Organizational Structure Running A Successful Business
Ch06s
1. Chapter 6
Problem Solutions
6.1
a.
I CQ 2
gm = = ⇒ g m = 76.9 mA/V
VT 0.026
β VT (180 )( 0.026 )
rπ = = ⇒ rπ = 2.34 kΩ
I CQ 2
VA 150
r0 = = ⇒ r0 = 75 kΩ
I CQ 2
b.
0.5
gm = ⇒ g m = 19.2 mA/V
0.026
(180 )( 0.026 )
rπ = ⇒ rπ = 9.36 kΩ
0.5
150
r0 = ⇒ r0 = 300 kΩ
0.5
6.2
(a)
I CQ 0.8
gm = = = 30.8 mA/V
VT 0.026
β VT (120 )( 0.026 )
rπ = = = 3.9 K
I CQ 0.8
VA 120
ro = = = 150 K
I CQ 0.8
(b)
0.08
gm = = 3.08 mA/V
0.026
(120 )( 0.026 )
rπ = = 39 K
0.08
120
ro = = 1500 K
0.08
6.3
I CQ I CQ
gm = ⇒ 200 = ⇒ I CQ = 5.2 mA
VT 0.026
β VT (125)( 0.026 )
rπ = = ⇒ rπ = 0.625 kΩ
I CQ 5.2
VA 200
r0 = = ⇒ r0 = 38.5 kΩ
I CQ 5.2
6.4
2. I CQ I CQ
gm = ⇒ 80 = ⇒ I CQ = 2.08 mA
VT 0.026
β VT β ( 0.026 )
rπ = ⇒ 1.20 = ⇒ β = 96
I CQ 2.08
6.5
(a)
2 − 0.7
I BQ = = 0.0052 mA
250
I C = (120 )( 0.0052 ) = 0.624 mA
0.624
gm = ⇒ g m = 24 mA / V
0.026
(120 )( 0.026 )
rπ = ⇒ rπ = 5 k Ω
0.624
ro = ∞
⎛ r ⎞ ⎛ 5 ⎞
(b) Av = − g m RC ⎜ π ⎟ = − ( 24 )( 4 ) ⎜ ⎟ ⇒ Av = −1.88
⎝ rπ + RB ⎠ ⎝ 5 + 250 ⎠
v v
(c) vS = O = O ⇒ vS = −0.426sin100t V
Av −1.88
6.6
I CQ
gm = , 1.08 ≤ I CQ ≤ 1.32 mA
VT
1.08 1.32
≤ gm ≤ ⇒ 41.5 ≤ g m ≤ 50.8 mA/V
0.026 0.026
β VT (120 )( 0.026 )
rπ = ; rπ ( max ) = = 2.89 kΩ
I CQ 1.08
(80 )( 0.026 )
rπ ( min ) = = 1.58 kΩ
1.32
1.58 ≤ rπ ≤ 2.89 kΩ
6.7
a.
β VT (120 )( 0.026 )
rπ = 5.4 = = ⇒ I CQ = 0.578 mA
I CQ I CQ
1 1
VCEQ = VCC = ( 5 ) = 2.5 V
2 2
VCEQ = VCC − I CQ RC ⇒ 2.5 = 5.0 − ( 0.578 ) RC ⇒ RC = 4.33 kΩ
I CQ 0.578
I BQ = = = 0.00482 mA
β 120
VBB = I BQ RB + VBE ( on )
= ( 0.00482 )( 25 ) + 0.70 ⇒ VBB = 0.820 V
b.
3. β VT (120 )( 0.026 )
rπ = = = 5.40 kΩ
I CQ 0.578
I CQ 0.578
gm = = = 22.2 mA/V
VT 0.026
VA 100
r0 = = = 173 kΩ
I CQ 0.578
⎛ r ⎞
V0 = − g m ( r0 RC ) Vπ , Vπ = ⎜ π ⎟ VS
⎝ rπ + RB ⎠
⎛ r ⎞ β ( r0 RC )
Av = − g m ⎜ π ⎟ ( r0 RC ) = −
⎝ rπ + RB ⎠ rπ + RB
(120 ) ⎡173
⎣ 4.33⎤
⎦ (120 )( 4.22 )
Av = − =− ⇒ Av = −16.7
5.40 + 25 30.4
6.8
a.
1
VECQ = VCC = 5 V
2
VECQ = 10 − I CQ RC ⇒ 5 = 10 − ( 0.5 ) RC ⇒ RC = 10 kΩ
I CQ 0.5
I BQ = = = 0.005
β 100
VEB ( on ) + I BQ RB = VBB = ( 0.70 ) + ( 0.005 )( 50 ) ⇒ VBB = 0.95 V
b.
I CQ 0.5
gm = = ⇒ g m = 19.2 mA/V
VT 0.026
β VT (100 )( 0.026 )
rπ = = ⇒ rπ = 5.2 kΩ
I CQ 0.5
VA ∞
r0 = = ⇒ r0 = ∞
I CQ 0.5
c. Av = −
β RC
=−
(100 )(10 ) ⇒ A = −18.1
v
rπ + RB 5.2 + 50
6.9
10 − 4
I CQ = = 1.5 mA
4
1.5
I BQ = = 0.015 mA
100
(100 )( 0.026 )
rπ = = 1.73 K
1.5
v 5sin ω t ( mV )
ib = be = = 2.89sin ω t ( μ A )
rπ 1.73 kΩ
So
4. iB ( t ) = I BQ + iEb = 15 + 2.89sin ω t ( μ A )
iC1 ( t ) = β iB ⇒ iC1 ( t ) = 1.5 + 0.289sin ω t ( mA )
vC ( t ) = 10 − iC1 ( t ) RC = 10 − [1.5 + 0.289sin ω t ] (γ )
vC1 ( t ) = 4 − 1.156sin ω t ( v )
vC ( t ) −1.156
Av = = ⇒ Av = −231
vbe ( t ) 0.005
6.10
vo = 1.2sin ω t ( V )
−1.2sin ω t
iC ( t ) RC + vo = 0 ⇒ iC ( t ) =
2
iC ( t ) = −0.60sin ω t ( mA )
iC ( t )
ib ( t ) = = −6sin ω t ( μ A )
β
vbe ( t ) = ib ( t ) ⋅ rπ g m rπ = β
100
rπ = =2K
50
vbe ( t ) = −12sin ω t ( mV )
6.11
a.
I CQ ≈ I EQ
VCEQ = 5 = 10 − I CQ ( RC + RE )
= 10 − I CQ (1.2 + 0.2)
I CQ = 3.57 mA
3.57
I BQ = = 0.0238 mA
150
R1 R2 = RTH = ( 0.1)(1 + β ) RE
= ( 0.1)(151)( 0.2 ) = 3.02 kΩ
1
VTH = ⋅ RTH ⋅ (10) − 5
R1
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5
1
(3.02)(10) − 5 = ( 0.0238)(3.02) + 0.7 + (151)( 0.0238)( 0.2) − 5
R1
1
( 30.2 ) = 1.50 ⇒ R1 = 20.1 k Ω
R1
20.1R2
= 3.02 ⇒ R2 = 3.55 kΩ
20.1 + R2
b.
(150 )( 0.026 )
rp = = 1.09 kΩ
3.57
3.57
gm = = 137 mA/V
0.026
5. V0
ϩ
V r gmV
Ϫ
ϩ
VS R1͉͉R2 RC
Ϫ
RE
2 β RC (150 )(1.2 )
Av = =2 ⇒ Av = 2 5.75
rp + (1 + β ) RE 1.09 + (151)( 0.2 )
6.12
a.
⎛ R2 ⎞ ⎛ 50 ⎞
VTH = ⎜ ⎟ VCC = ⎜ ⎟ (12 ) = 10 V
⎝ R1 + R2 ⎠ ⎝ 50 + 10 ⎠
RTH = R1 R2 = 50 10 = 8.33 kΩ
12 − 0.7 − 10
I BQ = = 0.0119 mA
8.33 + (101)(1)
I CQ = 1.19 mA, I EQ = 1.20 mA
VECQ = 12 − (1.20 )(1) − (1.19 )( 2 )
VECQ = 8.42 V
iC
4
1.19
8.42 12 EC
b.
V0
Ϫ
V r gmV
ϩ
ϩ
VS R1͉͉R2 RC
Ϫ
RE
6. (100 )( 0.026 )
rp = = 2.18 kΩ
1.19
V0 = g mVp RC
⎛V ⎞
VS = 2 Vp − ⎜ p + g mVp ⎟ RE
⎝ rp ⎠
⎡ rπ + (1 + β ) RE ⎤
= −Vp ⎢ ⎥
⎣ rp ⎦
2 β RC 2 (100 )( 2 )
Av = = ⇒ Av = 2 1.94
rp + (1 + β ) RE 2.18 + (101)(1)
c. Approximation: Assume rp does not vary significantly.
RC = 2 kΩ ± 5% = 2.1 kΩ or 1.9 kΩ
RE = 1 kΩ ± 5% = 1.05 kΩ or 0.95 kΩ
For RC ( max ) = 2.1 kΩ and RE ( min )
− (100 )( 2.1)
Av = = −2.14
2.18 + (101)( 0.95 )
For RC ( min ) = 1.9 kΩ and RE ( max ) = 1.05 kΩ
− (100 )(1.9 )
Av = = −1.76
2.18 + (101)(1.05 )
So 1.76 ≤ Av ≤ 2.14
6.13
(a)
VCC = ⎜ 1+ β ⎟ I CQ RE + VECQ + I CQ RC
⎛ ⎞
⎜ ⎟
β
⎝ ⎠
⎛ 101 ⎞
12 = ⎜ ⎟ I CQ (1) + 6 + I CQ ( 2 )
⎝ 100 ⎠
so that I CQ = 1.99 mA
1.99
I BQ = = 0.0199 mA
100
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)(1) = 10.1 k Ω
⎛ R2 ⎞ 1 1
VTH = ⎜ ⎟ V = ⋅ R ⋅ V = (10.1)(12 )
⎝ R1 + R2 ⎠ CC R1 TH CC R1
VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
121.2
12 = (101)( 0.0199)(1) + 0.7 + ( 0.0199)(10.1) +
R1
which yields R1 = 13.3 k Ω and R2 = 41.6 k Ω
2 β RC 2 (100 )( 2 )
(b) Av = = ⇒ Av = 2 1.95
rp + (1 + β ) RE 1.31 + (101)(1)
6.14
7. I CQ = 0.25 mA, I EQ = 0.2525 mA
I BQ = 0.0025 mA
I BQ RB + VBE ( on ) + I EQ ( RS + RE ) − 5 = 0
( 0.0025)( 50 ) + 0.7 + ( 0.2525)( 0.1 + RE ) = 5
RE = 16.4 kΩ
VE = − ( 0.0025 )( 50 ) − 0.7 = −0.825 V
VC = VCEQ + VE = 3 − 0.825 = 2.175 V
5 − 2.175
RC = ⇒ RC = 11.3 kΩ
0.25
− β RC
Av =
rπ + (1 + β ) RS
(100 )( 0.026 )
rπ = = 10.4 kΩ
0.25
− (100 )(11.3)
Av = ⇒ Av = −55.1
10.4 + (101)( 0.1)
Ri = RB ⎡ rπ + (1 + β ) RS ⎤
⎣ ⎦
= 50 ⎡10.4 + (101)( 0.1) ⎤
⎣ ⎦
Ri = 50 20.5 ⇒ Ri = 14.5 kΩ
6.15
(a)
VCC > I CQ ( RC + RE ) + VCEQ
9 = I CQ ( 2.2 + 2 ) + 3.75 So that
I CQ = 1.25 mA
Assume circuit is to be designed to be bias stable.
RTH = R1 R2 = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2 ) = 24.2 Ω
1.25
I BQ = = 0.01042 mA
120
1
VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I BQ (121)( RE )
R1
1
( 24.2 )( 9 ) = ( 0.01042 )( 24.2 ) + 0.7 + ( 0.01042 )(121)( 2 )
R1
= 0.2522 + 0.7 + 2.5216
= 3.474
62.7 R2
R1 = 62.7 K = 24.2
62.7 + R2
R2 = 39.4 K
(b)
8. 1.25
gm = = 48.08 mA / V
0.026
(120 )( 0.026 )
rp = = 2.50 k Ω
1.25
100
ro = = 80 k Ω
1.25
Vo
ϩ
IS V r ro RC RL
R1͉͉R2Ϫ gmV
Vo = 2 g mVp ( ro RC RL )
Vp = I S ( R1 R2 rp )
Then
Vo
Rm = = 2 g m ( R1 R2 rp )( ro RC RL )
Is
Rm = 2 48.08 ( 24.2 2.5 )( 80 2.2 1) = 2 48.08 ( 2.266 )( 0.6816 )
or
Vo
Rm = = −74.3 k Ω = −74.3 V / mA
Is
6.16
a.
0.80
I EQ = 0.80 mA, I BQ = = 0.0121 mA
66
I CQ = 0.788 mA
0.3
VB = I BQ RB ⇒ RB = ⇒ RB = 24.8 kΩ
0.0121
V − ( −5 ) 5 − 3
RC = C = ⇒ RC = 2.54 kΩ
I CQ 0.788
b.
