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Chapter 14
Exercise Solutions

EX14.1
                           −50
a.       ACL =                             ⇒ ACL = −49.949
                  ⎡ ⎛ 1 ⎞                ⎤
                  ⎢1 + ⎜ 5 × 104 ⎟ ( 51) ⎥
                  ⎣ ⎝            ⎠       ⎦
         dACL               51        dA
b.               = 10 ×          ⇒ CL = 0.0102%
          ACL            5 × 104       ACL
                       −50
         ACL =                    ⇒ ACL = −49.943
                 ⎡       51 ⎤
                 ⎢1 + 4.5 × 104 ⎥
                 ⎣              ⎦

EX14.2
a.      For R0 = 0

     = + (1 + 104 ) = 0.1 + 103
  1    1 1
 Ri f 10 10
⇒ Ri f = 10−3 kΩ = 1 Ω
b.       For R0 = 10 kΩ
 1    1 1 ⎡1 + 104 + 1 ⎤          104
    = + ×⎢             ⎥ ≅ 0.1 +
Ri f 10 10 ⎣ 1 + 1 + 1 ⎦         3 (10 )
Rif = 3 × 10−3 kΩ ⇒ Ri f = 3 Ω

EX14.3
                          ⎛ 40 ⎞
       40 (1 + 104 ) + 99 ⎜ 1 + ⎟
Ri f =                    ⎝    1 ⎠
                      99
                 1+
                       1
       4 × 10 + 4.059 × 103
             5
     ≅
                100
Ri f = 4.04 × 103 kΩ ⇒ Ri f = 4.04 MΩ

EX14.4
   R
1 + 2 = 100
   R1
          1    1 ⎡ 105 ⎤
a.           =    ⎢    ⎥ = 10 ⇒ R0 f = 0.1 Ω
         R0 f 100 ⎣100 ⎦


          1    1 ⎡ 105 ⎤
b.           = ⎢       ⎥ = 10
                              2

         R0 f 10 ⎣100 ⎦
R0 f = 10−2 kΩ ⇒ R0 f = 10 Ω

EX14.5
From Equation (14.43)
ACL 0
ACL ( f ) =
                               f
              1+ j ⋅
                       f PD ( A0 /ACL 0 )
                          25                         25
         =                                  =
                              f                             f
              1+ j ⋅                            1+ j ⋅
                     ( 50 ) (104 / 25)                   2 × 104
a.         f = 2 kHz
v0
   = 25 ⇒ v0 ( peak ) = 1.25 mV
vI
b.         f = 20 kHz
 v0   1
    =    ⋅ 25 ⇒ v0 ( peak ) = 0.884 mV
 vI    2
c.        f = 100 kHz
v0         25              25
   =                   =       = 4.90
vI   1 + (100 / 20 )     5.099
                     2


           ⇒ v0 = 0.245 mV

EX14.6
Full-scale response = 1× 5 = 5 V
     5
t=     ⇒ t = 2.5 μ s
     2

EX14.7
                              SR          0.63 × 106
a.        FPBW =                        =
                          2π V0 ( max )     2π (1)
          FPBW = 1.0 × 105 ⇒ FPBW = 100 kHz
                          0.63 × 106
b.        FPBW =                     = 1.0 × 104
                           2π (10 )
                   ⇒ FPBW = 10 kHz

EX14.8
             ⎛I ⎞                 ⎛ 1.85 × 10−14 ⎞
V0 S = VT ln ⎜ S 2 ⎟ = (0.026) ln ⎜         −14 ⎟
             ⎝ I S1 ⎠             ⎝ 2 × 10       ⎠
         ⇒ V0 S = 2.03 mV

