CHAPTER 13
Exercises
E13.1
The emitter current is

v
iE = I ES exp BE
 VT

given by the Shockley equation:
 
 − 1
 
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 v BE , we have
 i
v BE ≅ VT ln E
 I ES
v BC = v BE − v CE

 10 −2 
 = 26 ln −14  = 718.4 mV

 10 

= 0.7184 − 5 = −4.2816 V
β
50
=
= 0.9804
β + 1 51
iC = αiE = 9.804 mA
α=
iB =
E13.2
β=
iC
β
= 196.1 µA
α
1−α
α
0.9
0.99
0.999
β
9
99
999
E13.3
iB = iE − iC = 0.5 mA
α = iC / iE = 0.95
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
431
β = iC / iB = 19
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.
432
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.
433
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
434
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 I C < βI B .
(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 B = CC
= 71.5 µA
I C = βI B = 3.575 mA
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 βI B > I C , 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
(a) I B = I C / β = 20 µA RB = CC
= 965 kΩ
IB
V − 0. 7
(b) I B = I C / β = 6.667 µA RB = CC
= 2.985 MΩ
IB
435
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
= 19.3 µA
I C = βI B = 0.965 mA
IB =
VCE
RB
= 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
RB
= 19.3 µA
IC =
20 − 0.2
RC
= 1.98 mA
Because we have βI B > I C , we are assured that the transistor operates
in the active region.
E13.13
VB = VCC
R2
= 5V
IB =
VB −VBE
R1 + R2
RB + ( β + 1)RE
I C = βI B
VCE = VCC − RC I C − RE (I C + I B )
β
IB
100
300
32.01
12.86
(µA)
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 β 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 β 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 β.
E13.14
RB =
I BQ
rπ =
1
= 3.333 kΩ
1 R1 + 1 R2
VB −VBE
=
= 14.13 µA
RB + ( β + 1)RE
βVT
I CQ
=
VB = VCC
R2
R1 + R2
= 5V
I CQ = βI BQ = 4.239 mA
300(26 mV )
= 1840 Ω
4.238 mA
436
1
= 666.7 Ω
1 RL + 1 RC
RL ′ =
Av = −
βRL ′
rπ
= −108.7
RL β
1
= −163.0
= 1186 Ω
Z in =
1 R1 + 1 R2 + 1 rπ
rπ
Z
G = Av Ai = 7004
Ai = Av in = −64.43
RL
Z o = RC = 1 kΩ
Z in
v o = Avv in = Avv s
= −76.46 sin(ωt )
Z in + Rs
Avoc =
E13.15
First, we determine the bias point:
R2
1
= 50.00 kΩ
VB = VCC
= 10 V
RB =
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.
rπ =
βVT
=
300(26 mV )
= 1823 Ω
4.279 mA
I CQ
RL′ ( β + 1)
Av =
= 0.9910
rπ + ( β + 1)RL′
RL′ =
1
= 666.7 Ω
1 RL + 1 RE
RE (β + 1)
= 0.9970
rπ + (β + 1)RE
Z
1
= 40.10 kΩ Ai = Av in = 39.74
Z in =
1 RB + 1 [rπ + ( β + 1)RL′]
RL
Avoc =
Rs′ =
G = Av Ai = 39.38
Zo =
1
( β + 1) + 1 = 33.18 Ω
R ′ + r RE
s
π
Problems
P13.1
437
1
= 8.333 kΩ
1 RB + 1 Rs
P13.2
The sketch should resemble Figure 13.1a in the book. In normal
operation, current flows into the base, into the collector, and out of the
emitter.
P13.3
To forward bias a pn junction, the p-side of the junction should be
connected to the positive voltage. In normal operation of a BJT, the
emitter-base junction is forward biased and the collector-base junction
is reverse biased.
P13.4
The emitter current is

v
iE = I ES exp BE
 VT

P13.5
For a BJT, the parameters are defined as:
given by:
 
 − 1
 
iC
iE
i
α
β= C =
iB 1 − α
α=
It is assumed that the base-emitter junction is forward biased and that
the collector-base junction is reverse biased.
P13.6*
iE = iC + iB
= 9 + 0.3 = 9.3 mA
i
9
α= C =
= 0.9677
iE 9.3
β=
P13.7*
iC
9
=
= 30
iB 0.3
The emitter current is

v
iE = I ES exp BE
 VT

given by the Shockley equation:
 
