# Microelectronic Circuit Design Third Edition - Part III Solutions to

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Microelectronic Circuit Design
Third Edition - Part III
Solutions to Exercises
CHAPTER 10
Page 509
Vo 25.3V
Vo = 2Po RL = 2(20W )(16Ω) = 25.3 V          Av =     =       = 5.06x103
Vi 0.005V
Vo 25.3V                   Vi       0.005V                 I     1.58 A
Io =      =     = 1.58 A I i =         =            = 0.167µA Ai = o =         = 9.48x106
RL   16Ω                RS + Rin 10kΩ + 20kΩ                I i 0.167µA
Po   25.3V (1.58 A)
AP =         =
PS 0.005V (0.167µA)
= 4.79x1010                          (   )(        )
Checking : Ap = 5.06x103 9.48x106 = 4.80x1010

Page 510

€                                                   (          )                     (         )
(i ) AvdB = 20 log(5060) = 74.1 dB AidB = 20 log 9.48x106 = 140 dB APdB = 10 log 4.80x1010 = 107 dB
(ii ) AvdB          (      )                        (          )                     (         )
= 20 log 4x10 4 = 92.0 dB AidB = 20 log 2.75x108 = 169 dB APdB = 10 log 1.10x1013 = 130 dB

Page 511

€
(i) The constant slope region spanning a maximum input range is between 0.4 ≤ vI ≤ 0.65,
0.4 + 0.65
and the bias voltage VI should be centered in this range : VI =          V = 0.525 V.
2
vi ≤ 0.65 − 0.525 = 0.125 V and v i ≤ 0.525 − 0.40 = 0.125 V. For vI = 0.8V, the slope is 0. A v = 0.
(ii) vO = VO + vo For vi = 0, v I = VI = 0.6, VO = 14V and Av = +40. Vo = AvVi = 40(0.01V ) = 4V
vO = (14.0 + 4.00sin1000πt ) volts      VO = 14 V

€

1
Page 519
1
g11 =                     = 0.262 µS g21 = 0.262µS (76)(50kΩ) = 0.995
20kΩ + 76(50kΩ)
1    1   75                                          1            1
g22 =         +    +     = 3.82 mS               g12 = −              =−               = -0.0131
50kΩ 20kΩ 20kΩ                                   g22 (20kΩ)    3.82mS (20kΩ)
1                                               1
Rin =         = 3.82 MΩ        A = g21 = 0.995   Rout =       = 262 Ω
g11                                             g22

Page 520
2   RS + Rin
€   (i ) P = A A = A
v    i   v
RL
 50kΩ            8Ω     
(ii ) V o = 2(100W )(8Ω) = 40 V           40 = 0.001             A           → A = 46,800
 5kΩ + 50kΩ   0.5Ω + 8Ω 
2                2
I o RL 0.5Ω  40V                    40V  5kΩ + 50kΩ              8
P=           =           = 6.25 W Ai =                      = 2.75 x 10
2     2  8Ω                      8Ω  0.001V 
 5kΩ   8Ω 
(iii ) 40 = 0.001 5kΩ + 5kΩ  A 8Ω + 8Ω  → A = 160,000
                      
2              2
I o RL 8Ω  40V                              40V  5kΩ + 5kΩ              7
P=       =         = 100 W!              Ai =                  = 5.00 x 10
2    2  8Ω                               8Ω  0.001V 

Page 521
300s
€   Av ( s) =                             Zeros at s = 0 and s = ∞; Poles at s = -5000 and s = −100.
(s + 5000)(s + 100)
Page 523
2π x 106
−400                         5000π
€   Av ( s) = −                =      → Amid = −400 f H =       = 2.50 kHz
s + 5000π   s                          2π
1+
5000π
BW = f H − f L = 2.50 kHz − 0 = 2.50 kHz GBW = (400)(2.50kHz) = 1.00 MHz

€

2
Page 524
52 − 4
(i ) A ( j5) = 50
v
2
= 41.87 20 log(41.87) = 32.4 dB
(5 − 2)
2
+4 5    ( ) 2

 −2(5) 
(
∠Av ( j5) = ∠ 52 − 4 − tan−1 2  )   = 0 − −23.5o = 23.5o                            (        )
5 − 2
       
12 − 4
Av ( j1) = 50                                                        = 67.08 20 log(41.87) = 36.5 dB
2
(1 − 2)
2
+41         ( ) 2

 −2(1) 
(
∠Av ( j1) = ∠ 12 − 4 − tan−1 2

)
1 − 2
 = 180 o − −63.43o = 243o = −117 o

(        )


Page 524
20
€   (ii ) A ( jω ) =
v
0.1ω
1+ j
1− ω 2
20                                                                         0.1(0.95) 
Av ( j0.95) =
2
= 14.3       ∠Av ( j0.95) = ∠20 − tan 
−1
2
 1− 0.95 
(       )
 = 0 − 44.3o = −44.3o

12 +
(0.1) (0.95 )                    2
           
2
(1− 0.95 )                  2

20                                                                       0.1(1) 
Av ( j1) =
2
=0              ∠Av ( j1) = ∠20 − tan−1
 1−1 
2           ( )
 = 0 − 90 o = −90.0 o

12 +
( )
(0.1) 12                                                                        
2
(1−1 )       2

20                                                                             0.1(1.1) 
Av ( j1.1) =
2
= 17.7           ∠Av ( j1.1) = ∠20 − tan−1
 1−1.1 
2       (         )
 = 0 − −27.6 o = 27.6 o
2
1 +
(0.1) (1.1 )                      2
          
2
(1−1.1 )                 2

Page 526
1         1
€    fH =                        = 804 kHz
2π 1kΩ 100kΩ (200 pF )
(                         )
Page 527
250                                                                 250π
€   Av ( s) =                                Ao = 250                            fL =        = 125 Hz      fH = ∞    BW = ∞ −125 = ∞
250π                                                                2π
1+
s

3
€
Page 528
1         1
fL =                       = 15.8 Hz
2π 1kΩ 100kΩ (0.1µF )
(             )
Page 529
−400
€   (i ) A (s) = 
v
100        s 
Ao = 400 or 52 dB
1+      1+        
     s  50000 
100                     50000
fL =      = 15.9 Hz f H =           = 7.96 kHz BW = 7960 −15.9 = 7.94 kHz
2π                        2π
(ii ) ∠Av ( j0) = −90 − 0 − 0 = −90o
 100           100 
∠Av ( j100) = −90 o − tan−1      − tan 
−1
 = −90 − 45 − 0.57 = −136
o

 100           50000 
 50000           
−1 50000

∠Av ( j50000) = −90 o − tan−1          − tan            = −90 − 89.9 − 45 = −225
o

 100              50000 
∠Av ( j∞) = −90 − 90 − 90 = −270 o

Page 531
The numerator coefficient should be 6 x106.
€
2x105 s
Av ( s) = 30                            Ao = 30
s2 + 2x105 s + 1014
1                                   10 7                   1.59 MHz
fo =    1014 = 1.59 MHz               Q=       = 50      BW =             = 31.8 kHz
2π                                  2x105                      50

Page 533
6.4x1012 π 2 s
€   The transfer fucntion should be Av (s) =                                     2
.
(s + 200π )(s + 80000π )
1000
Av ( s) =                    2
Ao = 1000 or 60 dB
 200π    s 
1+    1+     
   s  80000π 
200π                                 80000π 
fL =        = 100 Hz            f H = 0.644         = 25.8 kHz         BW = 25800 −100 = 25.7 kHz
2π                                  2π 

€

4
CHAPTER 11
Page 545
10V                                   10V                                     10V
vid =     = 0.100V = 100 mV        v id =        = 0.001 V = 1.00 mV        vid =        = 1.00x10−5 V = 10.0 µV
100                                   10 4                                    10 6

Page 547
360kΩ                                          0.5V
€   Av = −       = −5.29 vO = −5.29(0.5V ) = −2.65 V iS =      = 7.35 µA iO = −i2 = −iS = −7.35 µA
68kΩ                                          68kΩ

Page 549
2V                                                  24kΩ
€   IS =       = 426 µA        I2 = I S = 426 µA      Av = −         = −5.11     VO = −5.11(2V ) = −10.2 V
4.7kΩ                                                 4.7kΩ

Page 551
36kΩ                                                             −3.80V
€   Av = 1+      = +19.0        vO = 19.0(−0.2V ) = − 3.80 V        iO =              = −100 µA
2kΩ                                                           36kΩ + 2kΩ

Page 552
39kΩ
€   (i )   Av = 1+
1kΩ
= +40.0   AvdB = 20log(40.0) = 32.0 dB           Rin = 100kΩ ∞ = 100kΩ

10.0V
vO = 40.0(0.25V ) = 10.0 V      iO =          = 250 µA
39kΩ + 1kΩ
54
R          R2               vO                    10
(ii ) Av = 1020 = 501     1+ 2 = 501
R1         R1
= 500 iO =
R2 + R1               R2 + R1
≤ 0.1 mA

R1 + R2 ≥ 100kΩ        501R1 ≥ 100kΩ → R1 ≥ 200 Ω            There are many possibilities.
( R1 = 200 Ω, R2 = 100 kΩ), but ( R1 = 220 Ω, R2 = 110 kΩ) is a better solution since
resistor tolerances could cause iO to exceed 0.1 mA in the first case.

Page 554
30kΩ
€   Inverting Amplifier : Av = −      = −20.0 Rin = R1 = 1.5 kΩ
1.5kΩ
v    −3.00V
vO = −20.0(0.15V ) = −3.00 V iO = O =           = −100 µA
R2    30kΩ
30kΩ                         v S 0.15V
Non - Inverting Amplifier : Av = 1+           = +21.0        Rin =      =      =∞
1.5kΩ                        iS   0A
vO       3.15V
vO = 21.0(0.15V ) = 3.15 V      iO =          =             = 100 µA
R2 + R1 30kΩ + 1.5kΩ

€

5
Page 555
 3kΩ                       3kΩ 
Vo1 = 2V  −    = −6V      Vo2 = 4V  −    = −6V vO = (−6sin1000πt − 6sin 2000πt ) V
 1kΩ                       2kΩ 
v1                         v2
The summing junction is a virtual ground : Rin1 =      = R1 = 1 kΩ    Rin2 =      = R2 = 2 kΩ
i1                         i2
Vo1 −6V              V     −6V
I o1 =      =    = −2mA I o2 = o2 =     = −2mA iO = (−2sin1000πt − 2sin 2000πt ) mA
R3 3kΩ                R3 3kΩ

Page 559
3V
€   (i )   I2 =
10kΩ + 100kΩ
= 27.3 µA

100kΩ                                               VO − V− VO − V+
(ii ) Av = − 10kΩ = −10.0 VO = −10(3V − 5V ) = 20.0 V IO = 100kΩ = 100kΩ
R4         100kΩ                     20.0 − 4.545                    5V
V+ = V2          =5               = 4.545V I O =              = 155 µA I2 =              = 45.5 µA
R3 + R4    10kΩ + 100kΩ                   100kΩ                    10kΩ + 100kΩ
36kΩ                                                 V − V V − V+
(iii )   Av = −
2kΩ
= −18.0 VO = −18(8V − 8.25V ) = 4.50 V IO = O − = O
36kΩ  36kΩ
R2            36kΩ                    4.50 − 7.816
V+ = V2         = 8.25            = 7.816V I O =              = −92.1 µA
R1 + R2        2kΩ + 36kΩ                   36kΩ

Page 560
V −V  5.001V − 4.999V
€   I= A B =                 = 1.00 µA
2R1       2kΩ
VA = V1 + IR2 = 5.001V + 1.00µA(49kΩ) = 5.05 V
VB = V2 − IR2 = 4.999V −1.00µA(49kΩ) = 4.95 V
 R                10kΩ 
VO = − 4 (VA − VB ) =  −    (5.05 − 4.95) = −0.100 V
 R3               10kΩ 

Page 564
26
R
€   (i ) Av = − R2 = −1020 = −20.0    R1 = Rin = 10kΩ   R2 = 20R1 = 200kΩ
1

1
C=                        = 265 pF Closest values : R1 = 10kΩ        R2 = 200kΩ C = 270 pF
2π (3kHz)(200kΩ)

€

6
Page 564
I                         5V  1 
(ii )   Rin = R1 = 10 kΩ                  ΔV = −
C
ΔT             C=               (1ms ) = 0.05µF
10kΩ 10V 
vO
2                           4                   6                 8        t (msec)
€

-10V

Page 567
dv S
vO = −RC              = −(20kΩ)(0.02µF )(2.50V )(2000π )(cos2000πt ) = −6.28cos2000πt V
dt

Page 569
R2    68kΩ
€   (i ) A  vA   = AvB = AvC = −
R1
=−
2.7kΩ
= −25.2                       RinA = RinB = RinC = R1 = 2.7 kΩ

The op - amps are ideal : RoutA = RoutB = RoutC = 0
3
(ii ) A  v   = AvA AvB AvC = (−25.2) = −16,000                            Rin = RinA = 2.7 kΩ     Rout = RoutC = 0

Page 570
 2.7kΩ 2
3                    3                                   2.7kΩ 2
€   Av = (−25.2)                ≥ 0.99(25.2)                                                   ≥ 0.99
 Rout + 2.7kΩ                                                   Rout + 2.7kΩ 
2.7kΩ
≥ 0.9950      Rout ≥ 13.6 Ω
Rout + 2.7kΩ

Page 574
ω2                                           ω2
€   (i) Av (0) = +1               Av ( s) =             o
Av ( jω ) =           o

s2 + s 2ω o + ω 2
o                              jω 2ω o + ω 2 − ω 2
o

1                            ω2                          1        4     4     4
Av ( jω H ) =                 →                     o
=        → 2ω o = ω o + ω H → ω o = ω H
2
2                                                    2
(ω   2
o   − ω2
H   )       + 2ω 2 ω 2
H o

