N4 Industrial Electronics

Industrial Electronics

i

N4

Industrial Electronics Learner Book Johann Kraft

© Future Managers 2012 All rights reserved. No part of this book may be reproduced in any form, electronic, mechanical, photocopying, or otherwise, without prior permission of the copyright owner. To copy any part of this publication, you may contact DALRO for information and copyright clearance. Any unauthorised copying could lead to civil liability and/or criminal sanctions.

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Industrial Electronics

N4 Industrial Electronics

iii

CONTENTS

Chapter 1 – Kirchhoff ’s Laws ............................................................................................1 Chapter 2 – Superposition Theorem...............................................................................21 Chapter 3 – Thevenin’s Theorem.....................................................................................29 Chapter 4 – Series RLC-networks...................................................................................41 Chapter 5 – Parallel RLC-networks................................................................................69 Chapter 6 – Q-factor, Bandwidth and Complex Notation...........................................89 Chapter 7 – Basic Atomic Theory.................................................................................105 Chapter 8 – PN–Junction Theory..................................................................................115 Chapter 9 – Semi-conductor Diodes............................................................................127 Chapter 10 – Diode Applications..................................................................................139 Chapter 11 – Special Diodes and Applications...........................................................183 Chapter 12 – Transistors.................................................................................................199 Chapter 13 – Amplification Classes, Coupling Methods and Feedback..................231 Chapter 14 – Hybrid-Parameters .................................................................................249 Chapter 15 – Uni-Junction- and Field Effect Transistors .........................................263 Chapter 16 – Power Control .........................................................................................271 Chapter 17 – Operational Amplifiers ..........................................................................289 Chapter 18 – Function Generator and Oscilloscope..................................................309 Chapter 19 – Transducers..............................................................................................331

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KIRCHHOFF’S LAWS

KIRCHHOFF’s laws


CHAPTER

1 Kirchhoff’s Laws Learning Outcomes On completion of this module you will be able to: • • • • •

Describe using applicable network diagrams the concept of Kirchhoff ’s: – Current Law; and – Voltage Law. Describe using suitable expressions and network diagrams the concept of: – Voltage division; and – Current division. Solve current magnitudes in a given network containing one power source; Solve current magnitudes in a given network containing two power sources; and Solve voltage magnitudes, power magnitudes and resistive magnitudes in a given network diagram using Ohm’s Law as well as Kirchhoff ’s voltage and current laws.

1.1 Introduction The concept of Kirchhoff ’s Laws is not a new concept since you have studied the module on Ohm’s Law and the basis was laid there. It would be true to say that these laws (Kirchhoff ’s Laws) have as origin Ohm’s Law. You will find that as you proceed with further network theorems that all the theorems have as origin Ohm’s Law.

1.2 Kirchhoff’s Laws 1.2.1 Kirchhoff’s Current Law (KCL)

Definition 1.1 Kirchhoff ’s Current Law (KCL)

Kirchhoff ’s Current Law states that the algebraic sum of currents entering a point will be equal to the algebraic sum of the currents leaving that point.

1

2

KIRCHHOFF’S LAWS

Figure 1.1(a) and (b) will illustrate this principle. i1

i3

i2

i4

i2

i3

i1 (a)

i5

(b)

Figure 1.1

According to figure 1.1 (a): IT = I1 + I2 + I3, and According to figure 1.1 (b): I1 + I2 = I3 + I4 + I5. Important to note is that this current law is only applicable to a parallel network since current has to divide.

Example 1.1 Consider the network below and determine by calculation all the unknown values. All calculations must be shown. I1 = 0,6 A

I5

I9 = 0,4 A I10

I3 = 0,3 A I6 = 0,7 A

I11

I2 I4 = 0,4 A

I12 = 0,3 A I7

I13 = 0,9 A

I8

Solution: I2 = I3 + I4 + I6 I5 = I1 + I3 I7 = I12 + I13 = 0,3 + 0,4 + 0,7 = 0,6 + 0,3 = 0,3 + 0,9 = 1,4 ampere = 0,9 ampere = 1,2 ampere I8 = I7 - I4 I10 = I5 - I9 I11 = I6 + I10 +I12 = 1,2 - 0,4 = 0,9 - 0,4 = 0,7 + 0,5 + 0,3 = 0,8 ampere = 0,5 ampere = 1,5 ampere


1.2.2 Kirchhoff’s Voltage Law (KVL)

Definition 1.2 Kirchhoff ’s Voltage Law (KVL)

Kirchhoff ’s Voltage Law states that the algebraic sum of the individual voltage drops in a closed network is equal to the algebraic sum of the applied voltage. Figure 1.2 will illustrate this principle.

R1

R2

R3

V1

V2

V3

VS

Figure 1.2

According to figure 1.2: VS = V1 + V2 + V3 Something that also need to be addressed here is that should we have a current in the network which happen to be in the opposite direction to that which we assume, it will still cause a voltage drop across that resistor but that it will be of the opposite polarity and should we then define Kirchhoff ’s Voltage Law in an expression we will find that: ΣVS - ΣIR = 0 This expression will be very useful when working on networks which have no supply present. Important to note is that the voltage law is only applicable to a series network since the voltage has to divide between the respective resistive values.

1.3 Application of Kirchhoff’s Laws We once again find that the easiest and most convenient way to apply Kirchhoff ’s Laws will be by doing a couple of activities. In doing these activities the basic concepts will be applied as well as expanded. It must however be noted that as with many concepts that we will be dealing with that we need to commence with the most basic of those concepts as will be the case with Kirchhoff ’s Laws.

3

4

KIRCHHOFF’S LAWS

1.3.1 Kirchhoff’s Laws with a single sourcer Consider the network illustrated in figure 1.3. There is more than one way in which to solve this activity and you are advised to seek more than one solution. The secret lies therein to set up two mathematical equations and then to solve them mathematically. In setting up the two equations you must understand the concepts that were explained in the previous section since that theory will form the basis of setting up the two equations you will require. Furthermore, the concepts of Ohm’s Law are equally applicable since Kirchhoff ’s Laws has as origin Ohm’s Law. The two equations will be set up using what is termed ‘loops’ and you will also notice that the network has been labelled from A through to H. R2

I1 C

D

3 ohm

R1 B

G

3 ohm R3

E I2

I1 + I2

F

6 ohm V1 A

V2

-

VS = 10 V

+

H

Figure 1.3

Example 1.2 Determine with the given information of the network the magnitude of the current produced by the battery (I1 + I2).

Solution: Loop ABCDGHA

Loop ABEFGHA

V1 + V2 = VS (KVL) V1 + V2 = VS (KVL) But V1 = R1 × (I1+I2) and But V1 = R1 × (I1+I2) V2 = R3 × I2 V2 = R2 × I1 R1 × (I1+I2) + R2 × I2 = VS R1× (I1 + I2) + R3 × I2 = VS But R1 = 3 ohm; R2 = 3 ohm; R3 = 6 ohm and VS = 10 V


5

3 × (I1 + I2) + 3 × I1 = 10 3 × (I1 + I2) + 6 × I2 = 10 = 10 3I1 + 3I2 + 6I2 = 10 3I1 + 3I2 + 3I1 = 10 (1) 3I1 + 9I2 = 10 (2) 6I1 + 3I2 We now have two equations with two unknowns and are solved in the following manner: (1) × 1: 6I1 + 3I2 (2) × 2: 6I1 + 18I2 (1) - (3) - 15I2 I2 Substitute I2 = 0,667 = 6I1 + 3I2 6I1 + 3 (0,667) = 6I1 + 3 (0,667) = = 6I1 + 2,001 6I1 = 6I1 = I1 =

= 10 (1) = 20 (3) = - 10 = 0,667 ampere

ampere in (1) 10 10 10 10 10 - 2,001 7,999 1,333 ampere

Current produced by the battery = I1 + I2 = 1,333 + 0,667 = 2 ampere Having obtained the magnitudes of the currents indicated it is of utmost importance that these answers be verified and this is done in the following manner. Use any one of the equations you set up and substitute the obtained values therein and you must then have the left hand side equal to the right hand side of the equation. = 10 (1) 6I1 + 3I2 Substitute I1=1,333 ampere and I2 = 0,667 ampere 6 (1,333) + 3 (0,667) = 10 7,998 + 2,001 = 10 9,999 = 10 10 = 10 It must be noted however that should you have set up your equations incorrectly from the onset, that, when substituting the current magnitudes in any of the equations will result in the left hand side being equal to the right hand side. Therefore, be sure that the equations you set up are correct!

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KIRCHHOFF’S LAWS

Example 1.3 R3

R1 A

1,2 k-ohm

B

I1

1 k-ohm

C

V1 I1 - i2

I1 V2

R2

VS = 30 V

F

4 k-ohm E

D

Consider the illustrated network and determine by calculation the: (a) Magnitude of the currents in the parallel section of the network by making use of Kirchhoff ’s Laws; (b) Magnitude of the voltage drop V2; and (c) Power dissipated by resistor R3.

Solution: (a) Loop ABEFA

Loop ABCDEFA

V1 + V2 = VS (KVL) V1 + V2 = VS (KVL) But V1 = R1 × I1 and V2 = R3 × I2 But V1 = R1 × I1 and V2 = R2 × (I1 - I2) R1 × I1 + R2 × (I1 - I2) = VS R1 × I1 + R2 × I2 = VS 3 3 3 3 1,2 × 10 I1 + 4 × 10 (I1 - I2) = 30 1, 2 × 10 I1 + 1 × 10 I2 = 30 (2) 1,2 × 103I1 + 4 × 103I1 -1 × 103I2 = 30 = 30 (1) 5,2 × 103I1 - 4 × 103I2 (1) × 1: 5,2 × 103I1 - 4 × 103I2 (2) × 4: 4,8 × 103I1 + 4 × 103I2 (1) + (3) 10 × 103I1 I1

= = = =

30 (1) 120 (3) 150 15 m-ampere

Substitute I1 = 15 × 10-3 ampere in (1) 5, 2 × 103 (15 × 10-3) - 4 × 103I2 78 – 4 × 103I2 4 × 103I2 I2

= = = =

30 30 - 48 12 m-ampere


7

Current through R3 = I2 = 12 m-ampere Current through R2 = I1 - I2 = 15 × 10-3 - 12 × 10-3 = 3 m-ampere (b) V1 = R1 × I1 = 1,2 × 103×15 × 10-3 = 18 volt (c) PR1 = I22 × R3 = (12 × 10-3)2 × 1 × 103 = 144 m-watt The question may now arise why we make use of Kirchhoff ’s Laws when Ohm’s Law may be applied in obtaining the same result. This statement is confirmed when you look at example done on the application of Kirchhoff ’s Laws. We do however find times when the application of Ohm’s Law is impossible and we then have to find a method to solve the current magnitudes and it is then that Kirchhoff ’s Laws must be applied. Up to now we have only really had a look at Kirchhoff ’s Voltage Law except for example 1.1. Our aim will now be to solve a network in which no source included.

1.3.2 Kirchhoff’s Laws without a source Up to now we have only really had a look at Kirchhoff ’s Voltage Law except for example 1.1. Our aim will now be to solve a network in which no source included.

Example 1.4

Determine, with the aid of Kirchhoff ’s Laws the magnitude of the currents I1 and I2 and I1 - I2. B 10 ohm

I1 - I2

I1 3A

A

Loop 1

20 ohm

Loop 2

20 ohm

3 - I1 30 ohm

24 ohm D

I2

C

3 - I2

3A

8

KIRCHHOFF’S LAWS

Solution: Before we commence with the solution to this network it is important to note that there is no supply source present and one would assume that it will be very easy to solve this type of network. Yes, you are correct that it is very easy to solve since VS = 0, but there is a more important factor that need to be taken into account. You will notice that the current direction is indicated by means of the arrow points on the network. Now, should you select the loops at random and you follow the loop you must indicate a minus when you move against the arrow indicated on the network. Please follow the steps in the solution. Loop 1 (ABDA)

Loop 2 (BCDB)

20I2 - 24 (3 - I2) - 20 (I1 - I2) = 0 10I1 + 20 (I1 - I2) - 30 (3 - I1) = 0 20I2 – 72 + 24I2 - 20I1 + 20I2 = 0 10I1 + 20I1 - 20I2 - 90 + 30I1 = 0 = 90 (1) 20I1 + 64I2 = 72 (2) 60I1 - 20I2 (1) × 1: 60I1 - 20I2 (2) × 3: 60I1 + 192I2 (1) + (3): 172I2 I2

= = = =

90 (1) 216 (3) 306 1,779 ampere

Substitute I2 =1,779 ampere in (1) = 90 60I1- 20I2 = 90 60I1- 20 (1,779) = 90 60I1 - 35, 58 = 125, 58 60I1 = 2,093 ampere I1 I1 - I2 = 2,093 - 1,779 = 0,314 ampere It should now be obvious why we need to be able to apply Kirchhoff ’s Laws and you are now advised to apply the same principles that you applied in activity 1.2 and see if you can solve all the current magnitudes through each resistor in the network.


Example 1.5

Calculate with the aid of Kirchhoff ’s Laws the magnitude of the currents I1 and I2. 0,2 A 40 ohm B

A

D C

I1 - 0,2 A I1

10 ohm

I2

10 ohm

I1 - I2 - 0,2 A

80 ohm

VS = 4 V F

E

Solution: Loop ABCEFA

Loop BDCB

10 (I1 - 0, 2) + 80I2 = 4 40 × 0, 2 - 10 (I1 - I2 - 0, 2) - 10 (I1 - 0, 2) = 0 = 4 8 - 10I1 + 10I2 + 2 - 10I1 + 2 = 0 10I1- 2 + 80I2 - 20I1 + 10I2 =-8 10I1 + 80I2 = 6 (1) = 8 (2) 20I1 - 10I2 (1) × 2: 20I1 + 160I2 (2) × 1: 20I1 - 10I2 (3) - (2) 170I2 I2

= = = =

Substitute I2 = 23, 53 × 10-3 10I1 + 80I2 10I1 + 80 (23, 53 × 10-3) 10I1 + 1,882 10I1 I1

ampere in (1) = 6 = 6 = 6 = 4, 12 = 0,412 ampere

12 (3) 8 (2) 4 23,53 m-ampere

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10

KIRCHHOFF’S LAWS

1.3.3 Kirchhoff’s Laws with two sources It was obvious that some networks cannot be solved by means of Ohm’s Law since there is no supply present but yet we were able to set up equations and then solve for the current magnitudes. The question may now arise of how to solve current magnitudes as well as direction of current flow should two or more voltage sources are present. Important to note here is that one very often has to assume certain conditions and it may be possible that these assumptions be proved incorrect once an answer had been obtained. One assumption that we can assume from the onset is that current will flow from the negative terminal of the voltage source toward the positive terminal of that voltage source. There may be arguments that this is not a valid assumption since some theories may indicate that current flow is from the positive terminal of the voltage source toward the negative terminal of that source. The latter theory is therefore termed conventional current flow and the first theory is termed current flow. The most productive method of explanation will once again be to solve a given network and such a network is illustrated in figure 1.4. In this given network the assumptions that we spoke about will mainly be indicated by the direction of the assumed current flow which of course may be incorrect but more than enough examples will be given in order to rectify our assumptions made at the onset. For purposes of clarity and understanding of the assumptions to be made more than one solution will be given and this will assist you as learner to fully understand the concepts. Important to note is that specific theoretical concepts are embedded and these should be fully understood in order to gain success. R1 A

R2

B

10 ohm

10 ohm I1 + I2

I1 R3

V1 = 10 V

F

10 ohm

E

Figure 1.3

C

I2

V2 = 20 V

D


Example 1.6

Solve for the given network the magnitude of the current I1 + I2 through the resistor R3.

Solution: Loop ABEFA R1 × I1 + R3 (I1 + I2) 10I1 + 20 (I1 + I2) 10I1 + 20I1 + 20I2 30I1 + 20I2

= = = =

Loop CBEDA

V1 R2 × I2 + R3 (I1 + I2) 10 15I2 + 20 (I1 + I2) 10 15I2 + 20I1 + 20I2 10 (1) 20I1 + 35I2

(1) × 20: 600I1 + 400I2 (2) × 30: 600I1 + 1050I2 (3) - (4) - 650I2 I2

= = = =

= = = =

V2 20 20 20 (2)

200 (3) 600 (4) - 400 0,615 ampere

Substitute I2 = 0,615 ampere in (1) = 10 30I1 + 20I2 = 10 30I1 + 20 (0,615) = 10 30I1 + 12,3 30I1 = - 2,3 I1 = - 0,0766 ampere Current through resistor R3 = I1 + I2 = - 0,0766 + 0,615 = 0,5384 ampere. This solution proves to us that assumptions we made were wrong! Initially we assumed that the current I1 flows from A to B but the negative value obtained indicates that the current in fact flows from B to A. Secondly we assumed that the currents will be additive between B and E where in fact they subtract from one another. So what does this prove? Source V2 is the stronger of the two sources and therefore negates the current produced by V1. The following solution to this network (figure 1.5) will illustrate that should we have made the correct assumptions initially that all values obtained will have a positive value. It must however be noted that the negative answer obtained is not incorrect but that it is an indication that the current actually flows in the opposite direction as was assumed. Also important is that when that negative value is substituted into any expression that the negative sign must be used.

11

12

KIRCHHOFF’S LAWS

R1 A

R2

B

10 ohm

10 ohm I2 - I1

I1 R3

V1 = 10 V

F

10 ohm

E

C

I2

V2 = 20 V

D

Loop CBEDC

Loop CBAFEDC

R2 × I2 + R3 (I2 - I1) 15I2 + 20 (I2 - I1) 15I2 + 20I2 - 20I1 35I2 - 20I1

= = = =

V2 20 20 20 (1)

R2 × I2 + R1 × I1 15I2 + 10I1 15I2 + 10I1

(1) × 1: 35I2 - 20I1 (2) × 2: 30I2 + 20I1 (1) + (3) 65I2 I2

= = = =

20 (1) 20 (3) 40 0,615 ampere

Substitute I2 = 0,615 35I2 - 20I1 = 20 35 (0,615) - 20I1 21,525 - 20I1 - 20I1 I1

20 20 - 1,525 0,07625 ampere

= V2 - V1 = 20 - 10 = 10 (2)

ampere in (1) = = = =

Current through resistor R3 = I2 - I1 = 0,615 - 0,07625 = 0,5388 ampere


Example 1.7 Make use of Kirchhoff ’s Laws and determine the magnitude of the current through RL. R2 A

B

15 ohm

I2 - I1 V1 = 10 V

C

I2 RL

20 ohm V2 = 20 V

I1 10 ohm

F

E

D

R1

Solution: Loop FEBAF R1 × I1 - RL (I2 - I1) 10I1 - 20 (I2 - I1) 10I1 - 20I2 + 20I1 30I1 - 20I2

= = = =

Loop CBEDC V1 R2 × I2 + RL (I2 - I1) 10 15I2 + 20 (I2 - I1) 10 15I2 + 20I2 - 20I1 10 (1) - 20I1 + 35I2

(1) × 20: 600I1 - 400I2 (2) × 30: - 600I1 + 1050I2 (3) + (4) 650I2 I2

= = = =

200 (3) 600 (4) 800 1,23 ampere

Substitute I2 = 1,23 ampere in (1) = 10 30I1 - 20I2 30I1 - 20 (1, 23) = 10 = 10 30I1 - 24, 6 = 34,6 30I1 = 1,15 ampere I1 Current through resistor RL = I2 - I1 = 1, 23 - 1,15 = 0,08 ampere

= = = =

V2 20 20 20 (2)

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14

KIRCHHOFF’S LAWS

Example 1.8

R1 A

R2 B

4 ohm

3 ohm

I1 + I2

I1

C

I2 R3

V1 = 15 V F

2 ohm E

V2 = 10 V D

Calculate with the given information the: (a) Magnitude of the current through resistor R3 using Kirchhoff ’s Laws; (b) Voltage drop across resistor R1; and (c) Power consumed by resistor R2.

Solution: (a) Loop ABEFA R1 × I1 + R3 (I1 + I2) 4I1 + 2 (I1 + I2) 4I1 + 2I1 + 2I2 6I1 + 2I2 (1) × 1: (2) × 3: (1) – (3)

6I1 + 2I2 6I1 + 15I2 -13I2 I2

= = = =

Loop CBEDC V1 R2 × I2 + R3 (I1 + I2) 15 3I2 + 2 (I1 + I2) 15 3I2 + 2I1 + 2I2 15 (1) 2I1 + 5I2 = = = =

= = = =

V2 10 10 10 (2)

15 (1) 30 (3) -15 1,154 ampere

Substitute I2 = 1,154 ampere in (1) = 15 6I1 + 2I2 = 15 6I1 + 2 (1,154) = 15 6I1 + 2,308 = 12,692 6I1 = 2,115 ampere I1

Current through resistor R3 = I1 + I2 = 2,115 + 1,154 = 3,269 ampere


(b)

Although this section is concerned with Kirchhoff ’s Laws the quantities that need to be calculated here must be done using Ohm’s Law.

VR1

= = =

I1 × R1 1,154 × 4 4,616 volt

(c) PR2

= = =

I22 × R2 (2,115)2 × 3 13,419 watt

Example 1.9

Determine from the following network diagram with the assistance of Kirchhoff ’s Laws: I1 - I2 D

4 ohm

I2

12 ohm

I1

R5 1,1 A R2

A

I2 - 1,1 A

VS = 20 V 10 ohm

R3

R1 4 ohm R4 C R6

(a) (b) (c)

12 ohm 4 ohm

B I1 - 1,1 A

The magnitude of the currents I1 as well as I2 (use the loops DABCD and DACD); and The potential difference (voltage drop) across points A and B. Which of the terminals in (b) above is positive?

15

16

KIRCHHOFF’S LAWS

Solution: (a)

Loop DABCD

R2 × I2 + R3 (I2 - 1,1) + R4 (I1 - 1,1) + R1 × I1 12I2 + 10 (I2 - 1,1) + 12 (I1 - 1,1) + 4I1 12I2 + 10I2 - 11 + 12I1 - 13,2 + 4I1 16I1 + 22I2 - 24,2 16I1 + 22I2

= = = = =

VS 20 20 20 44,2 (1)

Loop DACD

R2 × I2 + R6 (1,1) + R1 × I1 12I2 + 4 (1,1) + 4I1 12I2 + 4,4 + 4I1 4I1 + 12I2 + 4,4 4I1 + 12I2 (1) × 1: 16I1 + 22I2 (2) × 4: 16I1 + 48I2 (1) - (3) - 26I2 I2 Substitute I2 = 0, 7 16I1 + 22I2 = 16I1 + 22 (0,7) = = 16I1 + 15,4 16I1 = I1 =

= = = =

= = = = =

VS 20 20 20 15,6 (2)

44,2 (1) 62,4 (3) -18,2 0,7 ampere

ampere in (1) 44,2 44,2 44,2 28,8 1,8 ampere

(b) VAB = R3 (I1 - 1,1) = 10 (1, 8 - 1, 1) = 7 volt (c)

Observing the indicated direction point B will be positive.


Example 1.10

The following Wheatstone-bridge network is given. R1

B R3

20 ohm I1 - I2

I1 0,4 A

I2

A

G

C

0,4 A

0,4 - I1 0,4 - I2

20 ohm R2

D

RX

The variable resistor R3 is adjusted until the galvanometer reading indicates zero. It is then determined by measurement that R3 has a value of 180 ohm. Determine the value of the resistor RX by making use of Kirchhoff ’s Laws only. NB: You may not use Ohm’s Law!

Solution: Loop ABDA R1 × I1 - R2 (0,4 - I1) 20I1 - 20 (0,4 - I1) 20I1 - 8 + 20I1 40I1 I1 I2

= = = = = =

Loop BCDB

0 R3 × I2 - RX (0,4 - I2) 0 180 × 0,2 - RX (0,4 - 0,2) 0 90 - 0,4RX - 0,2RX 8 - 0,6RX 0,2 ampere RX 0,2 ampere

= = = = =

0 0 0 - 90 150 ohm

Exercise 1.1 1. Describe Kirchhoff ’s: 1.1 Voltage Law; and 1.2 Current Law in your own words. 2. Use applicable network diagrams and explain the concepts of: 2.1 Current division; and 2.2 Voltage division.

17

18

KIRCHHOFF’S LAWS

3. Determine by calculation the magnitude of all the unknown current magnitudes in the following diagram. I1

I5 = 0,9 A

I9 I10 = 0,5 A

I3 I6 = 0,7 A

I11 = 1,6 A

I2 = 1,4 A I4 = 0,4 A

I12 I7 = 1,2 A

I13

I8

4. Two resistors having values of 5 ohm and 15 ohm respectively are connected in parallel and this combination is connected in series with a 12 ohm resistor across a direct current source of 12 volts. 4.1 Draw a neat labelled network diagram that will illustrate the above information. 4.2 Make use of Kirchhoff ’s Laws to determine the current magnitudes through each of the parallel resistors. 4.3 Make use of Ohm’s Law to determine the voltage drop across the parallel section of the network. 4.4 Calculate the power consumed by the 12 ohm resistor. 5. B 4 ohm

I1 - I2

I1 2A

A

Loop 1

12 ohm

Loop 2

12 ohm

2 - I1

5.1 5.2 5.3 5.4

18 ohm

16 ohm D

I2

C

2A

2 - I2

Use Kirchhoff ’s Laws and determine the magnitude of the currents I1 and I2. Determine the current through each resistor of the network. Determine the voltage drop across each resistor of the network. Determine the power consumed by each resistor of the network.


6. I1 A

6.1 6.2 6.3 6.4

B I2

10 ohm

VS = 12 V

I1 - I2 - 0,128 A

I1 - I2

40 ohm

H

C 0,128 A

10 ohm

30 ohm

G

D

20 ohm

F

E

Use Kirchhoff ’s Laws and calculate the magnitude of the currents I1 and I2. Determine the current through each resistor of the network. All calculations must be shown. Determine the voltage drop across each resistor of the network. All calculations must be shown. Determine the power consumed by each resistor of the network. All calculations must be shown.

7. Consider the network below and determine by calculation the: 7.1 Current using Kirchhoff ’s Laws that is produced by the voltage sources V1 and V2; 7.2 Voltage drop across the 6 ohm resistor R3; and 7.3 Power consumed by R1 and R2. R1

A

R2

B

10 ohm

I2 - I1

I1

V1 = 10 V F

4 ohm

C

I2 R3

6 ohm E

V2 = 20 V D

19

20

KIRCHHOFF’S LAWS

8. 8.1 8.2

Consider the given network and determine the following quantities. Use Kirchhoff ’s Laws and determine the magnitude of the current I1 - I2. It was found that through a change in temperature that the value of the load resistor RL had been reduced to 1 ohm. Re-calculate the magnitude of the current I1 - I2. R2 A

B I1 - I2

C

I2

I1 RL

V2 = 20 V F

3 ohm

4 ohm R1

2 ohm E

V1 = 10 V D

SUPERPOSITION THEOREM


CHAPTER

2 Superposition Theorem Learning Outcomes On completion of this module you will be able to: • • • •

Define the Superposition Theorem; Use and explain by applicable network diagrams and calculations the principle of: – Voltage division; and – Current division. Solve current magnitudes and directions with the aid of the Superposition Theorem; and Solve voltage magnitudes, power magnitudes and resistive magnitudes in a given network using Ohm’s law.

2.1 Introduction

Definition 2.1 Superposition Theorem

The Superposition Theorem states that all current magnitudes and directions may be determined by considering each supply on its own.

Once we have solved the magnitudes and directions we superimpose the network on top of one another in order to obtain the final solution. When considering each supply on its own we will actually make use of Ohm’s Law from which this theorem also originates. Before we can actually commence with this theorem there are two very important aspects that need to be addressed namely voltage and current division. You will recall that this concept was also discussed in the previous module but we will now take this concept one step further.

21

22

Superposition Theorem

2.2 Current- and voltage division 2.2.1 Current division In the definition of Kirchhoff ’s current law (KCL) we found that it may only be applied to a parallel network. What is of further significance is that the way in which the current will divide is based on the ratio to one another of the resistive elements in that particular parallel network. We also know from Ohm’s Law that in order to determine the current through the respective resistors in a parallel section that we need to know the voltage drop across that parallel section. The current division rule bypasses this requirement. Consider the resistive network illustrated in figure 2.1. I1 IT

I2

R1 R2

Figure 2.1

We know that according to Kirchhoff ’s current law that the current will divide between the two resistors and a point that should also need to be mentioned is that current will always take the path of least resistance. The magnitudes of I1 and I2 are given by the following mathematical expressions: I1 =

R2 × IT I2 = R1 + R2

R1 × IT R1 + R2

Example 2.1 Determine from the given information of the network the magnitude of the currents I1 and I2. 10 ohm

10 A IT

I1

I2

R1 15 ohm R2


23

Solution: I1 =

R2 × IT I2 = R1 × IT R1 + R2 R1 + R2 = 15 × 10 = 10 × 10 15 + 10 15 + 10 = 6 ampere = 4 ampere

In the definition of Kirchhoff ’s voltage law (KVL) we found that it may only be applied to a series network. What is of further significance is that the way in which the voltage will divide is based on the ratio to one another of the resistive elements in that particular series network. We also know from Ohm’s Law that in order to determine the voltage drop across the respective resistors in a series section that we need to know the current flowing in the network. The voltage division rule bypasses this requirement. Consider the resistive network illustrated in figure 11.2. We know that according to Kirchhoff ’s voltage law that the voltage will divide between R1

R2

V1

V2 VS

Figure 2.2

the two resistors and a point that should also need to be mentioned is that the largest resistor will also have the largest voltage drop. The magnitudes of V1 and V2 are given by the following mathematical expressions: V1 =

R1 × VS V2 = R1 + R2

R2 R1 + R2

× VS

Example 2.2 Determine from the given information of the network the magnitude of the voltages V1 and V2 R1

R2

18 ohm

12 ohm

V1

V2 VS = 50 V

24


Solution: R1 × VS V2 R R 1 + 2 18 = × 50 18 + 12 = 30 volt

V1 =

R2 × VS R1 + R2 12 = × 50 18 + 12

=

= 20 volt

We have once again reached the point where we need to look at the application of the Superposition Theorem and we will once again do it by means of a number of activities. It does not matter where you start but most important is that you must short circuit one of the supplies and insert your assumed current directions. Important to note again is that you may safely assume that the current will flow from the negative terminal of the supply to the positive terminal of the supply. For the purpose of our explanation we will short circuit supply V2 and what is most important is to redraw your network as follows.

Example 2.3 Apply the Superposition Theorem to find the direction and current magnitudes through each resistor in the following network.

V1 = 10 V

R1

R2

10 ohm

15 ohm

R3

20 ohm

V2 = 20 V

Solution: Short circuit V2 and indicate current directions and calculate values.


R2

R1 IA

IB

10 ohm

V1 = 10 V

IC

15 ohm 20 ohm

R3

R × R3 VS RTA = R1 + 2 IA = R + R3 R 2 TA 15 × 20 = 10 + = 10 15 + 20 22,857 = 22,857 ohm = 0,438 ampere

IB

R2 = × IA IC R2 + R3 15 = × 0,438 15 + 20 = 0,188 ampere

R3 × IA R2 + R3 20 = × 0,438 15 + 20

=

= 0,25 ampere

Step 2 Replace V2 and short circuit V1 and indicate current directions and calculate values. R2

R1 IE

10 ohm

ID

15 ohm

IF R3

RTB = = = IE

20 ohm

R1 × R3 V ID = S R2 + R3 R TB 10 × 20 10 15 + = 10 + 20 21,667 21,667 ohm = 0,462 ampere

R2 +

R1 × ID IF R + R3 1 10 = × 0,462 10 + 20 = 0,154 ampere

=

V2 = 20 V

R3 × ID R1 + R3 20 = × 0,462 10 + 20

=

= 0, 308 ampere

25

26


Having obtained the current directions and magnitudes by considering each supply on its own we have now reached the stage where the two networks need to be superimposed onto one another in order to find the final current directions as well as the magnitudes thereof. Step 3

R2

R1 IF

10 ohm

IB

IC

15 ohm

IE IA

ID

V1 = 10 V

R3

V2 = 20 V

20 ohm

IR1 = IA - IF IR2 = ID - IC IR3 = IB + IE = 0,438 - 0,308 = 0,462 - 0, 25 = 0,188 + 0,154 = 0, 13 ampere = 0,212 ampere = 0,342 ampere

Example 2.4 Determine with the aid of the Superposition Theorem the magnitude of the current through the load resistor RL. R2

2 ohm

V1 = 10 V

RL

4 ohm

V2 = 20 V

3 ohm R1

Solution: R2 2 ohm V1 = 10 V

RL

4 ohm

IA

IC 3 ohm R1

IB


R2 × RL V R2 IA = S IB = × IA R2 + RL RTA R2 + RL 2×4 10 2 = 3 + = = × 2,307 2+4 4,333 2+4 = = 4,333 ohm = 2,307 ampere = 0,769 ampere

RTA = R1 +

R2 IF

2 ohm

RL

3 ohm

4 ohm

ID

V1 = 20 V

IE

R1

R × RL V R1 ID = S IE = × ID RTB = R2 + 1 R1 + RL R R + RL TB 1 3×4 = 2 + = 20 = 3 × 5,385 3+4 3,714 3 + 4 = 3,714 ohm = 5,385 ampere = 2,308 ampere

Current through the load resistor RL

= IE - IB = 2,308 - 0,769 = 1,539 ampere

Exercise 2.1 1.

Explain using suitable network diagrams and applicable mathematical expressions the concept of: 1.1 Current division; and 1.2 Voltage division.

2. Define the Superposition Theorem. 3. Two resistors having values of 15 ohm and 35 ohm respectively are used in a voltage divider network driven from a source of 100 volt. Determine using the voltage divider rule the magnitude of the voltage across each resistor. Use any other method (Ohm’s Law) to prove that the magnitudes you obtained are the correct values.

27

28


4. A constant current source supplies 15 ampere to a resistive network consisting of two parallel resistors having values of 25 ohm and 40 ohm respectively. Determine using the current division rule the magnitude of the current through each resistor. Use any other method (Ohm’s Law) to prove that the magnitudes you obtained are the correct values. 5. Give a step-by-step description to illustrate the application of the Superposition Theorem. Use a network and values for the resistors of your choice. 6. Solve for each of the networks below the current magnitude and direction through each resistor by making use of the Superposition theorem. V1 = 40 V R1

R2

85 ohm

R3

47 ohm

3 ohm

100 ohm

R3 5 ohm

V1 = 80 V

R2

RL 2 ohm

V2 = 50 V V2 = 60 V

2 ohm

2 ohm

R1

R2

V1 = 10 V

V2 = 6 V

R3 2 ohm

V3 = 4 V

THEVENIN’s THEOREM


29

CHAPTER

3 Thevenin’s Theorem Learning Outcomes On completion of this module you will be able to: • • •

Define Thevenin’s theorem; Solve current magnitudes with the aid of Thevenin’s theorem; and Solve voltage magnitudes, power magnitudes and resistive magnitudes in a given network using Ohm’s Law.

3.1 Introduction

Definition 3.1 Thevenin’s Theorem

Thevenin’s Theorem specifies that a complex network consisting of impedances and voltage sources may be replaced by a constant voltage source with a series impedance. Thevenin’s Theorem has the advantage that once the equivalent network of impedances and constant voltage source had been obtained that the current through a load resistor may be calculated by merely inserting the new load resistor.

3.2 Application of Thevenin’s Theorem 3.2.1 Thevenin’s Theorem with a single source As was the case in the preceding modules, Thevenin’s Theorem will once again be explained by means of a number of activities. It is however very important that you must have a thorough background of Ohm’s Law since Thevenin’s Theorem has that law as basis.

30

Thevenin’s Theorem

Example 3.1 Consider the network illustrated in figure 3.1 and determine with the aid of Thevenin’s Theorem the magnitude of the current through resistor RL. R1

RL

10 ohm

6 ohm

V1 = 20 V

R2

4 ohm

Figure 3.1

In order to solve for the current magnitude through resistor RL we have to follow the step-by-step approach in that we must determine VTHEVENIN as well as RTHEVENIN in order to obtain the Thevenin equivalent network.

Solution: Remove resistor RL and label the points AB. R1 A 10 ohm

V1 = 20 V

R2

4 ohm

B

Determine VTHEVENIN which is the voltage drop across resistor R2 by making use of the voltage divider rule. R2 = × V1 VTHEVENIN = VAB R + R2 1 4 = × 20 10 + 4 = 5,714 volt


Step 2 Short circuit the supply V1 and determine RTHEVENIN by looking in at points AB. R1 A 10 ohm R2

4 ohm

RTHEVENIN B

R1 × R2 RTH = R + R2 1 10 × 4 = 10 + 4 = 2,857 ohm

Step 3 Equivalent network Having obtained the values for VTH and RTH we can now draw the equivalent network and re-insert the load resistor RL and then use Ohm’s law to determine the current through the load resistor RL. RTH

A

2,857 ohm

VTH = 5,714 V

RL

B

IRL = = =

VTH RTH + RL 5,714 2,857 + 6

0,645 ampere

6 ohm

31

32


Example 3.2 R1

R2

4 k-ohm

1 k-ohm A

VS = 30 V B

Consider the given network and: (a) Obtain the equivalent Thevenin’s network; and (b) Determine the magnitude of the current through the following load resistors: (i) 10 kilo-ohm; (ii) 25 kilo-ohm; and (iii) 100 ohm.

Solution: R2 R × R2 (a) VTH = × VS RTH = 1 R1 + R2 R + R2 1 1 × 103 4 × 103 × 1 × 103 = 3 × 30 = 4 × 10 + 1 × 103 4 × 103 + 1 × 103 = 6 volt = 800 ohm

Equivalent network RTH

A

800 ohm

VTH = 6 V

RL

B


VTH (b)(i) IRL = R + RL TH 6 = 800 + 10 × 103 = 0,556 m-ampere VTH (ii) IRL = RTH + RL 6 = 800 + 25 × 103

= 0,233 m-ampere VTH (iii) IRL = R + RL TH 6 = 800 + 100 = 6,667 m-ampere

Example 3.3

VS = 60 V

R1

R2

36 ohm

6 ohm

12 ohm

R3

5 ohm

RL

Consider the given network and determine with the aid of Thevenin’s Theorem the: (a) Magnitude of the current through the 5 ohm resistor; and (b) Power dissipated by the 5 ohm resistor.

33

34


Solution: (a) R1

R2

36 ohm

6 ohm

A

VS = 60 V

R2

12 ohm

B

R3 VTH = × VS R1 + R3 12 = × 60 36 + 12 = 15 volt R1

R2

36 ohm

6 ohm

A

RTHEVENIN

12 ohm

R2

B

RTH

= = =

R1 × R3 R1 + R3 36 × 12 6 + 36 + 12

R2 +

15 ohm


A

15 ohm

VTH = 15 V

RL

B

5 ohm


VTH = IRL RTH + RL 15 = 15 + 5 = 0, 75 ampere

(c) PRL = IRL2 × RL = (0, 75)2 × 5 = 2,813 watt

3.2.2 Thevenin’s Theorem with two sources Example 3.4 Obtain for the network the Thevenin’s equivalent network and then calculate the magnitude of the current through the load resistor RL in figure 3.2.

V1 = 4 V

V2 = 6 V RL

20 ohm

R1

4 ohm

3 ohm

R2

Figure 3.2

A point that should be noted with this type of problem is that we now have two voltage sources which will both contribute to VTH and we have to consider each supply on its own initially since both sources will contribute to the total of VTH.

Solution: Step 1 Remove RL, mark terminals AB and short circuit any one of the voltage sources. In this instance we will short circuit V2 and calculate VTH1.

35

36


V1 = 4 V A B

R1

4 ohm

3 ohm

R2

R2 VTH1 = × V1 R1 + R2 4 = × 4 3+4 = 2,286 volts

Step 2 Short circuit V1, replace V2 and calculate VTH2

V2 = 6 V A

R1 3 ohm

VTH2 = = =

R1 × V2 R1 + R2 3 × 6 3+4

2,571 volt

VTHT = VTH1 + VTH2 = 2,286 + 2,571 = 4,857 volt

B

4 ohm R2


37

VTH1 and VTH2 may be added together since they will be supplying current in the same direction through RL. But, be aware that this will not always be the case and this will be dealt with in a later activity. Step 3 Short circuit both voltage supplies and calculate RTH by looking in at terminals AB.

A RTHEVENIN

R1

4 ohm

B 3 ohm

R2

R × R2 RTH = 1 R1 + R2 3×4 = 3+4 = 1,714 ohm

Step 4

Equivalent network

You can now draw the equivalent network, re-insert the load resistor RL and determine the magnitude of the current. RTH

A

1,714 ohm

VTH = 4,857 V

RL

B

IRL

= = =

VTHT RTH + RL 4,857 1,714 + 20

0,224 ampere

20 ohm

38


Example 3.5 Make use of Thevenin’s Theorem and determine the magnitude of the current through the load resistor RL. R2 2 ohm

RL

V1 = 10 V

4 ohm

V2 = 20 V

3 ohm R1

Solution: R2 2 ohm A V1 = 10 V

B 3 ohm R1

R2 VTH1 = × V1 R1 + R2 2 = × 10 2+3 = 4 volt R2 2 ohm A B 3 ohm R1

V2 = 20 V


R1 VTH2 = × V2 R1 + R2 = 3 × 20 2+3 = 12 volt

VTHT = VTH2 - VTH1 = 12 - 4 = 8V R2 2 ohm A RTHEVENIN B 3 ohm R1

R × R2 RTH = 1 R1 + R2 3×2 = 3+2 = 1,2 ohm


A

1,714 ohm

VTH = 4,857 V

RL

B

IRL

= = =

VTHT RTH + RL 8 1,2 + 4

1,538 ampere

20 ohm

39

40


Exercise 3.1 1. Give a suitable definition or description of Thevenin’s Theorem. 2. Use a network containing: 2.1 One voltage source; and 2.2 Two voltage sources and give a step-by-step description of the approach you would follow in order to apply Thevenin’s Theorem. 3. Consider each of the networks below and determine with the aid of Thevenin’s Theorem the magnitude of the current through the load resistor RL. V1 = 40 V R1

12 ohm

3 ohm

R2

80 ohm

RL

R1

R2 5 ohm

36 ohm VS = 80 V

V2 = 60 V

R1

2 ohm

V2 = 4 V 4 ohm

V1 = 10 V

R2

3 ohm

RL 2 ohm

RL

SERIES RLC-NETWORKS


CHAPTER

4 Series RLC-networks Learning Outcomes On completion of this module you will be able to: •

• •

• • •

• • •

Illustrate using circuit diagrams, graphical representation and phasor diagrams the effect that an alternating quantity has on a single: – Resistor; – Inductor; and – Capacitor. Illustrate with the aid of a circuit diagram and graphical representation a series circuit containing a resistor and an inductor; Determine by calculation the: – Impedance; – Current flow; – Phase angle; – Voltage drops; and Supply voltage in a series circuit containing a resistor and an inductor. Draw a phasor diagram for a series circuit containing a resistor and an inductor; Illustrate with the aid of a circuit diagram and graphical representation a series circuit containing a resistor and a capacitor; Determine by calculation the: – Impedance; – Current flow; – Phase angle; – Voltage drops; and – Supply voltage in a series circuit containing a resistor and a capacitor. Draw a phasor diagram for a series circuit containing a resistor and a capacitor; Illustrate with the aid of a circuit diagram and graphical representation a series circuit containing a resistor, an inductor and a capacitor; Determine by calculation the: – Impedance; – Current flow; – Phase angle;

41

42

Series RLC-networks

– Power factor; – True power; – Apparent power; – Voltage drops; and – Supply voltage in a series circuit containing a resistor, an inductor and a capacitor. Draw a phasor diagram for a series circuit containing a resistor, an inductor and a capacitor; Explain by means of a graphical representation the concept of resonance in a series circuit containing a resistor, an inductor and a capacitor; Explain the conditions for resonance in a series circuit containing a resistor, an inductor and a capacitor; and Calculate the resonant frequency in a series circuit containing a resistor, an inductor and a capacitor.

• • • •

4.1 The effect of an alternating quantity on a resistor Figure 4.1 illustrates a resistor connected across an alternating current supply (a) and (b) illustrates a graphical representation of the phase relationship between the current and the supply voltage whereas (c) illustrates a phasor diagram. It is of utmost importance that you grasp and understand, not only for the resistor but also for the inductor and capacitor, these graphical representations and phasor diagrams. VS R

IR

+

t

IR

IR

VS

VS (a)

(b)

(c)

Figure 4.1

Obvious is that the voltage and current are in phase and therefore makes it possible that Ohm’s Law as used in dc networks can also be used in this instance. These expressions will be given again just to refresh your memory. V where I = current in ampere I = S R = resistance in ohm R VS = supply voltage in volt


43

VS Also: R = and VS = I×R I Before we go any further it is important to remember that we are working with alternating current and that all values that will be given will be as RMS-values unless otherwise stated! Also of importance is that the frequency of the alternating supply has no effect on the resistor.

Example 4.1 A resistor having a value of 100 ohm is connected across a 380V/100Hz alternating current supply. Determine the current that will flow through this resistor.

Solution: V I = S R 380 = 100 = 3, 8 ampere

4.2 The effect of an alternating quantity on an inductor Figure 4.2 illustrates an inductor connected across an alternating current supply (a) and (b) illustrates a graphical representation of the phase relationship between the current and the supply voltage whereas (c) illustrates a phasor diagram.

