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Chapter 6 Exercise Problems EX6.1
(a) ( ) 2 0.7 2
650BB BE
BQB
V V onI A
Rμ
− −= = ⇒
Then ( )( )
( )( )100 2 0.20
5 0.2 15 2 CQ BQ
CEQ CC CQ C
I I A mAV V I R V
β μ= = == − = − =
(b) Now 0.20 7.69 /
0.026CQ
mT
Ig mA V
V= = =
and ( )( )100 0.026
13 0.20
T
CQ
Vr kIπβ= = = Ω
(c) We find
( )( ) 137.69 15 2.2613 650
v m CB
rA g R
r Rπ
π
⎛ ⎞= − ⎜ ⎟+⎝ ⎠
⎛ ⎞= − = −⎜ ⎟+⎝ ⎠
EX6.2 ( ) 0.92 0.7
100BB BE
BQB
V V onI
R− −= =
or 0.0022 BQI mA=
and ( )( )150 0.0022 0.33 CQ BQI I mAβ= = = (a)
( )( )
0.33 12.7 /0.026
150 0.02611.8
0.33
200 606 0.33
CQm
T
T
CQ
Ao
CQ
Ig mA V
V
Vr kI
Vr kI
πβ
= = =
= = = Ω
= = = Ω
(b)
( ) and o m o C sB
rv g v r R v v
r Rπ
π ππ
⎛ ⎞= − = ⋅⎜ ⎟+⎝ ⎠
so
( )
( ) ( )11.812.7 606 1511.8 100
which yields 19.6
ov m o C
B
v
v rA g r R
v r R
A
π
π π
⎛ ⎞= = − ⎜ ⎟+⎝ ⎠
⎛ ⎞= − ⎜ ⎟+⎝ ⎠= −
EX6.3 (a)
( ) 1.145 0.750
BB EBBQ
B
V V onI
R− −= =
or 0.0089 BQI mA=
( )( )Then 90 0.0089 0.801 CQ BQI I mAβ= = =
Now
( )( )
0.801 30.8 /0.026
90 0.0262.92
0.801
120 150 0.801
CQm
T
T
CQ
Ao
CQ
Ig mA V
V
Vr kIVr kI
πβ
= = =
= = = Ω
= = = Ω
(b) We have ( )o m o CV g V r Rπ=
and sB
rV V
r Rπ
ππ
⎛ ⎞= −⎜ ⎟+⎝ ⎠
so
( )
( ) ( )2.9230.8 150 2.52.92 50
which yields 4.18
ov m o C
s B
v
V rA g r R
V r R
A
π
π
⎛ ⎞= = − ⎜ ⎟+⎝ ⎠
⎛ ⎞= − ⎜ ⎟+⎝ ⎠= −
EX6.4 Using Figure 6.23 (a) For 0.2 ,CQI mA= 7.8 15 , 60 125,ie feh k h< < Ω < < 4 46.2 10 50 10 ,reh− −× < < × 5 13 oeh mhosμ< < (b) For 5 ,CQI mA= 0.7 1.1 , 140 210,ie feh k h< < Ω < < 4 41.05 10 1.6 10 ,reh− −× < < × 22 35 oeh mhosμ< < EX6.5
( )
1 2
2
1 2
35.2 || 5.83 5
5.83 55.83 35.2
TH
TH CC
R R R k
RV VR R
= = = Ω
⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
or 0.7105 THV V=
Then ( ) 0.7105 0.7
5TH BE
BQTH
V V onI
R− −= =
or 2.1 BQI Aμ=
and ( )( )
( )( )100 2.1 0.21
5 0.21 10CQ BQ
CEQ CC CQ C
I I A mAV V I R
β μ= = == − = −
and 2.9 CEQV V=
Now 0.21 8.08
0.026100 476 0.21
CQm
T
Ao
CQ
Ig mA
VVr kI
= = =
= = = Ω
And ( ) ( ) ( )8.08 476 10v m o CA g r R= − = −
so 79.1vA = −
EX6.6
( ) ( )
1 2
2
1 2
250 75 57.7
75 575 250
TH
TH CC
R R R k
RV VR R
= = = Ω
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
or
( )( ) ( )( )
1.154
1.154 0.71 57.7 121 0.6
TH
TH BEBQ
TH E
V VV V on
IR Rβ
=− −= =
+ + +
or
( )( )3.48
120 3.38 0.418 BQ
CQ BQ
I AI I A mA
μβ μ
== = =
(a) Now
( )( )
0 418 16 080 026
120 0 0267 46
0 418
CQm
T
T
CQ
I .g . mA / VV .
