Transcript
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: 100B
C
I
I .
Find:
The operating point and the state of the transistor.
Analysis:
V 6.0BEV and the BE junction is forward biased.
AVV
I
IIIIIVIV
BECC
B
BBCEEBEBCC
5.12911910
6.012
10191010820
101&91010820
3
3
mAIIBC
25.1
Writing KVL around the right-hand side of the circuit:
0 EECECCCC RIVRIV
V RIIRIVVEBCCCCCCE
1.8)910.0)(0125.025.1()2.2)(25.1(12
V VVVCEBEBC
5.71.86.0 : the BC junction is reverse biased
BECE VV
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 BEBCEC
IVIVP .
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:
mW IVIVPBEBCEC
97.46101.065.0109.595.7 33
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:
Applying KVL to the right-hand side of the circuit, A V
I BE
E480
30000
6.015
30000
15
Then, on the left-hand side, assuming >> 1:
V RIV
VRI
CCCB
CBCC
8.21015104801010
010
36
Chapter 10, Solution 6.
Known quantities:
The circuit of Figure P10.6, assuming the BJT has
V 6.0BEV 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
VIRIRVV
mAII
μARR
VVI
VVRR
RV
kΩRRR
EECCCCCE
BC
EB
BEBB
B
CCBB
B
857.7101515112001025.2330018
25.2
15)1(
5.3
078.12||
63
21
2
21
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
V6.0V .
Find:
The emitter current and the collector-base voltage.
Analysis:
Applying KVL to the right-hand side of the circuit,
0EBEEBB
VRIV
μAR
VVI
E
EBBB
E4.497
1039
6.0203
. Since 1 , μA4.497 EC II
VCC = 20V VBB = 20V
Applying KVL to the left-hand side: 0CCCCCB
VRIV
VVRIVCCCCCB
05.102010204.497 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 A 600 and A,200 A, 100 BI
b) VCE, assuming the maximum allowable
collector power dissipation is 0.5 W for
A 500 BI .
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 CCEBBECCE IVIVIVP , therefore:
VCE P
IC
0.5
79.5 103 6.29 V.
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: 030 111 BEBB VRI
μA07.3910750
7.03031
BI
mA907.311 BC II
V779.52.6907.33030 111 CCC IRV
V779.511 CCE VV .
b) Again, from KVL: 0779.5 222 EEBE RIV mA081.1107.4
7.0779.532
EI
and mA07.1101
100081.1
122
EC II .
Also, 0)(30 2222 CEECC VRRI V574.3)7.420()07.1(302 CEV .
Finally, V603.8)20()07.1(3030
22
22
C
C
CC V
R
VI .
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: 0500050 CEC VI .
Then, if 0CI , V50CEV ; and if 0CEV , mA10CI . The load line is shown superimposed on
the collector characteristic below:
The operating point is at the intersection of the
load line and the A 20BI line of the
characteristic. Therefore,
mA 5CQI and V 20CEQV .
Under these conditions, an A 5 increase in BI
yields an increase in CI of approximately
mA 156 . Therefore,
200105
1016
3
B
C
I
I
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
mA 9.111
1.012
CI
The base current is
IB IC
11.9
50 0.238 mA 238A
And since
mA 10
BEsatBBB
VVI
Therefore,
V V 98.26.0k10mA 238.0 BBV
Chapter 10, Solution 16.
Known quantities:
Collector characteristics of 2N3904 npn transistor; Transistor
circuits;
Find:
The operating point;
Analysis:
From KVL,
or
If 0CEV , mAk
IC 99.410
9.49 , and if 0CI , VVCE 9.49 . The load
line is shown superimposed on the collector characteristic below:
The operating point is at the intersection of the load line and the
AIB 20 line of the characteristic. Therefore, mAICQ 3 and
VVCEQ 8 .
Under these conditions, a A10 increase in BI yields an increase
in CI of approximately mAmAmA 235 . Therefore,
20010
2
A
mA
I
I
B
C
Addition of the emitter resistor effectively increased the current
gain by decreasing the magnitude of the slope of the load line.
0)20(5550 AIkVkI CCEC
9.491.05010 CCE kIV
Chapter 10, Solution 17.
Known quantities:
For the circuit shown in Figure 10.14 in the text:
mW100,mA10V,4.1
95,V,2.0V,7.0,V5,kΩ1,mA5,V5,V0
max
PIV
VVVRIVV
LEDLED
CEsatCCBBonoff
Find:
Range of RC.
Analysis:
34001.0
2.04.15
LED
CEsatLEDCCC
I
VVVR
From the maximum power
47
mA714.1
1.0
max
maxmax
LED
CEsatLEDCCC
LEDLED
I
VVVR
V
PI
Therefore, RC [47, 340]
Chapter 10, Solution 22.
Known quantities:
For the circuit shown in Figure 10.14 in the text:
A1
V,1V,7.0,V13,Ω12,kΩ1,mA1,V5,V0 max
C
CEsatCCBBonoff
I
VVVRRIVV
Find:
Minimum value of that will ensure the correct operation
of the fuel injector.
Analysis:
A112
113
R
VVI CEsatCCC
1000101
13
maxmin
B
C
I
I
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 mA4.401
EE II
.
