Chapter 8 Problem Solutions 8.1 a.
b. i. 80 VD DV =
Maximum power at 40 V2
25 0.625 A40
80 40 64 0.625
D DDS
TD
D S
D D
VV
PIV
R R
= =
= = =
−= ⇒ = Ω
ii. 50 VD DV =
Maximum power at 25 V2
25 1 A25
50 25 25 1
D DD S
TD
DS
D D
VV
PIV
R R
= =
= = =
−= ⇒ = Ω
8.2 a.
(max)2
2 (max) 2(20)So 1.67 A24
( / 2) 24 12 7.2 1.67
1.67 20.8 mA80
24 0.7 1.12 k20.8
C CQ C Q
QC Q
CC
CC CCL L
CQ
CQB
B B
VP I
PI
VV VR R
I
II
R R
β
= ⋅
= = =
− −= = ⇒ = Ω
= = ⇒
−= ⇒ = Ω
b. ( )( )
00
1.67 7.2462
0.026(max) 12(max) 12 V 26 mV
462
CQ Lv m L
T
P Pv
I RA g R
VVV V V
A
⋅= = = =
= ⇒ = = ⇒ ≅
8.3
a. For maximum power delivered to the load, set 2C C
C EQ
VV =
( )susSet 25 VC C CEV V= =
( ) ( )( )
,max
D,max
25Then 0.1
250 mA25 12.5 125 mA
0.1
max 0.125 12.52
1.56 W P125 1.25 mA10025 0.7 19.4 k
1.25
CCCm
L
Cm C
CQ
CCQ CQ
BQ
B B
VI
RI I
I
VP I
I
R R
= =
= <−= =
= ⋅ =
= <
= =
−= ⇒ = Ω
b. ( ) ( ) ( ) ( ) ( )221 1max 0.125 100 max 0.781 W rms2 2L CQ L LP I R P= ⋅ ⋅ = ⇒ =
8.4
Point (b): Maximum power delivered to load. Point (a): Will obtain maximum signal current output. Point (c): Will obtain maximum signal voltage output.
8.5 a.
b.
( )( )( )
( )
2
2
2
2
5 V, 0.25 5 4 0.25 A, 37.5 V, 9.375 W
6 V, 0.25 6 4 1.0 A, 30 V, 30 W
7 V, 0.25 7 4 2.25 A, 17.5 V, 39.375 W
8 V, 0.25 2 8 4
402.92
103.
GG D D S
GG D D S
GG D D S
GG D D S D S
D SD S
D
V I V P
V I V P
V I V P
V I V V
VV
I
= = − = = =
= = − = = =
= = − = = =
⎡ ⎤= = − −⎣ ⎦−
= ⇒ =
=
( ) 2
71 A, 10.8 W
9 V, 0.25 2 9 4
401.88 V
103.81 A, 7.16 W
GG D D S D S
D SD S
D
P
V I V V
VV
I P
=
⎡ ⎤= = − −⎣ ⎦−
= ⇒ =
= =
c. Yes, at ,max7 V, 39.375 W > 35 WGG DV P P= = = 8.6 a.
( )
( )
2
2
1 2
1 2
22
1
Set 25 V2
50 25 1.25 A20
1.25 4 6.5 V0.2
Let 100 kΩ
6.5 50 13 kΩ100
87 kΩ
DDDSQ
DQ
DQ n GS TN
GS
GS DD
VV
I
I K V V
V
RV VR R
R RR R
R
= =
−= =
= −
+ = =
⎛ ⎞= ⎜ ⎟+⎝ ⎠
+ =
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
=
b. ( )( )1.25 25 31.25 WD DQ DSQ DP I V P= = ⇒ = c.
,max ,max
,max ,max
,max
2 2.5 A
50 V
31.25 W
D DQ D
DS DD DS
D
I I I
V V V
P
= ⇒ =
= ⇒ =
=
d.
( )( )( )( )( )
( )
0
022
0
2 2 0.2 1.25 1 /
1 20 0.5 10 V
101 1 2.5 W2 2 20
31.25 2.5 28.75 W
m Li
m n DQ
L LL
Q Q
Vg R
V
g K I A V
V
VP PR
P P
=
= = =
= =
= ⋅ = ⋅ ⇒ =
= − ⇒ =
8.7 (a)
(b) ( )( ),max 25D D jP P Slope T= − −
,max ,max60At 0, 25 145 C0.5D j jP T T= = + ⇒ = °
(c) ,max,max
j caseD
dev amb
T TP
θ −
−= or 145 25 2
60dev amb dev amb C/Wθ θ− −−= ⇒ = °
8.8
,max,rated
dev case
,max ambdev case
,rated
or
150 25 2.5 C/W50
θ
θ
−
−
−=
−=
−= = °
j ambD
j
D
T TP
T TP
( )( ) ( )
dev amb dev case case amb
case amb case amb
Then 150 25 2.5 125 2.5
θ θθ θ
− −
− −
− = +− = + ⇒ = +
D
D D
T T PP P
8.9
( )( )( )
( )( )( )
( )( )
dev amb dev case case snk snk amb
dev dev
dev case dev case
case dev case
case sink case snk
sink c
4 5 20 W
25 20 1.75 0.8 3 111 136 C
20 1.75 3535 136 35 101 C
20 0.8 16 C
θ θ θ
θ
θ
− − −
−
−
= ⋅ = =− = + +− = + + = ⇒ = °
− = ⋅ = == − = − ⇒ = °
− = ⋅ = = °=
D D DS
D
D
D
P I VT T PT T
T T PT T T
T T PT T ase sink16 101 16 85 C− = − ⇒ = °T
8.10
( )( )
dev amb dev case case amb
case amb case amb200 25 25 3 4 C/Wθ θ
θ θ− −
− −
− = +− = + ⇒ = °
DT T P
8.11
,max ambdev case
,rated
,max amb
dev case case snk snk amb
175 25 10 C/W15
175 25 10 W10 1 4
θ
θ θ θ
−
− − −
− −= = = °
−=
+ +−= ⇒ =
+ +
j
D
jD
D
T TPT T
P
P
8.12
( )2
12 50%
L
S