0.788
gm = = 30.3 mA / V
0.026
( 65)( 0.026)
rπ = = 2.14 kΩ
0.788
75
r0 = = 95.2 kΩ
0.788
⎛ RC r0 ⎞
i0 = ⎜ g V , V = − vS
⎜R r +R ⎟ m π π⎟
⎝ C 0 L ⎠
i0 ⎛ RC r0 ⎞
Gf = = − gm ⎜ ⎟
vS ⎜R r +R ⎟
⎝ C 0 L ⎠
⎛ 2.54 95.2 ⎞
= − ( 30.3) ⎜
⎜ 2.54 95.2 + 4 ⎟
⎟
⎝ ⎠
G f = −11.6 mA/V
9. 6.17
(a)
I CQ = 0.8 mA ⇒ I BQ = 0.00667 mA
I BQ RS + 0.7 + (121) I BQ RE − 15 = 0
( 0.01667 )( 2.5) + 0.7 + (121)( 0.00667 ) RE = 15
RE = 17.7 K
VE = − ( 0.00667 )( 2.5 ) − 0.7 = −0.717 V
VC = −0.717 + 7 = 6.283 V
15 − 6.283
RC = = 10.9 K
0.8
0.8 (120 )( 0.026 )
gm = = 30.77 mA/V rπ = = 3.9 K
0.026 0.8
⎛ r ⎞
vo = − g m ( RC RL ) ⋅ vπ vπ = ⎜ π ⎟ vS
⎝ rπ + RS ⎠
− β ( RC RL ) − (120 ) (10.9 5 )
Av = =
rπ + RS 3.9 + 2.5
Av = −64.3
(b) For RS = 0
0.7 + (121)( 0.00667 ) RE = 15
RE = 17.7 K
VE = −0.7 ⇒ VC = −0.7 + 7 = 6.3
15 − 6.3
RC = ⇒ RC = 10.9 K
0.8
− β ( RC RL )
Av = = −30.77 (10.9 5 )
rπ
Av = −105
6.18
(a)
15 = ( 81) I BQ (10 ) + 0.7 + I BQ ( 2.5 )
15 − 0.7
I BQ = = 0.0176 mA
2.5 + ( 81)(10 )
I CQ = 1.408 mA
1.408 (80 )( 0.026 )
gm = = 54.15 mA/V rπ =
0.026 1.408
rπ = 1.48 K
RS
V0
IS
ϩ
VS ϩ V r RC Io RL
Ϫ gmV
Ϫ
10. − β ( RC RL ) − ( 80 ) ( 5 5 )
Vo = − g mVσ ( RC RL ) ⇒ Av = = ⇒ Av = −50.3
rπ + RS 1.48 + 2.5
⎛ RC ⎞
− g mVπ ⎜ ⎟
i
AI = o = ⎝ RC + RL ⎠ = − β ⎛ RC ⎞
V ⎜ ⎟
iS π ⎝ RC + RL ⎠
r
π
AI = −40
vo ( t ) = ( −50.3)( 4sin ω t )
vo ( t ) = −0.201sin ω t ( V )
4 sin ω t ( mV )
is = = 1.005sin ω t ( μA )
2.5 + 1.48
io = −40.2 sin ω t ( μA )
(b)
15 − 0.7
I EQ = = 1.43 mA
10
⎛ 80 ⎞
I CQ = ⎜ ⎟ (1.43) = 1.412 mA
⎝ 81 ⎠
1.412 (80 )( 0.026 )
gm = = 54.3 mA/V rπ = = 1.47 K
0.026 1.412
Av = − g m ( RC RL ) = − ( 54.3) ( 5 5 ) ⇒ Av = −136
⎛ RC ⎞ ⎛ 5 ⎞
AI = − β ⎜ ⎟ = −80 ⎜ ⎟ ⇒ AI = −40
⎝ RC + RL ⎠ ⎝5+5⎠
vo ( t ) = ( −136 )( 4sin ω t ) ⇒ vo ( t ) = −544sin ω t ⇒ vo ( t ) = −0.544sin ω t ( V )
4sin ω t ( mV )
is ( t ) = = 2.72sin ω t ( μA )
1.47 k
io ( t ) = ( −40 )( 2.72sin ω t )
io ( t ) = −109sin ω t ( μA )
6.19
RTH = R1 R2 = 27 15 = 9.64 K
⎛ R2 ⎞ ⎛ 15 ⎞
VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 9 ) = 3.214 V
⎝ R1 + R2 ⎠ ⎝ 15 + 27 ⎠
V − VBE ( on ) 3.214 − 0.7 2.514
I BQ = TH = =
RTH + (1 + β ) RE 9.64 + (101)(1.2 ) 130.84
I BQ = 0.0192 mA I CQ = 1.9214 mA
1.92 (100 )( 0.026 )
gm = = 73.9 mA/V rπ = = 1.35 K
0.026 1.92
11. RS
V0
IS
ϩ
VS ϩ RTH V r r0 RC I0 RL
Ϫ gmV
Ϫ
100
ro = = 52.1 K
1.92
⎛ r R ⎞
(
Vo = − g mVπ r0 RC RL ) Vπ = ⎜ π TH
⎜r R +R ⎟ VS
⎟
⎝ π TH S ⎠
rπ RTH = 1.35 9.64 = 1.184 K
⎛ 1.184 ⎞
Vπ = ⎜ ⎟ VS
⎝ 1.184 + 10 ⎠
= 0.1059VS
(
Av = − ( 73.9 ) ( 0.1059 ) 52.1 2.2 2 )
= − ( 73.9 ) ( 0.1059 ) ( 52.1 1.0476 )
= − ( 73.9 ) ( 0.1059 ) (1.027 )
Av = −8.04
⎛ ro RC ⎞
− g mVπ ⎜
⎜r R +R ⎟
⎟
I ⎝ o C L ⎠
AI = o =
IS Vπ
RTH rπ
⎛ ro RC ⎞
AI = − g m ( RTH rπ ) ⎜
⎜r R +R ⎟ ⎟
⎝ o C L ⎠
ro RC = 52.1 2.2 = 2.11 K
RTH rπ = 9.64 1.35 = 1.184 K
⎛ 2.11 ⎞
AI = − ( 73.9 ) (1.184 ) ⎜ ⎟
⎝ 2.11 + 2 ⎠
AI = −44.9
Ri = RTH rπ = 9.64 1.35
Ri = 1.184 K
6.20
a.
0.35
I E = 0.35 mA, I B = = 0.00347 mA
101
VB = 2 I B RB = 2 ( 0.00347 )(10 ) ⇒ VB = 2 0.0347 V
VE = VB − VBE ( on ) ⇒ VE = 2 0.735 V
b.
12. VC = VCEQ + VE = 3.5 − 0.735 = 2.77 V
⎛ b ⎞ ⎛ 100 ⎞
IC = ⎜ ⎟ IE = ⎜ ⎟ ( 0.35 ) = 0.347 mA
⎝ 1+ b ⎠ ⎝ 101 ⎠
V 1 − VC 5 − 2.77
RC = = ⇒ RC = 6.43 kΩ
IC 0.347
(c)
⎛ RB rp ⎞
⎜ R r + R ⎟( C o )
Av = 2 g m ⎜ R r
⎟
⎝ B π S ⎠
0.347 100
gm = = 13.3 mA/V , ro = = 288 k Ω
0.026 0.347
(100 )( 0.026 )
rp = = 7.49 k Ω
0.347
RB rp = 10 7.49 = 4.28 k Ω
⎛ 4.28 ⎞
Av = 2 (13.3) ⎜ ⎟ ( 6.43 288 ) ⇒ Av = 2 81.7
⎝ 4.28 + 0.1 ⎠
d.
⎛ RB rp ⎞
⎜ R r + R ⎟( C 0 )
Av = 2 g m ⎜ R r
⎟
⎝ B p S ⎠
RB rp = 10 7.49 = 4.28 kΩ
⎛ 4.28 ⎞
Av = 2 (13.3) ⎜ ⎟ ( 6.43 288 ) ⇒ Av = 2 74.9
⎝ 4.28 + 0.5 ⎠
6.21
a.
RTH = R1 R2 = 6 1.5 = 1.2 kΩ
⎛ R2 ⎞ + ⎛ 1.5 ⎞
VTH = ⎜ ⎟V = ⎜ ⎟ ( 5 ) = 1.0 V
⎝ R1 + R2 ⎠ ⎝ 1.5 + 6 ⎠
V − VBE ( on ) 1.0 − 0.7
I BQ = TH = = 0.0155 mA
RTH + (1 + β ) RE 1.2 + (181)( 0.1)
I CQ = 2.80 mA, I EQ = 2.81
VCEQ = V + − I CQ RC − I EQ RE
= 5 − ( 2.8 )(1) − ( 2.81)( 0.1) ⇒ VCEQ = 1.92 V
b.
(180 )( 0.026 )
rp = ⇒ rp = 1.67 kΩ
2.80
2.80
gm = ⇒ g m = 108 mA/V, r0 =`
0.026
(c)
⎛ R1 R2 rp ⎞
Av = 2 g m ⎜ ⎟ ( RC RL )
⎝ R1 R2 rp + RS ⎠
R1 R2 rp = 6 1.5 1.67 = 0.698 k V
⎛ 0.698 ⎞
Av = 2 (108 ) ⎜ ⎟ (1 1.2 ) ⇒ Av = 2 45.8
⎝ 0.698 + 0.2 ⎠
13. 6.22
a.
9 = I EQ RE + VEB ( on ) + I BQ RS
0.75
I EQ = 0.75 mA, I BQ = = 0.00926 mA
81
I CQ = 0.741 mA
9 = ( 0.75 ) RE + 0.7 + ( 0.00926 )( 2 ) ⇒ RE = 11.0 kΩ
b.
VE = 9 − ( 0.75 )(11) = 0.75 V
VC = VE − VECQ = 0.75 − 7 = −6.25 V
VC − ( −9 ) 9 − 6.25
RC = = ⇒ RC = 3.71 kΩ
I CQ 0.741
c.
⎛ rp ⎞
Av = 2 g m ⎜ ⎟ ( RC || RL || r0 )
⎝ rp + RS ⎠
(80 )( 0.026 )
rp = = 2.81 kΩ
0.741
80
r0 = = 108 kV
0.741
2 80
Av = ( 3.71||10 ||108 )
2.81 + 2
Av = 2 43.9
d.
Ri = RS + rp = 2 + 2.81 ⇒ Ri = 4.81 kΩ
6.23
4 − 0.7
I BQ = = 0.00647
5 + (101)( 5 )
I CQ = 0.647 mA
a. 80 ≤ h fe ≤ 120, 10 ≤ h0e ≤ 20 mS
2.45 kΩ ≤ hie ≤ 3.7 kΩ
low gain high gain
RS
V0
IS
ϩ
VS ϩ V r RC Io RL
Ϫ gmV
Ϫ
14. ⎛ 1 ⎞
V0 = 2 h fe I b ⎜ RC RL ⎟
⎝ hoe ⎠
RB
V
?S
R + RS
Ib = B
RTH + hie
RTH = RB RS = 5 1 = 0.833 kΩ
High-gain
⎛ 5 ⎞
⎜ ⎟ VS
5 +1⎠
Ib = ⎝ = 0.1838VS
0.833 + 3.7
Low-gain
⎛ 5 ⎞
⎜ ⎟ VS
Ib = ⎝ 5 +1⎠ = 0.2538VS
0.833 + 2.45
1 1
For hoe = 10 ⇒ || Rc || RL = || 4 || 4
hoe 0.010
= 100 || 2 = 1.96 kΩ
1
For hoe = 20 ⇒ || 4 || 4 = 50 || 2 = 1.92 kΩ
0.020
Av max
= (120 )( 0.1838 )(1.96 ) = 43.2
Av min
= ( 80 )( 0.2538 )(1.92 ) = 39.0
39.0 ≤ Av ≤ 43.2
b.
Ri = RB hie = 5 3.7 = 2.13 kV or Ri = 5 2.45 = 1.64 kΩ
1.64 ≤ Ri ≤ 2.13 kΩ
1 1
R0 = RC = 4 = 100 || 4 = 3.85 kΩ
hoe 0.010
1
or R0 = || 4 = 50 || 4 = 3.70 kΩ
0.020
3.70 ≤ R0 ≤ 3.85 kΩ
6.24
VCC ϭ 10 V
RC
R1
o
RS ϭ 1 k⍀
CC
s ϩ R2
Ϫ
RE CE
15. Assume an npn transistor with b = 100 and VA = ∞. Let VCC = 10 V .
0.5
Av = = 50
0.01
Bias at I CQ = 1 mA and let RE = 1 k Ω
For a bias stable circuit
RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)(1) = 10.1 k Ω
1 1 101
VTH = ⋅ RTH ⋅ VCC = (10.1)(10 ) =
R1 R1 R1
1
I BQ = = 0.01 mA
100
VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE
101
= ( 0.01)(10.1) + 0.7 + (101)( 0.01)(1)
R1
which yields R1 = 55.8 k Ω and R2 = 12.3 k Ω
Now
(100 )( 0.026 )
rp = = 2.6 k Ω
1
1
gm = = 38.46 mA/V
0.026
Vo = − g mVp RC
⎛ R1} R2 } rp ⎞ ⎛ 10.1} 2.6 ⎞
where Vp = ⎜ ⎟ ⋅ Vs = ⎜ ⎟ .Vs
⎝ R1} R2 } rp + RS ⎠ ⎝ 10.1} 2.6 + 1 ⎠
or Vp = 0.674 Vs
V
Then Av = o = − ( 0.674 ) g m RC = − ( 0.674 )( 38.46 ) RC = −50
Vs
which yields RC = 1.93 k Ω
With this RC, the dc bias is OK.
