Rizzoni, G. Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Chapter 10 Instructor Notes
Chapter 10 introduces bipolar junction transistors. The material on transistors has been reorganized in this 4th Edition, and is now divided into two independent chapters, one on bipolar devices, and one on field-effect devices. The two chapters are functionally independent, except for the fact that Section 10.1, introducing the concept of transistors as amplifiers and switches, can be covered prior to starting Chapter 11 if the instructor decides to only teach field-effect devices. Section 10.2 introduces the fundamental ideas behind the operation of bipolar transistors, and illustrates the calculation of the state and operating point of basic transistor circuits. The discussion of the properties of the BJT in Section 10.2 is centered around a description of the base and collector characteristics, and purposely avoids a detailed description of the physics of the device, with the intent of providing an intuitive understanding of the transistor as an amplifier and electronic switch. Example 10.5 introduces the use of Electronics Workbench (EWB) as a tool for analyzing electronics circuits; the CDROM contains an introduction to the software package and a number of solved problems. Section 10.3 introduces large-signal models of the BJT, and also includes the box Focus on Methodology: Using device data sheets (pp. 535-537). Example 10.7 (LED Driver) and the box Focus on Measurements: Large Signal Amplifier for Diode Thermometer (pp. 539-541) provide two application examples that include EWB solutions. Section 8.4 introduces the analysis of BJT switches and presents TTL gates. The end-of-chapter problems are straightforward applications of the concepts illustrated in the chapter. Learning Objectives 1. Understand the basic principles of amplification and switching. Section 1. 2. Understand the physical operation of bipolar transistors; determine and select the operating point of a bipolar transistor circuit; understand the principle of small signal amplifiers. Section 2. 3. Understand the large-signal model of the bipolar transistor, and apply it to simple amplifier circuits. Section 3. 4. Understand the operation of bipolar transistor as a switch and analyze basic analog and digital gate circuits. Section 4.
10.1
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Section 10.2: Operation of the Bipolar Junction Transistor
Problem 10.1 Solution:
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 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.
active region. (b)
(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. ______________________________________________________________________________________
Problem 10.2 Solution:
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.
10.2
G. Rizzoni, Principles and Applications of Electrical Engineering b)
Problem solutions, Chapter 10
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.
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= - 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.
d) With VBE
Problem 10.3 Solution:
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. V - VBE 12 - 0.6 = = 13.9A I B = CC R1 820 I C = I B = 1.39 mA
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 B + I C )RE
VBC = VBE VCE > VBE
= 12 - (1.39)(2.2) - (1.39 + 0.0139)(0.910) = 7.664 V + VCE = 0.6 + 7.664 = 8.264 V The transistor is in the active region.
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10.3
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Problem 10.4 Solution:
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, VCB = 7.3 V. Find: a) VCE. b) IC. c) The total power dissipated in the transistor, defined as Analysis: a) VCE = VCB c)
P = VCE I C + VBE I B .
- VEB = 7.3 - 0.65 = 6.65 V. b) IC = IE - IB = 6 - 0.1 = 5.9 mA.
The total power dissipated in the transistor can be found to be: P
VCEIC = 6.655.910-3 = 39
mW. ______________________________________________________________________________________
Problem 10.5 Solution:
Known quantities: The circuit of Figure P10.5, assuming the BJT has Find: The emitter current and the collector-base voltage. Analysis: Applying KVL to the right-hand side of the circuit, Then, on the left-hand side, assuming
V = 0.6 V .
>> 1 :
(0.6 + 15) ) = -520A V + 15 I E = - BE =- 30000 30000
VCB = 10 - I C RC
- 10 + I C RC + VCB = 0
= 10 - (-520) (15) 10 -3 = 17.8 V
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Problem 10.6 Solution:
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.
10.4
G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: Define RC
Problem solutions, Chapter 10
= 3.3k, RE = 1.2k, R1 = 62k, R2 = 15k, VCC = 18 V
By applying Thevenin's theorem from base and mass, we have
RB = R1 || R2 = 12.078 k VBB = IB = R2 VCC 3.5 V R1 + R2
VBB - VBE 15 A RB + RE (1 + )
I C = I B = 2.25 mA VCE = VCC - RC I C - RE I E = 18 - 3300 2.25 10 -3 - 1200 151 15 10 -6 = 7.857 V
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From the value of VCE it is clear that the BJT is in the active region.