EX14.9
We need
iC1 = iC 2 , vEC 3 = vEC 4 = 0.6 V, and vCE1 = vCE 2 = 10 V
By Equation (14.60(a))
            ⎡     ⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC1 = I S 1 ⎢exp ⎜ BE1 ⎟ ⎥ ⎜ 1 + ⎟
            ⎣     ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
            ⎡     ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
    = I S 3 ⎢ exp ⎜ EB 3 ⎟ ⎥ ⎜1 +    ⎟
            ⎣     ⎝ VT ⎠ ⎦ ⎝      50 ⎠
By Equation (14.60(b))
⎡     ⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC 2 = I S 2 ⎢ exp ⎜ BE 2 ⎟ ⎥ ⎜ 1 + ⎟
             ⎣     ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
             ⎡     ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
     = I S 4 ⎢ exp ⎜ EB 4 ⎟ ⎥ ⎜ 1 +    ⎟
             ⎣     ⎝ VT ⎠ ⎦ ⎝       50 ⎠
I S1 = I S 2, take the ratio:
    ⎛v −v ⎞ I
exp ⎜ BE1 BE 2 ⎟ = S 3
    ⎝    VT    ⎠ IS 4
                                ⎛I ⎞
vBE1 − vBE 2 = V0 S     = VT ln ⎜ S 3 ⎟
                                ⎝ IS 4 ⎠
                        = 0.026 ⋅ ln (1.05 )
                       ⇒ V0 S = 1.27 m V

EX14.10
           1   I Q ⎛ ΔK n ⎞
  VOS =      ⋅     ⋅⎜     ⎟
           2 2Kn ⎝ Kn ⎠
           1   150 ⎛ ΔK n ⎞
0.020 =      ⋅       ⋅
           2 2 ( 50 ) ⎜ 50 ⎟
                       ⎝   ⎠
       ⇒ ΔK n = 1.63μ A / V 2
           ΔK n 1.63
       ⇒       =     ⇒ 3.26%
            Kn   50

EX14.11
     ⎛ R5 ⎞ +
Want ⎜         ⎟V = 5 m V
     ⎝ R5 + R4 ⎠
               R5
R5     R4 so      × V + = 0.005
               R4
       ( 0.005)(100 )
R5 =                = 0.05 kΩ
            10
           ⇒ R5 = 50 Ω

EX14.12
R1′ = 25 1 = 0.9615 kΩ
 ′
R2 = 75 1 = 0.9868 kΩ
For I Q = 100 μ A ⇒ iC1 = iC 2 = 50 μ A
From Equation (14.75)
             ⎛ 50 × 10−6 ⎞
( 0.026 ) ln ⎜ −14 ⎟ + ( 0.050 )( 0.9615 )
             ⎝ 10        ⎠
                        ⎛i ⎞
         = ( 0.026 ) ln ⎜ C 2 ⎟ + ( 0.050 )( 0.9868 )
                        ⎝ IS 4 ⎠
0.58065 + 0.048075
                          ⎛i ⎞
           = ( 0.026 ) ln ⎜ C 2 ⎟ + 0.04934
                          ⎝ IS 4 ⎠
⎛i ⎞
ln ⎜ C 2 ⎟ = 22.284
   ⎝ IS 4 ⎠
50 × 10−6
            = 4.7625 × 109
    IS 4
            I S 4 ≅ 1.05 × 10−14 A

EX14.13
From Equation (14.79)
                        ⎛ R ⎞
v0 = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟
                        ⎝    R1 ⎠
For v0 = 0
                                              ⎛ 100 ⎞
0 = (1.1× 10−6 ) (100 kΩ ) − (1.0 × 10−6 ) R3 ⎜ 1 +    ⎟
                                              ⎝     10 ⎠
R3 (11) = (1.1)(100 kΩ ) ⇒ R3 = 10 kΩ

TYU14.1
v1CM ( max ) = V + − VSD1 ( sat ) − VSG1
v1CM ( min ) = V − + VDS 4 ( sat ) + VSD1 ( sat ) − VSG1
We have:
 I REF = 100 μ A, kn = 80 μ A / V 2 , k ′ = 40 μ A / V 2 ,
                   ′                    p

⎛W    ⎞
⎜     ⎟ = 25
⎝L    ⎠
For   M1 :
           ⎛ 40 ⎞
I D = 50 = ⎜ ⎟ ( 25 )(VSG1 + VTP )
                                   2

