Solution of Homework problems 2 in Section 10.2 Chapter 10, Solution 1. Known quantities: Transistor diagrams, as shown in Figure P10.1: (a) pnp, VEB = 0.6 V and VEC = 4.0 V (b) npn, VCB = 0.7 V and VCE = 0.2 V (c) npn, VBE = 0.7 V and VCE = 0.3 V (d) pnp, VBC = 0.6 V and VEC = 5.4 V Find: For each transistor shown in Figure P10.1, determine whether the BE and BC junctions are forward or reverse biased, and determine the operating region. Analysis: (a) VBE = - 0.6 V for a pnp transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 3.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. (b) VBC = - 0.7 V for a npn transistor implies that the CB junction is reverse-biased. VBE = VBC - VEC = -0.5 V. The BE junction is reverse-biased. Therefore, the transistor is in the cutoff region. (c) VBE = 0.7 V for a npn transistor implies that the BE junction is forward-biased. VBC = VEC - VEB = 0.4 V. The CB junction is forward-biased. Therefore, the transistor is in the saturation region. (d) VBC = 0.6 V for a pnp transistor implies that the CB junction is reverse-biased. VBE = VBC – VEC = - 4.8 V. The BE junction is forward-biased. Therefore, the transistor is in the active region. Chapter 10, Solution 2. Known quantities: Transistor type and operating characteristics: a) npn, VBE = 0.8 V and VCE = 0.4 V b) npn, VCB = 1.4 V and VCE = 2.1 V c) pnp, VCB = 0.9 V and VCE = 0.4 V d) npn, VBE = - 1.2 V and VCB = 0.6 V Find: The region of operation for each transistor.
Analysis: a) Since VBE = 0.8 V, the BE junction is forward-biased. VCB = VCE + VEB = - 0.4 V. Thus, the CB junction is forward-biased. Therefore, the transistor is in the saturation region. b) VBE = VBC + VCE = 0.7 V. The BE junction is forward-biased. VCB = 1.4 V. The CB junction is reverse-biased. Therefore, the transistor is in the active region. c) VCB = 0.9 V for a pnp transistor implies that the CB junction is forward-biased. VBE = VBC – VCE = - 1.3 V. The BE junction is forward-biased. Therefore, the transistor is in the saturation region. d) With VBE = - 1.2 V, the BE junction is reverse-biased. VCB = - 0.6 V. The CB junction is reverse-biased. Therefore, the transistor is in the cutoff region. Chapter 10, Solution 3. Known quantities: The circuit of Figure P10.3:
IC 100 . IB
Find: The operating point and the state of the transistor. Analysis: VBE 0.6 V and the BE junction is forward biased.
VCC 820 103 I B VBE 910 I E & I E I C I B 101I B VCC VBE 12 0.6 12.5A 3 820 10 910 101 911910 I C I B 1.25mA IB
Writing KVL around the right-hand side of the circuit: VCC I C RC VCE I E RE 0
VCE VCC I C RC I C I B RE 12 (1.25)(2.2) (1.25 0.0125)(0.910) 8.1V VBC VBE VCE 0.6 8.1 7.5 V : the BC junction is reverse biased VCE VBE
The transistor is in the active region.