Finish Design, Set RC = 2 K RE = 1 K
R1 = 56 K
R2 = 12 K
RTH = R1 R2 = 9.88 K
⎛ R2 ⎞ ⎛ 12 ⎞
VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 1.765 V
⎝ R1 + R2 ⎠ ⎝ 12 + 56 ⎠
1.765 − 0.7
I BQ = = 9.60 μ A
9.88 + (101)(1)
I CQ = 0.9605 mA
(100 )( 0.026 ) 0.9605
rπ = = 2.707 K gm = = 36.94
0.9605 0.026
RTH rπ = 2.125 K
⎛ RTH rπ ⎞ ⎛ 2.125 ⎞
Vπ = ⎜ ⎟ Vi = ⎜ ⎟ Vi = ( 0.680 ) Vi
⎝ RTH rπ + RS ⎠ ⎝ 2.125 + 1 ⎠
Av = − ( 0.680 ) g m RC = − ( 0.680 )( 36.94 )( 2 ) = −50.2
16. Design specification met.
6.25
a.
6 − 0.7
I BQ = = 0.0169 mA
10 + (101)( 3)
I CQ = 1.69 mA, I EQ = 1.71 mA
VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3)
VCEQ = 5.38 V
b.
1.69
gm = ⇒ g m = 65 mA/V
0.026
(100 )( 0.026 )
rp = ⇒ rp = 1.54 kV , r0 = ∞
1.69
(c)
− β ( RC RL ) RB Rib
Av = ⋅
rπ + (1 + β ) RE RB Rib + RS
Rib = rπ + (1 + β ) RE = 1.54 + (101)(3) = 304.5 k Ω
RB Rib = 10 304.5 = 9.68 k Ω
Then
− (100 )( 6.8 6.8 ) ⎛ 9.68 ⎞
Av = ⋅⎜ ⎟ ⇒ Av = −1.06
1.54 + (101)( 3) ⎝ 9.68 + 0.5 ⎠
⎛ RC ⎞
i0 = ⎜ ⎟ ( − β ib )
⎝ RC + RL ⎠
⎛ RB ⎞
ib = ⎜ ⎟ iS
⎝ RB + rπ + (1 + β ) RE ⎠
⎛ RC ⎞⎛ RB ⎞
Ai = − ( β ) ⎜ ⎟⎜⎜ R + r + (1 + β ) R ⎟ ⎟
⎝ RC + RL ⎠⎝ B π E ⎠
⎛ 6.8 ⎞ ⎛ 10 ⎞
= − (100 ) ⎜ ⎟⎜⎜ 10 + 1.54 + (101)( 3) ⎟ ⇒ Ai = −1.59
⎟
⎝ 6.8 + 6.8 ⎠ ⎝ ⎠
(d) Ris = RS + RB Rib = 0.5 + 10 304.5 = 10.2 k Ω
(e)
2 b ( RC RL )
Av =
rp + (1 + b ) RE
2 (100 ) ( 6.8} 6.8 )
Av = ⇒ Av = 2 1.12
1.54 + (101)( 3)
Ai = same as ( c ) ⇒ Ai = 2 1.59
6.26
17. ie
ϩ ϩ
vCE
ϩ vCE ris gmv o
vbe Ϫ
Ϫ
Ϫ
ϩ
vCE gmv
Ϫ
vCe 1
r= =
g m vCe g m
⎛ 1 ⎞
So re = rp ⎜ ⎟ r0
⎝ gm ⎠
6.27
Let b = 100, VA = ∞
VCC
RC
R1
o
RS ϭ 100 ⍀
CC
s ϩ R2
Ϫ
RE
Let VCC = 2.5 V
P = ( I R + I C ) VCC ⇒ 0.12 = ( I R + I C )( 2.5 ) ⇒ I R + I C = 48 mA, Let I R = 8mA, I C = 40 mA
VCC 2.5
R1 + R2 > = ⇒ 312.5 k Ω
IR 8
40
I BQ = = 0.4 mA
100
18. Let RE = 2 k Ω. For a bias stable circuit
RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)( 2 ) = 20.2 k Ω
1
VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE
R1
1
( 20.2 )( 2.5 ) = ( 0.0004 )( 20.2 ) + 0.7 + (101)( 0.0004 )( 2 )
R1
which yields R1 = 64 k V and R2 = 29.5 k Ω
(100 )( 0.026 )
rπ = = 65 k Ω Neglect RS
0.04
Vo 2 b RC
Av = >
Vs rπ + (1 + b ) RE
2 100 RC
−10 = ⇒ RC = 26.7 k Ω
65 + (101)( 2 )
With this RC , dc biasing is OK.
6.28
100
Need a voltage gain of = 20.
5
Assume a sign inversion from a common-emitter is not important. Use the configuration for Figure 6.31.
Let RS = 0. Need an input resistance of
5 × 102 3
Ri = = 25 × 103 = 25 k Ω
0.2 × 102 6
Ri = RTH Rib . Let RTH = 50 k Ω, Rib = 50 k Ω
Rib = rp + (1 + b ) RE > (1 + b ) RE
Rib 50
For b = 100, RE = = = 0.495 k Ω
1 + b 101
Let RE = 0.5 k V , VCC = 10 V , I CQ = 0.2 mA
0.2
Then I BQ = = 0.002 mA
100
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
1 1
⋅ RTH ⋅ VCC = ( 50)(10) = ( 0.002)(50) + 0.7 + (101)( 0.002 )( 0.5)
R1 R1
which yields R1 = 555 k Ω and R2 = 55 k Ω
− β RC (100)( 0.026)
Now Av = , rπ = = 13 k Ω
rπ + (1 + β ) RE 0.2
So
− (100 ) RC
−20 = ⇒ RC = 12.7 k Ω
13 + (101)( 0.5)
[Note: I CQ RC = ( 0.2 )(12.7 ) = 2.54 V. So dc biasing is OK.]
6.29
19. VCC ϭ 10 V
R1 RE
CC
o
s ϩ R2
Ϫ
RC
2 b RC
b = 80, Av =
rp + (1 + b ) RE
First approximation:
R
( Av ) ≈ C = 10 ⇒ RC = 10 RE
RE
Set RC = 12 RE
VEC ≈ VCC − I C ( RC + RE ) = 10 − I C (13RE )
1
For VEC = VCC = 5
2
5 = 10 − I C (13RE )
For I C = 0.7 mA
I E = 0.709, I B = 0.00875 mA ⇒ RE = 0.55 kΩ − RC = 6.6 kΩ
Bias stable ⇒
R1 R2 = RTH = ( 0.1)(1 + β ) RE
= ( 0.1)(81)( 0.55 ) = 4.46 kΩ
1
10 = ( 0.709 )( 0.55 ) + 0.7 + ( 0.00875 )( 4.46 ) + ( 4.46 )(10 )
R1
1
8.87 = ( 4.46 ) ⇒ R1 = 5.03 kΩ
R1
5.03R2
= 4.46 ⇒ R2 = 39.4 kΩ
5.03 + R2
10 10
= = 0.225 mA
R1 + R2 5.03 + 39.4
0.7 + 0.225 ≅ 0.925 mA from VCC source.
(80 ) ( 0.026 )
Now rπ = = 2.97 kΩ
0.7
(80 )( 6.6 )
Av = = 11.1
2.97 + ( 81)( 0.55 )
6.30
20. ϩ5V
R1 RC
CC2
CC1 o
RL ϭ 10 K
s ϩ
Ϫ R2 CE
RE
Ϫ5V
β = 120
Let I CQ = 0.35 mA, I EQ = 0.353 mA
I BQ = 0.00292 mA
Let RE = 2 kΩ. For VCEQ = 4 V ⇒ 10 = 4 + ( 0.35) RC + ( 0.353)( 2)
(120 )( 0.026 )
RC = 15.1 kΩ, rπ = = 8.91 kΩ
0.35
− β ( RC RL ) (120 ) (15.1 10 )
Av = =−
rπ 8.91
Av = −81.0
For bias stable circuit:
R1 R2 = RTH = ( 0.1)(1 + β ) RE
= ( 0.1)(121)( 2 ) = 24.2 kΩ
⎛ R2 ⎞ 1
VTH = ⎜ ⎟ (10) − 5 = ⋅ RTH ⋅ (10 ) − 5
⎝ R1 + R2 ⎠ R1
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5
1
( 24.2 )(10 ) − 5 = ( 0.00292 )( 24.2 ) + 0.7 + (121)( 0.00292 )( 2 ) − 5
R1
1
( 242 ) = 1.477, R1 = 164 kΩ
R1
164 R2
= 24.2 ⇒ R2 = 28.4 kΩ
164 + R2
10
= 0.052, 0.35 + 0.052 = 0.402 mA
164 + 28.4
So bias current specification is met.
6.31
From Prob. 6.12,
21. RTH = R1} R2 = 10}50 = 8.33 kΩ
⎛ R2 ⎞ ⎛ 50 ⎞
VTH = ⎜ ⎟ (12 ) = ⎜ ⎟ (12 ) = 10 V
⎝ R1 + R2 ⎠ ⎝ 50 + 10 ⎠
12 − 0.7 − 10
I BQ = = 0.0119 mA
8.33 + (101)(1)
I CQ = 1.19 mA, I EQ = 1.20 mA
VECQ = 12 − (1.19 )( 2 ) − (120 )(1) = 8.42 V
1.19
1 8.42 11 12
For 1 ≤ vEC ≤ 11
DvEC = 11 − 8.42 = 2.58
⇒ Output voltage swing = 5.16 V (peak-to-peak)
6.32
5 − 0.7
I BQ = = 0.00315 mA
50 + (101)( 0.1 + 12.9)
I CQ = 0.315 mA, I EQ = 0.319 mA
VCEQ = ( 5 + 5 ) − ( 0.315 )( 6 ) − ( 0.319 )(13)
VCEQ = 3.96 V
AC load line
Ϫ1
Slope ϭ
6.1 K
0.315
3.96 10
1
ΔiC = − Δv
6.1 eC
For ΔiC = 0.315 − 0.05 = 0.265 ⇒ ΔvEC = 1.62
vEC ( min ) = 3.96 − 1.62 = 2.34
Output signal swing determined by current:
22. Max. output swing = 3.24 V peak-to-peak
6.33
From Problem 4.18, I CQ = 1.408 mA, I EQ = 1.426 mA
(a) VECQ = 30 − (1.408 )( 5 ) − (1.426 )(10 ) = 8.7 V
IC (mA)
AC load line
Ϫ1
Slope ϭ
RC ͉͉RL
Ϫ1
1.408 ϭ
2.5 kΩ
8.7 EC (V)
vEC ( max ) = 8.7 + ΔI C ⋅ ( 2.5 ) = 8.7 + (1.408 )( 2.5 ) = 12.22
Set vEC ( max ) = 12 = 8.7 + ΔI C ( 2.5 ) ⇒ ΔI C = 1.32 mA
So ΔvEC (peak-to-peak) = 2(12 − 8.7) = 6.6 V
(b) ΔiC (peak-to-peak) = 2(1.32) = 2.64 mA
6.34
I EQ = 0.80 mA, I CQ = 0.792 mA
I BQ = 0.00792 mA
VE = 0.7 + ( 0.00792 )(10 ) = 0.779 V
VC = I CQ RC − 5 = ( 0.792 )( 4 ) − 5 = 2 1.83 V
VECQ = 0.779 − ( −1.83) = 2.61 V
Load line: Assume VE remains constant at ≈ 0.78 V
IC (mA)
AC load line
Ϫ1
Slope ϭ
RC ͉͉RL
Ϫ1
1.408 ϭ
2.5 kΩ
8.7 EC (V)
21
DiC = v
? ec
2 kV
Collector current swing = 0.792 − 0.08
= 0.712 mA
Dvec = ( 0.712 )( 2 ) = 1.424 V
Output swing determined by current.