Problem 10.7 Solution:
Known quantities: The circuit of Figure P10.7, assuming the BJT has Find: The emitter current and the collector-base voltage. Analysis: Applying KVL to the right-hand side of the circuit,
V = 0.6 V .
- VCC + I E RE + VEB = 0 VCC - VEB 20 - 0.6 = = 497.4A . Since >> 1 , I C I E = 497.4A RE 39 10 3 Applying KVL to the left-hand side: VCB + I C RC - VDD = 0 IE =
VCB = VDD - I C RC = 20 - 497.4 20 10 -3 = 10.05 V
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Problem 10.8 Solution:
Known quantities: The circuit of Figure P10.7, assuming the emitter resistor is changed to 22 k and the BJT has V = 0.6 V . Find: The operating point of the transistor. Analysis:
IE =
VCC - VEB 20 - 0.6 = = 881.8 A , I C I E = 881.8A RE 22 10 3
10.5
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
VCB = VDD - I C RC = 20 - 881.8 20 10 - 3 = 2.364 V
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Problem 10.9 Solution:
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 IB = 100 A, 200 A, and 600 A b) VCE, assuming the maximum allowable collector power dissipation is 0.5 W for IB Analysis: a) For IB
= 500 A.
= 100 A and VCE = 10 V, from the characteristics, we have IC = 17 mA. The ratio IC / IB is 170. For IB = 200 A and VCE = 10 V, from the characteristics, we have IC = 33 mA. The ratio IC / IB is 165. For IB = 600 A and VCE = 10 V, from the characteristics, we have IC = 86 mA. The ratio IC / IB is 143. -3 b) For IB = 500 A, and if we consider an average from a., we have IC = 159500 10 = 79.5 mA. The power dissipated by the transistor is P = VCE I C + VBE I B VCE I C , therefore: 0 .5 P = = 6.29 V . VCE I C 79.5 10 -3
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Problem 10.10 Solution:
Known quantities: Figure P10.10, assuming both transistors are silicon-based with Find: a) IC1, b) IC2, a)
= 100 .
VC1, VCE1. VC2, VCE2. - 30 + I B1 RB1 + VBE1 = 0
Analysis: From KVL:
I B1 =
I C1 = I B1 = 3.907 mA VC1 = 30 - RC1 I C1 = 30 - 3.907 6.2 = 5.779 V VCE1 = VC1 = 5.779 V .
b) Again, from KVL: and I C 2
30 - 0.7 = 39.07A 750 10 3
- 5.779 + VBE 2 + I E 2 RE 2 = 0
I E2 =
100 = I E2 + 1 = 1.081 101 = 1.07 mA .
10.6
5.779 - 0.7 = 1.081 mA 4.7 10 3
G. Rizzoni, Principles and Applications of Electrical Engineering Also,
Problem solutions, Chapter 10
- 30 + I C 2 ( RC 2 + RE 2 ) + VCE 2 = 0 VCE 2 = 30 - (1.07) (20 + 4.7) = 3.574 V . = 30 - VC 2 RC 2
VC 2 = 30 - (1.07) (20) = 8.603 V .
Finally, I C 2
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Problem 10.11 Solution:
Known quantities: Collector characteristics of the 2N3904 npn transistor, see data sheet pg. 536. 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 I C + 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:
load line
The operating point is at the intersection of the load line and the Therefore,
I B = 20A line of the characteristic.
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 10 -3 = = 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.
10.7
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
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Problem 10.12 Solution:
Known quantities: The circuit shown in Figure P10.12. With reference to Figure 10.20, assume
V = 0.6 V , Vsat = 0.2 V .