           ⎝ 2 ⎠
So 50 = 500 (VSG1 − 0.5 ) ⇒ VSG1 = 0.816 V
                             2


VSD1 ( sat ) = 0.816 − 0.5 = 0.316 V
Then
vCM ( max ) = V + − 0.316 − 0.816 = V + − 1.13 V
For M 4 :
            ⎛ 80 ⎞
I D = 100 = ⎜ ⎟ ( 25 )(VGS 4 − VTN )
                                     2

            ⎝ 2⎠
So 100 = 1000 (VGS 4 − 0.5 ) ⇒ VGS 4 = 0.816 V
                                 2


VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
VCM ( min ) = V − + 0.316 + 0.316 − 0.816 = V − − 0.184
So
V − − 0.184 ≤ vCM ≤ V + − 1.13 V

TYU14.2
vo ( max ) = V + − VSD 8 ( sat ) − VSG10
vo ( min ) = V − + VDS 4 ( sat ) + VDS 6 ( sat )
Now
50
VSG 8 = VSG10 =                  + 0.5 = 0.816 V
                    ( 40/ 2 )( 25)
VSD 8 ( sat ) = VSD10 ( sat ) = 0.316 V
So vo ( max ) = V + − 0.316 − 0.816 = V + − 1.13
Also
                50
VGS 6 =                  + 0.5 = 0.724 V
           (80/ 2 )( 25)
              100
VGS 4 =                  + 0.5 = 0.816 V
           (80/ 2 )( 25)
VDS 6 ( sat ) = 0.724 − 0.5 = 0.224 V
VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
So
vo ( min ) = V − + 0.316 + 0.224 = V − + 0.54
Then
V − + 0.54 ≤ vo ≤ V + − 1.13 V

TYU14.3
             500
ACL ( ideal ) = −= −25
              20
Within 0.1% ⇒ −25 + ( 0.001)( 25 ) ⇒ ACL = −24.975
                −25
−24.975 =
             ⎡    26 ⎤
             ⎢1 + A ⎥
             ⎣     0L ⎦

26     −25
    =        − 1 = 0.0010
A0 L −24.975
          A0 L = 25.974

TYU14.4
                     ACL ( ∞ )
a.         ACL =
                     ⎡ A (∞) ⎤
                 1 + ⎢ CL       ⎥
                     ⎣ A0 L ⎦
               R         495
ACL ( ∞ ) = 1 + 2 = 1 +        = 100
               R1          5
         100
ACL =           ⇒ ACL = 99.90
           100
       1+ 5
           10
                      ACL ( ∞ ) = 100
           dACL       100
b.              = 10 × 5 = 0.01%
            ACL       10
ACL = 99.90 − ( 0.0001)( 99.90 )
       ⇒ ACL = 99.89

TYU14.5
ACL ( ∞ ) − ACL                ACL                1
                      = 1−             = 1−
     ACL ( ∞ )               ACL ( ∞ )           A (∞)
                                              1 + CL
                                                   A0 L
              ACL ( ∞ )         ACL ( ∞ )
                 1+     −1
                A0 L              A0 L
So 0.001 =                 =
               A (∞)             A (∞)
           1 + CL            1 + CL
                  A0 L              A0 L
                      ACL ( ∞ )
0.001 = 0.999 ⋅
                        A0 L
              0.001          0.001
ACL ( ∞ ) =         ⋅ A0 L =       ⋅ (104 ) ⇒ ACL ( ∞ ) = 10.010
              0.999          0.999
or
 ACL = (1 − 0.001)(10.010 ) ⇒ ACL = 10.0

TYU14.6
 ii ⎛ Rif ⎞
   =⎜      ⎟
 I1 ⎝ Ri ⎠
          iI 0.1
a.           =   = 1× 10−5
          i1 104
            iI   10
b.             =    = 1× 10−3
            i1 104

TYU14.7
Voltage follower R2 = 0, R1 = ∞
Rif = Ri (1 + A0 L ) = 10 (1 + 5 × 105 )
     ≅ 5 × 106 kΩ ⇒ Rif = 5000 MΩ