.Vr . kI .π
= = =
β= = = Ω
We have o m CV g V Rπ= −
We find ( ) ( )( )1 7.46 121 0.6ib ER r Rπ β= + + = +
or 80.1 ibR k= Ω
Also 1 2
1 2
250 75 57.7 57.7 80.1 33.54 ib
R R kR R R k
= = Ω= = Ω
We find
1 2
1 2
33.5433.54 0.5
ibs s s
ib S
R R RV V V
R R R R⎛ ⎞ ⎛ ⎞′ = ⋅ = ⋅⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
or ( )0.985s sV V′ =
Now
( )1 1211 1 0.67.46s EV V R V
rπ ππ
β⎡ ⎤⎛ ⎞+ ⎡ ⎤⎛ ⎞′ = + = +⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥ ⎣ ⎦⎝ ⎠⎣ ⎦
or ( ) ( )( )0.0932 0.0932 0.985s sV V Vπ ′= =
So
( )( )( )( )16.08 0.0932 0.985 5.6ov
s
VA
V= = −
or 8.27vA = −
EX6.7
( )
1 2
2
1 2
15 85 12.75
85 1215 85
TH
TH CC
R R R k
RV VR R
= = = Ω
⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
or 10.2 THV V=
Now ( ) ( )1CC BQ E EB BQ TH THV I R V on I R Vβ= + + + +
so ( ) ( ) ( )12 101 0.5 0.7 12.75 10.2BQ BQI I= + + +
which yields 0.0174 BQI mA=
and ( )( )
( )( )
( )( ) ( )( )
100 0.0174 1.74 101 0.0174 1.76
12 1.76 0.5 1.74 4
CQ BQ
EQ
ECQ CC EQ E CQ C
I I mAI mA
V V I R I R
β= = == == − −= − −
or 4.16 ECQV V=
Now ( )( )100 0.026
1.49 1.74
T
CQ
Vr kIπβ= = = Ω
We have ( )π
ππ π
π
=
⎛ ⎞= − − +⎜ ⎟
⎝ ⎠
o m C L
s m E
V g V R || R
VV V g V Rr
Solving for Vπ and noting that ,mg rπβ = we find
( )( )
( )( )( )( )
1
100 4 21.49 101 0.5
C Lov
s E
R RVAV r Rπ
ββ
−= =
+ +
−=
+
or 2.56vA = − EX6.8 Dc analysis
( )( )10 0.7 0.00439
100 101 200.439 , 0.443
BQ
CQ EQ
I mA
I mA I mA
−= =+
= =
Now ( )( )100 0.026
5.92 0.439
0.439 16.88 /0.026100 228
0.439
m
o
r k
g mA V
r k
π = = Ω
= =
= = Ω
(a) Now
( )
100 5.92 5.59
o m o C
Bs
B S
B
V g V r R
R rV V
R r RR r k
π
ππ
π
π
= −
⎛ ⎞= ⋅⎜ ⎟+⎝ ⎠
= = Ω
( )5.59Then 0.9185.59 0.5 s sV V Vπ
⎛ ⎞= ⋅ =⎜ ⎟+⎝ ⎠
( ) ( ) ( )Then 16.88 0.918 228 10ov
s
VA
V= = −
or 148vA = − (b) 0.5 100 5.92 6.09 in S BR R R r kπ= + = + = Ω
and 10 228 9.58 o C oR R r k= = = Ω
EX6.9 (a)
( ) ( )( )
0.25 9.615 /0.026
100 400 0.25
9.615 400 100 769
CQm
T
Ao
CQ
v m o c
Ig mA V
VVr kI
A g r r
= = =
= = = Ω
= − = − = −
(b) ( ) ( )( )9.615 400 100 100
427v m o c L
v
A g r r rA
= − = −= −
EX6.10
( )( )
( )( ) ( )( )
5 0.7 0.00672 10 126 50.84 , 0.847 10 0.84 2.3 0.847 5
BQ
CQ EQ
CEQ
I mA
I mA I mAV
−= =+
= == − −
or 3.83 CEQV V=
dc load line ( ) ( )CE C C EV V V I R R+ −≅ − − +
or ( )10 7.3CE CV I= −
ac load line (neglecting ro) ( ) ( ) ( )2.3 5 1.58ce c C L c cv i R R i i= − = − = −
1.37
0.84
3.83 10
IC(mA)
AC load line
DC load line
Q-point
VCE(V) EX6.11 (a) dc load line
( ) ( )12 4.5EC CC C C E CV V I R R I≅ − + = − ac load line
( )ec c E C Lv i R R R≅ − + or ( ) ( )0.5 4 2 1.83ec c cv i i= − + = − Q-point values
( )( )12 0.7 10.2 0.0150
12.75 121 0.51.80 , 1.82
3.89
BQ
CQ EQ
ECQ
I mA
I mA I mAV V
− −= =+
= ==
2.67
1.80
3.89 12
AC load line
DC load line
IC(mA)
Q-point
VEC(V) (b)
( )( )( )
1.80 1.8 1.83 3.29
min 3.89 3.29 0.6
C CQ
EC
EC
i I mAv V
v V
Δ = =Δ = =
= − =
So maximum symmetrical swing 2 3.29 6.58 V= × = peak-to-peak EX6.12
(a)
( )( ) ( )( )( )
2
1 2 1
1
0.1 1 0.1 121 1
TH CC TH CC
TH E
RV V R VR R R
R Rβ
⎛ ⎞= ⋅ = ⋅ ⋅⎜ ⎟+⎝ ⎠= + =
so
( ) ( )1
112.1 , 12.1 12= Ω =TH THR k VR
We can write ( ) ( )1CC BQ E EB BQ TH THV I R V on I R Vβ= + + + +
We have 1.61.6 , 0.0133 120CQ BQI mA I mA= = =
Then
( )( )
( )( )1
112 0.7 12.1 120.0133