Find: The operating point of the transistor, by computing the ratio of collector current to base current. Analysis:
VCE = Vsat = 0.2 V , therefore I C =
VBE
10 - 0.2 = 9.8 mA RC 5 .7 - 0 .6 = V = 0.6 V , therefore I B = = 102A RB
I C 9.8 10 -3 = = 96.08 << I B 102 10 -6
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Problem 10.13 Solution:
Known quantities: For the circuit shown in Figure P10.13, Find: a) VB b) c)
VE = 1V and V = 0.6 V . IC e) f) .
d)
IB IE
Analysis: a) VEB = VE
- VB = V = 0.6 V VB = VE - VEB = 1 - 0.6 = 0.4 V
10.8
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
VB 0 .4 = = 20A RB 20 10 3 5 - VE 5 - 1 c) I E = = = 800A RE 5000 d) I C = I E - I B = 800 - 20 = 780A
b)
IB =
e) f)
I C 780 = = 39 IB 20 I 780 = C = = 0.975 . I E 800
=
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Problem 10.14 Solution:
Known quantities: For the circuit shown in Figure P10.14:
VCC = 20 V RL = 1k
= 130
R1 = 1.8 M
RS = 0.6 k
vS = cos 6.28 103 t mV . R1 , R2 , and VCC with respect to the
[
R2 = 300 k
]
RC = 3 k
RE = 1k
Find: The Thvenin equivalent of the part of the circuit containing terminals of
R2 . Redraw the schematic using the Thvenin equivalent.
Analysis: Extracting the part of the circuit specified, the Thvenin equivalent voltage is the open circuit voltage. The equivalent resistance is obtained by suppressing the ideal independent voltage source:
Note that
VCC must remain in the circuit because it supplies current to other parts of the circuit:
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10.9
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Problem 10.15 Solution:
Known quantities: For the circuit shown in Figure P10.14:
VCC = 15 V RL = 1.5 k
Find:
= 100
R1 = 68 k
RS = 0.9 k
vS = cos 6.28 10 t mV .
[
R2 = 11.7 k RC = 200
3
]
RE = 200
VCEQ and the region of operation.
Analysis: Simplify the circuit by obtaining the Thvenin equivalent of the biasing network in the base circuit:
VD : V BB = V TH = Suppress V CC :
15 11.7 V CC R 2 = = 2.202 V 68 + 11.7 R1 + R2 R1 R 2 = 68 11.7 = 9.982 k R B = Req = R1 + R 2 68 + 11.7
Redraw the circuit using the Thvenin 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 CQ = I BQ
Then:
I EQ = ( + 1) I BQ
10.10
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
KVL: - V BB + I BQ RB +V BEQ+ I EQ RE = 0 - V BB + I BQ RB +V BEQ+ [ +1 ] I BQ RE = 0 2.202- 0.7 V BB - V BEQ = 49.76 A = I BQ = RB + [ +1 ] RE 9982+ (100 + 1)(200)
-6 I CQ = I BQ = 100 49.76 10 = 4.976 mA -6 I EQ = ( + 1) I BQ = (100 + 1) 49.76 10 = 5.026 mA KVL : - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = V CC - I CQ RC - I EQ R E = 15 - 4.976 0.2 - 5.026 0.2 = 13.00 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. Notes: 1. If the collector-emitter voltage were less than its saturation value, the transistor would be operating in its "saturation" or "ohmic" region. In this case, the collector-emitter voltage is equal to its saturation value (0.3 V for Silicon). The solution for the base current remains valid. However, the parameter and the solution for the collector and emitter curents become invalid. In the saturation region, the base-emitter junction is still forward biased and its voltage remains the same. A solution for the collector and emitter currents is possible using the collector-emitter saturation voltage (0.3 V for Silicon) in a KVL. In the saturation region, the base current has no control over the collector or emitter currents (since is invalid). Therefore, amplification is impossible. ______________________________________________________________________________________ 2.
Problem 10.16 Solution:
Known quantities: For the circuit shown in Figure P10.14:
VCC = 15 V RL = 1.5 k
Find:
= 100
R1 = 68 k
RS = 0.9 k
vS = cos 6.28 103 t mV .
[
R2 = 11.7 k RC = 4 k
]
RE = 200
VCEQ and the region of operation.