TYU14.8

 f3dB =
           fT
                =
                  (105 ) (10 ) ⇒ 20 kHz
          ACL 0      50
                      SR
f max = f 3dB =
                  2π V0 ( max )
                  SR       0.8 × 106
V0 ( max ) =            =
                 2π f3dB 2π ( 20 × 103 )
           ⇒ V0 ( max ) = 6.37 V

TYU14.9
a.     v0 = I B1 R3 = (10−6 )( 200 × 103 )
                         ⇒ v0 = 0.20 V
b.          R4 = R1 R2 R3 = 100 50 200
                         ⇒ R4 = 28.6 kΩ

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Ch14p

  • 1. Chapter 14 Exercise Solutions EX14.1 −50 a. ACL = ⇒ ACL = −49.949 ⎡ ⎛ 1 ⎞ ⎤ ⎢1 + ⎜ 5 × 104 ⎟ ( 51) ⎥ ⎣ ⎝ ⎠ ⎦ dACL 51 dA b. = 10 × ⇒ CL = 0.0102% ACL 5 × 104 ACL −50 ACL = ⇒ ACL = −49.943 ⎡ 51 ⎤ ⎢1 + 4.5 × 104 ⎥ ⎣ ⎦ EX14.2 a. For R0 = 0 = + (1 + 104 ) = 0.1 + 103 1 1 1 Ri f 10 10 ⇒ Ri f = 10−3 kΩ = 1 Ω b. For R0 = 10 kΩ 1 1 1 ⎡1 + 104 + 1 ⎤ 104 = + ×⎢ ⎥ ≅ 0.1 + Ri f 10 10 ⎣ 1 + 1 + 1 ⎦ 3 (10 ) Rif = 3 × 10−3 kΩ ⇒ Ri f = 3 Ω EX14.3 ⎛ 40 ⎞ 40 (1 + 104 ) + 99 ⎜ 1 + ⎟ Ri f = ⎝ 1 ⎠ 99 1+ 1 4 × 10 + 4.059 × 103 5 ≅ 100 Ri f = 4.04 × 103 kΩ ⇒ Ri f = 4.04 MΩ EX14.4 R 1 + 2 = 100 R1 1 1 ⎡ 105 ⎤ a. = ⎢ ⎥ = 10 ⇒ R0 f = 0.1 Ω R0 f 100 ⎣100 ⎦ 1 1 ⎡ 105 ⎤ b. = ⎢ ⎥ = 10 2 R0 f 10 ⎣100 ⎦ R0 f = 10−2 kΩ ⇒ R0 f = 10 Ω EX14.5 From Equation (14.43)
  • 2. ACL 0 ACL ( f ) = f 1+ j ⋅ f PD ( A0 /ACL 0 ) 25 25 = = f f 1+ j ⋅ 1+ j ⋅ ( 50 ) (104 / 25) 2 × 104 a. f = 2 kHz v0 = 25 ⇒ v0 ( peak ) = 1.25 mV vI b. f = 20 kHz v0 1 = ⋅ 25 ⇒ v0 ( peak ) = 0.884 mV vI 2 c. f = 100 kHz v0 25 25 = = = 4.90 vI 1 + (100 / 20 ) 5.099 2 ⇒ v0 = 0.245 mV EX14.6 Full-scale response = 1× 5 = 5 V 5 t= ⇒ t = 2.5 μ s 2 EX14.7 SR 0.63 × 106 a. FPBW = = 2π V0 ( max ) 2π (1) FPBW = 1.0 × 105 ⇒ FPBW = 100 kHz 0.63 × 106 b. FPBW = = 1.0 × 104 2π (10 ) ⇒ FPBW = 10 kHz EX14.8 ⎛I ⎞ ⎛ 1.85 × 10−14 ⎞ V0 S = VT ln ⎜ S 2 ⎟ = (0.026) ln ⎜ −14 ⎟ ⎝ I S1 ⎠ ⎝ 2 × 10 ⎠ ⇒ V0 S = 2.03 mV EX14.9 We need iC1 = iC 2 , vEC 3 = vEC 4 = 0.6 V, and vCE1 = vCE 2 = 10 V By Equation (14.60(a)) ⎡ ⎛ v ⎞ ⎤ ⎛ 10 ⎞ iC1 = I S 1 ⎢exp ⎜ BE1 ⎟ ⎥ ⎜ 1 + ⎟ ⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠ ⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞ = I S 3 ⎢ exp ⎜ EB 3 ⎟ ⎥ ⎜1 + ⎟ ⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠ By Equation (14.60(b))
