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G. Rizzoni, 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
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.
______________________________________________________________________________________
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
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
b)
saturation region.
= - 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.
β=
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
=
= 13.9µA
I B = CC
R1
820
I C = β ⋅ I B = 1.39 mA
Writing KVL around the right-hand side of the circuit:
− VCC + I C RC + VCE + I E RE = 0
Ÿ
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
Ÿ The transistor is in the active region.
______________________________________________________________________________________
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
P = VCE I C + VBE I B .
Analysis:
a) VCE = VCB
- VEB = 7.3 - 0.65 = 6.65 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
≈ VCEIC = 6.65×5.9×10-3 = 39
mW.
______________________________________________________________________________________
Problem 10.5
Solution:
Known quantities:
The circuit of Figure P10.5, assuming the BJT has
Vγ = 0.6 V .
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
β >> 1 :
(0.6 + 15) ) = −520µA
§ V + 15 ·
I E = −¨ BE
¸=−
30000
© 30000 ¹
VCB = 10 − I C ⋅ RC
− 10 + I C RC + VCB = 0
Ÿ
= 10 − (−520) ⋅ (15) ⋅ 10 −3
= 17.8 V
______________________________________________________________________________________
Problem 10.6
Solution:
Known quantities:
The circuit of Figure P10.6, assuming the BJT has
VBE = 0.6 V and β =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
= 3.3kΩ, RE = 1.2kΩ, R1 = 62kΩ, R2 = 15kΩ, VCC = 18 V
By applying Thevenin’s theorem from base and mass, we have
RB = R1 || R2 = 12.078 kΩ
VBB =
IB =
R2
VCC ≅ 3.5 V
R1 + R2
VBB − VBE
≅ 15 µA
RB + RE (1 + β )
I C = β 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
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
Vγ = 0.6 V .
Find:
The emitter current and the collector-base voltage.
Analysis:
Applying KVL to the right-hand side of the circuit,
− VCC + I E RE + VEB = 0
VCC − VEB 20 − 0.6
=
= 497.4µA . Since β >> 1 , I C ≈ I E = 497.4µA
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
______________________________________________________________________________________
Problem 10.8
Solution:
Known quantities:
The circuit of Figure P10.7, assuming the emitter resistor is changed to 22 kΩ and the BJT has
Vγ = 0.6 V .
Find:
The operating point of the transistor.
Analysis:
IE =
VCC − VEB 20 − 0.6
=
= 881.8µ A , I C ≈ I E = 881.8µA
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
______________________________________________________________________________________
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 µA, 200 µA, and 600 µA
b) VCE, assuming the maximum allowable collector power dissipation is 0.5 W for IB
= 500 µA.
Analysis:
= 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.
-3
b) For IB = 500 µA, and if we consider an average β from a., we have IC = 159·500 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
a)
For IB
______________________________________________________________________________________
Problem 10.10
Solution:
Known quantities:
Figure P10.10, assuming both transistors are silicon-based with
Find:
a) IC1,
b) IC2,
β = 100 .
VC1, VCE1.
VC2, VCE2.
Analysis:
a)
From KVL:
− 30 + I B1 RB1 + VBE1 = 0
Ÿ
I B1 =
I C1 = β ⋅ I B1 = 3.907 mA
Ÿ
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
− 5.779 + VBE 2 + I E 2 RE 2 = 0 Ÿ
§ β ·
§ 100 ·
¸¸ = 1.081 ⋅ ¨
= I E 2 ¨¨
¸ = 1.07 mA .
© 101 ¹
© β +1¹
10.6
30 − 0.7
= 39.07µA
750 ⋅ 10 3
I E2 =
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 Ÿ VCE 2 = 30 − (1.07) ⋅ (20 + 4.7) = 3.574 V .
Finally, I C 2
=
30 − VC 2
RC 2
Ÿ VC 2 = 30 − (1.07) ⋅ (20) = 8.603 V .
