Preface This booklet on operational amplifiers has been compiled to fulfill the requirements of the Caribbean Advanced Proficiency Examinations (CAPE). It addresses a specific need in some of our high schools and is intended to augment the teachers’ material on the subject matter. Booklets are provided to students during the CAPE Physics Workshop offered annually by the Department of Physics at the University of the West Indies, Mona Campus. These workshops target problematic topics in the Physics Syllabus and use lectures and laboratory sessions to teach the material in a manner that enhances the students’ understanding.
The booklet starts with an overview of the CAPE Physics Electronics requirements then moves into the relevant subject matter. The author presents the content of each section in a way that relates to the targeted age group and facilitates quicker understanding. Worked examples are included throughout the booklet and additional CAPE type questions are added at the end.
The procedures for the practical sessions are added at the end of the booklet. In Lab 1, a chosen operational amplifier circuit is constructed and tested on the lab bench. The results of these measurements are plotted on a graph. In lab 2, the same circuit is simulated on a computer and its results are compared to that of Lab 1. This technique is intended to provide the students with a comparative feel for the hands-on experiments versus the simulated ones.
The Department of Physics hopes that the students find this booklet to be a valuable aid in their learning process and wishes each and every student success in their upcoming examinations.
Paul R. Aiken (PhD) Department of Physics UWI, Mona © 2014 1
Table of Contents
CAPE Requirements ……………………………… ……………………………………………………. …………………….
2
What are Operational Amplifiers? .. …………………………………….
3
Circuit Schematic Representation Representatio n …………………………………… ………………………………………. ….
4
Power Supply Requirements………………………………… Requirements …………………………………………….. …………..
4
The Ideal Operational Amplifier
5
……………………………………… …………………………………… …
The Typical (Real) Operational Amplifier
……………………………
9
The Op Amp as a Comparator
…………………………………… ……………………………… ……
10
Application of the Comparator
……………………………………
13
Potential Dividers and Variable Resistors ……………………….
13
Light Dependent Resistors (LDR) ……………………………… ………………………………..
14
Thermistors …………………………………… ……………………………………………………… …………………
14
Strain gauge ………………………… …………………………………………………….. …………………………....
14
Feedback in Op Amp Circuits
………………………………………… ……………………………… …………
17
…………………………………………… ……………………………… ……………
18
The Non-inverting Non-invertin g Amplifier …………………………………… ………………………………………….. ……..
21
Effect of Negative Feedback on Gain and Bandwidth …………………
22
Voltage Follower
…………………………………………………… ………………………………… …………………
26
Summing Amplifier ………………………… ………………………………………………….... ………………………....
27
Difference Amplifier
………………………………………………….. ……………………………… …………………..
29
An Operational Amplifier Circuit Example ……………………………. ……………………………...
29
Op Amp Problems ………………………………… ………………………………………………………. …………………….
32
Experiment 1
……..…………………………………………………. ……..………………… ………………………………... ..
34
Experiment 2
…………………………………………………………. …………………………………… …………………….
36
Positive Feedback Negative Feedback The Inverting Amplifier
2
CAPE Requirements
4. Operational Amplifiers Students should be able to: 4.1 describe the th e properties properties of the ideal operational amplifier; 4.2 compare the properties of the typical and the ideal operational amplifier; 4.3 use th e operational operati onal amplifi amplifier er as a comparator; 4.4 use the fact that magnitud e of the output voltage cannot exceed that of the power supply; 4.5 discuss the effect of positive and negative feedback in an amplifier; 4.6 explain the meaning of gain and bandwidth of an amplifier; 4.7 explain the gain-frequency gain-frequency curve for a typical operational amplifier;
Consider these effects in terms of whether they are advantages or disadvantages. Typical as well as ideal values for these quantities should be discussed. Include the fact that frequency is usually plotted on a logarithmic axis and explain the reason for this.
4. 8
determine bandwidth bandwidth from from a gain frequency curve;
Precise numerical value related to the response of the ear is not required.
4.9
draw the circuit diagram diagram for both the inverting and non-inverting amplifier with a single input;
Students should be familiar with several representations of the same circuit.
4.10 use the concept of virtual earth in the inverting amplifier;
Explain why the virtual earth cannot be connected directly to earth although it is `virtually" at earth potential.
4.11 4.1 1 derive and and use expressions for the the gain of both the inverting amplifier; 4.12 4.1 2 discuss the effect effect of negative feedback feedback on the gain and bandwidth of an operational amplifier; 4.13 perform calculations related to single-input inverting amplifier circuits; 4.1 4
State the two "Golden Rules' of Operational Amplifier circuit analysis and show how they lead to the results required here. Mention the effect of negative feedback on other op-amp characteristics.
perform calculations related to single-input, single-input, noninverting amplifier circuits;
4.15 4.1 5 describe the use of the inverting amplifier as a summing amplifier;
4.16
solve problems problems related to summing summing amplifier circuits;
4.17
describe the use of the operational amplifier as a voltage follower;
4.18
analyse simple simple operational operatio nal amplifi amplifier er circuits;
4.19
analyse the response of amplifier circuits to input signals, using timing diagrams.
Include the fact that it is also possible to configure the op-amp as a difference amplifier.
Mention the important practical use of the voltage follower as a buffer or matching amplifier. Refer to note 4.11
3
What are Operational Amplifiers? The term Operational Amplifier was originally used to describe an amplifier circuit which performed various mathematical operations such as differentiation, integration, summation and subtraction. Operational Amplifier, or Op Amp, is now more loosely applied to any high gain alternating current (ac) and direct current (dc) amplifier capable of operating in various configurations. Op Amps have extremely wide applications and may be found in all types of circuit and system designs.
