Ch06s
56
Chapter 6 Problem Solutions 6.1 a. ( )( ) 0 0 2 76.9 mA/V 0.026 180 0.026 2.34 k 2 150 75 k 2 CQ m m T T CQ A CQ I g g V V r r I V r r I π π β = = ⇒ = = = ⇒ = Ω = = ⇒ = Ω b. ( )( ) 0 0 0.5 19.2 mA/V 0.026 180 0.026 9.36 k 0.5 150 300 k 0.5 m m g g r r r r π π = ⇒ = = ⇒ = Ω = ⇒ = Ω 6.2 (a) ( )( ) 0.8 30.8 mA/V 0.026 120 0.026 3.9 K 0.8 120 150 K 0.8 CQ m T T CQ A o CQ I g V V r I V r I π β = = = = = = = = = (b) ( )( ) 0.08 3.08 mA/V 0.026 120 0.026 39 K 0.08 120 1500 K 0.08 m o g r r π = = = = = = 6.3 ( )( ) 0 0 200 5.2 mA 0.026 125 0.026 0.625 k 5.2 200 38.5 k 5.2 CQ CQ m CQ T T CQ A CQ I I g I V V r r I V r r I π π β = ⇒ = ⇒ = = = ⇒ = Ω = = ⇒ = Ω 6.4
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Transcript of Ch06s
Chapter 6 Problem Solutions 6.1 a.
( )( )
0 0
2 76.9 mA/V0.026
180 0.0262.34 k
2
150 75 k2
CQm m
T
T
CQ
A
CQ
Ig g
V
Vr rI
Vr rI
π πβ
= = ⇒ =
= = ⇒ = Ω
= = ⇒ = Ω
b.
( ) ( )
0 0
0.5 19.2 mA/V0.026180 0.026
9.36 k0.5
150 300 k0.5
m mg g
r r
r r
π π
= ⇒ =
= ⇒ = Ω
= ⇒ = Ω
6.2 (a)
( )( )
0.8 30.8 mA/V0.026
120 0.0263.9 K
0.8
120 150 K0.8
CQm
T
T
CQ
Ao
CQ
Ig
V
VrI
VrI
πβ
= = =
= = =
= = =
(b)
( )( )
0.08 3.08 mA/V0.026120 0.026
39 K0.08
120 1500 K0.08
m
o
g
r
r
π
= =
= =
= =
6.3
( )( )
0 0
200 5.2 mA0.026
125 0.0260.625 k
5.2
200 38.5 k5.2
CQ CQm CQ
T
T
CQ
A
CQ
I Ig I
V
Vr rI
Vr rI
π πβ
= ⇒ = ⇒ =
= = ⇒ = Ω
= = ⇒ = Ω
6.4
( )
80 2.08 mA0.026
0.0261.20 96
2.08
CQ CQm CQ
T
T
CQ
I Ig I
V
VrIπ
ββ β
= ⇒ = ⇒ =
= ⇒ = ⇒ =
6.5 (a)
( )( )
( )( )
2 0.7 0.0052 250
120 0.0052 0.624 0.624 24 /0.026120 0.026
5 0.624
BQ
C
m m
o
I mA
I mA
g g mA V
r r k
r
π π
−= =
= =
= ⇒ =
= ⇒ = Ω
= ∞
(b) ( )( ) 524 4 1.885 250v m C v
B
rA g R A
r Rπ
π
⎛ ⎞ ⎛ ⎞= − = − ⇒ = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
(c) 0.426sin100 1.88
O OS S
v
v vv v t VA
= = ⇒ = −−
6.6
, 1.08 1.32 mACQm CQ
T
Ig I
V= ≤ ≤
1.08 1.32 41.5 50.8 mA/V0.026 0.026m mg g≤ ≤ ⇒ ≤ ≤
( ) ( )( )
( ) ( )( )
120 0.026; max 2.89 k
1.08
80 0.026min 1.58 k
1.321.58 2.89 k
T
CQ
Vr rI
r
r
π π
π
π
β= = = Ω
= = Ω
≤ ≤ Ω
6.7 a.
( )( )
( )( )
( )( )( )
120 0.0265.4 0.578 mA
1 1 5 2.5 V2 2
2.5 5.0 0.578 4.33 k
0.578 0.00482 mA120
on0.00482 25 0.70 0.820 V
TCQ
CQ CQ
CEQ CC
CEQ CC CQ C C C
CQBQ
BB BQ B BE
BB
Vr II I
V V
V V I R R R
II
V I R VV
πβ
β
= = = ⇒ =
= = =
= − ⇒ = − ⇒ = Ω
= = =
= += + ⇒ =
b.
( )( )
( )
( ) ( )
( ) ( )( )
0
0 0
00
120 0.0265.40 k
0.578
0.578 22.2 mA/V0.026100 173 k
0.578
,
120 173 4.33 120 4.2216.7
5.40 25 30.4
T
CQ
CQm
T
A
CQ
m C SB
Cv m C
B B
v v
VrII
gVVrI
rV g r R V V V
r R
r RrA g r Rr R r R
A A
π
ππ π
π
π
π π
β
β
= = = Ω
= = =
= = = Ω
⎛ ⎞= − = ⎜ ⎟+⎝ ⎠
⎛ ⎞= − = −⎜ ⎟+ +⎝ ⎠
⎡ ⎤⎣ ⎦= − = − ⇒ = −+
6.8 a.
( )
( ) ( ) ( )( )
1 5 V210 5 10 0.5 10 k
0.5 0.005100
on 0.70 0.005 50 0.95 V
ECQ CC
ECQ CQ C C C
CQBQ
EB BQ B BB BB
V V
V I R R R
II
V I R V Vβ
= =
= − ⇒ = − ⇒ = Ω
= = =
+ = = + ⇒ =
b.
( )( )
0 0
0.5 19.2 mA/V0.026
100 0.0265.2 k
0.5
0.5
CQm m
T
T
CQ
A
CQ
Ig g
V
Vr rIVr rI
π πβ
= = ⇒ =
= = ⇒ = Ω
∞= = ⇒ = ∞
c. ( ) ( )100 1018.1
5.2 50π
β= − = − ⇒ = −+ +C
v vB
RA Ar R
6.9
( )( )
( ) ( )
10 4 1.5 mA4
1.5 0.015 mA100100 0.026
1.73 K1.55sin mV
2.89sin A1.73 k
CQ
BQ
beb
I
I
r
tvi tr
π
π
ωω μ
−= =
= =
= =
= = =Ω
So
( ) ( )( ) ( ) ( )( ) ( ) [ ]( )( ) ( )
( )( )
1 1
1
1
15 2.89sin A1.5 0.289sin mA
10 10 1.5 0.289sin4 1.156sin
1.156 2310.005
B BQ b
C B C
C C C
C
Cv v
be
i t I iE ti t i i t tv t i t R tv t t v
v tA A
v t
ω μβ ω
ω γω
= + = += ⇒ = += − = − += −
−= = ⇒ = −
6.10
( )
( ) ( )( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )
1.2sin V1.2sin0
20.60sin mA
6sin A
100 2 K50
12sin mV
o
C C o C
C
Cb
be b m
be
v tti t R v i t
i t ti t
i t t
v t i t r g r
r
v t t
π π
π
ωω
ω
ω μβ
β
ω
=−+ = ⇒ =
= −
= = −
= ⋅ =
= =
= −
6.11 a. CQ EQI I≈
( )( )
5 1010 1.2 0.2
CEQ CQ C E
CQ
V I R RI
= = − += − +
3.57 mA3.57 0.0238 mA150
CQ
BQ
I
I
=
= =
( )( )( )( )( )
1 2 0.1 10.1 151 0.2 3.02 kTH ER R R Rβ= = +
= = Ω
( )
( ) ( )( )( ) ( )( ) ( )( )( )
( )
1
1
11
22
2
1 10 5
on 1 51 3.02 10 5 0.0238 3.02 0.7 151 0.0238 0.2 5
1 30.2 1.50 20.1
20.1 3.02 3.55 k20.1
THTH
BQ BQ ETH TH BE
V RR
V I R V I R
R
R kR
R RR
β+
= ⋅ ⋅ −
= + + −
− = + + −
= ⇒ = Ω
= ⇒ = Ω+
b. ( )( )150 0 026
1 09 k3 57
3 57 137 mA/V0 026m
.r .
..g.
= = Ω
= =
p
RE
rV
V0
gmV
RCVS
R1R2
( )( )( )
( )( )150 1 2
5 751 1 09 151 0 2
Cv v
E
.RA A .r R . .
ββ
= = ⇒ =+ + +p
2 2 2
6.12 a.
( )
( )( )
( )( ) ( )( )
2
1 2
1 2
50 12 10 V50 10
50 10 8 33 k12 0 7 10 0 0119 mA
8 33 101 11 19 mA, 1 20 mA
12 1 20 1 1 19 28 42 V
TH CC
TH
BQ
CQ EQ
ECQ
ECQ
RV VR R
R R R ..I .
.I . I .
V . .V .
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= = = Ω
− −= =+
= =
= − −=
4
1.19
8.42 12
iC
EC b.
RE
rV
V0
gmV
RCVS
R1R2
( )( )
( )
( )( )( )
( )( )
0
100 0 0262 18 k
1 19
1
100 21 94
1 2 18 101 1
m C
S m E
E
Cv v
E
.r .
.V g V R
VV V g V Rr
r RV
r
RA A .r R .
π β
ββ
= = Ω
=
⎛ ⎞= − +⎜ ⎟
⎝ ⎠+ +⎡ ⎤
= − ⎢ ⎥⎣ ⎦
= = ⇒ =+ + +
p
p
pp p
p
pp
p
2
22 2
c. Approximation: Assume rp does not vary significantly.
( ) ( )( )( )
( )( )( ) ( )
( )( )( )( )
2 kΩ 5% 2.1 kΩ or 1.9 k1 kΩ 5% 1.05 kΩ or 0.95 k
For max 2.1 kΩ and min100 2.1
2.142.18 101 0.95
For min 1.9 kΩ and max 1.05 k100 1.9
1.762.18 101 1.05
C
E
C E
v
C E
v
RR
R R
A
R R
A
= ± = Ω= ± = Ω
=−
= = −+
= = Ω−
= = −+
So 1 76 2 14v. A .≤ ≤
6.13 (a)
( ) ( )
( )( ) ( )( )( )
( )( )
( ) ( )
2
1 2 1 1
1
10112 1 6 2100
so that 1 99
1 99 0 0199 1000 1 1 0 1 101 1 10 1
1 1 10 1 12
1
12
CQ E ECQ CQ C
CQ CQ
BQ
ETH
TH CC TH CC
CC BQ E EB BQ
CC
CQ
TH TH
V I R V I R
I I
I . mA
.I . mA
R . R . . k
RV V R V .R R R R
V I R V on I R V
ββ
β
β
⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠
+= + +
⎛ ⎞= + +⎜ ⎟⎝ ⎠
=
= =
= + = = Ω
⎛ ⎞= = ⋅ ⋅ =⎜ ⎟+⎝ ⎠= + + + +
= ( )( )( ) ( )( )1
1 2
121 2101 0 0199 1 0 7 0 0199 10 1
which yields 13 3 and 41 6
.. . . .R
R . k R . k
+ + +
= Ω = Ω
(b) ( )
( )( )( )( )
100 21 95
1 1 31 101 1C
v vE
RA A .r R .
ββ
= = ⇒ =+ + +p
22 2
6.14
( ) ( )( ) ( ) ( ) ( )
0.25 mA, 0.2525 mA0.0025 mA
on 5 00.0025 50 0.7 0.2525 0.1 5
16.4 k
= ==
+ + + − =+ + + =
= Ω
CQ EQ
BQ
BQ B BE EQ S E
E
E
I III R V I R R
RR
( )( )0.0025 50 0.7 0.825 V3 0.825 2.175 V
5 2.175 11.3 kΩ0.25
E
C CEQ E
C C
VV V V
R R
= − − = −= + = − =
−= ⇒ =
( )( )( )
( )( )( )( )
( )( )( )
1
100 0.02610.4 k
0.25100 11.3
55.110.4 101 0.1
1
50 10.4 101 0.1
50 20.5 14.5 k
Cv
S
v v
i B S
i i
RAr R
r
A A
R R r R
R R
π
π
π
ββ
β
−=
+ +
= = Ω
−= ⇒ = −
+
= + +⎡ ⎤⎣ ⎦
= +⎡ ⎤⎣ ⎦
= ⇒ = Ω
6.15 (a)
( )( )9 2 2 2 3 75 So that
1 25
CC CQ C E CEQ
CQ
CQ
V I R R VI . .
