Ch15p
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Transcript of Ch15p
Chapter 15 Exercise Solutions EX15.1 For the circuit shown in Figure 15.7
31
2dBfRCπ
=
or
( )6
33
1 1 3.979 102 2 40 10dB
RCfπ π
−= = = ××
For 75 KR = Then
115.31 10 53.1 pFC −= × =
3
4
We have 1.414C 75.1 pF0.707C 37.5 pF
CC
= == =
EX15.2
1C
eq
fCR
=
or
( )( )5 6
1 110 20 10
0.5 pFc eq
Cf R
C
= =×
=
EX15.3
Low-frequency gain: 1
2
30 65
CTC
= − = − = −
( )( )( )
3 122
3 312
100 10 5 106.63 kHz
2 2 12 10C
dB dBF
f Cf fCπ π
−
−
× ×= = ⇒ =
×
EX15.4
0
63
0
12 3
1 1 6.13 102 3 2 (15 10 ) 3
fRC
RCf
π
π π−
=
= = = ××
Let 0.001 μF 1 nFC = =
Then 6.13 kR = Ω so 2 8 49 kR R= = Ω EX15.5
( )( )( )
00
4
2 1 2
1 12 2
1 0.02 F2 800 10
2 2 10 20 k
f CRC f R
C C
R R R
π π
μπ
= ⇒ =
= ⇒ ≅
= = ⇒ = Ω
EX15.6
( )
1
1 2
1
1
1 1
1 1
2 (12)20
2 20 1240 10 4 k
TH HRV V
R R
RR
R RR R
⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞
= ⎜ ⎟+⎝ ⎠+ =
= ⇒ = Ω
EX15.7
( )
[ ]( )
( )
( )
1
1 2
1
1 2
2 2
1 1
2
1 2
1
2
0.10 10 10
201 200 1990.10
11 1 1 1.005 V199
on0.1
10 0.7 0.70.1 43 k0.2
42.9 k
TH TL H L
S REF
REF S REF
H BE
RV V V VR R
RR R
R RR R
RV VR R
RV V VR
V V VI
R
R
R
γ
⎛ ⎞− = −⎜ ⎟+⎝ ⎠
⎛ ⎞= − −⎜ ⎟+⎝ ⎠
+ = = ⇒ =
⎛ ⎞= ⎜ ⎟+⎝ ⎠
⎛ ⎞ ⎛ ⎞= + = + ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
− −=
+− −+ = = Ω
= Ω
EX15.8 At 0 ,t −= let 0 5v = − so 2.5.Xv = − For 0t >
( )10 2.5 10 exp XX
tvr
⎛ ⎞−= + − − ⎜ ⎟⎝ ⎠
When 5.0,Xv = output switches
( )
1
1
11 1
5.0 10 12.5 exp
10 5 5.0exp 12.5 12.5
12.5 12.5exp ln 0.9165.0 5.0
X
X
X XX
tr
tr
t t r t rr
⎛ ⎞= − −⎜ ⎟
⎝ ⎠⎛ ⎞ −− = =⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞+ = ⇒ = ⋅ ⇒ =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
During the next part of the cycle
[ ]( )5 5 5 exp XX
tvr
⎛ ⎞= − + − − −⎜ ⎟
⎝ ⎠
When 2.5,Xv = − output switches
( )
( ) ( )
( )( )
2
2
22 2
1 2
3 6 4
2.5 5 10 exp
5 2.5 2.5exp 10 10
10 10exp ln 1.392.5 2.5
1Period 0.916 1.39 2.31 Frequency2.31
50 10 0.01 10 5 10 s 866 Hz
Du
X
X
X XX
X XX
X
tr
tr
t t r t rr
t t T r rr
r f− −
⎛ ⎞− = − + −⎜ ⎟
⎝ ⎠⎛ ⎞ −− = =⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞+ = ⇒ = ⋅ ⇒ =⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠
= + = = + = ⇒ =⎡ ⎤⎣ ⎦
= × × = × ⇒ =
( )( ) ( )
1
1 2
0.916ty cycle 100% 100% Duty cycle 39.7%
0.916 1.39t
t t= × = × ⇒ =
+ +
EX15.9 a.