VS R

+

VS IL IL

t

IL

VS (a)

(b)

(c)

Figure 4.2

Upon careful inspection you will notice that there is a phase displacement between the alternating supply and the current through the inductor and is at a maximum of 90º with the voltage leading the current. In order to remember when voltage is leading or lagging we will look at the word CIVIL where I is used to identify current, C is used to

44

Series RLC-networks

identify a capacitor, V is used to identify voltage, and L is used to identify an inductor. Since we are presently working with an inductor and we underline part of the word CIVIL as indicated and we read it from left to right and we say: The voltage (V) will lead the current (I) in an inductor (L). This phase angle between the alternating supply voltage and the current will never be greater that 90º. Should we assume that we are working with a pure inductor we will find that it has no resistance (dc) but that it does offer an opposition to the flow of current which is termed the ‘inductive reactance’ and is given mathematically as: XL = 2 × π × f × L where

XL π f L

= inductive reactance of inductor in ohm = 3,142 = frequency of supply in hertz = value of inductor in Henry

Using Ohm’s Law as our basis we can also formulate the following expressions: V V IL = S and VS = IL × XL and XL = S XL I L

Example 4.2 A 250 mH inductor is connected across a 250V/100 Hz alternating current supply. Determine the: (a) Inductive reactance; (b) Current that will flow through the inductor.

Solution: (a) XL

= = =

2 × π × f × L 2 × 3,142 × 100 × 250 × 10-3 157, 1 ohm

V (b) IL = S X L 250 = 157, 1 = 1,591 ampere


Example 4.3 Re-calculate the value of the inductive reactance and current for example 4.2 above if it is given that the frequency of the alternating current supply changes to 50 Hz.

Solution: V XL = 2 × π × f × L IL = S X = 2 × 3,142 × 50 × 250 × 10-3 L 250 = 78, 55 ohm = 78, 55 = 3,182 ampere

Therefore, as our calculations indicate, the opposition to the flow of current is purely dependant upon the frequency of the supply. Also note that the inductive reactance of the inductor is directly proportional to the frequency of the supply.

4.3 The effect of an alternating quantity on a capacitor Figure 4.3 illustrates a capacitor connected across an alternating current supply (a) and (b) illustrates a graphical representation of the phase relationship between the current and the supply voltage whereas (c) illustrates a phasor diagram. IC VS

C IC

+

IC

t

VS (a)

(b)

VS

(c)

Figure 4.3

Upon careful inspection you will again notice that there is a phase displacement between the alternating supply and the current through the inductor and is at a maximum of 90º with the current leading the voltage. In order to remember when voltage is leading or lagging we will look at the word CIVIL where I is used to identify current, C is used to identify a capacitor, V is used to identify voltage, and L is used to identify an inductor.

45

46

Series RLC-networks

Since we are presently working with a capacitor and we underline part of the word CIVIL as indicated and we read it from left to right and we say: In a capacitor (C) the current (I) will lead the voltage (V). This phase angle between the alternating supply voltage and the current will never be greater that 90º. Should we assume that we are working with a pure capacitor we will find that it has no resistance (dc) but that it does offer an opposition to the flow of current and is termed ‘capacitive reactance’ and is given mathematically as: 1 XC = where XC 2×π×f×C π f C

= capacitive reactance of capacitor in ohm = 3,142 = frequency of supply in hertz = value of capacitor in Farad

Using Ohm’s Law as our basis we can also formulate the following expressions: V V IC = S and VS = IC × XC and XC = S XL I C

Example 4.4

A 250 µF capacitor is connected across a 250V/100 Hz alternating current supply. Determine the: (a) Capacitive reactance; (b) Current that will flow through the capacitor.

Solution: 1 (a) XC = 2×π×f×C 1 = 2 × 3,142 × 100 × 250 × 10-6

=

6,365 ohm

V (b) IC = S X C = 250 6,365 = 39,277 ampere


Example 4.5

Re-calculate the value of the capacitive reactance and current for example 4.4 above if it is given that the frequency of the alternating current supply changes to 50 Hz.

Solution: 1 XC = IC = 2×π×f×C = 1 = 2 × 3,142 × 50 × 250 × 10-6 = = 127,307 ohm

VS XC 250 127,307

1,963 ampere

Therefore, as our calculations indicate, the opposition to the flow of current is purely dependant upon the frequency of the supply. Also note that the capacitive reactance of the inductor is inversely proportional to the frequency of the supply. Another important factor that needs to be mentioned here is the opposition being offered to the flow of current when supplied from an alternating quantity is termed ‘reactance’ and not resistance as was the case with a direct current supply.

4.4 The series RL-network A network consisting of an inductor and resistor connected in series is illustrated in figure 4.4. Also indicated are all the variables that may be found in such a network.

R

L VL

VR I

VS

Figure 4.4

The graphic representation and phasor diagram for a series RL-network is given in figure 4.5 (a) and (b).

47

48

Series RLC-networks

VL

VR I

VS

VL

+

θ

t

I

(a)

VR

(b)

Figure 4.5

It is very important that when we draw phasor diagrams of series networks, irrespective of the combination of components in the network, that we use our current in the network as our reference. This may be done since the current in a series network is the same throughout the network. In order to derive our mathematical expressions for a series RL-network we also need to look at the reactance/impedance phasor diagram and that is illustrated in figure 4.6. XL

Z

θ

R I

Figure 4.6

For the purpose of obtaining mathematical expressions for series RL-networks we will refer to figure 4.5 (b) and figure 4.6. A new concept namely impedance now comes to the fore. This impedance of the network may be defined in this instance (RL-series networks) as the opposition offered to the flow of an alternating current and is the vector sum of the reactance of the inductor and the resistance of the resistor and is mathematically given by: where Z Z = (R2 + XL2) ½ R XL

= = =

impedance of the network in ohm resistance of resistor in ohm reactance of inductor in ohm

Observe figure 4.6 and you will notice that the above expression is derived by applying Pythagoras to the triangle. Similarly the supply voltage expression can be derived by


49

applying Pythagoras using figure 4.5 (b) and is mathematically given by: VS = (VR2 + VL2) ½ where VS VR VL

= = =

supply voltage in volt voltage drop across resistor in volt voltage drop across inductor in volt

where VL

=

I×R

=

I × XL

and VR

The current in the network is mathematically given by: V I = S where VS Z Z I

= = =

supply voltage in volt impedance of the network in ohm current flow in network in ampere

You will further notice that there is a phase angle θ between the current flowing in the network and the supply voltage. This is termed the phase angle and is mathematically given by: θ = tan-1XL or θ = tan-1VL where XL V L R VR VR R

= = = =

reactance of inductor in ohm voltage drop across inductor in volt voltage drop across resistor in volt resistance of resistor in ohm

Both these expressions are derived with trigonometry using figure 4.5 (b) and figure 4.6.

Example 4.6

Consider the network diagram given.

I

R

L

10 ohm

25 mH

VR

VL

VS = 250 V/ 50 Hz

50

Series RLC-networks

Determine from the given information on the network diagram the: (a) Inductive reactance of the inductor; (b) Impedance of the network; (c) Current flow in the network; (d) Voltage drops across the: (i) Resistor; and (ii) Inductor. (e) Supply voltage (f) Phase angle between the supply voltage and the current. (g) Draw a voltage phasor that will represent the quantities you have calculated above and insert all relevant values.

Solution: (a) XL = 2 × π × f × L = 2 × 3,142 × 50 × 25 × 10-3 = 7,855 ohm (b) Z = (R2 + XL2)½ = (102 + 7, 8552)½ = 12,716 ohm V (c) I = S Z 250 = 12,716 = 19, 66 ampere

(d)(i) VR (d)(ii) VL

= = =

I × R 19, 66 × 10 196, 6 volt

= = =

I × XL 19, 66 × 7,855 154, 43 volt

(e) VS = (VR2 + VL2)½ = (196, 62 + 154,432)½ = 250 volt The calculation done in (e) proves that the previous answers obtained are indeed correct since the vector sum of the individual voltage across the resistor and inductor is equal to the given supply voltage.


(f) θ = tan-1 XL or θ = tan-1 VL R VR = tan-1 7,855 = tan-1 154, 43 10 96, 6 = 38,15º = 38,15º What do these two answers mean? Since this is a series RL-network it will behave inductively and the supply voltage will lead the current by 38,15º, or alternatively the current will lag the supply voltage by 38,15º. This will become clearer when we draw the voltage phasor in 5. (g) VL = 154,43 V

VS = 250 V

38,15º I = 19,66 A

VR = 196,6 V

4.5 The series RC-network A network consisting a capacitor and resistor connected in series is illustrated in figure 4.7. Also indicated are all the variables that may be found in such a network. C

R VC

VR I

VS

Figure 4.7

The graphic representation and phasor diagram for a series RC-network is given in figure 4.8 (a) and (b).

51

52

Series RLC-networks

VR I

+

VC I

t

θ

VR

(a)

VC

VS (b)

Figure 4.8

It is very important that when we draw phasor diagrams of series networks, irrespective of the combination of components in the network, that we use our current in the network as our reference. This may be done since the current in a series network is the same throughout the network. In order to derive our mathematical expressions for a series RC-network we also need to look at the reactance/impedance phasor diagram and that is illustrated in figure 4.9.

I θ

XC

R

Z

Figure 4.9

For the purpose of obtaining mathematical expressions for series RC-networks we will refer to figure 4.8 (b) and figure 4.9. A new concept namely impedance now comes to the fore. This impedance of the network may be defined in this instance (RC-series networks) as the opposition offered to the flow of an alternating current and is the vector sum of the reactance of the capacitor and the resistance of the resistor and is mathematically given by: where Z = impedance of the network in ohm Z = (R2 + XC2)½ R = resistance of resistor in ohm XC = reactance of capacitor in ohm


53

Observe figure 12.9 and you will notice that the above expression is derived by applying Pythagoras to the triangle. Similarly the supply voltage expression can be derived by applying Pythagoras using figure 4.8 (b) and is mathematically given by: VS = (VR2 + VC2)½ where VS VR VC

= supply voltage in volt = voltage drop across resistor in volt = voltage drop across capacitor in volt

where VL = I × XC

= I × R

and

VR

The current in the network is mathematically given by: V I = S where VS Z Z I

= supply voltage in volt = impedance of the network in ohm = current flow in network in ampere

You will further notice that there is a phase angle θ between the current flowing in the network and the supply voltage. This is termed the phase angle and is mathematically given by: θ = tan-1 XC or θ = tan-1 VC R VR where XC = reactance of capacitor in ohm VL = voltage drop across capacitor in volt VR = voltage drop across resistor in volt R = resistance of resistor in ohm Both these expressions are derived with trigonometry using figure 4.8 (b) and figure 4.9.

Example 4.7 Consider the network diagram given. R

C

10 ohm

130 µF VC

VR I

VS = 250 V/50 Hz

54

Series RLC-networks

Determine from the given information on the network diagram the: (a) Capacitive reactance of the capacitor; (b) Impedance of the network; (c) Current flow in the network; (d) Voltage drops across the: (i) Resistor; and (ii) Capacitor. (e) Supply voltage; (f) Phase angle between the supply voltage and the current. (g) Draw a voltage phasor that will represent the quantities you have calculated above and insert all relevant values.

Solution: 1 (a) XC = 2×π×f×C 1 = 2 × 3,142 × 50 × 130 × 10-6 = 24, 48 ohm

(b) Z = (R2 + XC2)½ = (102 + 24, 482)½ = 26,443 ohm VS (c) I = Z 250 = 26,443 = 9,454 ampere

(d)(i) VR (d)(ii) VL

= = =

I × R 9, 454 × 10 94, 54 volt

= = =

I × XC 9,454 × 24, 48 231,434 volt

(e) VS = (VR2 + VC2)½ = (94, 54 + 231, 434)½ = 249,999 volt


55

The calculation done in (e) proves that the previous answers obtained are indeed correct since the vector sum of the individual voltage across the resistor and inductor is almost equal to the given supply voltage. or θ = tan-1 VC (f) θ = tan-1 XC R VR -1 -1 = tan 24, 48 = tan 231,434 10 94, 54 = 67,62º = 67,78º What do these two answers mean? Since this is a series RC-network it will behave capacitively and the supply voltage will lead the current by 67,62º, or alternatively the current will lag the supply voltage by 67,62º. This will become clearer when we draw the voltage phasor in (g). (g)

I = 9,454 A VR = 94,54 V

67,62º

VS = 250 V

VC = 231,434 V

4.6 The series RLC-network A network consisting a resistor, capacitor and inductor connected in series is illustrated in figure 4.10. Also indicated are all the variables that may be found in such a network. R

I

L

VR

VL VS

Figure 4.10

C

VC

56

Series RLC-networks

With all three components connected in the circuit it will either behave like an RLnetwork or an RC-network. This will depend on the frequency of the alternating supply in that either the inductive reactance (XL) or the capacitive reactance (XC) will be the greater. Should the inductive reactance be the greater then the network will behave inductively as if there was only an inductor and resistor in the network. On the other hand, should the capacitive reactance be the greater then the network will behave capacitively as if there was only a capacitor and resistor in the network. These two conditions will now be discussed in turn and you are referred to figure 4.10 and 4.11 (a) and (b).

4.6.1 Inductive behaviour The conditions for inductive behaviour include: • •

XL > XC; and VL > VC

The phasor diagram for inductive behaviour is illustrated in figure 4.11 (a) and (b).

XL

VL Z

XL - XC

θ I XC

VS

VL - VC

θ

R

I VC

(a)

VR

(b)

Figure 4.11

If you refer back to figure 4.5 (b) and figure 4.6 you will notice that they are identical. The only difference being is that VL and VC have been subtracted vectorially to yield VL VC. The reason for this is that they are 180º apart and cannot be subtracted algebraically.

4.6.2 Capacitive behaviour The conditions for capacitive behaviour include: • •

XC > XL; and VC > VL

The phasor diagram for capacitive behaviour is illustrated in figure 4.12 (a) and (b).


XL

VL I

I

R

θ

XC - XL

Z XC

57

θ

VC - VL

VS VC

(a)

VR

(b)

Figure 4.12

If you refer back to figure 4.8 (b) and figure 4.9 you will notice that they are identical. The only difference being is that VC and VL have been subtracted vectorially to yield VC VL. The reason for this is that they are 180º apart and cannot be subtracted algebraically. Referring to figure 4.11 and 4.12 we can once again use Pythagoras as basis and derive the following mathematical expressions: Z = [R2 + (XC ~ XL)2]½ where Z R XC XL ~

= impedance of the network in ohm = resistance of resistor in ohm = reactance of capacitor in ohm = reactance of inductor in ohm = difference between

VS = [VR2 + (VC ~ VL)2]½ where VS VR VL VC ~

= supply voltage in volt = voltage drop across resistor in volt = voltage drop across inductor in volt = voltage drop across capacitor in volt = difference between

where VC = I × XC VR =

I × R

VL

=

I × XL

The current in the network is mathematically given by: V I = S where Z

VS Z I

= supply voltage in volt = impedance of the network in ohm = current flow in network in ampere

You will further notice that there is a phase angle θ between the current flowing in the network and the supply voltage. This is termed the phase angle and is mathematically given by: θ = cos-1 R Z

58

Series RLC-networks

The power factor is mathematically given by: Cos θ = R Z The true power in a circuit containing an inductor and a capacitor is the power consumed taking into account the phase angle and is mathematically given by: True power

=

V × I Cos θ watt

The true power also termed apparent power in a circuit is the power in a pure resistive circuit and there is no phase angle and is mathematically given by: Apparent power = V × I watt A very important aspect concerning the behaviour of a series RLC-network behaving either inductive or capacitive needs to be noted. In a series RLC-network it is very easy to see whether XL or XC is the greater and that will determine whether the network is inductive or capacitive. It is further obvious that should XL be the greater that VL will also be the greater and vice versa. However, when we do parallel networks and we work in terms of current this direct proportion does not hold well. We will find there that should XL be the greater then IL will be smaller and the circuit actually behaves capacitive. Therefore, to determine whether a network is inductive in a series RLCnetwork we look at VL and VC and for a parallel RLC-network we look at IL and IC. The following two activities will serve to capture what we have discussed thus far. It should be noted that there is a lot of mathematical, as well as theoretical knowledge that need to be mastered. It is further noteworthy that a lot of ‘other’ work needs to be done before we can draw a phasor diagram. In order to illustrate this logical process the following two activities will be handled in this manner.

Example 4.8 A network consists of the following components connected in series across a 200V/200Hz alternating current supply: • R = 30 ohm • L = 5 mH • C = 10 µF Use this data and draw a fully labelled phasor diagram that will represent the conditions in this network. All relevant information must be indicated on the phasor diagram.


Solution: 1 XL = 2 x π x f x L XC = 2 x π xfxC = 2 × 3,142 × 200 × 50 × 10-3 = 62,84 ohm 1 = 2 × 3,142 × 200 × 10 ×10-6 = 79,56 ohm

Z = [R2 + (XC - XL)2]½ = [302 + (79, 56 - 62, 84)2]½ = 34, 34 ohm V R I = S θ = cos-1 Z Z 200 30 = = cos-1 34,34 34,34 = 5,82 ampere = 29,11º

VC = I × XC VR = = 5,82 × 79,56 = = 463 volt =

I × R VL 5,82 × 30 174,6 olt

= = =

VL = 365,73 V I = 5,82 A 29,11º

VC - VL = 97,27 V VC = 463 V

VR = 174,6 V

VS = 200 V

I × XL 5,82 × 62,84 365,73 volt

59

60

Series RLC-networks

Example 4.9 Consider the following network diagram.

I

R

L

10 ohm

14 ohm

150 V

180 V

C 8 ohm 90 V

100 Hz

Use the information in the above network diagram and determine the: (a) Supply voltage; (b) Impedance; (c) Line current; (d) Value of the inductor in mH; (e) Value of the capacitor in µF; and (f) Phase angle between the line current and supply voltage. (g) You are further required to answer the following questions. (i) Does the network behave inductive or capacitive? Motivate your answer! (ii) Is the line current leading or lagging the supply voltage? Motivate your answer!

Solution: (a) VS = [VR2 + (VC2 ~ VL2)]½ = [1502 + (180 - 90)2]½ = 174,9 volt (b) Z

= [R2 + (XC ~ XL)2]½ = [102 + (14 - 8)2]½ = 11,66 ohm

VS (c) I = Z 174, 9 = 11, 66 = 15 ampere


(d)

X L

=

61

2×π×f×L

XL L = 2×π×f 4 = 2 × 3,142 × 100 = 22,27 mH

(e) XC = = = =

1 2×π×f×C 1 2 × π × f × XC 1 2 × 3,142 × 100 × 8

198,9 µF

R (f) θ = cos-1 Z 10 = cos-1 11, 16 = 30,95º

(g)(i)

The circuit is behaving inductive since VL > VC.

(g)(ii) The line current is lagging the supply voltage since the circuit behaves inductive. Resonance is a condition that will occur in a series- or parallel RLC-network when the inductive reactance equals the capacitive reactance. In order to analyse the phenomena of resonance we need to look at the mathematical expressions we use to determine inductive- and capacitive reactance. It has already been mentioned that the inductive reactance is directly proportional to the supply frequency and we will find that as frequency increases so does the inductive reactance increase. The capacitive reactance however is indirectly proportional to the supply frequency and will have a decrease in capacitive reactance as the supply frequency increases. This concept is illustrated in figure 4.13. The graphic representation of resonance illustrated in figure 4.13 shows that the graph for inductive reactance (XL) is a straight line whereas the graph for the capacitive reactance (XC) is an exponential curve.

62

Series RLC-networks

There is the point where these two graphs intersect one another and is termed the resonant frequency and it is also this point where XL = XC. We will now use this information to derive a formula to determine this frequency.

R e a c t a n c e

XL

XC f fr

Figure 4.13

XL = XC ............... (1) But XL = 2 x π x f x L and XC = Substitute in (1)

1 2xπxfxC

2xπxfxL = 1 2xπxfxC = 1 f2 (2 x π)2 x L x C fr = 1 2 x π x (L x C)½ The expression derived above clearly indicates that only the inductor and capacitor plays a role in the resonance of a series RLC-network and that resistance plays no role at all.

4.7 Conditions for resonance At this point we will only consider the conditions for resonance in series RLC-networks. The following conditions will exist for resonance in a series RLC-network. • • • •

XL = XC VL = VC and is maximum VS = VR Z = R and is minimum


• θ = 0º • I is maximum • fr = 1 2 x π x (L x C)½ The explanation that will follow now we will prove mathematically as well as by means of graphic representation some of the conditions mentioned above. Z = [R2 + (XC ~ XL)2]½ but XL = XC θ = cos-1 R 2 2 ½ Z = [R + (0) ] -1 = R = cos R R -1 = cos 1 = 0º VS = [VR2 + (VC2 ~ VL2)½ = [VR2 + (0)2]½ = VR

but Z = R

but VL = VC

The following two graphs illustrated in figure 4.14 (a) and (b) illustrates two of the above conditions namely frequency versus impedance and frequency versus current. Also illustrated in figure 4.14 (c) is the phasor diagram for a series resonant network. R I m p e d a n c e

R XL > XC

XC > XL

RC

XC > XL C u r r e n t

RL

VL

XL > XC

RC

RL VR I

f

f

fr

fr

(a)

(b)

VC (c)

Figure 4.14

Example 4.10 A series RLC-network consists of the following components connected across a 300V alternating current supply: • R = 10 ohm • L = 20 mH • C = 75 µF

63

64

Series RLC-networks

Use this information to determine the: (a) Resonant frequency of the network; (b) Impedance of the network (prove your answer); (c) That the supply voltage VS = 300 V (prove your answer); and (d) That the phase angle θ = 0º (prove your answer).

Solution:

1 (a) fr = 2 x π x (L x C)½ 1 = 2 × 3,142 (20 × 10-3 × 75 × 10-6)½

= 129, 93 Hz

(b) Z = [R2 + (XC ~ XL)2]½ = [102 + (0)2]½ = 10 ohm

but XL = XC = 0

(c) XL = 2 × π × f × L = 2 × 3,142 ×129,93 × 20 × 10-3 = 16,33 ohm 1 XC = 2×π×f×C = 1 2 × 3,142 ×129, 93 × 75 × 10-6

= 16,33 ohm

V I = S VC = I × XC VR = I × R VL = I × XL Z = 30 × 16,33 = 30 × 10 = 30 × 16,33 300 = 489,9 volt = 300 volt = 489,9 volt = 10 = 30 ampere

VS = [VR2 + (VC2 ~ VL2 )]½ = [3002 + (0)2]½ = 300 volt

but VL = VC = 0


R (d) θ = cos-1 Z R = cos-1 R 10 = cos-1 10 = 0º

but

Z

=

R

=

10

Exercise 4.1 1. An alternating quantity will behave in a prescribed manner when applied to different components. Use applicable diagrams to illustrate this behaviour and draw the graphical- as well as phasor diagrams when an alternating quantity is applied to a (n): 1.1 Resistor; 1.2 Capacitor; and 1.3 Inductor. 2. Use a graph and give a detailed explanation of series resonance. You are further required to derive an expression that can be used to determine the resonant frequency of a series RLC-network. 3. Explain the concept of inductive- and capacitive behaviour of a series RLC-network by means of two fully labelled phasor diagrams. Use your own imaginary values. 4. A series network consists of a 25 mH inductor and a 15 ohm resistor connected across a 100V/60Hz alternating supply. Calculate the: 4.1 Impedance of the network; 4.2 Current that will flow in the network; 4.3 Voltage drop across the: 4.3.1 Resistor; and 4.3.2 Inductor. 4.4 Phase angle between the supply voltage and the current. 4.5 Draw a phasor diagram that will represent the quantities for the above network and insert all relevant values. 5.

A series network consists of a 25 µF capacitor and a 10 ohm resistor connected across a 150V/60Hz alternating supply. Calculate the: 5.1 Impedance of the network; 5.2 Current that will flow in the network;

65

66

Series RLC-networks

5.3 Voltage drop across the: 5.3.1 Resistor; and 5.3.2 Capacitor. 5.3.3 Phase angle between the supply voltage and the current. 5.3.4 Draw a phasor diagram that will represent the quantities for the above network and insert all relevant values. 6. Mention seven conditions that will exist in as series resonant RLC-network. 7. Consider the following phasor diagram. VL = 55 V

VS

VL - VC

I-3A

VR = 30 V

VC = 20 V

Determine using the available date the: 7.1 Magnitude of the supply voltage; 7.2 Capacitive value of the capacitor in µF; and 7.3 Inductive value of the inductor in mH. Note: Use a supply frequency of 50 Hz.

8. A series RLC-network consists of the following components: • Resistor = 2 ohm • Capacitor = 100 µF • Inductor = 20 mH This combination is connected across a 100mV alternating supply at a frequency of 500 radians/second. Calculate the: 8.1 Impedance of the network; 8.2 Current that will flow in the network; 8.3 Voltage drop across the: 8.3.1 Resistor; 8.3.2 Capacitor; and 8.3.3 Inductor. 8.4 Phase angle between the line current and the supply voltage. 8.5 Draw a phasor diagram that will represent the quantities for the above network and insert all relevant values.


9. Draw neat labelled graphs that will illustrate: 9.1 Current versus frequency; and 9.2 Frequency versus impedance for a series RLC-resonant circuit. 10. A series RLC-network draws a current of 1,6 mA from a 1,6 V alternating current supply when the network is at resonance at a frequency of 1 kHz. Should 16 volt be measured across the inductor at this frequency determine the: 10.1 Value of the inductor; 10.2 Value of the capacitor; and 10.3 Value of the resistor.

67

68

Series RLC-networks

PARALLEL RLC-NETWORKS


69

CHAPTER

5 Parallel RLC-networks Learning Outcomes On completion of this module you will be able to: • •

• • •

• • •

•

Illustrate with the aid of a circuit diagram and graphical representation a parallel circuit containing a resistor and an inductor; Determine by calculation the: – Impedance; – Current flow; – Phase angle; – Voltage drops; and – Supply voltage in a parallel circuit containing a resistor and an inductor. Draw a phasor diagram for a parallel circuit containing a resistor and an inductor; Illustrate with the aid of a circuit diagram and graphical representation a parallel circuit containing a resistor and a capacitor; Determine by calculation the: – Impedance; – Current flow; – Phase angle; – Voltage drops; and – Supply voltage in a parallel circuit containing a resistor and a capacitor. Draw a phasor diagram for a parallel circuit containing a resistor and a capacitor; Illustrate with the aid of a circuit diagram and graphical representation a parallel circuit containing a resistor, an inductor and a capacitor; Determine by calculation the: – Impedance; – Current flow; – Phase angle; – Voltage drops; and – Supply voltage in a parallel circuit containing a resistor, an inductor and a capacitor. Draw a phasor diagram for a parallel circuit containing a resistor, an inductor and a capacitor;

70

parallel RLC-networks

• • •

Explain by means of a graphical representation the concept of resonance in a parallel circuit containing a resistor, an inductor and a capacitor; Explain the conditions for resonance in a parallel circuit containing a resistor, an inductor and a capacitor; and Calculate the resonant frequency in a parallel circuit containing a resistor, an inductor and a capacitor.

5.1 The parallel RL-network A parallel network consisting of an inductor and resistor connected in parallel is illustrated in figure 5.1. Also indicated are all the variables that may be found in such a network. IR

R

IL

L

IT

VS

Figure 5.1

The graphic representation and phasor diagram for a parallel RL-network is given in figure 5.2 (a) and (b). VS IL

+

IR VS

t

IR

θ

IL IT (a)

(b)

Figure 5.2

It is very important that when we draw phasor diagrams of parallel networks, irrespective of the combination of components in the network, that we use our supply voltage to the network as our reference. This may be done since the voltage across a parallel network is the same across the network. In order to derive our mathematical


expressions for a parallel RL-network we also need to look at the reactance/impedance phasor diagram and that is illustrated in figure 5.3.

I

R

θ

Z

XL

Figure 5.3

For the purpose of obtaining mathematical expressions for parallel RL-networks we will refer to figure 5.2 (b) and figure 5.3. A new concept namely impedance now comes to the fore. This impedance of the network may be defined in this instance (RL-parallel networks) as the opposition offered to the flow of an alternating current and is the vector sum of the reactance of the inductor and the resistance of the resistor and is mathematically given by: where Z = R × XL (R2 + XL2)½

Z = impedance of the network in ohm R = resistance of resistor in ohm XL = reactance of inductor in ohm

V Z = S where Vs = supply voltage in volt I T IT = total current in ampere

The total current expression can be derived by applying Pythagoras using figure 5.2 (b) and is mathematically given by: IT = (IR2 + IL2)½ where IT = total current in ampere IR = current through resistor in ampere IL = current through inductor in ampere VS V where IL = and IR = S XL R The total current in the network is also mathematically given by: V IT = S where VS = supply voltage in volt Z Z = impedance of the network in ohm IT = current flow in circuit in ampere

71

72


You will further notice that there is a phase angle θ between the current flowing in the network and the supply voltage. This is termed the phase angle and is mathematically given by: I I θ = tan-1 L or θ = cos-1 R where IL = current through inductor in ampere IR IT IR = current through resistor in ampere IT = total current in circuit in ampere

Example 5.1 Consider the network and determine from the given information on the network diagram the: R IR

10 ohm L

IL

25 mH

IT 100 V/60 Hz

(a) (b) (c) (d) (e) (f)

Inductive reactance of the inductor; Impedance of the network; Current flow through the: (i) Resistor; and (ii) Inductor. Total current Phase angle between the supply voltage and the current. Draw a current phasor that will represent the quantities you have calculated above and insert all relevant values.

Solution: (a) XL

= = =

2 x π x f x L 2 × 3,142 × 60 × 50 × 10-3 18,852 ohm


73

(b) Z = R × XL Z = VS 2 2 ½ (R + XL ) IT = 10 × 18,852 = 100 11, 32 (102 + 18,8522)½ = 8,834 ohm = 8,834 ohm V (c)(i) IL = S X L 100 = 18,852 = 5,3 ampere V (c)(ii) IR = S R 100 = 10 = 10 ampere VS (d) IT = (IR2 + IL2)½ IT = Z = (102 + 5,32)½ 100 = 11,32 ampere = 8,834 = 11, 32 ampere

The calculation done in (d) proves that the previous answers obtained are indeed correct since the vector sum of the individual voltage across the resistor and inductor is equal to the total current flow.

I I (e) θ = tan-1 L or θ = cos-1 R IR IT 5,3 10 = tan-1 = cos-1 10 11,32 = 27,92º = 27,95º

What do these two answers mean? Since this is a parallel RL-network it will behave inductively and the supply voltage will lead the current by 27,92º, or alternatively the current will lag the supply voltage by 27,92 º. This will become clearer when we draw the voltage phasor in (f).

74


(f)

VS = 100 V

IR = 10 A

27, 92º

IT = 11,32 A

IL = 5,3 A

5.2 The parallel RC-network A network consisting a capacitor and resistor connected in parallel is illustrated in figure 5.4. Also indicated are all the variables that may be found in such a network. The graphic representation and phasor diagram for a parallel RC-network is given in figure 5.5 (a) and (b). IR

R

IC C IT

VS

Figure 5.4 IC

VS

+

IR

IT

IC t

θ VS

-

(a)

(b)

Figure 5.5

IR


75

It is very important that when we draw phasor diagrams of parallel networks, irrespective of the combination of components in the network, that we use our supply voltage to the network as our reference. This may be done since the supply voltage in a parallel network is the same across the network. In order to derive our mathematical expressions for a parallel RC-networks we also need to look at the reactance/impedance phasor diagram and that is illustrated in figure 5.6. XC

Z

θ I

R

Figure 5.6

For the purpose of obtaining mathematical expressions for parallel RC-networks we will refer to figure 5.5 (b) and figure 5.6. A new concept namely impedance now comes to the fore. This impedance of the network may be defined in this instance (parallel RC-networks) as the opposition offered to the flow of an alternating current and is the vector sum of the reactance of the capacitor and the resistance of the resistor and is mathematically given by: where Z = impedance of the network in ohm Z = R × Xc (R2 + XC2)½ R = resistance of resistor in ohm XC = reactance of inductor in ohm Z = VS where Vs = supply voltage in volt IT IT = total current in ampere The total current expression can be derived by applying Pythagoras using figure 13.2 (b) and is mathematically given by: IT = (IR2 + IC2)½ where IT = total current in ampere IR = current through resistor in ampere IC = current through inductor in ampere where IC

=

VS and IR = VS XC R

The total current in the network is also mathematically given by:

76


IT = VS where VS = supply voltage in volt Z Z = impedance of the network in ohm IT = current flow in circuit in ampere You will further notice that there is a phase angle θ between the current flowing in the network and the supply voltage. This is termed the phase angle and is mathematically given by: θ = tan-1 IC or θ = cos-1 IR where IC = current through inductor in ampere IT IR = current through resistor in ampere IR IT = total current in circuit in ampere

Example 5.2

Consider the network and determine from the given information on the network diagram the: R IR

100 ohm

IC

IT

C 10 µF 120 V/100 Hz

(a) (b) (c) (d) (e) (f)

Capacitive reactance of the capacitor; Impedance of the network; Current flow through the: (i) Capacitor; and (ii) Resistor. Current flow in the network; Phase angle between the supply voltage and the current. Draw a current phasor that will represent the quantities you have calculated above and insert all relevant values.


Solution: (a) XC = ½ x π x f x C = 1/2 × 3,142 × 60 × 10 × 10-6 = 159,134 ohm VS Z = (b) Z = R × XC IT (R2 + XC2)½ 120 100 × 159,134 = 1,417 = 2 (10 + 159,134)½ = 84,67 ohm = 84,67 ohm V V (c)(i) IC = S (ii) IR = S XC R 120 120 = = 159,134 10 = 0,754 ampere = 1,2 ampere VS (d) IT = (IR2 + IC2)½ IT = Z = (1,22 + 0,754)½ 100 = 1,417 ampere = 84,68 = 1,417 ampere

The calculation done in (d) proves that the previous answers obtained are indeed correct since the vector sum of the individual voltage across the resistor and inductor is equal to the total current flow. I I θ = cos-1 R (e) θ = tan-1 C or IR IT 0,754 1,2 = tan-1 = cos-1 1, 2 1,417 = 32,14º = 32,13º What do these two answers mean? Since this is a parallel RC-network it will behave capacitively and the current will lead the supply voltage by 32,14º, or alternatively the supply voltage will lag the current by 32,14 º. This will become clearer when we draw the voltage phasor in (f)

77

78


(f) IC = 0,754 A

IT = 1,417 A

32, 13º

IR = 1,2 A

VS = 120 V

5.3 The parallel RLC-network A network consisting a resistor, capacitor and inductor connected in series is illustrated in figure 5.7. Also indicated are all the variables that may be found in such a network. R IR C IC L IL IT VS

Figure 5.7

With all three components connected in the network it will either behave like an RLnetwork or an RC-network. This will depend on the frequency of the alternating supply in that either the inductive reactance (XL) or the capacitive reactance (XC) will be the greater. Should the inductive reactance be the greater then the network will behave capacitively as if there was only a capacitor and resistor in the network. On the other hand, should the capacitive reactance be the greater then the network will behave inductively as if there was only an inductor and resistor in the network. These two conditions will now be discussed in turn and you are referred to figure 5.8 (a) and (b) and 5.9 (a) and (b).

5.3.1 Capacitive behaviour The conditions for capacitive behaviour include:


• •

79

XL > XC; and IC > IL

The phasor diagram for capacitive behaviour is illustrated in figure 5.8 (a) and (b). XL

IC Z

XL - XC

θ

θ

R

I XC

IT

IC - IL

IL

(a)

IR VS (b)

Figure 5.8

If you refer back to figure 5.5 (b) and figure 5.6 you will notice that they are identical. The only difference being is that IC and IL have been subtracted vectorially to yield IC IL. The reason for this is that they are 180º apart and cannot be subtracted algebraically.

5.3.2 Inductive behaviour The conditions for inductive behaviour include: • •

XC > XL; and IL > IC

The phasor diagram for inductive behaviour is illustrated in figure 5.9 (a) and (b). the inductor is directly proportional to the frequency of the supply. XL

IC I θ

XC - XL

Z XC

VS

R

θ

IL - IC

IT IL

(a)

IR

(b)

Figure 5.9

If you refer back to figure 5.2 (b) and figure 5.3 you will notice that they are identical. The only difference being is that IL and IC have been subtracted vectorially to yield IL IC. The reason for this is that they are 180º apart and cannot be subtracted algebraically.

80


A new concept namely impedance now comes to the fore. This impedance of the network may be defined in this instance (parallel RLC-networks) as the opposition offered to the flow of an alternating current and is the vector sum of the reactance of the capacitor and the resistance of the resistor and is mathematically given by: Z = 1 where Z 2 ½ [1 + (1 ~ 1 ) ] R [R (XC XL ) ] XC XL

= = = =

impedance of the network in ohm resistance of resistor in ohm reactance of capacitor in ohm reactance of inductor in ohm

V Z = S where Vs = supply voltage in volt I T IT = total current in ampere The total current expression can be derived by applying Pythagoras using figure 5.5 (b) and is mathematically given by:

where IT IT = [IR2 + (IC ~ IL)2]½ or IR IT = [IR2 + IX2]½ IC IL IX where IC =

VS XC

= = = = =

total current in ampere current through resistor in ampere current through capacitor in ampere current through inductor in ampere difference between IC and IL

V V and IR = S and IL = S R X L

The current in the network is mathematically given by: V IT = S where VS = supply voltage in volt Z Z = impedance of the network in ohm IT = current flow in circuit in ampere

You will further notice that there is a phase angle º between the current flowing in the network and the supply voltage. This is termed the phase angle and is mathematically given by: θ = tan-1 IX or θ = cos-1 IR where IT IR

IC = current through capacitor in ampere IR = current through resistor in ampere IT = total current in network in ampere IX = difference between IC and IL


Example 5.3 A parallel RLC-network has the following components connected across a 150V/150 Hz alternating current supply: • Resistor = 12 ohm • Capacitor = 70 µF • Inductor = 15 mH Determine the: (a) Inductive reactance of the inductor; (b) Capacitive reactance of the capacitor; (c) Current flowing through the: (i) Inductor; (ii) Resistor; and (iii) Capacitor. (d) Total current flowing in the network; (e) The phase angle between the supply voltage and the line current (also indicate whether the network is capacitive or inductive and motivate your answer); and (f) Impedance of the network. (g) Draw a current phasor that will represent the quantities you have calculated above and insert all relevant values.

Solution: (a) XL

= = =

2×π×f×L 2 × 3,142 × 150 × 15 × 10-3 14,139 ohm

1 (b) XC = 2×π×f×C 1 = 2 × 3,142 × 150 × 70 × 10-6

=

15,155 ohm

V V V (c)(i) IL = S (ii)IR = S (iii)IC = S X R XC L 150 150 150 = = = 14,139 12 15,155 = 10, 61 ampere = 12, 5 ampere = 9,897 ampere

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82


(d) IT = [IR2 + (IL2 ~ IC2)]½ = 12,52 + (10,61 - 9,897)2]½ = 12,52 ampere or θ = cos-1 IR (e) θ = tan-1 IX IT IR = tan-1 0,713 = cos-1 12,5 12,5 12,52 = 3,2º = 3, 2º The phase angle will indicate that the network is inductive and the line current is lagging behind the supply voltage. (f) Z

=

1 Z = VS [ 1 + ( 1 ~ 1 )2] IT [R2 (XL XC ) ] = 150 = 1 12, 52 = 11,98 ohm [ 1 + ( 1 - 1 ) 2]½ [122 (14,139 15,155) ] = 11, 98 ohm

(g) IC = 9,897 A

VS = 150 V 3,2º

IL - IC IL = 10,61 A

IR =12,5 A

IT = 12,52 A


83

5.4 Resonance There that there is no difference between series- and parallel resonance and we can therefore use similar expressions and graphic representations as was illustrated for series resonance.

5.5 Conditions for resonance The following conditions will exist for resonance in a parallel RLC-network. • XL = • θ = • IL = • IT = • Z = • I is minimum • fr =

XC 0º IC and is maximum IR is minimum 1 2 × π × (L × C)½

The following three graphs in figure 5.10 (a), (b) and (c) shows two of the above conditions namely impedance versus frequency (a) and frequency versus current (b) whereas (c) illustrates the phasor diagram for a parallel resonant network. R I m p e d a n c e

XC > XL

R XL > XC

RC

C u r r e n t

RL

IC

XL > XC

XC > XL

VR IT and IR RC

RL

f

f

fr

fr

(a)

(b)

IL (c)

Figure 5.10

5.6 The tuned network (tank network) A more practical parallel network is illustrated in figure 5.11 and is also termed a ‘tuned network’ or a ‘tank circuit’.

84


R

L

IR IC

IT

C VS

Figure 5.11

It is generally accepted that a capacitor has no ‘dc resistance’ and is considered as a pure component. The inductor on the other hand contains ‘dc resistance’ since the inductor is wound with a conductor in the shape of a coil and this conductor has a resistance. However, should the inductive reactance of the inductor be ten times greater than the resistance of the inductor the resistance can be ignored. This will however depend on the frequency at which the network is operated. It could be generally accepted that when the inductor is indicated as is illustrated in figure 5.11 with a dotted block, that the resistance of the inductor can be ignored. Important to note is that although we say that the resistance of the inductor plays no role it will only be valid for the resonant frequency and it will play a role in the dynamic impedance and hence the total current that will be produced in the network. The following mathematical expressions are applicable for parallel resonant RLC-networks: 1 where fr = ½ 2 × π × (L × C)

f π L C

= frequency in Hertz = 3,142 = value of the inductor in Henry = value of the capacitor in Farad

V V and IC = S IL = S X X L C

where IL IC VS XL XC

= = = = =

current through the inductor in ampere current through the capacitor in ampere supply voltage in volt inductive reactance of the inductor in ohm capacitive reactance of the capacitor in ohm

L ZD = where ZD R×C R C L

= = = =

dynamic impedance of the network in ohm value of the resistance in ohm value of the capacitor in Farad value of the inductor in Henry


V IT = S where IT = total current in the network in ampere ZD VS = supply voltage in volt ZD = dynamic impedance of the network in ohm

Example 5.4 A parallel resonant tuned network consists of an inductor of 25 mH and a negligible resistance of 3 ohm and a capacitor of 100 µF connected across a 120V alternating supply. Use the given data to determine the: (a) Resonant frequency of the network; (b) Inductive reactance of the inductor; (c) Capacitive reactance of the capacitor; (d) Current through the inductor; (e) Current through the capacitor; (f) Dynamic impedance of the network; and (g) Total current produced by the network.

Solution: (a) 1 fr = 2 × π × (L × C)½ 1 = 2 × 3,142 × (25 × 10-3 × 100 × 10-6)½ = 100,645 Hz (b) XL = 2 × π × f × L = 2 × 3,142 × 100,645 × 25 ×10-3 = 15,811 ohm 1 = (c) XC 2×π×f×C 1 = 2 × 3,142 × 100 × 100 × 10-6

=

15,811 ohm

The question may now arise why these two values are the same? The reason for that is that the network is at resonance and at the resonant frequency XL = XC!

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86


V (d) IL = S X L 120 = 15,811 = 7,589 ampere V (e) IC = S X C 120 = 15,811 = 7,589 ampere

Note that the two current are also the same for the same motivation as given above! L (f) ZD = R×C 25 × 10-3 = 3 × 100 × 10-6

=

83,333 ohm

V (g) IT = S Z D 120 = 83,333 = 1, 44 ampere

Exercise 5.1 1. A parallel RL-network has the following components connected across a 200V/120 Hz alternating supply: • inductor = 60 mH • resistor = 15 ohm Use the given values and calculate the: 1.1 Impedance of the network; 1.2 Total current flowing in the network; and 1.3 Phase angle between the supply voltage and line current. 1.4 Draw a phasor diagram that will represent the quantities you calculated above and insert all relevant values.


2.

A parallel RC-network has the following components connected across a 150V/100 Hz alternating supply: • Capacitor = 60 µF • Resistor = 10 ohm Use the given values and calculate the: 2.1 Impedance of the network; 2.2 Total current flowing in the network; and 2.3 Phase angle between the supply voltage and line current. 2.4 Draw a phasor diagram that will represent the quantities you calculated above and insert all relevant values. 3.

Name seven conditions that may exist for RLC-parallel resonance.

4. Draw neat labelled graphs that will illustrate: 4.1 Current versus frequency; and 4.2 Frequency versus impedance for a parallel RLC resonant network. 5.

Consider the following network diagram and incomplete phasor diagram. R 4A C 4A L 6A IT VS

Complete the phasor diagram using the available data and then determine the total current flowing in the network.

6. Complete the following expressions with reference to a series resonant RLC-network. (i) XL = (ii) θ = (iii) VL = (iv) Z = (v) I = 7.

A parallel resonant tuned network consists of an inductor of 75 mH and a negligible resistance of 10 ohms and a capacitor of 150 µF connected across a 200V alternating supply. Use the given data to determine the:

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88


7.1 7.2 7.3 7.4 7.5 7.6 7.7

Resonant frequency of the network; Inductive reactance of the inductor; Capacitive reactance of the capacitor; Current through the inductor; Current through the capacitor; Dynamic impedance of the network; and Total current produced by the network.

8.

Consider the following network. R

L

10 ohm

80 mH

C

8.1 8.2 8.3

VS

Find the value of the variable capacitor should the dynamic impedance be equal to 800 ohm. Determine the value of the capacitor should the network resonate at 10 kHz. What will the phase angle between the supply voltage and line current be at 10 kHz?