I . mA
+ += + +
=
>
Assume circuit is to be designed to be bias stable. ( )( ) ( )( )( )
( ) ( )( )
1 2
1
0.1 1 0.1 121 2 24.2 1.25 0.01042 mA1201 on 121
TH E
BQ
TH TH CC BQ TH BE BQ E
R R R R
I
V R V I R V I RR
β= = + = = Ω
= =
= ⋅ ⋅ = + +
( )( ) ( )( ) ( )( )( )1
1 24.2 9 0.01042 24.2 0.7 0.01042 121 2
0.2522 0.7 2.52163.474
R= + +
= + +=
21
2
2
62.7 62.7 K 24.262.7
39.4 K
RRR
R
= =+
=
(b)
( )( )
1 25 48 08 /0 026120 0 026
2 50 1 25
100 80 1 25
m
o
.g . mA V.
.r . k
.
r k.
= =
= = Ω
= = Ω
p
RLRCror
Vo
V
gmV
IS
R1R2
( )
( )1 2
o m o C L
S
V g V r R R
V I R R r
=
=p
p p
2
Then
( ) ( )( )( ) ( )( )
1 2
48 08 24 2 2 5 80 2 2 1 48 08 2 266 0 6816
om m o C L
s
m
VR g R R r r R RI
R . . . . . . .
= =
= =
p2
2 2
or
74.3 74.3 /om
s
VR k V mAI
= = − Ω = −
6.16 a.
( )
0 800 80 mA, 0 0121 mA66
0 788 mA0 3 24 8 k
0 01215 5 3 2 54 k
0 788
EQ BQ
CQ
B BQ B B B
CC C
CQ
.I . I .
I ..V I R R R ..
VR R .
I .
= = =
=
= ⇒ = ⇒ = Ω
− − −= = ⇒ = Ω
b.
( )( )
0
0.788 30.3 mA / V0.02665 0.026
2.14 k0.788
75 95.2 k0.788
mg
r
r
π
= =
= = Ω
= = Ω
00
0
, C
SmC L
R ri g V V v
R r R π π
⎛ ⎞= = −⎜ ⎟⎜ ⎟+⎝ ⎠
( )
00
0
2.54 95.230.3
2.54 95.2 4
11.6 mA/V
Cf m
S C L
f
R riG g
v R r R
G
⎛ ⎞= = − ⎜ ⎟⎜ ⎟+⎝ ⎠
⎛ ⎞= − ⎜ ⎟⎜ ⎟+⎝ ⎠= −
6.17 (a)
( )( ) ( ) ( ) ( )
0.8 mA 0.00667 mA0.7 121 15 0
0.01667 2.5 0.7 121 0.00667 15
= ⇒ =+ + − =
+ + =
CQ BQ
BQ S BQ E
E
I II R I R
R
( )( )17.7 K
0.00667 2.5 0.7 0.717 V0.717 7 6.283 V
15 6.283 10.9 K0.8
E
E
C
C
RVV
R
== − − = −= − + =
−= =
( )( )
( )
120 0.0260.8 30.77 mA/V 3.9 K0.026 0.8m
o m C L SS
g r
rv g R R v v v
r R
π
ππ π
π
= = = =
⎛ ⎞= − ⋅ = ⎜ ⎟+⎝ ⎠
( ) ( ) ( )120 10.9 53.9 2.5
64.3π
β− −= =
+ += −
C Lv
S
v
R RA
r RA
(b) For 0SR =
( )( )
( ) ( )
0.7 121 0.00667 1517.7 K
0.7 0.7 7 6.315 6.3 10.9 K
0.8
30.77 10.9 5
105
E
E
E C
C C
C Lv
v
RRV V
R R
R RA
rA
π
β
+ === − ⇒ = − + =
−= ⇒ =
−= = −
= −
6.18 (a)
( ) ( ) ( )
( )( )
( )( )
15 81 10 0.7 2.515 0.7 0.0176 mA
2.5 81 101.408 mA
80 0.0261.408 54.15 mA/V0.026 1.408
1.48 K
BQ BQ
BQ
CQ
m
I I
I
I
g r
r
π
π
= + +−= =
+=
= = =
=
V0
RC
RS
RLrV
IS
Io
VS gmV
( ) ( ) ( )( )80 5 550.3
1.48 2.5
40
C Lo m C L v v
S
Cm
C Lo CI
S C L
I
R RV g V R R A A
r R
Rg VR Ri RA
Vi R Rr
A
σπ
π
β
βπ
π
− −= − ⇒ = = ⇒ = −
+ +
⎛ ⎞− ⎜ ⎟+ ⎛ ⎞⎝ ⎠= = = − ⎜ ⎟+⎝ ⎠
= −
( ) ( )( )( ) ( )
50.3 4sin0.201sin V
o
o
v t tv t t
ωω
= −= −
( ) ( )( )
4sin mV1.005sin μA
2.5 1.4840.2sin μA
ωω
ω
= =+
= −
s
o
ti t
i t
(b)
( )
15 0.7 1.43 mA10
80 1.43 1.412 mA81
EQ
CQ
I
I
−= =
⎛ ⎞= =⎜ ⎟⎝ ⎠
( )( )80 0.0261.412 54.3 mA/V 1.47 K0.026 1.412mg rπ= = = =
( ) ( )( )54.3 5 5 136
580 405 5
v m C L v
CI I
C L
A g R R A
RA A
R Rβ
= − = − ⇒ = −
⎛ ⎞ ⎛ ⎞= − = − ⇒ = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( )
136 4sin 544sin 0.544sin V
4sin mV2.72sin μA
1.47 k
o o o
s
v t t v t t v t t
ti t t
ω ω ω
ωω
= − ⇒ = − ⇒ = −
= =
( ) ( )( )( ) ( )
40 2.72sin109sin μA
o
o
i t ti t t
ωω
= −= −
6.19
( )
1 2
2
1 2
27 15 9.64 K
15 9 3.214 V15 27
TH
TH CC
R R R
RV VR R
= = =
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
( )( ) ( )( )
on 3.214 0.7 2.5141 9.64 101 1.2 130.84
0.0192 mA 1.9214 mA
TH BEBQ
TH E
BQ CQ
V VI
R RI I
β− −= = =
+ + += =
( )( )100 0.0261.92 73.9 mA/V 1.35 K0.026 1.92mg rπ= = = =
V0
RC
RS
RLRTH r0rV
IS
I0
VS gmV
100 52.1 K1.92or = =
( )0
1.35 9.64 1.184 K
1.1841.184 10
0.1059
THo m C L S
TH S
TH
S
S
r RV g V r R R V V
r R R
r R
V V
V
ππ π
π
π
π
⎛ ⎞= − = ⎜ ⎟⎜ ⎟+⎝ ⎠
= =
⎛ ⎞= ⎜ ⎟+⎝ ⎠=
( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )
73.9 0.1059 52.1 2.2 2
73.9 0.1059 52.1 1.0476
73.9 0.1059 1.0278.04
v
v
A
A
= −
= −
= −= −
( )
o Cm
o C LoI
S
TH
o CI m TH
o C L
r Rg V
r R RIA
VIR r
r RA g R r
r R R
π
π
π
π
⎛ ⎞− ⎜ ⎟⎜ ⎟+⎝ ⎠= =
⎛ ⎞= − ⎜ ⎟⎜ ⎟+⎝ ⎠
52.1 2.2 2.11 K
9.64 1.35 1.184 Ko C
TH
r R
R rπ
= =
= =
( ) ( ) 2.1173.9 1.1842.11 2
44.9
I
I
A
A
⎛ ⎞= − ⎜ ⎟+⎝ ⎠= −
9.64 1.35
1.184 Ki TH
i
R R r
Rπ= =
=
6.20 a.
( )( )( )
0 350 35 mA, 0 00347 mA101
0 00347 10 0 0347 V
on 0 735 V
E B
B B B B
E B BE E
.I . I .
V I R . V .
V V V V .
= = =
= = ⇒ =
= − ⇒ =
2 2 2
2
b.
( )
3 5 0 735 2 77 V
100 0 35 0 347 mA1 101
5 2 77 6 43 k0 347
C CEQ E
C E
CC C
C
V V V . . .
I I . .
V V .R R .I .
= + = − =
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠− −= = ⇒ = Ω
1
bb
(c)
( )
( )( )
( ) ( )
0 347 100 13 3 , 288 0 026 0 347100 0 026
7 49 0 34710 7 49 4 28
4 2813.3 6 43 288 81 74 28 0 1
Bv m C o
B S
m o
B
v v
R rA g R r
R r R
.g . mA/V r k
. ..
r . k.
R r . . k.A . A .
. .
π
⎛ ⎞= ⎜ ⎟⎜ ⎟+⎝ ⎠
= = = = Ω
= = Ω
= = Ω
⎛ ⎞= ⇒ =⎜ ⎟+⎝ ⎠
p
p
p
2
2 2
d.
( )
( ) ( )
0
10 7 49 4 28 k4 2813 3 6 43 288 74 9
4 28 0 5
Bv m C
B S
B
v v
R rA g R r
R r R
R r . ..A . . A .
. .
⎛ ⎞= ⎜ ⎟⎜ ⎟+⎝ ⎠
= = Ω
⎛ ⎞= ⇒ =⎜ ⎟+⎝ ⎠
p
p
p
2
2 2
6.21 a.
( )
( )( ) ( ) ( )
1 2
2
1 2
6 1.5 1.2 k
1.5 5 1.0 V1.5 6
on 1.0 0.7 0.0155 mA1 1.2 181 0.1β
+
= = = Ω
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠− −= = =
+ + +
TH
TH
TH BEBQ
TH E
R R R
RV VR R
V VI
R R
( ) ( ) ( ) ( )
2.80 mA, 2.81
5 2.8 1 2.81 0.1 1.92 V
CQ EQ
CEQ CQ C EQ E
CEQ
I I
V V I R I RV
+
= =
= − −= − − ⇒ =
b. ( )( )
0
180 0 0261 67 k
2 802 80 108 mA/V,
0 026m m
.r r .
..g g r.
= ⇒ = Ω
= ⇒ = =
p p
`
(c)
( )
( ) ( )
1 2
1 2
1 2 6 1 5 1 67 0 698 0 698108 1 1 2 45 8
0 698 0 2
v m C LS
v v
R R rA g R R
R R r R
R R r . . . k.A . A .
. .
⎛ ⎞= ⎜ ⎟+⎝ ⎠
= =
⎛ ⎞= ⇒ =⎜ ⎟+⎝ ⎠
p
p
p
2
V
2 2
6.22 a.
( )
( ) ( )( )
9 on0 750 75 mA, 0 00926 mA81
0 741 mA9 0 75 0.7 0 00926 2 11 0 k
EQ E EB BQ S
EQ BQ
CQ
E E
I R V I R.I . I .
I .. R . R .
= + +
= = =
== + + ⇒ = Ω
b. ( )( )
( )
9 0.75 11 0.75 V0 75 7 6.25 V
9 9 6 25 3 71 k0 741
E
C E ECQ
CC C
CQ
VV V V .
V .R R .I .
= − == − = − = −
− − −= = ⇒ = Ω
c.
( )
( )( )
( )
0
0
|| ||
80 0 0262 81 k
0 74180 108 k
0 74180 3 71||10 ||108
2 81 243 9
v m C LS
v
v
rA g R R rr R
.r .
.
r.
A ..
A .
⎛ ⎞= ⎜ ⎟+⎝ ⎠
= = Ω
= =
=+
=
p
p
p
2
V
2
2
d. 2 2 81 4 81 ki S iR R r . R .= + = + ⇒ = Ωp
6.23
( )( )4 0 7 0 00647
5 101 50 647 mA
BQ
CQ
.I .
I .
−= =+
=
a. 080 120, 10 20 Sfe eh h≤ ≤ ≤ ≤ m
low gain high gain
2 45 k 3 7 kie. h .Ω ≤ ≤ Ω
V0
RC
RS
RLrV
IS
Io
VS gmV
01
5 1 0 833 k
fe b C Loe
BS
B Sb
TH ie
TH B S
V h I R Rh
R VR R
IR h
R R R .
⎛ ⎞= ⎜ ⎟
⎝ ⎠
+=
+= = = Ω
2
?
High-gain 5
5 1 0 18380 833 3 7
S
b S
VI . V
. .
⎛ ⎞⎜ ⎟+⎝ ⎠= =
+
Low-gain 5
5 1 0 25380 833 2 45
S
b S
VI . V
. .
⎛ ⎞⎜ ⎟+⎝ ⎠= =
+
1 1For 10 || || || 4 || 40 010
100 || 2 1 96 k
oe c Loe
h R Rh .