1        C1 R2                           1
(ii )        2
=
C2 2R
→ C1 = 2C2 → C2 =
2 (2.26kΩ)(20000π )
= 4.98nF

C2 = 0.005 µF                 C1 = 0.01 µF

€

7
Page 574
(iii ) To decrease the cutoff frequency from 5kHz to 2 kHz, we must increase the
5kHz
resistances by a factor of              = 2.50 → R1 = R2 = 2.50(2.26kΩ) = 5.65 kΩ
2kHz
1C R1 R2
(iv )           =    → R12 + 2R1 R2 + R22 = 2R1 R2 → R12 = −R22 - - can' t be done!
2     C R1 + R2
R1 R2   dQ        1       R (R + R )                                    1
Q=               =          2
 1 1      2
− R1 R2  = 0 → R2 = R1 → Qmax =
R1 + R2   dR2 ( R + R )  2 R1 R2
                     
                          2
1    2

Page 575
Q        C1 dQ C1  1      R1 R2  C1 Q
€   SC1 =          =                 =      = 0.5
Q dC1 Q 2 C1C2 R1 + R2  Q 2C1
              
Q         R2 dQ                         1 C1    Q
SR 2 =                 R1 = R2 → Q =          → SR 2 = 0
Q dR2                         2 C2

Page 577
−ω 2              K                         K
€   (i )     Av ( jω o ) = K      2
o
2     2
=          Av ( jω o ) =        ∠90 o
−ω o + j (3 − K )ω o + ω o 3 − K                      3− K
1
(ii )    fo =                                                 = 5.19 kHz
2π 10kΩ(20kΩ)(0.0047µF )(0.001µF )
−1
                                        
 10kΩ 4.7nF + 1.0nF + 1− 2 20kΩ(1.0nF )  = 0.829
Q=
 20kΩ 4.7nF 1.0nF    ( ) 10kΩ 4.7nF 
            (      )           (      )
Q         K dQ     1            dQ    −1                                 K dQ
(iii )   SK =            Q=                  =        2 (
−1) = Q2         Q
SK =          = KQ
Q dK    3− K          dK (3 − K )                              Q dK

1                 Q
Q=            → KQ = 3Q −1 SK = 3Q −1 = 1.12
3− K

Page 578
1                                            1 82kΩ
€   Rth = 2kΩ 2kΩ = 1kΩ f o =                                                        = 879 Hz     Q=          = 4.53
2π 1kΩ(82kΩ)(0.02µF )(0.02µF )                                2 1kΩ

€

8
Page 582
R2                  294kΩ
(ii )    ABP ( jω o ) = KQ =
R1
10 =
R1
→ R1 = 29.4 kΩ

1          1                                               1          1
(iii )   fo =       =
2πRC 2π (40.2kΩ)(2nF )
= 1.98 kHz                   BW =        =
2πR2C 2π (402kΩ)(2nF )
= 198 Hz

R2    402kΩ
ABP ( jω o ) = −       =−        = −20.1
R1    20.0kΩ
(iv )    Blindly using the equations at the top of page 580 yields
1                 1
f omin =          =                                = 1946 Hz
2πRC 2π (1.01)(29.4kΩ)(1.02)(2.7nF )
1                 1
f omax =          =                                = 2067 Hz
2πRC 2π (0.99)(29.4kΩ)(0.98)(2.7nF )
1                 1
BW min =             =                               = 195 Hz
2πR2C 2π (1.01)(294kΩ)(1.02)(2.7nF )
1                 1
BW max =             =                               = 207 Hz
2πR2C 2π (0.99)(294kΩ)(0.98)(2.7nF )

min           R2     294kΩ(1.01)                        max       R2    294kΩ(0.99)
ABP = −           =−              = −20.4               ABP = −       =−              = −19.6
R1    14.7kΩ(0.99)                                  R1    14.7kΩ(1.01)
1
The W/C results are similar if R and C are not the same for example where ωo =                                   .
RA RB CACB

Page 583
0.02µF                     0.01µF
€   (i ) - (a )    R1 = R2 = 5(2.26kΩ) = 11.3 kΩ C1 =
5
= 0.004 µF   C2 =
5
= 0.002 µF

1
fo =                                                               = 4980 Hz
2π     (11.3kΩ)(11.3kΩ)(0.004µF )(0.002µF )
11.3kΩ (0.004µF )(0.002µF )
Q=                                    = 0.471
11.3kΩ 0.004µF + 0.002µF
0.02µF                  0.01µF
(b)      R1 = R2 = 0.885(2.26kΩ) = 2.00 kΩ               C1 =
0.885
= 0.0226 µF C2 =
0.885
= 0.0113 µF

1
fo =                                                                 = 4980 Hz
2π     (2.00kΩ)(2.00kΩ)(0.0226µF )(0.0113µF )
2.00kΩ (0.0226µF )(0.0113µF )
Q=                                       = 0.471
2.00kΩ 0.0226µF + 0.0113µF

€                                                                                                                        9
Page 583
1                                     82kΩ (0.02µF )(0.02µF )
(ii )   fo =                                        = 879 Hz    Q=
1kΩ 0.02µF + 0.02µF
= 4.53
2π   (1kΩ)(82kΩ)(0.02µF )(0.02µF )
0.02µF
The values of the resistors are unchanged. C1 = C2 =             = 0.005 µF
4
1                                         82kΩ (0.005µF )(0.005µF )
fo =                                             = 3520 Hz Q =                                   = 4.53
2π     (1kΩ)(82kΩ)(0.005µF )(0.005µF )                          1kΩ 0.005µF + 0.005µF

Page 585
€   The diode will conduct and pull the output up to vO = v S = 1.0 V. v1 = vO + vD = 1.0 + 0.6 = 1.6 V
For a negative input, there is no path for current through R, so vO = 0 V. The op - amp
sees a -1V input so the output will limit at the negative power supply : vO = −10 V.
The diode has a 10 - V reverse bias across it, so VZ > 10 V.

Page 587

€
(i ) vS = 2 V : Diode D1 conducts, and D2 is off. The negative input is a virtual ground.
v1 = −v D2 = −0.6 V.             The current in R is 0, so vO = 0 V.
vS = −2 V : Diode D2 conducts, and D1 is off. The negative input is a virtual ground.
R2        68kΩ
vO = −
R1
vS = −
22kΩ
(−2V ) = +6.18 V          v1 = vO + vD1 = 6.78 V.

15V
vS =        = −4.85 V v1 = vO + v D1 = 6.78 V. When vO = 15 V , v D2 = -15.6 V , so VZ = 15.6 V.
−3.09
        
(ii ) vO = 20kΩ  10.2kΩ  2V = 2.00 V
20kΩ  3.24kΩ  π

Page 589
R1              1kΩ                                                   1kΩ
€   V− = −         VEE = −             10V = −0.990 V                     V+ =               10V = +0.990 V
R1 + R2         1kΩ + 9.1kΩ                                           1kΩ + 9.1kΩ
Vn = 0.990V − (−0.990V ) = 1.98 V

Page 591
 1+ 0.5                                    1
€   T = 2(10kΩ)(0.001µF ) ln         = 21.97µs                   f =     = 45.5 kHz
 1− 0.5                                    T

€

10
Page 594
    0.7 
22kΩ                                      1+      
β =−               = 0.550 T = (11kΩ)(0.002µF ) ln      5  = 20.4 µs
22kΩ + 18kΩ                                 1− 0.550 
         
        5V  
1+ 0.55  
 5V  
Tr = (11kΩ)(0.002µF ) ln                = 13.0 µs
      0.7 
 1− 5         
              

€

11
CHAPTER 12
Page 612
1                    A       105
(i ) Aideal = β = 100       Av =      =
1+ Aβ 1+ 105 (0.01)
= 99.90

vo 9.99V
vo = 99.9(0.1V ) = 9.99 V         vid =
=         = 99.9 µV
A     105
(ii ) Values taken from OP - 27 specification sheet (www.jaegerblalock.com or www.analog.com)
(iii ) Values taken from OP - 27 specification sheet

Page 613
€                                    −R   −R Aβ 
2         2
       −        
−R2 Aβ                    R1   R1 1+ Aβ        Aβ     1
Av =                      FGE =                      = 1−      =
R1 1+ Aβ                          −R 
2
1+ Aβ 1+ Aβ
     
 R1 

Page 614
1                      1                                                  1
€   FGE =           =                             = 3.98x10−3 or 0.398 %      FGE ≅         = 0.40 %
1+ Aβ                     1kΩ                                            Aβ
1+ 10 4            
 1kΩ + 39kΩ 

Page 615

€
Values taken from OP - 77 specification sheet (www.jaegerblalock.com or www.analog.com)

Page 616
R         50Ω
€   A≅ o =                  = 20,000
βRout (0.025)(0.1Ω)

Page 617

€                   A        10 4
(i )   Av =        =
1+ Aβ 1+ 10 4 (0.025)
= +39.8

max       (39kΩ + 1kΩ)(1.05) = 44.2    Avmax =
(39kΩ + 1kΩ)(0.95) = 36.2
(ii ) A v     =
1kΩ(0.95)                              1kΩ(1.05)
4.20                                                  −3.80
GE = 44.2 − 40.0 = 4.20         FGE =          = 10.5 %     GE = 36.2 − 40.0 = −3.80    FGE =         = −9.5 %
40                                                    40

€

12
Page 618
R       200Ω
A≅ o =
βRout (0.01)(0.1Ω)
= 200,000                      (        )
AdB = 20 log 2x105 = 106 dB

Page 619
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)
€
Page 620
           10kΩ      
€   Rin = Rid (1+ Aβ ) = 1MΩ 1+ 10 4                = 251 MΩ
        10kΩ + 390kΩ 
v         1V                      βv     Aβ  v s   1V 
i− = − s = −           = −3.98 nA i1 = o =             ≅      = 100 µA i1 >> i−
Rin     251 MΩ                      R1 1+ Aβ  R1  10kΩ 
βvo   Aβ  v s   10 4 (0.025)  1V 
More exactly, i1 =     =      =                          = 99.6 µA
R1 1+ Aβ  R1  1+ 10 4 (0.025)  10kΩ 

Page 621
€                     R2              100kΩ                     ideal
Rin = R1 + Rid        = 1kΩ + 1MΩ        = 1001 Ω         Rin = R1 = 1000 Ω 1 Ω or 0.1 %
1+ A             1+ 105

Page 622
€   Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 626
        v ic 
€   vo = A vid +          
       CMRR 
min
        v ic                   5.000 
vo = A vid +             = 25000.002 −         = 3.750 V
     CMRR                       10 4 
max
         v ic                  5.000 
vo = A v id +            = 2500 0.002 +        = 6.250 V
     CMRR                       10 4 

Page 627
€               1                1                                            1 
A1+           10 41+     4
103 1+     3
 2CMRR           2x10                                         2x10 
Av =                =                 = 1.000                  Av =                  = 1.000
   1           4
   1                                       3
   1 
1+ A1−       1+ 10 1−      4
1+ 10 1−     3
 2CMRR            2x10                                         2x10 

€

13
Page 628
1   1
GE = FGE ( Av ) ≤ 5x10−5 (1) = 5x10−5            Worst case occurs for negative CMRR : GE ≅       +
A CMRR
1
If both terms make equal contributions: A = CMRR =                          = 4x10 4 or 92 dB
2.5x10 −5

−1                                            −1
        1                                         1 
For other cases : CMRR =  5x10 − 
−5
or   A = 5x10 −
−5

        A                                      CMRR 
−1
         1 
A = 100dB           CMRR =  5x10 − 5  = 2.5x10 4 or 88 dB
−5

        10 
−1
         1 
CMRR = 100dB                A =  5x10 − 5  = 2.5x10 4 or 88 dB
−5

        10 

Page 630
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)
€
Page 631

€   (i ) V ≤ 50(0.002V ) → −0.100 V ≤ V ≤ +0.100 V
O                                   O

(ii ) Values taken from op - amp specification sheets (www.jaegerblalock.com           or www.analog.com)

Page 633
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)
€
Page 634

€   (i )    R = 39kΩ 1kΩ = 975 Ω
VOS    I                    1.5mV        100nA
(ii )   v O (t ) = VOS +
RC
t + B2 t
C
1.5mV +                t+
10kΩ(100 pF ) 100 pF
t = 15V → t = 6.00 ms

Page 636

€   Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 637
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)
€

€

14
Page 638
−1
20V                1    1 
REQ     = RL ( R2 + R1 ) ≥         R1 + R2 ≥ 
= 4kΩ              −     = 20kΩ
5mA                4kΩ 5kΩ 
Including 5% tolerances, R1 + R2 ≥ 21kΩ A v = 10 → R2 = 9R1
A few possibilities : 27 kΩ and 3 kΩ, 270 kΩ and 30 kΩ, 180 kΩ and 20 kΩ, etc.