S CC Q
CCL P P Q
CC Q
CC Q
PP
P V IV
P V I I
V I
V I
η
η η
=
= ⋅
⎛ ⎞= ⋅ = ⎜ ⎟⎝ ⎠
⋅ ⋅= ⇒ =
⋅
8.13
( )( )
( ) ( )
( )
3 2
S
max 4.8 V0.7 5
4.3 mA1
min0.7 max 4.3 mA
1so 3.6 5.5 V min 4.3 V
o
C C
I o L
I o
v
i i
vv v i
v v
=− − −
= = =
= + = − =
− ≤ ≤ = −
8.14
( ) ( )
( )
( ) ( )( )( )
( )( )
( ) ( ) ( ) ( )
2 33 3
23 3
23
2
3
3 22
3 2
2 2
02 0
0 5K
12 0.5 52 11 2 0
11 11 4 12 22 12
1.072 V
12 1.072 0.5 3.93 mAsat 1.072 0.5 0.572 V
minmin : max 3.93 min 3.93 V
1mi
GSD GS TN
GS GS
GS GS
GS
GS GS
D D
DS GS TN
o
I
VI V V
RV V
V V
V
V V
I IV V V
Vv i V
v
− − −= − =
− = −− − =
± +=
= =
= = − == − = − =
= − = ⇒ = −
( ) ( )( )( ) ( )( )
( )
( )( ) ( )
1
21 1 1
1
n min 3.93 0.5min 3.43 V
max 5 sat 5 0.572max 4.43 V
4.43max 3.93 8.36 mA1
8.36 12 0.5 1.33 Vmax 4.43 1.33 max 5.76 V
o TN
I
o DS
o
D
D GS GS
I o GS I
v Vv
v Vv
I
I V Vv v V v
= + = − += −
= − = −=
= + =
= = − ⇒ == + = + ⇒ =
8.15 a. Neglect base currents.
( )
( )
( ) ( )( )
( )( )
0
1 1
1
max (sat) 10 0.2 9.8 V9.8 9.8(max) 9.8 mA
10 0.7 10
949 9.8
max 2 max 19.6 mA
min 0
max 9.8 mA
min 9.8 mA
CE
L Q QL
E Q E
E
L Q
L Q
v V V
i I IR
R R
i I i
i
i I
i I
+= − = − =
= = = ⇒ =
− − −= ⇒ = Ω
= ⇒ =
=
= =
= − = −
b.
( )( ) ( ) ( )
( ) ( )( ) ( )
2 21 1max 9.8 1 48.02 mW2 2
0
9.8 20 9.8 10 294 mW
48.02 16.3%294
L L L L
S Q Q
S
L
S
P i R P
P I V V I V
P
PP
η η
+ − −
= = ⇒ =
= − + −
= + ⇒ =
= = ⇒ =
8.16 a.
( ) ( ) ( )
( )
0 max 10min min 100 mA0.1
0 0.7 12113
100
Q QL
vI I
R
R R
= = ⇒ =
− − −= ⇒ = Ω
b.
( )( )( )
1 1 1100 12 1.2 W
(source) 2 12 2.4 WQ Q CE Q
Q
P I V P
P I
= ⋅ = ⇒ =
= =
c. ( )( )
22 101 0.5 W2 2 100
1.2 2.4 3.6 W
0.5 13.9%3.6
PL
L
S
L
S
VP
R
P
PP
η η
= ⋅ = =
= + =
= = ⇒ =
8.17
( ) ( )( )221
1
12 0 1.8
38.9 mAD n GS TN
D
I K V V
I
= − = − −
=
(a)
( )( )
( )
For max 4.8 V
sat 1.8 Vmin 5 1.8 3.2 V
0.7 2.5 5.5 V
L
o
DS GS TN
o
I o I
RvV V Vvv v v
= ∞== − == − + = −
= + ⇒ − ≤ ≤
(b) For ( )500 max 4.8 VL oR v= Ω =
For ( ) 23.20, min 3.2 V 6.4 mA
0.5o
o oL
vv v IR
−′< = − = = = −
2.5 5.5 VIv− ≤ ≤ (c)
( )
( ) ( )
( )
( )
2
2
22
For 2 , max 38.9 mA2min min 51.4
38.921 1 38.9 mW
2 2 51.438.910 38.9 389 mW % 10%389
o
L
oL L
L
L
v V I
R R
vP PR
P
′= − = −−= ⇒ = Ω
−
= ⋅ = ⋅ ⇒ =
= = = =
8.18
( )
( ) ( )
( )
22
2 2
2
1 1 , 2 2
So
100%η η
+
+ −− +
+
= =
= ⋅ + ⋅ = −
=
= ⇒ =
PL
L L
SL L
SL
L
S
VVPR R
V VP V V
R R
VP
R
PP
8.19 (a)
( ) ( ) ( )
( )
As maximum conversion efficiency
, 0.7854
4So max 0.785 5
max 5 V
πη
π
= =
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
P
CC
p
p
VV
V
V
(b) Maximum power dissipation occurs when ( )2 52
3.183 VCCp
VV
π π= = =
(c) ( )
( )
2
2
2
2
max
52 1.27
CC
L
LL
VPR
RR
θ π
π
=
= ⇒ = Ω
8.20
(a) 2
2
12
150 49 V 52 V, 52 V2 24
p
L
pp
VP
RV
V V V+ −
= ⋅
= ⋅ ⇒ = ⇒ = = −
(b) 49 2.04 A24
PP
L
VI
R= = =
(c) 494 4 5274.0%
P
CC
VV
π πη
η
⎛ ⎞= ⋅ = ⎜ ⎟⎝ ⎠
=
8.21 (a)
( )( ) ( )
( ) ( )
( )
( ) ( ) ( )( )( )
( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( ) ( )
2
2
2 00
020 0
20 0
2
0
10 max and max
max
max maxSo 10 max
5 0.4
max10 max
2max
100 20 max max2
max 20.5 max 100 0
20.5 20.5 4 100max
2
DS DS GS TN GS
DS o D L n GS
on GS
L
oGS
L n
o oo
L n
V V sat V V V
V V I I K VV
K VR
VV
R K
V VV
R K
VV
VV V
V V
V
≥ = − =
= − = =
=
=⋅
− = =⋅
− =⎡ ⎤⎣ ⎦
− + =
− + =
± −= ( )0 max 8 V
8 1.6 mA5
1.6 2 10 V0.4
L L
LGS I
n
V
i i
iV V VK
⇒ =
= ⇒ =
= = = ⇒ =
b.