Max. output swing = 2.85 V peak-to-peak
2.85
Swing in i0 current =
4
= 0.712 mA peak-to-peak
23. 6.35
6 − 0.7
I BQ = = 0.0169 mA
10 + (101)( 3)
I CQ = 1.69 mA, I EQ = 1.71 mA
VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3)
VCEQ = 5.38 V
AC load line
Ϫ1
Slope ϭ
3.4 ϩ 3
1.69 Ϫ1
ϭ
6.4 K
5.38 22
1
DiC = 2 Dvce
6.4
4.38
For vce ( min ) = 1 V, Dvce = 5.38 − 1 = 4.38 V ⇒ DiC = = 0.684 mA
6.4
Output swing limited by voltage:
Δvce = Max. swing in output voltage
= 8.76 V peak-to-peak
1
Δi0 = ΔiC ⇒ Δi0 = 0.342 mA
2
or Δi0 = 0.684 mA (peak-to-peak)
6.36
AC load line
Ϫ1
Slope ϭ
1.05 K
2.65
Q-point
ICQ
VCEQ 9
100
ro =
I CQ
Neglect ro as (E) approx. dc load line VCE = 9 − I C ( 3.4 )
24. ΔI C = I CQ − 0.1
ΔVCE = VCEQ − 1
Also ΔVCE = ΔI C ( RC RL ) = ΔI C (1.05 )
Or VCEQ − 1 = ( I CQ − 0.1) (1.05 )
Substituting the expression for the dc load line.
⎡9 − I CQ ( 3.4 ) − 1⎤ = ( I CQ − 0.1) (1.05 )
⎣ ⎦
8.105 = I CQ ( 4.45 ) ⇒ I CQ = 1.821 mA
VCEQ = 2.81 V
1.821
I BQ = = 0.01821
100
RTH = ( 0.1)(101)(1.2 ) = 12.12 K
1 1
VTH = ⋅ RTH ⋅ VCC = (12.12 ) ( 9 ) = ( 0.01821) (12.12 ) + 0.7 + (101)( 0.01821)(1.2 )
R1 R1
= 0.2207 + 0.7 + 2.20705
R1 = 34.9 K
R2 = 18.6 K
34.9 R2
= 12.12
34.9 + R2
6.37
dc load line
5
ϭ 4.55 mA
1 ϩ 0.1
AC load line
Ϫ1
Slope ϭ
ICQ 1͉͉1.2
Ϫ1
ϭ
0.545 K
VCEQ 5
For maximum symmetrical swing
ΔiC = I CQ − 0.25
1
ΔvCE = VCEQ − 0.5 and ΔiC = ⋅ | ΔvCE |
0.545 kΩ
VCEQ − 0.5
I CQ − 0.25 =
0.545
VCEQ = 5 − I CQ (1.1)
0.545 ( I CQ − 0.25 ) = ⎡5 − I CQ (1.1) ⎤ − 0.5
⎣ ⎦
( 0.545 + 1.1) I CQ = 5 − 0.5 + 0.136
I CQ = 2.82 mA, I BQ = 0.0157 mA
RTH = R1 R2 = ( 0.1)(1 + β ) RE
= ( 0.1)(181)( 0.1) = 1.81 kΩ
1
VTH = ⋅ RTH ⋅ V + = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
R1
25. 1
(1.81)( 5 ) = ( 0.0157 )(1.81) + 0.7 + (181)( 0.0157 )( 0.1)
R1
1
( 9.05 ) = 1.013 ⇒ R1 = 8.93 kΩ
R1
8.93R2
= 1.81 ⇒ R2 =2.27 kΩ
8.93 + R2
6.38
I CQ = 0.647 mA , VCEQ > 10 − ( 0.647 )( 9 ) = 4.18 V
DiC = I CQ = 0.647 mA
So DvCE = DiC ( 4} 4 ) = ( 0.647 )( 2 ) = 1.294 V
Voltage swing is well within the voltage specification. Then DvCE = 2 (1.294 ) = 2.59 V peak-to-peak
6.39
a.
RTH = R1} R2 = 10}10 = 5 kΩ
⎛ R2 ⎞ ⎛ 10 ⎞
VTH = ⎜ ⎟ (18 ) − 9 = ⎜ ⎟ (18 ) − 9 = 0
⎝ R1 + R2 ⎠ ⎝ 10 + 10 ⎠
0 − 0.7 − ( −9 )
I BQ = = 0.0869 mA
5 + (181)( 0.5 )
I CQ = 15.6 mA, I EQ = 15.7 mA
VCEQ = 18 − (15.7 )( 0.5 ) ⇒ VCEQ = 10.1 V
b.
AC load line
Ϫ1
Slope ϭ
0.5͉͉0.3
Ϫ1
15.6 ϭ
0.188 K
10.1 18
c.
(180 )( 0.026 )
rπ = = 0.30 kΩ
15.6
(1 + β )( RE RL ) ⎛ R1 R2 Rib ⎞
Av = ⋅⎜ ⎟
rπ + (1 + β ) ( RE RL ) ⎝ R1 R2 Rib + RS ⎠
Rib = rπ + (1 + β )( RE RL ) = 0.30 + (181)( 0.5 0.3) or Rib = 34.2 k Ω
R1 R2 Rib = 5 34.2 = 4.36 k Ω
(181)( 0.5 0.3) ⎛ 4.36 ⎞
Av = ⋅⎜ ⎟ ⇒ Av = 0.806
0.3 + (181)( 0.5 0.3) ⎝ 4.36 + 1 ⎠
d.
26. Rib = rp + (1 + b ) ( RE } RL )
Rib = 0.30 + (181)( 0.188 ) ⇒ Rib = 34.3 kΩ
rp + R1} R2 } RS 0.3 + 5}1
Ro = RE = 0.5 ⇒ Ro = 6.18 Ω
1+ b 181
6.40
a.
RTH = R1} R2 = 10}10 = 5 kΩ
⎛ R2 ⎞
VTH = ⎜ ⎟ ( −10 ) = 2 5 V
⎝ R1 + R2 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE − 10
2 5 − 0.7 − ( −10 )
I BQ = = 0.0174 mA
5 + (121)( 2 )
I CQ = 2.09 mA, I EQ = 2.11 mA
VCEQ = 10 − ( 2.09 )(1) − ( 2.11)( 2 ) ⇒ VCEQ = 3.69 V
b.
AC load line
Ϫ1
Slope ϭ
2͉͉2
Ϫ1
2.09 ϭ
1K
3.69 10
c.
(120 )( 0.026 )
rπ = = 1.49 kΩ
2.09
(1 + β ) ( RE RL ) ⎛ R1 R2 Rib ⎞
Av = ⋅⎜ ⎟
rπ + (1 + β ) ( RE RL ) ⎜ R1 R2 Rib + RS ⎟
⎝ ⎠
Rib = rπ + (1 + β ) ( RE RL ) = 1.49 + (121) ( 2 2)
Rib = 122.5 k Ω, R1 R2 Rib = 5 122.5 = 4.80 k Ω
(121) ( 2 2) ⎛ 4.80 ⎞
Av = ⋅⎜ ⎟ ⇒ Av = 0.484
1.49 + (121) ( 2 2) ⎝ 4.80 + 5 ⎠
d.
Rib = rπ + (1 + b ) ( RE } RL )
Rib = 1.49 + (121) ( 2} 2 ) 1 Rib = 122 kΩ
rπ + R1} R2 } RS 1.49 + 5}5
Ro = RE =2 1 Ro = 32.4 Ω
1+ b 121
6.41
a.
27. RTH = R1 R2 = 60 40 = 24 kΩ
⎛ R2 ⎞ ⎛ 40 ⎞
VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 5) = 2 V
⎝ R1 + R2 ⎠ ⎝ 40 + 60 ⎠
5 − 0.7 − 2
I BQ = = 0.0130 mA
24 + ( 51)( 3)
I CQ = 0.650 mA, I EQ = 0.663 mA
VECQ = 5 − I EQ RE = 5 − ( 0.663)(3) ⇒ VECQ = 3.01 V
b.
1.63
AC load line
Ϫ1
Slope ϭ 51
Θ 50 Ι Θ3͉͉4Ι
0.65 Ϫ1
ϭ
1.75 K
3.01 5
c.
( 50 )( 0.026 ) 80
rπ = = 2 kΩ, r0 = = 123 kΩ
0.650 0.65
′
Define RL = RE RL r0 = 3 4 123 = 1.69 kΩ
(1 + β ) RL
′ ( 51)(1.69 )
Av = = ⇒ Av = 0.977
rπ (1 + β ) RL′ 2 + ( 51)(1.69 )
⎛ RE r0 ⎞
Ai = (1 + β ) I b ⎜ ⎟
⎜R r +R ⎟
⎝ E 0 L ⎠
⎛ RTH ⎞
Ib = I S ⎜ ⎟
⎝ RTH + Rib ⎠
Rib = rπ + (1 + β ) RL = 2 + ( 51)(1.69 ) = 88.2
′
RE r0 = 3 r0 = 3 123 = 2.93
⎛ 2.93 ⎞ ⎛ 24 ⎞
Ai = ( 51) ⎜ ⎟⎜ ⎟ ⇒ Ai = 4.61
⎝ 2.93 + 4 ⎠ ⎝ 24 + 88.2 ⎠
d.
Rib = rπ + (1 + β ) RE RL r0 = 2 + ( 51)(1.69 ) ⇒ Rib = 88.2 kΩ
rπ ⎛ 2⎞
R0 = RE = ⎜ ⎟ 3 = 0.0392 3
1+ β ⎝ 51 ⎠
R0 = 38.7 Ω
e. Assume variations in rπ and r0 have negligible effects
R1 = 60 ± 5% R1 = 63 kΩ, R1 = 57 kΩ
R2 = 40 ± 5% R2 = 42 kΩ, R2 = 38 kΩ
RE = 3 ± 5% RE = 3.15 kΩ, RE = 2.85 kΩ
RL = 4 ± 5% RL = 4.2 kΩ, RL = 3.8 kΩ
28. ⎛ RE r0 ⎞ ⎛ RTH ⎞
Ai = (1 + β ) ⎜
⎜ R r + R ⎟⎜ R + R ⎟
⎟
⎝ E 0 L ⎠ ⎝ TH ib ⎠
Rib = rπ + (1 + β ) ( RE RL r0 )
RTH ( max ) = 25.2 kΩ, RTH ( min ) = 22.8 kΩ
Rib ( max ) = 92.5 kΩ, Rib ( min ) = 84.0 kΩ
RE ( max ) , RL ( min ) , Rib = 88.6 kΩ
RE ( min ) , RL ( max ) , Rib = 87.4 kΩ
RE ( max ) r0 = 3.07 kΩ
RE ( min ) r0 = 2.79 kΩ
For RE ( min ) , RL ( max ) , RTH ( min )
⎛ 2.79 ⎞ ⎛ 22.8 ⎞
Ai = ( 51) ⎜ ⎟⎜ ⎟ ⇒ Ai = 4.21
⎝ 2.79 + 4.2 ⎠ ⎝ 22.8 + 87.4 ⎠
For RE ( max ) , RL ( min ) , RTH ( max )
⎛ 3.07 ⎞⎛ 25.2 ⎞
Ai = ( 51) ⎜ ⎟⎜ ⎟ ⇒ Ai = 5.05
⎝ 3.07 + 3.8 ⎠⎝ 25.2 + 88.6 ⎠
6.42
(a)
0.5
I BQ = = 0.00617 mA
81
VB = I BQ RB = ( 0.00617 )(10 ) ⇒ VB = 0.0617 V
VE = VB + 0.7 ⇒ VE = 0.7617 V
(b)
⎛ 80 ⎞
I CQ = ( 0.5 ) ⎜ ⎟ = 0.494 mA
⎝ 81 ⎠
I CQ 0.494
gm = = ⇒ g m = 19 mA / V
VT 0.026
β VT (80 )( 0.026 )
rπ = = ⇒ rπ = 4.21 k Ω
I CQ 0.494
VA 150
ro = = ⇒ ro = 304 k Ω
I CQ 0.494
(c)
RS VЈ
S
Ϫ
IS
Vs ϩ RB V r ro
Ϫ
gmV
ϩ
Vo
RL
Io
For RS = 0
⎛V ⎞
Vo = − ⎜ π + g mVπ ⎟ ( RL ro )
⎝ rπ ⎠
29. −Vo
so that Vπ =
⎛1+ β ⎞
⎜ ⎟ ( RL ro )
⎝ rπ ⎠
Now Vs + Vπ = Vo
Vo
or Vs = Vo − Vπ = Vo +
⎛ 1+ β ⎞
⎜ ⎟ ( RL ro )
⎝ rπ ⎠
We find
Vo (1 + β )( RL ro ) (81)( 0.5 304 )
Av = = =
Vs rπ + (1 + β )( RL ro ) 4.21 + ( 81)( 0.5 304 )
(81)( 0.5 )
≅ ⇒ Av = 0.906
4.21 + ( 81)( 0.5 )
Rib = rπ + (1 + β )( RL ro ) ≅ 4.21 + ( 81)( 0.5 ) = 44.7 k Ω
⎛ RB ⎞ ⎛ ro ⎞
Ib = ⎜ ⎟ ⋅ I s and I o = ⎜ ⎟ (1 + β ) I b
⎝ RB + Rib ⎠ ⎝ ro + RL ⎠
Then
Io ⎛ RB ⎞⎛ ro ⎞
Ai = = (1 + β ) ⎜ ⎟⎜ ⎟
Is ⎝ RB + Rib ⎠⎝ ro + RL ⎠
⎛ 10 ⎞
Ai ≅ ( 81) ⎜ ⎟ (1) ⇒ Ai = 14.8
⎝ 10 + 44.7 ⎠
(d)
⎛ RB + Rib ⎞ ⎛ 10 44.7 ⎞
⎜ R R + R ⎟ s ⎜ 10 44.7 + 2 ⎟ s (
Vs′ = ⎜ ⋅V = ⋅ V = 0.803) Vs
⎟ ⎜ ⎟
⎝ B ib s ⎠ ⎝ ⎠
Then Av = ( 0.803)( 0.906 ) ⇒ Av = 0.728
Ai = 14.8 (Unchanged)
6.43
(a)
(100 )( 0.026 )
I CQ = 1.98 mA rπ = = 1.313 K
1.98
VA 100
ro = =
I CQ 1.98
= 50.5 K
rπ + RS 1.31 + 10
Ro = ro = 50.5 ⇒ Ro = 112 Ω
1+ β 101
0.112 50.5 ⇒ Ro ≅ 112 Ω
(b) From Equation 4.68
(1 + β ) ( ro RL ) 100
Av = ro = = 50.5 K
rπ + (1 + β ) ( ro RL ) 1.98
(i)
30. RL = 0.5 K
(101) ( 50.5 0.5)
Av =
1.31 + (101) ( 50.5 0.5 )
(101)( 0.4951)
Av = ⇒ Av = 0.974
1.31 + (101)( 0.4951)
(ii)
RL = 5 K ro RL = 50.5 5 = 4.5495
(101)( 4.55)
Av = ⇒ Av = 0.997
1.31 + (101)( 4.55 )
6.44
5 − 0.7
I EQ = = 1.303 I CQ = 1.293 mA
3.3
(125 )( 0.026 )
rπ = = 2.51 K
1.293
1.293
gm = = 49.73 mA/V
0.026
(a)
Rib = rπ + (1 + β ) ( RE RL ) = 2.51 + (126 ) ( 3.3 1)
Rib = 99.2 K
rπ 2.51
Ro = RE = 3.3 = 3.3 0.01992
1+ β 126
Ro = 19.8 Ω
(b)
v 2sin ω t
is = s = ⇒ is ( t ) = 20.2sin ω t ( μ A )
Rib 99.2
veb ( t ) = −is ( t ) rπ = ( −20.2 )( 2.51) sin ω t
veb ( t ) = −50.6sin ω t ( mV )
(1 + β ) ( RE RL ) (126 ) ( 3.3 1) (126 )( 0.7674 )
Av = = =
rπ + (1 + β ) ( RE RL ) 2.51 + (126 ) ( 3.3 1) 2.51 + (126 )( 0.7674 )
Av = 0.9747 ⇒ vo ( t ) = 1.95sin ω t ( V )
v (t )
io ( t ) = o ⇒ io ( t ) = 1.95sin ω t ( mA )
RL
6.45
a.