 − 1
 
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 v BE , we have
438
 i
v BE ≅ VT ln E
 I ES

 10 −2 
 = 26 ln −13  = 658.5 mV

 10 

v BC = v BE − v CE = 0.6585 − 10 = −9.341 V
β
100
α=
=
= 0.9901
β + 1 101
iC = αiE = 9.901 mA
iB =
P13.8
iC
β
= 99.01 µA
From Equation 13.9, we have
α
β=
1−α
Solving for α , we have
200
β
α=
=
= 0.9950
β + 1 201
P13.9
iC = βiB = 200 × (10 µA) = 2 mA
iE = iC + iB = 2.01 mA
P13.10
iB = iE − iC = 0.5 mA
i
10
α= C =
= 0.9524
iE 10.5
i
10
β= C =
= 20
iB 0.5
P13.11
Solving the Shockley equation, we have
I ES =
iE

 v BE  
 − 1
exp


 VT  
Also, we have VT = kT / q
Substituting values for 300 K, we obtain
VT = 25.875 mV
I ES = 8.500 × 10 −13 A
At 310 K, we have
VT = 26.74 mV
I ES = 3.794 × 10 −12 A
439
For the 10 K increase in temperature, we find that IES has increased by a
factor of 4.464.
P13.12
From the circuit, we can write:
15 − 7
15 − 0.7
iC =
= 0.8 mA
iB =
= 21.03 µA
10 kΩ
680 kΩ
Then, we have β =
P13.13
iC
= 38.04
iB

v  
We have iE = I ES exp BE  − 1
 VT  

Solving for the saturation current:
iE
10 × 10 −3
=
I ES =
= 2.030 × 10 −14
0
.
7
v




 −1
exp BE  − 1 exp
0
.
026


V
 T 
For iE = 1 mA , we have
 i 


10 −3
 = 0.6401 V
v BE ≅ VT ln E  = 0.026 ln
−14 
I
2.030 × 10
 ES 
For iE = 1 µA , we have
 i
v BE ≅ VT ln E
 I ES
P13.14




10 −6
 = 0.026 ln
−14

 2.030 × 10


 = 0.4605 V

Writing a current equation at the collector of Q1 , we have:
iC 1 + iB 1 + iB 2 = 1 mA
However, iB 2 = iB 1 and iC 1 = βiB 1 , so we have
βiB 1 + iB 1 + iB 1 = 1 mA
iB 1 = 9.804 µA
iB 2 = 9.804 µA
iC 1 = iC 2 = βiB 1 = 0.9804 mA
iE 1 = iE 2 = iB 1 + iC 1 = 0.9902 mA
As in the solution to Problem P13.7, we have:
 i 
 0.9902 × 10 −3 
 = 0.6583 V
v BE 1 = v BE 2 ≅ VT ln E  = 0.026 ln
10 −14


 I ES 
440
P13.15
The Shockley equation states:

v  
iE = I ES exp BE  − 1
 VT  

Thus, iE is proportional to I ES , and we can write:
iE 2 = 10iE 1
Because α and β are the same for both transistors, we also have:
iB 2 = 10iB 1 and iC 2 = 10iC 1
Writing a current equation at the collector of Q1 , we have
iC 1 + iB 1 + iB 2 = 1 mA
βiB 1 + iB 1 + 10iB 1 = 1 mA
iB 1 = 9.009 µA
iB 2 = 10iB 1 = 90.09 µA
iC 1 = βiB 1 = 0.9009 mA
iC 2 = βiB 2 = 9.009 mA
iE 1 = iB 1 + iC 1 = 0.9099 mA
iE 2 = 10iE 1 = 9.099 mA
As in the solution to Problem 13.13, we have
 i 
 0.9099 × 10 −3 
 = 0.6561 V
v BE 2 = v BE 1 ≅ VT ln E 1  = 0.026 ln
10 −14