( )
( )
281 6.4 mW2 520 1.6
10.2 mW
6.4 62.7%10.2
L
S
L
S
P
P
PP
π
η η
= ⋅ =
= =
= = ⇒ =
8.22
( )2 and O L L L D n GS TNv i R i i K v V= = = − or ( )2 and L n GS GS I Oi K v v v v= = −
( ) ( )
( )( )( )
( ) ( )
( )( )
2 2
0 00
00 0
00
0
Then
or 2
2 2 1
1 4 4
4or
1 4
O n L I O O I O
II I
I II
I
I I
v K R v v v v v
dv dvv v
dv dvdv
v v v vdv
v vdvdv v v
= − = −
⎛ ⎞= − −⎜ ⎟
⎝ ⎠
+ − = −⎡ ⎤⎣ ⎦
−=
+ −
For ( )
( )0 0
0
4 10 810 V, 8 V 0.889
1 4 10 8II I
dv dvv vdv dv
−= = ⇒ = ⇒ =
+ −
At 000, 0 0I
I
dvv vdv
= = ⇒ =
At 001, 0.5 0.667I
I
dvv vdv
= = ⇒ =
8.23 a.
( )
( )( )
3
13
5 10ln 0.026 ln5 10
0.5987 V 1.1973 V2
5 10 50 mW
CBE T
S
BBBE BB
Q C CE Q
iV VI
VV V
P i v P
−
−
⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠
= = ⇒ =
= ⋅ = ⇒ =
b.
( )
0
3
13
0
13
8 V8 80 mA
0.180 mA
80 10ln 0.026 ln5 10
0.6708 V
0.5987 0.6708 8 8.072 V2
1.1973 0.6708 0.5265 V
exp 5 10 exp
L L
Cp
CpEB T
S
EB
BBI EB I
BE BB EB
BECn S
T
v
i i
i
iv V
Iv
Vv v v v
V V v
vi IV
−
−
−
= −−= ⇒ = −
≈
⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠=
= − + = − − ⇒ = −
= − = − =
⎛ ⎞= = ×⎜ ⎟
⎝ ⎠
( ) ( )( ) ( )( )( )( )
22
0.5265 0.311 mA0.026
80 0.1 640 mW
0.311 10 8 5.60 mW
80 2 160 mW
Cn
L L L L
Qn Cn CE Qn
Qp Cp EC Qp
i
P i R P
P i v P
P i v P
⎛ ⎞ ⇒ =⎜ ⎟⎝ ⎠
= = ⇒ =
= ⋅ = − − ⇒ =
= ⋅ = ⇒ =
8.24 (a) ( )
( )( )
2
0.5 2 2.5 V 5.0 V2 2
0.5 10 5 mW
Dn n GSn TN
BBGSn BB
n n p
i K v V
Vv V
P P P
= −
+ = = = ⇒ =
= ⇒ = =
(b) ( )
( ) ( )( )( )
2 10 maxand
max max2 2
2 1
DS GS TN DS GS DS o
O OLGS TN
n L n
V V V V V V v
v viV VK R K
= − ⇒ = − = −
= + = + = +
so
( ) ( ) ( )0 00
max max10 max 2 2
2 2v v
v− = + − =
so ( )0 max 8 Vv =
8 8 mA1
8 2 4 V2
Dn L Dn L
GS GS
i i i i
V V
= = ⇒ = =
= + ⇒ =
( )
Then 8 4 2.5 9.5 V2
8 9.5 2.52
1 V cutoff 0
BBI o GS I
BBSGp o I
SGp p Dp
Vv v V v
Vv v v
v M i
= + − = + − ⇒ =
⎛ ⎞= − − = − −⎜ ⎟⎝ ⎠
= ⇒ ⇒ =
( ) ( )( )( )
22 8 1 64 mW
8 10 8 16 mW
0
L L L L
Mn Dn DS Mn
Mp Dp SD Mp
P i R P
P i v P
P i v P
= = ⇒ =
= ⋅ = − ⇒ =
= ⋅ ⇒ =
8.25 a.