Since the base-emitter junction voltage is assumed to be 0.6 V, then resistor R2 has a voltage:
V 56.06.52 VVV z , so the required value of R2 to be:
8.1230404.0
52
EI
VR
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,
59 CECC VV . 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 EI 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: V 11zV . That means:
V 11 ZB VV .
When the transistor is forward biased, according to KVL,
batteryBEBEZ VVRIV , where BER is the base resistance.
As the battery gets charged, the actual battery charging voltage batteryV will increase from 9.6 V to
10.4 V.
As batteryV increases gradually, ZV and V stay unchanged, then we can see that BEI will decrease
gradually.
So BEE II 1 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, CECC VV 11 . 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: V 12CCV 130 Ωk821 R Ωk222 R Ωk5.0ER
Ω16LR .
Find:
CEQV 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:
Ωk35.172282
2282Suppress
V538.22282
2212:VD
= = R + R
R R = R = R:V
= = R + R
R V = V = V = V
21
21eqBCC
21
2CCOCTHBB
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.
μA18.22
500113017350
7.0538.2
1 = =
R + + R
V - V = I
EB
BEQBBBQ
V55.105.0906.212
:KVL
mA906.21018.2211301 6
= = R I - V = V
0 = V + V - R I -
= + = I + = I
EEQCCCEQ
CCCEQEEQ
BQEQ
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: V12CCV 100 V 4EEV Ωk100BR
Ωk3CR Ωk3ER
Ωk6LR Ωk6.0SR mV )1028.6cos( 1 3tvS .
Find:
CEQV 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:
0 = R I] + [ + V + R I + V -
0 = R I + V + R I + V -
I] + [ = I [Si] 700 V
EBQBEQBBQBB
EEQBEQBBQBB
BQEQBEQ
1
:KVL
1mV
V 06.11
300010100.8273000109.818124
0 :KVL
A 0.82710189.8)1100()1(
A9.81810189.8)100(
A189.8)3000)(1100(100000
7.04
1
0 :KVL
)1(
][ mV 700
66
6
6
EEQCCQCCEECEQ
CCCCQCEQEEQEE
BQEQ
BQCQ
EB
BEQEEBQ
EEQBEQBBQEE
BQEQBQCQ
BEQ
RIRIVVV
VRIVRIV
II
II
RR
VVI
RIVRIV
IIII
SiV
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: V3Sv 100 Ωk60BR
Find:
a) The value of ER so that EI is 1 mA.
b) CR so that CV is 5 V.
c) The small-signal equivalent circuit of the amplifier
for k5LR
d) The voltage gain.
Analysis:
(a) With k 60BR and V 3BV , applying KVL, we have
EBBB RIRI )1(6.03
EB
RI
101k60
4.2
mARk
IE
E 110160
4.2101
Therefore,
k 81.1101
604.2101ER
(b) EECCCE RIRIV 15
From (a), we have mA 99.01
EC II
Therefore,
k 27.899.0
81.1515CR
(c) The small signal equivalent circuit is shown below
(d) iwB
SB
hR
VI
oeLCout
hRIv
1 Bfe
oe
outC Ih
h
VI
1
Since hoe is not given, we can reasonably assume that 1/hoe is very large. Therefore,
15.4100
ieB
L
s
outV
hR
R
v
vA
Chapter 10, Solution 36.
Known quantities:
For the circuit shown in Figure P10.36: Ωk200CR
Find:
e) The operating point of the transistor.
f) Voltage gain inout vv ; current gain inout ii
g) Input resistance ir
h) Output resistance or
Analysis:
(a) V 1.621
2
RR
RVV CCB
87.3749| | 21 RRRB
Assuming V 6.0BEV , we have
V 5.5 BEBEV VVV
kI
Vh
BQIB
BEie 6.60
100099.0
6.03
OUTvvS
+
-
IB
C
E
-
h ie
oeh
1
I C
R C
R L
+
BR B
I Bh fe
mA 22E
EE
R
VI
mA 088.01
b
II E
B
and
V 12.55.51021.912200-15
5.5
3-
.
CCCCECCE IRVVVV
(b) The AC equivalent circuit is shown on the
right:
kI
Vh
BQIB
BEie 82.6
10088.0
6.03
BCBEout IIIRv )1250(250)(
BieBoutieBin IhIvhIv 251250
Therefore, the voltage gain is
902.0in
outV
v
vA and
BCBout IIIi )1(
BBieBBB
inBin RIhII
R
vIi )251250(
and the current gain is
84.12)251250(
)1(
BBieBB
B
in
out
RIhII
I
i
i
(c) To find the input resistance we compute:
BieBin IhIv 251250
BBieBBin RIhIIi )251250(
Therefore. the input resistance is
3558in
ini
i
vr
(d) To find the output resistance we compute
BCBEout IIIRv )1250(250)(
BCBout IIIi )1(
Therefore, the output resistance is
250out
outo
i
vr
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 1ov .
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 2ov .
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:
mA4.22000
2.05
ci , therefore, iB = iC/ = 0.24 mA.
(vin)min = 2.0 V and (vin)max = 5.0 V, therefore, applying KVL: -vin +RB iB + 0.6 = 0
or B
inB
i
vR
6.0 . Substituting for (v
in)min and (v
in)max , we find the following range for R
B:
Ω333.18Ω833.5 kRk B
Chapter 10, Solution 46.
Known quantities:
For the circuit given in Figure P10.46:
Ωk27,kΩ10 2121 BBCC RRRR .
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.
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