89 Q-factor, Bandwidth and Complex Notation N4 Industrial Electronics

CHAPTER

6 Q-factor, Bandwidth and Complex Notation Learning Outcomes On completion of this module you will be able to: • • • • •

Define and do relevant calculations for Q-factor; Explain graphically the concept of bandwidth; Do relevant calculations to determine bandwidth; Explain the concept of complex notation and reasons why it is used; Perform by means of complex notation: – Addition; – Multiplication; and – Division. • Solve for series RLC-networks using complex notation; • Solve for parallel RLC-networks using complex notation;

6.1 Introduction In the previous two modules we had a look at series- and parallel networks as well as resonance for these two types of networks. There is however a number of other concepts regarding alternating current networks that need to be studied.

6.2 The Q-factor The Q-factor of a network is also termed the ‘magnification factor’ and is applicable to either a series- or parallel resonant network. This factor is mathematically expressed by: X 1 Q = L or Q = (L/C) ½ R R where X L = inductive reactance of inductor in ohm R = resistive value of resistor in ohm L = value of inductor in Henry C = value of capacitor in Farad

90

Q-factor, Bandwidth and Complex Notation

Since XL = XC at resonance it is obvious that: Q = XC or XL R R It will be found that since this phenomenon is termed the magnification factor that the following conditions will apply for series or parallel resonant networks: Series networks:

The voltage across the inductor and capacitor is mathematically given by:

VL or VC where VS

= VS × Q = value of supply voltage in volt

Parallel networks:

The current through the inductor and capacitor is mathematically given by:

IL or IC where IT

= IT × Q = value of total line current in ampere

6.3 Bandwidth The graphic representation in figure 6.1 illustrates the concept of bandwidth. V or I 0 dB

- 3 dB

0,707 VM or IM

f f1

fr Bandwidth

Figure 6.1

f2


91

Definition 6.1 Bandwidth

The bandwidth may be defined as that range of frequencies between f1 and f2 where the power has fallen or dropped to half its value. Assuming absolutely pure components (L and C), this bandwidth will lie symmetrically around the resonant frequency fr. Also of importance is that the bandwidth range of frequencies will have a current or voltage value which will be equal to 0,707 × VM or IM. In order to prove that this point where the bandwidth is taken is also termed the halfpower points or –3 dB points we need to substantiate this by means of the following explanation. Assume an amplifier has an output of 1 watt and we reduce the power by half to 0,5 watt. We know that: P dB = 10 log O Pi = 10 log 0, 5 1 = - 3,010

The bandwidth of a network is mathematically given by: BW =

R fr or BW = 2×π×L Q

A number of activities will now be given in order to explain these concepts. There are a number of approaches that may be used and you are requested to experiment with all approaches in solving the problems.

Example 6.1 A series RLC-network has the following component values and is connected across an alternating current supply. = 100 ohm R L = 0,5 H C = 40 µF Use the above data to determine the lower- and upper half-power frequencies.

92


Solution:

1 R fr = BW = 2 × π × (L × C)½ 2×π×L 100 = 1 2 × 3,142 × 0,5 = 2 × 3,142 × (0, 5 × 40 × 10-6)½ = 35,584 Hz = 31,826 Hz

Since we assume that the curve is symmetrical around the resonant frequency fr we can now calculate the lower- and upper half-power frequencies in the following manner: BW BW Lower frequency = fr Upper frequency = fr + 2 2 31,826 31,826 = 35,584 = 35,584 + 2 2 = 19,667 Hz = 51,493 Hz

Example 6.2

A series RLC-network has the following component values and is connected across an alternating current supply. R = 3 ohm; L = 0, 2 mH; C = 0,323 µF Use the above data to determine the bandwidth of the network if it is given that the network resonates at a frequency of 19, 802 kHz

Solution:

R BW = 2×π×L 3 = 2 × 3,142 × 0, 2 ×10-3

=

2,387 kHz


A second approach, somewhat more complex, will yield the same result. BW = fr Q But Q = XL R = 2 × π × f × L But XL Q = 2 × π × f × L R BW = fr 2×π×f×L R = fr × R 2×π×f×L = 19,803 × 103 × 3 2 × 3,142 × 19,802 × 103 × 0,2 × 10-3 = 2,387 kHz

Example 6.3

A series network consists of a resistance having a value of 30 ohm, an inductor having a value of 10 mH and a capacitor that has a capacitive reactance of 300 ohm. This combination is connected across an alternating supply of 30V and resonates at a frequency of 4,774 kHz. Determine the: (a) Q-factor of the network; (b) Bandwidth of the network; and (c) The voltage drop across the inductor and the capacitor.

Solution: XL (a) Q = R 300 = 30 = 10

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94


f (b) BW = r Q 4,774 × 103 = 10

=

477, 4 Hz

(c) VL = VS × Q VC = VS × Q = 30 × 10 = 30 × 10 = 300 volt = 300 volt

Example 6.4

A parallel resonant network resonates at an unknown frequency and consists of a 10 ohm resistor, a 30 mH inductor and a 100 µF capacitor. Determine the Q-factor of this network.

Solution: Q = 1 (L/C) ½ R = 1 (30 × 10-3 / 100 × 10-6)½ 10 = 1,732

Example 6.5 Consider the following RLC-network connected across a 600V alternating supply resonating at 2,387 kHz. 6A

R 600 V 2,387 kHz

100 ohms

L 10 mH

C 150 ohms


95

Use the above information and calculate the: (a) Q-factor of the network; (b) Lower- and upper half-power frequencies; (c) Bandwidth; (d) Current through the inductor and the capacitor; and; (e) Value of the capacitor.

Solution: XL (a) Q = R 150 = 100 = 1,5

f (b) BW = r Q = 2,387 × 103 1,5 = 1,591 kHz (c) Lf = = =

fr - BW Uf = fr + BW 2 2 = 2,387 × 103 + 1,591 × 103 2,387 × 103 - 1,591 × 103 2 2 1,592 kHz = 3,183 kHz

IC = IT × Q (d) IL = IT × Q = 6 × 1,5 = 6 × 1,5 = 9 ampere = 9 ampere (e) XC C

= = = =

½ ×π×f×C ½ × π × f × XC ½ × 3,142 × 2,387 × 103 × 150 444,45 nF

6.4 Complex notation Complex notation is a method used to calculate different quantities in alternating current networks in modulus and angle form which gives us a much easier method of calculation. Complex notation is very often also referred to as the ‘j’ notation.

96


The most common forms to be used include: α ± jβ referred to as the rectangular form and r/θº which is termed the polar form. Consider the following two networks as illustrated in figure 6.2 (a) and (b). -j R

R

j

Z

Z

(a)

(b)

Figure 6.2

According to figure 6.2 (a):

According to figure 6.2 (b):

Z = α + jβ (rectangular) Z = r/θº (polar)

= =

α – jβ (rectangular) r/θº (polar)

Rectangular to polar

Polar to rectangular

r2 = α2 + β2 θ = tan-1 jβ α

α jβ

(Moduli) (Arguments)

= =

r cos θ (Real value) r sin θ (Imaginary value)

Although the conversion can be done this way most modern pocket calculator makes provision for these conversions. All addition or subtractions of quantities are done using the rectangular form whereas division and multiplication are done using the polar form. •

Addition (Z1 + Z2)

Z1 + Z2 = (α1 + jβ1) + (α2 + jβ2) = (α1 + α2) + (jβ1 + jβ2)

All real values are added or subtracted whatever the case may be and all imaginary values are added or subtracted whatever the case may be.

α = jβ =

The sign of the ‘j’ term must always be taken into account.

real values imaginary values


•

97

Multiplication (Z1 × Z2)

Z1 × Z2 = r1/θ1º × r2/θ2º = r1 × r2/ θ1º + θ2º

All moduli values are multiplied and all argument values are added.

•

Division (Z1 ÷ Z2)

Z1 ÷ Z2 = r1/θ1º ÷ r2/θ2º = r1 ÷ r2/ θ1º - θ2º All moduli values are divided and all argument values are subtracted.

6.5 RLC-networks 6.5.1 Series RLC-networks Consider the network in figure 6.3 R

L

α

j

VR

VL

C -j VC

I VS

Figure 6.3

The impedance of the network is mathematically given by: Z = α ± jβ (rectangular form) = r/θº (polar form) where α = R jβ = reactance where

XL XC

V I = S where Z

VS = r/θº Z = r/θº

= =

j(2 × π × f × L) - j(1/2 × π × f × C)

You will notice that r/θº indicates the angle of the current in relation to the supply voltage and we can therefore draw a phasor diagram with much greater ease as was the case earlier.

98


Example 6.6 A series RLC-network has the following component values connected across a 250/θºV/50 Hz alternating quantity. R = 20 ohm; L = 150 mH; C = 100 µF Use this information to: (a) Calculate the impedance of the network; (b) Determine the current that will flow in the network; and (c) Determine the voltage drop across each component in the network. (d) Draw a phasor diagram indication all the relevant values calculated.

Solution: XL

= = = =

j(2 × π × f × L) XC j2 × 3,142 × 50 × 150 × 10-3 j47, 13 ohm 47, 13/90º ohm

= = = =

- j(1/2 × π × f ×C) - j(1/2 × 3,142 × 50 × 100 ×10-6) - j31, 83 ohm 31, 83/-90º ohm

(a) Z = α ± jβ = 20 + j47,13 - j31,83 = 20 + j15,3 ohm = 25,18/37,42º ohm V (b) I = S Z 250/θº = 25,18/37,42º = 9,929/-37,42º ampere (c)(i) VR = = =

I × R 9,929/-37,42º × 20/0º 198,58/-37,42º volt

(ii) VL = = =

I × XL 9,929/-37,42º × 47,13/90º 467,954/52,58º volt

(iii) VC = = =

I × XC 9,929/-37,42º × 31,83/-90º 316,04/-127,42º volt


(d) VL = 467,954/52, 58º V

VL - VC

VS = 250/θº V I =9,929 /-37, 42º A VC = 316, 04/-127, 42º V VR =198, 58/- 37, 42º

6.5.2 Parallel RLC-networks Example 6.7 The following parallel RLC-network is given. R

L

10 ohms

J 8 ohms

C - j 2 ohms 250 V

Determine with reference to the network above the: (a) Impedance of the network; and (b) Current that will flow in the network. (c) Use the information obtained in above and draw a phasor diagram indicating all relevant values.

Solution: (a) Z1

= α + jβ Z2 = 10 + j8 ohm = 12,8/38,66º ohm

= = =

α - jβ 0 - j2 ohm 2/90º ohm

99

100


Z × Z2 ZT = 1 Z + Z2 1 12,8/38,66º × 2/-90º = 10 + j8 + 0 - j2 25,6/-51,34º = 10 + j6 25,6/-51,34º = 11,66/ 0,96º = 2,196/-82, 3º ohm V (b) I = S Z 250/θº = 2,196/-82,3º = 113,843/82,3º ampere

(c)

I = 113,843/82, 3º A

VS = 250/θº V

Example 6.8 The following RLC-network is given. C

R

L

- j 2 ohms

5 ohms

j 20 ohms

L j 10 ohms

A

ZOUT

B

Determine the impedance ZOUT that will be measured across terminals AB and leave the answer you obtain in rectangular form.


Solution: Z1 = α + jβ Z2 = α - jβ Z3 = α + jβ = 5 + j20 ohm = 0 - j2 ohm = 0 + j10 ohm = 20,615/75,96º ohm = 2/-90º ohm = 10/90º ohm Z × Z3 ZOUT = Z1 + 2 Z2 + Z3 5 + j 20 + 2/-90º × 10/90º = 0 - j 2 + 0 + j 10 5 + j 20 + 20/0º = 0 + j8 5 + j 20 + 20/0º = 8/90º = 5 + j 20 + 2, 5/-90º = 5 + j 20 + 0 - j 2, 5 = 5 + j17, 5 ohm

Example 6.9 An RLC-network is constructed in the following manner. Z1 = 5 - j2 ohm; Z2 = 10 + j10 ohm; Z3 = 2 - j8 ohm. Z1 and Z2 are connected in parallel and this parallel combination is connected in series with Z3 across an alternating quantity of 100V. (a) Make a neat labelled network diagram of this combination and indicate all relevant values. (b) Calculate the: (i) Impedance of the network; (ii) Current that will flow in the network; and (iii) The voltage drop across the parallel section of the network. (c) Draw a neat labelled phasor diagram that will illustrate the relationship between the applied voltage and the current in the network.

101

102


Solution: (a) Z1 R C

5 ohms

- j 2 ohms

R

R

L

2 ohms

10 ohms

j 10 ohms

- j 8 ohms Z3

Z2 100 V

Z × Z2 + Z3 (b)(i) ZT = 1 Z + Z2 1 5 - j2 × 10 + j 10 = + 2 - j8 5 - j2 + 10 + j 10 5,385/-21, 8º × 14,142/45º = +2-j8 15 + j 8 76,155/23, 2º = +2-j8 17/28º = 4,479/-4,8º + 2 - j 8 = 4,463 - j0, 375 + 2 - j8 = 6,463 - j 8,375 = 10,578/-52,34º ohms V (c) I = S VZ1Z2 Z 100/θº = 10,578/-52,34º = 9,455/52,34º ampere

(d)

= I × Z Z1Z2 = 9,455/52,34º × 4,497/-4,8 º = 42,35/47,54º volt

I = 9,455 / 52,34º A

VS = 100 / θº V


Exercise 6.1 1. Define and then graphically represent the concept of bandwidth. 2. A series RLC-network has the following component values and is connected across an alternating current supply. R = 200 ohm; L = 0, 15 H; C = 60µF Use the above data to determine the lower- and upper half-power frequencies. 3.

A series network consists of a resistance having a value of 20 ohm, an inductor having a value of 15 mH and a capacitor that has a capacitive reactance of 400 ohm. This combination is connected across an alternating supply of 30V and resonates at a frequency of 4,774 kHz. Determine the: 3.1 Q-factor of the network; 3.2 Bandwidth of the network; and 3.3 The voltage drop across the inductor and the capacitor.

4.

Two impedances are given as (α1 + jβ1) and (α2 + jβ2). Explain how these values are converted to the polar form and then give the rules for: 4.1 Division; 4.2 Addition; and 4.3 Multiplication.

5.

A series RLC-network has the following component values connected across a 250/θº V/50 Hz alternating quantity. R = 20 ohm; L = 150 mH; C = 100 µF Use this information to: 5.1 Calculate the impedance of the network; 5.2 Determine the current that will flow in the network; and 5.3 Determine the voltage drop across each component in the network. 5.4 Draw a phasor diagram indication all the relevant values calculated.

6. An RLC-network has the following impedance values: Z1 = 6 - j2 ohm; Z2 = 8 + j4 ohm; Z3 = 2 + j6 ohm. Z1 and Z2 are connected in parallel and this parallel combination is connected in series with Z3 across an alternating quantity of 100V. 6.1 Make a neat labelled network diagram of this combination and indicate all relevant values.

103

104


6.2 Calculate the: 6.2.1 Impedance of the network; 6.2.2 Current that will flow in the network; and 6.2.3 The voltage drop across the parallel section of the network. 6.3 Draw a neat labelled phasor diagram that will illustrate the relationship between the applied voltage and the current in the network. 7. An alternating current network consists of the following components: 100 – j 26,25 ohm • Z1 = 120 + j 29,35 ohm • Z2 = 100 + j 35,85 ohm • Z3 = Z1 and Z2 are connected in parallel and then in series with Z3 across a 220V/50 Hz alternating current supply. Determine from this information: 7.1 The impedance of the parallel section; 7.2 The total impedance of the network; 7.3 The total current flow in the network; 7.4 The voltage that will be measured across the parallel section; and 7.5 The current that will flow in each branch of the parallel section.

BASIC ATOMIC THEORY


CHAPTER

7 Basic Atomic Theory Learning Outcomes On completion of this module you will be able to: • • •

• • • • • •

•

Describe and identify the nature of matter by naming forms and examples; Ascertain atomic weight and number of elements by using the Periodic Table of Elements; Define: – Compounds; – A molecule; and – An atom. Draw a two dimensional layout of any atom indicating all the particles as well as major- and sub-shells; Determine the number of particles in the major- and sub-shells; Discuss or explain the term valency and valency shell; Explain a covalent bond by means of a suitable sketch; Discuss the difference between electron flow and hole flow by means of an applicable sketch; Understand and be able to explain by means of simple sketches the energy bands which exist in atoms and the placing of these energy bands in: – A conductor; – An insulator; and – A semi-conductor. Define the electron volt.

7.1 The structure of matter

Definition 7.1 Matter

Matter may be defined as anything that has mass and that occupies space and can be composed of elementary substances that are found in nature.

105

106

BASIC ATOMIC THEORY

These compositions may be in pure form or it may be a compound or mixture of elements found in nature. Matter cannot be generated or destroyed but it may change from one form into another. Matter can be divided into the following groups: • • • •

Solids; Liquids; Gasses; and Plasma.

Examples of such elements found are: • • • • •

Iron (Fe) Silver (Ag) Carbon (C) Gold (Au) Aluminum (Al)

Classification of all elements in nature, are found in the Periodic Table of Elements and are classified by its atomic number and atomic weight. The atomic number of an element is given by the number of electrons orbiting the atom whereas the atomic weight is given by the number of protons and neutrons contained in the nucleus of an atom. All elements consist only of one type of atom as is indicated by the Periodic Table of Elements. Combining of different elements may form compounds by chemical reaction with one another and the most common one would be water. Water is a compound of Oxygen and Hydrogen.

Definition 7.2 Molecule

A molecule can be defined as the smallest part of an atom that will have the same characteristics of the original element without breaking up into atoms again.

7.2 Atoms

Definition 7.3 Atoms

An atom may be defined as the smallest part of an element that can participate in a normal chemical reaction.


107

All atoms consist of minute particles of electrical charges arranged in a set pattern and consist of: • • •

Electrons, moving around the nucleus of an atom and that has a negative electrical charge of 1,602 × 10-19 coulomb and are kept in its orbit by electrostatic forces of attraction; Protons in the nucleus of an atom that have a positive electrical charge; and Neutrons in the nucleus of an atom that have no electrical charge.

It will be found that all atoms have an equal number of protons and electrons, which indicates its atomic number, which makes all atoms electrically neutral. All electrons, protons and neutrons are identical and it is possible that they be replaced in other atoms that are not of the same element. A point of interest is that it is very easy for an atom to loose or gain outer orbit or valence electrons. Therefore, should an atom loose an electron it will loose some of its negative electrical charge and in the process become positively charged! This is then known as a positive ion or cation. The opposite however is also true in that when an atom gains an electron it will increase its negative electrical charge and will then be known as a negative ion or anion. The only difference between atoms is a numerical difference. Figure 7.1 illustrates a two dimensional layout of an atom.

Negatively charged orbiting electrons Protons (+) and neutrons (no charge)

K - shell L - shell Nucleus

Figure 7.1

7.3 Energy Shells All electrons are arranged around the nucleus of an atom in a pre-determined manner as mentioned earlier but we should however have a closer look at these arrangements. Together with figure 7.1 we should also refer to figure 7.2 to find out how the arrangement of the electrons, protons and neutrons in an atom is accomplished.

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BASIC ATOMIC THEORY

Letter designation of major shells M L K S

SP

SPD

Letter designation of sub - shells

2 8 18 Number of orbiting electrons per shell

Figure 7.2

In the representation of figure 7.2 the centre circle will represent the nucleus consisting of the protons and neutrons. The outer circle or circles will indicate the shells for the orbiting electrons. Although these representations are two dimensional, the shape of a tennis ball can almost represent an atom in that the major shells of an atom are also sub-divided into sub-shells. This concept is illustrated in figure 7.2. The major shells of an atom are designated the K, L M and N shells and the sub-shells are designated the S, P, D and F shells. Each shell, major or sub-shell, may only have a pre-determined number of electrons per shell and the number of electrons in the major shells is determined by the expression 2.n2 where n will represent the orbit number counting from the inner circle toward the outer circle. The number of electrons per major shell can thus be calculated in the following manner. K shell: 2.n2 = 2 ×12 L shell: 2.n2 = 2 × 22 M shell: 2.n2 = 2 × 32 N shell: 2.n2 = 2 × 42

= = = =

2 Thus, the K shell may only contain 2 electrons. 8 Thus, the L shell may only contain 8 electrons. 18 Thus, the M shell may only contain 18 electrons. 32 Thus, the N shell may only contain 32 electrons.

Since the number of electrons per major shell can be determined and predicted it follows that the number of electrons per sub-shell can also be determined and predicted. The number of electrons per sub-shell is given as follows: S shell P shell D shell F shell

= = = =

2 electrons 6 electrons 10 electrons 14 electrons


109

Electrons can be made to jump to other orbits by means of heat, light energy or an electric field at or near an atom. Whatever the form of energy used, it is referred to as ionization potential. This arrangement of electrons orbiting an atom will now allow one to determine the conducting properties of the element and especially the outer orbit or valence shell of an atom.

7.4 Valency The number of electrons in the outer shell of an atom, called the valence shell, will determine the valency of that element and may be defined as an indication of the ability of an atom to gain or loose electrons and will determine the electrical properties of that element. Should an atom not have the full complement of electrons in the valence shell, that element is deemed to be a conductor since that particular atom will always be on the lookout for electrons to fill the vacancies, which exist. Should an atom however have the full compliment of electrons in the valence shell it will not need to look for electrons to fill up vacancies since no vacancies exist and is then deemed to be an insulator. The above statements are not always true. A conductor may also be an insulator under certain specific conditions. Two or more atoms may combine to form a covalent bond. The best manner in which to explain this concept is to look at two hydrogen atoms as illustrated in figure 7.3. This is not a compound that will be formed but is the sharing of electrons by two or more nuclei of atoms. In this instance the only shell of electrons will contain two electrons, which make that particular shell complete. This phenomenon is particularly important in the semi-conductor field and will be explained in more detail when semi-conductors are discussed. Therefore, we will find that there are no free electrons available since each atom shares its electrons with one another and that they are firmly attached in a crystal lattice structure by this covalent bond. Orbiting electrons

Hydrogen atom 1

Hydrogen atom 2

Figure 7.3

7.5 Conduction A very important phenomena concerning conduction should however also be explained. In the incomplete valence shell of an element we refer to the spaces, which exist where, electrons should be, as holes. Strange as it may sound we will find that these holes also appear to be moving. This concept is illustrated in figure 7.4.

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BASIC ATOMIC THEORY

Movement of electrons or conduction can and will take place in any given conducting material, in a desired direction, should a source of power be applied across such material. This conduction process can be by one or both of the following processes namely hole flow or transfer or by electron motion. Free electrons in the conduction band will, under the influence of an applied potential, move around. The reason for this action is that since the electron has a negative charge it will be repelled by the negative charge of the applied potential but at the same time be attracted by the positive charge of the applied potential. In this manner we will find that electrons will always move from a negative charge toward a positive charge. This action is termed electron flow. Should we now observe figure 7.4 in more detail, it will be seen that as the electron moves from the negative charge of the applied potential toward the positive charge of the applied potential, that the hole moves from the positive charge of the applied potential toward the negative charge of the applied potential. This action is termed hole flow and both these concepts are of great importance and we will deal with these principles again when we discuss semi-conductors. Electron motion

Hole motion

-

+

-

+

-

+

Applied potential

Figure 7.4


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7.6 Energy bands In an isolated atom the electrons will be acted upon from within. Should we now however bring a number of atoms together, as in a solid material like a conductor; the electrons of the various atoms will come under the influence of forces from the other atoms within that solid. In this manner energy levels are created and electrons will merge into energy bands of distinct energy levels. This concept is illustrated in figure 7.5.

Conduction band

Forbidden gap

Energy level

Valence band

Figure 7.5

In any given material, conducting or insulating, there are two distinct energy bands in which electrons may exist, namely the conduction band and the valence band but they will be separated by the forbidden gap. The electrons in the conduction band are drifting around in the material since they have virtually become disconnected from the atom and they are easily moved around by the application of a very small amount of energy. The electrons in the valence band however, are tightly bound around the nucleus of the atom and a large amount of energy will be required to move these electrons.

Definition 7.4 Electron Volt

The energy required to move any electron, whether it is in the conduction band or in the valence band, is defined by the electron volt (eV) and is the amount of energy required to move one electron through a potential difference of one volt.

Between the conduction band and valence band we find an area termed the forbidden gap. This forbidden gap will vary in size for different materials, conductors or insulators, and may be small, large or non-existent as is indicated in figure 7.6 (a), (b) and (c).

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BASIC ATOMIC THEORY

Conduction band

Conduction band

Conduction band

Forbidden gap

Forbidden gap

Valence band

Valence band

Valence band

( a ) Insulator

( b ) Semi-conductor

( c ) Conductor

Figure 7.6

In the conductor we find that the forbidden gap is non-existent and that the conduction band and the valence band actually overlap one another creating a large number of electrons available for conduction. Very little energy is required for conduction to take place and will even conduct at the slightest variation in temperature. In the insulator we find a fairly large forbidden gap between the conduction band and the valence band and no electrons are found in the conduction band. Since all the electrons are found in the valence band only, which is tightly bound to the nucleus, a large amount of energy, in the region of 6 eV, is needed to cause an electron to move from the valence band to the conduction band. The third material shown is that of a semi-conductor and has a fairly narrow forbidden gap and requires less energy for conduction to take place. Here we have to differentiate between the two types of semi-conductor materials commonly used in semi-conductor manufacture namely, Silicon (Si) and Germanium (Ge). Typical energy levels required to cause conduction is 0,7 eV for Germanium and 1,19 eV for Silicon. Since this is termed a semi-conductor it is neither a conductor nor an insulator and has a typical resistance of 10 ohm/cm3. The forbidden gap will therefore determine whether a material can be termed as a conductor, insulator or semi-conductor.


Exercise 7.1 1. Name the four groups into which matter can be divided. 2. Define the following terms: 2.1 Molecules; and 2.2 An atom. 3. Indicate the charges that are found on the particles of an atom. 4. An atom contains 54 electrons. Calculate the number of electrons per major orbit and then draw a two dimensional layout indicating all relevant information. 5. The term valency will determine the behaviour of an element in that it may be a conductor or it may be an insulator. Express your opinion on this statement. 6. Explain the concept of a covalent bond and illustrate your explanation with a suitable labeled sketch. 7.

Use suitable sketches to illustrate the principle of conduction and differentiate between: 7.1 Electron motion; and 7.2 Hole motion.

8.

The conduction band and the valence band are separated by the forbidden gap in any given material and it makes no difference whether the material is a conductor or an insulator. Draw three neat labeled sketches that will illustrate this concept for: 8.1 An insulator; 8.2 A semi-conductor; and 8.3 A conductor.

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114

BASIC ATOMIC THEORY

PN–Junction Theory


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CHAPTER

8 PN–Junction Theory Learning Outcomes On completion of this module you will be able to: • • • • • •

Represent an atom as a crystal lattice structure; Use crystal lattice structures to explain the concept of: – Donor doping; and – Acceptor doping. Clarify the concept of the Fermi-levels; Explain using suitable sketches the formation of a PN-junction; Explain the principle of diffusion current and the formation of the depletion layer; and Explain using suitable sketches the following concepts of a PN-junction: – No bias; – Reverse bias; and – Forward bias.

8.1 Introduction In the field of electronics there are two main elements that are used in the manufacture of semi-conductor devices or components namely Silicon and Germanium. Although these two elements do not have the same atomic weight or number of orbiting electrons they do however have one aspect in common: there are four valence electrons in its valence shell although the orbit designation is different. In order to understand how the conduction will take place in semi-conductor devices you are advised to sharpen up on you’re already gained knowledge of Atomic Theory. As the name Semi-conductor suggests, it is not a very good conductor and something needs to be done in order to improve on its conducting capabilities. This process of improvement in conducting capabilities is termed doping but can be best explained by considering the elements (germanium and silicon) as a crystal lattice structure.

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PN–junction theory

8.2 Crystal lattice structures We will only deal with the two most common semi-conductor manufacturing elements namely Germanium and Silicon. Since all elements are electrically neutral we will only consider the valence electrons of each element since they have the same number of valence electrons. A simplified illustration of the crystal lattice structure of Germanium and Silicon is depicted in figure 8.1. This illustration is interpreted as follows: the circle in the centre represents the nucleus of the atom as well as the electrons contained in the electrons orbits or shells that have their full compliment of electrons for those shells. The squares indicate the valence electrons in the valence shell. Although only four atoms are indicated here you must realise and accept that there are many more atoms in a very small piece of Germanium or Silicon. Covalent bonding will now take place among adjoining atoms at random. This type of crystal lattice structure is found in all crystalline elements. Germanium and Silicon in its pure form is also known as an intrinsic material and as already mentioned does not have very good conducting properties. In order to improve the conducting properties the Germanium or Silicon is mixed with other elements in a manufacturing process termed doping.

Orbiting electrons of four nuclei (same for silicon and germanium)

Figure 8.1

8.3 Donor doping Donor doping is a mixing process that will generate a free (extra) electron in the conduction band of the atom as well as crystal lattice structure. The number of such free electrons will depend on the extent to which this doping is performed. The concept of


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donor doping is illustrated in figure 8.2. Plainly said, how much impurities are added to the Germanium or Silicon. The impurity added to the Germanium or Silicon has five valence electrons in its valence shell and is therefore referred to as Penta-valent atoms. Penta- is the Greek word for five. Typical elements used for donor doping include: • Antimony • Arsenic • Phosphorous

Penta-valent impurity atom

Orbiting electrons of three nuclei (same for silicon and germanium)

Extra free electron

Figure 8.2

Since covalent bonds are formed in the crystalline structure the impurities form part of this covalent bond and create an extra electron for conduction purposes for each impurity atom added and will enter the conduction band of the atom as a free electron. Donor doping creates what is termed an N-type material and this material is negatively charged since the electron number has increased by the introduction of the impurity atoms. This material is therefore no longer an intrinsic material but is referred to as an extrinsic material. What this process has now brought about is an extra or free electron which will contribute to the better conducting properties of the material.

8.4 Acceptor doping Acceptor doping is a mixing process that will generate a hole in the conduction band of the atom as well as crystal lattice structure. The number of such holes will depend on the extent to which this doping is performed. Plainly said, how much impurities are added to the Germanium or Silicon.

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The impurity added to the Germanium or Silicon has three valence electrons in its valence shell and is therefore referred to as tri-valent atoms. Tri- is the Greek word for three. Typical elements used for acceptor doping include: • Gallium • Boron • Aluminium The concept of acceptor doping is illustrated in figure 8.3.

Tri-valent impurity atom

Orbiting electrons of three nuclei (same for silicon and germanium)

Hole

Figure 8.3

Since covalent bonds are formed in the crystalline structure the impurities form part of this covalent bond and create a hole for conduction purposes for each impurity atom added and will enter the conduction band of the atom as a hole. Acceptor doping creates what is termed a P-type material and this material is positively charged since the electron number has decreased by the introduction of the impurity atoms. This material is therefore no longer an intrinsic material but is referred to as an extrinsic material. What this process has now brought about is a hole which will contribute to the better conducting properties of the material. Important to know is that although the pure Silicon or Germanium (intrinsic material) has been doped in order to obtain an N-type material or P-type material (extrinsic material) that the doped material is still electrically neutral in that the number of electrons will still equal the number of protons. Of further importance is the fact that in an N-type material conduction is by means of electrons as majority carrier and in a P-type material conduction is by means of holes as majority carrier. Should a suitable supply now be connected across this PN-junction a current flow will be measured.


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8.5 Fermi-levels Once the process of doping have been accomplished another interesting phenomena namely the Fermi-level comes to light which should also be investigated. It has been mentioned that when these materials are manufactured that the free electrons as well as the holes introduced move into the conduction band. This conduction band also has a band termed the valence band and is of great importance. But, the question may now arise of what this Fermi-level is?

Definition 8.1 Fermi-levels

The Fermi-level may be defined as the amount of energy the free electrons as well as the holes possess within the material.

8.5.1 The Fermi-level in an intrinsic material Since an extrinsic material is electrically neutral we find that the Fermi-level lies exactly in the middle between the conduction band and the valence band. This is brought about by the fact that the number of free electrons in the conduction band will be equal to the number of holes and the valence band and that the energy levels will decrease exponentially toward the centre between the conduction band and the valence band. This concept is illustrated in figure 8.4. Conduction band

-

-

-

-

-

-

Electron energy

Fermi-level

+

+

+

+

+

+

Hole energy

Valence band

Figure 8.4

8.5.2 The Fermi-level in an extrinsic material The position of this Fermi-level will however be influenced by the type of doping and material being obtained. This concept is illustrated in figure 8.5 (a) and (b).

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Conduction band

-

-

-

-

-

-

Conduction band Electron energy

-

-

-

-

-

-

Electron energy

Fermi-level Fermi-level

+

+

+

+

+

+

+

Hole energy

+

+

+

+

Valence band

Valence band

(a)

(b)

+

Hole energy

Figure 8.5

We find that in an N-type extrinsic material that more electrons have been added which makes the material negatively charged that there is no longer a balance between the conduction band and the valence band. Therefore the Fermi-level will move closer to the conduction band. This concept is illustrated in figure 8.5 (a). On the other hand we find the same phenomena with an extrinsic P-type material with the difference that more holes have been created which now makes the material positively charged. However, in this instance the Fermi-level will move closer to the valence band. This concept is illustrated in figure 8.5 (b).

8.5.3 The Fermi-level in a PN-junction Should a PN-junction be created by joining the two types of extrinsic material then the Fermi-level will be placed in this junction as illustrated in figure 8.6.

Conduction band

-

-

-

-

-

-

-

-

-

-

-

Electron energy

Fermi-level

Valence band

+

+

+

+

+ +

N-type

Figure 8.6

+

+

+ +

+

P-type

+ Hole energy

+


8.6 The PN-junction 8.6.1 The formation of a PN-junction A PN-junction is formed when a P-type material and an N-type material is joined together. This joining together is not an electrical junction but is a junction which is achieved through a manufacturing process. In this manner a single junction is created between the N-type and the P-type material. Electrons and holes are uniformly distributed in the two types of material provided they have been doped to the same extent. This concept is illustrated in figure 8.7. P-type Holes

+ + +

+ +

+ + +

+ +

N-type Electrons

+ + + -

-

-

-

-

Junction

Figure 8.7

During the manufacturing process this evenly distribution of electrons and holes are disturbed since some of these holes and electrons are very close to one another in the immediate vicinity of the junction. Electrons from the N-type material now cross the junction into the holes in the P-type material and holes from the P-type material also cross the junction toward the N-type material. This natural process is termed diffusion current and may be defined as a process whereby charge carriers in a region of high density will move to a region of low density until they are evenly balanced. However, this will only take place in the immediate vicinity of the junction. This results in the creation of negative ions on the P-side as well as positive ions on the N-side in the immediate vicinity of the junction. This concept is illustrated in figure 8.8.

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122


P-type Holes

+

+ + + + + + + + + +

+

N-type Electrons

+

+ + + + + -

Holes moved across

Electrons moved across Junction

Figure 8.8

This movement of charge carriers across the junction leaves on either side of the junction a region which is completely depleted or void of charge carriers and is termed the depletion layer. It is generally accepted that this depletion layer is of equal thickness around the junction provided that both extrinsic materials have been doped to the same extent. Therefore, the depletion layer on the N-side will consist of donor impurities which have lost electrons and have a positive charge and the depletion layer on the P-side will consist of donor impurities which have lost holes and have a negative charge. This concept is illustrated in figure 8.9. P-type Holes

+ + +

+ +

+ + +

+ +

N-type Electrons

-

+ + + + +

-

-

-

-

Positive charge Donor impurities Negative charge

Junction Depletion layer

Figure 8.9

The explanations discussed above constitute the behaviour of the PN-junction under no bias conditions. Bias is a method applied to semi-conductor components that will allow for it to either conduct or not to conduct.


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8.6.2 Reverse-biased junction A reverse-biased junction condition will not allow for any current to flow except for a leakage current which for all practical purposes may be ignored for our discussion. A voltage source is connected to the PN-junction in such a manner that the negative terminal of the source is connected to the P-type material of the junction and the positive terminal of the source is connected to the N-type material of the source. This concept is illustrated in figure 8.10. Junction P-type Holes

+ Negative potential

+ +

+ +

+ + +

N-type Electrons

-

+ + + + +

-

-

-

Positive potential

-

Depletion layer Depletion layer widened

Figure 8.10

Under these reverse-bias conditions no current will be flowing for the following reasons. Electrons from the N-type material will be attracted to the positive terminal of the source and holes from the P-type material will be attracted to the negative terminal of the source. The result is that the depletion layer will widen as is illustrated. Although it was mentioned earlier that under reverse-bias we have a leakage current which may be ignored it is of utmost importance that this reverse-bias not exceed a given value since the junction will then be destroyed. An interesting fact that need to be mentioned is that under reverse-bias conditions a PN-junction will act as a variable capacitor since the two doped materials are actually two conductors and since the junction is not an electrical junction the junction will act as a capacitor and the distance between the two conducting materials will be governed by the magnitude of the reverse bias.

8.6.3 Forward-biased junction A forward-biased junction condition will allow for current to flow. A voltage source is connected to the PN-junction in such a manner that the negative terminal of the source is connected to the N-type material of the junction and the positive terminal of the source is connected to the P-type material of the source. This concept is illustrated in figure 8.11.

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Junction P-type Holes

+ Positive potential

+ +

+ +

+ + +

N-type Electrons

-

+ + + + +

-

-

-

Negative potential

-

Depletion layer

Depletion layer narrowed

Figure 8.11

At a given point a voltage magnitude is reached when these charge carriers are able to cross the junction and the electrons originating from the N-type material are attracted by the positive terminal of the voltage source on the P-type material and at the same time holes originating from the P-type material are attracted toward the negative terminal of the voltage source on the N-type material. As long as this required supply voltage is maintained a current will flow for as long as required. The given points at which conduction will commence are defined as follows: • •

Germanium ± 0,3 volt Silicon ± 0,7 volt

These values are the approximate values of the supply voltage that need to be reached before any conduction will take place.


Exercise 8.1 1. Germanium and Silicon may be represented in a crystal lattice structure. Make use of a neat labelled sketch to illustrate this statement. 2.

Make use of two neat labelled crystal lattice structure sketches and explain the concept of: 2.1 Donor doping; and 2.2 Acceptor doping.

3. Differentiate between an intrinsic- and extrinsic material. 4.

Define the Fermi-level and then indicate by means of neat labelled sketches the placement of the Fermi-level in: 4.1 An intrinsic material; 4.2 An extrinsic N-type material; 4.3 An extrinsic P-type material; and 4.4 A PN-junction.

5. During the formation of a PN-junction a depletion layer is formed. Use suitable sketches and explain how this depletion layer is formed. 6. A PN-junction may either be forward- or reverse biased. Use applicable sketches to illustrate this biasing concept.

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126


Semi-conductor Diodes


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CHAPTER

9 Semi-conductor Diodes Learning Outcomes On completion of this module you will be able to: • • • • • • • • • •

Give the IEC circuit symbol and a suitable definition of a diode; Discuss the operation of a diode with the aid of a characteristic curve; Mention the basic rating factors for diodes; Explain the electrical characteristics of a diode with the aid of a specification sheet; Give the mathematical expression for a diode and explain the meaning of each symbol; Do calculations and derive conclusions using the standard expression as well as derivatives thereof; Give the mathematical expression for the forward resistance a diode and explain the meaning of each symbol; Do calculations and derive conclusions using the standard expression as well as derivatives thereof; Explain and draw a graphical representation of the diode loadline; and Explain the concept of a series limiting resistor when using diodes.

9.1 Introduction

Definition 9.1 Diode

A diode may be defined a single PN-junction two terminal device which will offer a low resistance when forward-biased and a high resistance when reverse-biased.

Therefore, should a diode be forward-biased it will act as a conductor and when it is reverse-biased it will act as an insulator (non-conductor).

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9.2 The diode and characteristic curve The IEC circuit symbol and characteristic curve for a diode is illustrated in figure 9.1 (a) and (b). Our first reference will be to figure 9.1 (a). The anode is always manufactured from a P-type material and the cathode from an N-type material. The electron flow through the diode is always against the arrow head indicated by the circuit symbol and for such electron flow (current flow) to take place the anode must always be positive with respect to the cathode. This does not necessarily mean that the anode must have a positive potential supplied to it. Remember that minus 10 volt is more positive than minus 30 volt. The characteristic curve in figure 9.1 (b) is typical of a diode and will describe its operation. The curve has two distinct areas in which it operates and these two areas have another two points which are very significant. Area A on the curve constitutes the forward bias area of the device and it is in this area that a diode is normally operated. Should a diode now be correctly forward-biased then current will flow freely for as long as this forward-bias is maintained. Point B which is also in the forward-bias area is termed the forward break over point and it is at this point that the diode starts conducting freely. This point differs between Germanium and Silicon diodes and is illustrated in figure 9.2 and has the following approximate values and only the forward characteristics are indicated in order to differentiate between the two types of diodes: • Germanium ± 0,3 volt • Silicon ± 0,6 volt This value also indicates the voltage drop across the diode during conduction. I A Forward bias area

Anode (P-type) Cathode (N-type)

IF VR

Electron motion (a)

-V

V VF

C IR

D Reverse bias area

-I (b)

Figure 9.1

B


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Point C (figure 9.1 (b) is probably the most important point on the characteristic curve and also falls in the reverse-bias area of the diode and is termed the reverse break down point or the peak-inverse-voltage point and this point is specified by the manufacturer and will differ between diodes. Should this point be reached the diode will be destroyed. Area D (figure 9.1 (b) on the curve constitutes the reverse-bias area of the device and the diode is normally not operated in this area. Should a diode be reverse-biased no current will flow except a small reverse current which may be ignored. I (mA)

20

Germanium

10

Silicon

Q-point Q-point V 1

2

Figure 9.2

9.3 Basic rating factors of a diode Rating factors will assist during the design phase of circuits so that reliable as well as satisfactory operation can be assured. Range Rating Low current Medium current High current

Up to 49 ampere 50 ampere to 199 ampere 200 ampere and higher

All these rating are supplied in the manufacturers’ specification sheets and are based on the absolute maximum rating which must never be exceeded.

9.4 Electrical characteristics of a diode The electrical characteristics of a diode are based on the absolute maximum rating system and provide information pertaining to the maximum values that may not be exceeded for a given diode. These specifications are always contained in the manufacturers’ specification sheets and only the most important specifications are listed below.

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• • • • • • • •

Peak reverse voltage (VRM) Also termed the peak-inverse-voltage (PIV) and is the maximum permissible reverse voltage that may be applied across the diode after which it will be destroyed. Reverse breakdown voltage (VBR) This is the minimum voltage that may be applied across the diode before it will break down. Steady-state forward current (F or IO) This specifies the maximum current that may be passed through the diode under normal forward bias conditions. Peak surge current (IFM (SURGE)) This is the maximum specified current which may pass through the diode for a specified time and is higher than the specified steady state forward current and will occur at switch-on or short circuit conditions. Static reverse current (IR) This is the reverse saturation current for a specified reverse bias voltage. Static forward voltage drop (VF) This is the voltage drop across the diode for a given steady-state forward current. Continuous power dissipation (P) This is the maximum power that the diode may dissipate on a continuous basis at a temperature or 25ºC. Reverse recovery time (trr) This is the time that it will take the diode to switch from the conducting state (on- state) to the non-conducting state (off-state).

9.5 The diode equation The electrical characteristics of a diode were given in 9.4 and some of these may be represented by the following mathematical expression termed the diode equation: i = IS (e qV / KT) – 1 where

i IS q V K T

= = = = = =

forward current in ampere reverse saturation current in ampere electron charge = 1, 6 ×10-19 coulomb potential difference across the diode in volt Boltzmann’s constant = 1, 38 ×10-23 J/K temperature in Kelvin

It must be noted that Kelvin is termed absolute temperature and is obtained in the following manner: K =

273 + º C


Example 9.1 Convert the following temperatures to Kelvin. (a) 28 º C (b) 22 º C (c) 36 º C

Solution: (a) K (b) K

= = =

273 + º C 273 + 28 301 Kelvin

= = =

273 + º C 273 + 22 295 Kelvin

(c) K

= = =

273 + º C 273 + 36 309 Kelvin

Example 9.2 Determine the forward current for a diode at 28 º C should the diode have a 0, 03 volts potential difference across it and a reverse saturation current of 0, 5 µ-ampere.

Solution: i = IS(e qV / KT) -1 = 0, 5 × 10-6 × e(1,6 × 10-19 × 0,03 / 1,38 × 10-23 × 301) - 1 = 1,088 µ-ampere

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132


Example 9.3 Determine the reverse saturation current for a rectifier diode given the following data: Potential difference across diode = 0, 25 volt Temperature = 17 º C Forward current = 42,6 m-ampere

Solution: i = IS(e qV / KT) -1 i IS = qV / KT e -1 42,6 × 10-6 = (1,6 × 10-19 × 0,25 / 1,38 × 10-23 × 290) e = 1,944 µ-ampere

-1

Example 9.4 A silicon diode produces a forward current of 6 m-ampere and a reverse saturation current of 21 µ-ampere. Determine the voltage across the junction at a temperature of 25 º C.