Op amps are a member of the family of linear integrated circuits. Integrated circuits (ICs) consist of many transistors and few resistors and capacitors. Transistors are a special form of semiconductors with properties similar to those of junction diodes. ICs are usually fabricated on a specially prepared material using extremely precise process control. The final product generally occupies areas less that one square centimeter, even for the most complex ICs.
(a) LM741 Single OP amp
(c) Pin labels for LM741
(b) Quad LM741 op amp (LM324)
(d) Pin labels for single packaged quad op amp
Fig. 1 The 1 The 741 Op amp single (8-pins) and quad packages (14-pins)
4
There are many different types of op amps designed for varying applications. The most popular of these is the LM741 op amp developed by National Semiconductor Company. Fig. 1 shows pictures of two package types of the LM741 Op amp and their corresponding internal schematic.
Circuit Schematic Representation An op amp is represented by the symbol shown in Fig 2. For the LM741 op amp, pins 2 and 3 are what we called the inverting and non-inverting inputs, respectively, because of the way the input signals are acted upon. Pin 6 is the output and pins 7 and 4 are the pins to which the positive and negative power supply inputs are connected. In some circuits, pin 4 may be connected to the circuit ground (or common) instead of a negative voltage supply. positive power supply voltage
2
Inverting input
7 6
3
Non-inverting input
Output
+ 4
Negative power supply voltage
Fig. 2 The Op amp schematic symbol
Power Supply Requirements All ICs have a minimum and maximum power supply voltage rating. The LM741 op amp may be operated from a ‘dual-rail supply voltage’ of ±5 V dc to ±18 V dc, or a ‘singlerail supply’ of 10Vdc to 36 V dc with respect to ground. The op amp will not work properly if smaller voltages (than the minimum in this range) are applied and will be damaged if greater voltages (than the maximum in this range) are applied. Before using
5
any IC in a circuit always check its manufacturer data sheets for its maximum ratings. This may be easily found by doing a Google search of the part number.
Fig. 3 show examples of single and dual rail power supplies. At this time, the most common type available to high school students are those made from connecting batteries in series aiding arrangements (Fig.3(c)). Other types of DC power supplies are available.
+ 9 Vdc +
+ 9 Vdc
9V
+
Ground
9V
+
-
Ground
9V (b) A 9-Volt battery
(a) Single rail 9 V supply
- 9 Vdc
(c) A dual rail ±9 V supply
Fig. 3 Power supply configuration
The Ideal Operational Amplifier The output voltage of an op amp is proportional to the difference of the voltages at its -
+
inverting (V ) and non-inverting (V ) inputs. This is represented by Equation 1
V out
+
−
= A0 (V − V
)
(1)
where A0 is called the Open-Loop gain of the op amp.
Ok…let’s pause here for a minute. What are these ‘open-loop’ and ‘gain’ things we have been talking about???. .. Well, the typical connection for an op amp in a circuit is one where some other component (usually a resistor) is connected between the output pin and the inverting input pin. This causes the signal that goes into the op amp to get loop back from its output to its input. This setup is called a ‘closed-loop’ configuration. Therefore, if there are no components connecting the output back to the input, then we can say that
6
we have an ‘open-loop’ configuration. So Eq.1 describes the gain of the op amp when the loop is open…. OK …so here we go with that ‘gain’ word again.
The gain of an amplifier is the amount of amplifications that is given to the input signal to get the output signal. In other words, it is how many times the input signal gets ‘multiplied’ to equal the output signal. The gain is always found by dividing the voltage value of the output signal by that of the input signal. The gain is always ‘just a number’ and has no units. So ‘open-loop gain’ means how many times the input signal gets amplified when there is nothing connected between the output and input pins. By the way, this connection we are talking about, the one between the output and the input, it is called ‘ feedback ’. That is, a portion of the output signal gets fed back to the input whenever we have the connection in place. We will talk some more on this later on.
We will now attempt to describe some of the main properties of an ideal op amp. Don’t be frightened by all the new terminologies (weird words). We will list the properties first then go through line-by-line and try to provide additional explanations. So here goes … the ideal operational amplifier may be assumed to have the following properties:
+
-
(a) An infinite open-loop gain. The slightest difference in V and V will caused the output to go to ‘saturation’. Saturation voltage cannot exceed the power supply voltage. (b) An infinite input impedance (resistance). This ensures that no current flows into +
-
the input terminals (V and V ). However, voltages may be present. (c) An infinite bandwidth. This assumes that it amplifies any input range of frequencies. (d) Zero output impedance (resistance). This ensures that the amplifier is unaffected whatever output circuit it is connected to. (e) An infinite slew rate. The means that the input and output frequency changes are always exactly in synch. (f) Zero voltage and current offsets. This ensures that when the input signal voltages are zero the output will also be zero regardless of the input source resistance.
7
If your head is spinning at this point, just stop, take a deep breath, get some water or something…. Now, read these explanations below, then go back and re-read the properties. It is very important that you understand these concept and terminologies.
Let’s start with the first one, +
-
(a) An infinite open-loop gain. The slightest difference in V and V will caused the output to go to ‘saturation’. Saturation voltage cannot exceed the power supply voltage.
By now, we all understand this open-loop gain thing. If not, re-read the top of this section. The difference now is that we put the word ‘infinite’ in front of it! Infinite just means very, very large, … countless … So we are just saying that when the op amp is configured without any ‘feedback’ it has a very, very, very large gain.