I EQ = 1 mA , VCEQ = VCC − I EQ RE
5 = 10 − (1)( RE ) ⇒ RE = 5 kΩ
1
I BQ = = 0.0099 mA
101
10 = I BQ RB + VBE ( on ) + I EQ RE
10 = ( 0.0099 ) RB + 0.7 + (1)( 5 ) ⇒ RB = 434 kΩ
b.
32. RTH = R1 R2 = 40 60 = 24 kΩ
⎛ 60 ⎞
VTH = ⎜ ⎟ (10 ) = 6 V
⎝ 60 + 40 ⎠
6 − 0.7
β = 75 I BQ = = 0.0131 mA
24 + ( 76 )( 5 )
I CQ = 0.984 mA
6 − 0.7
β = 150 I BQ = = 0.00680 mA
24 + (151)( 5 )
I CQ = 1.02 mA
( 75 )( 0.026 )
β = 75 rπ = = 1.98 kΩ
0.984
β = 150 rπ = 3.82 kΩ
β = 75 Rib = rπ + (1 + β )( RE RL ) = 65.3 kΩ
β = 150 Rib = 130 kΩ
(1 + β )( RE RL ) R1 R2 Rib
Av = ⋅
rπ + (1 + β )( RE RL ) R1 R2 Rib + RS
For β = 75, R1 R2 Rib = 40 60 65.3 = 17.5 k Ω
( 76 )( 0.833) 17.5
Av = ⋅ ⇒ Av = 0.789
1.98 + ( 76 )( 0.833) 17.5 + 4
For β = 150, R1 R2 Rib = 40 60 130 = 20.3 k Ω
(151)( 0.833) 20.3
Av = ⋅ ⇒ Av = 0.811
3.82 + (151)( 0.833) 20.3 + 4
So 0.789 ≤ Av ≤ 0.811
⎛ RE ⎞ ⎛ RTH ⎞
Ai = (1 + β ) ⎜ ⎟⎜ ⎟
⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠
β = 75
⎛ 5 ⎞⎛ 24 ⎞
Ai = ( 76 ) ⎜ ⎟⎜ ⎟ ⇒ Ai = 17.0
⎝ 5 + 1 ⎠ ⎝ 24 + 65.3 ⎠
β = 150
⎛ 5 ⎞ ⎛ 24 ⎞
Ai = (151) ⎜ ⎟ ⎜ ⎟ ⇒ Ai = 19.6
⎝ 6 ⎠ ⎝ 24 + 130 ⎠
17.0 ≤ Ai ≤ 19.6
6.47
(a)
33. ⎛ I ⎞
9 = ⎜ E ⎟ (100 ) + VBE ( on ) + I E RE
⎝ 1+ β ⎠
9 − 0.7
IE =
⎛ 100 ⎞
⎜ ⎟ + RE
⎝ 1+ β ⎠
8.3
β = 50 I E = = 2.803 mA
⎛ 100 ⎞
⎜ ⎟ +1
⎝ 51 ⎠
8.3
β = 200 I E = = 5.543 mA
⎛ 100 ⎞
⎜ ⎟ +1
⎝ 201 ⎠
2.80 ≤ I E ≤ 5.54 mA
VE = I E RE , β = 50, VE = 2.80 V
β = 200, VE = 5.54 V
(b) β = 50, I CQ = 2.748 mA, rπ = 0.473 K
β = 200, I CQ = 5.515 mA, rπ = 0.943 K
Ri = RB ⎡ rπ + (1 + β ) RE
⎣ RL ⎤
⎦
β = 50 ⇒ Ri = 100 ⎡ 0.473 + ( 51)(1 1) ⎤ = 100 25.97 = 20.6 K
⎣ ⎦
β = 200 ⇒ Ri = 100 ⎡0.943 + ( 201)(1 1) ⎤ = 100 101.4 = 50.3 K
⎣ ⎦
From Fig. (4.68)
(1 + β ) ( RE RL ) ⎛ Ri ⎞
Av = ⋅⎜ ⎟
rπ + (1 + β ) ( RE RL ) ⎝ Ri + RS ⎠
( 51) (1 1) ⎛ 20.6 ⎞
= ⋅⎜ ⎟
0.473 + ( 51) (1 1) ⎝ 20.6 + 10 ⎠
β = 50 ⇒ Av = 0.661
( 201) (1 1) ⎛ 50.3 ⎞
β = 200 ⇒ Av = ⎜ ⎟
0.943 + ( 201) (1 1) ⎝ 50.3 + 10 ⎠
Av = 0.826
6.48
Vo = (1 + β ) I b RL
Vs
Ib =
rπ + (1 + β ) RL
(1 + β ) RL
so Av =
rπ + (1 + β ) RL
For β = 100, RL = 0.5 k Ω
(100 )( 0.026 )
rπ = = 5.2 k Ω
0.5
34. (101)( 0.5 )
Then Av ( min ) = = 0.9066
5.2 + (101)( 0.5 )
Then β = 180, RL = 500 k Ω
(180 )( 0.026 )
rπ = = 9.36 k Ω
0.5
(181)( 500 )
Then Av ( max ) = = 0.9999
9.36 + (181)( 500 )
6.49
Rib
IS Ib ϩ
V r gmV ϭ Ib
Ϫ
S ϩ R1͉͉R2
Ϫ
RE RL
I0
⎛ RE ⎞
I 0 = (1+ β ) I b ⎜ ⎟
⎝ RE + RL ⎠
⎛ R1 R2 ⎞
Ib = I S ⎜ ⎟
⎝ R1 R2 + Rib ⎠
Rib = rπ + (1 + β )( RE RL )
VCC = 10 V, For VCEQ = 5 V
⎛1+ β ⎞
5 = 10 − ⎜ ⎟ I CQ RE
⎝ β ⎠
β = 80, For RE = 0.5 kΩ
I CQ = 9.88 mA, I EQ = 10 mA, I BQ = 0.123 mA
(80 )( 0.026 )
rπ = = 0.211 kΩ
9.88
Rib = 0.211 + ( 81)( 0.5 0.5 ) ⇒ Rib = 20.46 kΩ
I0 ⎛ RE ⎞ ⎛ R1 R2 ⎞
Ai = = (1 + β ) ⎜ ⎟⎜ ⎟
IS ⎝ RE + RL ⎠⎝ R1 R2 + Rib ⎠
⎛ 1 ⎞⎛ R1 R2 ⎞
8 = ( 81) ⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝ R1 R2 + 20.46 ⎠
0.1975 ⎡ R1} R2 + 20.46 ⎤ = R1} R2
⎣ ⎦
R1} R2 ⇒ 5.04 kΩ
35. VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE
1
( 5.04 )(10 ) = ( 0.123)( 5.04 ) + 0.7 + (10 )( 0.5 ) ⇒ R1 = 7.97 kΩ
R1
7.97 R2
= 5.04 ⇒ R2 = 13.7 kΩ
7.97 + R2
rπ 0.211
Now Ro = RE = 0.5 or Ro = 2.59 Ω
1+ b 81
(b)
Rib = 0.211 + (81) ( 0.5 2) = 32.6 k Ω
⎛ 0.5 ⎞ ⎛ 5.04 ⎞
Ai = ( 81) ⎜ ⎟⎜ ⎟ = ( 81)( 0.2 )( 0.134 )
⎝ 0.5 + 2 ⎠ ⎝ 5.04 + 32.6 ⎠
Ai = 2.17
6.50
Ri = RTH Rib where Rib = rπ + (1 + β ) RE
5 − 3.5
VCEQ = 3.5, I CQ = 0.75 mA
2
(120 )( 0.026 )
rπ = = 4.16 k Ω
0.75
Rib = 4.16 + (121) ( 2 ) = 246 k Ω
Then Ri = 120 = RTH 246 ⇒ RTH = 234 k Ω
0.75
I BQ = = 0.00625 mA
120
VTH = I BQ RTH + VBE ( on ) + (1+ β ) I BQ RE
1 1
⋅ RTH ⋅ VCC = ( 234)(5) = ( 0.00625) ( 234) + 0.7 + (121)( 0.00625 )( 2 )
R1 R1
which yields R1 = 318 k Ω and R2 = 886 k Ω
6.51
a.
12
Let RE = 24 Ω and VCEQ = 1 VCC = 12 V ⇒ I EQ = = 0.5 A
2 24
I CQ = 0.493 A, I BQ = 6.58 mA
( 75)( 0.026 )
rπ = = 3.96 Ω
0.493
Reb
Is Ib ϩ
V r gmV ϭ Ib
Ϫ
VS ϩ R1 ͉͉ R2
Ϫ
ϭ Rrn
RE RL
Io
36. ⎛ RE ⎞
I 0 = (1 + β ) I b ⎜ ⎟
⎝ RE + RL ⎠
⎛ RTH ⎞
Ib = I S ⎜ ⎟
⎝ RTH + Rib ⎠
Rib = rπ + (1 + β ) ( RE RL )
= 3.96 + ( 76 )( 24 8 ) ⇒ Rib = 460 Ω
I0 ⎛ RE ⎞ ⎛ RTH ⎞
Ai = = (1 + β ) ⎜ ⎟⎜ ⎟
IS ⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠
⎛ 24 ⎞ ⎛ RTH ⎞
8 = ( 76) ⎜ ⎟⎜ ⎟
⎝ 24 + 8 ⎠ ⎝ RTH + 460 ⎠
RTH
0.140 = ⇒ RTH = 74.9 Ω (Minimum value)
RTH + 460
dc analysis:
1
VTH = ⋅ RTH ⋅ VCC
R1
= I BQ RTH + VBE ( on ) + I EQ RE
1
( 74.9 )( 24 ) = ( 0.00658)( 74.9 ) + 0.70 + ( 0.5 )( 24 )
R1
= 13.19
136 R2
R1 = 136 Ω, = 74.9 ⇒ R2 = 167 Ω
136 + R2
b.