  • 3. ⎛ v ⎞ ⎤ ⎛ 10 ⎞ iC 2 = I S 2 ⎢ exp ⎜ BE 2 ⎟ ⎥ ⎜ 1 + ⎟ ⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠ ⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞ = I S 4 ⎢ exp ⎜ EB 4 ⎟ ⎥ ⎜ 1 + ⎟ ⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠ I S1 = I S 2, take the ratio: ⎛v −v ⎞ I exp ⎜ BE1 BE 2 ⎟ = S 3 ⎝ VT ⎠ IS 4 ⎛I ⎞ vBE1 − vBE 2 = V0 S = VT ln ⎜ S 3 ⎟ ⎝ IS 4 ⎠ = 0.026 ⋅ ln (1.05 ) ⇒ V0 S = 1.27 m V EX14.10 1 I Q ⎛ ΔK n ⎞ VOS = ⋅ ⋅⎜ ⎟ 2 2Kn ⎝ Kn ⎠ 1 150 ⎛ ΔK n ⎞ 0.020 = ⋅ ⋅ 2 2 ( 50 ) ⎜ 50 ⎟ ⎝ ⎠ ⇒ ΔK n = 1.63μ A / V 2 ΔK n 1.63 ⇒ = ⇒ 3.26% Kn 50 EX14.11 ⎛ R5 ⎞ + Want ⎜ ⎟V = 5 m V ⎝ R5 + R4 ⎠ R5 R5 R4 so × V + = 0.005 R4 ( 0.005)(100 ) R5 = = 0.05 kΩ 10 ⇒ R5 = 50 Ω EX14.12 R1′ = 25 1 = 0.9615 kΩ ′ R2 = 75 1 = 0.9868 kΩ For I Q = 100 μ A ⇒ iC1 = iC 2 = 50 μ A From Equation (14.75) ⎛ 50 × 10−6 ⎞ ( 0.026 ) ln ⎜ −14 ⎟ + ( 0.050 )( 0.9615 ) ⎝ 10 ⎠ ⎛i ⎞ = ( 0.026 ) ln ⎜ C 2 ⎟ + ( 0.050 )( 0.9868 ) ⎝ IS 4 ⎠ 0.58065 + 0.048075 ⎛i ⎞ = ( 0.026 ) ln ⎜ C 2 ⎟ + 0.04934 ⎝ IS 4 ⎠
  • 4. ⎛i ⎞ ln ⎜ C 2 ⎟ = 22.284 ⎝ IS 4 ⎠ 50 × 10−6 = 4.7625 × 109 IS 4 I S 4 ≅ 1.05 × 10−14 A EX14.13 From Equation (14.79) ⎛ R ⎞ v0 = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟ ⎝ R1 ⎠ For v0 = 0 ⎛ 100 ⎞ 0 = (1.1× 10−6 ) (100 kΩ ) − (1.0 × 10−6 ) R3 ⎜ 1 + ⎟ ⎝ 10 ⎠ R3 (11) = (1.1)(100 kΩ ) ⇒ R3 = 10 kΩ TYU14.1 v1CM ( max ) = V + − VSD1 ( sat ) − VSG1 v1CM ( min ) = V − + VDS 4 ( sat ) + VSD1 ( sat ) − VSG1 We have: I REF = 100 μ A, kn = 80 μ A / V 2 , k ′ = 40 μ A / V 2 , ′ p ⎛W ⎞ ⎜ ⎟ = 25 ⎝L ⎠ For M1 : ⎛ 40 ⎞ I D = 50 = ⎜ ⎟ ( 25 )(VSG1 + VTP ) 2 ⎝ 2 ⎠ So 50 = 500 (VSG1 − 0.5 ) ⇒ VSG1 = 0.816 V 2 VSD1 ( sat ) = 0.816 − 0.5 = 0.316 V Then vCM ( max ) = V + − 0.316 − 