The next word is saturation, what does it mean for the op amp to be saturated? What is it saturated with? Suppose we were to plot the gain of the op amp from its definition, i.e. gain =
V out V in
, where Vout and Vin are the output and input signal voltages, respectively.
We will get the plot shown in Fig. 4(a). Just a nice straight line with a constant gradient! Agreed!!! Ok.. let’s move on. Vout (Volts) Vsupply
Vout (Volts)
Region of Fixed output
Region of reducing gain
Constant Gain
Region of Constant gain
0
Vin (mV)
(a) Constant gain
0
V in (mV)
(b) saturation limit
Fig. 4 Op amp gain and voltage saturation
8
What is the maximum possible output voltage of the op amp?? No idea!!! Think about it for a while … let’s ask another question then. Where does the op amp get its voltage from? Yes, this is easier… The op amp gets its voltage from the voltage of the power supply that is connected to it. Therefore, it stands to reason that the maximum output
voltage of the op amp cannot exceed its power supply voltage. In other words, the power supply voltage sets the voltage output limit of the amplifier. Now, let’s get back to Fig. 4. Look especially at Fig.4(b)… Whenever the output voltage starts getting close to the value of the power supply voltage, Vsupply, something strange starts happening. The gain starts decreasing, i.e. Vout/Vin is getting smaller. As a result, the gradient of the slope starts decreasing and is getting flatter and flatter. By the time it reaches the power supply voltage value, it becomes a flat straight line which will never exceed the power supply voltage value. Just as it reaches this point, the op amp is said to be saturated . That is, the output signal remains at a constant voltage irrespective of any increases in the input signal voltage.
Moving on to the second property: (b) An infinite input impedance (resistance). This ensures that no current +
-
flows into the input terminals (V and V ). However, voltages may be present.
What is impedance anyway? It is the term used to describe the combined resistances of all the circuit elements, including elements that you will not study at the CAPE level. Impedance values depend on the frequency of the signal. Its unit is the Ohm, same as that for resistances. So, for the sake of simplification, let’s think of impedance as resistance.
Why is it that no current flows into an infinitely high resistance? It goes back to Ohm’s law, which state that V = IR. If V is 10 volts and R is 10 M Ω, then the current I = 10V ÷ -6
10,000,000Ω. Therefore, I = 1 µA or 10 A. In these kinds of circuit, this current is considered negligible. This is the same principle behind the operation of an ideal voltmeter. It measures the voltage while having negligible current flow through it.
Moving on …..
9
(c) An infinite bandwidth. This assumes that it amplifies any input range of frequencies.
Bandwidth is the range of frequencies over which the op amp operates with a constant gain. This will be explained in more detail later on.
(d) Zero output impedance (resistance). This ensures that the amplifier is unaffected whatever output circuit it is connected to.
This is self explanatory by now. It just means there is no output resistance when the op amp is connected to a load. In this state it is capable of ‘driving’ any load.
Points (e) and (f) are self explanatory. Also, understanding these features is not a CAPE requirement at this time.
So let’s now restate the assumptions of an ideal Op amp: (a) An infinite open-loop gain. (b) An infinite input impedance (resistance). (c) An infinite bandwidth. (d) Zero output impedance (resistance). (e) An infinite slew rate. (f) Zero voltage and current offsets.
The Typical (Real) Operational Amplifier Real op amps have characteristics that approach those of ideal op amp, but never quite attained them. They deviate from the ideal op amp in the following ways: 5
6
(a) The open loop gain is usually in the range of 10 – 10 . Although this is high, it is not infinite. 6
12
(b) They have large but finite input impedances usually in the range of 10 Ω – 10 Ω. Thus, drawing very small, but measurable currents at their input terminals.
10
(c) They have a finite bandwidth which is dependent on the gain. The higher the gain the smaller the bandwidth. This is usually described in its frequency response characteristics or the Gain-Bandwidth product. (d) The output impedance is usually about 100 Ω. (e) They have finite slew rate and voltage and current offsets.
Note that while the ideal op amp does not exist, its properties serve as a valuable starting point for preliminary circuit analysis.
The Op Amp as a Comparator As discussed earlier, an op amp has an inverting (V-) and a non-inverting (V+) input. These inputs may be connected as single-ended inputs or as a differential input. In singleended input mode, only one of the inputs has a voltage signal while the other is grounded. In differential mode, both inputs have voltages with respect to ground. Equation 1 may be applied to both of these cases to create these three open-loop gain scenarios:
+
+
1. V out
= A0V
2. V out
= A0 ( −V
3. V out
= A0 (V − V
-
Input signal (Vin) is on V input and V terminal is grounded −
-
)
+
+
Input signal (Vin) is on V input and V terminal is grounded −
) for cases where a differential input signal is applied
This may be represented schematically as shown in Fig. 5. + Vsupply
+ Vsupply
-
-
+ Vin
- Vsupply ground
(a) Case 1
V out
= A0V
+
Vin
+ - Vsupply
V out
−
= A0 (−V
ground
(b) Case 2
11
)
+ Vsupply
+
Vin
Vin
- Vsupply
V out
+
−
= A0 (V − V
)
ground
(c) Case 3 Fig. 5 Open-loop gain scenarios
Some important observations: 1. In case 1, the output signal will always be of the same polarity (or phase) of its input signal. 2. In case 2, the output signal will always be of the opposite polarity (or phase) of its input signal. 3. In case 3, the polarity of the output will depend on which of the inputs has the larger voltage: a. If V +