AC load line
Ϫ1
Slope ϭ
24͉͉8
Ϫ1
0.493 ϭ
6⍀
12 24
1
ΔiC = − Δvce
6
For ΔiC = 0.493 ⇒ Δvce = ( 0.493)( 6 ) ⇒ Max. swing in output voltage for this design
= 5.92 V peak-to-peak
c.
rπ 3.96
R0 = RE = 24 = 0.0521 24 ⇒ R0 = 52 mΩ
1+ β 76
6.52
The output of the emitter follower is
⎛ RL ⎞
vo = ⎜ ⎟ ⋅ vTH
⎝ RL + Ro ⎠
37. Ro
ϩ
TH ϩ
O RL
Ϫ
Ϫ
For vO to be within 5% for a range of RL , we have
RL ( min ) RL ( max )
= ( 0.95 )
RL ( min ) + Ro RL ( max ) + Ro
4 10
= ( 0.95 ) which yields Ro = 0.364 k Ω
4 + Ro 10 + Ro
⎛ r + R1 R2 RS ⎞
We have Ro = ⎜ π ⎟ RE ro
⎝ 1+ β ⎠
The first term dominates
Let R1 R2 RS ≅ RS , then
rπ + RS r +4
Ro ≅ ⇒ 0.364 = π
1+ β 1+ β
rπ 4 β VT 4
or 0.364 = + = +
1 + β 1 + β I CQ (1 + β ) 1 + β
VT 4
0.364 ≅ +
I CQ 1 + β
4 4 4 V
The factor is in the range of = 0.044 to = 0.0305. We can set Ro ≅ 0.32 = T
1+ β 91 131 I CQ
Or I CQ = 0.08125 mA. To take into account other factors, set I CQ = 0.15 mA,
0.15
I BQ = = 0.00136 mA
110
5
For VCEQ ≅ 5 V , set RE = = 33.3 k Ω
0.15
Design a bias stable circuit.
⎛ R2 ⎞ 1
VTH = ⎜ ⎟ (10) − 5 = ( RTH )(10) − 5
⎝ R1 + R2 ⎠ R1
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(111)(33.3) = 370 k Ω
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5
1
So ( 370 )(10 ) − 5 = ( 0.00136 )( 370 ) + 0.7 + (111)( 0.00136 )( 33.3) − 5
R
1
which yields R1 = 594 k Ω and R2 = 981 k Ω
(1 + β ) ( RE RL ) ⎛ RTH Rib ⎞
Now Av = ⋅⎜ ⎟
rπ + (1 + β ) ( RE RL ) ⎝ RTH Rib + RS ⎠
β VT
Rib = rπ + (1 + β ) ( RE RL ) and rπ =
I CQ
For β = 90, RL = 4 k Ω,
rπ = 15.6 k Ω, Rib = 340.6 k Ω
( 91)( 33.3 4 ) 370 340.6
Av = ⋅ ⇒ Av = 0.9332
15.6 + ( 91)( 33.3 4 ) 370 340.6 + 4
38. For β = 90, RL = 10 k Ω
Rib = 715.4 k Ω
( 91)( 33.3 10 ) 370 715.4
Av = ⋅ ⇒ Av = 0.9625
15.6 + ( 91)( 33.3 10 ) 370 715.4 + 4
For β = 130, RL = 4 k Ω
rπ = 22.5 k Ω, Rib = 490 k Ω
(131)( 33.3 4 ) 370 490
Av = ⋅ ⇒ Av = 0.9360
22.5 + (131)( 33.3 4 ) 370 490 + 4
For β = 130, RL = 10 k Ω
Rib = 1030 k Ω
(131)( 33.3 10 ) 370 1030
Av = ⋅ ⇒ Av = 0.9645
22.5 + (131)( 33.3 10 ) 370 1030 + 4
Now vO ( min ) = Av ( min ) .vS = 3.73sin ω t
vO ( max ) = Av ( max ) .vS = 3.86sin ω t
ΔvO
= 3.5%
vO
6.53
PAVG = iL ( rms ) RL ⇒ 1 = iL ( rms )(12 )
2 2
so iL ( rms ) = 0.289 A ⇒ iL ( peak ) = 2 ( 0.289 )
iL ( peak ) = 0.409 A
vL ( peak ) = iL ( peak ) ⋅ RL = ( 0.409 )(12 ) = 4.91 V
4.91
Need a gain of = 0.982
5
With RS = 10 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 RS and CC1 .
1
Set I EQ = 0.5 A, VCEQ = (12 − ( −12 ) ) = 8 V
3
24 = I EQ RE + VCEQ = ( 0.5 ) RE + 8 ⇒ RE = 32 Ω
50
Let β = 50, I CQ = ( 0.5 ) = 0.49 A
51
β VT ( 50 )( 0.026 )
rπ = = = 2.65 Ω
I CQ 0.49
Rib = rπ + (1 + β ) ( RE RL ) = 2.65 + ( 51) ( 32 12 )
Rib = 448 Ω
(1 + β ) ( RE RL ) ( 51) ( 32 12 )
Av = = = 0.994
rπ + (1 + β ) ( RE RL ) 2.65 + ( 51) ( 32 12 )
So gain requirement has been met.
39. 0.49
I BQ = = 0.0098 A = 9.8 mA
50
24
Let I R ≅ ≅ 10 I B = 98 mA
R1 + R2
So that R1 + R2 = 245 Ω
R2
VTH = ( 24 ) − 12 = I BQ RTH + VBE ( on ) + I EQ RE − 12
R1 + R2
⎛ R2 ⎞ ( 0.0098) R1 R2
⎜ 245 ⎟ ( 24 ) = 245
+ 0.7 + ( 0.5 )( 32 )
⎝ ⎠
Now R1 = 245 − R2
So we obtain
4 × 10−5 R2 + 0.0882 R2 − 16.7 = 0 which yields R2 = 175 Ω and R1 = 70 Ω
2
6.54
(a)
RTH = R1 R2 = 25.6 10.4 = 7.40 k Ω
⎛ R2 ⎞ ⎛ 10.4 ⎞
VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (18 ) = 5.2 V
⎝ R1 + R2 ⎠ ⎝ 10.4 + 25.6 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
5.2 − 0.7
I BQ = = 0.0117 mA
7.40 + (126 )( 3)
Then I CQ = 1.46 mA and I EQ = 1.47 mA
VCEQ = VCC − I CQ RC − I EQ RE
VCEQ = 18 − (1.46 )( 4 ) − (1.47 )( 3) ⇒ VCEQ = 7.75 V
(b)
(125) ( 0.026 )
rπ = = 2.23 k Ω
1.46
1.46
gm = = 56.2 mA / V
0.026
Re
Vo
Ie r Ib
Is RS RE RC RL
Ib
RTH
40. rπ + RTH 2.23 + 7.40
Re = = = 0.0764 k Ω
1+ β 126
− ( RS RE ) − (100 3)
Ie = ⋅ Is = ⋅ Is
(R S RE ) + Re (100 3) + 0.0764
or I e = − ( 0.974 ) I s
⎛ β ⎞
Vo = − I c ( RC RL ) = − ⎜ ⎟ I e ( RC RL )
⎝ 1+ β ⎠
Vo ⎛ β ⎞ ⎛ 125 ⎞
Then = −⎜ ⎟ ( −0.974 )( RC RL ) = ⎜ ⎟ ( 0.974 )( 4 4 )
Is ⎝ 1+ β ⎠ ⎝ 126 ⎠
V
Then Rm = o = 1.93 k Ω = 1.93 V / mA
Is
(c)
( ) (
Vs = I s RS R E Re = I s 100 3 0.0764 = I s ( 0.0744 ))
Vs
or I s =
0.0744
V V
which yields o = o ( 0.0744 ) = 1.93
I s Vs
Vo
or Av = = 25.9
Vs
6.55
(a)
β ( RC RL )
Av = , RL = 12 k Ω, β = 100
rπ + R1 R2
Let R1 R2 = 50 k Ω, I CQ = 0.5 mA
VTH = I BQ RTH + VBE ( on ) + (1+ β ) I BQ RE
0.5 (100 )( 0.026 )
I BQ = = 0.005 mA, rπ = = 5.2 k Ω
100 0.5
1 1
⋅ RTH ⋅ VCC = ( 50 )(12 ) = ( 0.005 )( 50 ) + 0.7 + (101)( 0.005 )( 0.5 )
R1 R1
which yields R1 = 500 k Ω
and R2 = 55.6 k Ω
(100 )(12 12 )
Av = = 10.9, Design criterion is met.
5.2 + 50
(b)
I CQ = 0.5 mA, I EQ = 0.505 mA
VCEQ = 12 − ( 0.5)(12) − ( 0.505)( 0.5) ⇒ VCEQ = 5.75 V
0.5
Av = g m ( RC RL ) , g m = = 19.23 mA / V
0.026
Av = (19.23) (12 12 ) ⇒ Av = 115
6.56
a. Emitter current
41. I EQ = I CC = 0.5 mA
0.5
I BQ = = 0.00495 mA
101
VE = I EQ RE = ( 0.5 )(1) ⇒ VE = 0.5 V
VB = VE + VBE ( on ) = 0.5 + 0.7 ⇒ VB = 1.20 V
VC = VB + I BQ RB = 1.20 + ( 0.00495 )(100 ) ⇒ VC = 1.7 V
b.
(100 )( 0.026 )
rπ = = 5.25 kΩ
(100 )( 0.00495 )
(100 )( 0.00495 )
gm = = 19.0 mA/V
0.026
Ri
gmV
V0
RS Ϫ
VS ϩ RE V r RB RL
Ϫ
ϩ
gmV
RS ͉͉RE Ϫ
RE ϩ
V Ϫ V r
RE ϩ RS S
ϩ
Vo = − g mVπ ( RB RL )
RE Rie
Vπ = − ⋅ VS = − ( 0.4971) VS
RE Rie + RS
Vo = (19 )( 0.4971) VS (100 1)
Av = 9.37
c.
gmV
IX Ϫ
VX ϩ
Ϫ RE V r
ϩ
VX VX
IX = + − g mVπ , Vπ = −VX
RE rπ
IX 1 1 1
= = + + gm
VX Ri RE rπ
1 1
or Ri = RE rπ = 1 5.253
gm 19
Ri = 0.84 0.05252 ⇒ Ri = 49.4 Ω
6.57
(a) I EQ = 1 mA, I CQ = 0.9917 mA
42. VC = 5 − ( 0.9917 )( 2 ) = 3.017 V
VE = −0.7 V
VCEQ = 3.72 V
(b)
Av = g m ( RC RL )
0.9917
gm = = 38.14 mA/V
0.026
Av = ( 38.14 ) ( 2 10 ) ⇒ Av = 63.6
6.58
(a)
10 − 0.7
I EQ = = 0.93 mA
10
I CQ = 0.921 mA
VECQ = 20 − ( 0.93)(10 ) − ( 0.921)( 5 )
VECQ = 6.10 V
(b)
0.921
gm = = 35.42 mA/V
0.026
Av = g m ( RC RL ) = ( 35.42 ) ( 5 50 )
Av = 161
6.59
(a) I EQ = 0.93 mA, I CQ = 0.921 mA
VECQ = 6.10 V
0.921
(b) gm = = 35.42 mA/V rπ = 2.82 K
0.026
From Eq. 6.90
Av = g m
( RC RL ) ⎡ rπ R R ⎤
RS ⎢1 + β E S ⎥
⎣ ⎦
( 35.42 ) ( 50 5 ) ⎡ 2.82 ⎤
= ⎢ 101 10 0.1⎥
0.1 ⎣ ⎦
( 35.42 )( 4.545 )
Av = [0.0218]
0.1
Av = 35.1
6.60
(a)
⎛ 60 ⎞
I CQ = ⎜ ⎟ (1) ⇒ I CQ = 0.984 mA
⎝ 61 ⎠
⎛ 1⎞
VCEQ = I BQ RB + VBE ( on ) = ⎜ ⎟ (100 ) + 0.7
⎝ 61 ⎠
VCEQ = 2.34 V
(b)
43. Av = g m
(RB RL ) ⎡ rπ ⎤
RS ⎥
⎢
RS ⎣1 + β ⎦
0.984
gm = = 37.85 mA/V
0.026
rπ = 1.59 K
( 37.85) (100 2 ) ⎡1.59
⎤
Av = ⎢ 61 0.05⎥
0.05 ⎣ ⎦
= 1484 ⎡ 0.0261 0.05⎤
⎣ ⎦
Av = 25.4
6.61
is ( peak ) = 2.5 mA, Vo ( peak ) = 5 mV
vo 5 × 102 3
So we need Rm = = = 2 × 103 = 2 k Ω
is 2.5 × 102 6
From Problem 4.54
Vo ⎛ β ⎞ ⎛ RS RE ⎞
=⎜ ⎟ ( RC RL ) ⎜
⎜R R +R ⎟ ⎟
Is ⎝ 1+ β ⎠ ⎝ S E ie ⎠
Let RC = 4 k Ω, RL = 5 k Ω, RE = 2 k Ω
Now β = 120, so we have
⎛ 120 ⎞ ⎛ RS RE ⎞ ⎛ RS RE ⎞
2=⎜ ⎟ ( 4 5) ⎜
⎜R R +R ⎟ = 2.204 ⎜
⎟ ⎜R R +R ⎟ ⎟
⎝ 121 ⎠ ⎝ S E ie ⎠ ⎝ S E ie ⎠
RS RE
Then = 0.9075
RS RE + Re
RS RE = 50 2 = 1.923 k Ω, so that Rie = 0.196 k Ω
Assume VCEQ = 3 V
VCC ≅ I CQ ( RC + RE ) + VCEQ
5 = I CQ ( 4 + 2 ) + 3 ⇒ I CQ = 0.333 mA
(120 )( 0.026 )
rπ = = 9.37 k Ω
0.333
r + RTH 9.37 + RTH
Rie = π ⇒ 0.196 =
1+ β 121
which yields RTH = 14.35 k Ω
Now VTH = I BQ RTH + VBE ( on ) + I EQ RE
1 ⎛ 121 ⎞
I BQ = = 0.00833 mA, I EQ = ⎜ ⎟ (1) = 1.008 mA
120 ⎝ 120 ⎠
1 1
VTH = ⋅ RTH ⋅ VCC = (14.35 )( 5 ) = ( 0.00833)(14.35 ) + 0.7 + (1.008 )( 2 )
R1 R1
which yields R1 = 25.3 k Ω
and R2 = 33.2 k Ω
6.62
a.
