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CHAPTER 13
Solutions for Exercises
E13.1
The emitter current is given by the Shockley equation:

v  
iE  I ES exp  BE   1
 VT  

v 
For operation with iE  I ES , we have exp  BE   1 , and we can write
 VT 
v 
iE  I ES exp  BE 
 VT 
Solving for vBE , we have
E13.2
 i 
 10 2 
v BE VT ln E   26ln 14   718.4 mV
 10 
 I ES 
v BC  v BE  v CE  0.7184  5  4.2816 V
β
50
α

 0.9804
β  1 51
iC  αiE  9.804 mA
i
iB  C  196.1 μA
β
α
β
1α
β
α
0.9
0.99
0.999
9
99
999
E13.3
iB  iE  iC  0.5 mA
α  iC / iE  0.95
β  iC / iB  19
E13.4
The base current is given by Equation 13.8:

v  

 v
 
iB  (1  α)I ES exp  BE   1  1.961  10 16 exp  BE   1
 0.026  

 VT  

which can be plotted to obtain the input characteristic shown in Figure
13.6a. For the output characteristic, we have iC  βiB provided that
1
v CE  approximately 0.2 V. Forv CE  0.2 V, iC falls rapidly to zero at
v CE  0. The output characteristics are shown in Figure 13.6b.
E13.5
The load lines for v in  0.8 V and - 0.8 V are shown:
As shown on the output load line, we find
VCE max  9 V,VCEQ  5 V, andVCE min  1.0 V.
2
E13.6
The load lines for the new values are shown:
As shown on the output load line, we have
VCE max  9.8 V,VCEQ  7 V, andVCE min  3.0 V.
3
E13.7
Refer to the characteristics shown in Figure 13.7 in the book. Select a
point in the active region of the output characteristics. For example, we
could choose the point defined by v CE  6 V and iC  2.5 mA at which we
find iB  50 μA. Then we have β  iC / iB  50. (For many transistors the
value found for β depends slightly on the point selected.)
E13.8
(a) Writing a KVL equation around the input loop we have the equation for
the input load lines: 0.8  v in (t )  8000iB  v BE  0 The load lines are
shown:
Then we write a KCL equation for the output circuit:
9  3000iC  v CE
The resulting load line is:
From these load lines we find
4
I B max  48 μA, I BQ  24 μA, I Bmin  5 μA,
VCE max  1.8 V,VCEQ  5.3 V,VCE min  8.3 V
(b) Inspecting the load lines, we see that the maximum of vin corresponds
to IBmin which in turn corresponds to VCEmin. Because the maximum of vin
corresponds to minimum VCE, the amplifier is inverting. This may be a
little confusing because VCE takes on negative values, so the minimum
value has the largest magnitude.
E13.9
(a) Cutoff because we have VBE  0.5 V andVBC VBE VCE  4.5 V which is
less than 0.5 V.
(b) Saturation because we have IC  βIB .
(c) Active because we have I B  0 andVCE  0.2 V.
E13.10
(a) In this case ( β  50) the BJT operates in the active region. Thus the
equivalent circuit is shown in Figure 13.18d. We have
V  0.7
I C  βIB  3.575 mA
I B  CC
 71.5 μA
RB
VCE VCC  RC I C  11.43 V
Because we have VCE  0.2, we are justified in assuming that the
transistor operates in the active region.
(b) In this case ( β  250) ,the BJT operates in the saturation region.
Thus the equivalent circuit is shown in Figure 13.18c. We have
V  0.7
V  0.2
VCE  0.2 V I B  CC
 71.5 μA I C  CC
 14.8 mA
RB
RC
Because we have βIB  IC , we are justified in assuming that the
transistor operates in the active region.
E13.11
For the operating point to be in the middle of the load line, we want
V VCE
VCE VCC / 2  10 V and I C  CC
 2 mA . Then we have
RC
V  0. 7
 965 k
(a) I B  I C / β  20 μA RB  CC
IB
V  0.7
(b) IB  IC / β  6.667 μA RB  CC
 2.985 M
IB
5
E13.12
Notice that a pnp BJT appears in this circuit.
(a) For β  50, it turns out that the BJT operates in the active region.
20  0.7
IB 
 19.3 μA
I C  βI B  0.965 mA
RB
VCE  RC I C  20  10.35 V
(b) For β  250, it turns out that the BJT operates in the saturation
region.
VCE  0.2 V I B 
20  0.7
 19.3 μA
RB
IC 
20  0.2
 1.98 mA
RC
Because we have βIB  IC , we are assured that the transistor operates
in the active region.
E13.13
VB VCC
IC  βIB
R2
 5V
R1  R2
IB 
VB VBE
RB  ( β  1)RE
VCE VCC  RC I C  RE (I C  I B )
β
IB
100
300
32.01
12.86
(A)
I C (mA)
VCE (V)
3.201
3.858
8.566
7.271
For the larger values of R1 and R2 used in this Exercise, the ratio of the
collector currents for the two values of  is 1.205, whereas for the
smaller values of R1 and R2 used in Example 13.7, the ratio of the
collector currents for the two values of  is 1.0213. In general in the
four-resistor bias network smaller values for R1 and R2 lead to more
nearly constant collector currents with changes in .
E13.14
RB 
I BQ
rπ 
1
 3.333 k
1 R1  1 R2
VB VBE