< V
b. If V +
> V
-
−
then polarity will be opposite to that of V
−
then polarity will be same as that of V
+
4. A negative output voltage can only be obtained if the op amp is connected to a dual rail supply (i.e ±Vsupply).
A Comparator may be made up of any of these three configurations. The signals at the input terminals are ‘compared’ and their difference is multiplied by the open-loop gain of the op amp to produce the output voltage. Let’s look at these three examples:
12
Example 1: What is the output Voltage for an op amp circuit with the following characteristics? +
-
V = 1V
V = 0 Volt (or grounded)
+Vsupply = +12Volts
-Vsupply = - 12 Volts
A0 = 10
5
Solution:
(Use circuit in case 1 of Fig. 4) +
V out = A0V
5
= 10 × 1 = +100,000 Volts
But stop right here…Remember, you cannot get an output voltage that is greater that your power supply voltages. Therefore, Vout cannot exceed +Vsupply, i.e. V out = + 12 Volts
Whenever this happens, the op amp is said to be SATURATED. Saturation voltages cannot exceed the power supply voltages.
+
-
Example 2: Same as Example 1 except V = 0, and V = 1 Volt.
Solution:
V out = -100,000 Volts
But by now we know this cannot exceed the -12 Vsupply. Therefore the real answer is:
V out = -12 Volts.
+
-
Example 3: Same as Example 1 except V = 1.3 V, and V = 1.1 V
Solution:
Applying Case 3 equation, +
V out = A0 (V
−
− V
= 0.2 x 105
) = 10 5 (1.3 − 1.1) = 20,000 Volts
Again the actual value is: V out = +12 Volts
13
Application of the Comparator The comparator is used in many circuit applications where two states need to be compared to produce a desired output signal, which is usually used to control some other circuit or to switch some state. A not so obvious use is that of a voltage-level-shifter . That is, use can be made of the fact that an output voltage that is equal to the power supply voltage can always be obtained, irrespective of how small the input voltage difference is.
Usually, when the op amp is used as a comparator, its input terminals are connected to some kind of potential divider circuit. One of these potential dividers creates a fixed voltage that is used as a reference, while the other varies in accordance with some other physical or electrical property that may either be internal or external to the circuit. It may be important at this point to say a little about potential dividers and variable resistors.
Potential Dividers and Variable Resistors A Potential Divider or voltage divider is a simple arrangement of two resistances across one voltage source. From Ohm’s and Kirchhoff’s laws, we know that the sum of the voltage drops across each resistance will be equal to the supply voltage. In other words, the voltage drop across each of the resistances is always less than that of the voltage source. By varying the resistances, we can vary the amount of voltage across them. We can collect an output voltage at the point just between the two resistances. In most cases this output is taken with respect to the ground. Note that we are referring to two resistances and not two resistors. There is a reason for this! While two resistors will work perfectly well as voltage dividers, there are a set of resistors that has variable resistances. They have three terminals, one of which is a slider that is used to vary the resistance that is seen at the output. Terminals 1 and 3 (see Fig. 6) may be connected across a voltage source and the output taken from terminal 2. An illustration of these resistors is shown in Fig. 6.
14
1
Variable resistor
Variable resistor
Fixed resistor
3
1
2
3
1 2
2
Fig. 6 Fixed and variable resistors
Fig. 7 shows actual voltage divider examples using these two types of resistors. The value of the output voltage (in Fig. 7) can always be determined from Equation 2.
V out =
R2V sup ply R1 + R2
(2)
Note that in Fig.7 (b) the variable resistor is divided into two resistances, the one above the slider, R1, and the other below, R2. Eq. (2) applies in all cases where the output is connected to a very high resistance (or impedance), for example, the input terminal of an op amp.
R1
+
+
R1
Vsupply
Vsupply R2
Vout
(a) Two fixed resistor divider
R2
Vout
(b) Variable resistor divider
Fig. 7 Voltage Dividers
15
Example 1: If Vsupply =12 V, R1 = R2 = 10 k Ω, what is the value of Vout?
Using Eq.2;
Solution:
V out =
10,000×12 10,000 + 10,000
=
12 2
=6
= 6 Volts
Example 2: If Vsupply =12 V, R1 = 10 k Ω, what value of R2 is required to let Vout = 4V? Modifying Eq.2 so that R2 becomes the subject:
Solution:
R2
=
R1V out V sup ply
− V out
=
10,000 × 4 12 − 4
=
5000Ω = 5k Ω
Other forms of variable resistors exist that have only two terminals, instead of three. Their resistances usually depend on some external physical condition, like temperature, light or strain. Resistors with resistances that depend on the amount of light present are called Light Dependent Resistor (LDR) and those that depend on temperature are called Thermistors. •
An LDR is made by sandwiching two metal electrodes by a film of cadmium sulphide. In complete darkness, it has a resistance of about 10M Ω, but in bright sunlight, its resistance falls to about 100Ω. Therefore, by varying the amount of light shining on the LDR, we can vary its resistance. In Example 1 above, Replace R2 with a LDR and calculate Vout for both darkness and sunlight cases.
•
A thermistor is a temperature dependent resistor which is manufactured from the oxides of various metals. They are made in all kinds of shapes and sizes. Negative temperature coefficient types have resistances which becomes smaller as temperature increases.
•
A strain gauge is made by sealing a length of very fine wire in a small rectangular of thin plastic sheet in such a way that if the plastic is stretched (i.e. under strain), the wire will be stretched, which in turn increases its resistance.