Page 640

€
(i )    Ao = 10
100
20
= 10   5
ωB =    =
(
ωT 2π 5x10
6

=
)
10 7 π
= 100π f B =
100π
= 50 Hz Av (s) =
10 7 π
Ao    105       105                 2π                   s + 100π
ωT   2π x 106  2π x 106
(ii ) Av (s) = ω = 2π x 106 = s + 10π
s+ T s+
Ao   2x105

Page 643

€
(
90

i ) Ao = 10 = 31600 f B =
20       f T 5x106
=      = 158 Hz Av (s) =
2π 5x106
=
10 7 π                    (           )
Ao 31600                    s + 2π (158) s + 316π

f H ≅ βfT = 0.01(5MHz) = 50 kHz                    Av ( s) =
(
2π 5x106    )           =
10 7 π
(
s + 2π 5x10 4        )       s + 105 π
ωT      βω T 1
(ii )   For ω H >> ω B : Aβ =
s + ωB
β≅     = = − j1 since ω H = βωT
jω H j

Page 645

€   (
90
f
i ) Ao = 10 = 31600 f B = T =
20           5x106
= 158 Hz                                Av (s) =
(
2π 5x106         )   =
10 7 π
Ao 31600                                                            s + 2π (158)             s + 316π

f H ≅ βfT =
5MHz
= 15.8 kHz   Av ( s) =
( ) = 10 π
2π 5x106                               7

50                                                                                    4
s + 2π (15.8x10 ) s + 3.16x10 π
3
1+ 10   20

1
(ii )   f H ≅ βfT = 1(5MHz) = 5 MHz                   f H ≅ βfT =
2
(5MHz) = 2.5 MHz

€

15
Page 647
1250
Av (0) = 50(25) = 1250     Av (ω H ) =           = 884
2
                          
1+    ω2       1+ ω2       = 2 → ω2            2


H
2
H
2          H     ( )        + 4.935x109ω 2 − 3.896x1018 = 0
H


(10000π )   (20000π ) 
26.3x103
ω 2 = 6.925x108 → ω H = 26.3x103 → f H =
H                                                        = 4190 Hz
2π

Page 648
5                    −3.34x105
€   (i )   Av (0) = −100(66.7)(50) = −3.34x10          Av (ω H ) =                = −2.36x105
2
                                    
1+  ω2H      1+ ω2H      1+ ω2H     =2
2             2             2
                                    
 (10000π )   (15000π )   (20000π ) 
4
ω 6 + 7.156x109 ω H + 1.486x1010 ω 2 − 8.562x1027 = 0
H                                H

21.7x103
Using MATLAB, ω H = 21.7x103 → f H =                       = 3450 Hz
2π
1
3
(ii ) Av (0) = (−30) = −2.70x104         f H = (33.3kHz) 2 3 −1 = 17.0 kHz

Page 653
SR 5x105V / s                             SR     5x105 V / s
€   VM ≤     =         = 3.98 V              fM =       =             = 7.96 kHz
ω 2π (20kHz)                            2πVFS    2π (10V )

Page 657
130kΩ                                                                              2
€   Av1 = 1+       = 6.909      vO1 = 0.001(6.909) = 6.91 mV              vO2 = 0.001V (6.909) = 47.7 mV
22kΩ
3                                         4                                 5
vO3 = 0.001(6.909) = 330 mV        vO 4 = 0.001V (6.909) = 2.28 V vO5 = 0.001V (6.909) = 15.7 V
vO5 > 15V → vO5 = 15 V      vO6 = 15V (6.909) = 104 V             vO6 > 15V → vO6 = 15 V

€

16
CHAPTER 13
Page 671
IC 1.5mA                           IC         1.5mA
(i )    (a) At the Q - point : β F =      =
I B 15µA
= 100      (b) I S =
 VBE 
=
 0.700V 
= 1.04 fA
exp      exp        
 VT       0.025V 
v be 8mV
(c) Rin =        =     = 1.6 kΩ          (d) With the given applied signal, the smallest value of vCE is
ib   5µA
min
v CE = 5V − 0.5mA(3.3kΩ) = 3.35 V which exceeds v BE = 0.708 V.

(ii )   (a) v DS = 10 − 3300iDS
vds 2.7 - 6.7 V
(b) Using the peak - to - peak voltage swings, Av =             =            = −4.0.
vgs 4.0 - 3.0 V
Note that there is some distortion in this amplifier since the negative output voltage excursion is
larger than the positive output change.
(c) v min = 2.7V with vGS − VTN = 4 −1 = 3V , so the transistor has entered the triode region.
DS

(d) Choose two points on the i - v characteristics. For example,
Kn
(3.5 − VTN ) and 1.0mA = Kn (3.0 − VTN ) .
2                           2
1.56mA =
2                           2
µA
Solving for Kn and VTN yields 500 2 and 1 V respectively.
V

Page 673
10kΩ
€   (i )    VEQ =
10kΩ + 30kΩ
12V = 3.00V REQ = 10kΩ 30kΩ = 7.5kΩ

VEQ − VBE                 3.0V − 0.7V
IC = β F I B = β F                    = 100                      = 1.45 mA
REQ + (β F + 1) R4       7.5kΩ + (101)(1.5kΩ)
 101 
VCE = 12 − 4300IC −1500I E = 12 − 4300(1.45mA) −1500     (1.45mA) = 3.57 V
 100 
(ii ) v (t ) = V
C        C   + vC = (5.8 −1.1sin 2000πt ) V vE (t ) = VE + 0 = 1.45mA(1.5kΩ) = 2.2 V
1.1V
ic =         = 0.256mA ∠ic = 180 o ic (t ) = −0.26sin 2000πt mA
4.3kΩ
1         1
(iii ) X C = ωC = 2000π 500µF = 0.318 Ω
(     )
Page 676
RB = 20kΩ 62kΩ = 15.1 kΩ               RL = 8.2kΩ 100kΩ = 7.58 kΩ
€

€                                                                                                               17
Page 680
VT             0.025V                        0.025V
(i )    rd =
ID + IS
rd =
1 fA
= 25.0 TΩ       rd =
50µA
= 500 Ω

0.025V               0.025V
rd == 12.5 Ω rd =        = 8.33 mΩ
2mA                   3A
0.025V           kT            V                      0.0322V
(ii) rd = 1.5mA = 16.7 Ω q = 8.62x10−5 K (373K ) = 0.0322 V rd = 1.5mA = 21.5 Ω
           

Page 685
βo   75
€   (i )    g m = 40IC = 40(50µA) = 2.00 mS         rπ =      =
g m 2mS
= 37.5 kΩ

VA + VCE 60V + 5V
ro =             =         = 1.30 MΩ            µ f = g mro = 2mS (1.30 MΩ) = 2600
IC      50µA
βo   50
(ii )    g m = 40IC = 40(250µA) = 10.0 mS        rπ =       =
g m 10mS
= 5.00 kΩ

VA + VCE 75V + 15V
ro =             =          = 360 kΩ           µ f = g mro = 10mS (360kΩ) = 3600
IC      250µA
(iii )   The slope of the output characteristics is zero, so VA = ∞ and ro = ∞.
βF          I   1.5mA                             ΔiC    0.5mA
β FO =           = βF = C =       = 100            gm =            =       = 62.5 mS
VCE        I B 15µA                              ΔvBE    8mV
1+
VA
ΔiC 500µA                        βo    100
βo =         =     = 100           rπ =      =       = 1.60 kΩ
ΔiB   5µA                        g m 62.5mS

Page 692
Avt = −g m RL = −9.80mS (18kΩ) = −176
€

€

18
Page 695
Assume the Q - point remains constant.
max                min                max              min
(a ) R   L      = 19.8kΩ   RL = 16.2kΩ        RE = 3.30kΩ      RL = 2.70kΩ
max                                             min
RiB = 10.2kΩ + 101(3.3kΩ) = 344kΩ               RiB = 10.2kΩ + 101(2.7kΩ) = 283kΩ
9.80mS (16.2kΩ)  104kΩ 344kΩ 
Avmin = −                                         = −4.70
1+ 9.80mS (3.30kΩ) 1kΩ + 104kΩ 344kΩ 
                   
9.80mS (19.8kΩ)  104kΩ 283kΩ 
Avmax    =−                                        = −6.98
1+ 9.80mS (2.70kΩ)  1kΩ + 104kΩ 283kΩ 
                   
125
( b) r  π   =
9.80mS
= 12.8kΩ RiB = 12.8kΩ + 126(3.0kΩ) = 391kΩ

9.80mS (18kΩ)  104kΩ 391kΩ 
Av = −                                        = −5.73
1+ 9.80mS (3.00kΩ)  1kΩ + 104kΩ 391kΩ 
                   
       101    
(c) VCE = 12V − 22kΩIC −13kΩI E = 12V − 0.275mA22kΩ + 100 13kΩ = 2.34 V
              
100
g m = 40(0.275mA) = 11.0mS rπ =        = 9.09kΩ
11.0mS
11.0mS (18kΩ)  104kΩ 312kΩ 
RiB = 9.09kΩ + 101(3.0kΩ) = 312kΩ Av = −                                        = −5.75
1+ 11.0mS (3.00kΩ)  1kΩ + 104kΩ 312kΩ 
                   

Page 697
max

€
(i ) RiB = 10.2kΩ + 101(1kΩ) = 111kΩ            RE 2 = 13kΩ −1kΩ = 12 kΩ
9.80mS (18kΩ)   104kΩ 111kΩ 
Av = −                                             = −16.0
1+ 9.80mS (1.00kΩ)  1kΩ + 104kΩ 111kΩ 
                   
(ii ) The reference to C    2   should be C3 . Av = −159 as calculated in the Ex. 13.4.
 1.38x10−23 V                                              IC           245µA
(iii ) VT =               (273K + 27K ) = 25.84mV            IS =              =               = 0.421 fA
 1.602x10 K                                                 VBE      0.700V 
−19
exp      exp          
 VT       0.02584V 
(iv ) From Ex. 13.3, R     iB   = 313 kΩ, and rπ = 10.2kΩ.

€

19
Page 700
          100(2kΩ)         
(i ) RiC = 320kΩ 1+                          = 4.85 MΩ                            µ f RE = 3140(2kΩ) = 6.28 MΩ

           (
1kΩ 104kΩ + 10.2kΩ + 2kΩ
)
RiC < µ f RE         Rout = 4.85MΩ 22kΩ = 21.9 kΩ
     β o RE           β R 
(ii )   lim RiC = lim ro 1+
R E →∞ R E →∞
 = ro 1+ o E  = (β o + 1)ro
 Rth + rπ + RE          RE 

Page 704

€   (i ) (a)                                          (             )           [
g m = 2Kn I D (1+ λVDS ) = 2 1mA /V 2 (0.25mA) 1+ 0.02(5) = 1.73 mS    ]
1
+ VDS 50V + 5V
ro = λ      =         = 220 kΩ                        µ f = g mro = 0.742mS (220kΩ) = 163
ID     250µA

(             )        [
g m = 2Kn I D (1+ λVDS ) = 2 1mA/V 2 (5mA) 1+ 0.02(10) = 3.46 mS                ]
1
+ VDS 50V + 10V
ro = λ      =          = 12 kΩ                        µ f = g mro = 3.46mS (12kΩ) = 41.5
ID      5mA
( b)    The slope of the output characteristics is zero, so λ = 0 and ro = ∞.
2.1V
ΔiD
For the positive change in vgs , g m =                     ≅ 3.3kΩ = 1.3 mS
ΔvGS        0.5V

2I D        2(25mA)
(ii ) v   gs   ≤ 0.2(VGS − VTN ) = 0.2
Kn
= 0.2
2.0mA/V 2
= 1.00 V             v be ≤ 0.005 V

Page 705
γ               0.75                               0.75
€   η=                       =                  = 0.48        η=               = 0.20
2 VSB + 2φ F         2 0 + 0.6                         2 3 + 0.6

€

20
Page 713
1.5MΩ
(i ) V
EQ    =
1.5MΩ + 2.2 MΩ
12V = 4.87V REQ = 1.5MΩ 2.2 MΩ = 892kΩ

Neglect λ in hand calculations of the Q - point.
 5x10−4          2
4.87 = VGS + 12000I D 4.87 = VGS + 12000           (VGS −1)
 2 
2
3VGS − 5VGS −1.87 = 0 → VGS = 1.981V I D = 241 µA
VDS = 12 − 22000I D −12000I D = 3.81 V                Q - po int : (241 µA, 3.81 V )

(ii ) A  vdB   = 20 log(4.50) = 13.1 dB

Page 714
VDS ≥ VGS − VTH = 1.981V −1V = 0.981 V
€
Page 717
2(241µA)                   CS       12V                   Kn2 2x10−3
€   VGS − VTN ≅                    = 0.491V    A   v    ≅−        = −24.4      M=    =       =4
2x10−3                            0.491V                 Kn1 5x10−4

Page 720
CS                                                    680kΩ
€   Rin = 680kΩ 1.0 MΩ = 405 kΩ               VEQ =                 12V = 4.86 V
680kΩ + 1MΩ

Page 725
 9.57kΩ 
€   The gain is proportional to RL : RL = 10kΩ 220kΩ = 9.57kΩ                   AvCE = −36.1          = −36.8
 9.540kΩ 

Page 726

€
(i ) µ f = gmro = 9.56mS (225kΩ) = 2150
 RCE         1                     0.33kΩ + 14.2kΩ 
(ii ) vbe = vi R +inRCE 1+ g R 
 I     in     m E
vi ≤ 5mV 
     14.2kΩ
 1+ 9.56mS (0.15kΩ) = 12.4 mV

[                   ]
vo ≤ 12.4mV (36.1) = 0.450 V
CE                            CE
(iv ) R   in   = 100kΩ 6.8kΩ = 6.37 kΩ Rout = 10kΩ 225kΩ = 9.57 kΩ
 RCE                                    6.37kΩ      
AvCE = −g m RL     in
CE
 RI + Rin 
(
 = -9.56mS 9.57kΩ 220kΩ            )      = −83.4
 0.33kΩ + 6.37kΩ 

€

21
Page 730
(i) The gain is proportional to RL.
CS
 25.9kΩ 
RL = 300kΩ Rout = 300kΩ 28.4kΩ = 25.9kΩ      AvCS = −6.85         = −6.50
 27.3kΩ 
The corrected gain agrees more closely with thevalue from SPICE.
(ii ) µ   f   = g mro = 0.515mS (258kΩ) = 133
 RCS           1     
(iii ) v gs = v i     in
CS            v gs ≤ 0.2(VGS − VTN )
 RI + Rin  1+ g m RS 
 10kΩ + 1MΩ 
vi ≤ 0.2(1V )
     1MΩ
[                    ]
 1+ 0.515mS (2kΩ) = 410 mV vo ≤ 410mV (6.85) = 2.81 V


CS
 RCS  g R                   1MΩ        0.515mS 30kΩ 300kΩ 
(             )
(iv ) Av = − R + RCS 1+ g R  = −10kΩ + 1MΩ 1+ 0.515mS 32kΩ  = −0.796
in         m L

 I       in      m S                  
            ( )    
CS
The calculation is slightly larger than the SPICE value since it neglects Rout .