 14.13 μA
RB   β  1RE
VB VCC
R2
 5V
R1  R2
I CQ  βI BQ  4.239 mA
βVT
30026 mV 

 1840 
I CQ
4.238 mA
6
βRL 
Av  
 108.7
rπ
1
 666.7 Ω
1 RL  1 RC
R β
Avo  L  163.0
rπ
Z
Ai  Av in  64.43
RL
RL  
Z o  RC  1 k
v o  Avv in  Avv s
E13.15
Z in 
1
 1186 
1 R1  1 R2  1 rπ
G  Av Ai  7004
Z in
 76.46 sin( ωt )
Z in  Rs
First, we determine the bias point:
R2
1
RB 
 50.00 k
VB VCC
 10 V
1 R1  1 R2
R1  R2
VB VBE
I BQ 
 14.26 μA I CQ  βI BQ  4.279 mA
RB   β  1RE
Now we can compute rπ and the ac performance.
βVT
30026 mV 

 1823 
I CQ
4.279 mA
RL  β  1
Av 
 0.9910
rπ   β  1RL
rπ 
1
 666.7 Ω
1 RL  1 RE
RE  β  1
Avo 
 0.9970
rπ   β  1RE
RL 
Z
1
 40.10 kΩ Ai  Av in  39.74
1 RB  1 [rπ   β  1RL]
RL
1
Rs 
 8.333 kΩ
G  Av Ai  39.38
1 RB  1 Rs
1
Zo 
 β  1  1  33.18 
RE
R  r
Z in 
s
π
Answers for Selected Problems
P13.6*
iE  9.3 mA
  0.9677
  30
7
P13.7*
v BE  658.5 mV
v BC  9.341 V
  0.9901
iC  9.901 mA
iB  99.01 A
P13.16* I ESeq  2  10 13 A
 eq  100
P13.18* At 180  C and iB  0.1 mA , the base-to-emitter voltage is approximately:
v BE  0.7  0.002180  30  0.4 V
P13.19*   400
  0.9975
P13.24* VCE max  18.4 V,VCEQ  15.6 V, and VCE min  12 V
Av  16
P13.28*
8
P13.29* (a) and (b)
(c)
I C min  0.5 mA, I CQ  1.0 mA, and I C max  1.5 mA
VCE min  15 V,VCEQ  10 V, VCE max  5 V
(d) and (e) The sketches of vCE(t) are:
P13.36* In the active region, the base-collector junction is reverse biased and
the base-emitter junction is forward biased.
In the saturation region, both junctions are forward biased.
In the cutoff region, both junctions are reverse biased. (Actually, cutoff
applies for slight forward bias of the base-emitter junction as well,
provided that the base current is negligible.)
9
P13.41* 1.
2.
Assume operation in saturation, cutoff, or active region.
Use the corresponding equivalent circuit to solve for currents and
voltages.
Check to see if the results are consistent with the assumption
made in step 1. If so, the circuit is solved. If not, repeat with a
different assumption.
3.
P13.44* The results are given in the table:
Circuit
β
Region of
operation
(a)
(a)
(b)
(b)
(c)
(c)
(d)
(d)
100
300
100
300
100
300
100
300
active
saturation
active
saturation
cutoff
cutoff
active
saturation
IC
(mA)
1.93
4.21
1.47
2.18
0
0
6.5
14.8
VCE
(volts)
10.9
0.2
5.00
0.2
15
15
8.5
0.2
P13.47* RB  31.5 k and RE  753 
P13.49* I C max  0.952 mA
I C min  0.6667 mA
P13.56* r 
1581
I CQ
For ICQ  1 mA, we obtain rπ  50 k.
10
P13.63*
ICQ
rπ
Av
Avoc
Z in
Ai
G
Zo
High impedance amplifier Low impedance amplifier
(Problem 13.57)
(Problem 13.56)
0.0393 mA
3.93 mA
66.2 kΩ
662 Ω
-75.5
-75.5
-151
-151
54.8 kΩ
548 Ω
-41.4
-41.4
3124
3124
100 kΩ
1 kΩ
P13.67* I CQ  βI BQ  6.41 mA
r  405 
Av  0.98
Avoc  0.996
Zin  4.36 k
Ai  8.61
G  8.51
Z o  12.1 
11