( )
[ ]
10
1 2
1
1 2
6
6
6
10 12 1.2 V10 90
0.10
0.711 / 12 ln ln 1 1 (0.10)
50 10 ln 1.18 (0.162)50 10 3.09 k
(0.1 10 )(0.162)
X X X
Y
PX X
X X
X X
r R C
Rv vR RR
R R
V VT r r
T r r
R R
γ
β
β
−
−
−
=
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
= =+
⎡ ⎤++ ⎢ ⎥⎡ ⎤= = ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎢ ⎥
⎢ ⎥⎣ ⎦= × = =
×= ⇒ = Ω×
b. Recovery time
( 1.2 ) exp X P PX
tv V Vr
⎛ ⎞= + − − −⎜ ⎟
⎝ ⎠
When ,Xv Vγ= 2t t=
( )
( )
( )( )
2
2
2
3 6 42
0.7 12 1.2 12 exp
12 0.7exp 0.85613.2
1 ln 0.1550.856
3.09 10 0.1 10 3.09 10 48.0 s
X
X
X X
X
tr
tr
t r r
r t μ− −
⎛ ⎞= + − − −⎜ ⎟
⎝ ⎠⎛ ⎞ −− = =⎜ ⎟⎝ ⎠
⎛ ⎞= =⎜ ⎟⎝ ⎠
= × × = × ⇒ =
EX15.10
6
1.1 RC75 10
TT −
== ×
Let 10 nFC = Then
( )( )6
9
75 10 6.82 K1.1 10 10
R−
−
×= =×
EX15.11
( ) ( ) ( ) ( )
( )
3 6
1 1 802 Hz0.693 2 0.693 20 2 80 10 0.01 10
20 80Duty cycle 100% 100% Duty cycle 55.6%2 20 2 80
A B
A B
A B
f fR R C
R RR R
−= = ⇒ =
+ + × × ×⎡ ⎤⎣ ⎦+ += × = × ⇒ =
+ +
EX15.12
a.
( )( )
212
2 2 8 1 4 V
4 0.5 A8
P
L
P L P
PP P
L
VPR
V R P V
VI IR
= ⋅
= = ⇒ =
= = ⇒ =
b. 12 4 8 V0.5 A
CE
C
VI
= − =≈
So ( )( )0.5 8 4 WC CEP I V P= ⋅ = ⇒ = EX15.13
( )( )
( ) ( )2
2 2 8 10 12.65 V
10 812.65
19.9 V
P L L
PS S
L
SS
P
S
V R P
VP VR
P RV
VV
πππ
= = =
⎛ ⎞= ⎜ ⎟
⎝ ⎠
= =
=
EX15.14
Line 0 0regulation Z
Z
dV dV dVdVdV dV+ += = ⋅
Now
0
1
101 210
10 0.0022710 4400
Z
Z Z
Z
dVdV
dV rr RdV +
⎛ ⎞= + =⎜ ⎟⎝ ⎠
⎛ ⎞= = =⎜ ⎟+ +⎝ ⎠
So Line ( )( )regulation 2 0.00227 0.00454= = 0.454% EX15.15
( )
( ) ( )( )
0 1 011
0 01 0 1 1
0 0 10 1 0
0
0 0 0 0 0 11
0 0 0
0 0 0
0
00 0 00
0
1 110 10 10 10 10
2 210 10 2
010
10 10
2(10) 21000 6.3 1000
10 0.5 0.5 20 2 0.5[0.10 2.0 0.05
L Z
L
L Z L
L
L
V V VV V
V VV V V V
V A V VV V VR R
V V V A V A VVR R R R
V A VR
VV V VI
V
− ⎛ ⎞= ⇒ + =⎜ ⎟⎝ ⎠
⎛ ⎞ = ⇒ = ⇒ =⎜ ⎟⎝ ⎠
− −−+ + =
+ + − = −
= −
+ + − = −
+ − + 0
0 0
1000] 12,600(1002.05) 12,600
IV I
+ =+ =
For 0 01 mA 12.5732I V= ⇒ = For 0 0100 mA 12.4744I V= ⇒ =
( ) ( )( )
0 0
0
NL FLLoad reg 100%
NL12.5732 12.4744 100%
12.5732Load reg 0.786%
V VV
−= ×
−= ×
=
EX15.16 a.