Chapter 10, Solution 4. Known quantities: The magnitude of a pnp transistor's emitter and base current, and the magnitudes of the voltages across the emitter-base and collector-base junctions: IE = 6 mA, IB = 0.1 mA and VEB = 0.65 V, VBC = 7.3 V. Find: a) VCE. b) IC. c) The total power dissipated in the transistor, defined as P VEC I C VEB I B . Analysis: a) VEC = VBC + VEB = 7.3 + 0.65 = 7.95 V. b) IC = IE - IB = 6 - 0.1 = 5.9 mA. c) The total power dissipated in the transistor can be found to be:
P VEC I C VEB I B 7.95 5.9 103 0.65 0.1 103 46.97 mW
Chapter 10, Solution 5. Known quantities: The circuit of Figure P10.5, assuming the BJT has V = 0.6 V. Find: Change 15 V to 15 V The emitter current and the collector-base voltage. Analysis:
15 VBE 15 0.6 480 A 30000 30000
Applying KVL to the right-hand side of the circuit, I E Then, on the left-hand side, assuming >> 1:
10 I C RC VCB 0
VCB 10 I C RC 10 480 10 6 15 103 2.8 V
Chapter 10, Solution 6. Known quantities: The circuit of Figure P10.6, assuming the BJT has VBE 0.6 V and =150. Find: The operating point and the region in which the transistor operates. Analysis: Define RC 3.3 k, RE 1.2 k, R1 62 k, R2 15 k, VCC 18 V By applying Thevenin’s theorem from base and mass, we have
R B R1 || R2 12.078 kΩ
V BB IB
R2 VCC 3.5 V R1 R2
V BB V BE 15 μA R B R E (1 )
I C I B 2.25 mA VCE VCC RC I C R E I E 18 3300 2.25 10 3 1200 151 15 10 6 7.857 V From the value of VCE it is clear that the BJT is in the active region. Chapter 10, Solution 7. Known quantities: The circuit of Figure P10.7, assuming the BJT has V 0.6 V . VCC = 20V Find: The emitter current and the collector-base voltage. Analysis: Applying KVL to the right-hand side of the circuit,
VBB I E RE VEB 0 V VEB 20 0.6 I E BB 497.4 μA . Since RE 39 103
1 , I C I E 497.4 μA
VBB = 20V
VCB I C RC VCC 0 VCB I C RC VCC 497.4 20 10 20 10.05V
Applying KVL to the left-hand side:
3
Chapter 10, Solution 9. Known quantities: The collector characteristics for a certain transistor, as shown in Figure P10.9. Find: a) The ratio IC/IB for VCE = 10 V and I B 100 A, 200A, and 600 A
b) VCE, assuming the maximum allowable collector power dissipation is 0.5 W for I B 500 A . Analysis: a) For IB = 100 µA and VCE = 10 V, from the characteristics, we have IC = 17 mA. The ratio IC / IB is 170. For IB = 200 µA and VCE = 10 V, from the characteristics, we have IC = 33 mA. The ratio IC / IB is 165. For IB = 600 µA and VCE = 10 V, from the characteristics, we have IC = 86 mA. The ratio IC / IB is 143. b) For IB = 500 A, and if we consider an average from a., we have IC = 159·500 10-3= 79.5 mA. The power dissipated by the transistor is P VCE I C VBE I B VCE I C , therefore: VCE
P 0.5 6.29 V. I C 79.5103
Chapter 10, Solution 10.
Known quantities: Figure P10.10, assuming both transistors are silicon-based with 100 . Find: a) IC1, VC1, VCE1. b) IC2, VC2, VCE2. Analysis: a) From KVL:
30 I B1RB1 VBE1 0 30 0.7 I B1 39.07 μA 750 103
I C1 I B1 3.907 mA
VC1 30 RC1I C1 30 3.907 6.2 5.779 V VCE1 VC1 5.779 V .
b) Again, from KVL: 5.779 VBE2 I E 2 RE 2 0
I E2
5.779 0.7 4.7 103
1.081 mA
100 1.081 1.07 mA . 1 101
and I C 2 I E 2
Also, 30 I C 2 ( RC 2 RE 2 ) VCE 2 0 VCE 2 30 (1.07) (20 4.7) 3.574 V . Finally, I C 2
30 VC 2 RC 2
VC 2 30 (1.07) (20) 8.603 V .
Chapter 10, Solution 11. Known quantities: Collector characteristics of the 2N3904 npn transistor, see data sheet pg. 560. Find: The operating point of the transistor in Figure P10.11, and the value of at this point. Analysis: Construct a load line. Writing KVL, we have: 50 5000 IC VCE 0 . Then, if I C 0 , VCE 50 V ; and if VCE 0 , I C 10 mA . The load line is shown superimposed on the collector characteristic below: The operating point is at the intersection of the load line and the I B 20 A line of the characteristic. Therefore, I CQ 5 mA and VCEQ 20 V . Under these conditions, an 5 A increase in I B yields an increase in I C of approximately 6 5 1 mA .