Page 734
239µA
€   (i )   PD = ICVCE + I BVBE = 239µA(3.67V ) +
65
(0.7V ) = 0.880 mW
   1
PS = ICVCC + I EVEE   = 239µA(5V ) + 239µA1+ (5V ) = 2.41 mW
 65 
(i )   PD = I DVDS = 250µA(4.5V ) = 1.13 mW
PS = I DVDD + I S VSS = 250µA(10V ) + 250µA(10V ) = 5.00 mW

Page 735

€
[                ]       [                          ]
(a) VM ≤ min IC RC , (VCE − VBE ) = min 239µA(10kΩ), (3.67 − 0.7)V = 2.39 V
Limited by the voltage drop across RC .
( b) V   M          [                    ]       [                       ]
≤ min I D RD , (VDS − VDSSAT ) = min 250µA(30kΩ), (4.50 −1)V = 3.50 V
Limited by the value of VDS .

€

22
CHAPTER 14
Page 754
RB = 160kΩ 300kΩ = 104 kΩ           RE = 3.00 kΩ    RL = 100kΩ 22kΩ = 18.0 kΩ
RG = 1.5MΩ 2.2 MΩ = 892 kΩ           RS = 2.00 kΩ   RL = 100kΩ 22kΩ = 18.0 kΩ

Page 761
CE       0.838V
€   (i ) A   v    =−
0.150
= −5.59

(ii ) R  iB   = 10.2kΩ + 101(1.0kΩ) = 111kΩ
9.80mS (18kΩ)  104kΩ 111kΩ 
AvCE = −                                          = −15.8   R4 = 13kΩ −1kΩ = 12 kΩ
1+ 9.80mS (1kΩ)  2kΩ + 104kΩ 111kΩ 
                   
0.491mS (18kΩ)  892kΩ 
AvCS = −                                     = −5.91 R4 = 12kΩ −1kΩ = 11 kΩ
1+ 0.491mS (1kΩ)  2kΩ + 892kΩ 
(iii )   RiB = 10.2kΩ
 104kΩ 10.2kΩ 
CE
A = −9.80mS (18kΩ)
v

 2kΩ + 104kΩ 10.2kΩ  = −145
                     
 892kΩ 
AvCS = −0.491mS (18kΩ)                   = −8.82
 2kΩ + 892kΩ 
 1.38x10−23 V                                      IC           245µA
(iv ) VT =                  (273K + 27K ) = 25.84mV I S =             =                = 0.421 fA
 1.602x10 K                                        VBE       0.700V 
−19
exp      exp           
 VT        0.02584V 
18kΩ
(v ) gm RL = -9.80mS(18kΩ) = −176, AvCE ≅ − 3kΩ = −9.00
18kΩ
g m RL = -0.491mS(18kΩ) = −8.84, AvCE ≅ −       = −9.00
2kΩ
(vi ) The exercise should refer to Fig. 14.2. RiB = 10.2kΩ + 101(3.0kΩ) = 313kΩ
CE                                              CS                   CE
Rin = RB RiB = 104kΩ 313kΩ = 78.1 kΩ            Rin = RG = 892kΩ     Rin > rπ

€

23
Page 761
2kΩ + 892kΩ                              (0.491ms) R       L
(i ) Avt = 892kΩ 0.956 = 0.958                    1+ (0.491ms ) R
= 0.958 → RL = 46.5kΩ
L

R6 100kΩ = 46.5kΩ → R6 = 86.9kΩ
CC
(ii )    RiB = 10.2kΩ + 101(13kΩ) = 1.32 MΩ          Rin = 104kΩ 1.32 MΩ = 96.4kΩ
9.80mS (13kΩ)  96.4kΩ 
AvCE = −                                   = +0.972
1+ 9.80mS (13kΩ)  2kΩ + 96.4kΩ 
0.491mS (12kΩ)  892kΩ 
AvCS = −                                   = +0.853
1+ 0.491mS (12kΩ)  2kΩ + 892kΩ 

(iv )    BJT : g m RL = 9.80mS (11.5kΩ) = 113 FET : g m RL = 0.491mS (10.7kΩ) = 5.25

Page 767
RI + RB RiB
€   vi ≤ 0.005V (1+ g m RL )
RB RiB
[
= 0.005V 1+ 9.8mS (11.5kΩ)                ] 2kΩ + 95.4kΩ = 0.580 V
95.4kΩ
RI + RG                                  2kΩ + 892kΩ
vi ≤ 0.2(VGS − VTN )(1+ g m RL )
RG
[
= 0.2(0.982) 1+ 0.491mS (10.7kΩ)
892kΩ
]
= 1.23 V

Page 769
1               1
€
CD
Rout = R6
gm
12kΩ
gm
= 120Ω → g m = 8.25mS                                  (               )
8.25mS = 2 500µA/V 2 I D → I D = 68.1 mA

2I D                       2(68.1mA)
gm =                      VGS − VTN =               = 16.5 V
VGS − VTN                      8.25mS
If one neglects R6 , I D = 69.4 mA and VGS − VTN = 16.8 V

Page 774
RI + R6                          2kΩ + 13kΩ 
€   BJT : v i ≤ 0.005V (1+ g m RI )
R6
[
= 0.005V 1+ 9.8mS (2kΩ) 
 13kΩ 
]
 = 119 mV

[
Neglecting R6 , v i ≤ 0.005V (1+ g m RI ) = 0.005V 1+ 9.8mS (2kΩ) = 103 mV           ]
RI + R6                               2kΩ + 12kΩ
FET : v i ≤ 0.2(VGS − VTN )(1+ g m RI )
R6
[
= 0.2(0.982) 1+ 0.491mS (2kΩ)
12kΩ
]
= 454 mV

[
Neglecting R6 , v i ≤ 0.2(VGS − VTN )(1+ g m RI ) = 0.2(0.982) 1+ 0.491mS (2kΩ) = 389 mV             ]

€

24
Page 775
   β R              100(1.73kΩ) 
RiC = ro 1+ o th  = 219kΩ 1+               = 3.40 MΩ
 Rth + rπ         1.73kΩ + 10.2kΩ
                

[
Or more approximately, RiC = ro [1+ g m Rth ] = 219kΩ 1+ 9.8mS (1.73kΩ) = 3.93 MΩ        ]
[
RiD = ro [1+ g m Rth ] = 223kΩ 1+ 0.491(1.71kΩ) = 4.10 MΩ]
Page 778
 1
€                          R6  
 gm                        R6
1                            1
R6 +                         R6 +
gm                           gm                       R6
(i ) AvCB = g m RL            1
= g m RL
R6
= g m RL
R6 (1+ g m RI ) + RI
R6                g m RI +
 gm                             1
RI +                                 R6 +
1                             gm
R6 +
gm
R6           1            g m RL  R6 
AvCB = g m RL                          =                      
R6 + RI       g R R 1+ g m Rth  R6 + RI 
1+ m I 6
R6 + RI
(ii ) The voltage gains are proportional to the load resistance
 22kΩ                                     22kΩ 
AvCE = +8.48        = +10.4              AvCG = +4.12        = +5.02
 18kΩ                                     18kΩ 
CB                                 RL   RL    18kΩ
(iii ) CB : A  v    ≤ g m RL = 176       AvCB ≅      =     =       = 10.4          8.48 < 10.4 << 176
Rth RI R6 1.73kΩ
RL   RL    18kΩ
CG : AvCG ≤ g m RL = 8.84            AvCB ≅      =     =       = 10.5         4.11 < 8.84 < 10.5
Rth RI R6 1.71kΩ

Page 786
CE

€
(i ) Since we need high gain, the emitter should be bypassed, and Rin = RB rπ = 250kΩ.
βo     100
If we choose RB ≅ rπ , IC =              ≅          = 5 µA
40rπ 40(500kΩ)
CG       1                1
(ii ) R
in   ≅
gm
IC ≅
40(2kΩ)
= 12.5 µA

€

25
Page 790
(i) Common − Emitter :
1
C1 >>                            = 8.07nF    Choose C1 = 82 nF = 0.082 µF
2π (250Hz)(1kΩ + 77.9kΩ)
1
C2 >>                                     = 0.269µF    Choose C2 = 2.7 µF
               1 
2π (250Hz)10kΩ  3kΩ +        

            9.80mS 

1
C3 >>                             = 6.13nF     Choose C31 = 68 nF = 0.068 µF
2π (250Hz)(21.9kΩ + 82kΩ)
Common − Source :
1
C1 >>                         = 713 pF       Choose C1 = 8200 pF
2π (250Hz)(1kΩ + 892kΩ)
1
C2 >>                                      = 0.221µF   Choose C2 = 2.2 µF
                1   
2π (250Hz)10kΩ  2kΩ +         

            0.491mS 

1
C3 >>                             = 6.15nF     Choose C3 = 68 nF = 0.068 µF
2π (250Hz)(21.5kΩ + 82kΩ)

Page 793
Common − Collector :
€
1
C1 >>                          = 6.60nF      Choose C1 = 68 nF = 0.068 µF
2π (250Hz)(1kΩ + 95.5kΩ)
1
C3 >>                           = 7.75nF     Choose C3 = 82 nF = 0.082 µF
2π (250Hz)(120Ω + 82kΩ)
Common − Drain :
1
C1 >>                           = 713 pF     Choose C1 = 8200 pF
2π (250Hz)(1kΩ + 892kΩ)
1
C3 >>                             = 7.60nF     Choose C3 = 82 nF = 0.082 µF
2π (250Hz)(1.74kΩ + 82kΩ)

€

26
Page 796
Common − Base :
1
C1 >>                           = 0.579µF       Choose C1 = 6.8 µF
2π (250Hz)(1kΩ + 0.1kΩ)
1
C2 >>                                                  = 12.2nF            Choose C2 = 0.12 µF
                                  
              [               (
2π (250Hz)160kΩ 300kΩ 10.2kΩ + 101 13kΩ 1kΩ 
       )]
1
C3 >>                             = 6.13nF       Choose C3 = 0.068 µF
2π (250Hz)(21.9kΩ + 82kΩ)
Common − Gate :
1
C1 >>                            = 0.232µF       Choose C1 = 2.2 µF
2π (250Hz)(1kΩ + 1.74kΩ)
1
C2 >>                                = 714 pF        Choose C2 = 8200 pF
(
2π (250Hz) 1.5MΩ 2.2 MΩ  )
1
C3 >>                             = 6.19nF       Choose C3 = 0.068 µF
2π (250Hz)(20.9kΩ + 82kΩ)

Page 797

€   (a ) Common − Source :
1
C2 =                                       = 55.3nF        Choose C2 = 0.056 µF
                1   
2π (1000Hz)10kΩ  2kΩ +         

            0.491mS 

(b) Common − Collector :
1
C3 >>                             = 795 pF       Choose C3 = 820 pF
2π (2000Hz)(120Ω + 100kΩ)

(c) Common − Gate :
1
C1 >>                             = 42.6nF       Choose C1 = 0.042 µF
2π (1000Hz)(2kΩ + 1.74kΩ)

€

27
Page 801
0.020                 2
(i ) 20V = V + 3600I 20 = V + 3600 2 (V −1.5) → V = 2.203V I
GS                 D            GS                GS                    GS                D   = 4.94 mA

V = 5 − (−V ) = 7.20 V Q - point : (4.94 mA, 7.20 V) R = R = 22 MΩ
DS                     GS                                                             in        G

CD     g R            2(4.94mA)
A =
v
m
g =    L
m= 14.2mS R = 3600Ω 3000Ω = 1630Ω       L                                      AvCD = 0.959
1+ g R      m (2.20 −1.50)V
L

1             1
(ii ) R CD
out   = 3.6kΩ
gm
= 3.6kΩ
0.0142
= 69.1 Ω                                           [
vgs ≤ 0.2(2.20 −1.50) 1+ 0.0142(1630) = 3.38 V    ]
1           1
+ VDS         + 5 + 2.21
(iii ) ro = λ       = 0.015            = 14.8kΩ                      RL = 3600Ω 3000Ω 14.8kΩ = 1470Ω AvCD = 0.954
ID          0.005
W Kn 2x10−2 400
(iv )    = ' =
L Kn 5x10−5
=
1

Page 802
g R                                   2(4.94mA)
€   A= m S                          gm =                     = 14.2mS    RS = 3600Ω AvCD = 0.981 Rin = RG = 22 MΩ
1+ g m RS                           (2.20 −1.50)V
CD                         1             1                                        3000Ω
Rout = 3.6kΩ                   = 3.6kΩ        = 69.1 Ω          AvCD = A                     = 0.959
gm         0.0142                                  69.1Ω + 3000Ω

Page 805

€
(i ) Reverse the direction of the arrow on the ermitter of the transistor as well
as the values of VEE and VCC .