Analysis:
10.11
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Simplify the circuit by obtaining the Thvenin equivalent of the biasing network in the base circuit:
Redraw the circuit using the Thvenin 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 CQ = I BQ
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 = 2.202 - 0.7 V BB - V BEQ = 49.76 A = 9982 + (100 + 1)(200) RB + [ +1 ] RE
-6 I CQ = I BQ = 100 49.76 10 = 4.976 mA -6 I EQ = ( + 1) I BQ = (100 + 1) 49.76 10 = 5.026 mA KVL : - I EQ R E - V CEQ - I CQ RC + V CC = 0
V CEQ = V CC - I CQ RC - I EQ R E = 15 - 4.976 4 - 5.026 0.2 = - 5.91 V
The collector-emitter voltage is less (more negative) than its saturation value (+ 0.3 V for Silicon). Therefore the initial assumption (operation in the active region) was incorrect and the solution is not valid. The device is operating in the saturation region with a saturation collector-emitter voltage equal to 0.3 V. The solutions for the collector and emitter currents are invalid. THE VALID SOLUTION:
KVL : - I EQ R E - vCE -SAT - I CQ RC + V CC = 0
I EQ = I BQ + I CQ
- [ I BQ + I CQ ] R E - vCE -SAT - I CQ RC + V CC = 0 15 - 49.76 0.2 10 -3 - 0.3 V CC - I BQ R E - vCE - SAT = = 3.498 mA 200 + 4000 R E + RC
I CQ
=
The solution for the base current is valid. The value of beta given is not valid. ______________________________________________________________________________________
10.12
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Problem 10.17 Solution:
Known quantities: For the circuit shown in Figure P10.17:
VCC = 12 V
Find:
= 130
R1 = 82 k
R2 = 22 k
RE = 0.5 k
RL = 16 .
VCEQ at the DC operating point.
Analysis: Simplify the circuit by obtaining the Thvenin equivalent of the biasing network (R1,, R2, VCC) in the base circuit:
VD : V BB = V TH = V OC = Suppress V CC :
12 22 V CC R2 = = 2.538 V R1 + R2 82 + 22 R1 R2 = 82 22 = 17.35 k R B = Req = 82 + 22 R1 + R2
Redraw the circuit using the Thvenin 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 = 2.538 - 0.7 V BB - V BEQ = = 22.18A 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. ______________________________________________________________________________________
Problem 10.18 Solution:
Known quantities: For the circuit shown in Figure P10.18:
10.13
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
VCC = 12 V RL = 6 k
Find:
VEE = 4 V Rs = 0.6 k
= 100
vS = cos 6.28 10 t mV .
[
RB = 100 k
3
]
RC = 3 k
RE = 3 k
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 Assume voltages. the transistor is operating in its active region; then, the base-emitter junction is forward biased and:
V BEQ 700 mV [Si] I CQ = I BQ I EQ = ( + 1) I BQ
KVL : - V EE + I BQ R B + V BEQ + I EQ R E = 0 4 - 0.7 V EE - V BEQ = 8.189 A = I BQ = 100000 + (100 + 1)(3000) RB + [ +1 ] RE
I CQ = I BQ = (100 ) 8.189 10 -6 = 818.9 mA I EQ = ( + 1) I BQ = (100 + 1) 8.189 10 -6 = 827.0 mA KVL : + V EE - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = V EE + V CC - I CQ RC - I EQ R E = 4 + 12 - 818.9 3 10 -3 - 827.0 3 10 -3 = 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. ______________________________________________________________________________________
Problem 10.19 Solution:
Known quantities: For the circuit shown in Figure P10.19:
VCC = 12 V
= 130
RB = 325 k RC = 1.9 k
RE = 2.3 k
10.14
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
RL = 10 k
Find:
Rs = 0.5 k
vS = cos 6.28 103 t mV .
[
]
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 base-emitter junction is forward biased. The base and collector currents both flow through the collector resistor in this circuit.