Therefore,
I C 1 103 200 I B 5 10 6
The same result can be obtained by checking the hFE gain from the data-sheets corresponding to 5 mA.
Load line
Chapter 10, Solution 14. Known quantities: The circuit of Figure P10.14, VCEsat=0.1V, VBEsat=0.6V, and β=50. Find: The base voltage required to saturate the transistor. Analysis: The collector current is 12 0.1 IC 11.9 mA 1
The base current is I 11.9 IB C 0.238 mA 238A 50
And since
V VBEsat I B BB mA 10
Therefore, V VBB 0.238 mA 10k 0.6 2.98 V
Chapter 10, Solution 16. Known quantities: Collector characteristics of 2N3904 npn transistor; Transistor circuits; Find: The operating point; Analysis: From KVL, 50 5kI C VCE 5k ( IC 20A) 0
or VCE 10kI C 50 0.1 49.9
If VCE 0 , I C
49.9 4.99mA , and if I C 0 , VCE 49.9V . The load 10k
line is shown superimposed on the collector characteristic below: The operating point is at the intersection of the load line and the I B 20A line of the characteristic. Therefore, I CQ 3mA and VCEQ 8V .
Under these conditions, a 10A increase in I B yields an increase in I C of approximately 5mA 3mA 2mA . Therefore,
I C 2mA 200 I B 10A
Addition of the emitter resistor effectively increased the current gain by decreasing the magnitude of the slope of the load line.
Chapter 10, Solution 17. Known quantities: For the circuit shown in Figure 10.14 in the text:
Voff 0 V, Von 5 V, I B 5 mA, RB 1 kΩ, VCC 5 V, V 0.7V, VCEsat 0.2 V, 95, V
LED
1.4 V, I LED 10 mA, Pmax 100 mW
Find: Range of RC. Analysis: RC
VCC V
LED
I LED
VCEsat
5 1.4 0.2 340 0.01
From the maximum power I LED max RC
Pmax 0.1 71 mA V LED 1.4
VCC V LED VCEsat 47 I LED max
Therefore, RC [47, 340]
Chapter 10, Solution 22. Known quantities: For the circuit shown in Figure 10.14 in the text: Voff 0 V, Von 5 V, I B max 1 mA, RB 1 kΩ, R 12 Ω, VCC 13 V, V 0.7V, VCEsat 1 V, IC 1A
Find:
Minimum value of that will ensure the correct operation of the fuel injector. Analysis: V VCEsat 13 1 I C CC 1A R 12 I 1 min C 1000 I B max 1 103
Chapter 10, Solution 25. Known quantities: The circuit of Figure P10.25: IC = 40 mA; Transistor large signal parameters. Find:
Design a constant-current battery charging circuit, that is, find the values of VCC, R1, R2 that will cause the transistor Q1 to act as a 40-mA constant current source. Assumptions: Assume that the transistor is forward biased. Use the large-signal model with = 100. Analysis: The battery charging current is 40 mA, IC = 40 mA. Thus, the emitter current must be I E
1 I 40.4mA . E
Since the base-emitter junction voltage is assumed to be 0.6 V, then resistor R2 has a voltage: V2 Vz V 5.6 0.6 5 V , so the required value of R2 to be: R2
V 5 123.8 I E 0.0404
Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply enough current fro the Zener to operate, for example R1 > 100 , so that there will be as little current flow through this resistance as possible. Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the sum of the battery voltage, the CE junction voltage and the voltage across R2. That is, VCC 9 VCE 5 . A collector supply of 24 V will be more than adequate for this task.
Chapter 10, Solution 26. Known quantities: The circuit of Figure of P10.26. Find:
Analyze the operation of the circuit and explain how I E is decreasing until the battery is full. Find the values of VCC, R1 that will result in a practical design.
Assumptions: Assume that the transistor is forward biased. Analysis: When the Zener Diode works in its reverse breakdown area, it provides a constant voltage: Vz 11 V . That means: VB VZ 11 V . When the transistor is forward biased, according to KVL, VZ I BE RBE V Vbattery , where RBE is the base resistance. As the battery gets charged, the actual battery charging voltage Vbattery will increase from 9.6 V to 10.4 V. As Vbattery increases gradually, VZ and V stay unchanged, then we can see that I BE will decrease gradually. So I E 1I BE will also decrease at the same time. Since the only purpose of R1 is to bias the Zener diode, we can select a value that will supply enough current fro the Zener to operate, for example R1 > 100 , so that there will be as little current flow through this resistance as possible.