( )
( )
31 2 3
3 3
34 4
4
44
3 on
5.6 3 0.6 3.8 0.482 mA3.9 3.4 0.576 7.88
ln
0.482(0.1) (0.026) ln
Z BEC
C C
CC T
C
CC
V VI
R R R
I I
II R VI
II
−=
+ +−
= = ⇒ =+ +
⎛ ⎞= ⎜ ⎟
⎝ ⎠⎛ ⎞
= ⎜ ⎟⎝ ⎠
By trial and error 4
7 7
0.213 mA
2(0.6) (0.482)(3.9) 3.08 VC
B B
I
V V
=
= + ⇒ =
b.
( )( ) ( )( ) ( )
130 8 7
13 12
12
12
12 12
2.23 (5) 3.082.232.23 5 3.08 2.23 3.08
11.15 6.868 3.08 1.39 k
B BR V V V
R R
RR
R R
⎛ ⎞= =⎜ ⎟+⎝ ⎠
⎛ ⎞=⎜ ⎟+⎝ ⎠
= += = ⇒ = Ω
TYU15.1
( )
3
54
3
12
1 1 1.59 102 2 10
dB
dB
fRC
RCf
π
π π−
=
= = = ×
Let 0.01 F 1.59 kC Rμ= ⇒ = Ω Then
1
2
3
0.03546 F0.01392 F0.002024 F
CCC
μμμ
===
6 6
3
1 1
2011 10
0.124or 18.1 dBd
Tf
f
T T
= =⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
= = −
TYU15.2
( )
33
63
1 12 2
1 3.18 102 50 10
dBdB
f RCRC f
RC
π π
π−
= ⇒ =
= = ××
Let 0.001 F 1 nF 3.18 kC Rμ= = ⇒ = Ω Then
1
2
3
4
2.94 k3.44 k1.22 k8.31 k
RRRR
= Ω= Ω= Ω= Ω
83
8 243
23 3
10.01
1
11 100.01
10 15.8 kHz10
dB
dB
dB dB
Tf
f
ff
f ff f
f
−
−
−
= =⎛ ⎞+ ⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞+ = =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
⎛ ⎞≅ ⇒ = ⇒ ≅⎜ ⎟
⎝ ⎠
TYU15.3
2
4
6
8
11-pole 3.87 dB12110
12-pole 4.88 dB12110
13-pole 6.0 dB12110
14-pole 7.24 dB12110
T
T
T
T
= ⇒ −⎛ ⎞+ ⎜ ⎟⎝ ⎠
= ⇒ −⎛ ⎞+ ⎜ ⎟⎝ ⎠
= ⇒ −⎛ ⎞+ ⎜ ⎟⎝ ⎠
= ⇒ −⎛ ⎞+ ⎜ ⎟⎝ ⎠
TYU15.4
1eq
C
Rf C
=
or 76
1 1 2 105 10C
eq
f CR
−= = = ××
If 10 pF 20 kHzCC f= ⇒ = TYU15.5
( )( )( )
0 04 12
42 2
1 1 65 kHz2 6 2 6 10 100 10
29 29 10 290 k
f fRC
R R R
π π −= = ⇒ ≅
×
= = ⇒ = Ω
TYU15.6
0 09 2
61 29
1 2
2
1
23
1
1 1 7.12 MHz(10 )2 (10 )22 10
1 1 0.25 mA / V4 10
m
m m
f fC CL
C C
C g RC
Cg gC R
ππ−
−−
= = ⇒ =⎛ ⎞ ⎡ ⎤
⋅⋅⎜ ⎟ ⎢ ⎥×+ ⎣ ⎦⎝ ⎠
=
= ⋅ = ⇒ =×
We have
( )
( )( )
2
3
6
22
20 A / V , 1 V0.25 10So 12.520 10 1
m GS Th
GS Th
k Wg V VL
k V VWL
μ−
−
′⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
′ ≅ − ≅×= =
×
and a value of / 12.5W L = is certainly reasonable. TYU15.7
( )
1
2
1 1
2 2
0.10 10 0.010
TH LRV VR
R RR R
⎛ ⎞= −⎜ ⎟
⎝ ⎠⎛ ⎞
= − − ⇒ =⎜ ⎟⎝ ⎠
Let 1 0.10 kR = Ω then 2 10 kR = Ω TYU15.8 a.