12.1 121 1BQR
I− −
= =+
which yields 1 15.24 R k= Ω
Since 1 2 12.1 ,THR R R k= = Ω we find 2 58.7 R k= Ω Also ( )( ) ( )( )12 1.6 4 1.61 1 3.99 ECQV V= − − =
(b) ac load line ( )ec c C Lv i R R= − Want 0.1 1.6 0.1 1.5 c CQi I mAΔ = − = − = Also 3.99 0.5 3.49 ecv VΔ = − =
Now 3.49 2.327 1.5
ecC L
c
v k R Ri
Δ= = Ω =
Δ
So 4 2.327 LR k= Ω which yields 5.56 LR k= Ω EX6.13
( )
1 2
2
1 2
25 50 16.7
50 525 50
TH
TH CC
R R R k
RV VR R
= = = Ω
⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
or
( )( ) ( )( )
3.33
3.33 0.71 16.7 121 1
TH
TH BEBQ
TH E
V VV V on
IR Rβ
=− −= =
+ + +
( )( )
or0.0191
Also120 0.0191 2.29
BQ
CQ
I mA
I mA
=
= =
Now
( )( )
2.29 88.1 /0.026
120 0.0261.36
2.29
100 43.7 2.29
CQm
T
T
CQ
Ao
CQ
Ig mA V
V
Vr kI
Vr kI
πβ
= = =
= = = Ω
= = = Ω
(a)
( ) ( ) ( ) ( )
1 2
1 2
1 1.36 121 1 43.7
ibs s
ib S
ib E o
R R RV V
R R R RR r R rπ β
⎛ ⎞′ = ⋅⎜ ⎟+⎝ ⎠= + + = +
or 1 2120 and 16.7 ibR k R R k= Ω = Ω
Then 1 2 16.7 120 14.7 ibR R R k= = Ω
Now
( )14.7 0.96714.7 0.5s s sV V V⎛ ⎞′ = ⋅ =⎜ ⎟+⎝ ⎠
and
( ) 1o m E o E o
VV g V R r V R r
r rπ
π ππ π
β⎛ ⎞ ⎛ ⎞+= + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
We have s oV V Vπ′ = +
then ( )0.967
1 11 1
ss
E o E o
VVV
R r R rr r
π
π π
β β′
= =⎛ ⎞ ⎛ ⎞+ ++ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( )
( ) ( )( )
We then obtain
10.967
1
1
0.967 11
E oo
vs
E o
E o
E o
R rrV
AV
R rr
R rr R r
π
π
π
β
β
ββ
⎛ ⎞+⎜ ⎟⎝ ⎠= =
⎛ ⎞++ ⎜ ⎟⎝ ⎠
+=
+ +
Now 1 43.7 0.978 E oR r k= = Ω
Then ( )( )( )
( )( )0.967 121 0.978
0.9561.36 121 0.978vA = =
+
(b) ( )( )1ib E oR r R rπ β= + +
or ( )( )1.36 121 0.978 120 ibR k= + = Ω
EX6.14 For RS = 0, then
1o E or
R R r π
β=
+
Using the parameters from Exercise 4.13, we obtain 1.361 43.7121oR =
or 11.1 oR = Ω
EX6.15 (a) For 1.25 and 100, we find
1.26 and 0.0125 CQ
EQ BQ
I mAI mA I mA
β= =
= =
Now 10CEQ EQ EV I R= − Or
( )4 10 1.26 ER= − which yields
4.76 ER k= Ω Then
( )( ) ( )( )( )0.1 1 0.1 101 4.76TH ER Rβ= + = or
48.1 THR k= Ω We have
( ) ( )2
1 2 1
110 5 10 5TH THR
V RR R R
⎛ ⎞= − = ⋅ −⎜ ⎟+⎝ ⎠
or
( )1
1 481 5THVR
= −
We can write ( )
( )0.7 51
THBQ
TH E
VI
R Rβ− − −
=+ +
Or
( )
( )( )1
1 481 5 0.7 50.0125
48.1 101 4.76R
− − +=
+
which yields 1 65.8 R k= Ω
Since 1 2 48.1 ,R R k= Ω we obtain
2 178.8 R k= Ω (b)
( )( )100 0.0262.08
1.25
125 100 1.25
T
CQ
Ao
CQ
Vr kI
Vr kI
πβ= = = Ω
= = = Ω
We may note that ( )m m b bg V g I r Iπ π β= =
Also ( ) ( )
( ) ( )1
2.08 101 4.76 1 100π β= + +
= +ib E L oR r R R r
or 84.9 ibR k= Ω
Now
( )1E oo b
E o L
R rI I
R r Rβ
⎛ ⎞= +⎜ ⎟⎜ ⎟+⎝ ⎠
where
1 2
1 2b s
ib
R RI I
R R R⎛ ⎞
= ⋅⎜ ⎟⎜ ⎟+⎝ ⎠
We can then write
( ) 1 2
1 2
1E ooI
s E o L ib
R r R RIA
I R r R R R Rβ
⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
We have 4.76 100 4.54 E oR r k= = Ω
so
( )4.54 48.11014.54 1 48.1 84.9IA ⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
or AI = 29.9
(c) 2.084.76 1001 101o E o
rR R r π
β= =
+
or 20.5 oR = Ω EX6.16 (a)
( ) ( )
1 2
2
1 2
70 6 5.53
610 5 10 570 6
TH
TH
R R R k
RVR R
= = = Ω
⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
or 4.2105 THV V= −
We find ( )
( )( )1
4.2105 0.7 52.91
5.53 126 0.2BQI Aμ− − − −
= ⇒+
and ( )( )
( )1 1
1 1
125 2.91 0.364
1 0.368 CQ BQ
EQ BQ
I I A mA
I I mA
β μβ
= = =
= + =
At the collector of Q1, ( )