02424 V 3 A8
3 73.2 mA41
L L N
Bn Bn
v i i i
i i
= ⇒ = ⇒ ≈ =
= ⇒ =
For 125 mA 25 73.2 98.2 mAD Ri i= ⇒ = + =
( ) 12
3ln 0.026 ln6 10
0.7004 V
NBE T
S
iV V
I −
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠=
( )
( )
( )
( )
1 11
3
12
12
30 24 0.7 5.3Then 98.2 53.97 98.2
25 100.026 ln 0.5759 V6 10
2 2 0.5759 0.70040.4514 V
0.4514exp 6 10 exp 0.208 mA0.026
−
−
−
− += ⇒ = ⇒ = Ω
⎛ ⎞×= =⎜ ⎟×⎝ ⎠= − = −
=
⎛ ⎞ ⎛ ⎞= = × ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
D
EB D BE
EBP S P
T
R RR
V
V V V
Vi I iV
b. Neglecting base current
( )1
12
30 0.6 30 0.6 545 mA53.970.5450.026 ln 0.656 V
6 10
D D
D
i iR
V −
− −≈ = ⇒ ≈
⎛ ⎞= =⎜ ⎟×⎝ ⎠
Approximation for Di is okay. Diodes and transistors matched 545 mAN Pi i⇒ = = 8.26 (a)
( )
( )( )
21 1 1
1
23 3 3
2 23 3 4
5 1 2 5
200 2 1 200 /
D GS TN
GS
D GS TN
n p
I K V V
V V
I K V V
K K K A Vμ
= −
= + =
= −
= − ⇒ = =
(b) 4 3 1I SG GS GS Ov V V V v+ + − =
( )
( )( )
21 1 1
11 1
0
00
For large,
So 2 2 10.5 5
32.5
o L n GS TN
oLGS TN TN
n L n
oI
I
v i i K V V
viV V VK R K
vv v
vv v
= = −
= + = +
⎛ ⎞+ + − + =⎜ ⎟⎜ ⎟
⎝ ⎠
= + −
0 0
0
0
0
1 112 2.5
11 12 2.5
I
I I I
I
dv dvdvdv dv dvv
dvdv v
= = + ⋅ ⋅
⎡ ⎤= +⎢ ⎥
⎢ ⎥⎣ ⎦
For 5 :Ov V=
( )( )0 0 011 1 1.1414 0.876
2 2.5 5I I I
dv dv dvdv dv dv
⎡ ⎤⎢ ⎥= + = ⇒ =⎢ ⎥⎣ ⎦
8.27
and 2
DnBBO I GS GS TN
n
IVv v V V VK
= + − = +
For 0, OO Dn DQ L DQ
L
vv I I i IR
≈ = + = +
Then ( )/
2DQ O LBB
O I TNn
I v RVv v VK
+= + − − or 1
2DQ OBB
O I TNn DQ L
I vVv v VK I R
= + − − ⋅ +
For Ov small, 112 2
DQ OBBO I TN
n DQ L
I vVv v VK I R
≅ + − − ⋅ + ⋅
1 112 2
DQ DQBBO I TN
n DQ L n
I IVv v VK I R K
⎡ ⎤+ ⋅ ⋅ = + − −⎢ ⎥
⎢ ⎥⎣ ⎦
Now 1 0.95
1 112
O
I DQ
n DQ L
dvdv I
K I R
= =⎡ ⎤
+ ⋅ ⋅⎢ ⎥⎢ ⎥⎣ ⎦
So 1 1 1 1 0.05262 0.95
DQ
n DQ L
IK I R
⋅ ⋅ = − =
For 10.1 , then 0.01052Ln DQ
R kK I
= Ω =
Or 95.1n DQK I =
We can write 2 190 m n DQg K I mA/V= = This is the required transconductance for the output transistor. This implies a very large transistor. 8.28
v m LA g R= −
So ( )12 2 6 mA/V= CQm m
T
Ig g
V− = − ⇒ =
( )( )6 0.026 0.156 mACQ CQI I= ⇒ =
But for maximum symmetrical swing, set 10 5 mA 122
CCCQ v
L
VI A
R= = = ⇒ >
Maximum power to the load:
( ) ( )( ) ( )
( )( )
22 101 max max 25 mW2 2 2
10 5 50 mWSo 50%
CCL L
L
S CC CQ
VP P
R
P V Iη
= ⋅ = ⇒ =
= ⋅ = ==
( ) ( )
( )( ) ( )( )( )
( )( )
1
1
11
2
5 0.0278 mA180
6 kon 1
Set 20 0.0278 6 0.7 181 0.0278 0.020
0.967 V1
10.967 6 10 62.0 k
6.64 k
CQBQ
TH
TH BQ TH BE BQ E
E
TH
TH
TH TH CC
II
R RV I R V I R
RVV
V R VR
RR
R
β
β
= = =
= = Ω= + + +
= Ω= + +=
= ⋅ ⋅
= ⇒ = Ω
= Ω
8.29
( ) ( )( ) ( )
( )( )( ) ( )( )( )
( )( )
22
1 1
1
2
15 15 mA1
15 0.15 mA100
151max max 112.5 mW2 2 1
Let 10 k1
0.15 10 0.7 101 0.15 0.11 13.715 10 15
40.4 k
13.3 k
β
= = =
= =
= ⋅ = ⇒ =
= Ω= + + += + +
= = ⋅ ⋅ = ⋅
= Ω
= Ω
CCCQ
L
BQ
CCL L
L
TH
TH BQ TH BE BQ E
TH TH CC
VIR
I
VP PR
RV I R V I R
V R VR R
R
R
8.30
( )
( ) ( )( )
( ) ( )
2
1 2
1 2
22
1.55 101.55 0.73
6.80 V0.73 1.55 0.496 k
6.80 0.701 0.496 26 0.02
6.0 mA, 150 mA
and 3 8 72
150 5.77 A/V0.026
TH CC
TH
TH BEBQ
TH E
BQ CQ
v m L L L
CQm
T
RV VR R
R R R
V VIR R
I I
A g R R a RI
gV
β
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=
= = = Ω
− −= =+ + +
= =
′ ′= − = = = Ω
= = ⇒
( )( )( )( )
5.77 72 415
415 0.017 7.06 V7.06 2.35 V
37.06 2.35 V
37.06 2.35 V
3
v
o v i
o
o
o
A
V A V
V
V
V
= − = −
′ = ⋅ = =
= =
= =
= =
( )( )0.15 10 1.5 W
0.345 23%1.5
S CQ CC
L
S
P I V
PP
η η
= ⋅ = =
= = ⇒ =
8.31 a. Assuming the maximum power is being delivered, then
( ) rms rms36 9peak 36 V 9 V 6.36 V4 2o oV V V V′ = ⇒ = = ⇒ = ⇒ =
b. 36 25.5 V2o oV V= ⇒ =
c. Secondary rms rmsrms
2 0.314 A6.36
LPI IV
= = ⇒ =
Primary 0.314 78.6 mA4P PI I= ⇒ =
d. ( )( ). 0.15 36 5.4 W
2 37%5.4
η η
= = =
= ⇒ =
S CQ CCP I V
8.32 a.