.
= ⇒ =
= = Ω
( )( )( )( )( )( )
max
min
1For 20 || 4 || 4 50 || 2 1 92 k0 020
120 0 1838 1 96 43 280 0 2538 1 92 39 0
oe
v
v
h ..
A . . .A . . .
= ⇒ = = Ω
= == =
39 0 43 2v. A .≤ ≤
b. 5 3 7 2 13 k or 5 2 45 1 64 k
1 64 2 13 ki B ie i
i
R R h . . R . .. R .
= = = = = Ω≤ ≤ Ω
V
0
0
0
1 1 4 100 || 4 3 85 k0 010
1or || 4 50 || 4 3 70 k0 020
3 70 3 85 k
Coe
R R .h .
R ..
. R .
= = = = Ω
= = = Ω
≤ ≤ Ω
6.24
RE
R2
CC
R1
s
o
RC
VCC 10 V
RS 1 k
CE
Assume an npn transistor with 100=b and . Let 10 .A CCV V V= ∞ = 0 5 500 01v.A.
= =
Bias at 1 CQI mA= and let 1 ER k= Ω For a bias stable circuit
( )( ) ( )( )( )
( )( )1 1 1
0 1 1 0 1 101 1 10 1 1 1 10110 1 10
TH E
TH TH CC
R . R . . k
V R V .R R R
= + = = Ω
= ⋅ ⋅ = =
b
( ) ( )
( )( ) ( )( )( )1
1 2
1 0 01 100
1101 0 01 10 1 0 7 101 0 01 1
which yields 55 8 and 12 3
BQ
TH BQ TH BE BQ E
I . mA
V I R V on I R
. . . .R
R . k R . k
= =
= + + +
= + +
= Ω = Ω
b
Now ( )( )100 0 026
2 6 1
1 38 46 0 026m
o m C
.r . k
g . mA/V.
V g V R
= = Ω
= =
= −
p
p
1 2
1 2
10 1 2 6where 10 1 2 6 1
or 0 674
s sS
s
R R r . .V V .VR R r R . .
V . V
⎛ ⎞ ⎛ ⎞= ⋅ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
=
pp
p
p
( ) ( )( )Then 0 674 0 674 38 46 50
which yields 1 93
ov m C C
s
C
VA . g R . . R
VR . k
= = − = − = −
= Ω
With this RC, the dc bias is OK. Finish Design, Set 2 KCR = 1 KER =
1
2
56 K12 K
RR
==
( )
1 2
2
1 2
9.88 K
12 10 1.765 V12 56
TH
TH CC
R R R
RV VR R
= =
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
( ) ( )1.765 0.7 9.60 A
9.88 101 10.9605 mA
BQ
CQ
I
I
μ−= =+
=
( )( )100 0.026 0.96052.707 K 36.940.9605 0.026
2.125 K
m
TH
r g
R r
π
π
= = = =
=
( )
( ) ( )( )( )
2.125 0.680 2.125 1
0.680 0.680 36.94 2 50.2
THi i i
TH S
v m C
R rV V V VR r R
A g R
ππ
π
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= − = − = −
Design specification met. 6.25 a.
( )( )
( ) ( )( ) ( )( )
6 0 7 0 0169 mA10 101 31 69 mA, 1 71 mA
16 6 1 69 6 8 1 71 35 38 V
BQ
CQ EQ
CEQ
CEQ
.I .
I . I .
V . . .V .
−= =+
= =
= + − −=
b.
( )( )0
1 69 65 mA/V0 026100 0 026
1 54 k , 1 69
m m.g g.
.r r . r
.
= ⇒ =
= ⇒ = = ∞p p V
(c) ( )( )( ) ( )( )11 1.54 101 3 304.5
10 304.5 9.68
C L B ibv
E B ib S
ib E
B ib
R R R RA
r R R R RR r R kR R k
π
π
βββ
−= ⋅
+ + += + + = + = Ω
= = Ω
Then ( )( )
( )( )
( )
( )
( ) ( )
( ) ( )( )
0
100 6.8 6.8 9.68 1.061.54 101 3 9.68 0.5
1
1
6.8 10100 1.596.8 6.8 10 1.54 101 3
vv
Cb
C L
Bb S
B E
C Bi
C L B E
i
A A
Ri i
R R
Ri iR r R
R RAR R R r R
A
π
π
β
β
ββ
− ⎛ ⎞= ⋅ ⇒ = −⎜ ⎟+ +⎝ ⎠
⎛ ⎞= −⎜ ⎟+⎝ ⎠⎛ ⎞
= ⎜ ⎟+ + +⎝ ⎠⎛ ⎞⎛ ⎞
= − ⎜ ⎟⎜ ⎟⎜ ⎟+ + + +⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞= − ⇒ = −⎜ ⎟⎜ ⎟⎜ ⎟+ + +⎝ ⎠⎝ ⎠
(d) 0 5 10 304 5 10 2 is S B ibR R R R . . . k= + = + = Ω (e)
( )( )
( ) ( )( )( )
( )
1
100 6 8 6 81 12
1 54 101 3same as c 1 59
C Lv
E
v v
i i
R RA
r R
. .A A .
.A A .
=+ +
= ⇒ =+
= ⇒ =
p
2 bb
22
2
6.26
vCE
vbe
vCE
vCE
ie
ris ogmv
gmv
0
1
1So
Ce
m Ce m
em
vrg v g
r r rg
= =
⎛ ⎞= ⎜ ⎟
⎝ ⎠p
6.27 Let 100, AV= = ∞b
RE
CC
R2
R1
s
o
RC
VCC
RS 100
Let 2 5 CCV . V=
( ) ( )( )
1 2
0 12 2 5 48 , Let 8 , 40 2 5 312 5
840 0 4
100
R C CC R C R C R C
CC
R
BQ
P I I V . I I . I I A I A I AV .R R . kI
I . A
= + ⇒ = + ⇒ + = = =
+ = ⇒ Ω
= =
m m m
m
>
( )( ) ( )( )( )
( ) ( )
( )( ) ( )( ) ( )( )( )
( )( )
1
1
1 2
Let 2 For a bias stable circuit0 1 1 0 1 101 2 20 2 1 1
1 20 2 2 5 0 0004 20 2 0 7 101 0 0004 2
which yields 64 and 29.5 100 0 026
65 0 04
E
TH E
TH TH CC BQ TH BE BQ E
R k .R . R . . k
V R V I R V on I RR
. . . . . .R
R k R k.
r k.π
= Ω= + = = Ω
= ⋅ ⋅ = + + +
= + +
= = Ω
= =
b
b
V
( )
( )( )
Neglect
1
10010 26 7
65 101 2
S
o Cv
s E
CC
R
V RA
V r RR
R . k
π
Ω
=+ +
− = ⇒ = Ω+
2 bb
2
>
With this ,CR dc biasing is OK. 6.28
Need a voltage gain of 100 205
.=
Assume a sign inversion from a common-emitter is not important. Use the configuration for Figure 6.31. Let 0SR .= Need an input resistance of
( ) ( )
33
6
5 10 25 10 25 0 2 10
Let 50 , 50 1 1
50For 100, 0 495 1 101
i
i TH ib TH ib
ib E E
ibE
R k.
R R R . R k R kR r R R
RR . k
×= = × = Ω×
= = Ω = Ω= + + +
= = = = Ω+
2
2
p b b
bb
>
Let 0 5 , 10 , 0 2 E CC CQR . k V V I . mA= = =V
( ) ( )
( )( ) ( )( ) ( )( )( )
( )( )( )
( )( )( )
11
1 2
0.2Then 0.002 100
1
1 1 50 10 0.002 50 0.7 101 0.002 0.5
which yields 555 and 55 100 0.026
Now , 13 1 0.2
So100
20 12.7 13 101 0.5
BQ
TH BQ TH BE BQ E
TH CC
Cv
E
CC
I mA
V I R V on I R
R VRR
R k R kR
A r kr R
RR k
ππ
β
ββ
= =
= + + +
⋅ ⋅ = = + +
= Ω = Ω
−= = = Ω
+ +
−− = ⇒ = Ω
+
[Note: ( )( )0 2 12 7 2 54 CQ CI R . . . V.= = So dc biasing is OK.] 6.29
RC
R2
R1
s
o
RE
VCC 10 V
CC
( )80,
1C
vE
RA
r R= =
+ +p
2 bbb
First approximation:
( ) 10 10Cv C E
E
RA R RR
≈ = ⇒ =
( ) ( )Set 12
10 13C E
EC CC C C E C E
R RV V I R R I R
=≈ − + = −
( )
1For 52
5 10 13
EC CC
C E
V V
I R
= =
= −
For 0 7 mA0 709, 0 00875 mA 0 55 k 6 6 kC
E B E C
I .I . I . R . R .
== = ⇒ = Ω − = Ω
Bias stable ⇒ ( )( )
( )( )( )
( )( ) ( )( ) ( )( )
( )
( )
1 2
1
11
22
2
1 2
0.1 10.1 81 0.55 4.46 k
110 0.709 0.55 0.7 0.00875 4.46 4.46 10
18.87 4.46 5.03 kΩ
5.03 4.46 39.4 kΩ5.03
10 10 0.225 mA5.03 39.4
0.7 + 0.225 0.925 mA from source.80 0.0
Now
TH E
CC
R R R R
R
RRR RR
R RV
rπ
β= = += = Ω
= + + +
= ⇒ =
= ⇒ =+
= =+ +
≅
=( )
( )( )( )( )
262.97 kΩ
0.780 6.6
11.12.97 81 0.55vA
=
= =+
6.30
RE
CE
RL 10 K
R2
R1
s
RC
5 V
5 V
CC1
CC2
o
( ) ( )( )( )( )
( ) ( )( )
120Let 0.35mA, 0.353 mA
0.00292 mALet 2 kΩ. For 4 V 10 4 0.35 0.353 2
120 0.02615.1 k , 8.91 k
0.35120 15.1 10
8.9181.0
CQ EQ
BQ
E CEQ C
C
C Lv
v
I I
IR V R
R r
R RA
rA
π
β
β
π
== =
== = ⇒ = + +
= Ω = = Ω
−= = −
= −
For bias stable circuit: ( )( )
( )( )( )
( ) ( )
( ) ( )
1 2
2
1 2 1
0.1 10.1 121 2 24.2 k
110 5 10 5
on 1 5
TH E
TH TH
BETH BQ TH BQ E
R R R R
RV RR R R
V I R V I R
β
β
= = += = Ω
⎛ ⎞= − = ⋅ ⋅ −⎜ ⎟+⎝ ⎠= + + + −
( )( ) ( )( ) ( )( )( )
( )1
1
1
22
2
1 24.2 10 5 0.00292 24.2 0.7 121 0.00292 2 5
1 242 1.477, 164 k
164 24.2 28.4 k164
10 0.052, 0.35 0.052 0.402 mA164 28.4
R
RR
R RR
− = + + −
= = Ω
= ⇒ = Ω+
= + =+
So bias current specification is met. 6.31 From Prob. 6.12,
( ) ( )
( )( )
( )( ) ( )( )
1 2
2
1 2
10 50 8 33 k
5012 12 10 V50 10
12 0 7 10 0 0119 mA8 33 101 11 19 mA, 1 20 mA12 1 19 2 120 1 8 42 V
TH
TH
BQ
CQ EQ
ECQ
R R R .
RVR R
.I ..
I . I .V . .
= = = Ω
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠− −= =+
= == − − =
1.19
8.42 12111 For 1 11ECv≤ ≤
11 8 42 2 58ECv . .= − =D ⇒ Output voltage swing 5 16 V.= (peak-to-peak) 6.32
( )( )
( ) ( )( ) ( )( )
5 0.7 0.00315 mA50 101 0.1 12.90.315 mA, 0.319 mA
5 5 0.315 6 0.319 13
3.96 V
BQ
CQ EQ
CEQ
CEQ
I
I I
V
V
−= =+ +
= =
= + − −
=
0.315
3.96 10
AC load line
Slope 1
6.1 K
( )
16.1
For i 0.315 0.05 0.265 1.62
min 3.96 1.62 2.34
C eC
ECC
EC
i v
v
v
Δ = − Δ
Δ = − = ⇒ Δ =
= − =
Output signal swing determined by current:
Max. output swing 3 24 V peak-to-peak.= 6.33 From Problem 4.18, 1.408 mA, 1.426 mACQ EQI I= =
(a) ( )( ) ( )( )30 1.408 5 1.426 10 8.7 VECQV = − − =
8.7
1.408
EC (V)
IC (mA)
AC load line
Slope 1
RCRL
1
2.5 kΩ
( ) ( ) ( )( )max 8.7 2.5 8.7 1.408 2.5 12.22EC Cv I= + Δ ⋅ = + =
Set ( ) ( )max 12 8.7 2.5 1.32 mAEC C Cv I I= = + Δ ⇒ Δ = So ECvΔ (peak-to-peak) 2(12 8.7) 6.6 V= − = (b) CiΔ (peak-to-peak) = 2(1.32) = 2.64 mA 6.34
( )( )( )( )
( )
0 80 mA, 0 792 mA0 00792 mA0 7 0 00792 10 0 779 V
5 0 792 4 5 1 83 V0 779 1 83 2 61 V
EQ CQ
BQ
E
C CQ C
ECQ
I . I .I .V . . .V I R . .