______________________________________________________________________________________
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 β 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 = 20µA line of the characteristic.
I CQ ≈ 5 mA and VCEQ ≈ 20 V .
Under these conditions, an
5 µA increase in I B yields an increase in I C of approximately
6 − 5 = 1 mA . Therefore,
∆I C 1 ⋅ 10 −3
β≈
=
= 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
_____________________________________________________________________________________
Problem 10.12
Solution:
Known quantities:
The circuit shown in Figure P10.12. With reference to Figure 10.20, assume
Vγ = 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:
10 − 0.2
= 9.8 mA
RC
5 .7 − 0 .6
= Vγ = 0.6 V , therefore I B =
= 102µA
RB
VCE = Vsat = 0.2 V , therefore I C =
VBE
I C 9.8 ⋅10 −3
=
= 96.08 << β
I B 102 ⋅ 10 −6
______________________________________________________________________________________
Problem 10.13
Solution:
Known quantities:
For the circuit shown in Figure P10.13,
Find:
a) VB
b)
c)
IB
IE
Analysis:
a) VEB = VE
VE = 1V and Vγ = 0.6 V .
d)
IC
e) β
f) α .
− VB = Vγ = 0.6 V Ÿ 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
=
= 20µA
RB 20 ⋅ 10 3
5 − VE 5 − 1
c) I E =
=
= 800µA
RE
5000
d) I C = I E − I B = 800 − 20 = 780µA
b)
e)
f)
IB =
I C 780
=
= 39
IB
20
I
780
α= C =
= 0.975 .
I E 800
β=
______________________________________________________________________________________
Problem 10.14
Solution:
Known quantities:
For the circuit shown in Figure P10.14:
VCC = 20 V
β = 130
R1 = 1.8 MΩ
RL = 1kΩ
RS = 0.6 kΩ
vS = cos 6.28 ×103 ⋅ t mV .
[
R2 = 300 kΩ
]
Find:
The Thèvenin equivalent of the part of the circuit containing
terminals of
RC = 3 kΩ
RE = 1kΩ
R1 , R2 , and VCC with respect to the
R2 . Redraw the schematic using the Thèvenin equivalent.
Analysis:
Extracting the part of the circuit specified, the Thèvenin 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:
______________________________________________________________________________________
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 kΩ
β = 100
R1 = 68 kΩ
RS = 0.9 kΩ
[
R2 = 11.7 kΩ RC = 200Ω
3
]
RE = 200Ω
vS = cos 6.28 ×10 ⋅ t mV .
Find:
VCEQ and the region of operation.
Analysis:
Simplify the circuit by obtaining the Thèvenin equivalent
of the biasing network in the base circuit:
15 ⋅11.7
V CC R 2
=
= 2.202 V
68 + 11.7
R1 + R2
R1 R 2 = 68 ⋅11.7 = 9.982 kΩ
R B = Req =
R1 + R 2 68 + 11.7
VD : V BB = V TH =
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 CQ = β ⋅ I BQ
I EQ = (β + 1) I BQ
Then:
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 Ÿ - V BB + I BQ RB +V BEQ+ [ β +1 ] I BQ RE = 0
2.202- 0.7
V BB - V BEQ
= 49.76 µA
=
I BQ =
RB + [ β +1 ] RE 9982+ (100 + 1)(200)
−6
I CQ = β I BQ = 100 ⋅ 49.76 ⋅10 = 4.976 mA
−6
I EQ = (β + 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 β
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 β is
invalid). Therefore, amplification is impossible.
______________________________________________________________________________________
2.
Problem 10.16
Solution:
Known quantities:
For the circuit shown in Figure P10.14:
VCC = 15 V
β = 100
R1 = 68 kΩ
RL = 1.5 kΩ
RS = 0.9 kΩ
vS = cos 6.28 ×103 ⋅ t mV .
[
R2 = 11.7 kΩ RC = 4 kΩ
]
Find:
VCEQ and the region of operation.