( )( )11
11 2
0.7 551
CCCQ
C E
VVI
R Rβ− − −−
= ++
or
( )( )( )11 0.7 55
0.3645 126 1.5
CC VV − − −−= +
which yields 1 2.99 CV V=
also ( )( )1 1 1 5 0.368 0.2 5E EQ EV I R= − = −
or 1 4 93 EV . V= −
Then ( )1 1 1 2 99 4 93 7 92 CEQ C EV V V . . . V= − = − − =
We find ( )1
2
0 7 54 86
1 5C
EQ
V .I . mA
.− − −
= =
and
( )2 1125 4 86 4 82
1 126CQ EQI I . . mAββ
⎛ ⎞ ⎛ ⎞= ⋅ = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠
We find 2 1 0 7 2 99 0 7 2 29 E CV V . . . . V= − = − =
and 2 25 5 2 29 2 71 CEQ EV V . . V= − = − =
(b) The small-signal transistor parameters are:
( )( )
( )( )
11
11
22
22
125 0.0268.93
0.364
0.364 14.0 /0.026125 0.026
0.674 4.82
4.82 185 /0.026
T
CQ
CQm
T
T
CQ
CQm
T
Vr kI
Ig mA V
V
Vr kII
g mA VV
π
π
β
β
= = = Ω
= = =
= = = Ω
= = =
We find ( ) ( )( )1 1 11 8 93 126 0 2ib ER r R . .π β= + + = + Or
1 34 1 ibR . k= Ω and
( )( )( )( )
2 2 210 674 126 1 5 10 165
ib E LR r R R. . k
π β= + += + = Ω
The small-signal equivalent circuit is:
Vs����
R1��R2
RE1 RE2 RL
RC1
�
�
�
�
Vo
V�1 r�1 gm1V�1 gm2V�2V�2 r�2
We can write
( ) ( )2 21o b E LV I R Rβ= + where
( )12 1 1
1 2
1 11
π
π π
⎛ ⎞= −⎜ ⎟+⎝ ⎠
= ⋅
Cb m
C ib
s
ib
RI g VR R
VV rR
Then
( ) ( ) 1 1 12
1 2 1
1 πβ
=
⎛ ⎞⎛ ⎞−= + ⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠
ov
s
C mE L
C ib ib
VAV
R g rR RR R R
so
( ) ( ) 5 125126 1 5 105 165 34 1
⎛ ⎞ ⎛ ⎞= − ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠vA .
.
or 17 7= −vA .
(c) 1 2 1 70 6 34 1 4 76 i ibR R R R . . k= = = Ω
and 2 12
0 676 51 51 126
Co E
r R .R R .π
β+⎛ ⎞ +⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠
or 43 7 oR .= Ω
Test Your Understanding Exercises TYU6.1
( )( )
0.25 9.62 /0.026120 0.026
12.5 0.25
150 600 0.25
CQm
T
T
CQ
Ao
CQ
Ig mA V
V
Vr kI
Vr kI
πβ
= = =
= = = Ω
= = = Ω
TYU6.2 75
200 A A
o CQCQ o
V Vr II r kΩ
= ⇒ = =
or ICQ = 0.375 mA
TYU6.3
As a first approximation, Cv
E
RA
R≅ −
Resulting gain is always smaller than this value. The effect of RS is very small. Set
10C
E
RR
=
Now ( )5 C C E CEQI R R V≅ + +
or ( )( )5 0 5 2 5C E. R R .= + +
which yields 5 C ER R k+ = Ω
So 10 RE + RE = 5 or
0 454 ER . k= Ω and 4 54 CR . k= Ω We have
0 5 0 005 100
CQBQ
I .I . mAβ
= = =
and ( )( ) ( )( )( )0 1 1 0 1 101 0 454TH ER . R . .β= + =
or 4 59 THR . k= Ω
also
2
1 2 1
1TH CC TH CC
RV V R V
R R R⎛ ⎞
= ⋅ = ⋅ ⋅⎜ ⎟+⎝ ⎠
or
( )( )1 1
1 234 59 5THV .R R
= =
We can write ( ) ( )1TH BQ TH BE BQ EV I R V on I Rβ= + + +
or
( )( ) ( )( )( )1
23 0 005 4 59 0 7 101 0 005 0 454. . . . .R
= + +
which yields 1 24 1 R . k= Ω
and since 1 2 4 59 R R . k= Ω we find 2 5 67 R . k= Ω TYU6.4 dc analysis
( )
1 2
2
1 2
15 85 12 75
85 1215 85
TH
TH CC
R R R . k
RV VR R
= = = Ω
⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
or VTH = 10.2 V We can write
( ) ( )( )12 0 7 12 0 7 10 2
1 12 75 101 0 5TH
BQTH E
. V . .IR R . .β
− − − −= =+ + +
or IBQ = 0.0174 mA
and 1 74 CQ BQI I . mAβ= =
ac analysis ( )o fe b C LV h I R R=
and
( )1s
bie fe E
VI
h h R−
=+ +
so ( )( )1
fe C Lov
s ie fe E
h R RVA
V h h R
−= =
+ +
For ICQ = 1.74 mA, we find ( ) ( )( ) ( )max 110, min 70max 2 , min 1.1
fe fe
ie ie
h hh k h k
= == Ω = Ω
We obtain
( ) ( )( )( )
110 4 2max 2 59
1 1 111 0 5vA .. .−
= = −+
and
( ) ( )( )( )
70 4 2min 2 49
2 71 0 5vA ..