CD                  1             1                           CB          75.1Ω
(ii ) R out   = RE
gm
= 13kΩ
40(331µA)
= 75.1 Ω             AV =
75Ω + 75.1Ω
(13.2mS )(7.58kΩ) = 50.1
(iii ) For v    CB      ≥ 0, we require v C ≥ 0. VC = 5 − IC RC = 2.29 V ∴ v c ≤ 2.29 V
vo ≤ 5mV ( g m RL ) = 5mV (13.2mS )(7580Ω) = 0.500 V
6.8V
(iv ) R   E                 [                   ]
= 75Ω 1+ 40(7.5 − 0.7) = 20.5 kΩ (a standard 1% value)                             IC ≅
20.5kΩ
= 332 µA

75
50 = 40(332µA) RL                         → RL = 7.53kΩ → RC = 8.14 kΩ → 8.06 kΩ (a standard 1% value)
75 + 75

€

28
Page 806
max       VEE − 0.7V 5(1.05) − 0.7V
max

(i ) 5% tolerances I              C     ≅
REmin
=
13kΩ(0.95)
= 368µA

min  max max
VCmin = VCC − IC RC = 5V (0.95) − 368µA(8.2kΩ)(1.05) = 1.58 V 1.58 ≥ 0, so active region is ok.

max     VEE − 0.7V 5(1.1) − 0.7V
max
10% tolerances I              C     ≅       min
=              = 410µA
RE      13kΩ(0.9)
min  max max
VCmin = VCC − IC RC = 5V (0.90) − 410µA(8.2kΩ)(1.1) = 0.802 V                         0.802 ≥ 0, so active region is ok.
CB           CB
(ii ) A = g        m   RC = (13.2mS )(8200Ω) = 108          Rout = 8.2kΩ Rin = 75 Ω
CB                                  100kΩ        
Rin        100kΩ       75Ω
AvCG =                CB
A CB         =          (108) 8200 + 100kΩ  = 49.9
75Ω + Rin  Rout + 100kΩ 75Ω + 75Ω                    

Page 811
1         1
€   (ii ) r
o    ≅ ro =              =
λI D 0.015 2x10−4
= 333kΩ
(        )
5
or more exactly VDS = 25 −10 I D − 9.1x103 I D = 25 −1.09x105 (0.2mA) = 3.18 V
1           1
+ VDS         + 3.18
ro = λ       = 0.015 −4     = 349kΩ                       RL = 100kΩ 100kΩ 349kΩ = 43.7kΩ
ID         2x10
Rin      2(0.2mA)          75Ω 
AvCS = −( g m RL )                    =−          (43.7kΩ) 75Ω + 75Ω  = −43.7
RI + Rin      0.2V                      
0.01       2
(iii ) I 2
D (0.25) = 0.3125mA VGS − VTN = 0.25V VGS = 0.25V − 2 = −1.75V
=

−V        1.75V                             A 50(0.25V )
RS = GS =             = 5.60kΩ → 5.6 kΩ RL = 2 v =             = 40 kΩ RD 100kΩ = 40kΩ
ID    0.3125mA                             g m 0.3125mA
RD = 66.7kΩ → 68 kΩ C1 remains unchanged.
1
C2 >>                                           = 0.853nF → Choose C2 = 8200 pF

6          1 
10 π  5.6kΩ       
       2.5mS 
1
C3 >>           6
= 1.90 pF → Choose C3 = 20 pF
10 π (68kΩ + 100kΩ)

(iv ) A = −g           m   RD = −(2mS )(100kΩ) = −200        Rout = 100kΩ Rin = 75 Ω
Rin       100kΩ       75Ω               100kΩ     
AvCG =                  A            =          (−200)100kΩ + 100kΩ  = −50.0
75Ω + Rin Rout + 100kΩ 75Ω + 75Ω                     

€
29
Page 813
Kn             2                   0.01            2
(i ) M :   1   ID =
2
(VGS − VTN ) VGS = −RS1I D I D = 2 (−200I D + 2) → I D = 5.00mA
VDS       = 15 − 5mA(820Ω) = 10.9 V

1+ λVDS 1+ 0.02(10.9)
g m = 2Kn I D = 2(0.01)(0.005) = 10.0 mS           ro =          =              = 12.2 kΩ
λI D   0.02(5mA)
22kΩ
Q2 : VEQ =
22kΩ + 78kΩ
(15V ) = 3.30 V REQ = 22kΩ 78kΩ = 17.2kΩ
3.30 − 0.7                                        151      
IC = 150                     = 1.52 mA VCE = 15 −1.52mA 4.7kΩ +     1.6kΩ = 5.41 V
17.2kΩ + 151(1.6kΩ)                                    150      
150                 80 + 5.41
g m = 40(1.52mA) = 60.8 mS        rπ =          = 2.47 kΩ ro =           = 56.2 kΩ
60.8mS                1.52mA
120kΩ
Q3 : VEQ =
120kΩ + 91kΩ
(15V ) = 8.53 V REQ = 120kΩ 91kΩ = 51.8kΩ
8.53 − 0.7                                 81     
IC = 80                    = 1.96 mA VCE = 15 −1.96mA 3.3kΩ = 8.45 V
51.8kΩ + 81(3.3kΩ)                              80     
80                   60 + 8.45
g m = 40(1.96mA) = 78.4 mS        rπ =        = 1.02 kΩ ro =           = 34.9 kΩ
78.4mS                  1.96mA
(ii) A typical op - amp gain is at least 10,000 which exceeds the amplification factor
of a single transistor.

Page 815

€
(i ) RL1 = 478Ω 12.2kΩ = 460Ω            RL2 = 3.53kΩ 54.2kΩ = 3.31kΩ        RL3 = 232Ω 34.4kΩ = 230Ω
 79.6mS (230Ω)      1MΩ     
Av = −10mS (460Ω)(−62.8mS )(3.31kΩ)                                 = 898 20 log(898) = 59.1 dB
1+ 79.6mS (230Ω) 10kΩ + 1MΩ 
                  
   VDD                  15
(ii ) Av ≅ − V − V (−10VCC )(1) = − 1 (−10)(15)(1) = 2250
 GS    TN 

(iii ) A    v    = −10mS (2.39kΩ)(−62.8mS )(19.8kΩ)(0.95)(0.99) = 28000

Page 820
    1     3990 
€       (i ) R  out      = 3300         +
 0.0796S 90.1 
 = 55.9 Ω

(ii ) Note that these are obtained from SPICE.
(iii ) A = −g R = − 2(0.01)(0.001) (3kΩ 17.2kΩ 2.39kΩ) = 5.52
v1        m   L1

A = −5.52(−222)(3.31kΩ)(0.95)(0.99) = 1150
v

€       30
CHAPTER 15
Page 842
60  15 − 0.7 
(i ) IC = α F I E =                  = 93.8 µA
61  2 ( 75kΩ)
VCE = 15 − 93.8µA(75kΩ) − (−0.7V ) = 8.67 V
          
60  15 − VBE                            IC     
(ii ) IC = α F I E =                  and VBE = 0.025V ln 
61  2 ( 75kΩ)                               -15 
 0.5x10 A 
→ IC = 94.7 µA
           

Page 844
v1 + v 2 1.01+ 0.99
€   vid = v1 − v2 = 1.01− 0.990 = 0.020 V vic =                =          = 01.00 V
2         2
v +v    4.995 + 5.005
vid = v1 − v2 = 4.995 − 5.005 = −0.010 V v ic = 1 2 =                       = 5.00 V
2          2
vod = Add v id + Acd v ic v oc = Adcv id + Accvic

2.20                 0.02       1.00
     = [ Add Acd ]                    → [ Add Acd ] = [100 0.20]
 0                  −0.01       5.00
1.002                 0.02       1.00
       = [ Adc Acc ]                   → [ Adc Acc ] = [0.100 1.00]
5.001                −0.01       5.00

Page 848

€   Differential output : Adm = Add = −20VCC = −300           Acm = 0   CMRR = ∞
Single - ended output : Adm =           = +10VCC = 150     CMRR = 20VEE = 300      Acm = −       = −0.5
2                                                       300

Page 851
 100  R  15 − 0.7 
€             1−     C          
VIC = 15V  101  2RC  15  = 5.30 V
      100  RC     
   1+             
      101  2RC    

Page 853
VO             15V
€   I DC = I SS −       = 100µA −       = 80 µA

€

31
Page 855
I
I D = SS = 100 µA VDS = 12 − I D RD + VGS = 12 −100µA(62kΩ) + VGS = 5.8V + VGS
2
2I D
VGS = VTN +
Kn                                         (
= VTN + 0.2V VTN = 1+ 0.75 VSB + 0.6 − 0.6                   )
VSB = −VGS − (−12V )

VSB = 11.8 − VTN                         (                       )
VTN = 1+ 0.75 12.4 − VTN − 0.6 → VTN = 2.75V            VDS = 8.75 V

Page 858
VA     60V
€   Rod = 2ro ≅ 2          =2        = 3.20 MΩ         Roc ≅ 2µ f REE = 2( 40)(60)(1MΩ) = 4.80 GΩ
IC    37.5µA
v cm   v
idm = g mvdm = 40( 37.5µA)vdm = 1.5x10−3 vdm            icm ≅         = cm = 5.00x10−7 v cm
2REE 2 MΩ

Page 863
100  150µA                        15V
€   IC1 = IC 2 =                   = 74.3 µA    IC 3 =        = 750 µA        VCE 3 = 15 − 0 = 0 V
101  2                            20kΩ
VCE1 = 15 − 74.3µA(10kΩ) − (−0.7) = 15.0 V              VCE 2 = 15 − ( 74.3µA − 7.5µA)(10kΩ) − (−0.7) = 15.0 V
750µA
VEB 3 = ( 74.3µA − 7.5µA)(10kΩ) = 0.668 V              IS 3 =                 = 1.87x10−15 A
 0.668V 
exp        
 0.025 

Page 865
max
€   (i ) Adm        = 560 (15) = 8400
8400
(ii ) I   C1   ≤ 50(1µA) = 50 µA      Adm =
28  500µA 
= 2210
1+             
100  50µA 
8400
(iii ) I   C1   ≤ 50(1µA) = 50 µA     Adm =
28  5mA 
= 290
1+           
100  50µA 
max
(iv ) A    dm    = 560 (1.5) = 840
50                        15V
(v ) R    in   = 2rπ = 2
40(50µA)
= 50 kΩ   Rout ≅
0.5mA
= 30 kΩ

50                       15V
Rin = 2rπ = 2                  = 50 kΩ   Rout ≅       = 3.0 kΩ
40(50µA)                    5mA

€

32
Page 866
CMRR ≅ g m2 R1 = 40 (50µA)( 750kΩ) = 1500                      CMRRdB = 20 log(1500) = 63.5 dB

Page 868
g m2  RC rπ 3              40  IC 2 RC rπ 3                     0.7r                560VA 3
€   (i )   Adm =                  ( g m3ro3 ) = 
2  RC + rπ 3              2  RC + rπ 3 
( 40IC 3ro3 ) ≅ 800
π3
(VA 3 ) =
 RC + rπ 3              R
1+ C
rπ 3
560VA 3          560VA 3              560VA 3           560VA 3
Adm =                =                       =                  =
40IC 3 RC      40IC 2 RC  IC 3    40(0.7)  IC 3     28  IC 3 
1+             1+             1+               1+ β  I 
β o3           β o3  IC 2         β o3  IC 2       o3  C 2 

max
(ii ) A  dm    = 560 (75) = 42000                IC1 ≤ 50(1µA) = 50 µA
42000                                          42000
Adm =                      = 11000                Adm =                   = 1450
28  500µA                                     28  5mA 
1+                                            1+           
100  50µA                                     100  50µA 
50                                  75V + 15V
(iii )   Rin = 2rπ = 2
40(50µA)
= 50 kΩ        Rout ≅ ro3 =
0.5mA
= 180 kΩ

50                               90V
Rin = 2rπ = 2             = 50 kΩ           Rout ≅       = 18.0 kΩ
40(50µA)                            5mA

Page 871

€
Avt1 = −3.50                             (
Avt 2 = −22mS 150kΩ 162kΩ 203kΩ = −1238        )
Avt 3 =
(
0.198S 2kΩ 18kΩ       )       = 0.9971 Adm = −3.50(−1238)(0.9971) = 4320
(
1+ 0.198S 2kΩ 18kΩ             )
100                                    1    ro3 R2        1       162kΩ 150kΩ
Rin = 2rπ = 2                  = 101 kΩ             Rout ≅       +        =            +             = 776 Ω
40( 49.5µA)                               g m4 β o4 + 1 40( 4.95mA)       101
P ≅ ( I1 + I 2 + I 3 )(VCC + VEE ) = (100 + 500 + 5000)µA(30V ) = 168 mW

€

33
Page 872
 150                          5mA                     0.7V
(i )    IC = 50µA
 151
 = 49.7µA IC 3 = 500µA +
151
= 533 µA RC =
        533 
= 15.2kΩ
 49.7 −     µA
        150 
150
rπ 3 =
40(533µA)
= 7.04 kΩ                                 (
Avt 2 = −20( 49.7µA) 15.2kΩ 7.04kΩ = −4.68    )
150                                   150                              75 + 14.3
IC 4 =       5mA = 4.97mA         rπ 4 =               = 755 Ω          ro3 =             = 168 kΩ
151                               40( 4.97mA)                           533µA

[
Avt 2 = −40(533µA) 168kΩ 755 + 151(2kΩ) = −2304    ]
0.199S (2kΩ)
g m4 = 40(4.97mA) = 0.199S          Avt 3 =                         = 0.998
1+ 0.199S (2kΩ)
Adm = −4.78(−2304 )(0.998) = 11000

150                             1    ro3 R2       1       168kΩ
Rid = 2rπ 1 = 2                 = 151 kΩ      Rout ≅       +        =           +       = 1.12 kΩ
40( 49.7µA)                        g m4 β o4 + 1 40(4.95mA)    151
CMRR is set by the input stage and doesn't change since the bias current is the same.
50 + 14.3
(ii )                   (             )
Avt 2 = −22mS 117kΩ 203kΩ = −1630                ro3 =
550µA
= 117 kΩ

1    ro3 R2        1       117kΩ
Adm = −3.50(−1630)(0.998) = 5700              Rout ≅       +        =            +       = 1.16 kΩ
g m4 β o4 + 1 40( 4.95mA)    101
CMRR and input resistance are set by the input stage and don't change.
Adm    6920                             Ro    1.62kΩ
(iii ) A v   =         =
= 0.99986        Rout =         =
1+ Adm 1+ 6920
= 0.234 Ω