Analysis:
10.11
RE = 200Ω
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Simplify the circuit by obtaining the Thèvenin equivalent of the biasing network in the base circuit:
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 CQ = β ⋅ I BQ
I EQ = (β + 1) I BQ
KVL : - V BB + I BQ R B + V BEQ + I EQ R E = 0 Ÿ - V BB + I BQ R B + V BEQ + [ β + 1 ] I BQ R E = 0
I BQ =
2.202 - 0.7
V BB - V BEQ
= 49.76 µA
=
9982 + (100 + 1)(200)
RB + [ β +1 ] RE
−6
I CQ = β I BQ = 100 ⋅ 49.76 ⋅10 = 4.976 mA
−6
I EQ = (β + 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
I CQ
=
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
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
β = 130
R1 = 82 kΩ
R2 = 22 kΩ
RE = 0.5 kΩ
RL = 16Ω .
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 R2
=
= 2.538 V
R1 + R2 82 + 22
R1 R2 = 82 ⋅ 22 = 17.35 kΩ
R B = Req =
82 + 22
R1 + R2
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 = [ β +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 + [ β + 1 ] I BQ R E = 0
I BQ =
2.538 − 0.7
V BB - V BEQ
=
= 22.18µA
17350 + (130 + 1) ⋅ 500
R B + (β + 1) ⋅ R E
−6
I EQ = (β + 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
VCC = 12 V
RL = 6 kΩ
VEE = 4 V
Rs = 0.6 kΩ
β = 100
[
RB = 100 kΩ
3
]
Problem solutions, Chapter 10
RC = 3 kΩ
RE = 3 kΩ
vS = cos 6.28 ×10 ⋅ 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 base-emitter junction is
forward biased and:
V BEQ ≈ 700 mV [Si]
I CQ = β ⋅ I BQ
I EQ = (β + 1) I BQ
KVL : - V EE + I BQ R B + V BEQ + I EQ R E = 0 Ÿ
4 - 0.7
V EE - V BEQ
= 8.189 µ A
=
Ÿ I BQ =
100000 + (100 + 1)(3000)
RB + [ β +1 ] RE
I CQ = β ⋅ I BQ = (100 ) ⋅ 8.189 ⋅10 −6 = 818.9 mA
I EQ = (β + 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
β = 130
RB = 325 kΩ RC = 1.9 kΩ
10.14
RE = 2.3 kΩ
G. Rizzoni, Principles and Applications of Electrical Engineering
RL = 10 kΩ
Rs = 0.5 kΩ
[
Problem solutions, Chapter 10
]
vS = 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 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 = β I BQ
I EQ = [ β + 1 ] I BQ
KCL : I BQ + I CQ - I RC = 0
I RC = I CQ + I BQ = β I BQ + I BQ = [ β + 1 ] I BQ
KVL : - I EQ R E - V BEQ - I BQ R B - I RC RC + V CC = 0
- [ β + 1 ] I BQ [ R E + RC ] - V BEQ - I BQ R B + V CC = 0
12 − 0.7
V CC - V BEQ
=
= 12.91µA
[325 + (130 + 1) ⋅ (2.3 + 1.9) ]⋅103
R B + [ β + 1 ] [ R E + RC ]
= I EQ = [ β + 1 ] I BQ = (130 + 1) ⋅12.91 ⋅10 −6 = 1.691mA
I BQ =
I RC
KVL : - I EQ R E - V CEQ - I RC RC + V CC = 0 Ÿ
Ÿ 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
C = 0.5µF
β = 170
RL = 1.7 kΩ
Rs = 70Ω
vS = cos 6.28 ×103 ⋅ t mV .
[
RB = 22 kΩ
]
10.15
RC = 3.3 kΩ
RE = 3.3 kΩ
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Find:
VCEQ and the region of operation.