( )( )
2
1
2
10 0
21
2
1
1
1( )
11
where 1 11
so
e m E m E
E
i e i e
e i e E
EEe e
E Li E i
E
ee
Vv g V R V g R
r r
V Rr
v V v V v v
v v v Rr
RRv r v nR R
v r R v nRr
v nv v vnn
n
ππ π
π π
ππ
π π
π
π
π
π
β
β
ββ
β β
⎛ ⎞ ⎛ ⎞′ ′= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞+ ′= ⎜ ⎟⎝ ⎠
= + ⇒ = −
⎛ ⎞+ ′= − ⎜ ⎟⎝ ⎠
+ ′⋅ ′+ ⎛ ⎞′= = = = ⎜ ⎟+ ′+ + ⎝ ⎠′+ ⋅
⎛ ⎞= − ⎜⎛ ⎞ ⎝ ⎠⎜ ⎟⎝ ⎠
( )( )
0
1
2
11so 1
E
i E
Rvv r Rn
nπ
ββ
⎟
′+= ⋅
′+ +⎛ ⎞⎜ ⎟⎝ ⎠
b.
( )( )
2 2 2
2 22
2 2
1 1 1, , so .2 2 2
.For 50% :
120.5 so
2 0.1 50 5
η
= ⋅ = = =
==
⋅= = = = = ⇒ =
⋅ ⋅
PL P L CQ L CQ L
S CQ CC
CQ LCQ L CC CC CCL
CQ CC CC CQ LS
InP I R a I P a I Rn a
P I V
a I R a I R V V VP a aI V V I RP
c.
( )( )
( )( )0 0
49 0.0260.255
1 1 50 0.1T
CQ
r VR RI
π ββ β
= = = ⇒ = Ω+ +
8.33 a. With a 10:1 transformer ratio, we need a current gain of 8 through the transistor.
( ) 1 2
1 2
1 and e b b iib
R Ri i i i
R R Rβ
⎛ ⎞= + = ⎜ ⎟⎜ ⎟+⎝ ⎠
so we need ( ) 1 2
1 2
8 1e
i ib
R Rii R R R
β⎛ ⎞
= = + ⎜ ⎟⎜ ⎟+⎝ ⎠ where
( ) ( ) ( )( )1 1 101 0.8 80.8ib L LR r R Rπ β β′ ′= + + ≈ + = =
( )
( )( ) ( )( )
1 2
1 2
1 21 2
1 2
1
1 21
Then 8 10180.8
0.0792 or 6.95 k80.8
2 12Set 15 mA2 0.8
15 0.15 mA100
1
1 6.95 12 0.15 6.95 0.7 47.9 k then
⎛ ⎞= ⎜ ⎟⎜ ⎟+⎝ ⎠
= = Ω+
′= ⇒ = = =′
= =
= +
⋅ ⋅ = +
= + ⇒ = Ω
CC CCL CQ
CQ L
BQ
TH BQ TH BE
TH CC BQ TH BE
R RR R
R RR R
R RV VR II R
I
V I R V
R V I R VR
R RR
8.13 k= Ω
b.
( ) ( )
( )( )
2
0.9 13.5 mA 135 mA
1 0.135 8 72.9 mW2
12 15 180 mW
40.5%
Le CQ L
L L
S CC CQ S
L
S
II I I
a
P P
P V I P
PP
η η
= = = ⇒ =
= ⇒ =
= = ⇒ =
= ⇒ =
8.34 a.
( )( )
( )( )
2
2 8 2 5.66 V peak output voltage
5.66 0.708 A peak output current8
Set 0.9 to minimize distortion0.9 18
Then 2.865.66
=
= = =
= = = =
= =
= ⇒ =
P L L
P
PP
L
e CC P
V R P
V
VIR
V V aV
a a
b.
( )( )
1 1 0.708Now 0.275 A0.9 0.9 2.86
Then 18 0.275 4.95 W Power rating of transistor
⎛ ⎞ ⎛ ⎞= = ⇒ =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠= = ⇒ =
PCQ CQ
Q CC Q Q
II Ia
P V I P
8.35 a. Need a current gain of 8 through the transistor.