−= = −
+
TYU6.5
First approximation, Cv
E
RA
R≅ −
This predicts a low value, so set 9C
E
RR
=
Now ( )CC CQ C E ECQV I R R V≅ + +
or ( )( )7.5 0.6 9 3.75E ER R= + +
which yields 0.625 ER k= Ω and 5.62 CR k= Ω
We have ( )( ) ( )( )( )0.1 1 0.1 101 0.625TH ER Rβ= + =
or 6.31 THR k= Ω
Also
( )( )1 1
1 1 6.31 7.5TH TH CCV R VR R
= ⋅ ⋅ =
We have 0.6 0.006 100
CQBQ
II mA
β= = =
The KVL equation around the E-B loop gives ( ) ( )1CC BQ E EB BQ TH THV I R V on I R Vβ= + + + +
or
( )( )( ) ( )( ) ( )( )1
17.5 101 0.006 0.625 0.7 0.006 6.31 6.31 7.5R
= + + +
which yields 1 7.41 R k= Ω
Since 1 2 6.31 ,THR R R k= = Ω then 2 42.5 R k= Ω TYU6.6
We have ( ) ( )0.951
C Cv
E E
R RA
r R Rπ
ββ
⎛ ⎞−= = − ⎜ ⎟+ + ⎝ ⎠
or ( ) 20.95 4.750.4vA ⎛ ⎞= − = −⎜ ⎟
⎝ ⎠
Assume 1.2 r kπ = Ω from Example 4.6. Then
( )( )( )
24.75
1.2 1 0.4β
β−
= −+ +
or 76β =
TYU6.7 dc analysis: By symmetry, VTH = 0
1 2 20 20 10 THR R R k= = = Ω We can write
( )( )( )
( )( )
0 0.7 5 0.00672
10 126 5125 0.00672 0.84
BQ
CQ BQ
I mA
I I mAβ
− − −= =
+= = =
Small-signal transistor parameters: ( )( )125 0.026
3.87 0.84
0.84 32.3 /0.026
200 238 0.84
T
CQ
CQm
T
Ao
CQ
Vr kII
g mA VV
Vr kI
πβ
= = = Ω
= = =
= = = Ω
(a) We have
( )o m o C LV g V r R Rπ= − and sV Vπ = so
( )
( )( )32.3 238 2.3 5
ov m o C L
s
VA g r R RV
= = −
= −
or 50.5vA = −
(b) 238 2.3 2.28 o o CR r R k= = = Ω
TYU6.8 We find 0.418 ,CQI mA= then
( )( ) ( )( )1215 0.418 5.6 0.418 0.6120CEQV ⎛ ⎞= − − ⎜ ⎟⎝ ⎠
or VCEQ = 2.41 V So
( )2.41 0.5 2CEvΔ = − × or
3.82 CEv VΔ = peak-to-peak TYU6.9 dc load line
( ) ( )10 10CE CQ C EV I R R≅ + − + or
( )20 10CE C EV I R= − + ac load line
( )10ce c C cv i R i= − = − Now
( ) ( )0.7 10 10CE CEQ C CQv V i IΔ = − = Δ = so
( )0.7 10CEQ CQV I− = We have that
( )20 10CEQ CQ EV I R= − + then
( ) ( )10 0.7 20 10CQ CQ EI I R+ = − + or
( )20 19.3CQ EI R+ = (1) The base current is found from
( )10 0.7
100 101BQE
IR
−=+
so ( )( )
( )100 9.3
100 101CQE
IR
=+
Substituting into Equation (1), ( )( )
( ) ( )100 9.320 19.3
100 101 EE
RR
⎡ ⎤+ =⎢ ⎥+⎢ ⎥⎣ ⎦
which yields 16.35 ER k= Ω
Then ( )( )
( )( )100 9.3
0.531 100 101 16.35CQI mA= =
+
and ( )( )20 0.531 10 16.35 6.0 CEQV V≅ − + =
Now 0.7 6 0.7 5.3 CE CEQv V VΔ = − = − =
or, maximum symmetrical swing 2 5.3 10.6 V= × = peak-to-peak TYU6.10 We can write
( )( )( )
0 0.7 106.60
100 131 10BQI Aμ− − −
= ⇒+
and ( )( )130 6.60 0.857 CQI A mAμ= =
Assume nominal small-signal parameters of: 4 , 134
10, 12 83.3
ie fe
re oeoe
h k h
h h S kh
μ
= Ω =
= = ⇒ = Ω
We find
( )( ) ( )
11
4 135 10 ||10 || 83.3 641
ib ie fe E Loe
R h h R Rh
k
⎛ ⎞= + + ⎜ ⎟
⎝ ⎠= + = Ω
To find the voltage gain:
100 641100 641 10
B ibs s s
B ib S
R RV V V
R R R′ = ⋅ = ⋅
+ +
or ( )0.896s sV V′ =
Also ( )
( )1
1feo
s ie fe
h RVV h h R
′+=
′ ′+ +
where 1 10 10 83.3 4.72 E Loe
R R R kh
′ = = = Ω
Then ( )( )( )
( ) ( )0.896 135 4.72
0.8914 135 4.72
ov
s
VA
V= = =
+
To find the current gain:
( )
( )
1
11
10 83.3 10013510 83.3 10 100 641
⎛ ⎞⎜ ⎟
⎛ ⎞⎜ ⎟= = + ⎜ ⎟⎜ ⎟ +⎝ ⎠+⎜ ⎟⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
Eoeo B
i fei B ib
E Loe
RhI RA h
I R RR R
h
or 8.59=iA
To find the output resistance: 1
1
4 10 10010 83.3 96.0 135