Analysis:
When β is very high (say, greater than 50) the "high beta approximation" can be used, i.e.:
IC ≈ I E ≈ β I B
Using this approximation in this problem:
I BQ ≈ 12.5µA
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 = 710Ω
RL = 3 kΩ
•
•
C = 0.47µ F
R1 = 220 kΩ
R2 = 55 kΩ
Rs = 0.6 kΩ
vS = Vi 0 ⋅ sin (ωt )
RC = 3 kΩ
Vi 0 = 10 mV .
DC operating point:
I BQ = 19.9µ A
V CEQ = 7.61 V
Transfer characteristic and β of the npn silicon transistor:
v BE
V BEQ + vbe
IC ≈ I s eVT = I s e
VT
β = 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 Thèvenin equivalent of the biasing network in the base circuit.
VD : V BB = V TH =
15 ⋅ 55
V CC R2
=
= 3.00 V
R1 + R2 220 + 55
Suppress V CC :
3
R1 R2 = 220 ⋅ 55 ⋅10 = 44.0 kΩ
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 = β I BQ = 100 ⋅19.9 ⋅10 = 1.99 mA
−6
I EQ = [ β + 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 =
iE =
β +1
iC
β
V CC - vCE
V CC - vCE
=
β +1
3717
R E + RC
β
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:
v BE
iC = I s e V T = I s e
V BEQ + v be
VT
= [ Is e
V BEQ
VT
v be
v be
] eV T = I CQ eV T
v be
iC = I CQ vbe =
iB =
eV T
I BQ eV T
β
β
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
KVL : - vi + vc + vbe + ve = 0
iB =
vbe
I BQ eV T
Ÿ vbe = vi
vc = ve = 0
= [ 19.9 µA ]
Problem solutions, Chapter 10
vi
e 26 mV
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):
ωt [rad]
0,π,2π
π/2
3π/2
vi [mV]
0
+10
-10
iB [µ
µA]
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 Ÿ 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 kΩ, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, β = 95,
Vγ LED = 1.4 V, I LED ≥ 10 mA, Pmax = 100 mW
Find:
Range of RC.
Analysis:
RC =
VCC − Vγ LED − VCEsat
I LED
≤
5 − 1.4 − 0.2
= 340 Ω
0.01
From the maximum power
I LED max =
RC >
Pmax
0.1
=
= 71 mA
Vγ LED 1.4
VCC − Vγ LED − VCEsat
I LED max
= 47 Ω
Therefore, RC ∈[47, 340] Ω
______________________________________________________________________________________
Problem 10.23
Solution:
Known quantities:
For the circuit shown in Figure 10.26:
VD = 1.1 V, RB = 33 kΩ, VCC = 12 V, VBE = 0.75 V, VCEQ = 6 V, β = 188.5, RS = 500 Ω
Find:
The resistance RC.
Analysis:
The current through the resistance RB is given by
IB =
VD − VBEQ
RB
=
1.1 − 0.75
= 10.6 µA
33000
The current through RS is
IS =
VCEQ − VD
RS
=
6 − 1.1
= 9.8 mA
500
It follows that the current through the resistance RC is
I CQ = β I B + I S = 11.8 mA
Finally,
RC =
VCC − VCEQ
I CQ
=
12 − 6
= 508.5 Ω
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 Ω, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V, β = 95,
Vγ LED = 1.4 V, I LED ≥ 10 mA, Pmax = 100 mW
Find:
Range of RB.
Analysis:
If the BJT is in saturation
IC =
VCC − Vγ LED − VCEsat
RC
= 10 mA
In order to guarantee that the BJT is in saturation
RB ≤
RB ≥
Von − Vγ
IC / β
Von − Vγ
I B max
=
5 − 0.7
= 40.85 kΩ
0.01
95
= 860 Ω
______________________________________________________________________________________
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 kΩ, RC = 340Ω, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V,
Vγ LED = 1.4 V, I LED ≥ 10 mA, Pmax = 100 mW
Find:
Minimum value of β that will ensure the correct operation of the LED.