V . . .
= === + == − = − == − − =
2
Load line: Assume VE remains constant at 0 78 V.≈
8.7
1.408
EC (V)
IC (mA)
AC load line
Slope 1
RCRL
1
2.5 kΩ
( )( )
12 k
Collector current swing 0 792 0 080 712 mA0 712 2 1 424 V
C ec
ec
i v
. .
.v . .
=
= −== =
2D ?V
D
Output swing determined by current.
0
Max. output swing 2.85 V peak-to-peak
2 85Swing in current4
0 712 mA peak-to-peak
.i
.
=
=
=
6.35
( )( )
( ) ( )( ) ( )( )
6 0 7 0 0169 mA10 101 31 69 mA, 1 71 mA
16 6 1 69 6 8 1 71 35 38 V
BQ
CQ EQ
CEQ
CEQ
.I .
I . I .
V . . .V .
−= =+
= =
= + − −=
1.69
5.38 22
1 6.4 K
AC load line
Slope 1
3.4 3
1
6 4C cei v.
=D 2 D
For ( ) 4 38min 1 V, 5 38 1 4 38 V 0 684 mA6 4ce ce C.v v . . i ..
= = − = ⇒ = =D D
Output swing limited by voltage:
0 0
0
Max. swing in output voltage8.76 V peak-to-peak
1 0.342 mA2
or 0.684 mA (peak-to-peak)
ce
C
v
i i i
i
Δ ==
Δ = Δ ⇒ Δ =
Δ =
6.36
9VCEQ
ICQ
2.65
AC load line
Slope 1
1.05 K
Q-point
100
oCQ
rI
=
Neglect ro as (E) approx. dc load line ( )9 3.4CE CV I= −
( ) ( )( )( )
0.11
Also 1.05
Or 1 0.1 1.05
C CQ
CE CEQ
CE C C L C
CEQ CQ
I IV VV I R R I
V I
Δ = −Δ = −Δ = Δ = Δ
− = −
Substituting the expression for the dc load line. ( ) ( ) ( )
( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( )1 1
1
2
9 3.4 1 0.1 1.05
8.105 4.45 1.821 mA2.81 V
1.821 0.018211000.1 101 1.2 12.12 K1 1 12.12 9 0.01821 12.12 0.7 101 0.01821 1.2
0.2207 0.7 2.2070534.9 K18.6 K
34.
CQ CQ
CQ CQ
CEQ
BQ
TH
TH TH CC
I I
I IV
I
R
V R VR R
RR
⎡ ⎤− − = −⎣ ⎦= ⇒ =
=
= =
= =
= ⋅ ⋅ = = + +
= + +==
2
2
912.12
34.9RR
=+
6.37 dc load line
ICQ
VCEQ 5
5 1 0.1
4.55 mA
1 0.545 K
AC load line
Slope 1
11.2
For maximum symmetrical swing
( )( ) ( )
( )
0.2510.5 and | |
0.545 k0.5
0.250.545
5 1.1
0.545 0.25 5 1.1 0.5
0.545 1.1 5 0.5 0.1362.82 mA, 0.0157 mA
C CQ
CE CEQ C CE
CEQCQ
CEQ CQ
CQ CQ
CQ
CQ BQ
i I
v V i v
VI
V I
I I
II I
Δ = −
Δ = − Δ = ⋅ ΔΩ
−− =
= −
⎡ ⎤− = − −⎣ ⎦+ = − +
= =
( )( )( )( )( )
( ) ( )
1 2
1
0.1 10.1 181 0.1 1.81 k1 on 1
TH E
TH TH BQ TH BE BQ E
R R R R
V R V I R V I RR
β
β+
= = += = Ω
= ⋅ ⋅ = + + +
( )( ) ( )( ) ( )( )( )
( )1
11
22
2
1 1.81 5 0.0157 1.81 0.7 181 0.0157 0.1
1 9.05 1.013 8.93 k
8.93 1.81 2.27 k8.93
R
RR
R RR
= + +
= ⇒ = Ω
= ⇒ = Ω+
6.38
( )( )
( ) ( )( )
0 647 , 10 0 647 9 4 18 0 647
So 4 4 0 647 2 1 294
CQ CEQ
C CQ
CE C
I . mA V . . Vi I . mAv i . . V
= − == =
= = =D
D D
>
Voltage swing is well within the voltage specification. Then ( )2 1 294 2 59 CEv . . V= =D peak-to-peak
6.39 a.
( ) ( )
( )( )( )
( )( )
1 2
2
1 2
10 10 5 k
1018 9 18 9 010 10
0 0 7 90 0869 mA
5 181 0 515 6 mA, 15 7 mA18 15 7 0 5 10 1 V
TH
TH
BQ
CQ EQ
CEQ CEQ
R R R
RVR R
.I .
.I . I .V . . V .
= = = Ω
⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠− − −
= =+
= == − ⇒ =
b.
1 0.188 K
15.6
10.1 18
AC load line
Slope 1
0.50.3
c.
( )( )
( )( )( )( )( )( ) ( )( )
( )( )( )( )
1 2
1 2
1 2
180 0.0260.30 k
15.61
1
1 0.30 181 0.5 0.3 or 34.2 5 34.2 4.36
181 0.5 0.3 4.36= = 0.8060.3 + 181 0.5 0.3 4.36 1
E L ibv
ib SE L
ib E L ib
ib
v v
r
R R R R RAR R R Rr R R
R r R R R kR R R k
A A
π
π
π
βββ
= = Ω
+ ⎛ ⎞= ⋅⎜ ⎟++ + ⎝ ⎠= + + = + = Ω
= = Ω
⎛ ⎞⋅ ⇒⎜ ⎟+⎝ ⎠
d.
( )( )( )( )
1 2
10 30 181 0 188 34 3 k
0 3 5 10 5 6 18 1 181
ib E L
ib ib
So E o
R r R RR . . R .
r R R R .R R . R .
= + += + ⇒ = Ω
+ += = ⇒ = Ω+
p
p
b
b
6.40 a.
( )
( ) ( )
1 2
2
1 2
10 10 5 k
10 5 V
on 1 10
TH
TH
TH BQ TH BE BQ E
R R R
RVR R
V I R V I R
= = = Ω
⎛ ⎞= − =⎜ ⎟+⎝ ⎠= + + + −
2
b
( )( )( )
( )( ) ( )( )
5 0.7 100 0174 mA
5 121 22.09 mA, 2.11 mA
10 2.09 1 2.11 2 3.69 V
BQ
CQ EQ
CEQ CEQ
I .
I I
V V
− − −= =
+= =
= − − ⇒ =
2
b.
2.09
3.69 10
11 K
AC load line
Slope 1
22
c.
( )( )
( )( )( )( )( ) ( ) ( )( )
( )( )( )( )
1 2
1 2
1 2
120 0.0261.49 k
2.091
1
1 1.49 121 2 2122.5 , 5 122.5 4.80
121 2 2 4.80= 0.4844.80 51.49 + 121 2 2
E L ibv
ib SE L
ib E L
ib ib
v v
r
R R R R RA
R R R Rr R R
R r R RR k R R R k
A A
π
π
π
βββ
= = Ω
+ ⎛ ⎞= ⋅⎜ ⎟⎜ ⎟++ + ⎝ ⎠= + + = += Ω = = Ω
⎛ ⎞⋅ ⇒ =⎜ ⎟+⎝ ⎠
d. ( )( )
( )( )1 2
11 49 121 2 2 122 k
1 49 5 52 32 4 1 121
ib E L
ib ib
So E o
R r R RR . R
r R R R .R R R .
π
π
= + += + = Ω
+ += = = Ω+
b
b
1
1
6.41 a.
( )
( )( )
( )( )
1 2
2
1 2
,
60 40 24 k
40 5 2 V40 60
5 0.7 2 0.0130 mA24 51 30.650 mA 0.663 mA
5 5 0.663 3 3.01 V
TH
TH CC
BQ
CQ EQ
ECQ EQ E ECQ
R R R
RV VR R
I
I I
V I R V
= = = Ω
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠− −= =+
= =
= − = − ⇒ =
b.
1.63
0.65
3.01 5
1 1.75 K
AC load line
Slope 1
5150 34
c.
( )( )
( )( )
( )( )( )( )
( )
( ) ( )( )
0
0
0
0
0 0
50 0.026 802 k , 123 k0.650 0.65
Define 3 4 123 1.69 k1 51 1.69
0.9771 2 51 1.69
1
1 2 51 1.69 88.23 3 123 2
L E L
Lv v
L
Ei b
E L
THb S
TH ib
ib L
E
r r
R R R rR
A Ar R
R rA I
R r R
RI IR R
R r RR r r
π
π
π
ββ
β
β+
=
= = Ω = = Ω
′ = = = Ω′+
= = ⇒ =′+ +
⎛ ⎞= + ⎜ ⎟⎜ ⎟+⎝ ⎠
⎛ ⎞= ⎜ ⎟+⎝ ⎠
′= + = + == =
( )
.932.93 2451 4.61
2.93 4 24 88.2i iA A⎛ ⎞⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
d. ( ) ( )( )0
0
0
1 2 51 1.69 88.2 k
2 3 0.0392 31 5138.7
ib E L ib
E
R r R R r R
rR R
R
π
π
β
β
+= + = + ⇒ = Ω
⎛ ⎞= = =⎜ ⎟+ ⎝ ⎠= Ω
e. Assume variations in rπ and r0 have negligible effects 1 1 1
2 2 2
60 5% 63 k , 57 k40 5% 42 k , 38 k3 5% 3.15 k , 2.85 k4 5% 4.2 k , 3.8 k
E E E
L L L
R R RR R RR R RR R R
= ± = Ω = Ω= ± = Ω = Ω= ± = Ω = Ω= ± = Ω = Ω
( )
( )( )( ) ( )
( ) ( )( ) ( )( ) ( )
0
0
0
1
1
max 25.2 k , min 22.8 kmax 92.5 k , min 84.0 kmax , min , 88.6kmin , max , 87.4k
E THi
E L TH ib
ib E L
TH TH
ib ib
E L ib
E L ib
R r RAR r R R R
R r R R r
R RR RR R RR R R
π
β
β
⎛ ⎞⎛ ⎞= + ⎜ ⎟⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠= + +
= Ω = Ω= Ω = Ω
= Ω= Ω
( )( )
0
0
max 3.07 kmin 2.79 k
E
E
R rR r
= Ω= Ω
For ( ) ( ) ( )min , max , minE L THR R R
( ) 2.79 22.851 4.212.79 4.2 22.8 87.4i iA A⎛ ⎞⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
( ) ( ) ( )For max , min , maxE L THR R R
( ) 3.07 25.251 5.053.07 3.8 25.2 88.6i iA A⎛ ⎞⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
6.42 (a)
( )( )
0.5 0.00617 81
0.00617 10 0.0617
0.7 0.7617
BQ
B BQ B B
E B E
I mA
V I R V V
V V V V
= =
= = ⇒ =
= + ⇒ =
(b)
( )
( )( )
800.5 0.494 810.494 19 /0.026
80 0.0264.21
0.494
150 304 0.494
CQ
CQm m
T
T
CQ
Ao o
CQ
I mA
Ig g mA V
V
Vr r kIVr r kI
π πβ
⎛ ⎞= =⎜ ⎟⎝ ⎠
= = ⇒ =
= = ⇒ = Ω
= = ⇒ = Ω
(c)
Vo
RB
RS
Vs
VS
r roV
gmV
IS
RL
Io
( )
For 0S
o m L o
R
VV g V R r
rπ
ππ
=
⎛ ⎞= − +⎜ ⎟
⎝ ⎠
( )so that
1o
L o
VVR r
r
π
π
β−
=⎛ ⎞+⎜ ⎟⎝ ⎠
( )
Now
or 1
s o
os o o
L o
V V VVV V V VR r
r
π
π
π
β
+ =
= − = +⎛ ⎞+⎜ ⎟⎝ ⎠
( )( )( )( )
( )( )( )( )
( )( )( )( )
( )( ) ( )( )
( )
We find1 81 0.5 304
1 4.21 81 0.5 304
81 0.50.906
4.21 81 0.5
1 4.21 81 0.5 44.7
and 1
L oov
s L o
v
ib L o
oBb s o b
B ib o L
R rVAV r R r
A
R r R r k
rRI I I IR R r R
π
π
ββ
β
β
+= = =
+ + +
≅ ⇒ =+
= + + ≅ + = Ω
⎛ ⎞ ⎛ ⎞= ⋅ = +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
( )
( ) ( )
Then
1
1081 1 14.810 44.7
o oBi
s B ib o L
i i
I rRAI R R r R
A A
β⎛ ⎞⎛ ⎞
= = + ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠⎛ ⎞≅ ⇒ =⎜ ⎟+⎝ ⎠
(d)
( )
( )( )
10 44.70.803
10 44.7 2
Then 0.803 0.906 0.728
14.8 (Unchanged)
B ibs s s s
B ib s
v v
i
R RV V V VR R R
A A
A
⎛ ⎞ ⎛ ⎞+′ = ⋅ = ⋅ =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠= ⇒ =
=
6.43 (a)
( )( )100 0.0261.98 mA 1.313 K
1.981001.98
50.5 K1.31 10 50.5 112
1 1010.112 50.5 112
CQ
Ao
CQ
So o o
o
I r
VrI
r RR r R
R
π
π
β
= = =
= =
=+ += = ⇒ = Ω+
⇒ ≅ Ω
(b) From Equation 4.68 ( )( )
( )( )1 100 50.5 K
1.981o L
v oo L
r RA r
r r Rπ
ββ
+= = =
+ +
(i)
( )( )( )( )
( )( )( )( )
0.5 K
101 50.5 0.51.31 101 50.5 0.5
101 0.49510.974
1.31 101 0.4951
L
v
v v
R
A
A A
=
=+
= ⇒ =+
(ii)
( )( )( )( )
5 K 50.5 5 4.5495
101 4.550.997
1.31 101 4.55
L o L
v v
R r R
A A
= = =
= ⇒ =+
6.44
( )( )
5 0.7 1.303 1.293 mA3.3
125 0.0262.51 K
1.2931.293 49.73 mA/V0.026
EQ CQ
m
I I
r
g
π
−= = =
= =
= =
(a) ( )( ) ( )( )1 2.51 126 3.3 1
99.2 K
2.513.3 3.3 0.019921 126
19.8
ib E L
ib
o E
o
R r R R
R
rR R
R
π
π
β
β
= + + = +
=
= = =+
= Ω
(b)
( ) ( )
( ) ( ) ( )( )( ) ( )
( )( )( )( )
( )( )( )( )
( )( )( )( )
( ) ( )
( ) ( ) ( ) ( )
2sin 20.2sin A99.2
20.2 2.51 sin50.6sin mV
1 126 3.3 1 126 0.76742.51 126 0.76741 2.51 126 3.3 1
0.9747 1.95sin V
1.95sin mA
ss s
ib
eb s
eb
E Lv
E L
v o
oo o
L
v ti i t tR
v t i t r tv t t
R RA
r R R
A v t t
v ti t i t t
R
π
π
ω ω μ
ωω
ββ
ω
ω
= = ⇒ =
= − = −= −
+= = =
++ + +
= ⇒ =
= ⇒ =
6.45 a.