16
So let’s get back to our discussion of the op amp as a comparator. Fig 8 shows an example of a comparator application. Resistors R3 and R4 fixed the voltage at the noninverting input to half the +Vsupply voltage. The voltage at the inverting terminal will depend on the value of the resistance of the LDR, R2. In the presence of light (such as +
-
daylight) the LDR will have a resistance that is less than R1. This makes V less than V
and the output to be equal to –Vsupply. In this output state, no current will flow through the Light Emitting Diode (LED) and it will remain off. +9V R1 10kΩ
R3 10kΩ
LDR R2
R5
+
+ LED
R4 10kΩ
-
ground
9V
+
9V -9 V
Fig. 8 Comparator application: A light sensitive circuit
In the absence of light (such as at night) the LDR resistance will exceed that of R1, creating a voltage at V+ that is greater than that at V-. The output of the op amp will then switch to the positive saturation voltage which equals +Vsupply. In this state, a current will flow through the LED to turn it on. That is, the LED will be lit. Resistor R5 is required to limit the current through the LED. Usually, LED draws a current of 10 mA and has1.8 V drop across it, therefore R5 value is
+ V sup ply − 1.8
0.01
=
720Ω .
This is the basic principle of operation of a photosensitive light detector, like those installed outside your houses. Note that the LDR can be replaced by a thermistor to produce an ice-warning LED circuit.
17
Feedback in Op Amp Circuits Feedback, as the word implies, is the process of taking some, or all, of the output signal of an amplifier and adding it back to its input. The basic arrangement for this is shown in Fig. 9.
V in + βV out
V in
Amplifier Gain, A0
Add
V out
βV out
Feedback Fraction, β
Fig. 9 Basic feedback in amplifier
A fraction β of the output voltage is fed back and added to the input applied voltage. By inspection of Fig. 9, it is seen that the fraction of signal fed back to the input is β V out . This gets added to the input, V in, to give V in
+
β V out . This combined input is amplified by
the gain A0 of the amplifier to give an output of V out
= A0
(V in + β V out ) . This may be
rewritten as:
V out
= A0V in + A0 β V out
A0V in
= V out − A0 β V out
A0V in
= V out
(1 − A0 β )
The overall gain of this feedback arrangement may now be expressed as:
V out V in
=
A0
(1 − A0 β )
(3)
18
There are two kinds of feedback that depends on the polarity of β . They are positive
feedback and negative feedback.
Positive Feedback
Referring to Eq. 3 above, if β is positive, then the A0 β term can be made to be equal to 1, making the denominator term (1 − A0 β ) equal to zero. That is, the overall gain will be infinite. So we now have an amplifier system with an infinite gain, even without any inputs. This is the basis of the principle of operation of oscillator circuits. Note that positive feedback is only possible when the output signal is fed back in phase with the input signal, so in the case of the op amp, the feedback signal must be sent to the noninverting input terminal.
Negative Feedback
If the fraction β is negative, then the denominator term in Eq. 3 is greater than unity. The overall gain of the feedback amplifying system is now much smaller than the open-loop gain. At first glance, this may seem pointless, but there are some very great advantages for doing this: a. Lowering the gain significantly increases the bandwidth. That is, it allows amplification over a greater frequency range. b. There is less distortion of the output signal. The op amp doesn’t have to saturate anymore. c. Overall improvement in the operational stability of the circuit.
The Inverting Amplifier The inverting amplifier is shown in Fig. 10. For simplification, the power supply connections are not shown. Also, it is assumed that the op amp is not saturated. An input signal Vin is applied to the resistor R 1, which is connected to the inverting input terminal. Resistor Rf is connected across the output and the inverting input terminal to provide negative feedback. 19
Rf
R1
Vin
VG
+ V out
Fig. 10 The inverting amplifier
The non-inverting input is connected to ground so its potential is at exactly zero volt. The behavior of an op amp is such that when any of its input terminals is grounded it causes a virtual ground condition to exist at the other terminal. This point is labeled VG
in Fig.10. This is an important point and needs restating. When the amplifier output is fed back to the inverting input terminal, the output voltage will always take on that value required to drive the signal difference between the amplifier input terminals to zero. Another point to note is that because of the very large input impedance no current flows into the input terminals of the op amp.
Let us now derive an expression for the gain in terms of the R 1 and Rf . The circuits of Fig.11 (a) and (b) show the directions of current flow in an inverting amplifier. Since no currents flow into the input terminals, all currents must flow through the external resistors, R1 and Rf .
20
Rf
Rf
Current, I Current, I
R1
R1
Current, I VG
-
Current, I
-
Current, I
Current, I
+ve input
VG
+
-ve input
+
+ve output
-ve output
0-volt
(a) +ve flowing
(b) -ve flowing
Fig. 11 Direction of current flow in the inverting amplifier
The same current, I, that flows through R 1, also flows through R f . Therefore, Ohm’s law may be used to derive a simple expression of the voltage gain.
Current through R1
=
Current through R f
From Ohm’s law (V=IR), we can express the current in terms of voltage and resistance;
V R1
Therefore,
R1
=
V R f R f
Where, VR1 and VRf are the voltage drops across resistors R 1 and Rf , respectively. But, since the voltage at the inverting input is 0-volt (at the point VG) then:
V R1
= V in − 0
and
V R f
= 0 − V out
Substituting,
21
−0
V in
=
R1 V in
=
R1
V out V in
0 − V out R f
− V out
R f R f
=−
(4)
R1
That is, the voltage gain of an inverting amplifier is equal to the negative of the ratio of its feedback resistance to its input resistance.