Analysis:
IB =
β min
Von − Vγ
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 kΩ, RC = 340Ω, VCC = 5 V, Vγ = 0.7V, VCEsat = 0.2 V,
Vγ LED = 1.4 V, I LED ≥ 10mA, Pmax = 100mW
10.20
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Find:
Minimum value of β that will ensure the correct operation of the LED.
Analysis:
IB =
β min
Von − Vγ
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 kΩ, R = 12 Ω, VCC = 13 V, Vγ = 0.7V, VCEsat = 1 V,
IC ≥ 1A
Find:
Minimum value of β that will ensure the correct operation of the fuel injector.
Analysis:
IC =
β 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, β = 2000, R = 12 Ω, VCC = 13 V, Vγ = 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
Problem solutions, Chapter 10
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 ≤
RB ≥
Von − Vγ
IC / β
Von − Vγ
I B max
=
5 − 0.7
= 8.6 kΩ
1
2000
= 4.3 kΩ
______________________________________________________________________________________
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, β = 2000, R = 12 Ω, VCC = 13 V, Vγ = 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 ≤
RB ≥
Von − Vγ
IC / β
Von − Vγ
I B max
=
3.3 − 0.7
= 5.2 kΩ
1
2000
= 2.6 kΩ
______________________________________________________________________________________
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
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
______________________________________________________________________________________
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
off
off
on
0
5V
0
5V
off
on
on
0
5V
5V
0
on
off
on
0
5V
5V
5V
on
on
off
5V
0
______________________________________________________________________________________
Problem 10.33
Solution:
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 β of at least 10.
V. Assume that the
Find:
The range for resistor RB that can guarantee that the transistor is on.
Analysis:
5 − 0 .2
= 2.4 mA , therefore, iB = iC/β = 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 kΩ ≤ RB ≤ 18.333k Ω
______________________________________________________________________________________
Problem 10.35
Solution:
Known quantities:
For the circuit given in Figure P10.35:
R1C = R2C = 10 kΩ , R1B = R2 B = 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.
10.24
G. Rizzoni, Principles and Applications of Electrical Engineering
Problem solutions, Chapter 10
Analysis:
vin is low Ÿ Q1 is cutoff Ÿ vB = 5 V Ÿ Q2 is in saturation Ÿ vout = low = 0.2 V.
b) vin is high Ÿ Q1 is in saturation Ÿ vB = 0.2 V Ÿ Q2 is cutoff Ÿ vout = high = 5 V.
a)
______________________________________________________________________________________
Problem 10.36
Solution:
Known quantities:
For the inverter given in Figure P10.36:
RC1 = RC 2 = 2 kΩ , RB = 5 kΩ .
Find:
The minimum values of β1 and β2 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
β
− 5 + RB i B1 + 0.6 + 0.6 + 0.6 = 0
+ i or 0.64 ⋅ β1 = 1.2 + 2.5
Therefore, iB1= 0.64 mA. i E1 = β1 ⋅ i B1 = 600
500 B 2
β2
ic =
Choose β2 = 10 Ÿ β1 = 2.27.
______________________________________________________________________________________
Problem 10.37
Solution:
Known quantities:
For the inverter given in Figure P10.36:
RC1 = 2.5 kΩ ,
Find:
Show that Q1 saturates when vin is high. Find a condition for
RC 2 = 2 kΩ , β1 = β 2 = 4 .
RC 2 to ensure that Q2 also saturates.
Analysis:
3 .2
= 0.8 mA Ÿ iC1 = 3.2 mA
4000
600
Applying KCL:
+ i B 2 = 3.2 Ÿ i B 2 = 2 mA ; iC 2 = β ⋅ i B 2 = 8 mA
500
Applying KVL: 5 − 0.2 = 0.008 ⋅ RC 2 Ÿ RC 2 = 600Ω
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
v1 Q1 Q2
v0
L
L
L
L
L
L
H H
L
H
H
L
L
H
H
H
H H
H
H
L : Low; H : High.
______________________________________________________________________________________
10.26
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