ie S Bo E
oe fe
h R RR R
h h+
=+
+= ⇒ Ω
TYU6.11 We find
( )
1 2
2
1 2
50 50 25
1 5 2.5 2
TH
TH CC
R R R k
RV V VR R
= = = Ω
⎛ ⎞ ⎛ ⎞= ⋅ = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠
Now ( )
( )
( )( )
( )( )
15 0.7 2.5 0.00793
25 101 2and
100 0.00793 0.793
CC EB THBQ
TH E
CQ
V V on VI
R R
mA
I mA
β− −
=+ +
− −= =+
= =
The small-signal transistor parameters:
( )( )
0.793 30.5 /0.026100 0.026
3.28 0.793
125 158 0.793
CQm
T
T
CQ
Ao
CQ
Ig mA V
V
Vr kIVr kI
πβ
= = =
= = = Ω
= = = Ω
(a)
Define 2 0.5 158 0.40 E L oR R R r k′ = = ≅ Ω Now
( )( )
( )( )( )( )
1 101 0.41 3.28 101 0.4v
RA
r Rπ
ββ
′+= =
′+ + +
or Av = 0.925 (b)
( ) ( )( )1 3.28 101 0.4ibR r Rπ β ′= + + = + or
43.7 ibR k= Ω
Also 3.282 1581 101o E o
rR R r π
β= =
+
or 32.0 oR = Ω TYU6.12
For VECQ = 2.5, 5 2.5 5 0.5
CC ECQEQ
E
V VI mA
R− −= = =
then
( )75 5 4.93 76
5 0.0658 76
CQ
BQ
I mA
I mA
⎛ ⎞= =⎜ ⎟⎝ ⎠
= =
Small-signal transistor parmaters: ( )( )75 0.026
0.396 4.93
75 15.2 4.93
T
CQ
Ao
CQ
Vr kI
Vr kI
πβ= = = Ω
= = = Ω
Define the small-signal base current into the base, then m bg V Iπ β= −
Now,
( )1E oo b
E o L
R rI I
R r Rβ
⎛ ⎞= +⎜ ⎟+⎝ ⎠
and
1 2
1 2b i
ib
R RI I
R R R⎛ ⎞
= ⋅⎜ ⎟+⎝ ⎠
The current gain is
( ) 1 2
1 2
1o E oI
i E o L ib
I R r R RA
I R r R R R Rβ
⎛ ⎞ ⎛ ⎞= = +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
We have
( )( )( )( )
0.5 1
0.396 76 0.5 0.5 15.2 19.1
E L
ib E L o
R R kR r R R r
kπ β
= = Ω= + += + = Ω
and 0.5 15.2 0.484 E oR r k= = Ω
Then
( ) 1 2
1 2
0.48410 760.484 0.5 19.1I
R RA
R R⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
which yields 1 2 6.975 R R k= Ω
Now
2
1 2 1
1TH CC TH CC
RV V R V
R R R⎛ ⎞
= ⋅ = ⋅ ⋅⎜ ⎟+⎝ ⎠
or
( )( )1
1 6.975 5THVR
=
We can write ( )
( )1CC EB TH
BQTH E
V V on VI
R Rβ− −
=+ +
or
( )( )5 0.7
0.06586.975 76 0.5
THV− −=
+
which yields
( )( )1
11.34 6.975 5THVR
= =
or 1 26.0 R k= Ω and 2 9.53 R k= Ω
TYU6.13 (a) dc analysis:
( )
( )
10 0.7 0.93 10
100 0.93 0.921 1 101
EE EBEQ
E
CQ EQ
V V onI mA
R
I I mAββ
− −= = =
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠
( )( )( ) ( ) ( ) ( )10 0.93 10 0.921 5 10
ECQ EE EQ E CQ C CCV V I R I R V= − − − −= − − − −
or 6.1 ECQV V=
(b) Small-signal transistor parameters:
( )( )100 0.0262.82
0.921
0.921 35.42 /0.026
T
CQ
CQm
T
Vr kI
Ig mA V
V
πβ= = = Ω
= = =
Small-signal current gain: o mI g Vπ= and sV Vπ =
also 1s
i m s mE E
VI g V V g
R r R rππ π
⎛ ⎞= + = +⎜ ⎟⎜ ⎟
⎝ ⎠
Then ( )
( )
( )( )( )( )
11
35.42 10 2.821 35.42 10 2.82
m Eo mI
i m Em
E
g R rI g VA
I g R rV g
R r
ππ
ππ
π
= = =+⎛ ⎞
+⎜ ⎟⎝ ⎠
=+
or 0.987IA =
(c) Small-signal voltage gain:
o m C m s CV g V R g V Rπ= = or
( )( )35.42 5ov m C
s
VA g R
V= = =
or 177vA =
TYU6.14 (a)
( )( ) ( )( )
10 0.71 100 101 10
EE BEBQ
B E
V V onI
R Rβ− −= =
+ + +
or 8.38 BQI Aμ= and 0.838 CQI mA=
( )( )100 0.0263.10
0.838
0.838 32.23 /0.026
0.838
T
CQ
CQm
T
Ao
CQ
Vr kI
Ig mA V
VVrI
πβ
= = = Ω
= = =
∞= = = ∞
(b) Summing currents, we have
sm
E S
V V V Vg V
r R Rπ π π
ππ
⎛ ⎞⎛ ⎞− − −+ = + ⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
or 1 1 1 s
E S S
VV
r R R Rππ
β⎡ ⎤⎛ ⎞+ + + = −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
We can then write
1s
E SS