Finally, we need to select an appropriate supply voltage. VCC must be greater than or equal to the sum of the battery voltage, the CE junction voltage. That is, VCC 11 VCE . A collector supply of 12 V should be adequate for this task.
Chapter 10, Solution 32. Known quantities: For the circuit shown in Figure P10.32: VCC 12 V
130
R1 82 kΩ
R2 22 kΩ
RE 0.5 kΩ
RL 16 Ω .
Find: VCEQ at the DC operating point. Analysis: Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network (R1,, R2, VCC) in the base circuit: 12 22 V CC R 2 = = 2.538 V + 82 22 R1 R 2 R1 R 2 = 82 22 = 17.35 kΩ R B = R eq = 82 22 R1 + R 2
VD : V BB = V TH = V OC = Suppress V CC :
Redraw the circuit using the Thèvenin equivalent. The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages.
Assume the transistor is operating in its active region. Then, the base-emitter junction is forward biased. V BEQ 700 mV [Si]
I EQ = [ + 1 ] I BQ
KVL : - V BB + I BQ R B + V BEQ + I EQ R E = 0 - V BB + I BQ R B + V BEQ + [ + 1 ] I BQ R E = 0
I BQ =
V BB - V BEQ 2.538 0.7 = = 22.18 μA 17350 130 1 500 R B + +1 R E
6 I EQ = + 1 I BQ = 130 + 1 22.18 10 = 2.906 mA
KVL : - I EQ R E - V CEQ + V CC = 0 V CEQ = V CC - I EQ R E = 12 2.906 0.5 = 10.55 V
The collector-emitter voltage is greater than its saturation value (0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid.
Chapter 10, Solution 33. Known quantities: For the circuit shown in Figure P10.33: VCC 12 V 100 VEE 4 V RC 3 kΩ RE 3 kΩ RL 6 kΩ
RS 0.6 kΩ
RB 100 kΩ
vS 1 cos(6.28 103 t ) mV .
Find: VCEQ and the region of operation. Analysis: The "DC blocking" or "AC coupling" capacitors act as open circuits for DC; therefore, the signal source and load can be neglected since this is a DC problem. Specify directions of current and polarities of voltages. Assume the transistor is operating in its active region; then, the baseemitter junction is forward biased and:
VBEQ 700 mV [ Si ] I CQ I BQ
I EQ ( 1) I BQ
KVL : VEE I BQ RB VBEQ I EQ RE 0 I BQ
VEE VBEQ
RB 1RE
4 0.7 8.189A 100000 (100 1)(3000)
I CQ I BQ (100) 8.189 10 6 818.9A I EQ ( 1) I BQ (100 1) 8.189 10 6 827.0 A KVL : VEE I EQ RE VCEQ I CQ RC VCC 0
VCEQ VEE VCC I CQ RC I EQ RE 4 12 818.9 10 6 3000 827.0 10 10 6 3000 11.06 V
The collector-emitter voltage is greater (more positive) than its saturation value (+ 0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was correct and the solution is valid. Notes: 1. DC power may be supplied to an npn BJT circuit by connecting the positive terminal of a DC source to the collector circuit, or, by connecting the negative terminal of a DC source to the emitter circuit, or, as was done here, both. 2. In a pnp BJT circuit the polarities of the sources must be reversed. Negative to collector and positive to emitter. Chapter 10, Solution 35. Known quantities: For the circuit shown in Figure P10.35: vS 3 V
100
RB 60 kΩ
Find: a) The value of RE so that I E is 1 mA. b) RC so that VC is 5 V. c) The small-signal equivalent circuit of the amplifier for RL 5k d) The voltage gain. Analysis: (a) With RB 60 k and VB 3 V , applying KVL, we have 3 I B RB 0.6 (1 ) I B RE 2.4 IB 60k 101RE
I E 101
2.4 1mA 60k 101RE
Therefore, RE
101 2.4 60 1.81 k 101