The Non-inverting Amplifier The non-inverting amplifier is shown in Fig.12. An input voltage V in is applied directly to +
the non-inverting input terminal, V . Negative feedback is applied by means of resistors Rf and R1, as shown.
Rf
R1 + Vout
Vin
0V Fig. 12 The non-inverting amplifier
22
Assuming that the amplifier is not saturated, then the voltages at the two input terminal +
-
+
-
must effectively be the same. That is V = V . In this setup, V is equal to V in and V is equal to the voltage dividing of V out by Rf and R1. This may be expressed as:
V in
+
= V
−
= V =
R1V out R1 + R f
By rearranging, the overall voltage gain of the amplifier is given by
V out V in
=1+
R f R1
(5)
Exercise: Verify Eq.5 by using the current flow technique used to derive the gain
equation for the inverting amp.
Important Observations:
1. The non-inverting amplifier produces an output that is in phase with the input signal, hence the name non-inverting. 2. The inverting amplifier produces an output that is ‘out-of-phase’ (usually by 180 degrees) with its input. Look back at the circuits of Fig. 10 and Fig. 12. 3. The input signal of the inverting amplifier goes through the input resistors. This means that the impedance seen by the input signal is dependent on the values of the resistors. 4. In the case of the non-inverting amplifier, however, the input signal goes directly to the non-inverting input of the amplifier, which has infinitely high input impedance. Because of this, the current drawn from the signal source is negligible and this circuit configuration is extremely suitable for ‘BUFFER AMPLIFIER’ applications. That is, it does not load down the signal driving circuit.
Effect of Negative Feedback on Gain and Bandwidth Fig.13 shows a circuit that could be used to measure the open-loop gain of a real op amp at a number of different frequencies. During measurement of gain, the amplifier must not 23
5
be saturated. Because the open-loop gain is about 10 , if Vsupply is ±12V then the maximum output voltage would be 12V. This means that the maximum input signal must 5
be 12V ÷10 = 0.12V = 120mVpeak. + 12V
Vac
+
Vout
- 12V ground
Fig. 13 Measuring the gain and bandwidth of an open loop op amp
The sinusoidal input signal may be obtained from a frequency generator and an oscilloscope may be used to measure the input and output voltages (or waveforms). The frequency may be varied from 0 Hz (dc) to 1 MHz. A plot of the gain versus frequency for this open loop amplifier is shown in Fig.14. This is called the frequency response curve of the op amp.
gain
106 105 104 103 102 10 1 10
102
103
104
105
106
Frequency (Hz)
Fig. 14 Frequency response of the open-loop op amp circuit
24
The plot shows that in the open-loop mode, the op amp does not amplify all frequencies equally. The range of frequencies for which the gain is more or less constant is called the bandwidth of the amplifier. For this op amp in open loop mode, its bandwidth is about dc
to 10 Hz, or just 10 Hz. An important observation from the plot is that if a larger bandwidth is desired, the gain must be reduced. How can we reduce the gain? By using negative feedback of course! We just did that in the previous sections.
Using the circuits of Fig. 10 and Fig. 12, shown together in Fig. 15, and applying a variable frequency sinusoidal signal, Gain-Frequency plots may be obtained when both input and output are measured with an oscilloscope. Rf
Rf
R1
R1 Vac
+
+ Vout ground
(a) Inverting setup
Vout
Vac 0V
(b) Non-inverting setup
Fig. 15 Circuits to determine frequency response of negative feedback amplifiers
Note that the gain of both the inverting and non-inverting amplifiers may be easily adjusted by varying the values of one are both resistors (R 1 and/or Rf ). They may also be adjusted to be the same value. They have same frequency response, but different resistor values. Fig. 16 shows the frequency response curves for gains of 10, 100 and 1000.
The main observation from these plots is that as the gain decreases, the bandwidth increases. That is, as more and more negative feedback is applied, the bandwidth increases. Therefore, negative feedback improves bandwidth while reducing gain. But look more closely at the relationship between the product of the Gain and the bandwidth. Yes, it seems always to be constant. Let’s look at the three cases of Fig. 16 again.
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gain
106
Without feedback
105
With feedback
10 4 Overall gain = 1000 103 102 10 1 10
10 2
103
10 4
105
10 6
Frequency (Hz)
gain
106
Without feedback
105
With feedback
104 103 Overall gain = 100 102 10 1 10
102
103
104
105
106
Frequency (Hz)
gain
106
Without feedback
105
With feedback
104 103 102 Overall gain = 10
10 1 10
102
103
104
105
106
Frequency (Hz)
Fig. 16 Gain-Bandwidth for variable gain feedback amplifier
3
6
When the gain is 1000 the bandwidth is 10 making a Gain-Bandwidth product of 10 . 4
The second case has a gain of 100 and a bandwidth of 10 giving a gain-bandwidth 26
6
product of 10 . Likewise the third case! Go back to Fig. 14 for the case without feedback. 5
6
Its gain is 10 and its bandwidth is 10 making the gain-bandwidth product of 10 . This is a very important property of the op amp and is used in many design considerations.
Important observation:
Sometimes you may be required to design circuits with large gain and large 5
bandwidths. For example, suppose a gain of 100 and a bandwidth of 10 are desired. From the plots in Fig. 16, this cannot be obtained from one op amp circuit. So why not cascade (join in series) two op amp circuits, each with a gain of 10.
Voltage Follower What is a voltage follower? It is also called a unity gain buffer , that is, it has very high input impedance and low output impedance and is very applicable in cases where signal sources must not be loaded down. To load down a circuit means to connect a load that will cause too much current to be drawn. Fig. 17 shows the configuration of an op amp as a voltage follower.