V rV R R
Rπ
π β⎡ ⎤⎛ ⎞
= − ⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦
Now ( )o m C LV g V R Rπ= −
So ( )
( )( )( )
1
32.23 10 1 3.10 10 11 101
C Lov m E S
s S
R RV rA g R R
V Rπ
β⎡ ⎤⎛ ⎞
= = ⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦
⎡ ⎤= ⎢ ⎥⎣ ⎦
or 0.870vA =
Now
( )
( )
10 0.76310 3.10
1001 101
10 0.90910 1
Ee i i i
E
c e e
Co c c c
C L
RI I I IR r
I I I
RI I I IR R
π
ββ
⎛ ⎞ ⎛ ⎞= ⋅ = ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠⎛ ⎞ ⎛ ⎞= ⋅ = ⋅⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠⎛ ⎞ ⎛ ⎞= ⋅ = ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
So we have
( ) ( )1000.909 0.763101
oI
i
IA
I⎛ ⎞= = ⎜ ⎟⎝ ⎠
or 0.687IA =
(c) We have
3.10101 101i E
rR R π
β= =
+
or 30.6 iR = Ω
Also 10 o CR R k= = Ω
TYU6.15 dc analysis: We can write ( )5 BQ B BE EQ EI R V on I R= + + or
( ) ( )( )( )
( )
5 0.7 4.3101 101
100 4.3101
BQB E B E
CQB E
IR R R R
IR R
−= =+ +
=+
Also 5 5CQ C CEQ EQ EI R V I R= + + − or
10110100CEQ CQ C EV I R R⎡ ⎤⎛ ⎞= − + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
ac analysis: ( )o m C LV g V R Rπ= −
and
1 Bs B
V RV V R V
r rπ
π ππ π
⎛ ⎞= − − ⋅ = − +⎜ ⎟
⎝ ⎠
or
sB
rV V
r Rπ
ππ
⎛ ⎞= − ⋅⎜ ⎟+⎝ ⎠
Then
( )ov C L
s B
VA R R
V r Rπ
β= =+
where mg rπβ =
For 1 ,CQI mA= ( )( )100 0.026
2.6 1
T
CQ
Vr kIπβ= = = Ω
Now ( )( )100 2 2
202.6v
B
AR
= =+
which yields 2.4 BR k= Ω
Also ( )( )
( )100 4.3
12.4 101CQ
E
IR
= =+
which yields 4.23 ER k= Ω
TYU6.16 (a)
dc analysis: For 2 1 ,EQI mA=
( )2100 1 0.990 101CQI mA⎛ ⎞= =⎜ ⎟⎝ ⎠
21
1 0.0099 1 101
EQEQ
II mA
β= = =
+
and 1
10.0099 0.000098
1 101EQ
BQ
II mA
β= = =
+
and ( )( )1 100 0.000098 0.0098 CQI mA= =
and ( )( )1 1 0.000098 10B BQ BV I R= − = −
or 1 0.00098 0BV = − ≅
so 1 0.7 EV V= −
and 2 1.4 EV V= −
Now 1 1 2 0.0098 0.990 1 CQ CQI I I mA= + = + ≅
then ( )( )5 1 4 1 OV V= − =
and ( )( )
2
1
1 1.4 2.4
1 0.7 1.7 CEQ
CEQ
V V
V V
= − − =
= − − =
(b) Small-signal transistor parameters:
( )( )
( )( )
11
11
22
22
1 2
100 0.026265
0.0098
0.0098 0.377 0.026100 0.026
2.63 0.990
0.990 38.1 /0.026
T
CQ
CQm
T
T
CQ
CQm
T
o o
Vr kI
Ig mA
V
Vr kI
Ig mA V
Vr r
π
π
β
β
= = = Ω
= = =
= = = Ω
= = =
= = ∞
(c) Small-signal voltage gain: From Figure 4.73 (b)
( )1 1 2 2
1 2
o m m C
s
V g V g V RV V V
π π
π π
= − += +
and
12 1 1 2 1 2
1 1
1ππ π π π π
π π
β⎛ ⎞ ⎛ ⎞+= + ⋅ = ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
mVV g V r V rr r
Then
1 1 2 2 11
1π π π
π
β⎡ ⎤⎛ ⎞+= − + ⋅ ⋅⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
o m m CV g V g r V Rr
Also
( )
1 2 11
21
1
1
1 1
π π ππ
ππ
π
β
β
⎛ ⎞+= + ⋅⎜ ⎟⎝ ⎠
⎡ ⎤⎛ ⎞= + +⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦
sV V r Vr
rVr
or
( )1
2
1
1 1π
π
π
β=
⎛ ⎞+ + ⎜ ⎟
⎝ ⎠
sVVrr
Now
( )
( )
21 2
1
2
1
1
1 1
π
π
π
π
β
β
⎡ ⎤⎛ ⎞+ + ⋅⎢ ⎥⎜ ⎟
⎝ ⎠⎣ ⎦= = −⎛ ⎞
+ + ⎜ ⎟⎝ ⎠
m m Co
vs
rg g RrVA
V rr
so
( ) ( ) ( )
( )
2.630.377 38.1 101 4265
2.631 101265
⎡ ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦= −⎛ ⎞+ ⎜ ⎟⎝ ⎠
vA
or 77.0= −vA
(d) ( ) ( )( )1 21 265 101 2.63iR r rπ πβ= + + = +
or 531 iR k= Ω
TYU6.17 (a) dc analysis: For 1 2 3 100 R R R k+ + = Ω
11 2 3
12 0.12 100
CCVI mAR R R
= = =+ +
Now ( )( )1 2
1 1 1
0.5 0.5 0.25 0.25 4 4.25