( )( )
( )( ) ( )( )
1 mA 5 10 1 5 kΩ
1 0.0099 101
10 on10 0.0099 0.7 1 5 434 kΩ
EQ CEQ CC EQ E
E E
BQ
BQ B BE EQ E
B B
I , V V I RR R
I mA
I R V I RR R
= = −= − ⇒ =
= =
= + += + + ⇒ =
b.
0
b
RERB
( )( )
( )( )
( )( )( )( )
0
0
100 0.0262.63 kΩ
0.991 101 5
0.9951 2.63 101 5
4 4.02 V peak-to-peak at base0.995 0.995
E
b E
b b
r
Rvv r R
vv v
π
π
ββ
= =
+= = =
+ + +
⇒ = = ⇒ =
RS
RBRib
S
b
( )1 508 kΩ434 508 234 kΩ
234 234234 0.7 234.7
4.020.997 4.03 V peak-to-peak0.997
ib E
B ib
B ib Sb S S
B ib S
b S S S
R r RR R
R R vv v vR R R
v v v v
π β= + + == =
= ⋅ = =+ +
= ⇒ = ⇒ =
c. ( ) ( )
( ) ( )1
2.63 101 5 1 86.8 kΩ
434 86.8 72.3 kΩ
ib E L
ib
B ib
R r R R
R
R R
π β= + +
= + =
= =
( )( )
( )( )( )( )
( )( )( )( ) ( )
0
0
72.3 0.99 0.99 4.0372.3 0.7
3.99 V peak-to-peak1
1
101 0.8333.99
2.63 101 0.8333.87 V peak-to-peak
b S S
b
E Lb
E L
v v v
vR R
v vr R R
v
π
ββ
⎛ ⎞= = =⎜ ⎟+⎝ ⎠=
+= ⋅
+ +
=+
=
6.46
( )
( )( )
( )( )
( )( )
( )( )
1 2 40 60 24 kΩ60 10 6 V
60 406 0.775 0.0131 mA
24 76 50.984 mA
6 0.7150 0.00680 mA24 151 5
1.02 mA75 0.026
75 1.98 k0.984
150 3.82 k
75 1 65.3 k1
TH
TH
BQ
CQ
BQ
CQ
ib E L
R R R
V
I
I
I
I
r
r
R r R R
π
π
π
β
β
β
ββ ββ
= = =
⎛ ⎞= =⎜ ⎟+⎝ ⎠−= = =
+=
−= = =+
=
= = = Ω
= = Ω
= = + + = Ω=
( )( )( )( )
1 2
1 2
50 130 k1
1
ib
E L ibv
E L ib S
RR R R R RA
r R R R R R Rπ
ββ
= Ω+
= ⋅+ + +
( )( )( )( )
( )( )( )( )
1 2
1 2For 75, 40 60 65.3 17.5 76 0.833 17.5 0.789
1.98 76 0.833 17.5 4For 150, 40 60 130 20.3
151 0.833 20.3 0.8113.82 151 0.833 20.3 4
ib
v v
ib
v v
R R R k
A A
R R R k
A A
β
β
= = = Ω
= ⋅ ⇒ =+ +
= = = Ω
= ⋅ ⇒ =+ +
So 0.789 0.811vA≤ ≤
( )
( )
( )
1
755 2476 17.0
5 1 24 65.3150
5 24151 19.66 24 130
E TH
E L TH ib
i i
i i
iR RA
R R R R
A A
A A
β
β
β
⎛ ⎞⎛ ⎞= + ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠=
⎛ ⎞⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠=
⎛ ⎞⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠
17.0 19.6iA≤ ≤ 6.47 (a)
( ) ( )9 100 on1
9 0.7100
18.350 2.803 mA
100 1518.3200 5.543 mA
100 1201
EBE E E
E
E
E
E
I V I R
IR
I
I
β
β
β
β
⎛ ⎞= + +⎜ ⎟+⎝ ⎠
−=⎛ ⎞ +⎜ ⎟+⎝ ⎠
= = =⎛ ⎞ +⎜ ⎟⎝ ⎠
= = =⎛ ⎞ +⎜ ⎟⎝ ⎠
2.80 5.54 mAEI≤ ≤ , 50, 2.80 VE E E EV I R Vβ= = =
200, 5.54 VEVβ = = (b) 50, 2.748 mA, 0.473 KCQI rπβ = = =
200, 5.515 mA, 0.943 KCQI rπβ = = =
( )1i B E LR R r R Rπ β= + +⎡ ⎤⎣ ⎦
( )( )50 100 0.473 51 1 1 100 25.97 20.6 KiRβ = ⇒ = + = =⎡ ⎤⎣ ⎦
( )( )200 100 0.943 201 1 1 100 101.4 50.3 KiRβ = ⇒ = + = =⎡ ⎤⎣ ⎦ From Fig. (4.68)
( )( )( )( )( )( )
( )( )
( )( )( )( )
11
51 1 1 20.620.6 100.473 51 1 1
50 0.661
201 1 1 50.320050.3 100.943 201 1 1
0.826
E L iv
i SE L
v
v
v
R R RAR Rr R R
A
A
A
π
ββ
β
β
+ ⎛ ⎞= ⋅⎜ ⎟++ + ⎝ ⎠
⎛ ⎞= ⋅⎜ ⎟++ ⎝ ⎠
= ⇒ =
⎛ ⎞= ⇒ = ⎜ ⎟++ ⎝ ⎠
=
6.48
( )
( )( )
( )
( ) ( )
1
1
1so
1For 100, 0.5
100 0.0265.2
0.5
o b L
sb
L
Lv
L
L
V I RVI
r R
RA
r RR k
r k
π
π
π
β
ββ
ββ
= +
=+ +
+=
+ += = Ω
= = Ω
( ) ( )( )( )( )
( )( )
( ) ( )( )( )( )
101 0.5Then min 0.9066
5.2 101 0.5Then 180, 500
180 0.0269.36
0.5181 500
Then max 0.99999.36 181 500
v
L
v
A
R k
r k
A
π
β
= =+
= = Ω
= = Ω
= =+
6.49
RE RL
Rib
rV gmV Ib
R1R2S
IbIS
I0
( )
( )( )
1 2
1 2
110 V, For 5 V
15 10
80, For 0.5 k
E0 b
E L
b Sib
ib E L
CC CEQ
CQ E
E
RI 1+ β IR +R
R RI IR R +R
R r R RV V
I R
R
π β
ββ
β
⎛ ⎞= ⎜ ⎟
⎝ ⎠⎛ ⎞
= ⎜ ⎟⎝ ⎠
= + += =
⎛ ⎞+= − ⎜ ⎟⎝ ⎠
= = Ω
( )( )
( )( )
( )
( )
0 1 2
1 2
1 2
1 2
9.88 mA, 10 mA, 0.123 mA80 0.026
0.211 k9.88
0.211 81 0.5 0.5 20.46 k
1
18 812 20.46
CQ EQ BQ
ib ib
Ei
S E L ib
I I I
r
R R
I R R RAI R R R R +R
R RR R +
π
β
= = =
= = Ω
= + ⇒ = Ω
⎛ ⎞⎛ ⎞= = + ⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
1 2 1 20 1975 20 46. R R . R R+ =⎡ ⎤⎣ ⎦
1 2 5.04 kR R ⇒ Ω
( ) ( )
( )( ) ( )( ) ( )( ) 11
22
2
on 11 5 04 10 0 123 5 04 0 7 10 0 5 7 97 k
7 97 5 04 13 7 k7 97
0 211Now 0 5 or 2 59 1 81
TH BQ TH BE BQ E
o E o
V I R V I R
. . . . . R .R. R . R .. R
r .R R . R .π
= + + +
= + + ⇒ = Ω
= ⇒ = Ω+
= = = Ω+
b
b
(b) ( ) ( )
( ) ( ) ( )( )
0.211 81 0.5 2 32.6 0.5 5.0481 81 0.2 0.134
0.5 2 5.04 32.62.17
ib
i
i
R k
A
A
= + = Ω
⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠=
6.50
( )
( )( )
( )( )
( ) ( )
( )( ) ( )1 1
where 15 3.53.5, 0.75
2120 0.026
4.16 0.75
4.16 121 2 246 Then 120 246 234
0.75 0.00625 120
11 1 234 5 0.00625 2
i TH ib ib E
CEQ CQ
ib
i TH TH
BQ
TH BQ TH BE BQ E
TH CC
R R R R r R
V I mA
r k
R kR R R k
I mA
V I R V on I R
R VR R
π β
β
π
= = + +−= =
= = Ω
= + = Ω= = ⇒ = Ω
= =
= + + +
⋅ ⋅ = = ( ) ( )( )( )
1 2 and
34 0.7 121 0.00625 2
which yields 318 886 R k R k
+ +
= Ω = Ω
6.51 a.