 I ES 1 
P13.16* We have:
 v BE  
 − 1
V
T
 



v  
iE 2 = I ES 2 exp BE  − 1
 VT  

Adding the respective sides of these equations, we have:

v  
iEeq = iE 1 + iE 2 = (I ES 1 + I ES 2 )exp BE  − 1
 VT  

Thus, we have:
I ESeq = I ES 1 + I ES 2 = 2 × 10 −13 A

iE 1 = I ES 1 exp
Also, we have iB 1 = iB 2 , and we can write:
βeq =
iCeq
iBeq
=
iC 1 + iC 2 100iB 1 + 100iB 2
=
= 100
iB 1 + iB 2
iB 1 + iB 2
441
P13.17
We have iCeq = iC 1 + iC 2 = β 1iB 1 + β 2iE 1 = β 1iB 1 + β 2 ( β 1 + 1)iB 1 . Then we can write
β eq =
iCeq
iBeq
=
iC 1 + iC 2 β 1iB 1 + β 2 ( β 1 + 1)iB 1
=
= β 1 + β 2 ( β 1 + 1)
iB 1
iB 1
P13.18* At 180 o 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* We select a point on the output characteristics in the active region and
compute β = iC / iB . For example on the curve for iB = 20 µA, we have
iC = 8 mA in the active region. Thus, β = (8 mA)/(20 µA) = 400 . Then, we
have α = β /( β + 1) = 0.9975.
P13.20
In the active region (which is for v CE > 0.2 V ), we have iC = βiB . In
saturation, the current falls to zero. The sketches are:
442
P13.21
P13.22
Distortion occurs in BJT amplifiers because the input characteristic of
the BJT is curved (rather than straight) and because the output
characteristic curves are not uniformly spaced.
P13.23
The slope of the load line does not change as VCC changes.
443
P13.24* Following the approach of Example 13.2, we construct the load lines
shown. We estimate that VCE max = 18.4 V,VCEQ = 15.6 V, and VCE min = 12 V .
Thus, the voltage gain magnitude is Av = (18.4 − 12) 0.4 = 16
P13.25
Following the approach of Example 13.2, we construct the load lines
shown. From the input load line, we have I B min ≅ I BQ ≅ I B max ≅ 0 . Thus, we
have VCE max ≅ VCE min ≅ 20 . Thus, the voltage gain magnitude is:
444
Av = (20 − 20 ) 0.4 = 0
P13.26
The input load line is the same as in Problem P13.24. Thus, we again
have I B min = 2 µA, I BQ = 5.5 µA and I B max = 10 µA . The output load line is
shown below. Because VCEQ = VCE min , we know that the output waveform is
severely distorted. It is not appropriate to compute voltage gain for
such a severely distorted output waveform.
445
P13.27
P13.28*
446
P13.29* (a) and (b) The characteristics and the load line are:
(c) We are given iB (t ) = is (t ) = 10 + 5 sin(2000πt ) . From which we
determine that I B min = 5 µA, I BQ = 10 µA, and I B max = 15 µA. Then at the
intersections of the load line with the respective characteristics, we
determine that I C min = 0.5 mA, I CQ = 1.0 mA, and I C max = 1.5 mA and that
VCE min = −15 V,VCEQ ≅ −10 V, VCE max ≅ −5 V.
(d) The sketches of vCE(t) are:
(e)
We are given iB (t ) = is (t ) = 20 + 5 sin(2000πt ) . From which we
determine that I B min = 15 µA, I BQ = 20 µA, and I B max = 25 µA. Then at the
intersections of the load line with the respective characteristics, we
determine that I C min = 1.5 mA, I CQ = 2.0 mA, and I C max = 2.0 mA and that
VCE min = −5 V,VCEQ ≅ −0.2 V,VCE max ≅ −0.2 V. A sketch of vCE(t) is shown
above.
447
P13.30
α = iC / iE = 0.995
β = iC / iB = iC /(iE − iC ) = 199.
P13.31
The magnitude of v BE is decreased by about 2 mV for each o C increase in