V BEQ 700 mV [Si] I CQ = I BQ I EQ = [ + 1 ] I BQ
KCL : I BQ + I CQ - I RC = 0 I RC = I CQ + I BQ = I BQ + I BQ = [ + 1 ] I BQ
KVL : - I EQ R E - V BEQ - I BQ R B - I RC RC + V CC = 0 - [ + 1 ] I BQ [ R E + RC ] - V BEQ - I BQ R B + V CC = 0
I BQ = I RC
12 - 0.7 V CC - V BEQ = = 12.91A [325 + (130 + 1) (2.3 + 1.9) ]103 R B + [ + 1 ] [ R E + RC ] = I EQ = [ + 1 ] I BQ = (130 + 1) 12.91 10 -6 = 1.691mA
KVL : - I EQ R E - V CEQ - I RC RC + V CC = 0
V CEQ = V CC - I RC RC - I EQ R E = 12 - 1.691 1.9 - 1.691 2.3 = 4.896 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. ______________________________________________________________________________________
Problem 10.20 Solution:
Known quantities: For the circuit shown in Figure P10.19:
VCC = 15 V RL = 1.7 k
C = 0.5F
= 170
Rs = 70
vS = cos 6.28 103 t mV .
[
RB = 22 k
]
RC = 3.3 k
RE = 3.3 k
10.15
G. Rizzoni, Principles and Applications of Electrical Engineering Find:
Problem solutions, Chapter 10
VCEQ and the region of operation.
Analysis: When is very high (say, greater than 50) the "high beta approximation" can be used, i.e.:
IC I E I B
Using this approximation in this problem:
I BQ 12.5A
I CQ I EQ 2.125 mA
V CEQ 0.975 V
The collector-emitter voltage is greater than the saturation value (but not by much) so the original assumption (operation in the active region) and the solution are valid. However, when a signal is introduced into the circuit, the collector-emitter voltage will vary about its Q point value. Since the Q point is close to saturation, saturation and severe distortion due to clipping is likely. ______________________________________________________________________________________
Problem 10.21 Solution:
Known quantities: For the circuit shown in Figure P10.14:
VCC = 15 V RE = 710 RL = 3 k
C = 0.47 F
R1 = 220 k
R2 = 55 k
RC = 3 k
Rs = 0.6 k
vS = Vi 0 sin (t )
Vi 0 = 10 mV .
DC operating point:
V CEQ = 7.61 V Transfer characteristic and of the npn silicon transistor:
IC I s eVT = I s e
v BE V BEQ + vbe VT
I BQ = 19.9 A
= 100
Device i-v characteristic plotted in Figure P9.21. Find: a) The no-load large signal gain [v0/vi]. b) Sketch the waveform of the output voltage as a function of time. c) How the output voltage is distorted compared with the input waveform. Analysis: a) Simplify the circuit by determining the Thvenin equivalent of the biasing network in the base circuit.
VD : V BB = V TH = Suppress V CC :
15 55 V CC R2 = = 3.00 V R1 + R2 220 + 55
3 R1 R2 = 220 55 10 = 44.0 k R B = Req = 220 + 55 R1 + R2 First, determine the collector and emitter Q point:
10.16
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
KVL : - I EQ R E - V CEQ - I CQ RC + V CC = 0 V CEQ = - 2.010 0.710 - 1.990 3 + 15 = 7.603 V
-6 I CQ = I BQ = 100 19.9 10 = 1.99 mA -6 I EQ = [ + 1 ] I BQ = 101 19.9 10 = 2.01mA
Now, determine the DC load line (circuit characteristic).
KVL : - i E R E - vCE - iC RC + V CC = 0
iC = V CC - vCE V CC - vCE = +1 3717 R E + RC
iE =
+1 iC
If : vCE = V CC = 15 V [Intercept 1] = 0 = 15 = 3717 = 4.035 mA If : vCE = 0 [Intercept 2]
Plot the DC load line on the same plot as the transistor i-v characteristic. Note that the collector current is a linear function of the collector-emitter voltage. Only two points (the two intercepts shown as Points 1 and 2 on the plot) are required to plot a linear function. The DC load line should pass through the Q point. The AC load line can now be plotted. It will pass through the Q point as does the DC load line; however, its intercepts cannot be directly determined. Instead, its slope will be determined. The slope and the Q point can then be used to plot the AC load line.