( )( )
12
12Let 24 and 12 V 0.5 A240.493 A, 6.58 mA
75 0.0263.96
0.493
E CEQ CC EQ
CQ BQ
R V V I
I I
rπ
= Ω = = ⇒ = =
= =
= = Ω
RE RLIo
Is Ib
Reb
rV gmV Ib
R1 R2VS
Rrn
( )0 1 Eb
E L
THb S
TH ib
RI IR R
RI IR R
β ⎛ ⎞= + ⎜ ⎟+⎝ ⎠⎛ ⎞
= ⎜ ⎟+⎝ ⎠
( ) ( )( ) ( )
13.96 76 24 8 460
ib E L
ib
R r R RR
π β= + += + ⇒ = Ω
( )
( )
0 1
248 7624 8 460
0.140 74.9 (Minimum value)460
THEi
S E L TH ib
TH
TH
THTH
TH
RI RAI R R R R
RR
R RR
β⎛ ⎞⎛ ⎞= = + ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
= ⇒ = Ω+
dc analysis:
( )
( )( ) ( )( ) ( )( )
1
1
21 2
2
,
1
on1 74.9 24 0.00658 74.9 0.70 0.5 24
13.19136136 74.9 167
136
TH TH CC
BQ TH BE EQ E
V R VR
I R V I R
R
RR RR
= ⋅ ⋅
= + +
= + +
=
= Ω = ⇒ = Ω+
b.
0.493
12 24
1 6
AC load line
Slope 1
248
( )( )
16
For 0.493 0.493 6 Max. swing in output voltage for this design5.92 V peak-to-peak
C ce
ceC
i v
i v
Δ = − Δ
Δ = ⇒ Δ = ⇒=
c.
0 03.96 24 0.0521 24 52 m
1 76ErR R Rπ
β= = = ⇒ = Ω
+
6.52 The output of the emitter follower is
Lo TH
L o
Rv vR R
⎛ ⎞= ⋅⎜ ⎟+⎝ ⎠
Ro
RL
TH O
For Ov to be within 5% for a range of ,LR we have ( )
( ) ( ) ( )( )
( )
min max0.95
min max4 100.95 which yields 0.364
4 10
L L
L o L o
oo o
R RR R R R
R kR R
=+ +
= = Ω+ +
1 2We have 1
So E o
r R R RR R rπ
β⎛ ⎞+
= ⎜ ⎟+⎝ ⎠
The first term dominates 1 2Let , then
40.3641 1
S S
So
R R R Rr R rR π π
β β
≅+ +
≅ ⇒ =+ +
( )4 4or 0.364
1 1 1 1T
CQ
r VI
π ββ β β β
= + = ++ + + +
40.3641
T
CQ
VI β
≅ ++
The factor 41 β+
is in the range of 4 0.04491
= to 4 0.0305.131
= We can set 0.32 To
CQ
VRI
≅ =
Or 0.08125 .CQI mA= To take into account other factors, set 0.15 ,CQI mA=
0.15 0.00136 110BQI mA= =
For 5 ,CEQV V≅ set 5 33.3 0.15ER k= = Ω
Design a bias stable circuit.
( ) ( ) ( )
( )( ) ( )( )( )( ) ( )
2
1 2 1
110 5 10 5
0.1 1 0.1 111 33.3 370 1 5
TH
TH E
TH TH BE BQ E
TH
BQ
RV RR R R
R R kV I R V on I R
ββ
⎛ ⎞= − = −⎜ ⎟+⎝ ⎠= + = = Ω= + + + −
( )( ) ( )( ) ( )( )( )
( )( )( )( )( )( )
1 2
and
1So 370 10 5 0.00136 370 0.7 111 0.00136 33.3 5 1
which yields 594 and 981
1Now
1
1
E L TH ib
TH ib SE L
Tib E L
CQ
R
R k R k
R R R RAv R R Rr R R
VR r R R rI
π
π π
ββ
ββ
− = + + −
= Ω = Ω
+ ⎛ ⎞= ⋅⎜ ⎟++ + ⎝ ⎠
= + + =
For 90, 4 ,LR kβ = = Ω
( )( )( )( )
15.6 , 340.6 91 33.3 4 370 340.6 0.9332
15.6 91 33.3 4 370 340.6 4
ib
v v
r k R k
A A
π = Ω = Ω
= ⋅ ⇒ =+ +
For 90, 10 LR kβ = = Ω
( )( )( )( )
715.4 91 33.3 10 370 715.4 0.9625
15.6 91 33.3 10 370 715.4 4
ib
v v
R k
A A
= Ω
= ⋅ ⇒ =+ +
For 130, 4 LR kβ = = Ω
( )( )( )( )
22.5 , 490 131 33.3 4 370 490 0.9360
22.5 131 33.3 4 370 490 4
ib
v v
r k R k
A A
π = Ω = Ω
= ⋅ ⇒ =+ +
For 130, 10 LR kβ = = Ω
( )( )( )( )
1030 131 33.3 10 370 1030 0.9645
22.5 131 33.3 10 370 1030 4
ib
v v
R k
A A
= Ω
= ⋅ ⇒ =+ +
( ) ( )( ) ( )
Now min min . 3.73sinmax max . 3.86sin
3.5%
v
S
O S
O
O
O
v A v tv A v tv
vv
ωω
= == =
Δ=
6.53
( ) ( )( )( ) ( ) ( )
( )( ) ( ) ( )( )
2 21 12
so 0.289 2 0.2890.409
0.409 12 4.91 4.91Need a gain of 0.982
5
AVG L L L
L L
L
L L L
P i rms R i rms
i rms A i peaki peak Av peak i peak R V
= ⇒ =
= ⇒ === ⋅ = =
=
With 10 ,SR k= Ω we will not be able to meet this voltage gain requirement. Need to insert a buffer or an op-amp voltage follower (see Chapter 9) between SR and 1CC .
( )( )( )
( )
( )( )
1Set 0.5 , 12 12 8 3
24 0.5 8 32
50Let 50, 0.5 0.49 51
50 0.0262.65
0.49
EQ CEQ
EQ E CEQ E E
CQ
T
CQ
I A V V
I R V R R
I A
VrIπ
β
β
= = − − =
= + = + ⇒ = Ω
= = =
= = = Ω
( )( ) ( )( )
( )( )( )( )
( )( )( )( )
1 2.65 51 32 12
448
1 51 32 120.994
1 2.65 51 32 12
ib E L
ib
E Lv
E L
R r R R
R
R RA
r R R
π
π
β
ββ
= + + = +
= Ω
+= = =
+ + +
So gain requirement has been met.
( ) ( )
( ) ( ) ( )( )
1 2
1 2
2
1 2
1 22
1 2
0.49 0.0098 9.8 50
24Let 10 98
So that 245
24 12 12
0.009824 0.7 0.5 32
245 245Now 245
BQ
R B
TH BQ TH BE EQ E
I A mA
I I mAR RR RRV I R V on I R
R RR RR
R R
= = =
≅ ≅ =+
+ = Ω
= − = + + −+
⎛ ⎞ = + +⎜ ⎟⎝ ⎠
= −
So we obtain 5 2
2 2 2 14 10 0.0882 16.7 0 which yields 175 and 70 R R R R−× + − = = Ω = Ω 6.54 (a)
( ) ( )
( ) ( )
( )( )
( )( ) ( )( )
1 2
2
1 2
25.6 10.4 7.40
10.4 18 5.2 10.4 25.61
5.2 0.7 0.0117 7.40 126 3
Then 1.46 and 1.47
18 1.46 4 1.47 3
TH
TH CC
TH BQ TH BE BQ E
BQ
CQ EQ
CEQ CC CQ C EQ E
CEQ
R R R k
RV V VR R
V I R V on I R
I mA
I mA I mA
V V I R I RV
β
= = = Ω
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= + + +
−= =+
= =
= − −= − − ⇒ 7.75 CEQV V=
(b) ( ) ( )125 0.026
2.23 1.46
1.46 56.2 /0.026m
r k
g mA V
π = = Ω
= =
RLRC
Vo
r Ib
RS
RTH
REIs Ib
Ie
Re
( )( )
( )( )
( )
( ) ( )
( )( ) ( )( )
2.23 7.40 0.0764 1 126
100 3100 3 0.0764
or 0.974
1
125Then 0.974 0.974 4 41 126
Then 1.93 1.93 /
THe
S Ee s s
S E e
e s
o c C L e C L
oC L
s
om
s
r RR k
R RI I I
R R R
I I
V I R R I R R
VR R
IV
R k V mAI
π
β
ββ
ββ
+ += = = Ω+
− −= ⋅ = ⋅
++
= −
⎛ ⎞= − = −⎜ ⎟+⎝ ⎠
⎛ ⎞ ⎛ ⎞= − − =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠
= = Ω =
(c)
( ) ( ) ( )100 3 0.0764 0.0744E eS ss s s
V I R R R I I= = =
or0.0744
ss
VI =
( )which yields 0.0744 1.93
or 25.9
o o
s s
ov
s
V VI V
VAV
= =
= =
6.55 (a)
( )
( ) ( )( )( )
( )( ) ( )( ) ( )( )( )
1 2
1 2
1 1
1
2
, 12 , 100
Let 50 , 0.5 1
100 0.0260.5 0.005 , 5.2 100 0.5
1 1 50 12 0.005 50 0.7 101 0.005 0.5
which yields R 500 and 55
C Lv L
CQ
TH BQ TH BE BQ E
BQ
TH CC
R RA R k
r R RR R k I mA
V I R V on I R
I mA r k
R VR R
kR
π
π
ββ
β
= = Ω =+
= Ω == + + +
= = = = Ω
⋅ ⋅ = = + +
= Ω=
( )( ).6
100 12 1210.9, Design criterion is met.
5.2 50v
k
A
Ω
= =+
(b)
( )( ) ( )( )
( )( )( )
0.5 , 0.505
12 0.5 12 0.505 0.5 5.75
0.5, 19.23 /0.026
19.23 12 12 115
CQ EQ
CEQ CEQ
v m C L m
v v
I mA I mA
V V V
A g R R g mA V
A A
= =
= − − ⇒ =
= = =
= ⇒ =
6.56 a. Emitter current
( )( )( )
( )( )
0 5 mA0 5 0 00495 mA101
0 5 1 0 5 V
on 0 5 0 7 1 20 V
1 20 0 00495 100 1 7 V
EQ CC
BQ
E EQ E E
B E BE B
C B BQ B C
I I ..I .
V I R . V .
V V V . . V .
V V I R . . V .
= =
= =
= = ⇒ =
= + = + ⇒ =
= + = + ⇒ =
b. ( )( )
( )( )( )( )
100 0 0265 25 k
100 0 00495
100 0 0049519 0 mA/V
0 026m
.r .
.
.g .
.
π = = Ω
= =
V0
RE
gmV
V r
RS
RLRBVS
Ri
RSRE
gmV
V r
RE
RE RS VS
( )
( )
( ) ( ) ( )
0 4971
19 0 4971 100 19 37
o m B L
E ieS S
E ie S
o S
v
V g V R RR R
V V . VR R R
V . VA .
π
π
= −
= − ⋅ = −+
==
c.
VX
VRE r
IXgmV
,
1 1 1
1 1or 1 5 25319
0 84 0 05252 49 4
X XX m X
E
Xm
X i E
i Em
i i
V VI g V V VR r
I gV R R r
R R r .g
R . . R .
π ππ
π
π
= + − = −
= = + +
= =
= ⇒ = Ω
6.57 (a) 1 mA, 0.9917 mAEQ CQI I= =
( )( )5 0.9917 2 3.017 V0.7 V
3.72 V
C
E
CEQ
VV
V
= − == −=
(b) ( )
( )( )
0.9917 38.14 mA/V0.02638.14 2 10 63.6
v m C L
m
v v
A g R R
g
A A
=
= =
= ⇒ =
6.58 (a)
( )( ) ( )( )
10 0.7 0.93 mA10
0.921 mA
20 0.93 10 0.921 56.10 V
EQ
CQ
ECQ
ECQ
I
I
VV
−= =
=
= − −=
(b)
( ) ( )( )
0.921 35.42 mA/V0.026
35.42 5 50
161
m
v m C L
v
g
A g R R
A
= =
= =
=
6.59 (a) 0.93 mA, 0.921 mAEQ CQI I= =
6.10 VECQV =
(b) 0.921 35.42 mA/V 2.82 K0.026mg rπ= = =
From Eq. 6.90 ( )
( )( )
( )( ) [ ]
1
35.42 50 5 2.82 10 0.10.1 101
35.42 4.5450.0218
0.135.1
C Lv m E S
S
v
v
R R rA g R R
R
A
A
π
β⎡ ⎤
= ⎢ ⎥+⎣ ⎦
⎡ ⎤= ⎢ ⎥⎣ ⎦
=
=
6.60 (a)
( )
( ) ( )
60 1 0.984 mA61
1 100 0.761
2.34 V
CQ CQ
CEQ BQ B BE
CEQ
I I
V I R V on
V
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
⎛ ⎞= + = +⎜ ⎟⎝ ⎠
=
(b)
( )1
0.984 37.85 mA/V0.0261.59 K
B Lv m S
S
m
R R rA g R
R
g
r
π
π
β⎡ ⎤
= ⎢ ⎥+⎣ ⎦
= =
=
( ) ( )37.85 100 2 1.59 0.050.05 61
1484 0.0261 0.05
25.4
v
v
A
A
⎡ ⎤= ⎢ ⎥⎣ ⎦= ⎡ ⎤⎣ ⎦=
6.61
( ) ( )3
36
2 5 5 5 10So we need 2 10 2
2 5 10
s o
om
s
i peak . A, V peak mVvR ki .