+Vs
+ Vin
Vout
-Vs ground
Fig. 17 A voltage follower
27
In this configuration negative feedback is provided by directly connecting the output to the inverting input. V out will adjust itself in such a way as to ensure than the voltage at V +
-
+
is the same as that of V . The voltage at V is V in, therefore, V out will always be the same as Vin. Some characteristics of the voltage follower are:
1. Infinite input impedance 2. Low output impedance 3. Input and output are always in phase 4. Output voltage is the same as the input voltage. That is, Gain = 1
Question:
What is the bandwidth of an op amp voltage follower?
Answer:
Since the gain is always 1, look at the frequency response plots. At gain=1, we always have maximum bandwidth. In Fig. 14, the bandwidth at gain = 1 is 1 MHz.
Summing Amplifier The inverting amplifier configuration of the Op amp may be configured to operate with more than one input signals. Fig. 18 shows such a configuration where 3 input signals, V1, V2, and V3 are connected to resistors R1, R2 and R3, respectively. The other ends of the resistors are connected together at the inverting input terminal. I total = I 1 + I 2 + I 3
V1 V2
R1
I1
R2
I2
R3
I3
V3
Rf +
V ou
Fig. 18 The summing amplifier
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The analysis we did earlier for a single input inverting amplifier may be extended for this 3-input configuration. From Kirchhoff’s current law, we know that the algebraic sum of the currents flowing into a junction must be equal to the algebraic sum of the currents +
-
flowing out of that junction. Since the V input terminal is grounded, then the V terminal must be at a ground potential as well, i.e. 0 volts. Also, no current will flow into the terminals of the op amp (because of their infinitely high input impedance). Therefore, the currents I1, I2 and I3 produced by the input voltages V 1, V 2, and V3 can only combine to flow through the feedback resistor, R f , creating a voltage drop across R f that (from Ohm’s law) is equal to R f (I1+I2+I3). But, the voltage drop across R f is equal to V out. Therefore, Vout = - Rf [I1+I2+I3]. Note the minus sign!! If current flowing into the junction is positive, then current flowing out must be negative.
We can express the currents in terms of the input voltages. Once again, because of the virtual ground (or earth) at the inverting input terminal, I 1
=
V 1 R1
, I 2
=
V 2 R2
and I 3
=
V 3 R3
.
Substituting in expression for V out:
V out
V 1 V 2 R + R 2 1
= − R f
+
V 3
R3
(6)
This expression may be extended for many inputs system. Likewise, if there were only 2 inputs then the term with
V 3 R3
should be removed.
Summing circuits are used in the design of audio mixers. Input 1 may be from a microphone, input 2 from a CD player, and input 3 from a turn table (phono player), etc. The input resistors may be made variable for separate volume adjustment, while the feedback resistor may be made variable for the master volume adjustment. In most applications, however, the output of the summer is fed to a different amplifier circuit for master volume adjustments.
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Difference Amplifier The op amp may also be configured as a difference amplifier (or subtractor). Fig. 19 shows such a configuration. Notice the use of the same resistances, R 1 and R2! R2
R1
-
V2 R1
V1
+
Vout
R2
Fig. 19 The difference amplifier
V out =
R2 R1
(V 1 − V 2 )
(7)
An Operational Amplifier Circuit Example In this section we will examine a number of op amp circuits operations with respect to dc and ac signal inputs. Wherever appropriate, waveform diagrams (or timing diagrams) will be used to show the relationship between the input and output signals.
Response to a variable voltage 1 KHz input signal : Fig. 20 shows a typical inverting
amplifier with supply voltages ± 10V. The input signal is obtained from a 10Vpeak, 1Kz signal generator and is varied (from 0 -10V) by adjusting the slider arm of the variable resistor.
100 kΩ
Signal Generator (10V, 1kHz 100kΩ Variable resistor
Slider
+10 V
10 k Ω
+ -10 V
Vout
Fig. 20 An inverting amplifier circuit
30
Try to answer the following: 1. What is the voltage gain of the amplifier? 2. What is the maximum peak voltage output? 3. Draw the output waveform corresponding to input voltages of a. 0.1 volt b. 1 volt c. 5 volts d. 10 volts 4. How can the gain of the amp be increased by a factor of 10? 5. What is the maximum input voltage that will not saturate the amp? Solutions:
1. Gain is
− R f
Rin
=
− 100k
10k
= - 10
2. Max. peak voltage = V supply = 10 V (or -10V) 3. In drawing waveform diagrams. Always draw the input and the output on the same plot. a. Note the 180 degree phase change Voltage 0.1V input time
1V
output
b. 1 volt input will give 10 volts output. Waveform looks like that of (a). 31
c. 5 volts input will saturate the amplifier. The output signal will be clipped at the saturation voltage of ±10 V.
Voltage 5V input time
10 V output
-10 V
d. A similar waveform is obtained as in (c) when input is 10 V. Waveform will look more like a square wave.
4. Change the 100 k Ω resistor to 1 MΩ., or the 10k Ω to 1k Ω, or use any other resistor combination that gives a gain of 100.
5. The maximum input signal that will not saturate the amp is the Vsupply divided by the gain. V sup ply gain
=
± 10V
10
= ±1 volt
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OP Amp Problems 1. The circuit of Fig. 21 was used as a buffer amplifier for an audio signal.