Solution: i = IS(e qV / KT) -1 K×T i-1 V = × ln q IS 1,38 × 10-23 × 298 = 1,6 × 10-19

=

119,645 m-volt

x In

6 × 10-3 21 × 10-6

- 1


Example 9.5 The following data concerning a rectifier diode is given: Junction potential = 0,126 volt Forward current = 2 m-ampere Reverse saturation current = 15 µ-ampere Determine from this given information the temperature of this junction in º C.

Solution: i = IS(e qV / KT) -1 q ×V T = K × ln i + 1 IS = 1,6 ×10-19 × 0,126 × ln 1,38 × 10-23 = 298,136 K ºC = 289,136 - 273 = 25,136 ºC

2 × 10-3 15 × 10-6

+ 1

9.6 Forward resistance of a diode All diodes are manufactured from semi-conductor materials and will have a resistance caused by its atomic structure. This resistance is given by: K×T R = where K q×i T q

= = =

Boltzmann’s constant = 1,38 × 10-23 J/K temperature in Kelvin electron charge = 1,6 × 10-19 coulomb

Example 9.6 A rectifier diode has a current of 300 m-ampere flowing through it at a temperature of 26ºC. Determine the forward resistance of this diode.

133

134


Solution: K×T R = q×i 1,38 × 10-23 × 299 = 1,6 × 10-19 × 300 × 10-3

=

0,0859 ohms

Example 9.7 A rectifier diode has a forward resistance of 0,103 ohm at a temperature of 25ºC. Calculate the forward current that will flow through the diode.

Solution: K×T R = q×i K×T i = q×R 1,38 × 10-23 × 299 = 1,6 × 10-19 × 0,103

=

249,54 m-ampere

Example 9.8 At what temperature in ºC will a diode operate should it have a forward resistance of 1,428 ohm and allows a forward current of 18 m-ampere to flow through it?

Solution:

K×T R = q×i q×i×R T = K 1,6 × 10-19 × 18 × 10-3 × 1,428 = 1,38 × 10-23

= ºC = =

298 Kelvin 289 - 273 25 ºC


9.7 The diode load-line The load-line for a diode is obtained by considering the maximum values of forward current and the maximum value of the forward bias for a particular rectifier diode. This load-line is then drawn on the forward characteristic of a germanium or silicon diode and the point where these two intersect one another is termed the Q-point or operating point. Should we assume a 2 volt forward bias and a maximum forward current of 20 m-ampere then the loadline will be as indicated by the graphic representation in figure 9.3. Since the loadline is always constructed using the maximum parameters of the diode it will be required to limit and protect the diode by using a series resistor with the diode in order to limit the current to an acceptable level. The reason for this is that the diode should never be operated at its maximum value since it will reduce the life expectancy of the diode. The network in figure 9.4 illustrates this concept. The value of the limiting resistor RS is mathematically given by: RS = where

VS - VDF IDF

RS = value of series limiting resistor in ohm VS = forward bias voltage in volt VDF = forward voltage value of diode in volt IDF = maximum forward value of diode current in ampere I (mA)

20

10

Germanium

Silicon

Q-point Q-point V 1

Figure 9.3

2

135

136


RS

D

VR

VDF VS

Figure 9.4

Example 9.9 Silicon as well as a Germanium diode needs to be protected by a series limiting resistor. The forward bias voltage has a value of 3 volt and the maximum forward current for both types of diodes is 850 m-ampere. Determine the value of a series limiting resistor for both diodes.

Solution: Silicon Germanium VS - VDF VS - VDF RS = RS = IDF IDF 3 - 0,6 3 - 0,3 = = 850 × 10-3 850 × 10-3

=

2,824 ohm

=

3,176 ohm

Exercise 9.1 1.

A diode may be best described by its characteristic curve. This curve defines four distinct areas or points. Give such a characteristic curve, indicate these distinct areas and then describe in detail the essence of these areas.

2. Tabulate the range as well as rating commonly used for diodes.


3.

Clarify the following electrical characteristic abbreviations of a diode and then also give a brief description of each one: 3.1 VRM; 3.2 VBR; 3.3 IO or IF; 3.4 IFM (SURGE); 3.5 IR; 3.6 VF; 3.7 P; and 3.8 trr.

4. Give the expression for a diode and explain the meaning of each symbol used in the expression. 5. The following data concerning rectifier diodes are given. 5.1 Potential difference across the diode = 0,96 volt Temperature = 31ºC Reverse saturation current = 0,0045 m-ampere Determine the forward current of the diode. 5.2 Potential difference across the diode = 0, 96 volt Temperature = 31ºC Forward current = 55,56 m-ampere Calculate the reverse saturation current for the diode. 5.3 Potential difference across the diode = 0,96 volt Forward current = 1,6 m-ampere Reverse saturation current = 0,0045 m-ampere Calculate the temperature of this diode in ºC. 6. Give the expression for the forward resistance of a diode and explain the meaning of each symbol used in the expression. 7. The following data concerning rectifier diodes are given. 7.1 Temperature = Forward current = Determine the forward resistance of the diode. 7.2 Forward resistance = Temperature = Calculate the forward current for the diode. 7.3 Forward current diode = Forward resistance = Calculate the temperature of this diode in ºC.

21ºC 0,045 m-ampere 0,916 ohm 26ºC 25 m-ampere 1,625 ohm

137

138


8. Explain how you would draw the loadline for diode given the following values. • forward bias = 3 volt • maximum forward current = 15 m-ampere Draw your own forward characteristic for a diode and then indicate clearly the Q-point on your characteristic. 9. Explain using a suitable network diagram the importance of a series limiting resistor used in conjunction with a diode. 10. Give the expression for a series limiting resistor of a diode and explain the meaning of each symbol used in the expression. 11. The following data concerning a diode is given: • forward bias voltage = 6,2 volt • forward current for Ge diode = 24 m-ampere • forward current for Si diode = 30 m-ampere Determine a suitable value for a series limiting resistor for a: 11.1 Silicon diode; and 11.2 Germanium diode.

Diode Applications


CHAPTER

10 Diode Applications Learning Outcomes On completion of this module you will be able to: •

Explain using suitable circuit diagrams as well as input and output waveforms the application of a diode in: – Series clippers; and – Parallel clippers. • Explain using suitable circuit diagrams, calculations as well as input and output waveforms the application of a diode in: – Half-wave rectification and filters (C, RC and LC); – Full-wave rectification (two diodes) and filters (C, RC and LC); and – Full-wave rectification (bridge circuit - four diodes) and filters (C, RC and LC). • Explain using suitable circuit diagrams the application of a diode in: – Full-wave voltage doubling circuits; and – Other voltage multiplication circuits. • Explain the concept and do calculations of the following diode rectifier concepts: – Transformer ratio; – Average dc voltage; – Ripple voltage; – Ripple factor; and – Peak-inverse-voltage (PIV) ratings.

10.1 Introduction Diodes have many applications and in this module we will cover the most common applications. It is however very important that you must understand the principle of operation of a diode as was discussed in the pervious module otherwise the concepts that will be discussed in this module will be futile.

139

140

Diodes APPLICATIONs

10.2 Application of diodes 10.2.1 Clippers The term clipper is used to represent a diode-resistor circuit or network which will change or separate a given waveform into two components or parts and may be interchanged with the term limiter. These terms are interchangeable since it will depend on the point where the output is taken from. As a rule of thumb a limiter is referred to as a circuit whose output is below a certain value whereas a clipper refers to a circuit whose output is always above a certain value. In this module we will only refer to clippers as it is most commonly referred to and these clippers are divided into series and parallel clippers.

10.2.1.1 Series clippers A series clipper may be described as a circuit or network that will remove or clip one half of the input signal and are divided into two main groups. A negative series clipper is illustrated in figure 10.1 (a), the input waveform is illustrated in figure 10.1 (b) and the output waveform is illustrated in figure 10.1 (c). A positive series clipper is illustrated in figure 10.2 (a), the input waveform is illustrated in figure 10.2 (b) and the output waveform is illustrated in figure 10.2 (c). •

Negative series clipper

D R

Input

Output

(a)

+ -

t

+ -

(b)

t

(c)

Figure 10.1


141

The circuit is so constructed that the diode D is connected in series with the resistor R and that the output is taken across the resistor R. The input voltage varies centrally around the zero volt line and has equal positive and negative values. During the positive half of the input cycle diode D is forward-biased and will conduct producing a current to flow through resistor R causing a voltage drop across the resistor R. During the negative half of the input cycle diode D becomes reverse-biased and does not conduct and no current can flow through resistor R. Since no current can flow through resistor R during the negative half of the input cycle there can be no voltage drop across resistor R and no output can be obtained. In this fashion the negative half of the input cycle is removed. Now, should the input voltage have a magnitude of VS then the output voltage will have a magnitude of VO = VS - VF where VF is equal to the forward voltage drop across the diode during the conduction cycle. The output voltage will therefore be slightly less than the input voltage. Account must be taken of the type of diode used, Germanium or Silicon in order to determine the value that needs to be subtracted. Note again that these values are ± 0,3 volt for Germanium and ± 0,6 volt for Silicon. • Positive series clipper The circuit is so constructed that the diode D is again connected in series with the resistor R and that the output is again taken across the resistor R. The input voltage varies centrally around the zero volt line and has equal positive and negative values. During the positive half of the input cycle diode D is reverse-biased and will not conduct producing no current to flow through resistor R causing no voltage drop across the resistor R. During the negative half of the input cycle diode D becomes forward-biased and conducts and a current can flow through resistor R. Since no current can flow through resistor R during the positive half of the input cycle there can be no voltage drop across resistor R and no output can be obtained. In this fashion the positive half of the input cycle is removed. D R

Input

Output

(a)

+ -

t

+ -

(b)

t

(c)

Figure 10.2

142

Diodes APPLICATIONs

However, we now find that should the input voltage have a magnitude of VS then the output voltage will have a magnitude of VO = - [VS - VF] where VF is equal to the forward voltage drop across the diode during the conduction cycle. The output voltage will therefore be slightly less than the input voltage. Account must be taken of the type of diode used, Germanium or Silicon in order to determine the value that needs to be subtracted. Note again that these values are ± 0,3 volt for Germanium and ± 0,6 volt for Silicon. An interesting observation concerning series clippers, positive or negative, is that the output will always consist of that part of the input cycle during which the diode conducts.

10.2.1.2 Parallel clippers A parallel clipper may also be described as a circuit or network that will remove or clip one half of the input signal and are divided into two main groups. A negative parallel clipper is illustrated in figure 10.3 (a), the input waveform is illustrated in figure 10.3 (b) and the output waveform is illustrated in figure 10.3 (c). A positive parallel clipper is illustrated in figure 10.4 (a), the input waveform is illustrated in figure 10.4 (b) and the output waveform is illustrated in figure 10.4 (c). It should be noted that parallel clippers are also referred to as shunt clippers. • Negative parallel clipper The circuit is so constructed that the diode D is again connected in series with the resistor R but that the output is now taken across the diode D. The input voltage varies centrally around the zero volt line and has equal positive and negative values. During the positive half of the input cycle diode D is reverse-biased and will not conduct producing no current to flow through resistor R causing no voltage drop across the resistor R. But, diode D is in parallel with the input and will appear as such across the output. During the negative half of the input cycle diode D becomes forward-biased and conducts and current can flow through resistor R. But a short circuit is produced when diode D conducts and no output is obtained. In this fashion the negative half of the input cycle is removed. We will now find that should the input voltage have a magnitude of VS then the output voltage will have a magnitude of VO = [VS - (IO × R)] where IO is equal to the output current. The output voltage will therefore be slightly less than the input voltage. Account must be taken of the type of diode used, Germanium or Silicon in order to determine the value that needs to be subtracted. Note again that these values are ± 0,3 volt for Germanium and ± 0,6 volt for Silicon. Note that during the negative half of the input cycle a small output is obtained and that will be equal to the forward voltage of the diode since it will only become a short circuit once this value has been exceeded.


143

IO

R Input

D

Output

(a)

+ -

t

+ -

t VF

(b)

(c)

Figure 10.3

• Positive parallel clipper The circuit is so constructed that the diode D is again connected in series with the resistor R but that the output is now taken across the diode D. The input voltage varies centrally around the zero volt line and has equal positive and negative values. During the positive half of the input cycle diode D is forward-biased and will conduct producing current to flow through resistor R causing a voltage drop across the resistor R. But, diode D is in parallel with the input and no output will appear across the output. During the negative half of the input cycle diode D becomes reversed-biased and will not conduct and no current can flow through resistor R. But an open circuit is produced when diode D does not conduct and an output is obtained. In this fashion the positive half of the input cycle is removed. We will now find that should the input voltage have a magnitude of VS then the output voltage will have a magnitude of VO = - [VS – (IO × R)] where IO is equal to the output current. The output voltage will therefore be slightly less than the input voltage. Account must be taken of the type of diode used, Germanium or Silicon in order to determine the value that needs to be subtracted. Note again that these values are ± 0,3 volt for Germanium and ± 0,6 volt for Silicon. Note that during the positive half of the input cycle a small output is obtained and that will be equal to the forward voltage of the diode since it will only become a short circuit once this value has been exceeded. An interesting observation concerning parallel clippers is that the output will always consist of that part of the input cycle during which the diode does not conduct. The arrow point of the diode also serves as a good indication of the type of clipper. Should it points downward the positive half of the input is clipped and only the negative half of the input is obtained as an output. On the other hand, should it point upward then the negative half of the input is clipped and only the positive half of the input is obtained as an output.

144

Diodes APPLICATIONs

IO

R Input

D

Output

(a)

+ -

t

+ -

t VF

(b)

(c)

Figure 10.4

10.2.2 The diode as a rectifier One of the other main uses of a diode is as a rectifier is that it will convert an alternating quantity into a pulsating direct quantity and depending on the efficiency of the filter circuit may supply an almost ripple-free output.

10.2.2.1

Half-wave rectifier

A half-wave rectifier is a network so constructed using one diode plus additional components to perform the function of rectification and is illustrated in figure 10.5 (a) and (b). A half-wave rectifier derives its name from the fact that the output voltage VO only consist of one half of the input voltage Vi. This can be seen on the waveforms illustrated in figure 10.6 (a), (b) and (c). A half-wave rectifier circuit is illustrated in figure 10.5 and consists of a mains step-down transformer T1, a diode D1 and load resistor RL connected in series with the diode across the secondary side of the mains transformer T1. The input voltage Vi is supplied to the primary side of the mains transformer T1 and the output voltage Vo is taken across the load resistor RL. There is no fundamental difference between the two circuits except for the fact that different halves of the input cycle will be rectified and this is illustrated in figure 10.6 (a), (b) and (c). During the first half of the input voltage Vi, that is between A and B in figure 10.6(a), the diode D1 is reverse-biased and no current will flow through the load resistor RL as illustrated in figure 10.5. The output voltage Vo taken across the load resistor RL is equal to zero since there will be no current flowing. The above statements of operation will be reversed for figure 10.5 (b). During the second half of the input voltage Vi, that is between B and C in figure 10.6 (b), the diode D1 is forward-biased and a current will


145

flow through the load resistor RL as indicated by the dotted line with the polarities as indicated in figure 10.5. The output voltage Vo taken across the load resistor RL is almost equal to Vi. The only difference there is will be the forward voltage drop across the diode D1 since the circuit on the secondary side operates as a series voltage divider with the load resistor RL. The above statements of operation will be reversed for figure 10.5 (b). The same pattern will be repeated for every alternate positive and negative half cycle and the output voltage thus obtained is termed a pulsating direct quantity. But this is not really what is required since the magnitude of the output is not constant. The way in which this pulsating direct quantity is smoothed is by means of a smoothing capacitor which will remove the varying quantity and an output as constant as possible will be obtained. This aspect will be discussed under the section that will deal with filter networks. The applicable waveforms for the network in figure 10.5 are illustrated in figure 10.6 (a) and (b) where figure 10.6 (a) are those for figure 10.5 (a) and figure 10.6 (b) are those for figure 10.5 (b). Figure 10.6 [a (i)] and [b (i)] are the inputs Vi for both networks and figure 10.6 [a (ii)] and [b (ii)] are the outputs obtained from the secondary side of the mains transformer for both networks. Figure 10.6 [a (iii)] and [b (iii)] are the outputs obtained across RL for both networks and note must be taken of the polarity of the respective outputs.

10.2.2.2

Full-wave rectification (two diodes)

As the name would imply, full-wave rectification will make use of both halves of the input voltage Vi. Such a network diagram is illustrated in figure 10.7 (a), (b) and (c). The associated waveforms are given in figure 10.8 (a) and (b). A circuit diagram of a full-wave rectifier, using two diodes is illustrated in figure 10.7 (a), (b) and (c) and consists of a centre-tap mains step-down transformer T1, two diodes D1 and D2 and a load resistor RL connected in series with the diodes across the secondary side of the centre-tap mains step-down transformer T1. The input voltage Vi is supplied to the primary side of the centre-tap mains transformer T1 and the output voltage Vo is taken across the load resistor RL.

146

Diodes APPLICATIONs

D1

T1

+

D1

T1

-

-

+

+

VF

VF RL

Vi

-

Vi

VO

RL

+

+

VO = Vi - VF

-

-

(a)

D1

T1

+

D1

T1

-

-

-

+

VF

VF RL

-

VO = Vi - VF

Vi

+

+

RL

+

VO

-

(b)

Figure 10.5

(i)

+ -

t

A

( ii )

( iii )

B

C

+ -

D

+ -

E

t

A

t

B

C

+ -

t

t (a)

(b)

Figure 10.6

E

t

-

+

D


147

During the first half of the input voltage Vi, that is between A and B in figure 10.8, the diode D1 is reverse-biased and diode D2 is forward-biased and a current will flow through the load resistor RL as indicated by the dotted line with the polarities as indicated in figure 10.7 (b). The output voltage Vo taken across the load resistor RL is almost equal to Vi. The only difference there is will be the forward voltage drop across the diode D2 since the circuit on the secondary side operates as a series voltage divider with RL. T1

D1 Vi RL

VO = Vi - VF

D2

+

T1

+

+

D1

Vi

-

T1

+ -

D1

+ -

D2

+

Vi

-

+

RL

VO = Vi - VF

D2

+ -

RL

VO = Vi - VF

-

Figure 10.7

During the second half of the input voltage Vi, that is between B and C in figure 10.8, the diode D2 is reverse-biased and diode D1 is forward-biased and a current will flow through the load resistor RL as indicated by the dotted line with the polarities as indicated in figure 10.7 (c). The output voltage Vo taken across the load resistor RL is almost equal to Vi. The only difference there is will be the forward voltage drop across the diode D1 since the circuit on the secondary side operates as a series voltage divider with RL. The same pattern will be repeated for every alternate positive and negative half cycle and the output voltage thus obtained is termed a pulsating direct quantity. But this is not really what is required since the magnitude of the output is not constant. The way in which this pulsating direct quantity is smoothed is by means of a smoothing capacitor which will remove the varying quantity and an output as constant as possible will be obtained. This aspect will be discussed under the section that will deal with filter networks.

148

Diodes APPLICATIONs

The applicable waveforms for the network in figure 10.7 are illustrated in figure 10.8. Figure 10.8 (a) is the input Vi for the network and figure 10.7 (b) is the output obtained from the secondary side of the centre-tap mains transformer for the network. Figure 10.7 (c) is the output obtained across RL for the network.

10.2.2.3 Full-wave rectification (bridge circuit-four diodes) As the name would imply, full-wave rectification will also make use of both halves of the input voltage Vi. Such a network diagram is illustrated in figure 10.9 (a), (b) and (c). The associated waveforms for this type of network is given in figure 10.10 (a) (b) and (c). The only difference between a bridge rectifier and a rectifier using two diodes is that a bridge rectifier uses two diodes per half cycle which would increase your current delivering capabilities if compared to only using one diode per half cycle. A great advantage in using a bridge rectifier is that does not require a centre-tap mains transformer which is quite expensive and this type of circuit may be fed directly from the mains supply line but is not very often done since the transformer does supply some sort of protection.

(a)

+ -

t

A

(b)

B

C

D

E

+ -

F

G

H

I

t

+ (c)

t

Figure 10.8

A network diagram of a full-wave rectifier, using four diodes commonly known as the bridge rectifier is illustrated in figure 10.9 (a), (b) and (c) and consists of a mains stepdown transformer T1, four diodes D1 to D4 and a load resistor RL connected in series with the diodes across the secondary side of the mains step-down transformer T1. The input voltage Vi is supplied to the primary side of the mains transformer T1 and the output voltage Vo is taken across the load resistor RL. During the first half of the input voltage Vi, that is between A and B in figure 10.10, the diodes D1 and D4 are reverse-biased and diodes D2 and D3 are forward-biased and


149

a current will flow through the load resistor RL as indicated by the dotted line with the polarities as indicated in figure 3.9 (b). The output voltage Vo taken across the load resistor RL is almost equal to Vi. The only difference there is will be the forward voltage drop across the diodes D2 and D3 since the circuit on the secondary side operates as a series voltage divider with the load resistor RL. During the second half of the input voltage Vi, that is between B and C in figure 10.10, the diodes D2 and D3 are reverse-biased and diodes D1 and D4 are forward-biased and a current will flow through the load resistor RL as indicated by the dotted line with the polarities as indicated in figure 10.9 (c). The output voltage Vo taken across the load resistor RL is almost equal to Vi. The only difference there is will be the forward voltage drop across the diode D1 and D4 since the circuit on the secondary side operates as a series voltage divider with the load resistor RL. The same pattern will be repeated for every alternate positive and negative half cycle and the output voltage thus obtained is termed a pulsating direct quantity. But this is not really what is required since the magnitude of the output is not constant. The way in which this pulsating direct quantity is smoothed is by means of a smoothing capacitor which will remove the varying quantity and an output as constant as possible will be obtained. This aspect will be discussed under the section that will deal with filter networks. The applicable waveforms for the network in figure 10.9 are illustrated in figure 10.10. Figure 10.10 (a) is the input Vi for the network and figure 10.10 (b) is the output obtained from the secondary side of the centre-tap mains transformer for the network. Figure 10.10 (c) is the output obtained across RL for the network. T1 D1

(a)

Vi

+

(b)

D2

T1

D3

D4

D1

D2

Vo = Vi - 2VP

RL

Vo = Vi - 2VP

-

Vi

-

RL

+

D3

D4

150

Diodes APPLICATIONs

T1

-

+ D1

(c)

D2

Vi

Vo = Vi - 2VP

RL

+

D3

-

D4

Figure 10.9

(a)

+ -

t

A

(b)

B

C

D

+ -

E

F

G

H

I

t

+ (c)

t

Figure 10.10

10.3 Rectifier concepts It was mentioned earlier that the output obtained from a rectifier network is deemed to be a pulsating dc. The ideal output however should be a steady dc but that will be explained at a later stage. There are a number of concepts that will determine the magnitude of the output obtained from a rectifier. The first and most important factor is that of the transformer ratio. A very brief description of that aspect will be given.


151

10.3.1 Transformer ratio Although different types of transformers are available on the market today we will concentrate mainly on the step-down transformer which is the most common transformer used in conjunction with rectifier networks. Such a transformer is illustrated in figure 10.11.

VO Secondary side

Vi Primary side

Figure 10.11

A transformer consists of two sets of windings, a primary winding or side across which the input is applied (Vi) and a secondary winding or side across which the output is taken (VO) and these two sets of winding are electrically isolated from one another. Energy is transferred from the primary side to the secondary side by means of a magnetic coupling or more commonly known as electromagnetic coupling. Altough all transformers are specifically designed to perform a specific task we need to have a look at the different transformers that are found in industry and these are given below in figure 10.12 (a), (b), (c), (d) and (e). • Step-down transformer These are transformers that have more primary windings than secondary winding which will result in a lower secondary voltage but with an increase in secondary current. This type of transformer consists of a single primary winding as well as a single secondary winding.

(a)

152

Diodes APPLICATIONs

• Step-up transformer These are transformers that have more secondary windings than primary winding which will result in an increased secondary voltage but with a decrease in secondary current. This type of transformer consists of a single primary winding as well as a single secondary winding.

(b)

• Centre-tap transformer These are transformers that consist of a single primary winding and a split secondary winding which actually amounts to two secondary windings. Each set of secondary windings has its own specified output.

(c)

• Multi-tap transformer These are transformers much like the centre-tap transformer but actually has more than one tap on the secondary side in order to obtain different voltages from the secondary side of the transformer.

(d)

• Auto transformer The auto transformer consists of a single inductor with the facility to change the ratio of the primary to the secondary by means of a tapping on the transformer. These transformers may be used as step-up or a step-down transformer.


153

(e) Figure 10.12

The magnitudes of the different quantities are mathematically related to one another and are mathematically expressed by the transformer equation. NP V I = P = S where NS VS IP

NP VP IS NS VS IP

= = = = = =

number of primary windings magnitude of primary voltage in volt magnitude of secondary current in ampere number of secondary windings magnitude of secondary voltage in volt magnitude of primary current in ampere

Important to note is that the secondary current is inversely proportional to the primary number of winding as well as the primary voltage whereas the primary current is also inversely proportional to secondary number of windings as well as the secondary voltage. Also of great importance is that the above values for the primary and secondary current as well as voltage are always given in RMS-values unless otherwise stated. These RMS-values are converted to maximum or peak values as follows: VP (MAX) VS (MAX)

= =

1,414 × VP 1,414 × VS

IP (MAX ) IS (MAX)

= 1,414 × IP = 1,414 × IS

154

Diodes APPLICATIONs

Example 10.1 The following data concerning a transformer is given: • Primary voltage = 250 volt • Turns ratio = 10 : 1 • Secondary current = 5 ampere Determine from the above information the: (a) Peak primary voltage; (b) Peak primary current; and (c) Peak secondary voltage.

Solution: (a) VP (MAX) = 1,414 × VP = 1,414 × 250 = 352,5 volts NP IS = IP (MAX) (b) NS IP IS × NS IP = NP 5×1 = 10 = 0,5 ampere

= = =

NP VP NP (c) = or = NS VS NS VP × NS VS = VS (MAX) = N P 250 × 1 = = 10 = 25 volt =

VS (MAX) =

1,414 × VS = 1,414 × 25 = 35,35 volt

1,414 × IP 1,414 × 0,5 0,707 ampere

VP (MAX) VS (MAX) VP (MAX) × NS NP 353,5 × 1 10

35,35 volt


10.3.2 Average dc-voltage Refer to the two waveforms illustrated in figure 10.13 (a) and (b). Maximum value (a)

+

Average dc-value t

Maximum value (b)

Average dc-value

+

t

Figure 10.13

The waveform in figure 10.13 (a) illustrates the output obtained from a half-wave rectifier network and figure 10.13 (b) illustrates the output obtained from a full-wave rectifier network. Note the average dc value that will be obtained between the two types of rectifier networks. These values can be determined by using the following mathematical expressions for full-wave and half-wave rectification respectively. Half-wave

VDC (AVE) =

0,318 × VMAX

Full-wave

VDC (AVE)

=

0,637 × VMAX

Example 10.2 A rectifier has the following peak voltage outputs: (a) 55 volt (b) 16 volt Determine the average dc-output in each instance for a half-wave and fullwave rectifier.

Solution: (a) Half-wave VDC (AVE)

= = =

0,318 × VMAX 0,318 × 55 17,49 volt

155

156

Diodes APPLICATIONs

VDC (AVE) (b) Full-wave VDC (AVE)

= = =

0,318 × VMAX 0,318 × 16 5,088 volt

= = =

0,637 × VMAX 0,637 × 55 34,485 volt

VDC (AVE)

= = =

0,637 × VMAX 0,637 × 16 10,192 volt

10.3.3

Ripple voltage

Refer to the two waveforms illustrated in figure 10.14 (a) and (b). A B

+

(a)

Ripple t A

(b)

B

Ripple

+ t

Figure 10.14

During the period indicated by A the filter capacitor will charge and during the period indicated by B the capacitor will discharge and the output will have a ripple and not be a constant value. This will be discussed in more detail when we deal with filter networks. The ripple voltage is a variation of voltage values around the average dcvalue and is always measured as an RMS-value and can be determined by the following mathematical expressions. Half-wave

Full-wave

Vr (RMS) = 0,385 × VMAX Vr (RMS) = 0,305 × VMAX


157

Example 10.3 The maximum output on the secondary side of a power supply is measured at 10 volt. Determine the ripple voltage should this value be measured across a: (a) Half-wave rectifier; and (b) Full-wave rectifier.

Solution: It is very important that the maximum value given be converted to an RMSvalue before the calculation is done! VS (MAX) (a) Vr (RMS)

= = =

(b) Vr (RMS)

1,414 × VS 1,414 × 10 14,14 volt = 0,385 × VMAX = 0,385 × 14,14 = 5,444 volt = = =

0,305 × VMAX 0,305 × 14,14 4,313 volt

A very important observation concerning the above answers obtained is that it should be obvious that the ripple voltage will be greater for a half-wave rectifier than it would be for a full-wave rectifier.

10.3.4

Ripple factor

The ripple factor of a rectifier, half-wave as well as full-wave is mathematically given by: V r = r (RMS) V DC (AVE)

The ideal situation would be that the ripple factor should be as close as possible to zero and would therefore give us an indication of the performance of the rectifier.

158

Diodes APPLICATIONs

Example 10.4 Calculate the ripple factor for a: (a) Half-wave rectifier; and (b) Full-wave rectifier.

Solution: (a) r = But Vr (RMS) VDC (AVE) r = =

Vr (RMS) VDC (AVE) = 0,385 × VMAX and = 0,318 × VMAX 0,385 × VMAX 0,318 × VMAX 1,21 Vr (RMS) VDC (AVE)

(b) r = But Vr (RMS) VDC (AVE)

= =

0,305 × VMAX and 0,637 × VMAX

0,305 × VMAX 0,637 × VMAX

r = =

0,48

The above answers would therefore indicate to us that a full-wave rectifier is more suitable since the ripple factor is much closer to zero than that of the half-wave rectifier.

10.3.5 PIV-rating (peak-inverse-voltage) The PIV-rating for a diode is probably the most important aspect when diodes are used in rectifier networks. This value will specify the maximum reverse voltage that may be applied across the diode or diodes and exceeding this value may render the component useless since the junction will be damaged. Half-wave

PIV = VMAX

Full-wave PIV

=

2 × VMAX


Example 10.5 A value of 16,2 volt is measured across the secondary side of a rectifier. Determine the PIV-rating for the diodes should the rectifier be used as a: (a) Half-wave device; and (b) Full-wave device.

Solution: (a) PIV = VMAX = 16,2 volt (b) PIV

= = =

2 ×VMAX 2 × 16,2 32,4 volt

Example 10.6 A step-down transformer has a turns ratio of 30:1 and is fed from a 300 V/50 Hz supply. Determine for a half-wave rectifier network the: (a) Average dc voltage output; (b) Ripple voltage; (c) Ripple factor; and (d) PIV-rating for the diode.

Solution: NP V (a) VDC (AVE) = 0,318 × VMAX where = P NS V = 0,318 × 14,14 S V × NS = 4,497 volt VS = P NP 300 × 1 = 30 = 10 volt

VS (MAX)

= 1,414 × VS = 1,414 × 10 = 14,14 volt

159

160

Diodes APPLICATIONs

(b) Vr (RMS)

= = =

0,385 × VMAX 0,385 × 14,14 5,444 volt

V (c) r = r (RMS) V DC (AVE) 5,444 = 4,497 = 1,21

(d) PIV = VMAX = 14,14 volt

Example 10.7 A diode has a PIV rating of 17,675 volt. Determine the average dc output voltage for a half-wave rectifier.

Solution: VDC (AVE)

= = =

0,318 × VMAX 0,318 × 17,675 5,62 volt

where VMAX = PIV = 17,675 volt

Example 10.8 A centre-tap step-down transformer has a turns ratio of 10 : 1 and is fed from a 100V/50 Hz supply. Determine for this rectifier network the: (a) Average dc voltage output; (b) Ripple voltage; (c) Ripple factor; and (d) PIV-rating for the diodes.


Solution: A point that should be noted here is that the above rectifier network is that of a fullwave rectifier since use is made of a centre-tap transformer.

NP V (a) VDC (AVE) = 0,637 × VMAX where = P NS V = 0,637 × 14,14 S V × NS = 9 volt VS = P NP 100 × 1 = 10 = 10 volt

VS (MAX)

= 1,414 × VS = 1,414 × 10 = 14,14 volt

(b) Vr (RMS)

= = =

0,308 × VMAX 0,308 × 14,14 4,355 volt

V (c) r = r (RMS) V DC (AVE) 4,355 = 9 = 0,484

(d) PIV

= = =

2 × VMAX 2 × 14,14 28,28 volt

161

162

Diodes APPLICATIONs

10.4 Filter networks It was mentioned earlier that the output obtained from a rectifier circuit is a pulsating dc which means that it has a varying value but is not the type of output we require. The ideal output is a steady dc-value and that is obtained by using a filter with the rectifier.

Definition 10.1 Filter

A filter can therefore be defined as a component that will remove the ripple (pulsating) component from the output of a rectifier circuit.

There are various types of filters that can be used and each one has its own distinct characteristics and each type will be explained. Before we actually commence with the explanation of the filter circuits it is important that we look at the concept of filter circuits. You are therefore requested to refer to figure 10.15 (a) and (b) where this principle is illustrated.

Half-wave rectifier

Filter

Input

Output (a)

Full-wave rectifier

Filter

Input

Output (b)

Figure 10.15


163

10.4.1 The capacitor filter 10.4.1.1

Half-wave rectification

A half-wave rectifier circuit with a capacitor filter is illustrated in figure 10.16. Note that the capacitor used is an electrolytic type which has a fixed polarity. D1

T1

+

-

IDC VF

+

Vi C

-

+

-

RL

+

VO

Figure 10.16

The load voltage VDC across RL is mathematically expressed by: V VDC = Vp - r (p - p) or VDC = Vp 2

IDC Vm and IDC = 2×f×C RL

The RMS ripple voltage Vr (RMS) across RL is mathematically expressed by: V Vr (RMS) = r (p - p) or Vr (RMS) = 2√3

IDC 2√3 × f × C

The ripple r of the output voltage VDC across RL is mathematically given by: Vr (RMS) r = or VDC

IDC or VDC × 2√3 × f × C

1 RL × 2√3 × f × C

Consider the graphical representation in figure 10.17 for a clarification of the quantities used in the various expressions.

164

Diodes APPLICATIONs

Capacitor discharge time

VC

Vr (p-p)

VDC

VP

t Capacitor charge time

Figure 10.17

Although a capacitor filter is extensively used it does have the following disadvantages: • Its current delivering capabilities are limited in that it cannot supply a large current; • Should there be an increase in current the ripple voltage will increase likewise;

Example 10.9 A half-wave rectifier is fed from a 220V/50 Hz supply and the following additional data is given: • Peak secondary voltage = 15 volt • Value of filter capacitor = 30 µF • Load current = 15 mA Use the above data and determine the: (a) Output dc voltage (VDC); (b) Output ripple voltage (Vr (RMS)); and (c) Ripple factor expressed as a percentage (r %).

Solution: IDC (a) VDC = Vp 2×f×C 15 × 10-3 = 15 2 × 50 × 30 × 10-6 = 10 volt


IDC (b) Vr (RMS) = 2 x √3 × f × C 15 × 10-3 = 2 ×√3 × 50 × 30 × 10-6

=

2,887 volt

Vr (RMS) × 100 VDC 2,887 × 100 10

(c) r = = =

28,87 %

and • It does not have good regulation capabilities.

10.4.1.2

Full-wave rectification

T1 D1

+

Vi

-

D2

RL

VO = Vi - VF

C

Figure 10.18 T1 D1

D2

+ Vi

D3

Figure 10.19

D4

RL C

Vo = Vi - 2VP

165

166

Diodes APPLICATIONs

A full-wave rectifier circuit using two diodes with a capacitor filter is illustrated in figure 10.18 and a bridge rectifier using four diodes is illustrated in figure 10.19. The load voltage VDC across RL is mathematically expressed by: V IDC Vm VDC = Vp - r (p - p) or VDC = Vp and IDC = 2 4×f×C RL The RMS ripple voltage Vr (RMS) across RL is mathematically expressed by: V Vr (RMS) = r (p - p) or Vr (RMS) = 2√3

IDC 4√3 × f × C

The ripple r of the output voltage VDC across RL is mathematically given by: Vr (RMS) r = or VDC

IDC or VDC × 4√3 × f × C

1 RL × 4√3 × f × C

Example 10.10 Consider the following full-wave rectifier circuit diagram. Use the given information and determine the following quantities for a 20 µF and 35 µF capacitor. (a) Average value; (b) RMS-value; (c) Output dc voltage; (d) Output RMS ripple voltage; and (e) Ripple percentage. T1 D1

10 mA

D2

+ 220 V/50 Hz

15 V

D3

Solution: (a) VAVE = Vm × 0,637 = 15 × 0,637 = 9,555 volt

D4

RL C

1 500 ohms


(b) VRMS = Vm × 0,707 = 15 × 0,707 = 10,605 volt (c) 20 µF capacitor

167

(d)

I IDC Vr (RMS) = VDC = Vp - DC 4×f×C 4√3 × f ×C 10 × 10-3 10 × 10-3 = 15 = -6 4 × 50 × 20 × 10 3 × √3 × 50 × 20 × 10-6 = 12,5 volt = 1,925 volt

35 µF capacitor

IDC IDC VDC = Vp Vr (RMS) = 4×f×C 4√3 × f ×C 10 × 10-3 10 × 10-3 = 15 = 4 × 50 × 35 × 10-6 3 × √3 × 50 × 35 × 10-6

=

13,571 volt

= 1,099 volt

(e) 20 µF capacitor V IDC r = r (RMS) × 100 r = × 100 VDC VDC × 4√3 × f × C 1,925 = × 100 10 × 10-3 12,5 = × 100 12,5 × 4 × √3 × 50 × 20 × 10-6

= 15,4 = 11,55% 1 r = × 100 RL × 4√3 × f × C 1 × 100 = 1 500 × 4√3 × 50 × 20 × 10-6

= 9,623%

168

Diodes APPLICATIONs

35 µF capacitor

Vr (RMS) IDC r = × 100 r = × 100 VDC V × 4√3 × f × C DC 1,099 = × 100 10 × 10-3 13,571 = × 100 13,571 × 4 × √3 × 50 × 35 × 10-6

= 8,098 = 6,077% 1 r = × 100 RL × 4√3 × f × C = 1 × 100 1 500 × 4√3 × 50 × 35 × 10-6

= 5,499%

10.4.2 The RC-filter 10.4.2.1


A half-wave rectifier circuit with a resistor-capacitor (RC) filter is illustrated in figure 10.20. This type of filter is very often referred to as a π-type RC-filter. The RC-type filter has the following advantages when compared to the capacitor filter: • • •

A decrease in the ripple voltage; A slight increase in current delivering capabilities; and An increase in the voltage regulation capabilities. D1

T1

+

+

Vi C1

-

IDC

RS

VF

-

+

+

+ C2

-

RL

VO

Figure 10.20

10.4.2.2


A full-wave rectifier circuit using two diodes with a resistor-capacitor filter is illustrated in figure 10.21 and a bridge rectifier using four diodes is illustrated in figure 10.22.


+

T1

+

+

Vi

+

-

IDC

RS

D1

-

D2

+

+ RL C1

VO

C1

-

Figure 10.21

+

T1

D1

+

+ Vi

-

D3

+

+

RS

D2

RL C1

D4

Figure 10.22

The load voltage V’DC across RL is mathematically expressed by: RL V’DC = VDC × R + RS L

The RMS ripple voltage V’r (RMS) across RL is mathematically expressed by: V’r (RMS) = Vr (RMS) × XC 2 (RS + XC2)½ where XC =

1 2×π×f×C

1 XC = 4×π×f×C

for half-wave and for full-wave.

The ripple r of the output voltage VDC across RL is mathematically given by: V’ r = r (RMS) V’ DC

Vo

C2

-

169

170

Diodes APPLICATIONs

Example 10.11 The following data concerning a power supply is given: Supply voltage = 250 V/ 0 Hz Voltage across first capacitor = 120 V Ripple voltage across first capacitor = 10 V Series resistor = 1 k-ohm Second capacitor = 150 µF Load resistor = 5 k-ohm Determine for half-wave- and full-wave rectifier the: (a) DC output voltage; (b) Ripple voltage; and (c) Ripple factor.

Solution: RL (a) V’DC = VDC × RL + RS = 120 × 5 × 103 5 × 103 + 1 × 103 = 100 volt

(b) Half-wave

Full-wave

1 1 XC = XC = 2×π×f×C 4×π×f×C = 1 1 = -6 2 × 3,142 × 50 × 150 × 10 4 × 3,142 × 50 × 150 × 10-6

= 21,2 ohm

= 10,6 ohm

V’r(RMS) = Vr(RMS) × XC V’r(RMS) = Vr(RMS) × XC 2 2 ½ 2 (RS + XC2)½ (RS + XC ) = 10 × 21,2 = 10 × 10,6 3 2 2 ½ [(1 × 103)2 + 10, 62]½ [(1× 10 ) + 21, 22 ] = 0,212 volt = 0,106 volt


(c) Half-wave r = = =

171

Full-wave

V’r(RMS) V’r(RMS) r = V’DC V’ DC 0,212 0,106 = 100 100 2,212 × 10-3 = 1,06 × 10-3

10.4.3 The LC-filter 10.4.3.1


A half-wave rectifier circuit with an inductor-capacitor (LC) filter is illustrated in figure 3.23. This type of filter is very often referred to as a π-type LC-filter. The inductor has a fairly large value which will tend to smooth the current flowing through it. The LC-type filter has the following advantages when compared to the capacitor filter: • • •

A decrease in the ripple voltage; A slight increase in current delivering capabilities; and An increase in the voltage regulation capabilities. D1

T1

+

-

L

VF

+

Vi C1

-

IDC

RS

+

+

-

C2

+

-

RL

VO

Figure 10.23

10.4.3.2


A full-wave rectifier circuit using two diodes with an inductor-capacitor filter is illustrated in figure 10.24 and a bridge rectifier using four diodes is illustrated in figure 10.25.

172

Diodes APPLICATIONs

+

T1

+

D1

+

Vi

+

-

IDC

RS

L

D2

-

+

+ RL C1

VO

C1

-

Figure 10.24

+

T1

D1

L

D2

+

+ Vi

-

D3

+

+

RS

C2

D4

Figure 10.25

The load voltage V’DC across RL is mathematically expressed by: RL V’DC = VDC × R + RS L

The RMS ripple voltage V’r (RMS) across RL is mathematically expressed by: V’r(RMS) =

Vr(RMS) (2πf )2L ×C

for half-wave and

V’r(RMS) =

Vr(RMS) (4πf )2L × C

for full-wave.

The ripple r of the output voltage VDC across RL is mathematically given by: V’ r = r(RMS) V’DC

Vo

RL C1

-


Example 10.12 Consider the following filter circuit and waveform for a full-wave rectifier. L

+

RS

+

+ 100 V

-

-

C1

RL C2

V 10 V

t

Supply voltage to power supply = = 250 µF C1 = 250 µF C2 L = 3H = 10 ohm RS = 5 k - ohm RL Use this information and determine the: (a) DC output voltage; and (b) Ripple factor.

250 V/50 Hz

Solution: RL (a) V’DC = VDC × RL + RS 5 × 103 = 100 × 5 × 103 + 10

=

99,8 volt

173

174

Diodes APPLICATIONs

V (b) Vr(RMS) = r(p - p) V’r(RMS) = Vr(RMS) 2√3 (4 × π ×f )2 L × C 10 = 2,886 = 2√3 (4 × 3,142 × 50)2 × 3 × 250 × 10-6 = 2,886 volt = 9,745 × 10-3 volt V’ r = r ( RMS ) V’ DC 9,745 × 10-3 = 99,8 = 9,76 × 10-5

10.5 No-load voltage The no-load voltage of any power supply may be defined as that voltage which is supplied by the secondary winding of the transformer when the load to that power supply is not connected. Consider figure 10.26.

+

T1

+

-

S

+

Vi

+

+

D1

D2

-

VO C

RL

Figure 10.26

During the conduction half-cycles of the input cycle the capacitor C1 will charge up to a value of ± VP and will discharge during the period when the input cycle goes through zero degrees. Since there is no discharge path for the filter capacitor according to figure 10.26 the voltage is measured across C1 and is termed the no-load voltage since the switch S is open. This voltage contains for all practical purposes no ripple. Should we however close the switch S then the capacitor C1 will have a discharge path through the load resistor RL since the current will be flowing through it. This output will then contain a ripple and is termed the full-load voltage. An important aspect that should also be noted is that the no-load voltage will always have a higher value than the fullload voltage.


175

10.6 Voltage regulation

Definition 10.2 Voltage

Voltage regulation may be defined as that change in the output voltage (full-load) for varying load conditions. This concept is illustrated in figure 10.27.

O u t p u t

No-load voltage Voltage variation Full-load voltage

v o l t a g e

Load current

Figure 10.27

The following conclusions can be made concerning the graphical representation in figure 3.26: • •

The output voltage of a power supply will decrease with an increase in load current; and The smaller the change in no-load voltage and full-load voltage, the more effective the power supply will be.

This voltage regulation is given by the following mathematical expression: Voltage regulation =

No-load voltage - Full-load voltage Full-load voltage

176

Diodes APPLICATIONs

Example 10.13 A full-wave power supply delivers 15,6 volt under no-load conditions. Determine the percentage regulation should the following voltages be delivered under full-load conditions. (a) 13,4 volt; and (b) 12 volt.