Slope AC = -
1 RC
= -
1 mA 2mA iC = - 0.3333 = = 3000 V 6V vCE
For AC the capacitors act as short circuits. The bypass capacitor in parallel with the emitter resistor shorts out any AC voltage across it. The collector coupling capacitor acts as a short and connects the load resistor to the circuit. However, no-load conditions are specified, i.e., the load current is zero or the load resistor is replaced by an open circuit. With no signal, the circuit operates at its Q point. When a signal is introduced, the operating point changes along the AC load line as a function of the signal voltage. Unfortunately, the device i-v characteristic is plotted as a function of input (base) current instead of voltage. But, there is the static characteristic that relates collector current to base-emitter voltage:
iC = I s e V T = I s e
v BE
V BEQ + v be VT
= [ Is e
V BEQ VT
] eV T = I CQ eV T
v be
v be
v be iC = I CQ vbe = iB = eV T I BQ eV T
The base-emitter signal voltage is related to the input signal voltage. At sufficiently high frequencies (the "mid"-frequency range) the capacitors can be modeled as short circuits for AC. This means that there is no AC voltage across either the AC coupling (DC blocking) capacitor or the capacitor bypassing the emitter resistor. Using superposition to sum only AC voltages around the loop:
10.17
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
KVL : - vi + vc + vbe + ve = 0
iB =
vbe I BQ eV T
vc = ve = 0
vi e 26 mV
vbe = vi
= [ 19.9 A ]
This transfer function can now be used to give the base current as a function of the input voltage. The intersection of the base current curves with the AC load line give the corresponding values of the collector current and collector-emitter voltage (points 3, Q and 4 on the plot): t [rad] 0,,2 /2 3/2 vi [mV] 0 +10 -10 iB [A] 19.9 29.23 13.54 iC [mA] 1.99 2.92 1.35 vCE [V] 7.61 4.55 9.15 vo [V] 0 -3.06 +1.54
The capacitor connecting the load to the circuit is a DC blocking [and AC coupling] capacitor. Remember that for no load conditions, the load resistor is assumed to be open. Again using superposition to sum only AC voltages around the loop and modeling the capacitors as shorts:
- ve + vce + vc + vo = 0
ve = vc = 0 vo = vce = vCE - V CEQ
The no-load voltage gain can be determined using the maximum excursion in the input and output voltages:
Avo =
vo (- 1.54) - (+ 3.06) = -230 = vi 10 10 -3 - - 10 10 -3
When the input voltage becomes more positive, the output voltage becomes more negative and vice versa. This is called "phase inversion" and is why the gain is negative. In normal amplification this is not important. If the loading effect of the load resistance is included, the AC load line will be steeper and the gain will be reduced. b) A plot is a large number of calculated points plotted and connected by a smooth curve. In a sketch, only the most significant points [maxima, minima, intercepts, etc] are calculated, plotted, and connected by a carefully drawn smooth curve. c) The output waveform is very distorted, i.e., it is not a true linearly amplified copy of the input waveform. This is most obvious when the positive and negative peaks of the two curves are compared: Input voltage +10 mV -10 mV Output voltage -3.06 V +1.94 V The two peaks of the input voltage are equal; however, the positive peak is amplified less than the negative peak. The nonlinear (exponential) behavior of the transistor causes the amplification to be nonlinear and the output waveform to be very distorted. This is characteristic of large signal amplifiers. Even more serious distortion could occur if the input voltage (also called the "excitation" or "drive") were increased so that saturation or cutoff occurs. This causes "clipping" and a severely distorted output. ______________________________________________________________________________________
(
) (
)
10.18
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Section 10.3: BJT Large-Signal Model
Problem 10.22 Solution:
Known quantities: For the circuit shown in Figure 10.24:
Voff = 0 V, Von = 5 V, I B = 5 mA, RB = 1 k, 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 - VCEsat I LED
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 I LED max = 47
VCC - V LED - VCEsat
Therefore, RC [47, 340] ______________________________________________________________________________________
Problem 10.23 Solution:
Known quantities: For the circuit shown in Figure 10.26:
VD = 1.1 V, RB = 33 k, VCC = 12 V, VBE = 0.75 V, VCEQ = 6 V, = 188.5, RS = 500