(b) VCE 15 IC RC I E RE From (a), we have I C I E Therefore, RC
0.99 mA 1
15 5 1.81 8.27 k 0.99
(c) The small signal equivalent circuit is shown below I B
(d)
VS RB hiw
V 1 I C out h fe I B vout I C RL 1 hoe hoe V 0.6 hie BE I BQ 60.6k I B 0.0099 103
RB +
vS h ie -
B
C
RC
IB h fe I
IC
B
1 RL h oe
E
Since hoe is not given, we can reasonably assume that 1/hoe is very large. Therefore, AV
vout 100 RL 4.15 vs RB hie
Chapter 10, Solution 36. Known quantities: For the circuit shown in Figure P10.36: RC 200 kΩ
Find: e) The operating point of the transistor. f) Voltage gain vout vin ; current gain iout iin g) Input resistance ri h) Output resistance ro Analysis: (a) VB VCC
R2 6.1 V R1 R2
RB R1 || R2 3749.87
Assuming VBE 0.6 V , we have VEV VB VBE 5.5 V
+ v OUT -
V I E E 22 mA RE I I B E 0.088 mA b 1
and
VCE VC VE V.CC RC I C 5.5 15 - 200 21.912 10-3 5.5 5.12 V
(b) The AC equivalent circuit is shown on the right: hie
VBE I B
I BQ
0.6 0.088 103
6.82k
vout RE ( I B I C ) 250(250 1) I B vin I B hie vout I B hie 250 251 I B
Therefore, the voltage gain is AV
vout 0.902 vin
and
iout I B I C ( 1) I B v iin I B in I B ( I B hie 250 251 I B ) RB RB
and the current gain is iout ( 1) I B 12.84 iin I B ( I B hie 250 251 I B ) RB
(c) To find the input resistance we compute: vin I B hie 250 251 I B iin I B ( I B hie 250 251 I B ) RB
Therefore. the input resistance is ri
vin 3558 iin
(d) To find the output resistance we compute vout RE ( I B I C ) 250(250 1) I B
iout I B I C ( 1) I B
Therefore, the output resistance is ro
vout 250 iout
Chapter 10, Solution 41. Known quantities: The circuit given in Figure P10.41. Find: Show that the given circuit functions as an OR gate if the output is taken at v01. Analysis: Construct a state table. This table clearly describes an AND gate when the output is taken at vo1 . v1
v2
Q1
Q2
Q3
vo1
vo2
0
0
off
off
on
0
5V
0
5V
off
on
off
5V
0
5V
0
on
off
off
5V
0
5V
5V
on
on
off
5V
0
Chapter 10, Solution 42. Known quantities: The circuit given in Figure P10.41. Find: Show that the given circuit functions as a NOR gate if the output is taken at v02. Analysis: See the state table constructed for Problem 10.41. This table clearly describes a NOR gate when the output is taken at vo 2 .
Chapter 10, Solution 45. Known quantities: In the circuit given in Figure P10.45 the minimum value of vin for a high input is 2.0 V.
Assume that the transistor Q1 has a of at least 10. Find: The range for resistor RB that can guarantee that the transistor is on. Analysis: ic
5 0.2 2.4 mA , therefore, iB = iC/ = 0.24 mA. 2000
(vin)min = 2.0 V and (vin)max = 5.0 V, therefore, applying KVL:
-vin +RB iB + 0.6 = 0
vin 0.6 . Substituting for (vin)min and (vin)max , we find the following range for RB: iB 5.833 kΩ RB 18.333k Ω
or
RB
Chapter 10, Solution 46. Known quantities: For the circuit given in Figure P10.46: R1C R2C 10 kΩ , R1B R2B 27 kΩ .
Find: a) vB, vout, and the state of the transistor Q1 when vin is low. b) vB, vout, and the state of the transistor Q1 when vin is high. Analysis: a)
vin is low Q1 is cutoff vB = 5 V Q2 is in saturation vout = low = 0.2 V.
b) vin is high Q1 is in saturation vB = 0.2 V Q2 is cutoff vout = high = 5 V.