temperature. Thus at 180 o C, we have:
v BE = −0.7 + 0.002(180 − 30 ) = −0.4 V
P13.32
We can write
β eq =
iCeq
iBeq
=
iE 2 ( β 2 + 1)iC 1 ( β 2 + 1) β 1iB 1
=
=
= β 1 ( β 2 + 1)
iB 1
iB 1
iB 1
P13.33
See Figure 13.16 in the text.
P13.34
See Figure 13.16 in the text.
P13.35
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.)
448
P13.37
(a)
(b)
(c)
Active region.
Cutoff region.
Cutoff region (because I B ≅ 0 forVBE = 0.4 V at room
temperature).
(d)
Saturation region (because I C is less than βI B ).
P13.38
P13.39
(a)
Cutoff region (because I B ≅ 0 forVBE = −0.3 V at room
(b)
temperature).
Saturation region (because I C is less than βI B ).
(c)
Active region.
(a)
I B = I E − I C = 0.5 mA
The transistor is in saturation because we have IC < β I B
Thus, we have I C = 1 mA, I E = 1.5 mA, andVCE = 0.2 V.
(b)
The transistor is in the active region because we have VCE > 0.2
and I B > 0. Thus, we have
I C = βI B = 10 mA, I E = 10.1 mA, andVCE = 5 V.
(c)
This pnp transistor is in the active region because we have
VCE < −0.2 and I B > 0. Thus, we have
I C = βI B = 2 mA, I E = 2.02 mA, andVCE = −4.6 V.
(d)
P13.40
This pnp transistor is in cutoff because both junctions are reverse
biased. Thus, we have I C = I B = I E = 0, andVCE = −3 V.
For the Darlington pair, we have VBEeq = VBE 1 +VBE 2 = 1.2 V.
For the Sziklai pair, we have VBEeq =VBE 1 = 0.6 V.
P13.41* 1. Assume operation in saturation, cutoff, or active region.
2. Use the corresponding equivalent circuit to solve for currents and
voltages.
3. 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.
449
P13.42
The fixed base bias circuit is:
It is unsuitable for mass production because the value of β varies by
typically a factor of 3:1 between BJTs of the same type. Consequently,
the bias point varies too much from one circuit to another.
P13.43
See Figure 13.22a in the text.
P13.44* The results are given in the table:
IC
Region of
β
operation
Circuit
(mA)
(a)
(a)
(b)
(b)
(c)
(c)
(d)
(d)
P13.45
100
300
100
300
100
300
100
300
active
saturation
active
saturation
cutoff
cutoff
active
saturation
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
The results are given in the table:
Region of
I
V
β
operation
Circuit
(mA) (volts)
(a)
100 active
2.38
5.25
(a)
300 saturation
4.45
9.80
(b)
100 cutoff
0
10
(b)
300 cutoff
0
10
(c)
100 active
4.26 -10.74
(c)
300 active
4.29 -10.71
9.53 9.53
(d)
100 Q1 active
Q2 active
(d)
300 Q1 active
14.8
14.8
Q2 saturated
450
P13.46 VB = VCC
IB =
R2
R1 + R2
VB −VBE
= 5V
RB =
1
= 66.67 kΩ
1 / R1 + 1 / R2
= 2.071 µA
RB + ( β + 1)RE
I C = βI B = 0.4141 mA
VCE = VCC − RC I C − RE (I C + I B ) = 6.697 V
P13.47* The BJT operates in the active region. We can write the voltage
equation:
5 = RB I B + 0.7 + RE I E
However, we can substitute using the relations:
I
β +1
IE =
I C and I B = C
β
β
Thus, we have:
I
β +1
IC
4.3 = RB C + RE
β
β
For β = 100, we require I C = 4 mA. Furthermore, for β = 300, we require
I C = 5 mA. Thus, we have two equations:
4.3 = RB (0.04 × 10 −3 ) + RE (4.04 × 10 −3 )
4.3 = RB (0.01667 × 10 −3 ) + RE (5.017 × 10 −3 )
Solving, we find:
RB = 31.5 kΩ and RE = 753 Ω
P13.48
It turns out that for the values given the BJT operates in the saturation
region and the equivalent circuit is:
451
Enclosing the transistor in a closed surface (supernode) and writing a KCL
equation, we have
VE VE + 0.2 − 15 VE + 0.7 − 15 VE + 0.7
+
+
+
=0
RE
RC
R1
R2
Substituting the resistance values and solving we obtain VE = 7.533 V.
Then we have IC =
15 − (VE + 0.2)
RC
= 0.7267 mA and
I B = I E − I C = 0.0266 mA. As a check we note that we have βI B > I C as
required for operation in the saturation region.
P13.49* From Equations 13.20 through 13.23, we have:
I C = βI B = β
RB =
VB −VBE
RB + ( β + 1)RE
1
1 R1 + 1 R2
VB = VCC
R2
R1 + R2
Maximum I C occurs for β = βmax = 200, RE = RE min = 0.95 × 4.7 kΩ =
4465 Ω, R1 = R1 min = 0.95 × 100 kΩ = 95 kΩ, and R2 = R2 max = 1.05 × 47 kΩ =
49.35 kΩ. With these values, we calculate:
RB = 32.48 kΩ
VB = 5.128 V
I C max = 0.952 mA
Minimum I C occurs for β = βmin = 50, RE = RE max = 1.05 × 4.7 kΩ =
4.935 kΩ, R1 = R1 max = 1.05 × 100 kΩ = 105 kΩ, and R2 = R2 min =
0.95 × 47 kΩ = 44.65 kΩ. With these values, we calculate:
RB = 31.33 kΩ
VB = 4.475 V
I C min = 0.6667 mA
P13.50
15 −VBE 1
= 10 µA IC 1 = β I B 1 = 1 mA
1.43 × 10 6
15 = 10 4 (I C 1 + I B 2 ) +VBE 2 + 10 3 I E 2 = 10 4 (10 −3 + I B 2 ) + 0.7 + 10 3 (101I B 2 )
I B 2 = 38.74 µA I C 2 = βI B 2 = 3.874 mA I E 2 = (β + 1)I B 2 = 3.913 mA
IB 1 =
VCE 1 = 10 3 I E 2 +VBE 2 = 4.613 V
VCE 2 = 15 − 10 3 I C 2 − 10 3 I E 2 = 7.213 V
452
P13.51
The circuit is:
We have:
I
15 + 0.7
= 157 µA
I B = C = 20 µA
β
100 kΩ
I 1 = I 2 + I B = 177 µA
V −VBE
5 − 0. 7
R1 = CE
=
= 24.3 kΩ
I1
177 × 10 −6
I = I C + I 1 = 2.177 mA
15 −VCE
RC =
= 4.59 kΩ
I2 =
I
P13.52
15 + 0.7
= 0.104 mA
150 kΩ
15 = 4700(I C + I 2 + I B ) + 47 × 10 3 (I 2 + I B ) + 0.7
I2 =
Substituting I C = βI B and solving for I B , we obtain:
453
15 − 0.7 − (51700 )I 2
= 8.96 µA
4700( β + 1) + 51700
I C = βI B = 1.79 mA
VCE = 15 − 4700(I C + I 2 + I B ) = 6.06 V
IB =
P13.53
See Figure 13.26 in the text.
P13.54
rπ =
P13.55
rπ =
βVT
I CQ
βVT
I CQ
ICQ
1 µA
0.1 mA
1 mA
rπ
2.6 MΩ
26 kΩ
2.6 kΩ
P13.56* We use the same approach as in Section 13.7. We can write
I BQ
2
(t )
iB (t ) = 10 −5vBE
−5
+ ib (t ) = 10 [VBEQ + v be (t )]2
2
2
I BQ + ib (t ) = 10 −5VBEQ
+ 2 × 10 −5VBEQv be (t ) + 10 −5v be
(t )
(1)
2
However for the Q-point, we have IBQ = 10 − 5VBEQ
. Therefore, the first
term on each side of Equation (1) can be canceled. Furthermore, the last
term on the right hand side of Equation (1) is negligible for small signals
[i.e., v be (t ) much less than VBEQ at all times]. Thus, Equation (1) becomes
ib (t ) ≅ 2 × 10 −5VBEQv be (t )
We have defined
v (t )
rπ = be
ib (t )
and we have
rπ =
1
5 × 10 4
5 × 10 4
1581
=
=
=
−5
5
5
2 × 10 VBEQ
I CQ
10 I BQ
10 I CQ / 100
For ICQ = 1 mA, we obtain rπ = 50 kΩ.
454
P13.57
The equivalent circuit is:
rπeq =
P13.58
v beeq
ibeq
=
rπ 1ib 1 + rπ 2 (ib 1 + β 1ib 1 )
= rπ 1 + rπ 2 (1 + β 1 )
ib 1
The equivalent circuit is:
rπeq =
v beeq
ibeq
=
rπ 1ib 1
= rπ 1
ib 1
P13.59
Coupling capacitors are used to provide a path for ac signals while