Solution: VNL - VFL (a) % regulation = × 100 VFL = 15,6 - 13,4 × 100 15,6 = 14,1% VNL - VFL × 100 (b) % regulation = VFL 15,6 -12 = × 100 15, 6 = 23,07%

10.7 Voltage multiplication Rectifier circuits, full-wave or half-wave, are limited to an output that will never be greater than the peak input voltage to that circuit. It may sometimes be required that greater voltages are required and for this purpose we make use of voltage multiplication circuits which may, depending on the design, supply two or more times the peak input value as an output. Two such circuits will be discussed.

10.7.1 Voltage doubler The circuits in figure 10.28 (a) and (b) illustrates a voltage doubler where figure 10.28 (a) illustrates one half of the input cycle and figure 10.28 (b) illustrates the other half of the input cycle.


Electron flow D1

+

C1 Vi = 240 V

-

D1

-

-

340 V C1

Vi = 240 V

+

C2

680 V

+

+ D2

+

+

+

177

340 V

D2

-

Electron flow

(a)

C2

-

-

(b)

Figure 10.28

Assume the polarities of the input voltage Vi as illustrated in figure 10.28 (a) then diode D1 will conduct and capacitor C1 will charge up to a value equal to the peak voltage VP which in this instance will be equal to: VP

= 1,414 × VRMS = 1,414 × 240 = 340 V with polarities as indicated.

Now assume the polarities of the input voltage Vi as illustrated in figure 10.28 (b) then diode D2 will conduct and capacitor C2 will charge up to a value equal to the peak voltage VP which in this instance will be equal to: VP

= 1,414 × VRMS = 1,414 × 240 = 340 V with polarities as indicated.

10.7.2 Voltage multiplication The circuit illustrated in figure 10.28 only has the capability to double whatever the input to the circuit is. What are we going to do should we require multiples of the input? The circuit in figure 10.29 (a) illustrates such a voltage multiplier circuit. The circuit in figure 10.29 (b) illustrates the effect of the positive half cycle of the input whereas figure 10.29 (c) illustrates the effect of the negative half cycle of the input. Assume the polarities as indicated in figure 10.29 (b) then the diode D1 will be forward biased and the diode D2 will be reverse biased. Since diode D1 is forward biased it will conduct charging capacitor C1 to a value of VP and polarities as indicated.

178

Diodes APPLICATIONs

Now assume the polarities as indicated in figure 10.29 (c) then the diode D2 will be forward biased and the diode D1 will be reverse biased. Since diode D2 is forward biased it will conduct charging capacitor C2 to a value of VP and polarities as indicated. The sum of the combination of capacitors C1 and C2 will now result in a magnitude of 2 VP. In this manner we will be able to obtain magnitudes of 3 VP and 4 VP. 4 VP 2 VP

-

+

-

C2

C4

D1

-

C1

+

D2

+

D3

-

C3

D4

+

VP 3 VP (a)

+

-

-

C2 D1

-

-

C1

+

+ C2

D2

D1

+ (b)

-

C1

D2

+ (c)

Figure 10.29

Exercise 10.1 1.

Clippers may be sub-divided into series and parallel as well as positive and negative types. Draw the network diagrams as well as input and output waveforms that will describe the principle of operation of each of these types of clippers. The magnitude of the input waveform has a peak- to-peak value of 10 volt and is centrally clamped around the zero- volt line. In your output waveforms you are required to make provision for Germanium as well as Silicon diodes.


2.

Diodes are commonly used for rectification purposes. Make use of suitable network diagrams as well as input and output waveforms to illustrate the following types of rectification. 2.1 Half-wave; 2.2 Full-wave (two diodes); and 2.3 Bridge rectifier. Indicate on your network diagrams the direction of all currents that may flow in the network.

3.

The transformer of a bridge type rectifier has a turns ratio of 15:1 and is supplied from a 220 V/100 Hz alternating supply. Calculate the: 3.1 Average dc output voltage; 3.2 Ripple voltage; 3.3 Ripple factor; and 3.4 PIV-ratings for the diodes.

4. A 220 V/50 Hz supply rated at 15 ampere feeds a step-down transformer with 220 windings on the primary side. Should 10 volt be measured on the secondary side then determine for full-wave as well as half-wave the: 4.1 Average dc output voltage; 4.2 Ripple factor; and 4.3 RMS secondary current. 5. Explain in your own words what is meant by a filter network used in conjunction with rectifier circuits. Enhance your explanation with suitable sketches. 6. Draw the following fully labelled circuit diagrams: 6.1 Half-wave rectifier with: (a) An RC-filter; and (b) A capacitor filter. 6.2 Full-wave rectifier (two diodes) with: (a) LC-filter; and (b) An RC-filter. 6.3 Full-wave rectifier (bridge circuit) with: (a) LC-filter; and (b) A capacitor filter. 7. A half-wave rectifier is fed from a 220 V/60 Hz supply and the following additional data is given: • Peak secondary voltage = 25 volt • Value of filter capacitor = 100 µF • Load current = 30 mA Use the above data and determine the:

179

180

Diodes APPLICATIONs

7.1 Output dc voltage (VDC); 7.2 Output ripple voltage (Vr(RMS)); and 7.3 Ripple factor expressed as a percentage (r %).

8. Use a graphical representation to illustrate the concept of voltage regulation and then use your own values to show how voltage regulation can be determined. 9. Use suitable circuit diagrams and illustrate the concept of voltage doubling as well as voltage multiplication. 10.

Use suitable sketches or graphic representations to illustrate the concept of voltage regulation. Use a graphical representation to illustrate the concept of voltage regulation and then use your own values to show how voltage regulation can be determined.

11. Use suitable circuit diagrams and illustrate the concept of voltage doubling as well as voltage multiplication. 12. A full-wave bridge rectifier is constructed and a capacitor filter is used. The following additional data concerning the rectifier is given: • Input to transformer = 280 V/60 Hz • Voltage measured across secondary winding = 18 volt • Value of load resistor = 1,2 k-ohm • Load current = 15 mA Use the above information and determine the following quantities for a 150 µF and 90 µF capacitor. 12.1 Average value; 12.2 RMS value; 12.3 Output dc voltage; 12.4 Output RMS ripple voltage; and 12.5 Ripple percentage. 13. The following data concerning a power supply with an RC-filter is given: • Supply voltage = 250 V/100 Hz • Voltage across first capacitor = 90 V • Ripple voltage across first capacitor = 10 V • Series resistor = 1,2 k-ohm • Second capacitor = 100 µF • Load resistor = 2,5 k-ohm Determine for half-wave and full-wave the: 13.1 dc-output voltage; 13.2 Ripple voltage; and 13.3 Ripple factor.


14.

An LC-filter consists of an inductor having a value of 1,4 H and a resistance of 120 ohm. The capacitor has a value of 120 µF and the load resistor has a value of 80 k-ohm. A voltage of 12 volt is measured across the first capacitor and has a ripple of 1,6 volt. The primary of the transformer is fed from a 180 V/50 Hz supply. Use the given information to calculate the: 14.1 dc-output voltage; and 14.2 Ripple factor. Before you commence with the calculations you are required to draw a neat labelled circuit diagram of the power supply inserting all relevant values as is given.

181

182

Diodes APPLICATIONs

Special Diodes and Applications

183

CHAPTER

11 Special diodes and applications Learning Outcomes On completion of this module you will be able to: •

•

Discuss with the aid of circuit diagrams the operation, characteristics and application of the following diodes: – Zener diodes; – Varactor diodes; – Tunnel diodes; – Photo-diodes: – Photo-sensitive diodes; and – Photo-emissive diodes. Calculate: – Current values; – Voltage values; – Resistor values; and – Wattage ratings with reference to a Zener diode.

11.1 The Zener diode Should an ordinary diode be reverse-biased then a small leakage current will flow which for all practical purposes are normally disregarded. Should we however increase the magnitude of the reverse-bias then the diode will be destroyed since it will not be able to handle this increase since the PIV of the diode will be exceeded. When this PIVpoint is controlled with the aid of a series resistor we end up with a Zener diode voltage regulator. In order to understand the operation of this type of diode we need to have a look at the characteristic curve for a Zener diode. A point that must be mentioned is that a Zener diode operates in the reverse characteristic of the curve and very good voltage regulation capabilities are obtained and will behave like a normal diode when forward-biased. The point at which the Zener diode displays the voltage regulation capabilities are specifically manufactured as such.

184


11.1.1 Electrical characteristics of a Zener diode The following characteristics are the most common and can be found in all specification sheets and are indicated on the characteristic curve illustrated in figure 11.1 (b) whereas figure 11.1 (a) illustrates the IEC circuit symbol. Zener current (IZM) This is the maximum allowable current when the diode is used as a reference source at the Zener point. Reference voltage (VZ) This is the Zener breakdown voltage and specifies the maximum voltage when the diode is used as a reference source. (Also termed the Zener knee) Average forward current (IF (AV)) This is the average forward current permissible when the diode is forward biased but is not important when the Zener diode is used as a reference source. Repetitive reverse current (IZ (RM)) This is the maximum specified reverse current that is permissible when the Zener diode is used as a reference source. Zener dissipation rating (PZ) This is the maximum power dissipation rating of the Zener diode. Junction temperature (Tj) This is the maximum operating temperature of the Zener diode. Zener current near the knee (IZK) This is the minimum current required to sustain breakdown. Important to note that this characteristic curve of the Zener diode only indicates the reverse bias characteristics since it is in this region that the diode is used.

IF Forward bias region VZ

VR

VF IS IZK

Cathode

IR IZT

Anode

Reverse bias region (a)

(b)

Figure 11.1

IZM

185

11.1.2 Application of the Zener diode All Zener diodes are manufactured in order to conduct at a given voltage rating in the reverse-bias area. Zener diodes are mainly used as voltage regulators and are available from 1, 5 volt up to 30 volt but higher values can be obtained. The voltage rating for which it is designed is termed the Zener voltage and has thát value as voltage reference. What this basically means is that the design value is thát value that will be measured across the Zener diode when it is conducting and is used as a voltage reference source. An elementary circuit diagram in figure 11.2 illustrates the typical application of a Zener diode. The circuit is termed a voltage regulator since the Zener diode will supply a constant voltage (regulated) to the load depending on what the design value is. The resistor RS is connected in series with the Zener diode to limit the current flow through the Zener diode as well as the load RL which is connected in parallel across the Zener diode and will therefore supply a constant voltage to the load. Using the concepts of Kirchhoff ’s current and voltage laws we may formulate the following expressions: VO = VZ

VRS = Vi - VZ

IRS = IZ + IL

Should the load RL not be connected then IZ = 0 and VRL = VZ Various component values can be determined by the following mathematical expressions:

+

VRS IRS

+

RS

Unregulated dc Vi

IZ

IL

VZ

RL DZ

Regulated dc VO

-

Figure 11.2

P VZ V IZ = Z RL = RS = RS VZ IL IRS

where VO = Output voltage VZ = Zener voltage

Vi = Input voltage IRS = Current through series resistor

186


VRS = Voltage across series resistor PZ = Zener power dissipation VRS = Voltage across series resistor

IZ = Zener current IL = Load current

Example 11.1 Consider the following circuit diagram. + IRS

RS

10 V

+

IZ

IL

20 mA

RL

6 V at 3 W

VO

DZ

-

Use the available data and determine the: (a) Value of the load resistor; (b) Value of the series resistor; and (c) Current magnitude through the Zener diode.

Solution: VZ (a) RL = IL 6 = -3 20 × 10

=

300 ohm

V (b) RS = RS VRS = Vi - VZ IRS = IZ + IL I RS = 10 – 6 = 0, 5 + 20 × 10-3 4 = = 4 volt = 0,52 ampere 0,52 = 7,69 ohm PZ (c) IZ = VZ 3 = 6 = 0,5 ampere

187

A more practical circuit diagram of a Zener diode voltage regulator is illustrated in figure 11.3. This circuit consists of two sections as indicated. The rectifier section will not be discussed since it was done in the previous chapter. The output from the rectifier section is applied to the regulator (voltage reference) section and ZD and RS form a voltage divider section. The voltage drop across the Zener diode will always be its design value and the rest of the voltage will appear across the series resistor. This will only be so if the voltage applied to the regulator section exceeds the Zener diode design value. Should this value be less than the Zener diode design value there will be no output.

+

-

T1

+

D1

+

D2

+ RS

+ -

+

C1 DZ

RL

VOB

VOA

Rectifier section

Regulator section

Figure 11.3

Since the load resistor RL is in parallel with the Zener diode, the Zener diode voltage will appear across the load as well as VOA which is termed the regulated output. Should the output be taken across the Zener diode and the series resistor an output VOB will be obtained which is termed the unregulated output.

11.2 The varactor diode These diodes are also known as epicaps, varicaps and/or voltage variable capacitance diodes and are also operated in the reverse-bias area of the characteristic curve. You will recall from the PN-junction theory that when such a junction is reverse-biased that the depletion layer increases. Since a PN-junction is not an electrical junction this depletion layer will have the characteristics of a capacitor.

188


11.2.1 Circuit symbol and characteristic curve The IEC circuit symbol and characteristic curve is illustrated in figure 11.4 (a) and (b). Important to note that the characteristic curve in figure 11.4 (b) is not current versus voltage as with many other characteristic curves but capacitance versus voltage and in this instance it is reverse voltage.

11.2.2 Application of a varactor diode The main application for a varactor diode is that of tuning circuits in FM-receivers as well as television receivers and replaces the variable tuning capacitor. Such an application is illustrated in figure 11.5.

Anode

C a p a c I t a n c e

Cathode

Reverse bias (a)

(b)

Figure 11.4

Since the magnitude of the reverse-bias is proportional to the width of the depletion layer a high reverse-bias will result in result in a wider depletion layer and a lower reverse bias will result in a narrower depletion layer. This will then in turn determine the capacitance CPN of the varactor diode which is inversely proportional to the magnitude of the reverse-bias.

189

T1

+ VCC

f out

L1

D1

R2 CC2

CC1

R3

Q1

RV1 Bias control for D1 capacitance

R1

- VCC

Figure 11.5

In the diagram illustrated in figure 11.5 a tuned circuit is formed by an inductor L of the transformer T1 and the capacitance of the varactor diode D1 which is further dependant on the magnitude of the reverse-bias which is controlled by the setting of RV1. The range of frequencies that the tuned circuit will operate is mathematically given by: where 1 fr = 2 × π (L × C) ½

fr π L C

= = = =

frequency in hertz 3,142 value of inductor in henry value of capacitance in farad

It should therefore be obvious that should the reverse-bias on D1 be changed that its capacitance will also change and therefore the frequency of the tuned circuit will likewise change.

190


Example 11.2 A variation in reverse bias will cause a capacitance variance of 15 pF to 100 pF in the circuit diagram illustrated in figure 11.5. Determine the frequency range of the tuned circuit if it is given that the inductor L has a constant value of 100 µH.

Solution: fr = 1 2 × π (L × C) ½ = 1 2 × 3,142 (100 × 10-6 × 15 × 10-12)½ = 4,1 MHz fr = 1 2 × π (L × C) ½ = 1 2 × 3,142 (100 × 10-6 × 100 × 10-12)½ = 1,6 MHz The frequency fout will therefore vary between 1,6 MHZ and 4,1 MHz.

11.3 The tunnel diode The tunnel diode is also termed an Esaki diode. It is also a two-terminal device and is almost exclusively used as a high-frequency component in the following applications: • • •

An amplifier; An oscillator; and A switch.

Its very fast response it displays makes it ideal for use in the above applications. The main advantage of this type of diode is that it requires a smaller bias voltage in the region of mV if compared to other types of diodes.

11.3.1 Circuit symbol and characteristic curve The IEC circuit symbol and characteristic curve is illustrated in figure 11.6 (a) and (b).

191

The tunnel diode is manufactured from heavily doped semi-conductor materials to form a PN-junction and is used in the forward-bias region as indicated in figure 11.6. The large number of impurities in the material ensures that the depletion layer to be so thin that electrical charges moves through the junction very easily and is termed ‘tunnelling’, hence the name Tunnel diode. The operating principle and characteristic curve can best be described in the following manner. At switch-on there is a sudden and steep increase in current ID up to the point IP then there is a sudden decrease in current ID up to the point IV with an increase in forwardbias voltage. This area in which the sudden decrease takes place is termed the ‘negative resistance area’ of the diode and the negative resistance RD of the diode is the most important characteristic of the Tunnel diode. This is completely opposite to a ‘resistance’ in which there is an increase in current when the supply voltage is increased. A further increase in forward-bias will have a corresponding increase in ID and will almost follow the characteristic of an ordinary diode. mA

Negative resistance region

IP

ID

Anode

Cathode IV mV VR (a)

VM

VV

VF

(b)

Figure 11.6

11.3.2 Application of a tunnel diode The main application of a Tunnel diode is that of a parallel amplifier and such a practical circuit diagram is illustrated in figure 11.7.

192


IB

IL ID

VB D

RL

VR

VS ac

Figure 11.7

This circuit must, for operation as an amplifier, be biased to the centre of its negative resistance region. This parallel diode amplifier has the following characteristics: • • •

Current gain; Power gain; and No voltage gain.

11.4 Photo-diodes Photo-diodes are classified in two main categories namely photo-sensitive components and photo-emissive components.

11.4.1 Photo-sensitive diodes The photo-diode is a semi-conductor component that will depend on light energy in order to become effective. But it is important to note that the photo-diode be connected to a supply voltage and reverse-biased in order to become operative. Should the PNjunction be exposed to a source of light electron-hole pairs will be created which will cause a small reverse current to flow. The magnitude of this current will be determined by the intensity of the light source. The more intense this light source the greater the magnitude of the reverse current.

11.4.1.1 Circuit symbol and characteristic curve The IEC circuit symbol and characteristic curve is illustrated in figure 11.8 (a) and (b). Photo-diodes are always used in the reverse-bias region of the characteristic curve and the incident falls onto the PN-junction through a window in the casing of the diode. This source of light energy may be from the sun or it may be artificially generated light. A small current, termed the ‘dark current’ will flow in the absence of a light source and typical values are 10 µA for germanium diodes and 1 µA for silicon diodes. It can be seen that as the light intensity increases so does the reverse current also increase.

193

I 1 200 lumen

Radiant light energy

800 lumen

IR

400 lumen 0 lumen V

Anode

VR

Cathode (a)

(b)

Figure 11.8

11.4.1.2 Application of a photo-diode A photo-diode may switch a device ‘on’ or ‘off ’ depending on how it is connected in the circuit. Two such applications are illustrated in figure 11.9 (a) and (b). Figure 11.9 (a) will switch the device ‘on’ whereas figure 11.9 (b) will switch the device ‘off ’. The circuit is constructed in such a manner that the device being controlled is in the ‘off ’ state (figure 11.9 (a)) and need to be energised. In the absence of incident light the current through diode D is very small (dark current) and the transistor Q1 cannot conduct. However, should incident light now fall onto the PN-junction a current will flow dependant on the intensity of the incident light. Typical values of sensitivity lie between 10 mA and 50 mA/lumen. Since the current will increase because of the presence of incident light the transistor will switch on and start conducting. The transistor is also an amplifier and should the collector current increase sufficiently the relay contact will be energised, providing that the holding current value of the relay is exceeded, into the N/C position and the device coupled to it will commence operation. The circuit is constructed in such a manner that the device being controlled is in the ‘on’ state (figure 11.9 (b)) and need to be de-energised. Should incident light fall onto the PN-junction producing current to flow the diode will become a virtual short circuit and the transistor will switch off.Since there will be no collector current the relay will go into the N/O position because the holding current value of the relay has dropped below its specified value. In this manner the device to be controlled will be switched ‘off ’.

194


N/O (off)

N/O (off)

Relay contacts N/C (on)

Relay contacts

To supply and device to be controlled

To supply and device to be controlled + VCC

N/C (on)

+ VCC Incident light Relay coil

RB

Relay coil D QI

Q1

Incident light RB

RE

RE D

- VCC

- VCC (a)

(b)

Figure 11.9

11.4.2 Photo-emissive diodes Electrons and holes will continually combine in a PN-junction when there is a flow of electrical current through the device. What is however important is that the electrons are in the conduction band and the holes are in the valence band. When an electrical current now flows through the diode the electrons have to give off energy in order to drop from the conduction band into the valence band and this energy is given off in the form of an electromagnetic radiation of different colours depending on the doping level of the semi-conductor material. Silicon and Germanium however radiate mainly in the Infra-red spectrum which is not detectable by the human eye. To overcome this problem Gallium Arsenide Phosphide (Cds) and Gallium Phosphide is used and the radiation will now take place within the visible spectrum. This radiation takes place at the area of re combination and the PN-junction is placed in a translucent encapsulation and the energy given off is now visible as light. This component is termed a ‘Light Emitting Diode’ (LED) and the colours available include red, green, amber, yellow and blue. For this radiation to take place the LED need to be forward-biased. There are however dual-colour LED’s available where forward-bias produces one colour and reverse-bias produces another colour.

11.4.2.1 Circuit symbol and characteristic curve The IEC circuit symbol, construction and characteristic curves are illustrated in figure 11.10 (a), (b), (c) and (d).

195

Figure 11.10 (a) illustrates the IEC circuit symbol and figure 11.10 (b) illustrates the physical construction of an LED. Figure 11.10 (c) illustrates the characteristic curve for an LED and it must be noted that these diodes normally operate at a voltage of between 1,4 V and 1,7 V and at currents ranging from 50 mA to 100 mA. The curve illustrated in figure 4.10 (d) indicates the intensity or brightness of the radiated light depending on the magnitude of the forward current. The higher the forward current the more intense the radiation is. I

Radiant light energy IF

Anode

V

Cathode

VF

(a)

(c)

I n t e n s I t y

Cathode

Lumen

I

Anode

IF (b)

(d)

Figure 11.10

11.4.2.2 Application of an LED Light emitting diodes are mainly utilised as indicating components or devices such as seven-segment displays, dot-matrix displays or panel indicators. A seven-segment LED display and a 7 × 5 dot-matrix are illustrated in figure 11.11 (a) and (b).

196


a f

b g

e

c d Decimal points (a)

(b)

Figure 11.11

• Seven-segment display The seven-segment display (figure 4.11 (a)) consists of seven LED’s arranged in a figure eight. Should the applicable segments be energised the decimal numbers from 0 to 9 can be displayed as well as a limited number of alphabetical letters. Typical applications include the screen of your pocket calculator. • 7 × 5 dot-matrix display The main disadvantage of not being able to display all the alphabetical letters using a seven-segment display is overcome by using a 7 × 5 dot-matrix (figure 4.11 (b)) display. By applying bias to the correct LED’s any alpha-numerical display can be obtained. A single LED can be mounted offset to the matrix which is utilised for the decimal point when displaying numerical numbers.

11.4.2.3 Circuit diagram for the application of an LED A typical circuit diagram for use with an LED is illustrated in figure 11.12. Since LED’s are current operated devices one is required to limit the current through the LED and that is acquired by inserting resistor R2 in the given circuit.

197

+ VCC LED

R2

Q1 Vi

R1 - VCC

Figure 11.12

Exercise 11.1 1. Describe with the aid of a characteristic curve and circuit symbol the operation and electrical characteristics of a Zener diode. 2. You are supplied with the following components: • A step-down mains transformer. • Four diodes. • Two resistors. • One electrolytic capacitor. • One other component. Use the above components and draw the circuit diagram of a device that will be able to deliver a constant 6,4 volts to a load of 150 ohm. 3. The following values were measured in a rectifier and regulator circuit: • Load current = 50 mA • Zener diode specifications = 12 V/3 W • dc input to regulator section = 16 V Use this data and determine the: 3.1 Value of the series resistor; 3.2 Current magnitude through the Zener diode; and 3.3 Value of the load resistor. You are now further required to draw a neat labelled circuit diagram of this regulator section and insert all relevant values that were supplied and calculated. 4. Draw the IEC circuit symbol and then describe in your own words the difference between a photo-sensitive component and a photo-emissive component.

198


5. A device can be controlled by using a photo-diode which can cause the device to be controlled to either switch ‘on’ or ‘off ’. Draw two neat labelled circuit diagrams and explain the operation thereof in detail. 6. Explain with the aid of the characteristic curves the operation of a photo- emissive component. 7. Explain in your own word why certain semi-conductor material emits light in the visible spectrum. 8. What advantage does the 7 × 5 dot-matrix display have over the seven- segment display? 9. Show by means of a neat labelled circuit diagram how an LED can be protected against excessive current.

TRANSISTORS

199

CHAPTER

12 Transistors Learning Outcomes On completion of this module you will be able to: • • •

• •

• •

Describe the operation of a transistor with the aid of schematic block diagrams and circuit symbols; Understand the concept of transistor voltages; Calculate the values of: – Base currents; – Collector currents; and – Emitter currents. Understand and explain the transistor as an amplifier and switching device taking note of the areas of operation as well as the switching time of a transistor; Explain the characteristics and the basic operation of a transistor as a: Switch; – Basic common emitter amplifier; – Basic common collector amplifier; and – Basic common base amplifier. Do development of a common emitter amplifier, fixed bias and automatic bias amplifiers; and Do the graphical analysis of a common emitter amplifier.

12.1 Introduction The name transistor is derived from the term ‘transfer resistor’ and you will notice that the parts of the words printed in Italics spells out the word transistor. The principle of operation of the transistor is an extension of the PN-theory dealt with when the diode was discussed. The transistor is a three-terminal two junction component and is commonly referred to as a ‘junction transistor’ but it must be noted that there are other types of transistors also available but will be covered at a later stage.

200

TRANSISTORS

12.2 The transistor The junction transistor consists of two types of extrinsic or doped semi-conductor material. Since it contains only two junctions it however has three terminals and one type of doped material (N- or P-type) is sandwiched between two types of the other type of doped material (N- or P-type). This arrangement provides us with the opportunity of obtaining two types of transistors namely an NPN- or PNP transistor. Their basic operation is virtually the same but their bias arrangement is different. The schematic block diagrams and IEC circuit symbols are illustrated in figure 12.1 (a) and (b). Figure 12.1 (a) illustrates an NPN transistor and figure 12.1 (b) illustrates a PNP transistor. Take note of the voltage polarities indicated on the IEC circuit symbols since this will be your guide for the future in electronics. It must also be noted, and this is a very important aspect, that when we talk about current flow, that we refer to electron flow and that is from negative to positive. We will further refer to hole flow which is the opposite to electron flow, from positive to negative. It may not be correct to refer to hole flow as conventional current flow. Emitter

Base

Collector

Emitter

Base

Collector

N

P

N

P

N

P

Emitter-base junction

Collector-base junction

Emitter-base junction

VCE Emitter

VCE Collector

-

Emitter

++

Base

Collector

+

+ VBE

Collector-base junction

--

VCB

VBE

(a)

Base

VCB

(b)

Figure 12.1

In order to grasp and understand the operation of a transistor we need to familiarise ourselves with the concept of the depletion layers again. This is illustrated in figure 12.2 (a) and (b).

201

Emitter

Base

Collector

Emitter

Base

Collector

N

P

N

P

N

P

Depletion layers (a)

(b)

Figure 12.2

The following points need to be taken note of. • • • •

The base region is much thinner than the two outer regions. The two outer reasons are more heavily doped than the base region. Because of the doping level of the two outer regions the depletion layer at both junctions will penetrate deeper into the base region. This deeper penetration of the depletion layers at both junctions therefore minimises the distance between the two junctions.

These are the conditions that will exist when there is no voltage supply connected to the transistors. Should we now connect a voltage supply to the transistor a number of significant and important changes will take place. This concept is illustrated in figure 12.3 (a) and (b).

IE

+

-

Forward bias

IB

Emitter

+

Reverse bias

+ IE

-

+

Forward bias

IC

Collector

IB

Emitter

-

Reverse bias

IC

Collector

Holes emitted

N

P

N

P

Base

N

P Base

Depletion layers (a)

(b)

Figure 12.3

In both transistors the emitter-base is forward-biased and the collector-base is reversebiased. This is a fundamental condition to exist should a transistor be used as an amplifier. This will be further explained when the operating areas of a transistor is discussed. (Refer to the section on PN-junctions since the transistor may be seen as two PN-junctions connected back-to-back).

202

TRANSISTORS

You will notice that for both types of transistors the current directions are indicated to be in the same direction but there is a subtle difference. In the NPN transistors we have a current flow which is by means of electron flow as majority carriers. However, in the PNP transistors we have a current flow which is by means of hole flow as majority carriers. Important to note is that the width of the depletion layers have changed since we added the supply voltages. It would therefore appear that the width will have an influence on the conduction properties but this is not the case. Should you refer back to figure 12.1 (a) and (b) that in the case of the NPN transistor the high positive potential will pull the electrons through the base region and in the case of the PNP transistor the high negative potential will pull the holes through the base region. Note that the current directions are indicated as external direction. The current magnitudes for both transistors are given by the following mathematical expression: where IE = emitter current in ampere IE = IB + IC = base current in ampere IB = collector current in ampere IC Important to note is that the magnitude of the base current will seldom exceed ± 2% of the emitter current. To obtain a clearer view of the above mathematical expression refer to figure 12.4 (a), (b) and (c). You will recall the transistor is derived from ‘transfer resistor’ and we can now apply Ohm’s Law to this schematic diagram. A resistor variable R, acting as the base, is placed between the two semi-conductor materials and we will find that the current between the emitter and collector will be regulated by the variable resistor R. Should we now vary the resistance of the variable resistor R the electrons that will be emitted will not be the same as those collected by the collector. The resistance of R will thus have an influence on the magnitude of the current flow. We can therefore safely assume that the conduction of a transistor is mainly dependant upon the increase and decrease of current between the base-emitter. It will be found that the higher the current between the base-emitter junctions the lower the resistance of R and the lower the current between base-emitter junctions the higher the resistance of R. We can therefore safely assume that the collector current is controlled by the base current and the collector current is almost equal to the emitter current.

203

Base Emitter

R

NPN

Collector

C

IC

B Electron or hole flow

IE IE

VS (a)

C

IC

B

IB

Amperemeter

PNP

E

IB

IE IE

E

Electron flow

Hole flow

(b)

(c)

Figure 12.4

Example 12.1 It was found by measurement that the current flow in the emitter of a transistor is 80 mA and that causes a current flow of 0,8 mA in the base of the transistor. Determine the magnitude of the collector current.

Solution: IE = IC = = =

IB + IC IE - IB 80 × 10-3 - 0,8 × 10-3 79,2 mA

12.3 Operating regions of a transistor There are three regions in which a transistor is normally operated and that is illustrated in the graphical representation in figure 12.5.

204

TRANSISTORS

IC

Saturation region IB IB IB

Active region

IB IB Cut-off region VCE

Figure 12.5

Three main operating regions can be identified for a transistor as illustrated in figure 12.5. The output characteristics illustrated is that for the common emitter configuration in which most transistor applications are used. Important to note is that these three regions are termed ‘static characteristics’ since they are determined by dc conditions which are static. There are also another condition which will be covered later namely ‘dynamic characteristics’ which takes into consideration and ac (signal) application which on the other hand is a changing quantity. • The cut-off region The cut-off region is thát region where the emitter-base as well as the collector-base junctions is reverse biased. There is no major current that will flow besides a minute leakage current and the transistor is ‘off ’ since there is no collector current that can flow. • The active region The active region is thát region where the emitter-base junction is forward biased and the collector-base junction is reverse biased. Majority carriers, electrons or holes, are allowed to flow as a current. The transistor is now ‘on’ and typical voltage values that will be measured are: • ± 0,7 V between emitter and base for silicon transistors; • ± 0,3 V between emitter and base for germanium transistors; and • ± 20 V between collector and base and may in some instances even exceed this value. In this region a transistor may be used as an amplifier.

205

• The saturation region The saturation region is thát region where the emitter-base as well as the collectorbase junctions is forward biased. The transistor is conducting heavily and the voltage between collector and emitter is less than 0, 6 V. In this region a transistor may be used as a switch.

12.4 The switching speed of a transistor The switching time of a transistor is a very important characteristic. In order to explain this concept we need to study the circuit diagram and waveforms illustrated in figure 12.6 (a) and (b). On Base current Off + VCC 90 % IC

RC Collector current Q1

IB 10 %

- VCC

(a)

td

tr

ts

ton

tf toff

(b)

Figure 12.6

Like may things in life response is not always immediate and when an input current is applied to the base of a transistor it does not respond immediately since the electrons have, what is termed a ‘transit time’, to move across the junction as well as the junction capacitance to overcome. This time delay is the time it takes between the applications of the base current and the instance that the collector current starts flowing. This is indicated on the graphical representation as td and is termed the delay time. However, even if the current flow commences the collector current does not reach maximum value immediately and a time period lapses termed the rise time. This rise time tr may be defined as the time it takes for the collector current to rise from 10% of its maximum value to 90% of its maximum value. It should be obvious that the time taken for the collector current to reach 90% of its maximum value is the sum of the delay time td and the rise time tr which gives us a total time ton as illustrated in figure 12.6.

206

TRANSISTORS

Should the input base current now be removed from the transistor will not switch off immediately. The collector current will only cease flowing once the turn-off time toff has lapsed. This turn-off time also consists of two time periods namely the storage time ts and the fall time tf and these times may be defined in the following manner. Storage time is thát time that lapses since the majority charge carriers are trapped in the depletion layer of the junction when the junction polarity is reversed. The fall time on the other hand is thát time taken for the collector current to reduce from 90% of its maximum value to 10% of its maximum value. These times are clearly indicated on the graphical representation in figure 12.6. It must be noted that these time delays mentioned, ton and toff are in the order of micro-seconds and even nano-seconds for fast switching transistors.

12.5 The transistor as a switch A transistor can be utilised as an electronic switch by merely applying the principles that we have discussed up to this point. We have seen the effect that the base current IB has on the operation of a transistor and this will be the principle that we will use to operate a transistor as an electronic switch. Such a circuit diagram of a transistor used as an electronic switch is illustrated in figure 12.7. Supply voltage to device or load to be controlled + VCC N/C ( on )

D1

N/O ( off ) Relay coil and contacts

Q1

Battery

S1

- VCC

Figure 12.7

Device or load to be controlled

207

The circuit is illustrated with the switch S1 as well as the relay contacts in the N/O (off) position. The load, which could for instance be a borehole pump, is also off. When switch S1 is activated a high positive voltage will appear on the base of transistor Q1 and the transistor will start conducting heavily which will pull the relay contacts into the N/C (on) position, provided that the holding current value of the relay is exceeded, and the remote load will also switch on. Please note that the load to be controlled must have its own supply. In order to protect the transistor when switch S1 is de-activated a diode D1 is placed in parallel with the relay coil. The reason for this is that during the time that the transistor is conducting there will be a current flowing through the relay coil which will keep the relay contacts in the N/C (on) position and a magnetic field is set up around the relay coil. At switch-off the magnetic field around the relay coil will collapse causing a back-emf which could damage the transistor. During this period of back-emf the diode will conduct thus short-circuiting the relay coil and no damage will be done to the transistor.

12.6 The transistor as an amplifier By virtue of a transistor having three terminals, emitter, base and collector, will allow us to use a transistor in three configuration or modes as an amplifier. But we first need to clarify the term amplifier. An amplifier may de defined as an electronic device that will increase the value of whatever is fed into the device. The three configurations or modes of amplifier operation are: • • •

Common emitter; Common base; and Common collector (emitter follower).

These amplifiers are termed common on the grounds of reference to the terminal which is common to both the input as well as the output. In the case of the common emitter the emitter is common to the input as well as the output, in the common base the base is common to the input as well as the output, in the common collector the collector is common to the input as well as the output. Each configuration is unique having its own characteristics as well as application which will be dealt with in this section. Only the very basic configurations will be dealt with and a more detailed approach will follow in latter sections.

12.6.1

The common emitter amplifier

The circuit diagram in figure 12.8 illustrates a transistor configured in the common emitter mode.

208

TRANSISTORS

+ VCC

RC CC2 CC1

Q1 Output

Input - VCC

Figure 12.8

The common factor, emitter and other terminals are circled with a dotted line. The input signal is applied between the base and emitter and the output signal is obtained between the emitter and collector and the common denominator here being the emitter. This is the most frequently and commonly used configuration and has the following characteristics: • • • • •

12.6.2

High current gain; High voltage gain; High input impedance; High output impedance; and 180º phase shift between input and output.

The common collector amplifiere

The circuit diagram in figure 12.9 illustrates a transistor configured in the common collector mode. The common factor, collector and other terminals are circled with a dotted line. The input signal is applied between the base and collector and the output signal is obtained between the emitter and collector and the common denominator here being the collector. Note that the -VCC line becomes the collector under dynamic conditions.

209

+ VCC CC1

Q1 CC2 C

Input RE

Output

- VCC

Figure 12.9

This configuration is mainly used for matching purposes and has the following characteristics: • • • • •

12.6.3

High current gain; Low voltage gain; High input impedance; Low output impedance; and 0º phase shift between input and output.

The common base amplifier

The circuit diagram in figure 12.10 illustrates a transistor configured in the common base mode. + VCC

RC

CC2 CC1

Q1 Output

Input

- VCC

Figure 12.10

210

TRANSISTORS

The common factor, base and other terminals are circled with a dotted line. The input signal is applied between the base and emitter and the output signal is obtained between the base and collector and the common denominator here being the base. This configuration is also mainly used for matching purposes and has the following characteristics: • • • • •

Low current gain; High voltage gain; Low input impedance; High output impedance; and 0º phase shift between input and output.

All the circuit diagrams given up to this point have made use of NPN transistors but PNP transistors may also be used. All that need to be done is to substitute the NPN transistor with a PNP transistor as well change the +VCC line and -VCC line around. The rest of the circuit including the input and output stay exactly the same. The only real difference between the two circuits is the majority carriers for the different types of transistors. All of these characteristics mentioned here, especially those for the common emitter and common base will be proved when the h-parameters are discussed at a later stage in your career. For simplicity a summary of these characteristics is given in table 12.1.

Characteristics Current gain Voltage gain Input impedance Output impedance Phase shift

Type of amplifier Common emitter Common base High High High High High 180º

Low High Low High 0º

Common Collector High Low High Low 0º

Table 12.1

12.7 Common emitter amplifier development The first concept that needs to be discussed is that of transistor biasing since this concept will determine the operation of the transistor. This concept is illustrated by the schematic block diagram (a) and circuit diagram (b) in figure 12.11.

211

VCE IE

IC

VBE

VCB IB

IC

Emitter

IB

VCE

Collector VBE

N

P

IE

N Base (a)

(b)

Figure 12.11

Two important concepts are identified in figure 12.11. • •

The emitter-base is forward-biased by VBE; and The collector-base is reverse-biased by VCE.

Upon careful inspection of figure 12.11 (a) you will notice that two batteries are used in order to obtain a voltage on the base of the transistor. Since these batteries are in series forming a voltage divider they may be replaced by a voltage divider resistive network and the transistor may then be operated off a single voltage source. The following two methods of obtaining transistor biasing are widely used for a single voltage source.

12.7.1

Fixed bias common emitter amplifier

The common emitter amplifier circuit diagram illustrated in figure 12.12 is that of a fixed bias configuration. This means that only one supply is used bias both the base and the collector. + VCC

RB

RC CC2

Q1 Vi

CC1

VBE

Figure 12.12

VCE

VO

- VCC

212

TRANSISTORS

This circuit will operate as an amplifier when and only when the base-emitter is forward-biased and the base-collector junction is reverse-biased. In order to function as an amplifier the collector-emitter voltage VCE must always be positive with respect to ground. Should this not be the case and the collector-emitter voltage VCE reduces to a fraction of the base-emitter voltage VBE the base-collector junction will become forward biased and the transistor will then not function as an amplifier since the transistor will then move into the saturation area of its characteristic. To ensure satisfactory operation as an amplifier RB and RC must be so calculated so as to prevent the transistor from going into saturation which means that the collector-emitter voltage VCE must always remain greater than the base-emitter voltage VBE. The biasing of the amplifier will also be dependant upon the current - or β-gain (beta) of the transistor and for the transistor to always operate in the linear region of the characteristic the following principle shall apply: = β × IB where IC = collector current in ampere IC β = current gain of transistor IB = base current in ampere This type of fixed bias however does produce problems with stability since the collector current IC will vary with a change in temperature because of the following factors: • • •

Leakage current will double for every 10ºC change in temperature; The base-emitter voltage VBE will decrease by 2,5mV/º C; and The current gain β which increases with an increase in temperature.

The following mathematical expressions can be used for the design of a fixed bias amplifier. V -V RC = CC CE where IC V -V RB = CC BE IB

RC VCC VCE RB VBE

= = = = =

value of collector resistor in ohm supply voltage in volt collector-emitter voltage in volt value of base resistor in ohm emitter-base voltage in volt

213

Example 12.2 The following data for the fixed bias amplifier in figure 12.12 is given: = 10 V • VCC • β = 100 = 10 mA • IC = 5V • VCE Use the above information and calculate the resistive values of the base resistor and the collector resistor. Use is made of a Silicon resistor and VBE may be used as 0,7 volts.

Solution: V -V RC = CC CE IC 10 - 6 RC = 10 × 10-6

= 400 k-ohm

V -V RB = CC BE IC = β × IB IB IB = IC = 10 - 0,7 β -6 = 10 × 10-6 0, 1 × 10 = 9,3 M-ohm 100 = 0,1 µA

The stability for the amplifier may be improved by adding an emitter resistor RE as well as an emitter by-pass capacitor CE and for the calculation of these component values one may assume that IE is approximately equal to IC and VE is equal to 1/10 of VCC. The value of the emitter by-pass capacitor must have a reactance value of 1/10 of the emitter resistor value. Such a circuit diagram is illustrated in figure 12.13. Therefore, for a fixed bias amplifier with emitter resistor and emitter by-pass capacitor the component value expressions are mathematically given by: VE = IE = CE =

1/10 × VCC IC 10 2 × π × f × RE

214

TRANSISTORS

RE = RC = RB =

VE V = E IC I E VCC - VCE - VE IC VCC-VBE - VE IB + VCC

RB

RC CC2

Q1 CC1 VO Vi RE CE - VCC

Figure 12.13

Example 12.3 The following data for the fixed bias amplifier in figure 12.13 is given: = 15 V • VCC • β = 200 = 5 mA • IC = 6V • VCE • f = 60 Hz Use the above information and calculate the resistive values of the base resistor, collector resistor, emitter resistor and the capacitance value of the emitter bypass capacitor. Use is made of a Silicon resistor and VBE may be used as 0,7 volts.

Solution: VE = 1/10 × VCC = 1/10 × 15 = 1,5 volt

215

V V 10 RE = E = E CE = IC IE 2 × π × f × RE 10 = 1,5 1,5 2 × 3,142 × 60 × 300 = = 5 × 10-3 5 × 10-3 = 88,4 µF = 300 ohm V -V -V RC = CC CE E IC = β × IB IC IB = IC β 15 – 6 – 1,5 = 5 × 10-3 5 × 10-3 = 200 = 1, 5 k-ohm = 25 µA V -V - V RB = CC BE E IB 15 – 0,7 – 1,5 = 25 × 10-6

12.7.2

= 512 k-ohm

Voltage divider bias common emitter amplifier

The common emitter amplifier circuit diagram illustrated in figure 12.14 is that of a voltage divider bias configuration. This means that only one supply is used to bias both the base and the collector. + VCC

RB1

RC CC2

CC1

Q1

Output Input RB2

RE CE - VCC

Figure 12.14

216

TRANSISTORS

The fixed bias amplifier discussed has a serious disadvantage in that the values of bias current as well as voltage is dependant upon the β-gain of the transistor. Since this gain is extremely sensitive to temperature changes especially with Silicon transistors it is required that we design a dc-bias network which will be independent of the β-gain and thus not so sensitive to temperature variations. It should be mentioned that our greatest enemy with amplifier circuits is that of temperature variations. The use of a voltage divider network for our dc-bias arrangement overcomes to some extent this serious problem and use is made of such a network indicated as RB1 and RB2 as illustrated in figure 12.14. The values of the voltage divider resistors RB1 and RB2 is mathematically given by: RB1

=

RB2 × (VCC – VB) VB

and RB2

=

1/10 (RE × β)

The magnitude of the base voltage VB is mathematically given by: RB2 VB = × VCC where RB1 + RB2

VB RB1 RB2 VCC

= = = =

value of base voltage in volt resistive value of base resistor in ohm resistive value of base resistor in ohm supply voltage in volt

Referring to the mathematical expression it must be noted that the voltage drop across RB2 is the voltage that will be measured on the base of the transistor. The value of the collector load resistor RC is mathematically given by: RC

=

VRC where IC

RC IC

= =

resistive value of load resistor in ohm magnitude of collector current in ampere

Also

VCC = VRC + VCE

where

VCC = magnitude of supply voltage in volt VRC = magnitude of voltage drop across load resistor in volt VCE = magnitude of voltage between emitter and collector in volt

Having obtained an output from the transistor we now encounter a more serious problem termed ‘thermal runaway. It was mentioned at an earlier stage that a small leakage current flows across the reverse biased base-collector junction and that it could be ignored. But this is not altogether true. An increase in this leakage current causes

217

an increase in collector current which increases the temperature of the transistor which decreases its resistance which increases the leakage current which increases the collector current and in this manner it will snowball until such time that the transistor base-collector junction is damaged and the transistor becomes unserviceable. Once an amplifier comes into operation this leakage current affects the stability of the amplifier if not countered. In order to counter this thermal runaway and keep the transistor at a stable operating point we add to the circuit an emitter resistor as illustrated in figure 12.14. Although the addition of an emitter resistor RE has a stabilising affect on the transistor it causes negative feedback which will affect the gain of the amplifier which is an unwanted situation. To counter this reduction in gain we add an emitter capacitor CE in parallel with the emitter resistor RE as illustrated in figure 12.15. A further precaution that can be used is to mount the transistor on a ‘heat sink’ which will dissipate a lot of the heat generated by the current flowing through the transistor. This is often done with power transistors. If you compare the circuit diagram illustrated in figure 12.14 with the circuit diagram illustrated in figure 6.8 you will notice quite a difference although the characteristics and operation is exactly the same. The circuit diagram in figure 12.15 illustrates an automatic bias common base amplifier.