E CQ E
C E CEQ
V I R VV V V V
= = == + = + =
and
2 1 2 4.25 4 8.25 C C CEQV V V V= + = + =
So 2 12 8.25 7.5 0.5
CC CC
CQ
V VR k
I− −= = = Ω
Also ( )1 1 0.25 0.7 0.95 B E BEV V V on V= + = + =
And
( )3 31
1 2 3
0.95 12100B CC
R RV V
R R R⎛ ⎞
= ⋅ ⇒ =⎜ ⎟+ +⎝ ⎠
which yields 3 7.92 R k= Ω
We have ( )2 1 4.25 0.7 4.95 B C BEV V V on V= + = + =
and
2 32
1 2 3B CC
R RV V
R R R⎛ ⎞+
= ⋅⎜ ⎟+ +⎝ ⎠
or
( )2 7.924.95 12
100R +⎛ ⎞= ⎜ ⎟
⎝ ⎠
which yields 2 33.3 R k= Ω
Then 1 100 33.3 7.92 58.8 R k= − − = Ω
(b) Small-signal transistor parameters:
( )( )1 2
100 0.0260.5
T
CQ
Vr rIπ πβ
= = =
or 1 2 5.2 r r kπ π= = Ω
and
1 20.5
0.026CQ
m mT
Ig g
V= = =
or 1 2 19.23 mA/V= =m mg g
and 1 2o or r= = ∞
(c) Small-signal voltage gain: We have
( )2 2o m C LV g V R Rπ= − Also
22 2 1 1
2m m
Vg V g V
rπ
π ππ
+ =
or 2
2 1 1 1 and 1m s
rV g V V Vπ
π π πβ⎛ ⎞
= =⎜ ⎟+⎝ ⎠
We find
( )
( )
22 1
2
1
1
ov m C L m
s
m C L
V rA g R R g
V
g R R
π
β
ββ
⎛ ⎞= = − ⎜ ⎟+⎝ ⎠
⎛ ⎞= − ⎜ ⎟+⎝ ⎠
We obtain
( )( ) 10019.23 7.5 2101vA ⎛ ⎞= − ⎜ ⎟⎝ ⎠
or 30.1= −vA
TYU6.18 (a) dc analysis: with vs = 0
( )
1 2
2
1 2
125 30 24.2
30 12125 30
TH
TH CC
R R R k
RV VR R
= = = Ω
⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
or 2.32 THV V=
Now ( ) ( )1TH BQ TH BE BQ EV I R V on I Rβ= + + +
or
( )( )2.32 0.7 0.0250
24.2 81 0.5BQI mA−= =+
and ( )( )80 0.025 2.00 CQI mA= =
Also
( ) ( )
1
8112 2 2 0.580
ββ
⎡ ⎤⎛ ⎞+= − +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦
⎡ ⎤⎛ ⎞= − + ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
CEQ CC CQ C EV V I R R
or 6.99 CEQV V=
Power dissipated in RC: ( ) ( )22 2.0 2 8.0 RC CQ CP I R mW= = =
Power dissipated in RL: 0 0LQ RLI P= ⇒ =
Power dissipated in transistor:
( )( ) ( )( )0.025 0.7 2.0 6.99 14.0 Q BQ BEQ CQ CEQP I V I V
mW= += + =
(b) With ( )18cossv t mVω=
( )( )80 0.0261.04
2.0T
CQ
Vr kIπβ= = = Ω
We can write
( ) cosce C L Pv R R V trπ
β ω=
Power dissipated in RL:
( ) ( )
( )( )
2 2
2
3
1 12
1 1 80 2 2 0.0182 1.042 10
ceRL C L P
L L
v rmsp R R V
R R rπ
β⎡ ⎤= = ⋅ ⋅ ⎢ ⎥
⎣ ⎦
⎡ ⎤= ⋅ ⋅ ⎢ ⎥× ⎣ ⎦
or 0.479 RLp mW= Power dissipated in RC: Since 2 ,C LR R k= = Ω we have 8.0 0.479 8.48 RCp mW= + = Power dissipated in transistor: From the text, we have
( )
( )( ) ( )
2 2
223 3 3
3
2
80 0.0182 10 6.99 2 10 21.04 10 2
π
β
−
⎛ ⎞ ⎛ ⎞≅ − ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞= × − × ×⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠
PQ CQ CEQ C L
Vp I V R Rr
or 13.0 Qp mW=
TYU6.19 (a) dc analysis:
( )
1 2
2
1 2
53.8 10 8.43
10 553.8 10
TH
TH CC
R R R k
RV VR R
= = = Ω
⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
or 0.7837 THV V=
Now 0.7837 0.7 0.00993
8.43BQI mA−= =
and ( )( )100 0.00993 0.993 CQI mA= =
We have CEQ CC CQ CV V I R= −
or ( )2.5 5 0.993 CR= −
which yields 2.52 CR k= Ω
(b) Power dissipated in RC:
( ) ( )22 0.993 2.52RC CQ CP I R= = or
2.48 RCP mW= Power dissipated in transistor:
( )( )0.993 2.5Q CQ CEQP I V≅ = or
2.48 QP mW= (c) ac analysis: Maximum ac collector current:
( ) ( )0.993 cosci t mAω= average ac power dissipated in RC:
( ) ( ) ( )2 21 10.993 0.993 2.522 2RC Cp R= =
or 1.24RCp mW=
Now 1.24Fraction 0.25
2.48 2.48= = =
+ +RC
RC Q
pP P