44. 20 − 0.7
I EQ = = 1.93 mA
10
I CQ = 1.91 mA
VECQ = VCC + VEB ( on ) − I C RC
= 25 + 0.7 − (1.91)( 6.5 ) ⇒ VECQ = 13.3 V
b.
Rie
RS h fe Ib
V0
IS Ie
VS ϩ RE RC RL
Ϫ hie
Ib
Neglect effect hoe
From Problem 6-16, assume
2.45 ≤ hie ≤ 3.7 kΩ
80 ≤ h fe ≤ 120
Vo = ( h fe I b ) ( RC RL )
hie ⎛ RE ⎞
Rie = , Ie = ⎜ ⎟ IS
1 + h fe ⎝ RE + Rie ⎠
⎛ I ⎞ VS
Ib = ⎜ e ⎟ , I S =
⎜ 1+ h ⎟ RS + RE Rie
⎝ fe ⎠
⎛ h fe ⎞ ⎛ RE ⎞ ⎛ 1 ⎞
⎜ 1+ h ⎟( C
Av = ⎜ R RL ) ⎜ ⎟×⎜ ⎟
⎟
⎝ fe ⎠ ⎝ RE + Rie ⎠ ⎝ RS + RE Rie ⎠
High gain device: hie = 3.7 kΩ, h fe = 120
3.7
Rie = = 0.0306 kΩ
121
RE Rie = 10 0.0306 = 0.0305
⎛ 120 ⎞ ⎛ 10 ⎞⎛ 1 ⎞
Av = ⎜ ⎟ ( 6.5 5 ) ⎜ ⎟⎜ ⎟ ⇒ Av = 2.711
⎝ 121 ⎠ ⎝ 10 + 0.0306 ⎠ ⎝ 1 + 0.0305 ⎠
Low gain device: hie = 2.45 kΩ, h fe = 80
2.45
Rie = = 0.03025 kΩ
81
RE Rie = 10 0.03025 = 0.0302
⎛ 80 ⎞ ⎛ 10 ⎞⎛ 1 ⎞
Av = ⎜ ⎟ ( 6.5 5 ) ⎜ ⎟⎜ ⎟ ⇒ Av = 2.70 So Av ≈ constant
⎝ 81 ⎠ ⎝ 10 + 0.03025 ⎠ ⎝ 1 + 0.0302 ⎠
2.70 ≤ Av ≤ 2.71
c.
Ri = RE Rie
We found 0.0302 ≤ Ri ≤ 0.0305 kΩ
Neglecting hoe , Ro = RC = 6.5 kΩ
6.63
a. Small-signal voltage gain
45. Av = g m ( RC RL ) ⇒ 25 = g m ( RC 1)
For VECQ = 3 V ⇒ VC = −VECQ + VEB ( on ) = −3 + 0.7 ⇒ VC = −2.3
5 − 2.3 2.7
VCC − I CQ RC + VC = 0 ⇒ I CQ = = = I CQ
RC RC
For I CQ = 1 mA, RC = 2.7 kΩ
1
gm = = 38.5 mA/V
0.026
Av = ( 38.5 )( 2.7 1) = 28.1
Design criterion satisfied and VECQ satisfied.
⎛ 101 ⎞
IE = ⎜ ⎟ (1) = 1.01 mA
⎝ 100 ⎠
5 − 0.7
VEE = I E RE + VEB ( on ) ⇒ RE = ⇒ RE = 4.26 kΩ
1.01
b.
β VT (100)( 0.026)
rπ = = ⇒ rπ = 2.6 kΩ, g m = 38.5 mA/V, ro = ∞
I CQ 1
6.64
a.
⎛ R2 ⎞ ⎛ 20 ⎞
VTH 1 = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) ⇒ VTH 1 = 2.0 V
⎝ R1 + R2 ⎠ ⎝ 20 + 80 ⎠
RTH 1 = R1 R2 = 20 80 = 16 kΩ
2 − 0.7
I B1 = = 0.0111 mA
16 + (101)(1)
1.11
I C1 = 1.11 mA ⇒ g m1 = ⇒ g m1 = 42.74 mA/V
0.026
(100)( 0.026)
rπ 1 = ⇒ rπ 1 = 2.34 kΩ
1.11
∞
r01 = ⇒ r01 = ∞
1.11
⎛ R4 ⎞ ⎛ 15 ⎞
VTH 2 = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 1.50 V
⎝ R3 + R4 ⎠ ⎝ 15 + 85 ⎠
RTH 2 = R3 R4 = 15 85 = 12.75 kΩ
1.50 − 0.70
IB2 = = 0.01265 mA
12.75 + (101)( 0.5 )
1.265
I C 2 = 1.265 mA ⇒ g m 2 = ⇒ g m2 = 48.65 mA/V
0.026
(100 )( 0.026 )
rπ 2 = ⇒ rπ 2 = 2.06 kΩ
1.26
r02 = ∞
b.
Av1 = − g m1 RC1 = − ( 42.7 )( 2 ) ⇒ Av1 = −85.48
Av 2 = − g m 2 ( RC 2 RL ) = − ( 48.5) ( 4 4) ⇒ Av 2 = −97.3
c. Input resistance of 2nd stage
46. Ri 2 = R3 R4 rπ 2 = 15 85 2.06
= 12.75 2.06 ⇒ Ri 2 = 1.773 kΩ
Av′1 = − g m1 ( RC1 Ri 2 ) = − ( 42.7 ) ( 2 1.77B)
Av′1 = −40.17
Overall gain: Av = ( −40.17 )( −97.3) ⇒ Av = 3909
If we had Av1 ⋅ Av 2 = ( −85.48)( −97.3) = 8317
Loading effect reduces overall gain
6.65
a.
⎛ R2 ⎞ ⎛ 12.7 ⎞
VTH 1 = ⎜ ⎟ VCC = ⎜ ⎟ (12) ⇒ VTH 1 = 1.905 V
⎝ R1 + R2 ⎠ ⎝ 12.7 + 67.3 ⎠
RTH 1 = R1 R2 = 12.7 67.3 = 10.68 kΩ
1.905 − 0.70
I B1 = = 0.00477 mA
10.68 + (121)( 2 )
I C1 = 0.572 mA
0.572
g m1 = ⇒ g m1 = 22 mA/V
0.026
(120 )( 0.026 )
rπ 1 = ⇒ rπ 1 = 5.45 kΩ
0.572
∞
r01 = ⇒ r01 = ∞
0.572
⎛ R4 ⎞ ⎛ 45 ⎞
VTH 2 = ⎜ ⎟ VCC = ⎜ ⎟ (12) ⇒ VTH 2 = 9.0 V
⎝ R3 + R4 ⎠ ⎝ 45 + 15 ⎠
RTH 2 = R3 R4 = 15 45 = 11.25 kΩ
9.0 − 0.70
I B2 = = 0.0405 mA
11.25 + (121)(1.6)
I C2 = 4.86 mA
4.86
gm2 = ⇒ g m 2 = 187 mA/V
0.026
(120 )( 0.026 )
rπ 2 = ⇒ rπ 2 = 0.642 kΩ
4.86
r02 = ∞
b.
I E1 = 0.577 mA
VCEQ1 = 12 − ( 0.572 ) (10 ) − ( 0.577 ) ( 2 ) ⇒ VCEQ1 = 5.13 V
I E 2 = 4.90
VCEQ 2 = 12 − ( 4.90 )(1.6 ) ⇒ VCEQ 2 = 4.16 V
47. Q1
AC load line
Ϫ1
Slope ϭ
10͉͉7.92
Ϫ1
ϭ
4.42 K
0.572
5.13 12
Q2
AC load line
Ϫ1
Slope ϭ
1.6͉͉0.25
Ϫ1
4.86 ϭ
0.216 K
4.16 12
Ri 2 = R3 R4 Rib
Rib = rπ 2 + (1 + β ) ( RE 2 RL )
= 0.642 + (121) (1.6 0.25 )
Rib = 26.8
Ri 2 = 15 45 26.8
Ri 2 = 7.92 kΩ
c.
Av1 = − g m1 ( RC1 Ri 2 ) = − ( 22 )(10 7.92 ) ⇒ Av 2 = −97.2
(1 + β )( RE 2 RL )
Av 2 =
rπ 2 + (1 + β )( RE 2 RL )
(121)( 0.216 )
= = 0.976
0.642 + (121)( 0.216 )
Overall gain = ( −97.2 )( 0.976 ) = −94.9
d.
RiS = R1 R2 rπ 1 = 67.3 12.7 5.45 ⇒ RiS = 3.61 kΩ
rπ 2 + RS
Ro = RE 2 where
1+ β
RS = R3 R4 RC1
= 15 45 10 ⇒ RS = 5.29 kΩ
0.642 + 5.29
Ro = 1.6 ⇒ 0.049 1.6 ⇒ Ro = 47.6 Ω
121
e.
−1
ΔiC = ⋅ Δvce , ΔiC = 4.86
0.216 kΩ
Δvce = ( 4.86 )( 0.216 ) = 1.05 V
Max. output voltage swing = 2.10 V peak-to-peak
6.66
(a)
48. 5 − 2 ( 0.7 )
I R1 = = 72 mA
0.050
0.7
IR2 = = 1.4 mA
0.5
⎛ β ⎞
IC 2 =⎜ ⎟ ( 72 − 1.4 ) ⇒ I C 2 = 69.9 mA
⎝ 1+ β ⎠
69.9
IB2 = = 0.699 mA
100
⎛ β ⎞
I C1 =⎜ ⎟ (1.4 + 0.699 ) ⇒ I C1 = 2.08 mA
⎝ 1+ β ⎠
(b)
ϩ
Vs ϩ
Ϫ V1 r1
gm1V1 gm2V2
Ϫ
r2 ϩ
0.5 k⍀ V2
Ϫ
Vo
50 Ω
Vs = Vπ 1 + Vπ 2 + Vo
(1)
⎛V V ⎞
Vo = ⎜ π 2 + π 2 + g m 2Vπ 2 ⎟ ( 0.05 )
⎝ 0.5 rπ 2 ⎠
(100 )( 0.026 )
rπ 2 = = 0.0372 k Ω
69.9
69.9
gm2 = = 2688 mA / V
0.026
⎛ 1 1 ⎞ V
Vo = Vπ 2 ⎜ + + 2688 ⎟ ( 0.05 ) so that (1) Vπ 2 = o
⎝ 0.5 0.0372 ⎠ 135.8
(2)
49. Vπ 1 V V
+ g m1Vπ 1 = π 2 + π 2
rπ 1 0.5 rπ 2
(100 )( 0.026 )
rπ 1 = = 1.25 k Ω
2.08
2.08
g m1 = = 80 mA / V
0.026
⎛ 1 ⎞ ⎛ 1 1 ⎞
Vπ 1 ⎜ + 80 ⎟ = Vπ 2 ⎜ + ⎟
⎝ 1.25 ⎠ ⎝ 0.5 0.0372 ⎠
⎛ V ⎞
Vπ 1 ( 80.8 ) = Vπ 2 ( 28.88 ) = ⎜ o ⎟ ( 28.88 ) or (2) Vπ 1 = Vo ( 0.00261)
⎝ 136.7 ⎠
V V
Then Vs = Vo ( 0.00261) + o + Vo = Vo (1.00993) or Av = o = 0.990
136.7 Vs
(c)
Rib = rπ 1 (1 + β ) [ Rx ]
Ix
ϩ
Vx ϩ
Ϫ 0.5 k⍀ V2 r2
gm2V2
Ϫ
Vo
50 ⍀
Vπ 2 Vπ 2 ⎛ 1 1 ⎞
Ix = + = Vπ 2 ⎜ + ⎟
0.5 rπ 2 ⎝ 0.5 rπ 2 ⎠
Vo V − Vπ 2
= x = I x + g m 2Vπ 2
0.05 0.05
⎛ 1 ⎞
Ix ⎜ + gm2 ⎟
Vx ⎛ 1 ⎞ ⎝ 0.05 ⎠
− I x = Vπ 2 ⎜ + gm2 ⎟ =
0.05 ⎝ 0.05 ⎠ ⎛ 1 1 ⎞
⎜ + ⎟
⎝ 0.5 rπ 2 ⎠
Vx
We find = Rx = 4.74 k Ω
Ix
Then Rib = 1.25 + (101) ( 2.89 ) ⇒ Rib = 480 k Ω
50. ϩ
V1 r1
gm1V1 gm2V2
Ϫ
r2 ϩ
0.5 k⍀ V2
Ϫ
Ix
ϩ
50 ⍀ Ϫ
Vx
To find Ro:
Vx V
(1) Ix = − g m 2Vπ 2 − π 2
0.05 0.5 rπ 2
⎛V ⎞ ⎛ 1 ⎞
(2) Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ( 0.5 rπ 2 ) = Vπ 1 ⎜ + 80 ⎟ ( 0.5 0.0372 ) or Vπ 2 = ( 2.77 ) Vπ 1
⎝ rπ 1 ⎠ ⎝ 1.25 ⎠
(3) Vπ 1 + Vπ 2 + Vx = 0 ⇒ Vπ 1 + ( 2.77 ) Vπ 1 + Vx = 0
so that Vπ 1 = − ( 0.2653) Vx
and Vπ 2 = ( 2.77 ) ⎡ − ( 0.2653) Vx ⎤ = − ( 0.735 ) Vx
⎣ ⎦
Vx ⎛ 1 ⎞
Now I x = − Vπ 2 ⎜ g m 2 + ⎟
0.05 ⎜ 0.5 rπ 2 ⎟
⎝ ⎠
Vx ⎡ 1 ⎤ Vx
So that I x = + ( 0.735 ) Vx ⎢ 2688 + ⎥ which yields Ro = = 0.496 Ω
0.05 ⎢
⎣ 0.5 0.0372 ⎥
⎦ Ix
6.67
a.