+ VCC

RB1

RC

CC1 Q1 CC2 Output RB2

CB

RE

Figure 12.15

Input

- VCC

The circuit diagram in figure 12.16 illustrates an automatic bias common collector amplifier.

218

TRANSISTORS

+ VCC

RB1

RC

C

Q1

CC1

CC2

Input

RB2

RE

Figure 12.16

Output

- VCC

12.8 Common emitter graphical analysis The common emitter has three main characteristic curves which defines its behaviour. It will be possible to determine the transistor operation for static as well as dynamic conditions.

12.8.1

Input characteristics

The input characteristic curve is obtained by plotting the obtained base current IB against the base-emitter voltage VBE and this is illustrated in the graphical representation in figure 12.17. The input characteristic curve will allow us to calculate the static impedance ZS as well as the dynamic impedance ZD. Before we continue we need to differentiate between the terms dynamic and static.

219

IB (µ-ampere)

6

4

2

VB (mV) 0,3

0,4

0,5

Figure 12.17

• Static Static conditions imply that we only consider fixed parameters such as dc-values. • Dynamic Dynamic conditions imply that we only consider variable parameters such as ac-values. The static- and dynamic impedances are given by: V ΔVBE ZS = BE ZD = where IB ΔIB

ZS ZD IB VBE Δ

= = = = =

static impedance in ohm dynamic impedance in ohm base current in ampere base voltage in volt difference between

Example 12.4 Use the data in the graph in figure 6.18 and determine the: (a) Static impedance; and (b) Dynamic impedance.

220

TRANSISTORS

Solution: V (a) ZS = BE IB 0,4 × 10-3 = 4 × 10-6

=

100 ohm

ΔVBE (b) ZD = ΔIB (0,5 - 0,3) × 10-3 = (6 - 2) × 10-6

12.8.2

=

50 ohm

Transfer characteristics

The transfer characteristic curve is obtained by plotting the obtained base current IB as input against the collector current IC as output and this is illustrated in the graphical representation in figure 12.18. The transfer characteristic gives a clearer indication that a small change in base current IB causes a substantial change in collector current IC and since the base current IB is our input current and the collector current IC our output current it is now possible that we can calculate our current gain of the amplifier. Note that the collector current IC is controlled by the base current IB and this is termed the beta (β) gain. IC (mA)

30

20

10

IB (µ-ampere) 10

Figure 12.18

20

30

221

The static- and dynamic current gain is given by: I I ΔI ΔIC AS = O = C AD = O = Ii IB ΔIi ΔIB where AS = static current gain AD = dynamic current gain IB = base current in ampere IC = collector current in ampere IO = output current in ampere Ii = input current in ampere Δ = difference between

Example 12.5 Use the data in the graph in figure 12.18 and determine the: (a) Static current gain; and (b) Dynamic current gain.

Solution: I (a) AS = C I B 30 × 10-3 = 20 × 10-6 = 1 500

(b) AD = = =

12.8.3

ΔIC ΔIB (50 - 10) × 10-3 (30 - 10) × 10-6

2 000

Output characteristics and dc-loadline

The output characteristics are obtained by plotting the collector current IC to the emitter-collector voltage VCE and also indicating the base current IB and this is illustrated in the graphical representation in figure 12.19. In order to get a clearer

222

TRANSISTORS

picture of the output characteristic curve we need to use a circuit with this curve since we need to draw a dc-loadline on the graph and then analyse the quantities. Refer to figure 12.20. IC (mA) IB IB IB

µ-ampere

IB IB VCE

Figure 12.19

We now need to draw the dc-loadline on the graph in figure 12.19. This dc-loadline follows the equation y = mx + c, which is a straight line and only two points need to be plotted. Point A VCE = VCC - IC × RC Should the transistor not be conducting, that is, no signal is being applied, then IC = 0. VCE

= 20 - 0 = 20 volt

VCC Point B IC = R C 20 = 10 × 103

=

2 mA

223

+ 20 V

RB1

RC

10 k-ohms CC2

Q1

CC1

Output Input

RB2

RE

- 20 V

Figure 12.20

These two points A and B can now be plotted on the graphical representation of the output characteristic curve illustrated in figure 12.21. IC (mA) Point B IB

2

IB IB 1 IB

dc bias point (Q-point)

IB Point A VCE 10

20

Figure 12.21

On this dc-loadline a point is now selected as the dc-bias point or Q-point and must indicate the collector current IC and collector-emitter voltage VCE when no signal is applied. Should a signal now be applied the base current IB will vary causing the collector current IC to also vary according to the amplitude of the input signal resulting in an output voltage variation across the collector resistor RC. This concept is illustrated in figure 12.22.

224

TRANSISTORS

2

90

1,8

80

1,6

70

1,4 1,2

Collector output current

60

Base input current

50

1

40

0,8

30

0,6

20

0,4

10

0,2

0

2

4

6

8

10

12

14

16

18

IB (µ-ampere)

20

Collector output voltage

Figure 12.22

The dc-operating or Q-point in this instance is taken at 10 volt. But how will this point be determined. Refer back to the mathematical expression of: RB2 VB = R + RB2 B1

× VCC

and this is the expression we will use to determine the dc-operating or Q-point on the base of the transistor. This is a very important concept since this dc-operating or Q-point will determine our class of amplification which will be dealt with at a later stage. The use and interpretation of these graphical representations allow us to make quite a number of observations and calculations concerning the amplifier. These will now be discussed.

225

12.8.3.1 Gain

Definition 12.1 Gain

The gain of an amplifier may be defined as that capability of an amplifier to increase the magnitude of the input signal ideally without distortion.

The output must be a replica but larger in amplitude than the input signal.

12.8.3.1.1 Current gain The current gain has been discussed but it will be repeated here. The static- and dynamic current gain is given by: I I ΔI ΔI AS = O = C AD = O = C I I ΔIi ΔI B i B = static current gain where AS = dynamic current gain AD = base current in ampere IB = collector current in ampere IC = output current in ampere IO = input current in ampere Ii Δ = difference between

12.8.3.1.2 Voltage gain The static- and dynamic voltage gain is given by: V I × ZO ΔV ΔI × ZO AVS = O = O AVD = O = O V I × Zi ΔV ΔIi × Zi i i i

where

AVS AVD ZO Zi VO Vi

= = = = = =

static voltage gain dynamic voltage gain output impedance in ohm input impedance in ohm output voltage in volt input voltage in volt

226

TRANSISTORS

12.8.3.1.3 Power gain AP = VO = IO2 × ZO Ii2 × Zi Vi

where AP

=

power gain in watt

12.8.3.1.4 Decibel gain

It is more than often required to express the gain of an amplifier in decibels (dB). All the previous expressions have no unit and is merely a figure and the dB is a convenient method of expressing gain. Current gain Ai = Voltage gain AV = Power gain AP =

I 20 log O where PO = output power in watt Ii Pi = input power in watt V 20 log O Vi PO 10 log Pi

Example 12.6 Refer to figure 12.22 and use the indicated information to determine the: (a) Static current gain; (b) Magnitude of the input signal; (c) Voltage magnitude of the output signal; and (d) Dynamic current gain.

Solution: I (a) AS = C I B 1 × 10-3 = 40 × 10-6 = 25

(b)

20 µA (p-p)

(c)

4 V (p-p)

227

(d) AD = = =

ΔIC ΔIB 0,4 × 10-3 20 × 10-6

20

Example 12.7 An amplifier has the following specifications: • Input power = 3 500 watt • Output power = 1 000 watt • Input current = 200 ampere • Input voltage = 200 volt • Output impedance = 20 ohm Use the above data and determine in dB the: (a) Current gain; (b) Voltage gain; and (c) Power gain.

Solution: I (a) Ai = 20 log O PO = I2O × ZO Ii P I2O = O 50 ZO = 20 log 200 1000 = 20 = - 12,04 dB = 50 ampere

(b) AV = = = (c) AP

= = =

VO VO = Vi = 1 000 = 20 log 200

20 log

13,979 dB PO Pi 1 000 10 log 3 500

10 log

- 5,44 dB

IO × ZO 50 × 20 1 000 volt

228

TRANSISTORS

An interesting point to note here is the minus sign in front of the current gain and the power gain. What does it mean? It merely means that the amplifier operates at a loss and not a gain!

Exercise 12.1 1. Use suitable schematic diagrams that will illustrate a: 1.1 Biased NPN-transistor; and 1.2 Biased PNP-transistor. 2. Determine the magnitude of the base current of a transistor if it was found by measurement that a current of 80 mA flowed from the emitter and 78,5 mA was measured in the collector of the transistor. 3. Draw a neat labelled graphical representation that will illustrate the three regions of operation of a transistor and then give a detailed description of the three regions with a possible application. 4. Draw a neat labelled graphical representation that will illustrate the switching speed of a transistor and then give a detailed description of the times indicated. 5. Complete the following table with reference to three amplifier configurations. Characteristics Current gain Voltage gain Input impedance Output impedance Phase shift

Type of amplifier Common emitter Common base Common Collector High High High Low 0º

6. Explain with the aid of a suitable circuit diagram the operation of a transistor when used as a switch. 7.

Draw neat labelled circuit diagrams with input - and output waveforms of a PNP transistor used as an amplifier in the following configurations: 7.1 Common collector; 7.2 Common emitter; and 7.3 Common base.

229

8.

Use suitable graphs to explain the following characteristics of a transistor: 8.1 Input characteristics; 8.2 Transfer characteristics; and 8.3 Output characteristics.

9. Consider the following circuit diagram and output characteristic. 9.1 Use the available information on the circuit diagram and output characteristic curve and draw the dc loadline showing all relevant calculations. 9.2 Draw the collector output current waveform as well as the base input current waveform. IC (mA) IB

2

IB IB 1 IB IB VCE 10

20

30

+ 30 V

RB1

RC

15 k-ohms

CC2 CC1

Q1

Output Input

RB2

RE

- 30 V

µ-ampere

230

TRANSISTORS

9.3

Should the amplifier have an input impedance of 300 ohm and an output impedance of 400 ohm then calculate the: 9.3.1 Static current gain; 9.3.2 Dynamic current gain; 9.3.3 Static voltage gain; 9.3.4 Dynamic voltage gain; 9.3.5 Power gain; 9.3.6 Current gain in dB; 9.3.7 Voltage gain in dB; and 9.3.8 Power gain in dB.

Amplification Classes, Coupling Methods and Feedback


231

CHAPTER

13


Learning Outcomes On completion of this module you will be able to: • •

Understand and explain the concept of amplification classes; Understand and explain with the aid of circuit diagram and input- and output waveforms: – RC-coupling of transistors as well as frequency response; – Transformer coupling of transistors as well as frequency response; – Direct coupling of transistors as well as frequency response; – Push-pull amplification; and – Complementary symmetric amplification; • Explain using waveforms the concept of distortion and methods of overcoming such distortion; and • Understand and explain the concept of negative feedback.

13.1 Introduction All amplifiers fall into a specific category of amplification which is dependant upon the application of the amplifier as well as the coupling method used. The class of amplification refers to the conduction period of the output signal compared to the input signal and there are four types of amplification classes that will be discussed.

13.2 Classes of amplification 13.2.1 Class A amplification The graph in figure 13.1 illustrates the concept of class A amplification

232


IC (mA)

Output collector current

Base current

VCE

dc-bias point

Input voltage

Figure 13.1

Definition 13.1 Class A Amplifiction

Class A amplification may be defined as a process whereby an output collector current will flow for the full 360 º of the input cycle.

The advantage of this type of amplification is that it gives a fairly distortion free output in that the output is an almost exact replica of the input signal. The disadvantage is that it has a very low efficiency, in the region of 25%. ` Typical application includes pre-amplifiers and general purpose amplifiers. Studying figure 13.1 one can see that the transistor is biased in the linear part of the output characteristic on the dc-loadline.

13.2.2

Class B amplification

The graph in figure 13.2 illustrates the concept of class B amplification.


IC (mA)


Base current

VCE dc-bias point

Input voltage

Figure 13.2

Definition 13.2 Class B Amplifiction

Class B amplification may be defined as a process whereby an output collector current will flow for only 180º of the input cycle.

The advantage is that it has a higher efficiency if compared to class A, in the region of 78,5 %. Typical applications include power amplifiers. Studying figure 13.2 one can see that the transistor is biased at cut-off on the output characteristic on the dc-loadline.

13.2.3

Class AB amplification

The graph in figure 13.3 illustrates the concept of class AB amplification.

Definition 13.3 Class AB Amplifiction

Class AB amplification may be defined as a process whereby an output collector current will flow for more than 180º but less than 360º of the input cycle.

233

234


IC (mA)


Base current

VCE dc-bias point

Input voltage

Figure 13.3

The efficiency is also quite high and is often used in the push-pull configuration of amplification which will be dealt with later. Studying figure 13.3 one can see that the transistor is biased above cut-off on the output characteristic on the dc-loadline.

13.2.4

Class C amplification

The graph in figure 13.4 illustrates the concept of class C amplification. IC (mA)


Base current

VCE Input voltage

dc-bias point

Figure 13.4


235

The efficiency is very high and typical applications include waveform shaping circuits used in television receivers as well as oscillator circuits. Studying figure 13.4 one can see that the transistor is also biased beyond cut-off on the output characteristic on the dcloadline.

Definition 13.4 Class C Amplifiction

Class C amplification may be defined as a process whereby an output collector current will flow for less than 180º of the input cycle.

13.3 Amplifier coupling methods It is sometimes required to connect two or more transistors in cascade in order to increase the gain since one transistor may not supply the required gain on its own. There are several methods that may be employed and they will now be discussed.

13.3.1

RC-inter-stage coupling

An RC (resistive-capacitive) inter-stage coupling is illustrated in figure 13.5.

+ VCC RC1 RB1

RC2 RB3

CC2

CC3 Q1

CC1

Q2

RB2

RB4 RE1

CE1

RE2

CE1 - VCC

Stage 1

Stage 2

Figure 13.5

236


Two common emitter amplifiers are connected in cascade (series) by means of RCcoupling compromising of RC1 and CC2 which are enclosed in dotted lines in figure 13.5. The coupling from stage 1 to stage 2 is by means of CC2 which also serves to isolate the dc-biasing between the two stages from one another. Also of importance is the frequency response of an amplifier. All amplifiers are designed to cover a pre-determined range of frequencies and this response is given by the frequency response curve for RC-coupled amplifiers which is illustrated in figure 13.6. All frequency response curves are interpreted in the same manner and it does not matter whether we have a single stage or more than one stage. The bandwidth, between fLCO (lower cut-off frequency) and fHCO (higher cut-off frequency), is that range of frequencies that will be reproduced by the amplifier in a fairly constant manner. You will notice that the bandwidth is taken at a point termed the half-power points. That is when the output is at its RMS-value and is also termed the -3 dB point. Also important to note is that the frequency is indicated on a logarithmic scale and not a linear scale since it will be rather difficult to fit such a scale. VO 0,707 or - 3 dB (half-power points) Bandwidth

100 Hz

1 kHz

10 kHz

100 kHz

f

fHCO

fLCO

Figure 13.6

13.3.2 Direct inter-stage coupling A direct inter-stage coupling is illustrated in figure 13.7. Two common emitter amplifiers are connected in cascade (series) by means of direct coupling which are enclosed in dotted lines in figure 13.7. A simpler direct coupled amplifier is illustrated in figure 13.8


+ VCC RC1

RC2

RB1

RB3 CC2 Q1

CC1

Q2

RB2

RB4 RE1

CE1

CE1

RE2

- VCC Stage 1

Stage 2

Figure 13.7 + VCC RB Q2 Q1

RE

- VCC

Figure 13.8

Another form of direct coupling, the Darlington pair is illustrated in figure 13.9

237

238


+ VCC RB Q1

CC

Darlington pair Q2

RE

- VCC

Figure 13.9

The latter two applications of direct coupling are normally associated with low frequency applications. The dc-levels are related to one another for the complete circuit and the bias arrangement is so designed to cater for that. All variations in bias voltages will affect both transistors which will have an effect on the stability of the amplifier as a whole. The advantage of the Darlington pair is that it is an extremely good current amplifier and is available in a single package. Again the bandwidth, between fLCO (lower cut-off frequency) and fHCO (higher cut-off frequency), is that range of frequencies that will be reproduced by the amplifier in a fairly constant manner. Again it is important to note is that the frequency is indicated on a logarithmic scale and not a linear scale. The frequency response curve for a direct coupled amplifier is illustrated in figure 13.10. VO 0,707 or - 3 dB (half-power points) Bandwidth

100 Hz fLCO

Figure 13.10

1 kHz

10 kHz

100 kHz fHCO

f


239

13.3.3 Transformer inter-stage coupling A transformer inter-stage coupling is illustrated in figure 13.11. Two common emitter amplifiers are connected in cascade (series) by means of transformer coupling which are enclosed in dotted lines in figure 13.11. Important to note is the phase shifts that takes place.

+ VCC

T1 RC1 RB1

RB3 CC2 Q1

CC1

Q2

RB2

RB4 RE1

CE1

CE1

RE2

- VCC Stage 1

Stage 2

Figure 13.11

The frequency response curve for a transformer coupled amplifier is illustrated in figure 13.12. VO 0,707 or - 3 dB (half-power points)

Bandwidth

100 Hz

1 kHz

10 kHz

100 kHz fHCO

fLCO

Figure 13.12

f

240


Again the bandwidth, between fLCO (lower cut-off frequency) and fHCO (higher cut-off frequency), is that range of frequencies that will be reproduced by the amplifier in a fairly constant manner. Again it is important to note is that the frequency is indicated on a logarithmic scale and not a linear scale since it will be rather difficult to fit such a scale in on a linear scale. The disadvantage of this type of coupling is a poorer frequency response and increased size and cost. We do however have the advantage that power consumption is improved in that the transformer can be used for impedance matching as well.

13.3.4 Push-pull amplifiers 13.3.4.1 Two similar transistors In a previous section we discussed the classes of amplification and there we mentioned that class B has a very high efficiency. This concept is often used in push-pull amplifiers but it must be noted that the amplifier does not use class B but rather use class AB as will be discussed at a later stage. There are two methods employed for push-pull amplifiers and the first method is illustrated in figure 13.13.

+ VCC RB1 T2

Q1

T1

Q2 RB2 RE

Figure 13.13

- VCC

The input transformer T1 has a centre-tapped secondary winding which will act as a phase splitter for the transistors Q1 and Q2. The transistors Q1 and Q2 will only conduct when their respective bases become positive with respect to their collectors. It is therefore obvious that only one transistor can conduct at any given time. The output transformer T2 which has a centre-tapped primary will now combine the two halves of the signal to produce an output. Worthy of note is that the output


241

transformer T2 also acts as impedance matching. The resistors RB1 and RB2 are so chosen that the amplifier operates in class AB operation to minimise distortion. That means that before one transistor switches off the other transistor already switches on and in this manner cross-over distortion is minimised or in some cases negated. A different circuit may also be utilised as a phase splitter and such a circuit is illustrated in figure 13.14. + VCC RC Output 1 Q1 Output 2 RE - VCC

Figure 13.14

This particular circuit has its own advantages and disadvantages in that should RC and RE have the same resistive values then the magnitude of the two outputs will be exactly the same. The problem however is that the output impedance from the collector is high and the output impedance from the emitter is low which will cause a mismatch to the two transistors of the push-pull amplifier.

13.3.4.2

Complementary pair

The complementary pair is also termed a complementary symmetric push-pull amplifier and is illustrated in figure 13.15.

242


+ VCC RB1

Q1

CC1 RV1

CC3

CC2

Q2

RB2

Figure 13.15

- VCC

In this type of circuit use is made of an NPN- and a PNP transistor. The NPN transistor Q1 will amplify only the positive half of the input cycle to it whereas the PNP transistor Q2 will only amplify the negative half of the input cycle to it. These two halves are then combined to give an output. The base biasing resistors RB1 and RB2 are once again so chosen so as to provide for class AB amplification thereby minimising cross-over distortion.

13.4 Cross-over distortion Cross-over distortion falls into the category of distortion and distortion may be defined as a condition that will occur when the output waveform is not an amplified version of the input waveform. This is particularly so with push-pull amplifiers since the two halves of the input waveform is amplified separately and then need to be put together again and it is here where the problem comes in. We have already discussed what need to be done to overcome this problem, that is selecting the base bias resistors in such a fashion that the switching times of the two transistors actually overlap. Thus, when one transistor is still conducting the other transistor already switches on before the other transistor switches off. The graphical representation in figure 13.16 illustrates the concept of cross-over distortion.


Cross-over distortion

Figure 13.16

13.5 The transistor as a switch As can be seen in figure 13.16 there is a distortion of the waveform as it approaches the zero volt line. The desired output is that indicated by the dotted line on the graphical representation.

13.6 Feedback

Definition 13.5 Feedback

Feedback can be defined as a process whereby a part of the output signal is fed back to the input in anti-phase so as to stabilise the gain of the amplifier or similar circuit.

This process is termed negative feedback. In order to derive at an expression for negative feedback you are referred to the illustration in figure 13.17 (a) and (b).

243

244


Vi

VS

VO

Amplifier

Vf

Feedback network (Beta network) (a)

+

= Vf

VS

VO (b)

Figure 13.17

Figure 13.17 (a) illustrates and amplifier with a feedback network Beta. A portion of the output waveform VO is fed back in anti-phase to the input waveform VS and is termed Vf. In this manner the input waveform Vi to the amplifier is reduced in that Vi = VS Vf which results in the overall gain of the amplifier stage being reduced and kept at a constant level. This can be observed from the waveforms illustrated in figure 13.17 (b). The gain with feedback of an amplifier is mathematically given by: A where A’ = 1–A×β

A’ = A = β =

overall gain with feedback overall gain without feedback feedback fraction

The product A is termed the loop gain or feedback factor and when this factor is negative we have achieved negative feedback. Important to note that negative feedback is also termed de-generative feedback.


Example 13.1 An amplifier has an open loop gain of -50 with no feedback. Should a negative feedback fraction of 0, 04 be applied to the amplifier determine what the new gain would be.

Solution: A A’ = 1-A×β - 50 = 1 - (- 50 × 0,04) = - 16,667

13.6.1 Types of feedback Although feedback is a very important concept we will not go into detail with circuit description but merely investigate such feedback by means of types of feedback and block diagram descriptions.

13.6.1.1 Series voltage feedback It should be mentioned that our greatest enemy with amplifier circuits is that of temperature variations. A block diagram for series voltage feedback is illustrated in figure 13.18.

Vi

Amplifier

VO

Figure 13.18

13.6.1.2 Shunt (parallel) voltage feedback A block diagram for shunt (parallel) voltage feedback is illustrated in figure 13.19.

245

246


Amplifier

Vi

VO

Figure 13.19

13.6.1.3 Series current feedback A block diagram for series current feedback is illustrated in figure 13.20.

Vi

Amplifier

VO

Figure 13.20

13.6.1.4 Shunt (parallel) current feedback A block diagram for shunt (parallel) current feedback is illustrated in figure 13.21.

Vi

Amplifier

Figure 13.21

VO


247

Worthy of note concerning the effects of the different types of negative feedback includes: • • • •

Series feedback will tend to increase the input impedance to the amplifier; Shunt (parallel) feedback will tend to decrease the input impedance to the amplifier; Voltage feedback will tend to decrease the output impedance to the amplifier; and Current feedback will tend to increase the output impedance to the amplifier

The effects of negative feedback will lead to: • • • • •

Reduced but stabilised gain; Reduction in frequency distortion; Reduction of noise generation particularly in the early stages of amplification; Reduction of non-linear distortion; and Increase in bandwidth particularly in the higher frequency band.

Exercise 13.1 1.

Describe with the assistance of waveforms and applicable descriptions: 1.1 Class A amplification; 1.2 Class AB amplification 1.3 Class B amplification; and 1.4 Class C amplification. Mention at least one application for each class of amplification.

2. An amplifier has a feedback fraction of 0,02 and an open loop gain of -80. Determine the gain should negative feedback be applied to the amplifier. 3.

Two common emitter amplifier stages using PNP transistors need to be connected in tandem so as to increase the gain. Draw neat labelled circuit diagrams with waveforms of: 3.1 Transformer coupling; 3.2 Direct coupling; and 3.3 RC-coupling.

4. Draw neat fully labelled graphic representations of the frequency response of: 4.1 Transformer coupling; 4.2 Direct coupling; and 4.3 RC-coupling. 5. Explain using a suitable circuit diagram what is meant by a Darlington pair.

248


6.

Draw neat labelled circuit diagrams of a push-pull amplifier using: 6.1 Complementary symmetry; and 6.2 Two similar resistors. Indicate on both circuit diagrams all the relevant waveforms.

7.

Phase splitting required for push-pull amplifiers may be obtained from a centre-tap transformer. A transistor may also be used but has its own advantages and disadvantages. Draw a neat labelled circuit diagram using an NPN transistor and indicate the input- and the output waveforms. What is the main disadvantage of this circuit?

8. Push-pull amplifiers suffer from phenomena termed cross-over distortion. Use an applicable sketch to illustrate this concept and indicate how this problem can be overcome. 9. Use a suitable block diagram, waveforms and mathematical expression that will illustrate the concept of negative feedback. 10.

Negative feedback can either be series or parallel, current or voltage. Draw neat labelled block diagrams that will illustrate: 10.1 Parallel current feedback; 10.2 Series voltage feedback; 10.3 Series current feedback; and 10.4 Parallel voltage feedback.

11. Negative feedback can either be series or parallel, current or voltage but will have an effect on the input- or output impedance of the amplifier. Mention these effects. 12. Mention five effects negative feedback will have on the characteristics of an amplifier.

Hybrid-Parameters


249

CHAPTER

14 Hybrid-Parameters Learning Outcomes On completion of this module you will be able to: • • •

• • •

Explain using a simple sketch what is meant by matching of amplifier circuits; Convert an NPN and PNP transistor to a two-port device in the: – Common emitter mode; and – Common base mode. Clarify the meaning of the following h-parameters: – hi; – ho; – hr; and – hf. Explain the subscript of ‘e’ and ‘b’. Convert common base and common emitter amplifier circuits to an equivalent circuit; Do calculations using the approximate method the: – Input impedance; – Reverse voltage gain; – Forward current gain and; – Output conductance for common emitter and common base amplifiers.

14.1 Introduction Any transistor has a relation to current, voltage and impedance and is very often referred to the parameters of the transistor. In the previous module a number of activities were done with reference to these parameters, but was not a very accurate of determining specific gains of an amplifier and in most cases only took into consideration the dc-conditions of the amplifier. The question may now be raised whether the amplifier will display the same characteristics when a signal is applied to the amplifier. We therefore have to investigate the operation and characteristics of a transistor amplifier when we apply ac-conditions

250

Hybrid-Parameters

and for that we need to look at the principle of ‘hybrid-parameters’ which literally means ‘of mixed origin’ which will now include the ac-conditions for that specific amplifier. It is of utmost importance that circuits need to be ‘matched’ to one another in order to obtain maximum power transfer. This matching of circuits specifically refers to the impedances to be matched. This concept of matching is illustrated in figure 14.1 Amplifiers Z A

H I g h

Z B

L o w

Mismatch. Input and output impedances not the same

Z

Z

H I g h

H I g h

Z C

H I g h

Matched. Input and output impedances the same

Figure 14.1

According to figure 14.1 there is a mismatch of impedances between amplifiers A and B and maximum power transfer will not take place. On the other hand the impedances between amplifiers B and C are matched and maximum power transfer will take place. For the purpose of this section only the common emitter and common base amplifiers will be discussed. Our first step will therefore be to reduce an amplifier circuit to its equivalent ac-format and use a ‘black-box’ approach or more commonly known in electronic terms ‘a two-port device’ in order to eliminate the dccircuit and to allow for small signal acanalysis to be done.

14.2 The transistor as a two-port device The treatment of ac-analysis will be done in a manner that makes no distinction between an NPN and PNP transistor. The transistor in figure 14.2 illustrates a transistor as a two-port device in the common emitter configuration.


IC

IB C

B

B

C

VBE

VCE

E

IB

E

E

(a) (b)

Figure 14.2

The transistor in figure 14.3 illustrates a transistor as a two-port device in the common base configuration. E

IE

C

IC E

C

VBE B

B

VBC

IB B

(a) (b)

Figure 14.3

Figure 14.3 (a) illustrates an NPN transistor and figure 14.3 (b) illustrates the same transistor as a four-terminal network. The hybrid equivalent network for the input and output is illustrated in figure 14.4. Important to note is that the input circuit is represented by a Thevenin equivalent circuit and the output circuit by a Norton equivalent circuit. The complete hybrid equivalent circuit is illustrated in figure 14.5 (a) and (b) where figure 14.5 (a) is a common emitter network and figure 14.5 (b) is a common base network. I1

I2 hi hr

V1

Input

hf

Figure 14.4

ho

Output

V2

251

252

Hybrid-Parameters

IC

IB B

hi

V1

IE C

hr

ho

IC

E

V2

C

hi

V1

hr

ho

hf

IE

V2

hf

IB

E

(a)

B

(b)

Figure 14.5

What do the different h-parameters mean? Refer to table 14.1

Parameter hi

Meaning input resistance

e Common emitter

b Common base

hr

reverse transfer voltage ratio

Common emitter

Common base

hf

forward transfer current ratio

Common emitter

Common base

ho

output conductance

Common emitter

Common base

Therefore, hie would mean: input resistance, common emitter and hib would mean: input resistance, common base.

14.3 Small signal analysis (common emitter) All amplifiers are basically two-port devices in that it will consist of two input terminals and two output terminals. In order to arrive at mathematical expression for the solving of the above quantities you are referred to the illustration in figure 14.6 I2

I1 B RS Z1

V1

C

hi hr

hf

VS E

Figure 14.6

ho

V2

Z2

ZL


The following mathematical expressions are applicable: Δ Vin hie = = Δ Iin

Δ Vbe Δ Ib

Δ Vin hre = = Δ Vo

Δ Vbe Δ Vce

Δ Io Δ Ic hfe = = Δ Iin Δ Ib Δ Io Δ Ic hoe = = Δ Vo Δ Vce

Important to note is that 1/hoe = output impedance The best way of explaining small signal analysis is by doing a number of activities and then to indicate the different and various approaches that need to be followed.

Example 14.1 Consider the fixed bias amplifier and obtain the equivalent h-parameter network. + VCC RB

RC CC2

CC1

Q1 ZO

Vi

VO

Zi - VCC

253

254

Hybrid-Parameters

Solution: Step 1 Short circuit all dc-supplies and capacitors and obtain the equivalent network for the amplifier. + VCC RB

RC

CC2 Q1

CC1 Vi

ZO

VO

Zi

- VCC

Step 2 Equivalent network Ii

I1

I2

B

RB

Zi

C

hie

V1

IO

hre

Z1

hfe

hoe

E

Example 14.2 Consider the following characteristics and determine: (a) The input impedance hie; (b) The output impedance 1/hoe; and (c) The forward current gain Ai = hfe.

V2

Z2

RC

ZO


IB (µ-ampere)

8

6

4

VB (mV) 0,2

0,3

0,4

Ic (mA)

8 6 4

Vce 4

6

Solution: Δ Vin (a) hie = Δ Iin 0,2 = 4 × 10-6 = 50 k-ohm

8

255

256

Hybrid-Parameters

Δ Vo Δ Io

(b) 1/hoe = =

4 4 × 10-3

=

1 k-ohm Δ Io Δ Iin

(c) Ai = hfe = = =

4 × 10-3 4 × 10-6

1000

14.4 Small signal analysis (common base) All amplifiers are basically two-port devices in that it will consist of two input terminals and two output terminals. In order to arrive at mathematical expression for the solving of the above quantities you are referred to the illustration in figure 14.7. All the expressions already given are also applicable to a common base amplifier. A few subtle changes however to occur but will be taken care of with the activities that will be done. I1

I2 E

RS Z1

C

hi

V1

hr

hf

VS B

Figure 14.7

The following mathematical expressions are applicable: Δ Vin hib = = Δ Iin

Δ Vbe Δ Ie

Δ Vin hrb = = Δ Vo

Δ Vbe Δ Vcb

ho

V2

Z2

ZL


257

Δ Io Δ Ic hfb = = Δ Iin Δ Ie Δ Io Δ Ic hob = = Δ Vo Δ Vcb

Important to note is that 1/hob = output impedance When observing figure 14.7 you will notice that it is similar to figure 14.6 with the exception that the base and emitter terminals have changed around. This implies that the same process that was followed to obtain the equivalent circuit for a common emitter amplifier is followed for a common base amplifier. To further note is that for all the quantities that need to be calculated that the ‘e’ is substituted with a ‘b’.

Example 14.3 Consider the amplifier and then obtain the equivalent h-parameter network. Ii CC1

Q1 RE

Vi

IO

Zi

CC2

RC ZO

RL

VO

Solution: Step 1 Short circuit all dc-supplies and capacitors and obtain the equivalent network for the amplifier.

258

Hybrid-Parameters

Ii

IO

CC1

CC2

Q1 RE

Vi

RC

Zi

ZO

RL

VO

Step 2 Equivalent network Ii

I1 E

RE

C

hib hrb

Vi

hfb

hob

V1 Zi

Z1

IO

I2

RC

RL

V2 B

VO Z2

ZO

ZL

Example 14.4 Consider the following characteristics and determine: 14.4.1 The input impedance hib; 14.4.2 The output impedance 1/hob; and 14.4.3 The forward current gain Ai = hfb.


Ic (mA)

8 6 4

VCE 4

6

8

IB (µA) 8 6 4

0,2

Solution: (a) hib = = =

Δ Vin Δ Iin 0,2 × 10-6 4 × 10-6

50 ohm

Δ Vo (b) 1/hob = Δ Io = 4 4 × 10-3 = 1 k-ohm

0,4

VBE (mV)

259

260

Hybrid-Parameters

(c) Ai = hfb = = =

Δ Io Δ Iin 4 × 10-3 4 × 10-6

1000

Exercise 14.1 1. Explain the meaning of ‘h-parameters’. 2. 2.1 You are given the following amplifier circuit. You are required to obtain the equivalent circuit for this amplifier circuit. + VCC RB

RC CC2

CC1

Q1 ZO

Vi

VO

Zi - VCC

2.1 2.3

Give the mathematical expression you would use to calculate the following quantities: 2.2.1 The input impedance hie; 2.2.2 The output impedance 1/hoe; and 2.2.3 The forward current gain Ai = hfe. You are given the following amplifier circuit. You are required to obtain the equivalent circuit for this amplifier circuit. Ii CC1

Q1 RE

Vi

IO

Zi

CC2

RC ZO

RL

VO


2.3 3.

Give the mathematical expression you would use to calculate the following quantities: 2.3.1 The input impedance hib; 2.3.2 The output impedance 1/hob; and 2.3.3 The forward current gain Ai = hfb.

The following characteristics for an amplifier are given. Determine from the characteristics the following: 3.1 The input impedance hie; 3.2 The output impedance 1/hoe; and 3.3 The forward current gain Ai = hfe. IB (µ-ampere)

10

8

6

VB (mV) 0,3

0,4

0,5

Ic (mA)

7 5 3

Vce 5

7

9

261

262

Hybrid-Parameters

4.

The following characteristics for an amplifier are given. Determine from the characteristics the following: 4.1 The input impedance hib; 4.2 The output impedance 1/hob; and 4.3 The forward current gain Ai = hfb. Ic (mA)

7 5 3

VCE 2

4

6

IB (µA) 6 4 2

0,2

0,4

VBE (mV)

263 Uni-Junction- and Field Effect Transistors N4 Industrial Electronics

CHAPTER

15 Uni-Junction- and Field Effect Transistors Learning Outcomes On completion of this module you will be able to: • • • • • • •

Discuss the Uni-Junction Transistor with the aid of suitable sketches indicating the construction, IEC symbol and characteristic curve; Explain the operation of the Uni-Junction Transistor with the aid of its equivalent circuit; Give a circuit diagram with output waveforms and operation for the application of the Uni-Junction Transistor; Calculate the frequency of operation of a Uni-Junction Transistor; Mention the characteristics of Field Effect Transistors; Give with the aid of suitable sketches the construction, IEC symbol and characteristic curve of an N-channel and P-channel Junction Field Effect Transistors; and Give with the aid of suitable sketches the construction, IEC symbol and characteristic curve of an N-channel and P-channel Metal Oxide Semi-conductor Junction Field Effect Transistors.

15.1 Introduction The Uni-Junction Transistor is mainly used in digital circuits and for the firing circuits in SCR-Control (Silicon Controlled Rectifier). Before we can commence on the application of a UJT as a relaxation oscillator we need to explain the principle of operation of a UJT.

15.2 The UJT-Transistor The construction of a UJT-transistor is illustrated in figure 15.1 (a) and consists of a piece of N-type silicon material on which a heavily doped P-type material is attached and is termed the emitter. The UJT is a three-terminal device but only contains one PNjunction. The IEC-circuit symbol is illustrated in figure 15.1 (b) and the characteristic curve for a UJT-transistor is illustrated in figure 15.1 (c). The two ends of the N-type

264

Uni-Junction- and Field Effect Transistors

material are termed base 1 and base 2. During normal operation base 2 is held positive with respect to base 1 and is reverse biased. Should the emitter voltage now increase current will flow from the emitter to the base 1 region and the potential of the emitter will be reduced and goes into the negative resistance area of the characteristic curve. This is mainly caused by the reduction in resistance between the emitter and the base 1 region. B2

Negative resistance region

Cut-off region Base 2

Saturation region

Emitter E

P

Valley point

N

Base 1 IE

0

B1 (a)

(b)

(c)

Figure 15.1

In order to fully understand the operation of a UJT is required that we familiarise ourselves with an equivalent circuit thereof and this is illustrated in figure 15.2. The silicon bar possesses a relatively high resistance when no forward bias is applied to the emitter resulting in a small current that will flow between base 1 and base 2. When a forward bias is now applied on the emitter-base 1 region the internal resistance between base 1 and the emitter will decrease. This in turn causes an increase in current between base 1 and base 2. Should a signal now be applied the voltage across the device will vary in sympathy with the input signal. The intrinsic stand-off ratio of the UJT is taken as the ratio of the resistance RB1/RB2. + B2 RB1

IE E

VBB D1 VE

VBE RB2

Figure 15.2

B1

A typical UJT-transistor relaxation oscillator is illustrated in figure 15.3.


265

The output from the UJT-transistor is obtained via base 1 and base 2 as indicated on the circuit diagram. When the UJT fires a surge of current through base 1 causes a voltage drop across resistor R1 and a positive-going spike is produced. At the same time a reduction in the voltage across the emitter-base 2 junction causes the current flow towards base 2 to increase rapidly and produces a negative-going pulse from base 2 across resistor R2. The circuit compromising resistor RE and the capacitors C1, C2 and C3 allows for a variable frequency outputs to be obtained. These three capacitors may also be replaced by a single variable capacitor in order to get an adjustable or variable output. + VBB R2

RE

Negative going pulses

UJT

Output spikes

S1

C1

C2

C3

Positive going pulses

R1

- VBB

Figure 15.3

The main application of a UJT-transistor is for the firing circuits in power control which will be dealt with at a later stage. The frequency of oscillation of a UJT-transistor oscillator is mathematically given by: f = 1,5 where RE × C

f = C = RE =

frequency of oscillator in hertz value of capacitor (C1, C2 or C3) value of emitter resistor in ohms

Example 15.1 The following component values for the circuit diagram illustrated in figure 15.3 are given: = 10 k-ohm • RE = 5 nF • C1 = 100 nF • C2 = 25 pF • C3 Determine in each instance the frequency of operation of the oscillator.

266


Solution: 1,5 1,5 f = f = R × C1 R × C2 E E 1,5 1,5 = = 3 -9 10 × 10 × 5 ×10 10 × 103 × 100 × 10-9

= 30 kHz = 1,5 kHz

1,5 f = RE × C3 = 1,5 10 × 103 × 25 × 10-12

= 6 MHz

15.3 Field Effect Transistors The main difference between the bi-polar transistors and Field Effect Transistors is that bi-polar transistors are voltage operated and that Field Effect Transistors are current operated. Field Effect Transistors are also three terminal devices like bi-polar transistors with two PN-junctions. The Field Effect Transistor is a channel of semi-conductor material of which the resistance can be controlled which results in the current through the device being controlled. Field Effect Transistors however have the following characteristics over bi-polar transistors: • • • • • •

No off-set voltage when used as a switch; Small gain-bandwidth; Low noise level; Relatively immune to radiation; Extremely high input impedance; and Good thermal stability.

Having indicated the characteristics of the Field Effect Transistor we need to mention that it has one major disadvantage when compared to bi-polar transistors in that it is very susceptible when it comes to handling them in that the acid present in the human hand may lead to damage. Field Effect Transistors are mainly divided into Junction Field Effect Transistors (JFET) and Metal Oxide Semi-conductor Field Effect Transistor (MOSFET).


15.3.1

267

Junction Field Effect Transistors (JFET)

The JFET is constructed either of N-type silicon or P-type silicon. The N-type has two P-type regions diffused into it to form the gate region whereas the P-type has two N-type regions diffused into it to form the gate region. These concepts are illustrated in figure 15.4 (a) which is an N-type channel with its IEC circuit symbol and characteristic curves and figure 15.4 (b) is a P-type channel with its IEC circuit symbol and characteristic curves. The terminals associated with a FET are termed the gate, drain and source.

G

ID VGS = 0 V

D

P

VGS = -1 V S

N

D

G VGS = -2 V

P

S

VDS

(a)

G

ID

D

N

VGS = 0 V VGS = 1 V

S

P

D

G VGS = 2 V

N

S

VDS

(b)

Figure 15.4

The question may now be raised on how the JFET operates. In order to do this we need to look at the illustration in figure 15.5 (a), (b) and (c).

268


D

ID

ID

D

Pinch-off point VGS = 0 V

IDSS G

P

N

P

VDD

G

P

N

P

VGS = -1 V VGS = -2 V

VGS S

ID

(a)

S VP (b)

VDS

(c)

Figure 15.5

The most important factor that need to taken into account is that the correct polarities must be maintained when connecting a JFET into a circuit. Current will be able to flow from source to drain in the absence of no gate-source reverse bias. Should we now apply a reverse bias across the gate-source a depletion layer will start forming around the two P-type materials (figure 15.5 (b) and will extend into the N-type material thereby increasing the resistance resulting in that less area for current to flow and the current flow will decrease. Should the reverse bias increase further a point will be reached when the current will stop flowing and a cut-off point will be reached and current will cease flowing. This cut-off point is also termed the pinch-off point and is clearly indicated in figure 15.5 (c).

15.3.2 Metal Oxide Semi-conductor Field Effect Transistors (MOSFET’s) Metal Oxide Semi-conductor Field Effect Transistors (MOSFET’s) are found in two main types namely the enhancement mode type and the depletion mode type. Worthy to take note of is that MOSFET’s are often referred to as IGFET’s (Insulated Gate Field Effect Transistors). The main difference between a JFET and a MOSFET lies therein that the gate terminal is isolated from the channel which is created between the source and the drain in the absence of a gate potential. Current will only flow when a potential is applied to the gate.

15.3.2.1 The Enhancement Mode (MOSFET) The channel is not part of the original construction but is obtained when a potential is applied across the gate-source terminals for both the N-and P-channels. Figure 15.6 (a) (b) and (c) illustrates the construction of a P-channel enhancement MOSFET (a), the IEC circuit symbol (b) and the characteristic curve (c).


S

G

D

ID Silicon Oxide layer VGS = 0 V

D N

N P-substrate

269

Induced Nchannel

P-substrate

G

VGS = -1 V

S

VGS = -2 V VDS

(a)

(b)

(c)

Figure 15.6

Figure 15.7 (a) (b) and (c) illustrates the construction of an N-channel enhancement MOSFET (a), the IEC circuit symbol (b) and the characteristic curve (c). S

G

D


D P

P N-substrate

Induced Pchannel

N-substrate

G

VGS = 1 V

S

VGS = 2 V VDS

(a)

(b)

(c)

Figure 15.7

15.3.2.2 The Depletion Mode (MOSFET) In the depletion mode the channel already exists between the drain and the source and a current will be produced when a potential is applied across the source and drain terminals. Figure 15.8 (a) (b) and (c) illustrates the construction of a P-channel depletion MOSFET (a), the IEC circuit symbol (b) and the characteristic curve (c).

S

G

D


D P

P P-channel

N-substrate

G

N-substrate

VGS = 1 V

S

VGS = 2 V VDS

(a)

(b)

Figure 15.8

(c)

270


Figure 15.9 (a) (b) and (c) illustrates the construction of an N-channel depletion MOSFET (a), the IEC circuit symbol (b) and the characteristic curve (c). S

G

D


D N

N N-channel

P-substrate

G

P-substrate

VGS = -1 V

S

VGS = -2 V VDS

(a)

(b)

Figure 15.9

(c)

Exercise 15.1 1. Explain by means of an equivalent circuit the operation of a Uni-Junction Transistor. 2. The following component values for the circuit diagram illustrated in figure 15.3 are given: = 20 k-ohm • RE = 3 nF • C1 = 80 nF • C2 = 40 pF • C3 Determine in each instance the frequency of operation of the oscillator. 3. Refer to figure 15.3 and discuss its operation. 4. What is the main difference between a FET and a bi-polar transistor? 5. Mention six characteristics that a FET possesses over a bi-polar transistor. 6. Mention the main difference between an enhancement MOSFET and depletion MOSFET.