Rin = Rid (1+ Adm ) = 101kΩ(6921) = 699 MΩ

€

34
Page 873
2(500µA)                      1.63V
(i )   VGS 3 = 1+
2.5mA
= 1.63 V   RD =
100µA
= 16.3 kΩ

1
Avt1 = −     2(0.005)(100µA) (16.3kΩ) = −8.16
2
            
1
Avt 2 = −g m3ro3 = − 2(0.0025)(0.0005)             
 0.01 0.5mA  = −316
     (     )
7.07mS (2kΩ)
g m4 = 2(0.005mA)(0.005mA) = 7.07mS                 Avt 3 =                      = 0.934
1+ 7.07mS (2kΩ)
1      1
Adm = −8.16(−316)(0.934 ) = 2410             Rid = ∞ Ro =          =       = 141 Ω
g m4 7.07mS
CMRR = g m1 R1 = 1.00mS ( 375kΩ) = 375 or 51.5 dB
(ii ) P ≅ ( I1   + I 2 + I 3 )(VDD + VSS ) = (5.7mA)(24V ) = 137 mW

Page 879

€
(i ) VGS1 + VSG2 = 0.5mA(4.4kΩ) = 2.2 V Since the device parameters are the same,
0.025        2
VGS1 = VSG 2 = 1.1 V       ID =
2
(1.1−1) = 125 µA
0.5mA(2.4kΩ)
(ii ) Since the device parameters are the same,          VBE1 = VEB 2 =
2
= 0.6 V
 0.6 
(
IC = 10−14 A exp )      = 265 µA
 0.025 

Page 883
v                                          2
€                                (           )
Av1 = d = −g m n 2 RL = − 50mA/V 2 (2V −1V )(10) (8Ω) = −40.0
vg
1    40.0
Avo = Av1     =−      = −4.00         v g ≤ 0.2(2 −1)V = 0.200 V          v d ≤ 0.2V ( 40) = 8.00 V
n     10
8V
vo ≤       = 0.800 V
10

€

35
Page 890
   150(18.4kΩ) 
(i ) R  bb → 0 Rout = 432kΩ1+               = 32.5 MΩ
 18.8kΩ + 18.4kΩ
                
270kΩ
(ii )   VEQ = −15V
110kΩ + 270kΩ
= −10.66 V REQ = 110kΩ 270kΩ = 78.2 kΩ

−10.66 − 0.7 − (−15)               (75 + 11.5)V = 446 kΩ r = 150 = 19.3 kΩ
IC = 150                      = 195 µA ro =                        π
78.2kΩ + 151(18kΩ)                    195µA                  40(195µA)
       150(18kΩ)       
Rout = 446kΩ1+                      = 10.9 MΩ
 78.2kΩ + 19.3kΩ + 18kΩ
                       
15V
(iii ) R + R
1       2
20µA
≅ = 750kΩ Using a spreadsheet with I o = 200 µA yields VBB = 9V.

                      
 9V                               150 9 − 0.7 −1.33µA(180kΩ)
R1 = 750kΩ         = 450 kΩ R1 = 300 kΩ RE =                               = 40.0 kΩ
 15V                              151         200µA         

 75 + 15 − 8.3        150(40.0kΩ)        
Rout =                1+                         = 10.7 MΩ
 2x10−4  180kΩ + 18.75kΩ + 40.0kΩ
                          

Page 893
2(0.2mA)
€   (i ) V        ≥ VGS − VTN = 1+                   = 1.40 V       VD = VS + 1.40 = −15 + 0.2mA(18.2kΩ) + 1.40 = −9.96V
DS
2.49mA/V 2
W Kn 2.49mA/V 2 99.6
(ii )     = ' =
L Kn   25µA /V 2
=
1
2                                      15V
(iii ) P RS    = (0.2mA) 18.2kΩ = 0.728 mW           I BIAS =
499kΩ + 249kΩ
= 20.1 µA
2                                        2
PR 4 = (20.1µA) 499kΩ = 0.202 mW                 PR 3 = (20.1µA) 249kΩ = 0.101 mW
510kΩ
VGG = −15V                 = −10.2V −10.2 − VGS −18000I D = −15V
510kΩ + 240kΩ
2.49mA         2
4.8 − VGS −18000
2
(VGS −1) = 0 VGS = 1.390 V I D = 189 µA
1            (
2 2.49x10−3      )
Rout    ≅ µ f RS ≅
0.01   189x10−6
[              ]
1+ 0.01(11.6) (18kΩ) = 10.3 MΩ

Page 894
Ravg = 10kΩ(1+ 0.2) = 12 kΩ             12kΩ(1− 0.01) ≤ R ≤ 12kΩ(1+ 0.01)         11.88 kΩ ≤ R ≤ 12.12 kΩ
€

€

36
Page 896
2I REF                                2(150µA)
(i ) V  DS1   = VTN +
Kn (1+ λVDS1 )
VDS1 = 1+
250µA /V 2 [1+ 0.0133VDS1 ]
→ VDS1 = 2.08 V

1+ 0.0133(10)
IO = 150µA                         = 165 µA
1+ 0.0133(2.08)

2I D                      2(150µA)
(ii )   VDS ≥ VGS − VTN      VD − (−10V ) ≥
Kn
VD ≥ −10V +
250µA/V 2
= −8.91 V

Page 897
25 /1                                2(50µA)                                1+ 0.02(15)
€   MR =            = 8.33      VDS1 = 1V +                      = 2.16 V    MR = 8.33                   = 10.4
3/1                             (
3 25µA/V 2     )                           1+ 0.02(2.16)

2 /1                                 2(50µA)                               1+ 0.02(10)
MR =           = 0.400      VDS1 = 1V +                      = 1.89 V    MR = 8.33                   = 0.463
5 /1                             (
5 25µA/V 2     )                           1+ 0.02(1.89)

Page 899
 V  V         2                                             V       2 
€   I REF = I S exp BE1 1+ BE1 +                100µA = (0.1 fA) exp( 40VBE1 )1+ BE1 +    → VBE1 = 0.690
 VT  VA1 β FO                                               50V 100 
VCE ≥ VBE → VC ≥ −VEE + 0.690 V

Page 900
15
€                                                                       1+
0.5 A                      0.5                                  60
MR =       = 0.5          MR =        = 0.490       MR = 0.5              = 0.606
A                          1.5                             0.7 1.5
1+                             1+      +
75                              60 75
15
1+
5A                      2.50                              60
MR =         = 2.50       MR =          = 2.39      MR = 2.5              = 2.95
2A                        3.5                         0.7 3.5
1+                          1+      +
75                          60 75

€

37
Page 901
10 /1                                       20 /1
(i ) IO 2 = 100µA       = 200 µA
 5 /1 
IO 3 = 100µA       = 400 µA
 5 /1 
 40 /1                                      2.5 /1
IO4 = 100µA        = 800 µA                 IO5 = 100µA         = 50 µA
 5 /1                                       5 /1 
1+ 0.02(10)                                 1+ 0.02(5)
(ii ) I   O2   = 200µA
1+ 0.02(2)
= 231 µA IO 3 = 400µA
1+ 0.02(2)
= 423 µA

1+ 0.02(12)                                  1+ 0.02(8)
IO4 = 800µA                         = 954 µA IO5 = 50µA                   = 55.8 µA
1+ 0.02(2)                                 1+ 0.02(2)

Page 902
1
€   (i )   IO 2 = 10µA
17
= 7.46 µA IO3 = 5( 7.46µA) = 37.3 µA IO4 = 10( 7.46µA) = 74.6 µA
1+
50
10                                  10
1+                                  1+
IO2 = 10µA        50     = 8.86 µA IO 3 = 50µA       50    = 44.3 µA
0.7 17                              0.7 17
1+      +                           1+     +
50 50                               50 50
10
1+
IO4 = 100µA         50     = 88.6 µA
0.7 17
1+       +
50 50
10                     10 − 9.957
(ii ) MR =      11
= 9.957 FE =
10
= 4.3x10−3 VCE 2 = VBE1 + VBE 3 = 1.4 V
1+
50(51)

Page 903
1+ 0.02(10)                            50V + 10V
€   MOS : IO 2 = 200µA                             = 231 µA Rout 2 =              = 260 kΩ
1+ 0.02(2)                           231µA
1+ 0.02(5)                             50V + 5V
IO3 = 400µA                        = 423 µA Rout 3 =               = 130 kΩ
1+ 0.02(2)                              423µA
10
1+
50                            50V + 10V
BJT : IO 2 = 10µA                             = 10.1 µA Rout 2 =              = 5.94 MΩ
0.7 17                              10.1µA
1+      +
50 100
10
1+
50                                 50V + 10V
IO3 = 50µA                            = 50.7 µA     Rout 3 =             = 1.19 MΩ
0.7 17                                   50.7µA
1+      +
50 100

38
€
Page 904
0.7V                                   10V
1+                                   1+
IC1 = 100µA        50V     = 89.4 µA IC 2 = 500µA         50V     529µA
0.7V    6                              0.7V     6
1+      +                              1+      +
50V 50V                                50V 50V
1        1                    529µA                   50V + 10V
Rin ≅     =            = 280 Ω β =            = 5.92 Rout =            = 113 kΩ
g m1 40(89.4 µA)               89.4 µA                   529µA

Page 906
10V
€                           2(100µA)                          1+
VDS1 = VGS1 = 0.75V +           = 1.20 V I D 2 = 100µA 50V = 117 µA
1mA /V 2
1.2
1+
50V
1            1                    117µA                   50V + 10V
Rin ≅      =              = 2.24 kΩ β =            = 1.17 Rout =           = 513 kΩ
g m1                               100µA                     117µA
( )( )
−3
2 10 10 −4

Page 907
V  I REF A  0.025V  100µA 
€   (i ) R = I T ln I AE 2  = 25µA ln 25µA 5 = 3000 Ω
 O E1                       
O

100µA                    75V 
K = 1+ ln        5 = 4.00 Rout = 4       = 12.0 MΩ
 25µA                    25µA 
VT  I REF A         0.025V  1000µA 
(ii ) IO = R ln I AE 2  IO = 100Ω ln I  → IO = 300.54 µA
 O E1                             
O

 100µA                              75V 
K = 1+ ln            10 = 2.202 Rout = 2.202            = 550 kΩ
 300.54 µA                          300.54µA 

Page 909
                  
€        1 2I REF        IO (W/L)1           1  2(200µA)       IO 1 
IO =           1−                    IO =               1−          
R Kn1         I REF (W/L)          2kΩ 25µA /V 2 
   200µA 10 

                2 

        IO 
IO = 2.00mA1−
                → IO = 764 µA
     2.00mA   
50V + 10V                               
Rout =
764µA
(        )(         )
1+ 2000 2 2.5x10−4 7.64 x10−4  = 176 kΩ
                              

Page 912
β r 150  50V + 15V                               50V + 15V
€   Rout ≅ o o =              = 97.5 MΩ      Rout = ro =             = 1.30 MΩ
2     2  50µA                                     50µA

€
39
Page 913

VDS 2 = VGS 2 = 0.8V +
(
2 5x10−5     ) = 1.43 V   VDS 4 = 15 −1.43 = 13.6 V
−4
2.5x10
 1                1               
       V + 13.6V        V + 1.43V 
(
Rout ≅ µ f 4 ro2 = 2 2.5x10−4 5x10−5  )(        )[1+ 0.015(13.6)  0.015
]               0.015            = 379 MΩ
      50µA             50µA       
                                  
66.7V + 15V
Rout = ro =                   = 1.63 MΩ
50µA

Page 914
β r 100  67V + 14.3V                                                67V + 15V
€   Rout ≅ o o =                = 81.3 MΩ                       Rout = ro =             = 164 kΩ
2     2    50µA                                                      50µA

Page 916
10V − 20V
€   (i ) I  O      = 25.015µA +
1.66GΩ
= 25.009 µA

(ii ) V DS 4     ≥ VGS 4 − VTN = 0.2 V VD 4 ≥ VS 4 + 0.2 VD4 ≥ 0.95 + 0.2 = 1.15 V
20V
(iii ) I   O   = 50µA ± 0.1%        ΔI O ≤ 50nA
50nA
= 400 MΩ Choose Rout = 1 GΩ.
Rout ≥
2
50V                              1 2Kn           0.01         50µA    mA
ro ≅
50µA
= 1 MΩ → µ f = 1000 µ f ≅
λ ID
Kn = 
 V
(1000) 2 = 2.5 V 2

2.5x10−3 50                        1  50  25
(W / L)2 = (W / L)4 = 5x10−5 = 1 (W / L)3 = (W / L)1 = 2  1  = 1
 

Page 917
5V − 0.7V                    7.5V − 0.7V
€   (i )   I REF =
43kΩ
= 100 µA I REF =
43kΩ
= 158 µA

V − (−V )
(ii ) Since the transistors have the same parameters, VGS1 = DD 3 SS
4 x10−4           2                          4 x10−4     2
I D2 = I D1 =
2
(1.667 −1) = 89.0 µA I D 2 = I D1 = 2 (2.5 −1) = 450 µA
Page 918
0.025V     5 −1.4                                        0.025V    7.5 −1.4
€   IO ≅        ln −16     = 101 µA                      IO ≅            ln −16      = 103 µA
6.8kΩ 10 (39kΩ)                                          6.8kΩ 10 (39kΩ)

€

40
Page 919
VT  I A                          0.025V
(i ) IO = R ln I C1 AE 2 
 C 2 E1 
IO =
1000Ω
[       ]
ln 10(10) = 115 µA

(ii)   VCC + VEE ≥ VBE1 + VBE 4 ≅ 1.4 V

Page 921
2                  5
€   R=                              1−
       = 8.65 kΩ
(           )( )
5 25x10−6 10−4             50 


Page 923
V  I A  0.02588V  25 
€   (ii ) R = I T ln I C1 AE 2  = 45µA ln 5  = 926 Ω
 C 2 E1               
O