= =×= = = × = Ω×
2
2
m
From Problem 4.54
( )
( )
1
Let 4 , 5 , 2 Now 120, so we have
1202 4 5 2 204121
Then 0 9075
50 2 1 923 , so tha
S EoC L
s S E ie
C L E
S E S E
S E ie S E ie
S E
S E e
S E
R RV R RI R R R
R k R k R k
R R R R.
R R R R R R
R R.
R R RR R . k
ββ
β
⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
= Ω = Ω = Ω=
⎛ ⎞ ⎛ ⎞⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠
=+
= = Ω
( )( )
( )( )
( )
t 0 196 Assume 3
5 4 2 3 0 333 120 0 026
9 37 0 333
9 370 1961 121
which yields 14 35 Now
1 0 00833 , 120
ie
CEQ
CC CQ C E CEQ
CQ CQ
TH THie
TH
TH BQ TH BE EQ E
BQ EQ
R . kV V
V I R R VI I . mA
.r . k
.r R . RR .
R . kV I R V on I R
I . mA I
π
π
β
= Ω=
≅ + += + + ⇒ =
= = Ω
+ += ⇒ =+
= Ω= + +
= = ( )
( )( ) ( )( ) ( )( )1 1
1
2
121 1 1 008 120
1 1 14 35 5 0 00833 14 35 0 7 1 008 2
which yields 25 3 and 33 2
TH TH CC
. mA
V R V . . . . .R R
R . kR . k
⎛ ⎞= =⎜ ⎟⎝ ⎠
= ⋅ ⋅ = = + +
= Ω= Ω
6.62 a.
( )( )( )
20 0.7 1.93 mA10
1.91 mA
on25 0.7 1.91 6.5 13.3 V
EQ
CQ
ECQ CC EB C C
ECQ
I
I
V V V I RV
−= =
=
= + −= + − ⇒ =
b.
V0
RE
IS
RS
Ie
Ib
hieRLRCVS
RiehfeIb
Neglect effect oeh From Problem 6-16, assume 2 45 3 7 k
80 120ie
fe
. h .h
≤ ≤ Ω≤ ≤
( )( )
( )
, 1
, 1
11
o fe b C L
ie Eie e S
fe E ie
e Sb S
fe S E ie
fe Ev C L
fe E ie S E ie
V h I R R
h RR I Ih R R
I VI I
h R R R
h RA R Rh R R R R R
=
⎛ ⎞= = ⎜ ⎟+ +⎝ ⎠
⎛ ⎞= =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟+ + +⎝ ⎠ ⎝ ⎠⎝ ⎠
High gain device: 3 7 k , 120ie feh . h= Ω =
( )
3 7 0 0306 k121
10 0 0306 0 0305120 10 16 5 5 2 711121 10 0 0306 1 0 0305
ie
E ie
v v
.R .
R R . .
A . A .. .
= = Ω
= =
⎛ ⎞ ⎛ ⎞⎛ ⎞= ⇒ =⎜ ⎟ ⎜ ⎟⎜ ⎟+ +⎝ ⎠ ⎝ ⎠⎝ ⎠
Low gain device: 2 45 k , 80ie feh . h= Ω =
( )
2 45 0 03025 k81
10 0 03025 0 030280 10 16 5 5 2 70 So constant81 10 0 03025 1 0 0302
ie
E ie
v v v
.R .
R R . .
A . A . A. .
= = Ω
= =
⎛ ⎞ ⎛ ⎞⎛ ⎞= ⇒ = ≈⎜ ⎟ ⎜ ⎟⎜ ⎟+ +⎝ ⎠ ⎝ ⎠⎝ ⎠
2 70 2 71v. A .≤ ≤ c. i E ieR R R=
We found 0 0302 0 0305 ki. R .≤ ≤ Ω
Neglecting , 6 5 koe o Ch R R .= = Ω 6.63 a. Small-signal voltage gain
( ) ( )( )
( )( )
25 1For 3 on 3 0 7 2 3
5 2 3 2 70
For 1 mA, 2 7 k
1 38 5 mA/V0 02638 5 2 7 1 28 1
v m C L m C
ECQ C ECQ EB C
CC CQ C C CQ CQC C
CQ C
m
v
A g R R g RV V V V V . V .
. .V I R V I IR R
I R .
g ..
A . . .
= ⇒ == ⇒ = − + = − + ⇒ = −
−− + = ⇒ = = =
= = Ω
= =
= =
Design criterion satisfied and ECQV satisfied.
( )
( )
101 1 1.01 mA100
5 0.7on 4.26 k1.01
E
EE E E EB E E
I
V I R V R R
⎛ ⎞= =⎜ ⎟⎝ ⎠
−= + ⇒ = ⇒ = Ω
b. ( )( )
100 0.026
2.6 k , 38.5 mA/V,1 m
To
CQ
Vr r g rIπ πβ= = ⇒ = Ω = = ∞
6.64 a.
( )
( )( )
( )( )
21 1
1 2
1 1 2
1
1 1 1
1 1
01 01
20 10 2.0 V20 80
20 80 16 k2 0.7 0.0111 mA
16 101 11.111.11 mA 42.74 mA/V0.026
100 0.0262.34 k
1.11
1.11
TH CC TH
TH
B
C m m
RV V VR R
R R R
I
I g g
r r
r r
π π
⎛ ⎞ ⎛ ⎞= = ⇒ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= = = Ω
−= =+
= ⇒ = ⇒ =
= ⇒ = Ω
∞= ⇒ = ∞
( )42
3 4
42 3
15 10 1.50 V15 85
15 85 12.75 k
TH CC
TH
RV V
R RR R R
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= = = Ω
( )( )
( )( )
2
2 2
2 2
02
1.50 0.70 0.01265 mA12.75 101 0.5
1.2651.265 mA 48.65 mA/V0.026
100 0.0262.06 k
1.26
B
C m m2
I
I g g
r r
r
π π
−= =+
= ⇒ = ⇒ =
= ⇒ = Ω
= ∞
b. ( )( )
( ) ( )( )1 1 1 1
2 2 2 2
42.7 2 85.48
48.5 4 4 97.3v m C v
v m C L v
A g R A
A g R R A
⇒= − = − = −
= − = − ⇒ = −
c. Input resistance of 2nd stage
( ) ( )( )
2 3 4 2
2
1 1 1 2
1
15 85 2.0612.75 2.06 1.773 k
42.7 2 1.77B40.17
i
i
v m C i
v
R R R rR
A g R RA
π= == ⇒ = Ω
′ = − = −′ = −
( )( )( )( )1 2
Overall gain: 40.17 97.3 3909
If we had 85.48 97.3 8317v v
v v
A A
A A
= − − ⇒ =
⋅ = − − =
Loading effect reduces overall gain 6.65 a.
( )
( )( )
( )( )
21 1
1 2
1 1 2
1
1
1 1
1 1
01 0
12.7 12 1.905 V12.7 67.3
12.7 67.3 10.68 k1.905 0.70 0.00477 mA
10.68 121 20.572 mA
0.572 22 mA/V0.026120 0.026
5.45 k0.572
0.572
TH CC TH
TH
B
C
m m
RV V VR R
R R R
I
I
g g
r r
r r
π π
⎛ ⎞ ⎛ ⎞= = ⇒ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= = = Ω
−= =+
=
= ⇒ =
= ⇒ = Ω
∞= ⇒ 1 = ∞
( )
( )( )
( )( )
42 2
3 4
2 3 4
2
2 2
2 2
02
45 12 9.0 V45 15
15 45 11.25 k9.0 0.70 0.0405 mA
11.25 121 1.64.86 mA
4.86 187 mA/V0.026120 0.026
0.642 k4.86
TH CC TH
TH
B
C2
m m
RV V VR R
R R R
I
I
g g
r r
r
π π
⎛ ⎞ ⎛ ⎞= = ⇒ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠= = = Ω
−= =+
=
= ⇒ =
= ⇒ = Ω
= ∞
b.
( ) ( ) ( ) ( )
( ) ( )
1
1 1
2
2 2
0.577 mA12 0.572 10 0.577 2 5.13 V
4.9012 4.90 1.6 4.16 V
E
CEQ CEQ
E
CEQ CEQ
IV V
IV V
== − − ⇒ =
== − ⇒ =
125.13
0.572
AC load lineQ1
Slope 1
107.92
1
4.42 K
124.16
4.86
Q2
AC load line
Slope 1
1.60.25
1
0.216 K
( ) ( )( ) ( )
2 3 4
2 2
2
2
10.642 121 1.6 0.2526.815 45 26.87.92 k
i ib
ib E L
ib
i
i
R R R RR r R R
RRR
π β== + += +=== Ω
c. ( ) ( )( )
( )( )( )( )
( )( )( )( )( )( )
1 1 1 2 2
22
2 2
22 10 7.92 97.21
1
121 0.2160.976
0.642 121 0.216
Overall gain 97.2 0.976 94.9
v m C i v
E Lv
E L
A g R R AR R
Ar R Rπ
ββ
= − = − ⇒ = −+
=+ +
= =+
= − = −
d. 1 2 1
22
3 4 1
67.3 12.7 5.45 3.61 k
where1
15 45 10 5.29 k0.642 5.29 1.6 0.049 1.6 47.6
121
iS iS
So E
S C
S
o o
R R R r R
r RR R
R R R RR
R R
π
π
β
= = ⇒ = Ω
+=
+== ⇒ = Ω
+= ⇒ ⇒ = Ω
e.
( )( )
1 , 4.860.216 k
4.86 0.216 1.05 VMax. output voltage swing 2.10 V peak-to-peak
C ce C
ce
i v i
v
−Δ = ⋅ Δ Δ =Ω
Δ = ==
6.66 (a)
( )
( )
( )
1
2
2 2
2
1 1
5 2 0.772
0.0500.7 1.4 0.5
72 1.4 69.9 1
69.9 0.699 100
1.4 0.699 2.08 1
R
R
C C
B
C C
I mA
I mA
I I mA
I mA
I I mA
ββ
ββ
−= =
= =
⎛ ⎞= − ⇒ =⎜ ⎟+⎝ ⎠
= =
⎛ ⎞= + ⇒ =⎜ ⎟+⎝ ⎠
(b)
gm1V1 gm2V2
Vo
r2
r1
50 Ω
0.5 k V2
Vs V1
1 2s oV V V Vπ π= + +
(1)
( )
( )( )
( ) ( )
2 22 2
2
2
2
2 2
0.050.5
100 0.0260.0372
69.969.9 2688 /
0.0261 1 2688 0.05 so that 1
0.5 0.0372 135.8
o m
m
oo
V VV g Vr
r k
g mA V
VV V V
π ππ
π
π
π π
⎛ ⎞= + +⎜ ⎟⎝ ⎠
= = Ω
= =
⎛ ⎞= + + =⎜ ⎟⎝ ⎠
(2)
( ) ( )
( ) ( ) ( ) ( )
( )
1 2 21 1
1 2
1
1
1 2
1 2 1
0.5100 0.026
1.25 2.08
2.08 80 /0.0261 1 180
1.25 0.5 0.0372
80.8 28.88 28.88 or (2) 0.00261136.7
Then 0.00261 1.136.7
m
m
oo
os o o o
V V Vg V
r r
r k
g mA V
V V
VV V V V
VV V V V
π π ππ
π π
π
π π
π π π
+ = +
= = Ω
= =
⎛ ⎞ ⎛ ⎞+ = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞= = =⎜ ⎟⎝ ⎠
= + + = ( )00993 or 0.990ov
s
VA
V= =
(c) ( )[ ]1 1ib xR r Rπ β= +
Vo
gm2V2
r2Vx
Ix
V20.5 k
50
( ) ( )
2 22
2 2
22 2
2
2 2
2
1 10.5 0.5
0.05 0.051
1 0.050.05 0.05 1 1
0.5
We find 4.74
Then 1.25 101 2.89 480
x
o xx m
x mx
x m
xx
x
ib ib
V VI V
r rV V V
I g V
I gV
I V g
rV
R kI
R R k
π ππ
π π
ππ
π
π
⎛ ⎞= + = +⎜ ⎟
⎝ ⎠−
= = +
⎛ ⎞+⎜ ⎟⎛ ⎞ ⎝ ⎠− = + =⎜ ⎟ ⎛ ⎞⎝ ⎠+⎜ ⎟
⎝ ⎠
= = Ω
= + ⇒ = Ω
gm1V1 gm2V2r1
r2
Ix
0.5 k
50
V1
V2
Vx
To find Ro:
(1) 22 2
20.05 0.5x
x mV V
I g Vr
ππ
π
= − −
(2) ( ) ( ) ( )12 1 1 2 1 2 1
1
10.5 80 0.5 0.0372 or 2.771.25m
VV g V r V V V
rπ
π π π π π ππ
⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
(3) ( )1 2 1 10 2.77 0 x xV V V V V Vπ π π π+ + = ⇒ + + =
( )( ) ( ) ( )
( )
1
2
2 22
so that 0.2653 and 2.77 0.2653 0.735
1Now 0.05 0.5
1So that 0.735 2688 which yields 0.496 0.05 0.5 0.0372
x
x x
xx m
x xx x o
x
V VV V V
VI V gr
V VI V R
I
π
π
ππ
= −= − = −⎡ ⎤⎣ ⎦
⎛ ⎞= − +⎜ ⎟⎜ ⎟
⎝ ⎠⎡ ⎤
= + + = = Ω⎢ ⎥⎢ ⎥⎣ ⎦
6.67 a.