Find: The resistance RC. Analysis: The current through the resistance RB is given by
IB =
VD - VBEQ RB
=
1.1 - 0.75 = 10.6 A 33000 6 - 1.1 = 9.8 mA 500
The current through RS is
IS =
VCEQ - VD RS
=
It follows that the current through the resistance RC is
I CQ = I B + I S = 11.8 mA RC = VCC - VCEQ I CQ =
Finally,
12 - 6 = 508.5 0.0118
______________________________________________________________________________________
10.19
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Problem 10.24 Solution:
Known quantities: For the circuit shown in Figure 10.24:
Voff = 0 V, Von = 5 V, I B max = 5 mA, RC = 340 , 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 RB. Analysis: If the BJT is in saturation
IC =
VCC - V LED - VCEsat RC Von - V IC / Von - V I B max =
= 10 mA
In order to guarantee that the BJT is in saturation
RB
5 - 0.7 = 40.85 k 0.01 95
RB
= 860
______________________________________________________________________________________
Problem 10.25 Solution:
Known quantities: For the circuit shown in Figure 10.24:
Voff = 0 V, Von = 5 V, I B max = 5 mA, RB = 10 k, RC = 340, VCC = 5 V, V = 0.7V, VCEsat = 0.2 V, V LED = 1.4 V, I LED 10 mA, Pmax = 100 mW
Find: Minimum value of that will ensure the correct operation of the LED. Analysis:
IB =
Von - V
min
RB I = LED min IB
=
4.3 = 0.43 mA 10000 0.01 = = 23.25 0.43 10 -3
______________________________________________________________________________________
Problem 10.26 Solution:
Known quantities: For the circuit shown in Figure 10.24:
Voff = 0 V, Von = 3.3 V, I B max = 5 mA, RB = 10 k, RC = 340, VCC = 5 V, V = 0.7V, VCEsat = 0.2 V,
V LED = 1.4 V, I LED 10mA, Pmax = 100mW
10.20
G. Rizzoni, Principles and Applications of Electrical Engineering Find: Minimum value of that will ensure the correct operation of the LED. Analysis:
Problem solutions, Chapter 10
IB =
Von - V
min
RB I = LED min IB
=
3.3 - 0.7 = 0.26 mA 10000 0.01 = = 38.5 0.26 10 -3
______________________________________________________________________________________
Problem 10.27 Solution:
Known quantities: For the circuit shown in Figure 10.24:
Voff = 0 V, Von = 5 V, I B max = 1 mA, RB = 1 k, R = 12 , 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:
IC =
min
VCC - VCEsat 13 - 1 = = 1A 12 R I 1 = C = = 1000 I B max 1 10 -3
______________________________________________________________________________
Problem 10.28 Solution:
Known quantities: For the circuit shown in Figure 10.24:
Voff = 0 V, Von = 5 V, I B max = 1 mA, = 2000, R = 12 , VCC = 13 V, V = 0.7V, VCEsat = 1 V, IC 1A
Find: The range of RB that will ensure the correct operation of the fuel injector.
10.21
G. Rizzoni, Principles and Applications of Electrical Engineering Analysis: If the BJT is in saturation
Problem solutions, Chapter 10
IC =
VCC - VCEsat = 1A R
Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation
RB
Von - V IC /
=
5 - 0.7 = 8.6 k 1 2000
RB
Von - V I B max
= 4.3 k
______________________________________________________________________________________
Problem 10.29 Solution:
Known quantities: For the circuit shown in Figure 10.24:
Voff = 0 V, Von = 3.3 V, I B max = 1 mA, = 2000, R = 12 , VCC = 13 V, V = 0.7V, VCEsat = 1 V, IC 1A
Find: The range of RB that will ensure the correct operation of the fuel injector. Analysis: If the BJT is in saturation
IC =
VCC - VCEsat = 1A R
Because this is the minimum value allowed for the current to drive the fuel injector, it is necessary to guarantee that the BJT is in saturation. In order to guarantee that the BJT is in saturation
RB
Von - V IC /
=
3.3 - 0.7 = 5.2 k 1 2000
RB
Von - V I B max
= 2.6 k
______________________________________________________________________________________
10.22
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Section 10.4: BJT Switches and Gates
Problem 10.30 Solution:
Known quantities: The circuit given in Figure P10.30. 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
vo 2
0 0 5V 5V
0 5V 0 5V
off off on on
off on off on
on off off off
0 5V 5V 5V
5V 0 0 0
______________________________________________________________________________________
Problem 10.31 Solution:
Known quantities: The circuit given in Figure P10.30. 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 10.30. This table clearly describes a NOR gate when the output is taken at vo 2 . ______________________________________________________________________________________
Problem 10.32 Solution:
Known quantities: The circuit given in Figure P10.32. Find: Show that the given circuit functions as an AND 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 .