R2 = 400kΩ
+12V
R1 = 100kΩ
Vaudio
+
Vout -12V
Fig. 21
Buffer Amplifier
Answer the following questions:
(a) What amplifier configuration is this (inverting or non-inverting)? (b) What is the theoretical gain? (c) Sketch of graph of the output vs input voltage when the input voltage is varied from 100mV to 5V. (d) What is the saturation voltage of this circuit? Show this on your sketch in (c) above. (e) What is the maximum input before saturation? Show this on your sketch in (c) above. (f) Are there any advantages of using this buffer amp over other configuration? Explain.
2. Using the circuit of Fig. 21, but replacing the audio input with variable frequency generator. It was found that the gain was different for different frequencies. Explain!
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3. Explain the effect of negative feedback on (a) The gain of an op amp (b) The bandwidth of an op amp (c) The gain-bandwidth product (d) The saturation voltage
4. Design a 2-input audio mixer circuit that has gain of 10.
5. Fig. 22 shows a 2-stage op amp circuit. 600kΩ
+15V 0.2V
-
+12V
100kΩ
-
Va
+
Vout + -15 V
-12V
Fig. 22 Two stage amplifier
Determine the following: (a) The value of the voltage V a. (b) The output voltage, V out (c) The total gain of the circuit. (d) Which of the two op amp configuration above determines the bandwidth of the circuit?
6. Use any amplifier configuration to explain the concept of virtual ground.
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EXPERIMENT #1
Inverting Amplifier
AIM:
Determine the effect of negative feedback on the gain and bandwidth of an Operational Amplifier.
APPARATUS:
Dual voltage power supply ( ± 12 V), 741 Operational Amplifier, Cathode Ray Oscilloscope, Resistors (1 k Ω, 10 k Ω, 100 k Ω) and function generator (0 – 1 MHz).
DIAGRAM:
1.
Choose two resistors R f = 10kΩ and R i = 1kΩ, such that the Gain of the amplifier is 10 and connect the circuit as shown in the diagram. The pin labels for the Op amp may be obtained from page 3 of this workbook. Place the input signal on Channel 1 and the output signal on channel 2 of the scope.
2.
Set your signal generator to produce a sinusoidal waveform and set voltage amplitude to any value between) 0.5V and 1.0 V.
3.
Adjust the frequency of signal generator to 100 Hz. Remember the Cathode Ray Oscilloscope must be used to check/verify the output (frequency and voltage) of the signal generator.
35
4.
Measure the output voltage of the Operational Amplifier using the cathode ray oscilloscope.
5.
Repeat steps 3 & 4 at frequencies 1 kHz, 10 kHz, 100 kHz, 1 MHz.
6.
Tabulate your results in the table below (in Part 2).
7.
Plot graph of log of Gain vs. log of frequency.
PART 2 1. Repeat the experiment in Part 1, but set R f = 100kΩ and R i = 1kΩ to produce voltage gains of 100. Tabulate your results below:
Input Voltage Output Voltage Volatge Gain (Vin) (Vout) (Vout/Vin)
Log of Gain
Frequency (Hz)
10
100
10
1000
10
10000
10
100000
10
1000000
100
100
100
1000
100
10000
100
100000
100
1000000
Log of Frequency
2.
Make a plot of log(Gain) vs. log(frequency) on the same graph created in Part 1.
3.
Determine the Gain-Bandwidth product and the maximum bandwidth of the OpAmp..
DISCUSSION:
Discuss your result as they relate to the aim of the experiment.
36
EXPERIMENT #2
Simulation of Experiment 1
AIM:
Using PSPICE simulator to determine the effect of negative feedback on the gain and bandwidth of an Operation Amplifier.
APPARATUS:
Computer with student version of PSPICE loaded.
DIAGRAM:
Draw the circuit of Experiment 1 in PSPICE Schematic Editor as shown:
Procedure: The lab instructor will guide you through the simulation setup and measurement via ‘ac analysis’ and ‘transient analysis’. You will use PSPICE to obtain the output voltages and make the plots as required by Experiment 1.
37
1. Click “START” , select “ALL PROGRAMS”, then “PSPICE STUDENT”, then Schematics (as shown).
2. The PSPICE window should open up and looks like
38
3. You will now draw the circuit inside the SPICE window by selecting and placing the components. Click on the “looking glass” icon ( part browser ) to open the component selection window. Select and place each component as follows: uA741 (for the op amp), r (for resistor x 3), V dc for voltage supply times 2, Vsin for signal, and “GND_EARTH” for the ground connections.
4. Place the components to match the layout in the circuit diagram (Exit 1) then make all wire connections. This is done by clicking on the “THIN PENCIL”, place the mouse at one component, make one click, then place the mouse at the other component, and make another click. Repeat this procedure the components are connected as shown:
39
5. Double click on each resistor to set its value. 6. Double click on the V dc parts and set values to 12V 7. Double click on Vsin to open a window. Set values as DC = 0 AC = 0.1 (this is the desired input voltage level: 100mV) Voff = 0 Vamp = 0.1 (same as AC) Freq = 100 (this set it to 100 Hz) Then click ok to save values. During the experiment you must re-select this window to change the frequency value to 1k, 10k, 100k, and 1MHz. :
8. Insert Vin and Vout labels by double clicking on the input and output wires:
40
9. You are now ready to start your simulations. Click on “Analysis” then “Setup” and select “Transient” box. Enter the values shown in the transient window.
10. Close ‘Analysis Setup’ Window. 11. Click on Analysis and select “Simulate” to start your simulations. A percentage completion bar will be displayed on the bottom right of the screen. 12. Display the input and output waveform (Vin and Vout) and take the relevant measurements. The lab instructors will show you how to do this. 13. Record your data in the table below:
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