( )
( )
( )
2
1 2
1
1 2
1
1 2
10 21 10
1.82 V
11.82 101 10
2.73 V
11.82 101 10
0.91 V
S REF
S
TH S H
TH
TL S L
TL
RV VR R
V
RV V VR R
V
RV V VR R
V
⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=
⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=
⎛ ⎞ ⎛ ⎞= + = + −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=
b.
TYU15.9
1
2
1 1
2 2
1
and
S REF
TH S L TL S H
RV VR
R RV V V V V VR R
⎛ ⎞= +⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
Hysteresis 1
2
Width ( )TH TL H LRV V V VR
⎛ ⎞= − = −⎜ ⎟
⎝ ⎠
[ ]( )1 1
2 2
2.5 5 5 10R RR R
⎛ ⎞ ⎛ ⎞= − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
So 1
2
0.25RR
=
Then
1
2
1 1 (1 0.25) 0.8 VS REF REF REFRV V V VR
⎛ ⎞= − = + = + ⇒ = −⎜ ⎟
⎝ ⎠
Then ( )( )( )( )
1 0.25 5 0.25 V
1 0.25 5 2.25 VTH TH
TL TL
V V
V V
= − − − ⇒ =
= − − ⇒ = −
TYU15.10
10 0 0
1 2
10 110 20 3
100,3
1010 10 exp 3
X
X
XX
Rv v v vR R
t v
tvr
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
= = −
⎛ ⎞⎛ ⎞= + − − −⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
Output switches when 103Xv =
( )( )
1
1
1
1
4 6 3
10 10 13.33 exp 3
10 3.33 6.67exp 13.33 13.33
13.33exp 26.67
ln (2) (0.693)2(0.693)
12(0.693)
10 0.1 10 1 10 722 Hz Duty cycle 50%
X
X
X
X X
X
X
X X X
tr
tr
tr
t r rT r
fr
r R C f− −
⎛ ⎞= − −⎜ ⎟
⎝ ⎠⎛ ⎞ −− = =⎜ ⎟⎝ ⎠⎛ ⎞
+ = ≅⎜ ⎟⎝ ⎠
= ==
=
= = × = × ⇒ = ⇒ =
TYU15.11
( )( )
( )
1
1 2
4 6 4
4
20 0.33320 40
10 0.01 10 1 10
0.711 / 8 ln 1 10 ln 48.9 s1 1 0.333
X X X
PX
RR R
r R C
V VT r Tγ
β
μβ
− −
−
⎛ ⎞= = =⎜ ⎟+ +⎝ ⎠
= = × = ×
⎡ ⎤++ ⎢ ⎥⎛ ⎞= = × ⇒ =⎢ ⎥⎜ ⎟− −⎝ ⎠ ⎢ ⎥
⎢ ⎥⎣ ⎦
Recovery time
( )
10
1 2
2
2
2 2
20 (8) 2.667 V20 40
0.7 8 2.667 8 exp
8 0.7exp 0.684410.66
1 ln 37.9 s0.685
Y
X
X
X
Rv vR R
tr
tr
t r t μ
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠⎛ ⎞
= + − − −⎜ ⎟⎝ ⎠
⎛ ⎞ −− = =⎜ ⎟⎝ ⎠
⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠
TYU15.12
( )( )
( )
10.693
10.693
Let 0.01 F, 1kHz
A B
A B
fR R C
R RfC
C fμ
=+
+ =
= =
( )( )( )
( )( )( )( )( )
53 6
5
5
5
1 1.443 100.693 10 0.01 10
Duty cycle 55 100%2
1.443 10 10055
1.443 10
1.443 10 100 55118 k so 26.2 k
55
A B
A B
A B
B
B B A
R R
R RR R
R
R R R
−+ = = ×
×
+= = ×+
×=
× +
× −= ⇒ = Ω = Ω
TYU15.13
a. 01 2
1
02 4
3
301 1 2.520
50 2.520
I
I
v Rv Rv Rv R
⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠
= − = − = −
(b) 2 21 1 [12 ( 12)] 240
2 2 1.2L
L
VP mWR
− −= ⋅ = ⋅ =
Or 0.24 P W=
c. 12 4.8 V2.5 piV= =