POWER CONTROL


CHAPTER

16 Power Control Learning Outcomes On completion of this module you will be able to: •

• •

•

•

• •

Give the IEC circuit symbols of: – A Silicon Controlled Rectifier (SCR); – A Diac; – A Triac; and – A Quadrac. Explain the operating principle of an SCR by means of: – Two transistor analogy; and – Characteristic curve. Explain using graphical representation where applicable: – Conduction angle; – Delay angle; – Phase control; – Cycle control; – Cyclotronic control; – Commutation; and – Duty cycle. Draw a circuit diagram with the applicable waveforms to illustrate the application of an SCR in: – Half-wave control; – Full-wave control; and – dc-circuits. Explain using characteristic curves the operating principle of: – A Diac; – Triac; and – A Quadrac. Draw a circuit diagram with the applicable waveforms to illustrate the application of a Diac and Triac in a speed control circuit; and Draw a circuit diagram with the applicable waveforms to illustrate the application of a Quadrac in a speed control circuit.

271

272

Power Control

16.1 Introduction Power control forms an integral part in the electronic as well as electric field in modern industry today. This module will explain the concepts of different electronic components in use today to accomplish power control.

16.2 The Silicon Controlled Rectifier (SCR)

Definition 16.1 SCR

An SCR may be defined as an ordinary diode with a control element namely the gate.

The forward current that the device will experience will only flow once the control element (gate) has been activated. Once the gate has been activated the voltage on the gate can be removed and the SCR will stay in conduction provided certain parameters are adhered to. The SCR will therefore act as a high-speed switch. An SCR may be represented by a four-layer, three junction PNPN device and by two transistor analogy. This concept is illustrated in figure 16.1 (a), (b), (c) and (d). Figure 16.1 (a) illustrates a four-layer, three junction PNPN device and figure 16.1 (b) is an equivalent representation of an SCR. Figure 16.1 (c) illustrates the two transistor analogy around which the operating principle will be discussed. Figure 16.1 (d) is the IEC circuit symbol for an SCR. The four-layer PNPN device in figure 16.1 (a) has three junctions: J1 between P1 and N1, J2 between N1 and P2 and J3 between P2 and N2 and has three terminals namely an anode connected to P1, a cathode connected to N2 and the control element (gate) connected to P2. This four layer PNPN device is now split according to figure 16.1 (b) and now forms two transistors Q1 and Q2 of which Q1 is a PNP-transistor and Q2 is an NPN-transistor and is illustrated in figure 16.1 (c). It is shown in figure 16.1 (c) that the gate of the SCR is connected to the base of transistor Q2 and the emitter of transistor Q1 and the collector of transistor Q2 is connected to the base of transistor Q1. The anode is taken from the emitter of transistor Q1 and the cathode is taken from the emitter of transistor Q2. This concept is termed two transistor analogy and is in most cases used to explain the principle of operation of an SCR. Should the device now be forward-biased, anode positive with respect to the cathode no forward current will flow since the gate needs to be triggered by means of a positive pulse. This positive pulse will forward-bias the base of transistor Q2 thereby causing a current to flow in the collector of Q2 which in turn is connected to the base of


Anode

Anode

P1

P1 J1

N1

N1

N1

P2

P2

J2 Gate

Gate

P2 J3 N1

N2

Cathode

Cathode

(a)

(b)

Anode

Q1 Gate

Anode Q2

(c)

Cathode

Cathode Gate

(d)

Figure 16.1

transistor Q1 which will cause transistor Q1 to also switch on. The collector current of both transistors will provide more than enough base current and both transistors will therefore remain in a state of conduction even if the gate pulse is removed and only a small voltage across the anode-cathode of the SCR. Under these conditions the SCR is said to have latched. The characteristics of an SCR may be summarised in the following fashion: • • • • •

A positive pulse is required to switch an SCR on. Once the SCR is conducting the gate has no further control provided that the forward current remains above the holding current value for the device. The SCR will continue conducting until the anode-cathode voltage is reduced to near-zero. An SCR is an open-circuit device and will only conduct should the gate be triggered. It therefore behaves like an on-off switch. The gate may be triggered by either an ac or dc source.

The characteristics of the SCR, as with most semi-conductor components, are best described by means of a characteristic curve which is illustrated in figure 16.2.

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Power Control

Forward conduction region

IF Holding currents IH

+V IG1

IG2

IG3

Figure 16.2

The characteristic curve illustrated in figure 16.2 only indicates the forward conduction characteristics. The reverse characteristics are the same as that of an ordinary diode. Also indicated is the different gate triggering values as well as the holding current values.

Definition 16.2 Holding Current

The holding current may be defined as that minimum value of current required to keep the SCR in the state of conduction and should this value be reduced below the specified value the SCR will switch off. There is also another type of SCR available namely the Light Activated Silicon Controlled Rectifier (LASCR) and will operate in exactly the same manner as an ordinary SCR with the exception the SCR is triggered by means of a light source instead of a positive voltage on the gate. The characteristics are the same as that illustrated in figure 16.2. Figure 16.3 illustrates the IEC-circuit symbol of a LASCR. Anode

Cathode Gate

Figure 16.3

Another concept that must be understood is that of commutation.


275

Definition 16.3 Commutation

Commutation is a process that will switch an SCR from its conducting state to its non-conducting state.

This in short means it is a method to switch an SCR off and there are two methods in which this can be accomplished. • Line commutation An SCR is switched off every time that the input waveform moves through zero going negative and this is termed line commutation and line commutation may therefore be defined as switching an SCR off by means of the input alternating supply. • Forced commutation In dc circuits a problem arises in that there is no alternating supply that can switch an SCR off and the SCR must be forced to switch off. This is done by placing a capacitor in series with the load and thereby forcing a current through the SCR in the opposite direction of normal conduction hence the term forced commutation.

16.3 Delay angle and conduction angle

Definition 16.4 Delay Angle

The delay angle may be defined as that part of the waveform for which no conduction will take place whereas the conduction angle may be defined as that part of the waveform for which conduction will take place. Consider the waveforms illustrated in figure 16.3 (a) and (b). The waveform in figure 16.3 (a) illustrates that of half-wave control and the waveform in figure 16.3 (b) illustrates that of full-wave control. The concept of delay angle and conduction angle is the same whether it is half-wave or full-wave. You will further notice that half-wave only utilises only one half of the waveform whereas full-wave utilises both halves of the waveform. In both waveforms the area indicated by A is termed the delay angle and the area indicated by B is termed the conduction angle.15.3 Field Effect Transistors.

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Power Control

+ (a)

t

-

+ (b)

t

A

B

A

B

A

B

A

B

Figure 16.4

A point that must be noted is that once an SCR conducts that it then behaves like an ordinary diode which will actually deliver a pulsating dc and when averaged will have a value proportional to the conduction angle. Since an SCR could be used for controlling the speed of a motor, drilling machine or as a light dimmer it will be obvious that the conduction angle will determine either the speed of the motor, drilling machine or brightness of the lamp. Having this in mind the following conclusions may be made: • •

The smaller the delay angle the more of the input waveform is made available therefore the higher the voltage will be across the load. Should we therefore apply this voltage to a motor for instance it will have a relative high speed. The larger the delay angle less of the input waveform is made available therefore the lower the voltage will be across the load. Should we therefore apply this voltage to a motor for instance it will have a relative low speed.

Consider a basic circuit diagram of a half-wave SCR control and waveforms illustrated in figure 16.4 (a) and (b). • • •

The gate of the SCR is triggered by the leading edge of the train of pulses. The delay angle and therefore the conduction angle is the same for every positive half cycle of the input waveform. Should you graphically add the shaded areas of the voltage across the load and the voltage across the SCR then it will equal the input waveform.

The waveforms illustrated in figure 16.5 indicate the concept of full-wave control.

277


Input voltage

t VSCR Gate trigger pulses t

VIN

VL

RL

(b)

Voltage across load

t

Voltage across SCR

(a)

t

Figure 16.5

Input voltage

t

Gate trigger pulses

t

Voltage across load

t

Voltage across SCR

t

Figure 16.6

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Power Control

16.4 SCR applications It has already been mentioned that an SCR may be used for speed control or for a light dimmer. No matter what application is used, the operating principle will be the same.

16.4.1

Half-wave control

The circuit diagram illustrated in figure 16.6 indicates the concept of half-wave control. Only the circuit diagram is given and the waveforms are the same as that illustrated in figure 16.4 (b). During the positive half cycle of the input waveform the anode of the SCR is positive with respect to the cathode but the SCR will not conduct until such time that the gate is triggered. This is determined by the delay angle to the gate terminal of the SCR by the time constant of RV1 and C. This part of the circuit is also termed the triggering circuit. As soon as the time constant lapses a positive pulse will be applied to the gate of the SCR via the diode D1 and the SCR will commence conduction. The main purpose of the diode D1 is to prevent a pulse on the gate when the input cycle goes negative since repeated negative triggering may damage the SCR. Should the polarity of the input waveform change the anode of the SCR will now be negative with respect to the cathode and the SCR will switch off. This process of changing polarity will continue for as long as there is an input waveform. Note the position of the different loads. Universal load R

SCR VIN RV1 D dc load C

Figure 16.7

16.4.2

Full-wave control

The circuit diagram illustrated in figure 16.7 indicates the concept of full-wave control.


279

Only the circuit diagram is given and the waveforms are the same as that illustrated in figure 16.5. In order to obtain full-wave control two SCR’s are connected in parallel (back-to-back) as illustrated in figure 16.7. During the positive half of the input waveform SCR1 is forward-biased and will commence conduction as soon as the delay angle has been reached. At the same time SCR2 is reversed biased and will not commence conduction. During the negative half of the input waveform the roles will be reversed in that SCR1 becomes reverse-biased and will not conduct whereas SCR2 now becomes forwardbiased and will commence conduction as soon as the delay angle has lapsed.

Universal load

SCR1

SCR2

Triggering circuit

VIN

dc load

Figure 16.8

16.4.3

An SCR in a DC-circuit

The circuit diagram illustrated in figure 16.8 indicates the concept of an SCR in a dccircuit. The SCR is connected across the output terminals of a full-wave bridge rectifier and in this manner a single SCR can be used to obtain operation from both halves of the input waveform. The current through the SCR will be in the same direction for both halves of the input waveform.

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Power Control

Universal load D1

D2 R1

SCR VIN

D3

D4

dc load

R2

Figure 16.9

16.5 SCR control methods 16.5.1

Phase control

Since an alternating quantity has a cycle of 360º it is possible to trigger such a waveform at any desired angle and thereby controlling the amount of voltage applied to the SCR and hence the load. The waveforms in figure 16.9 (a) and (b) illustrate this principle.

+ (a)

t

-

+ (b)

t

-

Figure 16.10

The waveform in figure 16.9 (a) illustrates half-wave control and the waveform in figure 4.9 (b) illustrates full-wave control. It must however be noted that the angle for conduction (shaded area in both waveforms) are determined by the triggering circuit. Also to be noted is that the larger the delay angle the smaller the conduction angle and the smaller the conduction angle the lower the voltage and hence the speed of the motor or brightness of the lamp whatever our application is. It must however be indicated that this type of control generates unwanted transients which is not desirable.


281

Definition 16.5 Transient

A transient may be defined as unwanted voltage and/or current peaks which occur at the point (angle) where the triggering takes place.

16.5.2

Cycle control

+ t

On time

Off time

On time

Figure 16.11

The waveform in figure 16.10 illustrates the principle of cycle control. Cycle control is obtained by applying a number of full cycles to the load for a number of pre-determined periods of time and then removing the supply for a pre-determined period of time. Cycle control has the advantage that very little or no transients are generated since the control takes place at 0º.

16.5.3

Cyclotronic control

The waveform in figure 16.11 illustrates the principle of cyclotronic control. Cyclotronic control is a combination of the previous two methods discussed and is obtained by applying phase control for a number of cycles where the conduction angle is increased by a pre-determined amount until full cycle control is obtained where it is then kept for a number of pre-determined cycles. The conduction angle is then decreased by a predetermined amount until such time that zero is reached. The supply is then switched of completely for a pre-determined number of cycles and the whole process starts all over again. Typical applications include for instance a situation where a motor has to start running but at a very low speed with a gradual increase in speed until full speed is reached and then needs to be slowed down again.

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Power Control

t

Phase control

Fully on

Phase control

Off

Phase control

Figure 16.12

16.5.4

Duty cycle

The waveforms in figure 16.12 (a), (b), (c) and (d) illustrate the principle of duty cycle. The waveform illustrated in figure 16.12 (a) has a reference time of 2 seconds, whereas the reference time in figures 16.12 (b), (c) and (d) has a duty cycle of 75%, 50% and 25% compared to that of figure 16.12 (a).

Definition 16.6 Duty Cycle

Duty cycle may therefore be defined as the ratio of on time to the reference time and the reference time being the sum of the on time and the off time.

+

100 %

(a)

t 1

2

3

4

5

6

+

7

8

9

10

75 %

(b)

t

+

50 %

(c)

t

+

25 %

(d)

t

Figure 16.13


283

16.6 The Diac A Diac is a two terminal bi-directional semi-conductor component which is normally used in conjunction with a Triac. The IEC circuit symbol is illustrated in figure 16.13 (a) and the characteristics in figure 16.13 (b). A Diac can conduct in both directions, hence the term bi-directional meaning two, and may be seen as two diodes connected backto-back. The use of the Diac will be discussed under the next section that will deal with the Triac. +I

Forward bias Conducting region Anode 1

Anode 2

-V

+V

Reverse bias Conducting region (a) -I (b)

Figure 16.14

16.7 The Triac A Triac is a three terminal bi-directional gate-controlled semi-conductor component which is normally used in conjunction with a Diac. The IEC circuit symbol is illustrated in figure 5.14 (a) and the characteristics in figure 16.14 (b). This component has the advantage that it may be triggered by a positive pulse on the gate for one half of the input waveform as well as a negative pulse on the other half of the input waveform thereby giving us full-wave control in a single component. +I Forward bias Conducting region Holding currents IH MT1

MT2

-V

+V

Holding currents IH

Gate (a)

Reverse bias Conducting region -I (b)

Figure 16.15

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Power Control

16.8 Power control using a Diac and Triac The circuit diagram illustrated in figure 16.15 is that of a Diac and Triac used for speed control or a light dimmer. The waveforms are not given since it will be the same as that for figure 16.5. Universal load R

Triac VIN RV1 Diac

dc load

C

Figure 16.16

Using a Diac and a Triac gives us the opportunity to make use of full-wave control. The purpose of the diac is mainly to allow a negative pulse on the gate during the negative half of the input waveform and to allow a positive pulse on the gate during the positive half of the input waveform.

16.9 The Quadrac A Quadrac is a three terminal bi-directional semi-conductor component and consists of a Diac and a Triac in one package and has the characteristics of each component on its own. The characteristic curve of the Quadrac will therefore not be given and neither will a circuit diagram illustrating its application be given. The application is the same as that for figure 16.15 with the exception that a Quadrac is used as a single component instead of a Diac and Triac as indicated. The IEC circuit symbol is illustrated in figure 16.16. Anode

Gate

Cathode

Figure 16.17


16.10

285

Control systems

There are two types of main control systems available in industry today but it must be mentioned that both types are not always used.

16.10.1

Open-loop control systems

In this type of control the result cannot be controlled since the output is not monitored. The following scenario will illustrate this concept. Assume a borehole is used to fill a water tank. The borehole pump will pump water from the borehole into the tank for as long as the pump is activated. Since there is no indication when the tank is full the pump will keep on pumping which will result in a waste of a very scares commodity. The only time when the flow of water will be interrupted is when the pump is manually deactivated. This action can be rectified by using a closed-loop system.

16.10.2

Closed-loop control systems

Output Comparator Input

Quantity to be controlled

Sensor

Reference section supplies a constant signal

Correcting signal Feedback signal Correcting circuit

Amplifier

Figure 16.18

Error signal

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Power Control

A quantity need to be controlled to alleviate the scenario that was mentioned in 16.10.1. A sensor is used to determine the state of the output to comply with the reference we supply. As soon as this reference is exceeded a signal will be sent to the comparator which will compare the received signal and the reference signal and if they are different it will generate an error signal. This error signal will be amplified and sent as a feedback signal to the correction section which will send a correcting signal which will adjust the quantity to be controlled to the required level in order to obtain a pre-set output specified.

Exercise 16.1 1. Define an SCR. 2. An SCR may be represented by two transistor analogy in order to explain its operating principle. Draw such a neat labelled circuit diagram and explain the operating principle of an SCR. 3. Use a suitable characteristic curve to illustrate the principle of operation of an SCR and then tabulate its characteristics. 4. Differentiate between line commutation and forced commutation. 5. Use a graphic representation and illustrate the difference between a delay angle and conduction angle with reference to an SCR. In your illustration you must make reference to full-wave as well as half-wave control. 6.

Draw a neat fully labelled circuit diagram with input and output waveforms to illustrate the application of an SCR in: 6.1 Full-wave control; 6.2 A dc circuit; and 6.3 Half-wave control.

7. Use suitable graphic representation to illustrate the principle of: 7.1 Cyclotronic control; 7.2 Phase control; and 7.3 Cycle control, and mention at least one advantage or disadvantage of each. 8. Use your own reference time and a square wave to explain the concept of duty cycle. 9. Compare the characteristics by means of a characteristic curve of a Diac and a Triac.


10.

Full-wave control may be obtained by using a Quadrac. Draw such a neat labelled circuit diagram and indicate the following waveforms: 10.1 An input waveform; 10.2 An output waveform across the load; and 10.3 An output waveform across the Quadrac.

11. Explain the difference between an open-loop control system and a closed- loop control system. 12. Draw a neat labelled sketch of a closed-loop control system and then give a detailed description of the operation thereof.

287

288

Power Control

Operational Amplifiers


289

CHAPTER

17 Operational Amplifiers Learning Outcomes On completion of this module you will be able to: • • •

•

•

Describe the characteristics of an operational amplifier; Draw the IEC circuit symbol and pin-layout of an operational amplifier; Give the uses for the following operational amplifier applications: – Inverting amplifier; – Non-inverting amplifier; – Summing amplifier (inverting); – Voltage follower; – Integrator; and – Differentiator. Draw the circuit diagram indicating the input and output waveforms of an operational amplifier used as a (n): – Inverting amplifier; – Non-inverting amplifier; – Summing amplifier (inverting); – Voltage follower; – Integrator; and – Differentiator. Do calculations of resistor values, capacitor values, input and output voltages for the following operational amplifier applications: – Inverting amplifier; – Non-inverting amplifier; – Summing amplifier (inverting); and – Integrating amplifier.

17.1 Introduction In the previous modules on transistor applications it was found that all circuits had to be constructed using discreet components (transistors, resistors, capacitors, inductors, diodes, etc.) and that it is quite a cumbersome effort. The operational amplifier

290


solves this problem to quite an extent in that an amplifier is now available in a single integrated circuit (IC) package. Other types of IC-packages are also available but are not covered by the scope of this module. These IC-packages are fairly cheap to manufacture and has the following characteristics: • • • • •

High input impedance; Low output impedance; A high voltage gain in the region of 106; Wide bandwidth; and The ability to handle both ac and dc signals.

17.2 The operational amplifier Various modes of operation can be obtained from an operational amplifier and the operation is based on the concept of a differential amplifier which we will not discuss at this point in time. The IEC-circuit symbol and pin-layout for a 741 dual-in-line (DIL) 8-pin operational amplifier is illustrated in figure 17.1 (a) and (b). 8 -

7

6

5

3

4

+V

Vi

+

-V

Vi

+

VO 1

(a)

2 (b)

Figure 17.1

The following list supplies the purpose of each pin on the 741 dual-in-line (DIL) 8-pin operational amplifiers: 1 Null offset This pin is utilised together with pin 5 in order to obtain a zero output when no signal is present on any of the inputs. 2 Inverting input This is the pin on which an input may be applied and that will give an amplified but inverted output which means that the output is 180º out of phase with the input. 3 Non-inverting input The pin on which an input may be applied and that will give an amplified but an in phase output and the phase angle is 0º.


291

4 -V The pin on which the negative polarity of the bias supply is connected to the operational amplifier. 5 Null offset This pin is utilised together with pin 1 in order to obtain a zero output when no signal is present on any of the inputs. 6 Output This is the pin from which the amplified output is taken. 7 +V The pin on which the positive polarity of the bias supply is connected to the operational amplifier. 8 NC This pin is unconnected. In order to get a clearer view of an operational amplifier and how it functions you are referred to the following illustrations. An operational amplifier has two input terminals namely an inverting input marked - and a non-inverting input marked +. A single output can be obtained depending on which of the above terminals have been utilised as the input. Figure 17.2 illustrates the concept of a signal being applied to the inverting input terminal of an operational amplifier and the non-inverting input grounded. +V -

+ -V

Figure 17.2

The terminal marked - is the inverting input and a signal is applied between the inverting input and ground. The output will therefore be the inverted (180 ºout of phase) with the input. Figure 17.3 illustrates the concept of a signal being applied to the non-inverting input terminal of an operational amplifier and the inverting input grounded.

292


+V +

-V

Figure 17.3

The terminal marked + is the non-inverting input and a signal is applied between the non-inverting input and ground. The output will therefore be in phase (0 º) with the input. Figure 17.4 illustrates the concept of a signal being applied to both the input terminals of an operational amplifier. +V -

+ -V

VO = 0

Figure 17.4

Here we have a situation where two signals, equal in magnitude and in phase with one another are applied to both the inverting input as well as the non-inverting input and the output is zero. But why is the output zero? The input on the inverting input will cause a phase shift of 180º at the output and the input on the non-inverting input will have a 0º phase shift at the output and since these outputs are out of phase with one another they will cancel one another out and no output will be obtained. It must however be noted that this condition will only exist when the null-offset has been set to zero and that the operational amplifier have no output in the absence of any input signal. Figure 17.5 and figure 17.6 illustrates the concept of a signal being applied to both the input terminals of an operational amplifier but is different from the circuit illustrated in figure 17.4. Please observe the phases of the signals to the inverting input and noninverting input of the operational amplifier since it will have an effect on the output of the operational amplifier.


293

+V -

+ -V

Figure 17.5

Here we have a situation where two signals, equal in magnitude and out phase with one another are applied to both the inverting input as well as the non-inverting input and the output is double the magnitude of a single input. But why is the output double? The input on the inverting input will cause a phase shift of 180º at the output and the input on the non-inverting input will have a 0º phase shift at the output and since these outputs are in phase with one another they will add together out and double the output will be obtained. It must however be noted that this condition will only exist when the null-offset has been set to zero and that the operational amplifier have no output in the absence of any input signal. +V -

+ -V

Figure 17.6

17.3 Modes of operation The good characteristics, size and cost of operational amplifiers makes them very versatile in a number of applications.

17.3.1 The inverting amplifier An operational amplifier connected in the inverting mode is illustrated in figure 17.7. This type of configuration is normally used for amplification purposes since the main characteristic of an operational is its good gain that can be obtained. It is also very useful in applications where a phase inversion is required.

294


Rf RIN

Vi

+

VO

Figure 17.7

The output of this mode of operation is mathematically given by: -R × Vi VO = f where VO RIN Rf Vi RIN

= = = =

magnitude of output voltage in volt value of feedback resistor in ohm magnitude of input voltage in volt value of input resistor in ohm

The gain of the operational amplifier in this particular mode is mathematically given by: - Rf A = R IN

VO

=

A × Vi

Example 17.1 The values for an inverting operational amplifier are given as follows: • Input voltage = 3 volt • Input resistor (RIN) = 150 ohm Determine the value of the output voltage when the following resistance values are inserted for the feedback resistor: (a) 300 ohm (b) 500 ohm (c) 1 500 ohm (d) Having obtained the output voltage values for the different feedback resistors, what conclusions can be made concerning the answers?


Solution: - R × Vi (a) VO = f RIN - 300 × 3 = 150 = - 6 volt - R × Vi (b) VO = f RIN - 500 = × 3 150 = - 10 volt - R × Vi (c) VO = f RIN -1 500 = × 150 = - 30 volt

3

(d) The higher the resistive value of the feedback resistor the higher the output voltage. This is so because the portion of the voltage which is fed back negatively becomes smaller and therefore the gain of the amplifier increases. A point that needs to be mentioned is that for the inverting mode of operation the answer is preceded by a minus sign and this minus sign indicates that there is a 180º phase shift between the input and the output. However, should any other value be determined it is important that this minus sign then be used as a mathematical sign since it would be impossible to determine the value of a resistor and ending up with a minus value.

17.3.2 The non-inverting amplifier An operational amplifier connected in the inverting mode is illustrated in figure 17.8. This type of configuration is normally used for amplification purposes since the main characteristic of an operational is its good gain that can be obtained. The overall gain of this mode (non-inverting) is higher than that of the inverting mode of operation. The output of this mode of operation is mathematically given by:

295

296


R VO = 1 + f × Vi where RIN

VO Rf Vi RIN

= = = =

magnitude of output voltage value of feedback resistor in ohm magnitude of input voltage value of input resistor in ohm

The gain of the operational amplifier in this particular mode is mathematically given by: Rf RIN

A =

1+

VO

A × Vi

=

+

Vi Rf

VO

RIN

Figure 17.8

Example 17.2 The values for a non-inverting operational amplifier are given as follows: • Input voltage = 3 volt • Input resistor (RIN) = 150 ohm Determine the value of the output voltage when the following resistance values are inserted for the feedback resistor: (a) 300 ohm (b) 500 ohm (c) 1 500 ohm (d) Having obtained the output voltage values for the different feedback resistors, what conclusions can be made concerning the answers? In your conclusion you are also to refer to the answers you obtained in example 17.1 above.


297

Solution: Rf (a) VO = 1 + × Vi RIN 300 × 3 = 1 + 150 = 9 volt

(b) VO = = =

Rf × Vi RIN 500 1+ ×3 150

1+

13 volt

Rf (c) VO = 1 + × Vi RIN 1 500 = 1 + ×3 150 = 33 volt

(d)

The higher the resistive value of the feedback resistor the higher the output voltage. This is so because the portion of the voltage which is fed back negatively becomes smaller and therefore the gain of the amplifier increases. However, should we make use of an operational amplifier in the non-inverting mode the gain increases further.

Worthy to take not of is that whether we are connecting an operational amplifier in the inverting mode or non-inverting mode, that the input resistor and the feedback resistor form a voltage divider network and the amount of feedback is determined by the ratio between these two resistors.

17.3.3 The inverting summing amplifier (adder) An operational amplifier connected in the inverting summing mode (adder) is illustrated in figure 17.9. Typical application is that of data collection where it is often required to add several output voltages from various sources such as sensors etc.

298


Rf RIN1 RIN2 +

Vi1

VO

RIN3

Vi2 Vi3

Figure 17.9

The output of this mode of operation is mathematically given by: -R -Rf -Rf VO = f × Vi1 + × Vi2 + × Vi3 RIN1 RIN2 RIN3

where VO Rf Vi RIN =

= = = =

magnitude of output voltage in volt value of feedback resistor in ohm magnitude of input voltage in volt value of input resistor in ohm

V Vi2 V Rf ( i1 + + i3 ) ( RIN1 RIN2 RIN3)

should

RIN1 ≠ RIN2 ≠ RIN3

Should RIN1 = RIN2 = RIN3 = R then: R VO = f (Vi1 + Vi2 + Vi3) R

The gain of the operational amplifier in this particular mode is mathematically given by: Rf Rf Rf A1 = A2 = A3 = R RIN2 R IN3 IN1

VO = AT × (Vi1 + Vi2 + Vi3)

where

AT

=

A1 + A2 + A3


Example 17.3 The values for a non-inverting summing operational amplifier are given as follows: 10 k-ohm Vi1 = 2,5 m-volt RIN1 = 15 k-ohm Vi2 = 7,5 m-volt RIN2 = 25 k-ohm Vi3 = 12,5 m-volt RIN3 = Determine the output voltage for the following value feedback resistors. (a) 100 k-ohm; (b) 50 k-ohm; and (c) 150 k-ohm.

Solution: (a)VO = = = =

-Rf -Rf × Vi1 + × Vi2 + RIN1 RIN2

-Rf × Vi3 RIN3

100 × 103 (2,5 × 10-3 + 7,5 × 10-3 + 12,5 × 10-3) 25 × 103 ) (10 × 103 15 × 103 3 -6 -6 100 × 10 (0,25 × 10 + 0,5 × 10 + 0,5 × 10-6 ) -125 m-volt

-R -Rf -Rf (b) VO = f × Vi1 + × Vi2 + × Vi3 R RIN2 RIN3 IN1 = 50 × 103 (2,5 × 10-3 + 7,5 × 10-3 + 12,5 × 10-3) 25 × 103 ) (10 × 103 15 × 103 = 50 × 103 (0,25 × 10-6 + 0,5 × 10-6 + 0,5 × 10-6 ) = -62,5 m-volt -R -Rf -Rf (c) VO = f × Vi1 + × Vi2 + ×V RIN1 RIN2 RIN3 i3 = 150 × 103 (2,5 × 10-3 + 7,5 × 10-3 + 12,5 × 10-3) 25 × 103 ) (10 × 103 15 × 103 3 -6 -6 = 150 × 10 (0,25 × 10 + 0,5 × 10 + 0,5 × 10-6 ) = -187,5 m-volt

299

300


Example 17.4 The values for a non-inverting summing operational amplifier are given as follows: = R = 15 k-ohm RIN1 = RIN2 = RIN3 Vi2 = 15 m-volt; Vi3 = 20 m-volt Rf = 60 k-ohm; Determine the output voltage for this operational amplifier.

Solution: (a) VO = -Rf (Vi1 + Vi2 + Vi3) R = 60 × 103 (12,5 × 10-3 + 15 × 10-3 + 20 × 10-3) 15 × 103 = -190 m-volt

Example 17.5 The output of an operational amplifier is given as: - (A + B + C) volts VO = Draw a neat fully labelled circuit diagram of an operational amplifier that will conform to the given output expression. NO component values are required but input and output wave forms must be indicated.

Solution: Rf RIN1 RIN2 +

A

RIN3

B C

VO


301

17.3.4 The voltage follower An operational amplifier connected as a voltage follower in the inverting mode is illustrated in figure 12.10. It has a gain of unity and VO = Vi and it is sometimes referred to an amplifier where the output follows the input hence the name ‘voltage follower’. Its main application is that of and impedance matching circuit since it has a high input impedance and a low output impedance. It is further used as a ‘buffer circuit’ when two circuits need to be isolated from one another and the first circuit need to be protected.

17.4 Mathematical functions using operational amplifiers + VO

Vi

Figure 17.10

Over and above the application already mentioned for operational amplifiers it may also be used for mathematical decision making applications.

17.4.1 The differentiator An operational amplifier used as a differentiator and will be applied where differentiation of an input waveform is required and is illustrated in figure 17.11 (a), the input waveform is illustrated in figure 17.11 (b) and the output waveform is illustrated in figure 17.12 (c).

302


Rf CIN VIN

+

VO

(a)

Input

Output

(b)

(c)

Figure 17.10

17.5.2 The integrator An operational amplifier used as an integrator and will be applied where integration of an input waveform is required and is illustrated in figure 17.12 (a), the input waveform is illustrated in figure 17.12 (b) and the output waveform is illustrated in figure 17.12 (c).

Cf

RIN

VIN

+

VO

(a)

Input

Output

(b)

(c)

Figure 17.11


303

The rate of change of the output voltage is mathematically given by: Vi Vo = V/sec where Vo C×R Vi IN C RIN

= = = =

Rate of change of output voltage Value of input voltage in volt Value of feedback capacitor in Farad Value of input resistor in ohm

Example 17.6 The following data concerning an integrating operational amplifier is given: Feedback capacitor = 100 µF Input resistor = 10 k-ohm Input voltage = 5 volt Use this information to determine the rate of change of the output voltage and then represent this change determined graphically.

Solution: Vi Vo = V/sec C × RIN Vo = 5 100 ×10-6 × 10 ×103 = - 5 V/sec 1 sec t

Slope

-

Output voltage

5 V

304


The following set of waveforms in figure 17.13 illustrates the input- as well as different output waveforms of the different applications of operational amplifiers that have been discussed.

+ -

t

Square wave input to all circuits

+ -

t

Output for inverting amplifier and voltage follower

+ -

t

Output for noninverting amplifier

t

Output for integrator circuit

t

Output for differentiator circuit

-

+ -

Figure 17.13

Exercise 17.1 1. Name five characteristics of an operational amplifier. 2. Draw a neat labelled sketch of an 8-pin dual-in-line representation of a 741 operational amplifier and describe the purpose of each pin. 3.

Draw neat labelled circuit diagrams of an operational amplifier complete with input- and output waveforms for the following conditions: 3.1 Input on non-inverting input with inverting input grounded; 3.2 Equal magnitude in-phase input on both inputs;


3.3 Input on inverting input with non-inverting input grounded; and 3.4 Equal magnitude out of phase inputs on both inputs.

4. An operational amplifier is marked with a positive and a negative sign. What is the meaning of these signs? 5. The output of an operational amplifier can either be in-phase or out of phase with the input signal. Draw neat labelled circuit diagrams that will illustrate this principle. 6. The output from three sensors is given as: = 4,5 m-volt • Vi1 = 6,5 m-volt • Vi2 = 2,5 m-volt • Vi3 and is applied to the non-inverting input of an operational amplifier. Determine the output voltage of this amplifier if it uses a 50 k-ohm feedback resistor and the following input resistors: 6.1 15 k-ohm; 6.2 25 k-ohm; and 6.3 40 k-ohm. 7. Consider the following input and output waveforms.

(a)

(b)

(c)

+ -

+ -

t

t

t

Waveform (a) is an input waveform and waveforms (b) and (c) are output waveforms. Draw neat labelled circuit diagrams of an operational amplifier that will have as an output: 7.1 Waveform (b): and 7.2 Waveform (c).

305

306


8. The following data is given: • Feedback resistor = 100 k-ohm • Input voltage = 125 m-volt Determine the output voltage for: 8.1 A non-inverting operational amplifier; and 8.2 An inverting operational amplifier. The following resistors are connected as input resistors for the two modes of operation. (a) 15 k-ohm; (b) 50 k-ohm; and (c) 75 k-ohm. 9. A network with a high input impedance need to be matched to a network having low impedance by means of an operational amplifier. Draw a neat labelled circuit diagram of an operational amplifier that will conform to this specification. 10. The following data concerning an integrating operational amplifier is given: • Feedback capacitor = 200 µF • Input resistor = 10 k-ohm • Input voltage = 4 volt Use this information to determine the rate of change of the output voltage and then represent this change determined graphically.

Function generator and oscilloscope N4 Industrial Electronics

307

CHAPTER

18 Function Generator and Oscilloscope Learning Outcomes On completion of this module you will be able to: •

Explain by means of a block diagram a(n): – Function generator; and – Oscilloscope. • Indicate the uses of a(n): – Function generator; and – Oscilloscope. • Explain using suitable sketches the operating principle of a cathode ray oscilloscope (CRO); • Discuss the following focusing concepts: – Electrostatic; and – Electromagnetic. • Discuss the following deflection concepts: – Electrostatic; and – Electromagnetic. • Explain using suitable sketches the concept of synchronisation; • Explain, describe and do calculations of signal analysis for: – Amplitude measurement; – Time measurement; and – Frequency. In the field of Electronics it is required that use is made of different test instruments to perform our task. Some of these instruments are used to provide specific quantities ie function generator and other instruments are used to be able to measure, but more so to observe ie oscilloscope. A function generator is an instrument that is capable of delivering a choice of waveforms and whose frequencies are adjustable over a fairly large range. The output wave forms available include:

308

Function Generator and Oscilloscope

• • • •

Sine waves; Triangular waves; Square waves; and Saw-tooth waves from terminals A, B and C.

Depending on where and how the function generator is utilised it is of utmost need and importance in an electronic workshop. The block diagram in figure 18.1 illustrates such a function generator. It must be noted however that the function generator is not deemed a measuring instrument but deemed a test instrument and will be used together with certain measuring instruments, digital voltmeter and oscilloscope, to perform specific tasks. The purpose of each block illustrated in figure 18.1 may be summarised as follows: •

Frequency control network This section is controlled from the front panel of the function generator and will control the upper current source and the lower current source.

Upper current source C

Frequency control network

Voltage comparator

A

Intrgrator

Lower current source

Output amplifier 1

B

Shaping circuit

C

Output

Output amplifier 2

Figure 18.1

• • •

Lower current source The lower current source supplies a constant negative current to the integrator of which the output voltage will increase in a linear fashion with time. Upper current source The upper current source supplies a constant positive current to the integrator of which the output voltage will increase in a linear fashion with time. Integrator A triangular wave is obtained from the integrator output and its magnitude is determined by the constant current source and this triangular waveform is supplied to the comparator and wave-shaping circuits.


• • •

309

Comparator The main purpose of the comparator is to deliver a square wave at the same frequency of the triangular waveform. Wave shaping circuit This circuit will use the triangular waveform slope changes to convert the waveform to a sine-wave. Output amplifiers Two amplifiers supplies individually and independent outputs simultaneously.

An oscilloscope is a measuring instrument which is not only capable of giving the magnitude of a measurement but also the shape of the variable. Such a block diagram is illustrated in figure 18.2 and an oscilloscope may be used in the following applications: • • • • • •

Studying waveforms; Indicating phase relationships of waveforms; ac- and dc-voltage measurement; Amplification gain; Frequency determination; and/or Distortion indication.

The block diagram in figure 18.2 indicates all the main controls of an oscilloscope. The following list will give an indication of the main purpose of the indicated controls.

Vertical position

V/div

VIN

Attenuator

Vertical amplifier

CRT

Focus

Z

Trigger circuit External trigger

Time-base

Horisontal amplifier

+ Level

-

Slope

T/div

Figure 18.2

CRT control

Horisontal position

Intensity

310


Focus and intensity Ensures a well-focused display at a convenient intensity. Vertical- and horisontal position

The vertical- and horisontal position will allow for movement of the trace left and/or right and/ or up and/or down.

V/div

The V/div will indicate the voltage per division on the oscilloscope display screen.

T/div

The T/div will indicate the time per division on the oscilloscope display screen.

Slope and level The slope and level controls will select the point on the waveform where the display is to commence. The main component in an oscilloscope is the cathode ray tube (CRT) and is illustrated in figure 18.3 and consists of an electron gun, electron lens, Y-deflection plates, X-deflection plates and an oscilloscope screen. The operation is quite elementary and can be described in the following manner: • • • • •

The cathode termed the electron gun which is at a negative potential and emits electrons; A focusing and accelerating section which provides for acceleration of the electron beam as well as a well defined focus; A deflection system which may be electrostatic or magnetic controlling the movement of the electron beam onto the screen; A screen at a very high positive potential covered in phosphor which will glow when the electron beam strikes the screen; and The above components are covered by an evacuated glass tube.


311

X-deflecting plates

Electron gun

CRT screen Anode

Cathode Electron lens

Y-deflecting plates

Electron beam

Figure 18.3

Focusing of the electron beam can either be done magnetically or electrostatically. Figure 18.4 illustrates the concept of electromagnetic focusing. There are two coils placed around the neck of the glass tube and the magnetic thus created will extend into the tube and it will affect the focusing of the electron beam. The magnitude of the current in the external coils will determine the magnetic field intensity and therefore the sharpness or focus of the spot produced on the CRT screen. Coil Electron gun

CRT screen Anode

Cathode Coil

Figure 18.4

Figure 18.5 illustrates the concept of electrostatic focusing. This is the most commonly used concept of focusing used in oscilloscopes and employs the use of intermediate anodes. The most acceleration takes place between the cathode and the first anode which is at a fairly high positive potential. The second anode is at a lower positive potential than the first and third and controls the focusing. The third anode will provide for further acceleration providing enough energy to produce a spot on the screen since the screen anode is at a higher positive potential than the preceeding anodes.

312


Deflection of the electron beam can either be done electromagnetically or electrostatically. Anodes

CRT screen Anode Electric field

Electron beam

Figure 18.5

Two pairs of coils are placed symmetrically on the outside of the glass tube. The X-deflector coils are mounted horizontally below and above the tube and the Y-deflector coils are mounted vertically one on either side of the tube. Electrostatic deflection uses two pairs of parallel plates which operate independently from one another. Should a potential difference between the two horizontally mounted Y-deflection plates the electrons are deflected in the vertical plane. Should a potential difference between the two vertically mounted X-deflection plates the electrons are deflected in the horisontal plane. The latter of the above two concepts are illustrated in figure 18.6. Electrons will move from X1 towards X2 at a constant speed and in the middle of the screen provided that the voltages on all four plates (X1, X2, Y1 and Y2) are equal in magnitude. The velocity of the electrons will be determined by the time base signal. The waveform to be studied is applied to the vertical amplifier and from there to the vertical deflection plates (Y1 and Y2) and is controlled by the V/div control whereas the


313

Y1 Beam movement

X1

X2 Input signal

Y2

Time-base signal

Figure 18.6

time base signal is applied to the horisontal amplifier and from their to the horisontal deflection plates (X1 and X2) and is controlled by the T/div control and it is this signal that will display a trace across the screen. When all four plates have the same magnitude voltage the electron beam will be displayed exactly in the middle of the screen. Increasing the voltage on the Y1-plate will cause the beam to move up. The converse to this is also true. Increasing the voltage on the X1-plate will cause the beam to move left. The converse to this is also true. Note that this will be the case whether we apply a signal or not. Should we now apply a signal to the Y-plates we want our display to be as stable as possible. This concept is termed synchronisation.

18.1 Synchronisation The signal to be studied and the time base signal must be synchronised in order to ensure a stable signal on the oscilloscope screen. The graphical representation in figure 18.7 illustrates this concept. Should the signal to be viewed and the time-base frequency be synchronised then the display on the oscilloscope screen will be stable as illustrated in figure 18.7 (a). However, should the time-base frequency be too low then the display will drift to the left as illustrated in figure 18.7 (b). The converse is also true in that should the time-base frequency be too high then the display will drift to the right as illustrated in figure 18.7 (c).

314


+ -

(a)

+ +

+ + + -

+ -

(c)

Vertical input signal

t

Horisontal input signal

t

-

(b)

t

+ + -

Stable display

t


t


t

Display drifts to the left

t


t


t

Display drifts to the right

Figure 18.7

18.2 Signal analysis In order to analyse signals we need to understand the layout of the CRT-screen and this is illustrated in figure 18.8.


Vertical axis

1 cm

1 cm

Horisontal axis

Figure 18.8

18.2.1 Amplitude measurement The amplitude (peak-value) of the displayed signal is governed by the V/div control. This control is further calibrated in terms of V/cm and since the screen is normally divided into 1 cm squares it converts to each block representing a given voltage. The amplitude (VP) measurement is determined by the following mathematical expression. VP

=

height × V/div

and VP – P

=

2 × VP

Example 18.1 The following waveform is displayed on an oscilloscope screen and the V/div control is set at 2,5 V/div. Determine the peak-to-peak value of the quantity.

Solution: VP = = =

height × V/div VP – P 3 × 2,5 7,5 volt

= = =

2 × VP 2 × 7,5 15 volt

315

316


Example 18.2 Calculate the peak-to-peak value of an alternating quantity for a V/div setting of 1,5 V/cm when the amplitude of this alternating quantity is 4 cm on the oscilloscope screen.

Solution: VP

= = =

height × V/div VP – P 4 × 1,5 6 volt

= = =

2 × VP 2 × 6 12 volt

18.2.2 Time Measurement The time measurement, also termed periodic time of the displayed signal is governed by the T/div control. This control is further calibrated in terms of seconds/cm and since the screen is normally divided into 1 cm squares it converts to each block representing a given time period. The time measurement is determined by the following mathematical expression. t

=

length of wave × T/div

Example 18.3 The following waveform is displayed on an oscilloscope screen and the T/div control is set at 2, 5 µ-sec/div. Determine the time duration of the signal.

Solution: t

= = =

length of wave × T/div 8 × 2,5 × 106 10 µ-sec


317

Example 18.4 Calculate the periodic time of an alternating quantity for a T/div setting of 2 m-sec/cm when the distance of this alternating quantity is 10 cm on the oscilloscope screen.

Solution: t

18.2.3

= = =

length of wave × T/div 10 × 2 × 103 20 m-sec

Frequency measurement

Although we term this section frequency measurement it is actually the calculation of frequency since frequency can only be measured by means of a frequency counter. Since frequency may be related to time we may define frequency calculation by the following mathematical expression. 1 f = t

where

t

=

periodic time

Example 18.5 Determine the frequencies of example 18.3 and 18.4 above.

Solution: f = 1 f = 1 t t = 1 = 1 6 20 × 103 10 × 10 = 400 MHz = 50 Hz

318


Example 18.6 Refer to the following oscilloscope display.

The oscilloscope is adjusted to the following settings. V/div = 7, 5 V/cm and V/div = 25 µ-sec/cm Determine the: (a) Peak-to-peak value; and (b) Frequency of the alternating quantity.