RTH = R1 R2 = 335 125 = 91.0 kΩ
⎛ R2 ⎞
VTH = ⎜ ⎟ VCC
⎝ R1 + R2 ⎠
⎛ 125 ⎞
=⎜ ⎟ (10 ) = 2.717 V
⎝ 125 + 335 ⎠
VTH = I B1 RTH + VBE1 + VBE 2 + I E 2 RE 2
I E 2 = (1 + β ) I E1 = (1 + β ) I B1
2
2.717 − 1.40
I B1 = ⇒ I B1 = 0.128 μΑ
91.0 + (101) (1)
2
I C1 = 12.8 μΑ
I C 2 = β I E1 = β (1 + β ) I B1 = (100 )(101)( 0.128 μΑ )
I C 2 = 1.29 mΑ, I E 2 = 1.31 mΑ
I RC = I C 2 + I C1 = 1.29 + 0.0128 = 1.30 mΑ
51. VC = 10 − I RC RC = 10 − (1.30 )( 2.2 ) = 7.14 V
VE = I E 2 RE 2 = (1.30 )(1) = 1.30 V
VCE 2 = 7.14 − 1.30 = 5.84 V
VCE1 = VCE 2 − VBE 2 = 5.84 − 0.7
VCE1 = 5.14 V
Summary:
I C1 = 12.8 μΑ I C 2 = 1.29 mΑ
VCE1 = 5.14 V VCE 2 = 5.84 V
b.
0.0128
g m1 = = 0.492 mΑ / V
0.026
1.292
gm2 = = 49.7 mΑ / V
0.026
Rib
V0
ϩ
Ib
V1 r1
gm1V1
ϩ Ϫ
VS ϩ RC
Ϫ
gm2V2
R1͉͉ R2 V2 r2
Ϫ
V0 = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC
VS = Vπ 1 + Vπ 2 , Vπ 1 = VS − Vπ 2
⎛V ⎞
Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2
⎝ rπ 1 ⎠
⎛1+ β ⎞
Vπ 2 = Vπ 1 ⎜ ⎟ rπ 2
⎝ rπ 1 ⎠
V0 = − ⎡ g m1 (VS − Vπ 2 ) + g m 2Vπ 2 ⎤ RC
⎣ ⎦
V0 = − ⎡ g m1VS + ( g m 2 − g m1 ) Vπ 2 ⎤ RC
⎣ ⎦
⎛r ⎞
Vπ 2 = (VS − Vπ 2 )(1 + β ) ⎜ π 2 ⎟
⎝ rπ 1 ⎠
⎡ ⎛ r ⎞⎤ ⎛r ⎞
Vπ 2 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥ = VS (1 + β ) ⎜ π 2 ⎟
⎣ ⎝ rπ 1 ⎠ ⎦ ⎝ rπ 1 ⎠
52. ⎧ ⎛r ⎞⎫
⎪ VS (1 + β ) ⎜ π 2 ⎟ ⎪
⎪
V0 = − ⎨ g m1VS + ( g m 2 − g m1 ) ⋅ ⎝ rπ 1 ⎠ ⎪ R
⎬ C
⎪ ⎛r ⎞
1 + (1 + β ) ⎜ π 2 ⎟ ⎪
⎪ ⎝ rπ 1 ⎠ ⎪
⎩ ⎭
V0
Av =
VS
⎧ 2.01 ⎞ ⎫
⎪ ( 49.7 − 0.492 )(101) ⎛
⎜ ⎟⎪
⎪ ⎝ 203 ⎠ ⎪ 2.2
= − ⎨( 0.492 ) + ⎬
⎪ ⎛ 2.01 ⎞ ⎪
1 + (101) ⎜ ⎟
⎪
⎩ ⎝ 203 ⎠ ⎪
⎭
Av = −55.2
c.
Ris = R1 R2 Rib
Rib = rπ 1 + (1 + β ) rπ 2
= 203 + (101)( 2.01) = 406 kΩ
Ris = 91 406 = 74.3 kΩ = Ris
R0 = RC = 2.2 kΩ
6.68
R0
Ix
ϩ
ϩ Vx
V1 r1 ro1
Ϫ
gm1V1
Ϫ
VA
ϩ
V2 r2 ro2
gm2V2
Ϫ
Vx Vx − VA
(1) I x = g m 2Vπ 2 + + + g m1Vπ 1
ro 2 ro1
Vx − VA VA
(2) + g m1Vπ 1 =
ro1 rπ 1 rπ 2
(3) Vπ 2 = VA = −Vπ 1
Then from (2)
Vx ⎛ 1 1 ⎞
= VA ⎜ + g m1 + ⎟
ro1 ⎜r rπ 1 rπ 2 ⎟
⎝ o1 ⎠
Vx Vx VA ⎛ 1 1 ⎞ ⎛ 1 ⎞
(1) I x = g m 2VA + + − − g m1VA or I x = Vx ⎜ + ⎟ + VA ⎜ g m 2 − − g m1 ⎟
ro 2 ro1 ro1 ⎝ ro1 ro 2 ⎠ ⎝ ro1 ⎠
Solving for VA from Equation (2) and substituting into Equation (1), we find
53. 1 1
+ g m1 +
V ro1 rπ 1 rπ 2
Ro = x =
Ix 1 ⎛ 1 1 ⎞ 1 ⎛ 1 ⎞
⎜ + g m1 + ⎟+ ⎜ + gm2 ⎟
ro 2 ⎝ ro1 rπ 1 rπ 2 ⎠ ro1 ⎝ rπ 1 rπ 2 ⎠
For β = 100, VA = 100 V , I C1 = I Bias = 1 mA
100
ro1 = ro 2 = = 100 k Ω
1
(100 )( 0.026 )
rπ 1 = rπ 2 = = 2.6 k Ω
1
1
g m1 = g m 2 = = 38.46 mA/V
0.026
1 1
+ 38.46 +
100 2.6 2.6
Then Ro =
1 ⎛ 1 1 ⎞ 1 ⎛ 1 ⎞
⎜ + 38.46 + ⎟+ ⎜ + 38.46 ⎟
100 ⎜ 100
⎝ 2.6 2.6 ⎟ 100 ⎜ 2.6 2.6
⎠ ⎝
⎟
⎠
or Ro = 50.0 k Ω
Now I C 2 = 1 mA, I Bias = 0
IC 2 β I
Replace I Bias by ⋅ = C 2 , I C1 ≅ 0.01 mA
β 1+ β 1+ β
100 100
ro 2 = = 100 k Ω, ro1 = = 10, 000 k Ω
1 0.01
1
gm2 = = 38.46 mA/V , g m1 = 0.3846 mA/V
0.026
(100 )( 0.026 )
rπ 2 = = 2.6 k Ω, rπ 1 = 260 k Ω
1
Then Ro = 66.4 k Ω
6.69
a.
RTH = R1 R2 = 93.7 6.3 = 5.90 k Ω
⎛ R2 ⎞
VTH = ⎜ ⎟ VCC
⎝ R1 + R2 ⎠
⎛ 6.3 ⎞
=⎜ ⎟ (12 ) = 0.756 V
⎝ 6.3 + 93.7 ⎠
0.756 − 0.70
I BQ = = 0.00949 mA
5.90
I CQ = 0.949 mA
VCEQ = 12 − ( 0.949 )( 6 ) ⇒ VCEQ = 6.305 V
Transistor:
PQ ≈ I CQVCEQ = ( 0.949 )( 6.305 ) ⇒ PQ = 5.98 mW
RC : PR = I CQ RC = ( 0.949 ) ( 6 ) ⇒ PR = 5.40 mW
2 2
b.
54. 2 AC load line
Ϫ1
Slope ϭ
6͉͉105
Ϫ1
ϭ
0.949 5.68 K
6.31 12
100
r0 = = 105 kΩ
0.949
Peak signal current = 0.949 mA
V0 ( max ) = ( 5.68 )( 0.949 ) = 5.39 V
1 V0 ( max ) 1 ⎡ ( 5.39 ) ⎤
2 2
PRC = ⋅ = ⎢ ⎥ ⇒ PRC = 2.42 mW
2 RC 2⎢ 6 ⎥
⎣ ⎦
6.70
(a)
10 = I BQ RB + VBE ( on ) + (1 + β ) I BQ RE
10 − 0.7
I BQ = = 0.00369 mA
100 + (121)( 20 )
I CQ = 0.443 mA, I EQ = 0.447 mA
For RC : PRC = ( 0.443) (10 ) ⇒ PRC = 1.96 mW
2
For RE : PRE = ( 0.447 ) ( 20 ) ⇒ PRE = 4.0 mW
2
(b)
ΔiC = 0.667 − 0.443 = 0.224 mA
1 1
( ΔiC ) RC = ( 0.224 ) (10 )
2 2
Then P RC =
2 2
P RC = 0.251 mW
6.71
a.
10 − 0.7
I BQ = = 0.00596 mA
50 + (151)(10 )
I CQ = 0.894 mA, I EQ = 0.90 mA
VECQ = 20 − ( 0.894 )( 5 ) − ( 0.90 )(10 ) ⇒ VECQ = 6.53 V
PQ ≅ I CQVECQ = ( 0.894 )( 6.53) ⇒ PQ = 5.84 mW
PRC ≅ I CQ RC = ( 0.894 ) ( 5 ) ⇒ PRC = 4.0 mW
2 2
PRE ≅ I EQ RE = ( 0.90 ) (10 ) ⇒ PRE = 8.1 mW
2 2
b.
55. AC load line
Ϫ1
Slope ϭ
5͉͉2
Ϫ1
ϭ
0.894 1.43 K
6.53 20
−1
ΔiC = ⋅ Δvec
1.43 kΩ
ΔiC = 0.894 ⇒ Δvec = ( 0.894 )(1.43) = 1.28 V
⎛ 5 ⎞
Δi0 = ⎜ ⎟ ΔiC = 0.639 mA
⎝5+2⎠
1
PRL = ( 0.639 ) ( 2 ) ⇒ PRL = 0.408 mW
2
2
1
PRC = ⋅ ( 0.894 − 0.639 ) ( 5 ) ⇒ PRC = 0.163 mW
2
2
PRE = 0
PQ = 5.84 − 0.408 − 0.163 ⇒ PQ = 5.27 mW
6.72
10 − 0.70
I BQ = = 0.00838 mA
100 + (101)(10 )
I CQ = 0.838 mA, I EQ = 0.846 mA
VCEQ = 20 − ( 0.838 )(10 ) − ( 0.846 )(10 ) ⇒ VCEQ = 3.16 V
AC load line
Ϫ1
Slope ϭ
RE ͉͉RL͉͉r0
0.838
3.16 20
100
r0 = = 119 kΩ
0.838
Neglecting base currents:
a.
RL = 1 kΩ
−1 −1
slope = =
10 1 119 0.902 kΩ
−1
ΔiC = ⋅ ΔVce
0.902 kΩ
ΔiC = 0.838 ⇒ ΔVce = ( 0.902 )( 0.838 ) = 0.756 V
1 ( 0.756 )
2
PRL = ⇒ PRL = 0.286 mW
2 1
b.