0.816 = V + − 1.13 V For M 4 : ⎛ 80 ⎞ I D = 100 = ⎜ ⎟ ( 25 )(VGS 4 − VTN ) 2 ⎝ 2⎠ So 100 = 1000 (VGS 4 − 0.5 ) ⇒ VGS 4 = 0.816 V 2 VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V VCM ( min ) = V − + 0.316 + 0.316 − 0.816 = V − − 0.184 So V − − 0.184 ≤ vCM ≤ V + − 1.13 V TYU14.2 vo ( max ) = V + − VSD 8 ( sat ) − VSG10 vo ( min ) = V − + VDS 4 ( sat ) + VDS 6 ( sat ) Now
  • 5. 50 VSG 8 = VSG10 = + 0.5 = 0.816 V ( 40/ 2 )( 25) VSD 8 ( sat ) = VSD10 ( sat ) = 0.316 V So vo ( max ) = V + − 0.316 − 0.816 = V + − 1.13 Also 50 VGS 6 = + 0.5 = 0.724 V (80/ 2 )( 25) 100 VGS 4 = + 0.5 = 0.816 V (80/ 2 )( 25) VDS 6 ( sat ) = 0.724 − 0.5 = 0.224 V VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V So vo ( min ) = V − + 0.316 + 0.224 = V − + 0.54 Then V − + 0.54 ≤ vo ≤ V + − 1.13 V TYU14.3 500 ACL ( ideal ) = −= −25 20 Within 0.1% ⇒ −25 + ( 0.001)( 25 ) ⇒ ACL = −24.975 −25 −24.975 = ⎡ 26 ⎤ ⎢1 + A ⎥ ⎣ 0L ⎦ 26 −25 = − 1 = 0.0010 A0 L −24.975 A0 L = 25.974 TYU14.4 ACL ( ∞ ) a. ACL = ⎡ A (∞) ⎤ 1 + ⎢ CL ⎥ ⎣ A0 L ⎦ R 495 ACL ( ∞ ) = 1 + 2 = 1 + = 100 R1 5 100 ACL = ⇒ ACL = 99.90 100 1+ 5 10 ACL ( ∞ ) = 100 dACL 100 b. = 10 × 5 = 0.01% ACL 10 ACL = 99.90 − ( 0.0001)( 99.90 ) ⇒ ACL = 99.89 TYU14.5
  • 6. ACL ( ∞ ) − ACL ACL 1 = 1− = 1− ACL ( ∞ ) ACL ( ∞ ) A (∞) 1 + CL A0 L ACL ( ∞ ) ACL ( ∞ ) 1+ −1 A0 L A0 L So 0.001 = = A (∞) A (∞) 1 + CL 1 + CL A0 L A0 L ACL ( ∞ ) 0.001 = 0.999 ⋅ A0 L 0.001 0.001 ACL ( ∞ ) = ⋅ A0 L = ⋅ (104 ) ⇒ ACL ( ∞ ) = 10.010 0.999 0.999 or ACL = (1 − 0.001)(10.010 ) ⇒ ACL = 10.0 TYU14.6 ii ⎛ Rif ⎞ =⎜ ⎟ I1 ⎝ Ri ⎠ iI 0.1 a. = = 1× 10−5 i1 104 iI 10 b. = = 1× 10−3 i1 104 TYU14.7 Voltage follower R2 = 0, R1 = ∞ Rif = Ri (1 + A0 L ) = 10 (1 + 5 × 105 ) ≅ 5 × 106 kΩ ⇒ Rif = 5000 MΩ TYU14.8 f3dB = fT = (105 ) (10 ) ⇒ 20 kHz ACL 0 50 SR f max = f 3dB = 2π V0 ( max ) SR 0.8 × 106 V0 ( max ) = = 2π f3dB 2π ( 20 × 103 ) ⇒ V0 ( max ) = 6.37 V TYU14.9 a. v0 = I B1 R3 = (10−6 )( 200 × 103 ) ⇒ v0 = 0.20 V b. R4 = R1 R2 R3 = 100 50 200 ⇒ R4 = 28.6 kΩ