Solution: (a) VP = height × V/div VP - P = 5 × 7,5 = 37,5 volt

= 2 × VP = 2 × 37,5 = 75 volt

(b) t = length of wave × T/div f = 1 -6 = 12 × 25 × 10 t = 300 µ-sec = 1 300 × 10-6 = 3,333 kHz

18.3

Waveforms

During the process of utelising test instruments ie oscilloscopes and function generators one will meet up with various forms of waves which need to be described and understood.

18.3.1 Sine-waves The development and obtaining of a sine wave was done in previous studies and only a brief background will be given so as to recap the principles and concepts that were discussed at that stage.


A sine-wave is illustrated in figure 18.9 with all important aspects noted. The most important concepts will be recapped. V or I

Pos

Peak-to-peak value

Peak value

t

Neg

One cycle V or I

(b)

Figure 18.9

Frequency The frequency of this alternating quantity is related to time as well as the number of times the reversal of direction takes place. Since the same repetitive action takes place every 360º and one cycle is completed in this time. In order to determine the frequency we make use of the expression:

Definition 18.1 Frequency

Frequency may therefore be defined as the number of full cycles that moves past a given point in one second.

f =

1 where t

t f

= =

time in seconds frequency in hertz

319

320


Example 18.7 Determine in each instance the frequency for the alternating quantity: (a) 1 second; (b) 1 m-second; and (c) 10 m-seconds.

Solution: 1 (a) f = t 1 = 1 = 1 Hz 1 (b) f = t = 1 1 × 10-3 = 1 kHz

(c) f = 1 t = 1 10 × 10-3 = 100 Hz

Figure 18.10 (a) and (b) illustrates the frequency of an alternating quantity over the same time period. Figure 18.10 (a) illustrates a frequency of one cycle over a given time period and figure 18.10 (b) illustrates a frequency of two cycles over the same time period.

+

+ t

-

t

(a)

Figure 18.10

(b)


Periodic time

Definition 18.2 Periodic Time

The periodic time of an alternating quantity is that time it takes to complete one full cycle of 360°.

Example 18.8 The following frequencies of an alternating quantity were measured: (a) 150 kHz; (b) 10 Hz (c) 1 MHz Calculate in each instance the periodic time.

Solution: 1 (a) f = t 1 t = f = 1 150 × 103 = 6,667 µ-seconds 1 (b) f = t 1 t = f = 1 10 = 100 m-seconds 1 (c) f = t 1 t = f = 1 1 × 106 = 1 µ-second

321

322


Peak value The maximum- or peak value will be reached when the conductor lies at right angles (90º) to the magnetic field. You will notice that this value is reached twice during a full cycle, at 90º and once at 270º where the first is during the positive half-cycle and the second during the negative half-cycle. Peak-to-peak value The peak-to-peak value is the total value measured between the two peak values. This is mathematically expressed as: VP-P = 2 × VP

where

VP = peak value of alternating quantity

Example 18.9 Calculate the peak-to-peak value of the following alternating quantities: (a) 35 V; (b) 60 V; and (c) 120 V

Solution: (a) VP-P = 2 × VP = 2 × 35 = 70 volt = 2 × VP (b) VP-P = 2 × 60 = 120 volt (c) VP-P = 2 × VP = 2 × 120 = 240 volt


323

Example 18.10 An alternating quantity has the following peak-to-peak values: (a) 240 V; (b) 100 V; and (c) 180 V Determine in each instance the peak value of the alternating quantity.

Solution: (a) VP-P

=

2 × VP

V VP = P-P 2 240 = 2 = 120 volt

(b) VP-P

=

2 × VP

V VP = P-P 2 100 = 2 = 50 volt

(c) VP-P

=

2 × VP

V VP = P-P 2 180 = 2 = 90 volt

RMS- and Average values The root mean square (RMS) value of an alternating quantity is also termed the effective value and is that value that actually does the work.

324


Definition 18.3 RMS-value

The RMS-value may be defined as that value of the alternating quantity that will produce the same heat energy as that of a direct quantity when applied to the same resistive component.

This RMS-value is given by the following mathematical expression: VRMS = 0,707 × VP where VRMS = 0,707 × IP IRMS IRMS VP IP

= = = =

RMS-value in volt RMS-value in ampere peak value of voltage peak value of current

The average value of an alternating quantity is mathematically given by: VAVE = 0,637 × VP where IAVE = 0,637 × IP

VAVE IAVE VP IP

= = = =

average value in volt average value in ampere peak value of voltage peak value of current

Example 18.11 An alternating quantity has a maximum voltage value of 120 V at a peak-topeak current of 10 A. Determine in each instance the: (a) RMS-value of the voltage and current; and (b) Average value of the voltage and current.

Solution: (a) VRMS = 0,707 × VP IRMS = 0,707 × IP = 0,707 × 120 = 0,707 × 10 = 84,84 volt 2 = 3,535 ampere = 0,637 × IP (b) VAVE = 0,637 × VP I AVE = 0,637 × 120 = 0,637 × 10 = 76,44 volt 2 = 3,185 ampere


Figure 18.11 illustrates a square wave. A square wave can either be positive going or negative going and are mainly used in computer circuitry as well as logic circuits since it will either be on or off. On

Off Peak value

Leading edge

Mark

Space

Trailing edge

One cycle

Figure 18.11

Figure 18.12 illustrates a sawtooth wave. The sawtooth waveform is also termed a triangular wave corresponding to its shape. It is mainly used in oscilloscope time-base circuits in order to deflect the electron beam across the oscilloscope screen. You will further notice that the positive ramp and the Positive ramp

Negative ramp

Rise time Fall time

Figure 18.12

negative ramp have different slopes and that the rise time is greater than the fall time which is also termed the fly-back time,

Exercise 18.1 1.

Draw neat labelled block diagrams of a (n): 1.1 Function generator; and 1.2 Oscilloscope and indicate all the main controls. Give in each instance as many uses as possible for all the measuring instruments and test equipment above.

325

326


2. There are eight major controls on an oscilloscope. Discuss each of these controls. 3. Use a suitable sketch to describe the operation of a cathode ray tube with specific reference to the display of a signal. 4. Use suitable graphic representations to illustrate the concept of synchronisation. 5. Use suitable sketches to illustrate signal analysis with special reference to: 5.1 Amplitude measurement; and 5.2 Frequency determination. 6.

The following data concerning an oscilloscope display is given: Vertical height = 4 cm Horisontal width = 3 cm V/div = 25 µV/cm T/div = 20 µsec/cm Determine the: 6.1 Peak-to-peak voltage of the signal; and 6.2 Frequency of the signal.

7. Consider the following displayed oscilloscope signal.

7.1 How many cycles are displayed? 7.2 Calculate the: 7.2.1 Peak-to-peak value of the displayed signal; and 7.2.2 Frequency of the displayed signal. The following additional data is given. V/div = 10 mV/cm T/div = 10 m-sec/div


8.

Draw neat labelled graphic representation with all relevant characteristics of the following waveforms: 9.1 Square wave; 9.2 Sine wave; and 9.3 Triangular wave.

327

328


TRANSDUCERS


CHAPTER

19 Transducers Learning Outcomes On completion of this module you will be able to: •

• • •

•

•

•

Define: – A transducer; – Temperature coefficient of resistance; – A Thermistor; and – A strain gauge. Discuss the selection of transducers; Discuss transducer applications; Describe with the aid of sketches and graphic representations where applicable the following mechanical transducers: – Pressure transducer; and – Temperature transducer. Describe with the aid of sketches and graphic representations where applicable the following electrical transducers: – Resistive transducers; – Potentiometer application of the Wheatstone bridge; – Strain gauge application of the Wheatstone bridge; and – Thermistor application of the Wheatstone bridge. Describe with the aid of sketches and graphic representations where applicable the following capacitive transducers: – Pressure measurement; – Flow measurement; – Displacement measurement; and – Level measurement. Describe with the aid of sketches and graphic representations where applicable the following inductive transducers: – Speed measurement; – Pressure measurement; – Displacement measurement; and – The Linear Variable Differential Transformer (LVDT).

329

330

TRANSDUCERS

•

Describe with the aid of sketches and graphic representations where applicable the following photosensitive transducers: – Photovoltaic cells; – Photoconductor (LDR); – Photodiode; – Phototransistor; – Light Emitting Diode (LED); • Describe with the aid of sketches the following: – Opto-isolators – Thermocouples

19.1 Introduction In the world of the modern Electronics we are at times required to observe specific changes that have taken place and in most instances we cannot do this should the changes be very small. Take for example the difference between 19 °C and 20 °C. Without the proper equipment or measuring device this will be an impossible task. Transducers are normally associated with measuring instruments in the instrumentation field in that it will enable us to measure quantities such as: • • • •

Liquid level; Pressure; Temperature; and Light intensity.

Definition 19.1 Transducer

A transducer may be defined as a device that will convert one form of energy into another form of energy. Although there are various types of transducers available and which will be impossible to mention here the following list is not exhaustive. • • • • • •

Mechanical. Electrical. Optical. Pressure. Thermal Chemical.


331

Transducers are mainly divided into mechanical and electrical transducers where the output of a mechanical transducer will be indicated on an analogue calibrated scale. A typical example of such a device is a bathroom scale. An electrical transducer will on the other hand supply us with an electrical signal which then need to be calibrated in terms of the change that had taken place and it is these principles that we will be discussing in this module. Electrical transducers are however divided into two distinct groups. Passive transducers These transducers will produce a variation in electrical parameters such as resistance, inductance, capacitance, etc which will be measured in current and/or voltage. Important to note is that these types of transducers will require an external excitation supply. Active transducers or self-generating transducers These are transducers which will generate a voltage and/or current when subject to some form of quantity and therefore do not require an external excitation supply. Table 18.1 will indicate the type of transducer, the principle of operation of the transducer and the typical application of that transducer. All transducers are classified under the headings of application, nature of the output signal obtained and the method used for the energy conversion. This module will be devoted to discussing the following types of transducers. • • • • •

Mechanical. Resistive. Capacitive. Inductive. Photosensitive.

19.2 Selection of transducers Before we discuss the criteria for the selection of a transducers we need to determine the major elements of a measuring system and this concept is illustrated in the block diagram in figure 19.1.

Input device

Process device

Figure 19.1

Output device

332

TRANSDUCERS

The input device will receive a quantity, normally non-electrical. This quantity under measurement is converted to an electrical quantity and will be passed onto the process device in the form of a proportional electrical signal. The process device will then modify the quantity to an acceptable format for display purposes. Since there are a multitude of transducers available on the market today that may be utelised for virtually any type of measurement that need to be observed it is always required that the following criteria need to be taken into account when selecting a transducer. • • •

Determine the physical quantity that needs to be measured. Which transducer principle is best suited to measure a particular quantity? What is the level of accuracy that will be required for this measurement?

Table 19.1 indicates the classification of a variety of transducers. This classification is according to electrical principles and is divided into two groups.

Type of transducer

Operating principle Passive transducers Resistive types Resistance strain gauge Resistance of wire changes when subjected to elongation or compression due to external stress being applied. Resistance Resistance wire with a high positive thermometer temperature coefficient changes with a variance in temperature. Thermistor Resistance of certain metal oxides with a negative temperature coefficient changes with a variance in temperature. Resistance Resistance of conductive material hygrometer changes with a variance in moisture content. Photoconductive cell Resistance of cell changes with a variance in incident light.

Application

Measurement of force, torque or displacement. Measurement of temperature and radiant heat. Measurement of temperature.

Relative humidity.

Photosensitive relay.


Type of transducer

Variable capacitance pressure gauge Capacitor microphone Dielectric gauge

Type of transducer

Differential transformer

Magnetostriction gauge

Type of transducer

Photoemissive cell

Photomultiplier

Operating principle Passive transducers Capacitive types Distance between two parallel plates is varied by an externally applied force. Sound pressure varies the capacitance between a fixed plate and a moveable diaphragm. Variation in capacitance by variation of the dielectric medium.

Operating principle Passive transducers Inductive types The difference voltage across two series opposed secondary transformer windings is varied by the positioning of an iron core. Magnetic properties are varied.

Operating principle Passive transducers Voltage and Current types Electron emission will occur due to incident light on photoemmissive surface. Secondary electron emission will occur due to incident light on photoemmissive surface.

Application

Displacement or pressure.

Music, noise or speech.

Liquid level and thickness gauging

Application

Force, direction, pressure, position and displacement.

Displacement and rotational speed.

Application

Radiant light, radiation and photosensitive relay. Radiant light, radiation and photosensitive relay.

333

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TRANSDUCERS

Type of transducer Thermocouple

Piezoelectric

Photovoltaic cells

Operating principle Active Transducers An emf will be generated across the junction of two dissimilar metals when the junction temperature is varied. An emf will be generated when an external force is applied across certain crystalline materials. An emf will be generated when incident

Application Temperature, heat flow and radiation. Pressure changes, vibration, acceleration and sound. Solar cell and light meter.

light strikes the surface of a semi-conductor junction. Table 19.1

Exercise 19.1 1. Define a transducer. 2. Differentiate between an active- and passive transducer. 3. Name five types of transducers in use. 4. What criteria need to be taken into account when selecting a transducer?

19.3 Transducer applications 19.3.1

Mechanical transducers

It must be noted that when we discuss mechanical transducers that no electrical variations take place. The input is of a mechanical nature and the output is also of a mechanical nature.

19.3.1.1

Pressure transducer

The transducer illustrated in figure 19.2 is one of many that are used for the measurement of pressure.


Pressure in

335

Calibrated scale

Movement

Bellows capsule

Figure 19.2

When pressure is exerted via the inlet the bellows capsule will be compressed and since the pointer is mechanically linked to the bellows capsule there will be a movement of the pointer on the calibrated scale. Using a suitable conversion ratio the scale will indicate the magnitude of the pressure exerted.

19.3.1.2

Temperature transducer

The transducer illustrated in figure 19.3 is one of many that are used for the measurement of temperature. Calibrated scale

Metal 1

Bi-metal strip

Metal 2

Figure 19.3

Temperature measurement is accomplished by using a bi-metal strip. This bi-metal strip is manufactured from two different materials that have different coefficients of expansion and this means that one material will expand at a quicker tempo than the other one causing the bi-metal strip to bend. Since the pointer is mechanically linked to the bi-metal strip there will be movement on the calibrated scale. Using a suitable conversion ratio the scale will indicate the magnitude of the temperature measured.

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TRANSDUCERS

19.3.2

Electrical transducers

19.3.2.1

Resistive transducers

As a basis for resistive transducers we will commence with an ordinary potentiometer as illustrated in figure 19.4.

V1 VS

RV1 V2

Figure 19.4

Assuming that the potentiometer is a linear device we know that VS = V1 + V2. Should the slider of the potentiometer be exactly in the middle we will find that V1 = V2. Movement of the slider away from the centre position will result in V1 < V2 or V1 > V2 depending on the direction of the movement. It is therefore now possible that the direction as well as magnitude of the change that had occurred be determined by suitable calibration. A second very popular resistive transducer is the strain gauge which can be used where space or inaccessible areas are at the order of the day. These strain gauges are very often used to measure stress on: • • •

Concrete pylons; Aircraft wings and structures; and Bridging structures.

Although these strain gauges are available in various patterns the principle of operation is exactly the same and the basic construction of a strain gauge is illustrated in figure 19.5.


337

Connecting wire manufactured from resistance wire

Paper

Figure 19.5

The principle of operation of a strain gauge is determined by the following mathematical expression. ρ×ℓ where a

R=

R ρ ℓ a

= = = =

resistance of strain gauge in ohms specific resistance of strain gauge in ohms/m length of conductor in m cross-sectional area of conductor in m2

Definition 19.2 Strain Gauge

A strain gauge is a component that will have a change in resistance due to dimensional changes of the conductor from which the strain gauge is manufactured.

When stress is applied to the strain gauge and according to the mathematical expression there are two variables that will undergo a change in that the length (ℓ) will increase and the cross-sectional area (a) will decrease. Since the length, which directly proportional to the resistance, will increase the resistance will also increase but at the same time the cross-sectional area, which is inversely proportional to the resistance, will decrease resulting in the resistance to increase. Important to note is that the specific resistance of the strain gauge will remain the same. This principle of operation of the strain gauge is best explained in the following example.

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TRANSDUCERS

Example 19.1 The following data concerning a strain gauge is given. Condition 1 Condition 2 Before stress application After stress application R = Unknown R = unknown −6 ρ = 0, 0177 × 10−6 ohm/m ρ = 0,0177 × 10 ohm/m −6 2 a = 2 × 10−6 m2 a = 4 × 10 m ℓ = 1,4 m ℓ = 1,5 m Determine in each instance the resistance of the strain gauge.

Solution: Condition 1

Condition 2

R = ρ × ℓ R = ρ × ℓ a a = 0,0177 × 10−6 × 1,4 = 0,0177 × 10−6 × 1,5 2 × 10−6 4 × 10−6 = 6,195 m-ohm = 13,275 m-ohm

It can be seen from the solutions that were obtained for condition 1 and 2 that the resistance increased for condition 2 since: • •

The length increased; The cross-sectional area decreased.

A third very popular resistive transducer is the thermistor. The first two types we discussed both had the ability to measure the magnitude of stress and the thermistor on the other hand has the ability to measure a change in temperature.

Definition 19.3 Thermistor

A thermistor is a component that will have a change in resistance due to temperature changes of the material from which the thermistor is manufactured. The IEC-circuit symbol and characteristic curve for a thermistor is illustrated in figure 19.6 (a) and (b).


-t°

339

R e s I s t a n c e

C

(a)

Temperature (b)

Figure 19.6

Not all materials, however, respond to temperature in the same manner. Some materials have a positive temperature coefficient (PTC) whereas other materials have a negative temperature coefficient (NTC). What does this mean in practice? Refer to figure 19.7. This characteristic illustrates the variation of resistance with temperature for conductors, insulators or non-conductors and semi-conductors. Conductors having a PTC or NTC have that particular characteristic since it is manufactured from a material which possesses that particular characteristic.

R e s I s t a n c e

Insulator

Conductor (NTC) Semi-conductor

Conductor (PTC) Temperature

Figure 19.7

This phenomenon is termed the temperature coefficient of resistance.

Definition 19.4 Temperature Coefficient of Resistance

Temperature coefficient of resistance may be defined as the relationship between the increases in resistance per °C temperature increase to the increase in the actual resistance. The temperature coefficient of resistance is given in ohm/ohm ˚C. Typical values include 0,00428 ohm/ohm ˚C for copper and 0,0072 ohm/ohm ˚C for mercury. For the purpose of our discussion we only need to look at the following two aspects: should the temperature increase then the resistance of the conductor will also increase and should

340

TRANSDUCERS

the temperature decrease the resistance of the conductor will also decrease. Sintered mixtures of metallic oxides that are used in the manufacturing of thermistors include: • • • • •

Manganese; Nickel; Cobalt; Iron; and Uranium.

The resistance of these materials varies between 0,5 ohms to 75 M-ohms. An important point that need to be mentioned is the fact that the abovementioned materials exhibit and excellent characteristic of NTC (negative temperature coefficient) in that the resistance will decrease with an increase in temperature as illustrated in the characteristic curve in figure 1.6 (b). The operation of a thermistor is mathematically given by: Rt = RO(1 + α∆t) where Rt RO α ∆t

= Resistance at the new temperature in ohms = Resistance at initial temperature in ohms = Temperature coefficient of material in ohm/ohm °C = Change in temperature in °C

Example 19.2 Calculate the difference in resistance of a copper conductor should it have a resistance of 28 ohm at 10 ˚C and a temperature coefficient of 0,00428 ohm/ ohm ˚C at a final temperature of: (a) 145˚C; and (b) 263˚C

Solution: (a) Rt1

= RO [1 + αΔt] = 28 [1 + 0,00428 × 35] = 32,19 ohm

(b) Rt2

= = =

RO [1 + αΔt] 28 [1 + 0,00428 × 53] 34,35 ohm

Difference in resistance

= Rt2 - Rt1 = 34,35 - 32,19 = 2,16 ohm


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Having discussed the operation of the different resistive transducers we need to also discuss the use of the Wheatstone bridge concept which is very widely used in transducer principles. This concept is illustrated in figure 19.8. The Wheatstone bridge is made up of two equal resistors R1 and R2 connected between AD and AB respectively. Resistor R3 is a variable resistor and resistor R4 may be a resistor of which the value is unknown connected between DC and BC respectively. The supply VS is connected across AC and a galvanometer is connected across BD. In order to determine the value of the unknown resistor R4 the variable resistor R3 is adjusted until there is no deflection on the galvanometer. Since there is no deflection on the galvanometer it results in the bridge being balanced and you will find that the currents I1 and I2 are equal since the resistance ADC = ABC and the voltage across ABC = ADC. To determine the value of the unknown resistor R4 we need to find a mathematical expression and we will derive such expression in the following manner. A

VS

I1

I2

R1

R2

R3

R4

D

B 0 Galvanometer

C

Figure 19.8

I1 × R1 = I2 × R2 I1 × R3 = I2 × R4

…… (1) …… (2)

Divide equation (1) by equation (2) I1 × R1 I2 × R2 = I1 × R3 I2 × R4 R1 R2 = R3 R4 R2 × R3 R4 = R1 Any deviation from the balanced condition will indicate a change that has occurred. This phenomenon can now be utilised with great ease in resistive transducers.

342

TRANSDUCERS

Example 19.3 The following Wheatstone bridge is constructed. A

VS

R1

R2

R3

R4

D

B

C

The bridge has the following parameters. = 100 volt R1 = 10 k-ohm VS = 10 k-ohm R3 = 10 k-ohm R2 R4 = Unknown The bridge is adjusted by means of R3 until it is balanced again and the value of R3 is then measured at 5 k-ohm. Use the data and determine the value of the unknown resistor R4.

Solution: R × R3 (a) R4 = 2 R1 = 10 × 103 × 5 × 103 10 × 103 = 5 k-ohm

19.3.2.2

Potentiometer application of the Wheatstone bridge

Assume we need to measure the pressure in a pipe by using resistive potentiometers. The layout in figure 19.9 illustrates such an application. Referring back to figure 19.4 you will recall that the voltage drops across the two parts of the potentiometers are equal when they are situated exactly in the centre of the potentiometers. The slider of RV2 is locked into the centre position by a set screw resulting in two equal voltage drops across RV2. The slider of RV1 is mechanically linked to the bellows capsule and also has two equal voltage drops across the potentiometer in the absence of any pressure being applied. When pressure is now applied the bellows capsule will expand thereby moving the slider of RV1 downward and the voltage drop across the two halves will no longer be the same and will now be proportional to the pressure applied and the bridge becomes imbalanced. This imbalance causes an output voltage to be obtained and can now be calibrated in terms of the magnitude of the pressure applied ie 1 V = 5 kPA, 1 V = 10 kPA, etc. Such a potentiometer as illustrated in figure 18.9 has the advantage that it can be utelised on ac- or dc-supplies.


343

Pressure

RV2

RV1 Bellows capsule

VS

VO

Figure 19.9

19.3.2.3

Strain gauge application of the Wheatstone bridge

Assume we need to measure the stress on a concrete pylon by using a strain gauge. The layout in figure 19.10 illustrates such an application. When using the Wheatstone bridge configuration for this application the following points need to be taken into account.

Active gauge

Temperature compensation gauge

VS

R1

R2

A

V

Figure 19.10

• Two strain gauges need to be used, one as the active gauge and the other one for temperature compensation; • The two strain gauges must be mounted at right angles (90°) to one another; and • The Wheatstone bridge must be balanced before stress is applied. The question may be raised on why two strain gauges are used when one would be sufficient to measure the stress applied. One of the main criteria for using the Wheatstone bridge configuration is that it needs to be balanced before measurements

344

TRANSDUCERS

are done and we need to compensate for any temperature changes that may occur since resistance will vary with changes in temperature. Should any stress now be applied to the active gauge it will have a change in resistance which will cause the bridge to become unbalanced and an output voltage will be obtained in proportion to the magnitude of the stress applied. An operational amplifier can now be used in conjunction with the bridge in order to obtain a readable output. The strain gauge has one major disadvantage in that it is not useable for a second time since it has undergone dimensional changes and will have a different resistance in comparison to its design value.

19.3.2.4

Thermistor application of the Wheatstone bridge

Assume we need to measure the temperature in an oven by using a thermistor. A thermistor is also termed a resistance thermometer hence the application of determining temperature changes. The layout in figure 19.11 illustrates such an application and the thermistor is placed in the area where the temperature needs to be measured like the inside of an oven. Before the oven is switched on you must ensure that the bridge is balanced. When the thermistor now heats up caused by an increase in the oven temperature an out of balance reading will be obtained on the measuring device which is caused by the resistance of the thermistor decreasing with the increase in temperature. This imbalance causes an output voltage to be obtained and can now be calibrated in terms of the magnitude of the temperature change ie 1 V = 5 °C, 1 V = 10 °C, etc.

R1

R2

VS R3

-t°

C

A

V

Figure 19.11

A more practical application for a thermistor is that of a control mechanism for a heater element of a geyser and is illustrated in figure 19.12.


N/C Relay -t°

N/O Heater element

C

Relay coil

Rheostat

Figure 19.12

Exercise 19.2 1. Differentiate by definition the difference between a strain gauge and a thermistor. 2. The temperature need to be measured in an oven and you are requested to design a transducer circuit that will be able to do such measurement. 3. Explain the construction of a thermocouple 4. An ordinary potentiometer may be used to illustrate the principle of operation of a transducer. Refer to the sketch given and explain the operation thereof. V1 VS

RV1 V2

5. Refer to the following sketch and describe the operation thereof. Pressure

RV1 Bellows capsule

RV2

VS

VO

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346

TRANSDUCERS

19.4 Capacitive transducers The basic construction of a capacitor is illustrated in figure 19.13. The capacitor is constructed by means of two conducting materials electrically isolated from one another by a dielectric material. This dielectric material is a non-conductive material and the most commonly used include the following: • Air; • Mica; • Paper; • Ceramic; and • Electrolytic and each material will have its own given dielectric constant.

+

S1

S1

+

+

Dielectricum

-

(a)

(b)

Figure 19.13

Before the switch S1 is closed (see figure 19.13 (a)) each plate of the capacitor has a net charge of zero, there is no electric field between the plates resulting in no voltage existing between the plates. The capacitor is therefore uncharged. Should the switch S1 be closed (see figure 19.13 (b)), free electrons will leave the upper plate and move to the positive side of the battery terminal and at the same time free electrons leave the negative battery terminal and will accumulate on the lower plate of the capacitor. Since there was a movement of electrons we now find that the upper plate has no free electrons and will be positively charged whereas the lower plate has an excess of electrons and will be negatively charged and this will produce an electric field which will in turn produce a voltage. This action will continue until such time that the voltage across the two capacitor plates equal the supply voltage. Only now will the capacitor be fully charged. Should the voltage source now be removed the charges will be trapped on the respective plates thus storing the energy accumulated. There are three main factors that influence the capacity of a capacitor. Dielectric constant It was mentioned earlier that the two plates constituting a capacitor is separated by a dielectric material and all such materials will have by virtue of the type of material their own dielectric constant. Important to note is that dielectric used is of a non-conductive nature.


347

Cross-sectional area of plates By virtue of the cross-sectional are being large or small will determine the amount of energy being stored. The smaller the cross-sectional area the less space there will be for storing energy and the smaller the capacitance will be . Should we however have a large cross-sectional area more space will be available for storing energy and the higher the capacitance will be. Distance between the plates The closer the two capacitor plates are to one another the greater its capacitance would be and the further they are apart the smaller the capacitance would be. The capacitance of a capacitor can be expressed by the following mathematical expression. C = 2,24 × 10-13 [k × A (N - 1)] d where

k A N d C

= = = = =

dielectric constant cross-sectional area of one plate in metres2 number of plates distance separation of plates in meters capacitance value in Farad

Example 19.4 The following information concerning a two-plate capacitor is given: Cross-sectional area of one plate = 0,5 m2 Distance apart = 900 mm Dielectric constant = 1,006 Applied voltage = 30 volt Determine from the given information the capacitance of the capacitor.

Solution: C = 2,24 × 10-13 [k × A (N - 1)] d = 2,24 × 10-13 × [1,006 × (0,5)2 × (2 -1)] 900 × 10-6 = 6,26 × 10-11 Farad = 62,6 Pico Farad

348

TRANSDUCERS

It will be found that the only variable that will change in capacitive transducers is the distance between the capacitor plates or the cross-sectional area of the capacitor plates depending on the application. Another important factor that needs to be noted is that capacitive transducer needs to be exited from an alternating current source. Capacitive transducers are suitable to measure the following quantities. • • • •

Pressure. Level. Flow rate. Displacement.

In order to fully grasp the operation of a capacitive transducer we need to look at the principle of frequency. It is quite normal for a capacitive transducer to form part of an oscillator circuit to obtain the required measurement. An oscillator network consists of an inductor and a capacitor normally arranged in parallel which will then determine the frequency at which the network will operate. This frequency of operation is given by the following mathematical expression. f = 1 2 π (L × C)½ where f = π = L = C =

Frequency of oscillator in Hz 3,142 Value of inductance in Henry value of capacitance in Farad

Observing the above expression you must realise that the only constants that will vary is L and C and since we are discussing capacitive transducers it should be obvious that only C will vary and L will stay constant. Therefore, with a variation in C a new frequency will be obtained which will be in relation to the quantity that was applied to the capacitive transducer. This concept will be illustrated by the following activity.

Example 19.5 The following information concerning a capacitive transducer is given. Condition 1 Condition 2 No pressure Pressure applied L = 120 mH L = 120 mH C = 100 pF C = 140 pF Determine in each instance the frequency of operation.


349

Solution: Condition 1 f = 1 2 π (L × C)½ = 1 2 × 3,142 (120 × 13−3 × 100 × 10−12)½ = 695,1 Hz Condition 2 f = 1 2 π (L × C)½ = 1 2 × 3,142 (120 × 13−3 × 140 × 10−12)½ = 2,554 kHz From the above it can be seen that there is a difference in frequency between no pressure applied and pressure applied and can now be calibrated in terms of the quantity being measured irrespective of the type of quantity being measured ie 100 Hz = 1 kPA; 300 Hz = 1 kPA, etc. Our next objective will be to discuss the different types of capacitive transducers available as well as their principle of operation.

19.4.1

Pressure measurement

The capacitive transducer illustrated in figure 19.14 is that for measuring pressure and is the basis on which a capacitor microphone also operates. Flexible metal foil diaphram

Fixed plate

Pressure in

d

Output

Dielectric

Figure 19.14

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TRANSDUCERS

When pressure is applied the flexible metal foil diaphragm will vibrate at the tempo at which the pressure is applied thereby decreasing the distance between the moveable plate and the fixed plate. This new value of capacitance can now be calibrated in terms of the pressure applied.

19.4.2

Flow measurement

The capacitive transducer illustrated in figure 19.15 is that for measuring flow rate or pressure. Gas flow Movement

d

To oscillator

Bellows capsule

Figure 19.15

This transducer consists of a bellows capsule attached to the pipe in which the flow rate or pressure needs to be measured. The capacitor plate attached to the bellows capsule is the moveable plate whereas the plate on the right-hand side is the fixed plate and is coupled to an oscillator network. As soon as a quantity is moved through the pipe the bellows capsule will decrease and the distance between the two plates will be reduced thereby causing a new value for the capacitor and hence a new frequency as was indicated in activity 19.5. This type of capacitive transducer is mainly used to measure magnitude of a quantity but not the direction of the quantity.

19.4.3

Displacement measurement

In order to measure direction and magnitude we make use of a capacitor transducer as illustrated in figure 19.16 (a) and (b).


Fixed plate

R1 Movement

Moveable plate

To oscillator

351

Supply

R2

Fixed plate

Rotation

(a)

Fixed plate R1 Movement Rotation

To oscillator

Supply

R2 Fixed plate

(b)

Figure 19.16

You will notice that the capacitive transducers illustrated in figure 19.16 (a) and (b) once again make use of the Wheatstone bridge configuration. Note that when the moveable plate is exactly in the middle then the bridge will be balanced and no output is obtained. Moving away from the centre will result in an out of balance output which could be positive or negative indicating the direction of displacement and the magnitude thereof will indicate the change that is taking place. It should also be noted that the resistive arm of the Wheatstone bridge can be replaced with inductors and the same effect will be obtained.

19.4.4

Level measurement

In order to measure level we make use of a capacitor transducer as illustrated in figure 19.17. It was noted that there are three effects that will determine the capacitance of a capacitor and in the previous discussions we had a change of capacitance caused by a change in the distance between the capacitor plates. In order to measure level we make use of the change in the capacitor plate area since the liquid level will determine the area of the plates that will be submersed in the liquid.

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TRANSDUCERS

Electronic circuitry and read-out devices

Container Liquid feeder pipe

Liquid level Liquid serving as dielectric medium

Capacitor plates

Figure 19.17

The graphic representation in figure 19.18 will clearly illustrate this concept. Capacitance

Capacitance of air as dielectric

Fluid level

Figure 19.18

Exercise 19.3 1.

Draw neat labelled diagram of a capacitive transducer that will be able to measure the following quantities. 1.1 Flow; and 1.2 Level. You are further required to discuss the operation of each one with the aid of a mathematical expression and example.


353

19.5 Inductive transducers The operation of inductive transducers is somewhat different from resistive- and capacitive transducers in that with the aforementioned transducers the resistance or capacitance changed with the application of a varying quantity. Inductive transducers make use of a change in magnetic field characteristics since it is impossible to alter the inductance value of the inductor. In order to more clearly understand this concept you are referred to the illustration in figure 19.19 (a) and (b).

(a)

(b)

Figure 19.19

Observing figure 19.19 (a) the soft iron core is outside the inductor and will have no effect on the magnetic field which exists around the inductor. When the soft iron core is now inserted into the inductor without coming into contact with the inductor the existing magnetic lines become more concentrated and it this effect which is utelised in inductive transducers as illustrated in figure 19.19 (b).

19.5.1

Speed measurement

The illustration in figure 19.20 shows a method used to determine the rotational speed of a gear, wheel or axle.

Secondary windings

&

A

7

7

7

7

Digital readout Toothed wheel

Direction of rotation

Primary windings

Figure 19.20

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TRANSDUCERS

The primary windings are exited from a dc-source thereby creating a steady magnetic field around the primary winding. As soon as the toothed wheel is rotated the tooth will cause an increase in the magnetic field strength across the secondary windings thereby creating a pulse every time a tooth penetrates deeper into the primary magnetic field. By suitable methods of calibration the number of pulses can now be counted for a given time period and the speed can be calculated. A second concept makes use of a tacho-generator and is illustrated in figure 19.21. Read-out unit

Direction of rotation Armature

Figure 19.21

A tacho-generator is basically a two-phase induction motor and has two coils mounted at right-angles to one another. An alternating current is supplied to the one winding and a read-out unit normally a calibrated voltmeter is connected to the other winding. When the armature is now rotated it will induce a voltage into the coil to which the calibrated voltmeter is connected and the reading obtained will be in proportion and relation to the speed at which the armature is rotated.

19.5.2

Pressure measurement

The inductive transducer illustrated in figure 19.22 is that for measuring flow rate or pressure. Pressure Movement

Soft iron core Inductor wound around a cardboard holder

Bellows capsule

Figure 19.22

Supply and measuring device


355

This transducer consists of a bellows capsule attached to the pipe in which the flow rate or pressure needs to be measured. Attached to the bellows capsule is the moveable piston that will move in and out of the inductor coil thereby creating a varying magnetic field. This type of inductive transducer is mainly used to measure magnitude of a quantity but not the direction of the quantity.

19.5.3

Displacement measurement

The inductive transducer illustrated in figure 19.23 will measure direction as well as magnitude of any variance applied. Iron core

L1

Displacement

R1

L2

R2 VS

Figure 19.23

You will once again notice that use is made of a Wheatstone bridge and this Wheatstone bridge consists of inductors L1 and L2 with resistors R1 and R2. When the iron core is situated in the middle of the two inductors L1 and L2 then the voltage drops across resistors R1 and R2 are equal and no output is obtained. Moving the core in either direction will cause an imbalance in the Wheatstone bridge and an output will be obtained directly proportional to the magnitude of the change.

19.5.4

The Linear Variable Differential Transformer (LVDT)

The operation and graphic representation of an LVDT is illustrated in figure 19.24 (a) and (b).

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TRANSDUCERS

The LVDT consists of a primary winding L1 and two secondary windings L2 and L3. The secondary windings are connected in series opposition and the output will be zero when the core is exactly in the centre position since the magnetic flux is distributed evenly between L2 and L3. Moving away from the centre position will result in an increase in magnetic coupling depending on the direction of displacement and either a positive voltage or negative voltage output will be obtained which will indicate direction of the quantity as illustrated in figure 19.24 (b). The voltage magnitude of the quantity applied will indicate the magnitude of the quantity.

+V L1

B

Displacement Iron core

A

Output across L3

B Output across L2

L2

A

L3 -V

Output (a)

(b)

Figure 19.24

Exercise 19.4 1.

The rotational speed of a wheel needs to be determined by means of an inductive transducer. Show by means of a neat labelled sketch how you would be able to perform this requirement and then discuss the operation of the circuit you had provided.

2. Discuss with the aid of a suitable sketch the operation of a tacho- generator.

19.6 Photo-sensitive transducers Semi-conductors exhibits the phenomena to change its characteristics when exposed to light whether it be natural light or artificial light hence the term ‘photosensitive’ which literally translates to being sensitive to light. These components may operate when they are subjected to a source of light and are then termed ‘photoconductive or


357

photosensitive’ components but in this instance they need to be connected in a network or circuit that has a supply voltage. On the other hand they may radiate light and is then termed ‘photo-emmissive’ devices and will radiate light when subjected to a supply voltage.

19.6.1

Photo-voltaic cells

Photovoltaic cells are also termed solar cells and having already discussed the meaning of the term ‘photo’ which means light we also need to discuss the term ‘voltaic’. Voltaic means voltage and we can therefore proclaim that ‘photovoltaic’ means the generation of a voltage by means of light which still conforms to our definition of a transducer given earlier. It was mentioned earlier that semi-conductors changes its characteristics when exposed to a source of light and this change is the combination of electron-hole pairs in the semi-conductor material by thermal agitation which constitutes a current flow in the external circuit coupled to the device. The voltage developed by these cells will be dependant upon the strength and intensity of the light falling onto the cell. The output obtained from these cells is in the region of ± 0,5 volt per cell and in order to get a reasonable output depending on the application, these cells are connected in seriesparallel. These cells are used in what is termed a ‘solar panel’ to supply electricity to devices where the ESKOM network cannot reach. The construction of a selenium cell and a silicon cell is illustrated in figure 19.25 (a) and (b). Light

Light

Ring-shaped electrodes Glass cover

Glass cover

+

Gold Selenium Metal base

+

V

P-type layer

V -

-

(a)

N-type layer Metal base

(b)

Figure 19.25

19.6.2

Photo-conductor (LDR)

A photoconductor is a component that will have a change in resistance when subjected to a source of light. Important to note is that an LDR will not generate a voltage as was the case with the solar cells and will need a voltage source in the network where it is installed. The following materials used in the manufacture of LDR’s.

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• • • •

Selenium. Lead Sulphide. Cadmium Sulphide. Cadmium Selenite

The construction of an LDR is illustrated in figure 19.26.

Conductor

Figure 19.26

The circuit symbol and characteristic curve for an LDR is illustrated in figure 19.27 (a) and (b). R

Incident light

(a)

Light intensity in lumen (b)

Figure 19.27

In the absence of light onto the LDR it has a very high resistance termed the ‘dark resistance’ and the resistance will decrease when the LDR is exposed to a source of light. The amount by which the resistance will decrease is dependant upon the intensity of the light source as illustrated in figure 18.27 (b). The following applications of an LDR are illustrated in figure 19.28 (a) and (b). The circuit in figure 19.28 (a) is off in the absence of light and as soon as the LDR is illuminated then the circuit will switch on. The circuit in figure 19.28 (b) is on in the absence of light and as soon as the LDR is illuminated then the circuit will switch off.


359

+ VCC

+ VCC

LDR

RB Relay

Relay

Q

Q

RB

LDR

- VCC

- VCC

(a)

(b)

Figure 19.28

19.6.3 Photo-diode A photodiode operates in more or less the same principle as an LDR but the difference is that use is made of a PN-junction. The photodiode is connected in the reverse bias mode in the circuit and the junction is exposed to light through a window in the casing of the photodiode. The circuit symbol and characteristic curve for a photodiode is illustrated in figure 19.29 (a) and (b).

I

600 lumen 500 lumen 400 lumen

Anode

300 lumen

Cathode


Incident light

-V

(a)

(b)

Figure 19.29

In the absence of incident light a small current, termed dark current, will flow but is too small to make any significant change in the characteristics of the photodiode. This dark current is typically in the region of 10 μA for germanium diodes and 1 μA for silicon diodes.

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The following applications of a Photodiode are illustrated in figure 19.30 (a) and (b). The circuit in figure 19.30 (a) is off in the absence of light and as soon as the diode D is illuminated then the circuit will switch on. The circuit in figure 19.30 (b) is on in the absence of light and as soon as the diode D is illuminated then the circuit will switch off. + VCC

+ VCC

D

RB Relay

Relay

Q

Q

RB

D

- VCC

- VCC

(a)

(b)

Figure 19.30

19.6.4 Photo-transistor A photo-transistor has similar characteristics to a conventional bi-polar transistor. Should the emitter-base junction be exposed to a source of light a current in the base will be produced which in turn causes a collector current to flow. The magnitude of the collector current is therefore dependant upon the magnitude of the base current and is determined by the intensity of the light source. The circuit symbol and characteristic

IC


C

300 lumen

B

Incident light

200 lumen 100 lumen

E

0 lumen

(a)

V (b)

Figure 19.31


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A typical application of a phototransistor is illustrated in figure 19.32. In the absence of incident light the base current is too small to cause any significance change in collector current and the relay can not be energised since the collector current is below the holding current value of the relay. Any incident light on the base-emitter junction of the phototransistor causes an increase in base current which has a corresponding change in collector current and since the collector current now has a value above the holding current value of the relay, the relay will be energised. Figure 19.32 illustrates the application of a phototransistor. + VCC

RB1 Relay Q

RB2

Figure 19.32

19.6.5

- VCC

Light Emitting Diode (LED)

An LED falls into the category of photo-emmissive transducers since it will give off light when supplied with a suitable voltage source. Therefore it still conforms to the definition of a transducer. In any PN-junction there is a continuous combination of holes and electrons when a current is passed through the junction. The electrons of the junction are in the conduction band whilst the holes are in the valence band. In order to drop from the conduction band into the valence band the electrons have to give off energy for this transition, and are given off in the form of electromagnetic radiation of different colours depending on the level of doping of the junction as well as the material used for the junction. By placing this PN-junction in a translucent encapsulation the electromagnetic radiation energy is given off as light and the intensity thereof is determined by the magnitude of the forward current passed through the junction. These LED’s are then available in red, green, amber, blue and yellow. It is possible to obtain an LED in

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dual colours in that forward current will produce one colour and reverse current will produce a different colour. The construction and circuit symbol for an LED is illustrated in figure 19.33 (a) and (b).

Anode

Cathode

Cathode

Anode

Radiated light

(a)

(b)

Figure 19.33

Although there are various applications for an LED the most popular ones are those for digital read-out devices like you would find on the calculator you use. These LED’s are arranged either as a seven-segment display or as a 7 × 5 matrix arrangement. These two concepts are illustrated in figure 19.34 (a) and (b).

a f

b g

e

c d

(b)

(a)

Figure 19.34

19.6.6 Opto-isolators An opto-isolator or opto-coupler utilises an LED and phototransistor in one package as a six pin dual-in-line and is illustrated in figure 19.35. 1

6 5

2 3

NC

4

Figure 19.35


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A current is allowed to flow through the LED and light is emitted which is directed at the base-emitter junction of the phototransistor which is then switched on and the device that is to be controlled by the photo-transistor becomes operational. The opto-isolator is therefore used as a switch with both the LED and phototransistor initially off. The coupling is therefore optical which allows for a high degree of electrical isolation between the input and the output of the opto-isolator and this feature enables us to use the opto-isolator as an interface between a low voltage- and a high-voltage system.

19.7 Thermo-couples It is possible to generate a voltage by means of a thermo-couple. In analysing the word thermo-couple it will be noticed that ‘thermo’ has reference to temperature (heat) and it is on this principle that a thermo-couple operates. A thermo-couple consists of two dissimilar metal wires joined at one end termed the sensing or hot junction and terminated at the other end termed the reference or cold junction. This concept is illustrated in figure 19.36.

+

G

Hot junction

Cold junction

Figure 19.36

Should a temperature difference exist between the hot and the cold junction a voltage will be generated equal to the magnitude of the temperature difference. This magnitude of the voltage developed will also be depending on the type of material used in the manufacture of the thermocouple. Typical materials used in the manufacture of thermocouples include: • • • •

Chromel-Constantan; Iron-Constantan; Copper-Constantan; and Platinum-Platinum/Rhodium.

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Exercise 19.5 1.

The following components can be used to switch a device on or off. 1.1 Photo-diodes; and 1.2 Light Dependant Resistors. Draw two neat labelled circuit diagrams that will illustrate this concept.

2. Use suitable sketches that will illustrate the construction of: 2.1 A selenium cell; and 2.2 A silicon cell. 3. Illustrate using suitable sketches two applications of a light emitting diode. 4. Discuss the operation of a thermo-couple.

N4 Industrial Electronics

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