( ) 1 2
1 2
8 1b
i ib
R Rii R R R
β⎛ ⎞
= = + ⎜ ⎟⎜ ⎟+⎝ ⎠ where ( )( )1 0.9 90.9 kΩibR β≈ + =
( )( ) ( ) ( )
1 21 2
1 2
1 21
8 0.0792 or 7.82 k101 90.9
2 12Set 0.9 k 13.3 mA2 0.9
13.3 0.133 mA1001Then 7.82 12 0.133 7.82 0.7 53.9 k and R 9.15 k
CCCQ
CQ
BQ
R RR R
R R
V II
I
RR
⎛ ⎞= = = Ω⎜ ⎟⎜ ⎟+⎝ ⎠
= Ω ⇒ = =
= =
= + ⇒ = Ω = Ω
b.
( )
( ) ( )
( )( )
2
0.9 12 mA 120 mA
1 0.12 8 57.6 mW2
12 13.3 159.6 mW
57.6 36.1%159.6
η η
= = = ⇒ =
= ⇒ =
= = ⇒ =
= = ⇒ =
Le CQ L
L L
S CC CQ S
L
S
II I Ia
P P
P V I P
PP
8.36 a. All transistors are matched.
1 313 mA
61 13 2.90 mA60 60
ββ β
⎛ ⎞+= + = +⎜ ⎟⎝ ⎠
⎛ ⎞= + ⇒ =⎜ ⎟⎝ ⎠
CE B C
C C
ii i i
i i
b. For 6 Vov = , let 200 .LR = Ω
3
3
1
1 1
2 2
2 1
6 0.03 A 30 mA20030 0.492 mA613 0.492 2.508 mA2.508 41.11 A
6133 mA 49.18 A61
49.18 41.11 8.07 A
o E
B
E
B B
E B
I B B I
i i
i
i
i i
i i
i i i i
μ
μ
μ
= = = ≅
= =
= − =
= ⇒ =
≅ ⇒ = ⇒
= − = − ⇒ =
( )
( )
33
6
33
3 13
3
31
1 13
1
Current gain 30 10 3.72 10
8.07 1030 10ln 0.026 ln5 10
0.6453 V
2.508 10ln 0.026 ln5 10
0.5807 V
i i
EBE T
S
BE
EEB T
S
EB
A A
iV V
IV
iV VI
V
−
−
−
−
−
−
×= ⇒ = ××⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠
=
⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠=
0 3 1
0
6 0.6453 0.58076.0646 V
Voltage gain 6 0.989
6.0646
I BE EB
I
v vI
v v V Vv
vA Av
= + − = + −=
= = ⇒ =
8.37
a. For 0 311 A, 20 mA
50Bi I= ≅ ⇒
We can then write ( )0,max 31
1 1
1010 2 20BEEBv VV
R R
⎡ ⎤− +−= −⎢ ⎥
⎢ ⎥⎣ ⎦
If, for simplicity, we assume 1 3 0.7 V,EB BEV V= = then ,max
1 1
210 40oBE vVR R− = +
If we assume 0,max 4 V,v = then ( )
1 1
2 49.3 40R R
= + which yields 1 2 32.5 R R= = Ω
b. For 0,Iv = 1 1 29.3 0.286 A
32.5E E EI I I= ⇒ = =
Since 3,4 1,210 ,S SI I= then 3 4 2.86 AE EI I= = c.
( )( )
( )( )
13 1
10
3
33
3
11
1
0
0
We can write
112 1
50 0.026Now 0.4545
2.86120 0.026
10.91 0.286
So
10.910.4545 32.51 1212 51
10.9132.5 32.5 0.0902 0.0900121
1Then 2
T
C
T
C
rr RR
VrI
VrI
R
R
ππ
π
π
ββ
β
β
⎧ ⎫+⎪ ⎪+⎪ ⎪= ⎨ ⎬+⎪ ⎪
⎪ ⎪⎩ ⎭
= = = Ω
= = = Ω
⎧ ⎫+⎪ ⎪⎪ ⎪= ⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭
= =
= 00.4545 0.0900 or R 0.00534
51+⎧ ⎫ = Ω⎨ ⎬
⎩ ⎭
8.38
( ) ( )( )
( )( )
( ) ( )( )( )
1 1 3
1 3
1 1 1 227.2 mA and 7.2 mA
60 0.026Then 0.217 k
7.21So 0.217 61 2 0.217 61 0.221 0.217 61 2 12.4 or 52.6 k2
π π
π
β β⎡ ⎤= + + + +⎣ ⎦≈ ≈
= = Ω
⎡ ⎤= + +⎣ ⎦
= + ⎡ ⎤ = Ω⎣ ⎦
i L
C C
i
i
R r R r R
i i
r
R
R
8.39 a.
b.