maintaining an open circuit for dc. In connecting the signal source or
load, coupling capacitors prevent the source or load from affecting the
bias point of the amplifier. When it is desired to amplify dc signals,
coupling capacitors should not be used.
P13.60
The common-emitter amplifier is inverting. Both the voltage gain and
current gain magnitudes are potenitally greater than unity.
455
P13.61
See Figure 13.27a in the text.
P13.62
The dc circuit is:
The bias point calculations are:
RB =
1
= 3.197 kΩ
1 R1 + 1 R2
I BQ =
rπ =
VB −VBE
RB + ( β + 1)RE
βVT
I CQ
=
VB = VCC
= 39.3 µA
R2
R1 + R2
= 4.796 V
I CQ = βI BQ = 3.93 mA
100(26 mV )
= 662 Ω
3.93 mA
Then, we compute the amplifier performance:
RL′ =
1
= 500 Ω
1 RL + 1 RC
RC β
= −151.0
rπ
Z
Ai = Av in = −41.4
RL
Z o = RC = 1 kΩ
Avoc =
Av = −
Z in =
βRL′
rπ
= −75.5
1
1 R1 + 1 R2 + 1 rπ
= 548 Ω
G = Av Ai = 3124
P13.63* The solution is similar to that of Problem P13.62. The results are:
456
ICQ
rπ
Av
Avoc
Z in
Ai
G
Zo
P13.64
High impedance amplifier Low impedance amplifier
(Problem 13.63)
(Problem 13.62)
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
(a) The small-signal equivalent circuit is:
in which we have defined RL′ =
1
1 / RC + 1 / RL
(b) From the equivalent circuit we can write:
vin = rπib + ( β + 1)RE ib
(1)
v o = − βRL′ib
Then we have
Av =
− βRL′
βRL′ib
vo
=
=
v in rπ ib + ( β + 1)RE ib rπ + ( β + 1)RE
(c) From the equivalent circuit we can write:
iin =
vin
+i
RB b
Then solving Equation (1) for ib and substituting, we have
vin
vin
+
RB rπ + ( β + 1)RE
v
1
Rin = in =
iin 1 / RB + 1 /[rπ + ( β + 1)RE ]
iin =
457
(d) Because the coupling capacitors are open circuits for dc, we can
ignore the signal source and the load in the dc analysis. To attain ICQ = 1
mA, we must have IBQ = (1 mA)/β = 50 µA. Writing a voltage equation we
have
VCC = RB I BQ +VBEQ + ( β + 1)I BQ RE
Substituting values and solving, we obtain RB = 375.9 kΩ.
(e) First, we have rπ = βVt / ICQ = (100 × 0.026) / 0.005 = 520 Ω . Then
using the equations from parts (a) and (b) we determine that Av = -9.416
and Rin = 10.33 kΩ.
P13.65
See Figure 13.30a in the text.
P13.66
The voltage gain of an emitter follower is positive and less than unity in
magnitude. The current gain and power gains are potentially greater than
unity.
P13.67* The dc circuit is:
The bias point calculations are:
1
= 5 kΩ
RB =
1 R1 + 1 R2
VB = VCC
I BQ =
I CQ
R2
= 7. 5 V
R1 + R2
VB −VBE
= 64.1 µA
RB + ( β + 1)RE
= βI BQ = 6.41 mA
Now we can compute rπ and the ac performance.
rπ =
βVT
I CQ
=
100(26 mV )
= 405 Ω
6.41 mA
458
1
= 333 Ω
1 RL + 1 RE
RL′ ( β + 1)
Av =
= 0.98
r + ( β + 1)R ′
RL′ =
L
π
Avoc =
Z in =
RE (β + 1)
= 0.996
rπ + (β + 1)RE
1
= 4.36 kΩ
1 RB + 1 [rπ + ( β + 1)RL′ ]
Ai = Av
Zo =
Z in
= 8.61
RL
G = Av Ai = 8.51
1
1
in which Rs′ =
= 833 Ω
( β + 1) + 1
1 RB + 1 Rs
Rs′ + rπ RE
Z o = 12.1 Ω
P13.68
The solution is similar to that of Problem P13.67. The results are:
ICQ
rπ
Av
Avoc
Z in
Ai
G
Zo
High impedance amplifier Low impedance amplifier
(Problem 13.68)
(Problem 13.67)
64.1 µA
6.41 mA
40.5 kΩ
405 Ω
0.988
0.988
0.996
0.996
436 kΩ
4.36 kΩ
8.61
8.61
8.51
8.51
1210 Ω
12.1 Ω
459