AE1 = A AE 2 = 25 AE1 = 25 A AE 3 = A AE 4 = 5.58 AE 3 = 5.58 A

Page 927
250µA                          2(125µA)
€   I D3 = I D 4 = I D1 = I D 2 =         = 125 µA VGS1 = 0.75V +            = 1.75 V
2                           250µA /V 2
2(125µA)
VGS 3 = −0.75V −                        = −1.87 V
200µA/V 2
VDS1 = VD1 − VS1 = (5 −1.87) − (−1.75) = 4.88 V VSD 3 = VSG 3 = 1.87 V
M1 and M 2 : (125µA, 4.88V )                M 3 and M 4 : (125µA,1.87V )

(              )(
Gm = g m1 = 2 2.5x10−4 1.25x10−4 = 250 µS         )
75.2V + 4.88V 75.2V + 1.87V
Ro = ro2 ro4 =                               = 314 kΩ
125µA         125µA
Av = Gm Ro = 78.5

Page 928
 1 2K                      1 2K 
€   CMRR ≅ µ f 3 g m2 RSS = 
λ I 

n3

D3 
(      λ I  2Kn2 I D 2 RSS
 2Kn2 I D 2 RSS =      ) n3

      D3 
           (   )
1
Kn3 = Kn2 I D 2 = I D 3 CMRR =           2(0.005)10 7 = 5.99x10 6 or 136 dB
0.0167

€

41
Page 932
2VA
For the buffered current mirror, VEC 4 = VEB3 + VEB11 +
β FO 4 (β FO11 + 1)
2IC 4 2IC 4 IC 4                         I         
IC11 ≅          =    =             ΔVEB = 0.025ln C 4  = 80.5 mV
β FO 4   50   25                         IC 4 / 25 
2(60)                        2(60V )                        47mV
VEC 4 = 0.7 + (0.7 − 0.081) +               = 1.37 V ΔVEC =                 = 47.1mV    VOS =        = 0.47 mV
50(51)                         50(51)                         100

Page 933
    I  150
€   Adm ≅ Av1 Av 2 Av 3       Av1 ≅ β o5 C 2  =   = 50       Av 2 ≅ µ f 5 ≅ 40(75) = 3000 Av 3 ≅ 1
    IC 5    3
Adm ≅ 50( 3000)(1) = 150000 Note that this assumes R L = ∞.

Page 934
−1
                         
€           2       1             1      = 5.45x106 →135 dB
CMRR =                 −
100 100(40)(75) 2(40) 10 10 
−4 7
( )( )
                         

Page 941
22 + 22 −1.4                   0.025V  1.09mA 
€   (i ) I REF =     39kΩ
= 1.09 mA I1 =
5kΩ
ln
 I1 
 → I1 = 20.0 µA

23.4V                                       21.3V
1+                                           1+
I2 = 0.75(1.09mA)            60V    = 1.08 mA I2 = 0.25(1.09mA)          60V   = 351 µA
0.7V 2                                      0.7V 2
1+          +                                 1+      +
60V 50                                       60V 50
      I     A  60V + 13.5V           733µA 
(ii ) Ro = ro211+ ln C 20 E 20  =
      IC 21 AE 21   18.4µA 
1+ ln       1 = 18.7 MΩ
18.4µA 

Page 945
2VA6          2(60V )
€   (ii )   VCE6 = VCE 5 +
β FO6
= 0.7 +
100
= 1.90 V

Page 948
 60 + 13   60 + 1.3
€   Rth = 2ro4 1.3ro6 = 2          1.3       = 20.1MΩ 11.1MΩ = 7.15 MΩ
 7.25µA   7.16µA 

Page 950

€                         (                )               (
Av1 = −1.46x10−4 6.54 MΩ Rin11 = −1.46x10−4 6.54 MΩ 20.7kΩ = −3.01             )

€
42
Page 952
50(0.025)
(i )   Req2 = rπ 16 + (β o16 + 1) RL =
2mA
+ 51(2kΩ) = 103 kΩ

0.025V  0.025V           
Req1 = rd14 + (rd13 + R3 ) Req2 =         +         + 344kΩ 103kΩ = 79.4 kΩ
0.216mA  0.216mA          
50(0.025V )
Rin12 =                 + 51(79.4kΩ) = 4.06 MΩ
0.216mA
        rπ 12 + y22   0.025V
−1                     0.025V    5.79kΩ + 89.1kΩ 
Req3 = (rd13 + R3 )  rd14 +              =       + 344kΩ          +                  = 1.97 kΩ
         β o12 + 1   0.216mA          0.216mA         51        
0.025               625 + 1970
rπ 16 = 50         = 625Ω Rout =             + 22 = 75.9 Ω
2mA                     51
0.7V                     0.7V
(ii ) I SC + ≅ 27Ω = 25.9 mA I SC − ≅ − 22Ω = −31.8 mA

€

43
CHAPTER 16
Page 988
1                    2      2                               1                      2      2
fL ≅    102 + 10002 − 2(50) − 2(0) = 159 Hz               fL ≅      1002 + 10002 − 2(500) − 2(0) = 114 Hz
2π                                                          2π

Page 989
200s            200s( s + 100)                       ω 2 + 1002         ω 2 + 102
€              ≥ 0.9                              1 ≥ 0.9              → 0.81 ≤ 2        → ω ≥ 205 rad/s
(s + 1000)       (s + 10)(s + 1000)                      ω 2 + 102         ω + 1002

Page 996

€
(i ) The value of C2 does not change A mid , ω P1, ω P3 , ω z1, or ω Z3.
1                                                1
ω P2 = −                           = −1000 rad/s        ω Z3 = −               = −385 rad/s
         1                                        2µF (1.3kΩ)
2µF 1.3kΩ        
      1.23mS 
1
fL =
2π
(               )
41.02 + 95.92 + 10002 − 2 02 + 02 + 3852 = 135 Hz
13.5
4.732
(ii ) Amid = 10 20 = 4.732 4.3kΩ 100kΩ ro =
1.23mS
→ ro = 57.5kΩ

Note that the SPICE value of gm probably differs from 1.23mS as well.
1
(iii )   ωP2 = −
         1         
10µF 1.3kΩ        57.5kΩ
      1.23mS       
1
fL =
2π
(                   )
41.02 + 95.92 + 2022 − 2 02 + 02 + 76.92 = 31.8 Hz

Page 999
140(0.025V )
€                                                    (                        )
(ii ) rπ = 175µA = 20.0 kΩ R1S C1 = 1kΩ + 75kΩ 20.0kΩ 2µF = 33.6 ms Rth = 75kΩ 1kΩ = 987 Ω
      20.0kΩ + 987Ω 
R2S C2 = 13kΩ                 10µF = 1.47 ms R3S C3 = (43kΩ + 100kΩ)0.1µF = 14.3 ms
           141      
1  1           1        1 
fL ≅             +        +        = 124 Hz
2π  33.6ms 1.47ms 14.3ms 

€

44
Page 1001
Rin  β o   1260  100 
(i ) Av = −          RL  ≅ −    
RI + Rin  rπ   2260  1.51kΩ 
(            )
 4.3kΩ 100kΩ = −157

Rin  β o   1260  100 
Av = −           RL  ≅ −     
RI + Rin  rπ   2260 1.51kΩ 
(
 4.3kΩ 100kΩ 46.8kΩ = −140 )
ro is responsible for most of the discrepancy. rπ and βo will also be differ from our hand calculations.
Note that 45% of the gain is lost because of the amplifier's low input resistance.
2(1.5mA)
(ii )   gm =
0.5V
= 6.00 mS   R1S C1 = (1kΩ + 243kΩ)0.1µF = 24.4 ms
          1 
R2S C2 = 1.3kΩ         10µF = 1.48 ms R3S C3 = (4.3kΩ + 100kΩ)0.1µF = 10.4 ms
      6.00mS 
1  1          1         1 
fL ≅             +        +         = 129 Hz
2π  24.4ms 1.48ms 10.4ms 

Page 1003
                 1 
€   (i )    g m = 40(0.1mA) = 4.00 mS
R1S C1 = 100Ω + 43kΩ         4.7µF = 1.64 ms
4.00mS 

1  1          1 
R2S C2 = (22kΩ + 75kΩ)1µF = 97.0 ms f L ≅              +         = 98.7 Hz
2π  1.64ms 97.0ms 

Page 1004
2(1.5mA)                                        1 
€   gm =          = 6.00 mS    R1S C1 = 100Ω + 1.3kΩ         1µF = 0.248 ms
0.5V                                      6.00mS 
1     1         1 
R2S C2 = (4.3kΩ + 75kΩ)0.1µF = 7.93 ms f L ≅               +         = 662 Hz
2π  0.248ms 7.93ms 

€

45
Page 1005
100
(i )    g m = 40(1mA) = 40.0 mS
.04S
= 2.50 kΩ
rπ =
                                     
[                  (
R1S C1 = 1kΩ + 100kΩ 2.5kΩ + 101 3kΩ 47kΩ 0.1µF = 7.52 ms
                                                           )]
            2.5kΩ + 100kΩ 1kΩ 
R2S C2 =   47kΩ + 3kΩ                         (            )
100µF = 4.70 s f ≅ 1  1 + 1  = 21.2 Hz
             
L
                    101                           2π  7.52ms 4.7s 
                                

Amid ≅
(β   o   + 1) RL    RI 
         =
101 3kΩ 47kΩ           (                        )              100kΩ 
              = +0.978
Rth + rπ + (β o + 1) RL  RI + RB  990Ω + 2.5kΩ + 101 3kΩ 47kΩ                    (                 )    1kΩ + 100kΩ 

(ii )    R1S C1 = (1kΩ + 243kΩ)0.1µF = 24.4 ms
               1                                                  1  1        1 
R2S C2 =  24kΩ + 1.3kΩ     47µF = 1.15 s                           fL ≅             +       = 6.66 Hz
              1mS                                                2π  24.4ms 1.15s 
 RG  g m RL
Amid = +         
 243kΩ  1mS 1.3kΩ 24kΩ
= +       
(
= +0.550
)
 RI + RG  1+ g m RL     244kΩ  1+ 1.3kΩ 24kΩ        (                       )
Page 1009
g                                Cµo
€   Cπ = m − Cµ             Cµ =
ωT                                 VCB
1+
ϕ jc

(100µA,8V ) :           Cµ =
2 pF
= 0.551 pF        Cπ =
( )
40 10−4
− 0.551x10−12 = 0.722 pF
7.3V                                 2π (500 MHz)
1+
0.6V

(2mA,5V ) :     Cµ =
2 pF
= 0.700 pF        Cπ =
(
40 2x10−3            )       − 0.700x10−12 = 24.8 pF
4.3V                                2π (500 MHz)
1+
0.6V

(50mA,8V ) :        Cµ =
2 pF
= 0.551 pF       Cπ =
(
40 5x10−2               )    − 0.551x10−12 = 636 pF
7.3V                                 2π (500 MHz)
1+
0.6V

€

46
Page 1012
1
(i )    CGS = CGD = CISS = 0.5 pF
2
gm                        2(0.01)(0.01)
(ii )    CGS + CGD =
ωT
5CGD + CGD =
2π (200 MHz)
= 11.3 pF   CGD = 1.88 pF CGS = 9.38 pF

Cµo            2 pF                          gm          40(20µA)
(iii ) C  µ   =
VCB
=
7.3V
= 0.551 pF Cπ =
ωT
− Cµ =
2π (500 MHz)
− 0.551pF = −0.296 pF
1+              1+
ϕ jc            0.6V

Page 1015
Rin  β o          
€   Av = −                    RL  Rin = 7.5kΩ (1.51kΩ + 250Ω) = 1.43kΩ
RI + Rin  rx + rπ 
 1430  100 
≅ −       
 2430 1.76kΩ 
(
 4.3kΩ 100kΩ = −139    )
Rin  β o   1260  100 
Av = −           RL  ≅ −        
RI + Rin  rπ   2260 1.51kΩ 
(
 4.3kΩ 100kΩ = −157      )
Page 1023
RL
€   (ii )    The term CL
rπo
is added to the value of CT .

R          4120                                1
CL L = 3 pF        = 18.8 pF f P1 =                          = 1.39 MHz
rπo        656                    2π (656Ω)(156 + 18.8) pF
gm            0.064S
fP2 =                  =                 = 445 MHz
2π (Cπ + CL ) 2π (19.9 + 3) pF
0.064
(iii )   Cπ =
2π (500 MHz )
−10−12 = 19.4 pF

                 4120                         1
CT = 19.4 + 11+ 0.064(4120) +       = 290 pF f P1 =                 = 837 kHz
                  656                  2π (656Ω)290 pF
gm    0.064S                                      gm   0.064S
fP2 =         =              = 525 MHz               fZ =       =          = 10.2 GHz
2πCπ 2π (19.4 pF )                                2πCµ 2π (1pF )
Amid = −135 is not affected by the value of f T .