( )
( )
21 1
12
1 21
5 10 2 2.707 V10 2.7075 1.46 k
+ −= + =
= − ⇒ =−= ⇒ = = Ω
SGSG TP
SG SG
V VI K V VR
V V
R RR
c. 100 LR = Ω For a sinusoidal output signal:
( ) ( )
( ) ( )
( )
2 2
3 3
3
1 1
51 1 125 mW2 2 0.1
550 mA
0.1
50 2 4.236 V10
10 4.236 50.523 mA
1.46
oL L
L
oD D
L
GS
D
vP P
Rv
i iR
V
I I
= ⋅ = ⋅ ⇒ =
≈ = ⇒ =
= + =
− += ⇒ =
10.523 2 2.229 V
105 4.236 2.229 7.007 V
SG
I I
V
v v
= + =
= + − ⇒ =
( ) ( ) ( )
( )
( ) ( )( )( )
22
2
2
2
1010 2
1.4617.007
10 4 41.46
14.6 57.4 41.4 0
57.4 57.4 4 14.6 41.42 14.6
I GSD GS
GSGS GS
GS GS
GS
V VI V
VV V
V V
V
− − −= = −
−= − +
− + =
± −=
( )2
22 2
4 2
4 4
2.98 V
10 2.98 2 9.60 mA
7 2.98 4.02 V5 4.02 0.98 V 0
GS
D D
G I GS
SG D
V
I I
V v VV I
=
= − ⇒ =
= − = − == − = ⇒ =
8.40 For 0 0v =
( )
( )
( )
( )( ) ( )
3 2 1
33 2 2
3 2
22 1 1
2 1
3 1
1 1 1
1
1
1
1
1
11
11 1
10 1051 50 5011
Q C C E
n CB E C
n n
C n C
CPB C E
P n
PC n E
P
PC n n E
p
P PQ n n E n E E
P P
E
I I I I
II I I
I I
II I I
I I
I I
I I I I
I
ββ β
ββ
β β
βββ
ββ ββ
β ββ β ββ β
= + +
⎛ ⎞+= = =⎜ ⎟
⎝ ⎠= +
⎛ ⎞= = =⎜ ⎟+⎝ ⎠
⎛ ⎞= ⎜ ⎟+⎝ ⎠
⎛ ⎞= + ⎜ ⎟⎜ ⎟+⎝ ⎠
⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
⎛ ⎞= +⎜ ⎟⎝ ⎠
( ) ( )
( )( ) ( )
1 1
1 1 1
1 1
2 2
3 3
112318.18 45.451.692 A 1.534 A
1050 1.692 76.9 A11
1051 50 1.692 3.92 mA11
E E
Q E E E
E C
C C
C C
I I
I I I II I
I I
I I
μ μ
μ
⎛ ⎞ +⎜ ⎟⎝ ⎠
= + += ⇒ =
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
Because of 1rπ and Z, neglect effect of r0. Then neglecting r01, r02 and r03, we find
3 3 2 2 1 11
XX m m m
VI g V g V g V
r Zπ π ππ
= + + ++
( )( )
[ ]
( )
11 2 1 1 2
1
3 1 1 2 2 3
1 1 2 1 1 2 3
13 1 1 2 2 3
1
1 1 2 33
1
1 1 22 1 2
1 1
Now
,
and
and
Then
X m
m m
m m m
m m m X
X
m X X
rV V V g V rr Z
V g V g V rg V g g V r r
rV g g g r r Vr Z
rV V
r Z
r rV g r V Vr Z r Z
ππ π π π
π
π π π π
π π π π
ππ π π
π
ππ
π
π ππ π
π π
β β β
β
⎛ ⎞= ≅⎜ ⎟+⎝ ⎠
= += +⎡ ⎤⎣ ⎦⎛ ⎞
= + ⋅⎜ ⎟+⎝ ⎠+
= ⋅+
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
( )
( )( )( )
( ) ( )( ) ( )( ) ( )
1 1 2 3 1 2 1
1 1 1 1
10
1 1 2 1 1 2 3
1
0
0
Then
1
10 0.0260.169 MΩ
1.53425 kΩ
Then169 25
1 10 10 50 10 10 50 50
194 0.00746 kΩ or 7.46 26,011
XX X X X
X
X
o
VI V V Vr Z r Z r Z r Z
r ZVRI
r
Z
R
R R
π π π π
π
π
β β β β β β β
β β β β β β β
+= ⋅ + ⋅ + ⋅ +
+ + + +
+= =
+ + + +
= =
=
+=+ + + +⎡ ⎤⎣ ⎦
= = = Ω
8.41 a Neglect base currents.
( )
Bias
3
13
2 2 ln
5 102 0.026 ln 1.281 V10
BB D TS
BB
IV V VI
V−
−
⎛ ⎞= = ⎜ ⎟
⎝ ⎠⎛ ⎞×= ⇒ =⎜ ⎟⎝ ⎠
1 3
1 3 2
2 3 3
2 2 3
1
1
BE EB BB
E E C
PB C E
P
PC n B n E
P
V V VI I I
I I I
I I I
ββ
ββ ββ
+ == +
⎛ ⎞= = ⎜ ⎟+⎝ ⎠
⎛ ⎞= = ⎜ ⎟+⎝ ⎠
1 3 3
1 3
1
11
PE E n E
P
PE E n
P
I I I
I I
βββ
βββ
⎛ ⎞= + ⎜ ⎟+⎝ ⎠
⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦
1 31 1 1
1n P P
C C nn P P
I Iβ β ββ
β β β⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +
= +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣ ⎦
( ) ( )
1 31 3
1 3
3 3
ln , ln
21 201.01 1 10020 21
21 100 101.0520
C CBE T EB T
S S
C C
C C
I IV V V V
I I
I I
I I
⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥
⎣ ⎦ ⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
⎡ ⎤= + =⎢ ⎥⎣ ⎦
1 3
3 3
23
2
23
2
3 C3
100.05
100.05ln ln
100.05ln
100.05exp
exp 0.4995 mA100.05
C C
C CT T BB
S S
CT BB
S
C BB
TS
S BBC
T
I I
I IV V V
I I
IV V
I
I VVI
I VI I
V
=
⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠⎛ ⎞
=⎜ ⎟⎝ ⎠
⎛ ⎞= ⎜ ⎟
⎝ ⎠
⎛ ⎞= = =⎜ ⎟
⎝ ⎠
( ) ( )
3
1 3 1
2 2
Then 0.5245 mANow 100.05 49.97 mA
20100 0.5245 49.95 mA21
E
C C C
C C
II I I
I I
== = =
⎛ ⎞= = =⎜ ⎟⎝ ⎠
31
1 13
33
3 13
1 3
49.97 10ln 0.026 ln10
0.70037
0.4995 10ln 0.026 ln10
0.58062Note: 0.70037 0.58062 1.28099
CBE T
S
CEB T
S
BE EB
BB
IV V
I
IV V
I
V VV
−
−
−
−
⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
=
⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
=+ = + =
=
b.