10.23
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
v1
v2
Q1
Q2
Q3
vo1
vo 2
0 0 5V 5V
Problem 10.33 Solution:
0 5V 0 5V
off off on on
off on off on
on on on off
0 0 0 5V
5V 5V 5V 0
______________________________________________________________________________________
Known quantities: The circuit given in Figure P10.32. Find: Show that the given circuit functions as a NAND gate if the output is taken at v02. Analysis: See the state table constructed for Problem 10.32. This table clearly describes a NAND gate when the output is taken at vo 2 . ______________________________________________________________________________________
Problem 10.34 Solution:
Known quantities: In the circuit given in Figure P10.34 the minimum value of vin for a high input is 2.0 transistor Q1 has a of at least 10. Find: The range for resistor RB that can guarantee that the transistor is on. Analysis:
V. Assume that the
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 v - 0 .6 or . Substituting for (vin)min and (vin)max , we find the following range for RB: RB = in iB ic =
5.833 k RB 18.333k
______________________________________________________________________________________
Problem 10.35 Solution:
Known quantities: For the circuit given in Figure P10.35:
R1C = R2C = 10 k , R1B = R2 B = 27 k .
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.
10.24
G. Rizzoni, Principles and Applications of Electrical Engineering Analysis:
Problem solutions, Chapter 10
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.
a)
______________________________________________________________________________________
Problem 10.36 Solution:
Known quantities: For the inverter given in Figure P10.36:
RC1 = RC 2 = 2 k , RB = 5 k .
Find: The minimum values of 1 and 2 to ensure that Q1 and Q2 saturate when vin is high. Analysis:
5 - 0 .2 2.5 = 2.4 mA , therefore, ic = mA . Applying KVL: 2000 - 5 + RB i B1 + 0.6 + 0.6 + 0.6 = 0 + i or 0.64 1 = 1.2 + 2.5 Therefore, iB1= 0.64 mA. i E1 = 1 i B1 = 600 500 B 2 2 ic =
Choose 2 = 10 1 = 2.27. ______________________________________________________________________________________
Problem 10.37 Solution:
Known quantities: For the inverter given in Figure P10.36:
RC1 = 2.5 k ,
RC 2 = 2 k , 1 = 2 = 4 .
Find: Show that Q1 saturates when vin is high. Find a condition for Analysis:
RC 2 to ensure that Q2 also saturates.
3 .2 = 0.8 mA iC1 = 3.2 mA 4000 600 Applying KCL: + i B 2 = 3.2 i B 2 = 2 mA ; iC 2 = i B 2 = 8 mA 500 Applying KVL: 5 - 0.2 = 0.008 RC 2 RC 2 = 600 i B1 =
______________________________________________________________________________________
Problem 10.38 Solution:
Known quantities: The basic circuit of a TTL gate, shown in Figure P10.38. Find: The logic function performed by this circuit. Analysis: The circuit performs the function of a 2-input NAND gate. The analysis is similar to Example 10.8.
10.25
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
______________________________________________________________________________________
Problem 10.39 Solution:
Known quantities: The circuit diagram of a three-input TTL NAND gate, given in Figure P10.39. Find: vB1, vB2, vB3, vC2, and vout, assuming that all the input voltages are high. Analysis: Q2 and Q3 conduct, while Q4 is cutoff. vB1
= 1.8 V, vB2 = 1.2 V, vB3 = 0.6 V, and vC2 = vout = 0.2
V. ______________________________________________________________________________________
Problem 10.40 Solution:
Known quantities: Figure P10.40. Find: Show that when two or more emitter-follower outputs are connected to a common load, as shown in Figure P10.54, the OR operation results; that is, v0 = v1 OR v2. Analysis:
v2 L L H H L : Low; H : High.
v1 Q1 Q2 L L L L L L H H H H H H
v0 L H H H
______________________________________________________________________________________
10.26