€

47
Page 1024
                     4120                          1
CT = 10 + 2 1+ 1.23mS (4.12kΩ) +       = 30.4 pF f P1 =                  = 5.26 MHz
                      996                   2π (996Ω)30.4 pF
gm    1.23mS                                     gm   1.23mS
fP2 =          =            = 19.6 MHz             fZ =        =           = 97.9 MHz
2πCGS 2π (10 pF )                               2πCGD 2π (2 pF )
gm         1.23mS
fT =                 =            = 16.3 MHz
2π (CGS + CD ) 2π (12 pF )

Page 1031

€
(ii ) 1+ gm RE = 1+ 0.064(100) = 7.40            RiB = 250 + 1560 + 101(100) = 11.9 kΩ
rπ 0 = 11.9kΩ (882 + 250) = 1030 Ω
10kΩ 30kΩ 11.9kΩ                                     264 
Ai =                                   = 0.821      Amid = −0.821      = −29.3
1kΩ + 10kΩ 30kΩ 11.9kΩ                                     7.4 
1
fH =                                                 = 6.70 MHz GBW = 196 MHz
19.9 pF           264 4120 
2π (1.03kΩ)        + 0.5 pF 1+  +     
 7.4              7.4 1030 

Page 1033
'
g m RL       '       βo              100                                 3.96mS (17.0kΩ)
€   (i ) Amid ≅        '
1+ g m RE
gm =
rx + rπ
=
100
= 3.96 mS        Amid ≅
1+ 3.96mS (100Ω)
= +48.2
250 +
40(0.1mA)
1
fH ≅                           = 18.7 MHz          GBW = 903 MHz
2π (17.0kΩ)(0.5 pF )
1           1                            265
(ii ) R iS   = R4
gm
= 1.3kΩ
3mS
= 265 Ω          Ai =
100 + 265
= 0.726

1
Amid = 0.726( g m RL ) = 0.726(3mS )(4.12kΩ) = 8.98             fH ≅                      = 9.66 MHz
2π (4.12kΩ)(4 pF )
3mS
fT ≅              = 43.4 MHz           GBW = 86.7 MHz
2π (11pF )

€

48
Page 1035
100
g m = 40(1.5mA) = 60 mS        rπ =        = 1.67 kΩ               RL = 3kΩ 47kΩ = 2.82kΩ
60mS

Ai =
[
100kΩ 150 + 1670 + 101(2820)          ]         = 0.987       Amid = 0.987
60mS (2.82kΩ)
= 0.980
[
1kΩ + 100kΩ 150 + 1670 + 101(2820)            ]                                1+ 60mS (2.82kΩ)

60mS                               1                                        1
Cπ =                − 0.5 pF = 18.6 pF f H ≅                                                               = 230 MHz
2π (500 MHz)                          2π                                               18.6 pF 
(                       )
(990 + 150) 286kΩ 0.5 pF + 1+ 169 
                 

Page 1036
430kΩ
€   RL = 1.3kΩ 24kΩ = 1.23kΩ            Ai =               = 0.998
1kΩ + 430kΩ
3mS (1.23kΩ)                               1                   1
Amid = 0.998                        = 0.785        fH ≅                                       = 50.9 MHz
1+ 3mS (1.23kΩ)                            2π                        10 pF 
(              )
1kΩ 430kΩ 1pF +


1+ 3.69 

Page 1039
1                                              1
€   fZ =                = 6.37 kHz             fP =                       = 6.33 MHz
2π (25MΩ)1pF                                2π (50.25kΩ)0.5 pF

Page 1041
Differential Pair : Adm = −g m RC = −40(99.0µA)(50kΩ) = −198
€
40(99.0µA)                                             100
Cπ =                     − 0.5 pF = 0.761 pF         rπ =              = 25.3 kΩ
2π (500 MHz)                                         40(99.0µA)
1
fH =                                              = 3.18 MHz
                          50kΩ 
2π (250Ω)0.761pf + 0.5 pF 1+ 198 +      
                          250Ω 
 1 
g m1      
 g m2           198                      1
 1 ( m C)
CC − CB Cascade : Av =                 g R =+      = +99 f H ≅                    = 6.37 MHz
2              2π (50kΩ)(0.5 pF )
1+ g m1      
 g m2 

€

49
Page 1042
100                          64.0mS
g m = 40(1.6mA) = 64.0 mS       rπ =        = 1.56 kΩ      Cπ =                − 0.5 pF = 19.9 pF
64mS                       2π (500 MHz)
rπ                           1.56 kΩ
Amid =
RI + rx + rπ
(−g m RL ) = 882Ω + 250Ω + 1.56 kΩ (−64.0mS )(4.12kΩ) = −153
1                  1
f P1 ≅                      =                       = 11.6 MHz
2π rπ 0 (Cπ + 2Cµ ) 2π (656Ω)(19.9 + 1) pF
1                  1
fP 2 ≅                   =                       = 7.02 MHz
2π RL (Cµ + CL ) 2π (4120Ω)(0.5 + 5) pF

Page 1043
1        0.02(100µA)                             1        0.02(25µA)
€   f P1 ≅               =             = 159 kHz     f P1 ≅               =            = 39.8 kHz
4 π CGGDro2     4 π (1pF)                        4 π CGGDro2    4 π (1pF)

Page 1050
1
€   (i )    XC =
2π (530Hz) 39 pF
= 7.69 MΩ >> 2.39 kΩ

1
XC =                   = 300 MΩ >> 51.8kΩ 19.8kΩ = 14.3 kΩ
2π (530Hz)1pF
1
(ii )   X1 =
2π (667kHz)0.01µF
= 23.9 Ω << 1.01 MΩ

1
X2 =                      = 5.07 mΩ << 66.7 Ω
2π (667kHz) 47µF
1
X3 =                    = 239 mΩ << 2.69 kΩ
2π (667kHz)1µF

Page 1052
1
€     ZC =                       = − j3.18 Ω
2π j (5 MHz)0.01µF

€

50
Page 1053
1                                         50V + 15V −1.6V
(i )   fo =                                      = 4.59 MHz       ro =
3.2mA
= 19.8kΩ
2π     (10µH )(100 pF + 20 pF )
100kΩ 100kΩ ro                         4.59 MHz
Q=                                  = 49.2   BW =             = 93.3 kHz
2π ( 4.59 MHz)(10µH )                            49.2

(                 )                         (
Amid = −g m 100kΩ 100kΩ ro = − 2(0.005)(0.0032) 100kΩ 100kΩ 19.8kΩ = −80.2             )
(ii ) f   o   is unchanged

€

51
CHAPTER 17
Page 1078
A                       10kΩ                                         10 4
(i ) Av = 1+ Aβ               β=
10kΩ + 91kΩ
= 0.0990                 Av =
1+ 10 4 (0.0990)
= 10.1

Ro        103
[               ]
Rin = Rid (1+ Aβ ) = 25kΩ 1+ 10 4 (0.0990) = 24.8 MΩ Rout =                                     =
1+ Aβ 1+ 10 4 (0.0990)
= 1.01 Ω

1kΩ + 25kΩ + 9.01kΩ                                      4870
(ii )    A = 4730
25kΩ + 9.01kΩ
= 4870                Av =
1+ 4870(0.0990)
= 10.1

Ro       662Ω
[                  ]
Rin = Rid (1+ Aβ ) = 34.0kΩ 1+ 4870(0.0990) = 16.4 MΩ Rout =                                       =
1+ Aβ 1+ 4870(0.0990)
= 1.37 Ω

A
(iii )   h12 = 0 (No reverse gain through the amplifier.)

A i
h = 2               =−
( ) = −2.5x10
25000 10 4           5         F
h21 =
i2
=−
10kΩ
= −0.0990
21
i1     v 2 =0
1000                                 i1   v 2 =0
10kΩ + 91kΩ

Page 1086
A   i                                     F      i2                  1
€   (i ) y21 = v2              = +g m = +39.1 mS    y21 =
v1 v
=−
RF
= −10−5 S
1 v    2 =0                                       2 =0

A
y = 0 (No reverse gain through the amplifier.)
12

4.76kΩ
(ii )    A=−
4.76kΩ + 3.84kΩ
(
(150) 10kΩ 100kΩ 5kΩ 52.6kΩ = −252 kΩ                        )
1                                   −252kΩ
β =−    = −10−5 S = −0.01 mS Atr =                     = −71.6 kΩ
RF                             1− 252kΩ(−0.01mS )
A
Rin   2.13kΩ                  Ro    10kΩ 100kΩ 5kΩ 52.6kΩ
Rin =      =        = 605 Ω Rout =       =                       = 863 Ω
1+ Aβ 1+ 2.52                1+ Aβ          1+ 2.52

€

52
Page 1096
vx

ib
ix
rπ                                   ro
+
β oib
-
Aov1 = 0

        β 
vx = (ix − β oib )ro + ix rπ       ib = −ix   v x = (ix − β oix )ro + ix rπ = ix ro  βo + 1+ o 

        µf 

vx             β 
Rout =        = ro βo + 1+ o  ≅ βoro
                          for βo >> 1 and µ f >> βo
ix             µf 


Page 1100
RE is now connected between the emitter and collector of Q2 in the upper half of
€                                                              F
Fig. 17.36. However, when the A - circuit is constructed, y22 is connected in parallel
with RE , so the resistance at the output of the A - circuit is still 901 Ω. The value of
F
y11 is unchanged, so overall the A - circuit is unchanged. Likewise, β is unchanged,
and so Atr is the same.

Page 1102
R1 Rid                                          vo         R1 Rid
€   ( i)   Using voltage division : v+ − v− = −vP1                                   v o = A(v + − v− )   T =−
v P1
=A
(
R2 + R1 Rid    )                                                (
R2 + R1 Rid   )
R1          Rid         R1          Rid                  R1 Rid
(ii) T = A R + R             Rid + R3
=A
R1 + R2          RR
=A
Rid ( R1 + R2 ) + R1 R2
1        2
Rid + 1 2
R1 + R2
R1 Rid
R1 Rid                R1 + Rid          R1 Rid
T=A                          =A                =A
R1 Rid + ( R1 + Rid ) R2     R1 Rid           R2 + R1 Rid
+ R2
R1 + Rid

€

53
Page 1108
                                        
RD = 50kΩ1+
100 5kΩ 25kΩ(         )            = 1.84 MΩ
 5kΩ + 2.5kΩ + 5kΩ 25kΩ
                        (           )   


TSC = 10 4
101(3.85kΩ)
= 9810        TOC = 10 4
(5kΩ 25kΩ)        = 3570
5kΩ + 2.5kΩ + 101(3.85kΩ)                                5kΩ + 2.5kΩ + (5kΩ 25kΩ)
 1+ 9810 
Rout = 1.84 MΩ          = 5.06 MΩ
 1+ 3570 

Page 1114
The analysis is the same with rπ and βo → ∞.
€
       β o3                       1                                    βo + 1   µ
RD = ro3 1+
 µ + β + 1 → 2ro3 TOC =
                           →0                  TSC =             → f
                                  βo + 1                                2β + 1   2
f2     o3
1+                                     1+ o
µf                                     µf
µf
1+
Rout = 2ro3      2 ≅µ r
f o3
1+ 0

Page 1116
f   10 7 Hz                             10 7 Hz
€    fB = T =         = 100 Hz      f H = βfT =           = 10 kHz
Ao   105                                 103

Page 1120
5
€   ∠T ( jω ) = −3tan −1 (ω ) = −π → ω = 1.732         T ( jω ) =              3
= 0.625
( ω + 1)
2

 1 
GM = 20 log        = 4.08 dB
 0.625 

Page 1122
From the Bode plot, the phase shift at the unity gain frequency (10 MHz) is -130o.
€   The phase margin is 180 o - 130 o = 50o.

54
Page 1124
1          1
( )(           )
Gm = g m2 = 2 10−3 5x10−5 = 0.316 mS             RO = ro4 ro2 ≅              =
2λI D 2(0.02)5x10−5
= 500 kΩ

1  Gm  1  0.316mS 
fT =       =               = 2.51 MHz
2π  CC  2π  20pF 
                            
                            
1         1         1          1           1               1

fB =                     =                      ≅                                = 158 Hz
2π  RO CC (1+ Av2 )  2π  RO CC (1+ µ f 2 )  2π                           
2(0.001) 
500kΩ(20pF)1+ 1
                                        

            0.02 5x10−4 
                          

Page 1126
1 g m5   1    2(0.001)5x10−4                                 1           1
€   fz ≅         =                          = 7.96 MHz           R=        =                = 1.00 kΩ
2π CC 2π             20x10   −12
g m5   2(0.001)5x10−4

Page 1127
100V
€                 (       )
Gm = g m2 = 40 5x10−5 = 2.00 mS         RO = rπ 5 ≅                           = 4.54 kΩ
(
40 5.5x10−4 A      )

fT =
1  Gm  1  2.00mS 
 =              = 10.6 MHz         fZ =
1  g m5  1  40 5x10
      =
−4
(     ) = 106 MHz

2π  CC  2π  30pF                           2π  CC  2π  3x10−11                
                     

1          1           1            1

fB =                        ≅                             = 584 Hz
2π  rπ 5CC (1+ µ f 2 )  2π 4.54kΩ(30pF) 1+ 40(50) 
                                         [                ]
                         

Page 1128
100µA            V        V
€   (ii ) SR = 20 pF = 5.00x106 s = 5.00 µs
Page 1129
100µA           V       V
€   SR =       = 5.00x106 = 5.00
20 pF           s       µs

Page 1132
1  Gm / 2  1  10.0mS 
€                 (           )
Gm = g m2 = 40 2.5x10−4 = 10.0 mS        fT =                =            = 15.7 MHz
2π  CC + Cµ3  4π  50.8pF 
          
 15.7 MHz         
−1 15.7 MHz
        
−1 15.7 MHz
        
−1 15.7 MHz
        o
φ M = 90 − tan−1           + tan            + tan            + tan            = 69.5
 142 MHz           173MHz           192 MHz          206 MHz 

€                                                                                                            55
Page 1137
I1         1mA               V       V
(i )   SR ≅           =         = 15.4x106 = 15.4
CC + CGD5 65 pF                    s      µs
    fT                 fT              fT   
(ii ) 30o = tan−1 49.2 MHz  + tan −1 82.1MHz  + tan −1100 MHz  → fT = 16.6 MHz
                                            
 8.5MHz 
CC = 65 pF               − 2 pF = 31.3 pF
 16.6 MHz 

Page 1144
1                                     3(0.6V )
€   fo =                   = 15.9kHz     vo =                         = 3.00 V
2π (10kΩ)(1nF )                       10kΩ  24kΩ  24kΩ
2 −   1+      −
 10kΩ  12kΩ  10kΩ

Page 1149
1                                                    1
€   fP =                                  = 5.016 MHz   fP =                                   = 5.008 MHz
 31.8 fF (7 pF )                                 31.8 fF (25 pF ) 
2π 31.8mH                                        2π 31.8mH                  
 7.0318 pF 
                                                  25.0318 pF 
                 

€

56

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