( )
( )
( )
0 1 1
1
3
13
1 13
3
133
220
1010 V 0.10 A100
100 1 mA100
4 102 0.026 ln 1.2694 V10
0.10.026 ln 0.718410
1.2694 0.7184 0.55099 V0.5509910 exp 0.1598 mA0.026
10100
E C
B
BB
BE
EB
C
L LL
v i i
i
V
V
V
I
VP P
R
−
−
−
−
= ⇒ ≈ = =
= =
⎛ ⎞×= =⎜ ⎟⎝ ⎠
⎛ ⎞= =⎜ ⎟⎝ ⎠
= − =
⎛ ⎞= =⎜ ⎟⎝ ⎠
= = ⇒ 1 W=
( )( )( ) [ ]( )
( )( ) ( )( )( ) [ ]( )
1 1 1 1
3 3 3 3
2 3
2 2 2 2
0.1 12 10 0.2 W
0.1598 10 0.7 12 3.40 mW
100 100 0.1598 15.98 mA15.98 10 12 0.352 W
Q C CE Q
Q C EC Q
C C
Q C CE Q
P i v P
P i v P
i iP i v P
= ⋅ = − ⇒ =
= ⋅ = − − ⇒ =
= = == ⋅ = − − ⇒ =
8.42 a.
( )
( )( )
3
12
1 2 3
2 21 3 2
1 2 3
32
3 3
32
12
10 103 0.026 ln 1.74195 V2 10
,
ln ln ln
ln
exp
1.720 20 10 exp
BB BB
BE BE EB BB
C CC C
n n
C C CT T T BB
S S S
CT BB
n S
BBC n S
T
V V
V V V VI II I
I I IV V V VI I I
IV VI
VI IV
β β
β
β
−
−
−
⎛ ⎞×= ⇒ =⎜ ⎟×⎝ ⎠+ + =
≈ ≈
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎡ ⎤
=⎢ ⎥⎣ ⎦
⎛ ⎞= ⎜ ⎟
⎝ ⎠
= ×
( )
( )
( )
3
2 1 3
3
1 112
2 212
3
3 312
41950.026
0.20 A, 10 mA, 0.5 mA
10 100.026 ln 0.58065 V2 10
0.20.026 ln 0.6585 V2 10
0.5 100.026 ln 0.50276 V2 10
C C C
BE BE
BE BE
EB EB
I I I
V V
V V
V V
−
−
−
−
−
⎛ ⎞⎜ ⎟⎝ ⎠
= ≈ ≈
⎛ ⎞×= ⇒ =⎜ ⎟×⎝ ⎠⎛ ⎞= ⇒ =⎜ ⎟×⎝ ⎠⎛ ⎞×= ⇒ =⎜ ⎟×⎝ ⎠
b.
( )2 2
0 00
1 110 W= max 20 V2 2 20L
L
V VP VR
= ⋅ = ⋅ ⇒ =
For ( )0 maxv :
( )
( )( )
220
0
5 4 3
5 45
5 55
5
2020 W
2020max 1 A20
max 1 A
11
1
11 11 1
1 20 1120 21 21
L LL
C C E o
pC n CC
n n n p
pC n C nC
n n n n n p
C
vP P
R
i
i i i i
i ii
i ii
i
βββ β β β
ββ ββ β β β β β
= = ⇒ =
= − = −
+ + = − =
⎛ ⎞+⎛ ⎞+ ⋅ + =⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠
⎡ ⎤⎛ ⎞+⎛ ⎞ ⎛ ⎞+ + =⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ + ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦⎡ ⎤⎛ ⎞ ⎛+ +⎜ ⎟⎢ ⎥⎝ ⎠ ⎝⎣ ⎦
1 6 120 5⎡ ⎤⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎢ ⎥⎠ ⎝ ⎠⎣ ⎦
( )5 5
4
3
3
1.05048 1 0.952 A
0.0453 A
0.00272 A50.002726
0.002267 A
C C
C
E
C
i i
i
i
i
= =
=
=
⎛ ⎞= ⎜ ⎟⎝ ⎠
=
( )3
3 12
1 2
2 2
2.267 100.026 ln 0.54206 V2 10
1.74195 0.54206 1.19989
ln ln 1.19989
EB
BE BE
C CT T
n S S
V
V V
I IV V
I Iβ
−
−
⎛ ⎞×= =⎜ ⎟×⎝ ⎠+ = − =
⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
( )
2
2
1.19989exp0.026
20 18.83 mA93.9 mA
C n S
C
i I
i
β ⎛ ⎞= ⋅ ⎜ ⎟⎝ ⎠
==
( )( ) ( ) ( )( ) ( )( )
21
2 2
5
93.9 4.47 mA1 2124 20 0.0939 44 4.13 W
0.952 10 24 13.3 W
C nC
n n
Q C
Q
ii
P I
P
ββ β
⎛ ⎞= = =⎜ ⎟+⎝ ⎠= − − = =
= − − − =
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