( )
( ) ( )
1 2
2
1 2
1 1 2 2 22
2 1 1
335 125 91.0 k
125 10 2.717 V125 335
1 1
TH
TH CC
TH B TH BE BE E E
E E B
R R R
RV VR R
V I R V V I R
I I Iβ β
= = = Ω
⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞= =⎜ ⎟+⎝ ⎠
= + + +
= + = +
( ) ( )
( ) ( )( )( )
1 12
1
2 1 1
2 2
2 1
2.717 1.40 0.128 91.0 101 112.8
1 100 101 0.128 1.29 m , 1.31 m
1.29 0.0128 1.30 m
B B
C
C E B
C E
RC C C
I I
I
I I II I
I I I
μ
μ
β β β μ
−= ⇒ = Α+
= Α
= = + = Α= Α = Α
= + = + = Α
( ) ( )( ) ( )2 2
2
1 2 2
1
10 10 1.30 2.2 7.14 V1.30 1 1.30 V
7.14 1.30 5.84 V5.84 0.7
5.14 V
C RC C
E E E
CE
CE CE BE
CE
V I RV I RVV V VV
= − = − == = =
= − == − = −=
Summary: 21
1 2
12.8 1.29 m5.14 V 5.84 V
C C
CE CE
I IV V
μ= Α = Α= =
b.
1
2
0.0128 0.492 m / V0.026
1.292 49.7 m / V0.026
m
m
g
g
= = Α
= = Α
V0
r2
r1
V2
V1
Ib
RC
VS
gm2V2
gm1V1
R1R2
Rib
( )0 1 1 2 2
1 2 1 2,m m C
S S
V g V g V RV V V V V V
π π
π π π π
= − += + = −
12 1 1 2
1
2 1 21
1
mV
V g V rr
V V rr
ππ π π
π
π π ππ
β
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞+= ⎜ ⎟
⎝ ⎠
( )( )
0 1 2 2 2
0 1 2 1 2
m S m C
m S m m C
V g V V g V R
V g V g g V Rπ π
π
= − − +⎡ ⎤⎣ ⎦= − + −⎡ ⎤⎣ ⎦
( )( )
( ) ( )
22 2
1
2 22
1 1
1
1 1 1
S
S
rV V V
r
r rV V
r r
ππ π
π
π ππ
π π
β
β β
⎛ ⎞= − + ⎜ ⎟
⎝ ⎠⎡ ⎤⎛ ⎞ ⎛ ⎞
+ + = +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦
( )( )
( )
( )( )( )
( )
2
10 1 2 1
2
1
0
1
1 1
2.0149.7 0.492 1012030.492 2.2
2.011 101203
55.2
S
m S m m C
vS
v
rVr
V g V g g Rrr
VAV
A
π
π
π
π
β
β
⎧ ⎫⎛ ⎞+⎪ ⎪⎜ ⎟
⎪ ⎪⎝ ⎠= − + − ⋅⎨ ⎬⎛ ⎞⎪ ⎪+ + ⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
=
⎧ ⎫⎛ ⎞− ⎜ ⎟⎪ ⎪⎪ ⎪⎝ ⎠= − +⎨ ⎬⎛ ⎞⎪ ⎪+ ⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭
= −
c.
( )( )( )
1 2
1 2
0
1203 101 2.01 406 k91 406 74.3 k
2.2 k
is ib
ib
is is
C
R R R R
R r r
R R
R R
π πβ=
= + += + = Ω= = Ω =
= = Ω
6.68
V2
Ix
R0
r1 ro1
r2 ro2
V1
Vx
VA
gm1V1
gm2V2
(1) 2 2 1 12 1
x x Ax m m
o o
V V VI g V g Vr rπ π
−= + + +
(2) 1 11 1 2
x A Am
o
V V Vg Vr r rπ
π π
−+ =
(3) 2 1AV V Vπ π= = − Then from (2)
11 1 1 2
2 1 2 12 1 1 1 2 1
1 1
1 1 1(1) or
xA m
o o
x x Ax m A m A x x A m m
o o o o o o
V V gr r r r
V V VI g V g V I V V g gr r r r r r
π π
⎛ ⎞= + +⎜ ⎟⎜ ⎟
⎝ ⎠⎛ ⎞ ⎛ ⎞
= + + − − = + + − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
Solving for AV from Equation (2) and substituting into Equation (1), we find
11 1 2
1 22 1 1 2 1 1 2
1
1 1
1 1 1 1 1
For 100, 100 , 1
mx o
ox
m mo o o
A C Bias
gV r r rRI
g gr r r r r r rV V I I mA
π π
π π π π
β
+ += =
⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠= = = =
( )( )1 2
1 2
1 2
100 100 1100 0 026
2 6 1
1 38 46 0 026
o o
m m
r r k
.r r . k
g g . mA/V.
π π
= = = Ω
= = = Ω
= = =
1 138 46100 2 6 2.6
Then 1 1 1 1 138 46 38 46
100 100 2 6 2.6 100 2 6 2.6
or 50.0
o
o
..
R. .
. .
R k
+ +=
⎛ ⎞ ⎛ ⎞+ + + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠= Ω
2Now 1 , 0C BiasI mA I= =
2 21Replace by , 0 01
1 1C C
Bias CI II I . mAββ β β
⋅ = ≅+ +
( )( )
2 1
2 1
2 1
100 100100 , 10,000 1 0 01
1 38 46 , 0 3846 0 026100 0 026
2 6 , 260 1
Then 66 4
o o
m m
o
r k r k.
g . mA/V g . mA/V.
.r . k r k
R . k
π π
= = Ω = = Ω
= = =
= = Ω = Ω
= Ω
6.69 a.
( )
( )( )
1 2
2
1 2
93 7 6 3 5 90
6 3 12 0 756 V6 3 93 7
0 756 0 70 0 00949 mA5 90
0 949 mA12 0 949 6 6 305 V
TH
TH CC
BQ
CQ
CEQ CEQ
R R R . . . k
RV VR R
. .. .. .I ..
I .V . V .
= = = Ω
⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞= =⎜ ⎟+⎝ ⎠
−= =
== − ⇒ =
Transistor: ( )( )0.949 6.305 5.98 mWQ CQ CEQ QP I V P≈ = ⇒ =
( ) ( )22: 0.949 6 5.40 mWC R CQ C RR P I R P= = ⇒ = b.
2
126.31
0.949
AC load line
Slope 1
6105
1
5.68 K
( ) ( )( )( ) ( )
0
0
220
100 105 k0 949
Peak signal current 0.949 mAmax 5 68 0 949 5 39 V
max 5 391 1 2 42 mW2 2 6RC RC
C
r.
V . . .
V .P P .
R
= = Ω
== =
⎡ ⎤= ⋅ = ⇒ =⎢ ⎥
⎢ ⎥⎣ ⎦
6.70 (a)
( ) ( )
( )( )
10 110 0 7 0 00369
100 121 200 443 0 447
BQ B BE BQ E
BQ
CQ EQ
I R V on I R.I . mA
I . mA, I . mA
β= + + +−= =
+= =
( ) ( )2For : 0 443 10 1 96 C RC RCR P . P . mW= ⇒ =
( ) ( )2For : 0 447 20 4 0 E RE RER P . P . mW= ⇒ = (b)
( ) ( ) ( )2 2
0 667 0 443 0 224 1 1Then 0 224 102 20 251
C
RC C C
RC
i . . . mA
P i R .
P . mW
Δ = − =
= Δ =
=
6.71 a.
( )( )
( )( ) ( )( )EQ
10 0 7 0 00596 mA50 151 100 894 mA, 0 90 mA20 0 894 5 0 90 10 6 53 V
BQ
CQ
ECQ ECQ
.I .
I . I .V . . V .
−= =+
= == − − ⇒ =
( )( )( ) ( )( ) ( )
22
22
0 894 6 53 5 84 mW
0 894 5 4 0 mW
0 90 10 8 1 mW
Q CQ ECQ Q
RC CQ C RC
RE EQ E RE
P I V . . P .
P I R . P .
P I R . P .
≅ = ⇒ =
≅ = ⇒ =
≅ = ⇒ =
b.
206.53
0.894
AC load line
Slope 152
1
1.43 K
( )( )
( ) ( )
( ) ( )
0
2
2
11.43 k0.894 0.894 1.43 1.28 V
5 0.639 mA5 2
1 0.639 2 0.408 mW21 0.894 0.639 5 0.163 mW20
5.84 0.408 0.163 5.27 mW
C ec
C ec
C
RL RL
RC RC
RE
Q Q
i v
i v
i i
P P
P P
P
P P
−Δ = ⋅ ΔΩ
Δ = ⇒ Δ = =
⎛ ⎞Δ = Δ =⎜ ⎟+⎝ ⎠
= ⇒ =
= ⋅ − ⇒ =
=
= − − ⇒ =
6.72
( )( )
( )( ) ( )( )EQ
10 0 70 0 00838 mA100 101 100 838 mA, 0 846 mA20 0 838 10 0 846 10 3 16 V
BQ
CQ
CEQ CEQ
.I .
I . I .V . . V .
−= =+
= == − − ⇒ =
203.16
0.838
AC load line
Slope 1
RERLr0
0100 119 k
0 838r
.= = Ω
Neglecting base currents: a.
( )( )( )2
1 k1 1slope
10 1 119 0 902 k1
0 902 k0 838 0 902 0 838 0.756V
0.7561 0.286 mW2 1
L
C ce
C ce
RL RL
R
.
i V.
i . V . .
P P
= Ω− −= =
Ω−Δ = ⋅ Δ
ΩΔ = ⇒ Δ = =
= ⇒ =
b.
10 k1 1slope
10 10 119 4 80
LR
.
= Ω− −= =
( )( )For 0 838 0 838 4 80 4 02C cei . v . . .Δ = ⇒ Δ = =
( )23 161Max. swing determined by voltage 0 499 mW2 10RL RL
.P P .= ⇒ =
6.73 a.
( )( )
( )( ) ( )( )EQ
10 0 7 0 00838 mA100 101 100 838 mA, 0 846 mA20 0 838 10 0 846 10 3 16 V
BQ
CQ
CEQ CEQ
.I .
I . I .V . . V .
−= =+
= == − − ⇒ =
( )( )( ) ( )22
0 838 3 16 2 65 mW
0 838 10 7 02 mW
Q CQ CEQ Q
RC CQ C RC
P I V . . P .
P I R . P .
≅ = ⇒ =
≅ = ⇒ =
b.
203.16
0.838
1 RCRL
1 101
1 0.909 K
AC load line
Slope
( )( )
( ) ( )
( ) ( )
0
2
2
10 909 k
For 0 838 0 909 0 838 0 762 V
10 0 762 mA10 1
1 0 762 1 0 290 mW21 0 838 0 762 10 0 0289 mW2
2 65 0 290 0 0289 2 33 mW
C ce
C ce
CC C
C L
RL RL
RC RC
Q Q
i v.i . v . . .
Ri i i .
R R
P . P .
P . . P .
P . . . P .
−Δ = ⋅ ΔΩ
Δ = ⇒ Δ = =
⎛ ⎞ ⎛ ⎞Δ = Δ = Δ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
= ⇒ =
= ⋅ − ⇒ =
= − − ⇒ =
