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Chapter 6
Exercise Problems

EX6.1
                        VBB − VBE ( on )       2 − 0.7
(a)            I BQ =                      =           ⇒ 2 μA
                              RB                650
Then
 I CQ = β I BQ = (100 )( 2 μ A ) = 0.20 mA
VCEQ = VCC − I CQ RC = 5 − ( 0.2 )(15 ) = 2 V
(b)        Now
       I CQ    0.20
 gm =       =        = 7.69 mA / V
       VT     0.026
and
      β VT (100 )( 0.026 )
rπ =        =               = 13 k Ω
      I CQ         0.20
(c)
We find
               ⎛ r        ⎞
Av = − g m RC ⎜ π ⎟
               ⎝ rπ + RB ⎠
                      ⎛ 13 ⎞
    = − ( 7.69 )(15 ) ⎜          ⎟ = −2.26
                      ⎝ 13 + 650 ⎠
EX6.2
       V − VBE ( on ) 0.92 − 0.7
I BQ = BB                =
              RB               100
or I BQ = 0.0022 mA
and I CQ = β I BQ = (150 )( 0.0022 ) = 0.33 mA
(a)
        I CQ        0.33
gm =           =         = 12.7 mA / V
        VT         0.026
        β VT        (150 )( 0.026 )
 rπ =           =                     = 11.8 k Ω
        I CQ              0.33
        VA    200
 ro =       =     = 606 k Ω
        I CQ 0.33
(b)
                                        ⎛ rπ ⎞
vo = − g m vπ ( ro        RC ) and vπ = ⎜         ⎟ ⋅ vs
                                        ⎝ rπ + RB ⎠
so
       vo           ⎛ rπ ⎞
Av =       = − gm ⎜           ⎟ ( ro RC )
       vπ           ⎝ rπ + RB ⎠
                 ⎛ 11.8 ⎞
     = − (12.7 ) ⎜             ⎟ ( 606 15 )
                 ⎝ 11.8 + 100 ⎠
 which yields Av = −19.6
EX6.3
(a)
        V − VEB ( on ) 1.145 − 0.7
 I BQ = BB                =
              RB                  50
or I BQ = 0.0089 mA
Then I CQ = β I BQ = ( 90 )( 0.0089 ) = 0.801 mA
Now
        I CQ       0.801
gm =           =         = 30.8 mA / V
        VT         0.026
        β VT       ( 90 )( 0.026 )
 rπ =          =                     = 2.92 k Ω
        I CQ           0.801
        VA    120
 ro =       =      = 150 k Ω
        I CQ 0.801
(b)          We have Vo = g mVπ ( ro RC )
           ⎛ r         ⎞
and Vπ = − ⎜ π ⎟ Vs
           ⎝ rπ + RB ⎠
so
     V           ⎛ rπ ⎞
 Av = o = − g m ⎜           ⎟ ( ro RC )
     Vs          ⎝ rπ + RB ⎠
                      ⎛ 2.92 ⎞
         = − ( 30.8 ) ⎜            ⎟ (150 2.5 )
                      ⎝ 2.92 + 50 ⎠
 which yields Av = −4.18
EX6.4
Using Figure 6.23
(a)      For I CQ = 0.2 mA, 7.8 < hie < 15 k Ω, 60 < h fe < 125, 6.2 × 10−4 < hre < 50 × 10−4 ,
5 < hoe < 13 μ mhos
(b)          For I CQ = 5 mA, 0.7 < hie < 1.1 k Ω, 140 < h fe < 210, 1.05 × 10−4 < hre < 1.6 ×10−4 ,
22 < hoe < 35 μ mhos
EX6.5
RTH = R1 R2 = 35.2 || 5.83 = 5 k Ω
       ⎛ R2 ⎞              ⎛ 5.83 ⎞
VTH = ⎜          ⎟ ⋅ VCC = ⎜             ⎟ (5)
       ⎝ R1 + R2 ⎠         ⎝ 5.83 + 35.2 ⎠
or
VTH = 0.7105 V
Then
       V − VBE ( on ) 0.7105 − 0.7
I BQ = TH               =
             RTH                5
or
I BQ = 2.1 μ A
and
 I CQ = β I BQ = (100 )( 2.1 μ A ) = 0.21 mA
VCEQ = VCC − I CQ RC = 5 − ( 0.21)(10 )
and
VCEQ = 2.9 V
Now
        I CQ        0.21
gm =           =         = 8.08 mA
        VT         0.026
        VA    100
 ro =       =     = 476 k Ω
        I CQ 0.21
And
Av = − g m ( ro      RC ) = − ( 8.08 ) ( 476 10 )
so
 Av = −79.1
EX6.6
RTH = R1 R2 = 250 75 = 57.7 k Ω
      ⎛ R2 ⎞               ⎛ 75 ⎞
VTH = ⎜         ⎟ (VCC ) = ⎜          ⎟ (5)
      ⎝ R1 + R2 ⎠          ⎝ 75 + 250 ⎠
or
VTH = 1.154 V
          VTH − VBE ( on )           1.154 − 0.7
I BQ =                        =
         RTH + (1 + β ) RE        57.7 + (121)( 0.6 )
or
I BQ = 3.48 μ A
I CQ = β I BQ = (120 )( 3.38 μ A ) = 0.418 mA
(a)
Now
         I CQ       0.418
gm =            =         = 16.08 mA / V
         VT         0.026
         βVT (120 )( 0.026 )
 rπ =         =              = 7.46 k Ω
         I CQ   0.418
We have
Vo = − g mVπ RC
We find
Rib = rπ + (1 + β ) RE = 7.46 + (121)( 0.6 )
or
Rib = 80.1 k Ω
Also
R1 R2 = 250 75 = 57.7 k Ω
R1 R2 Rib = 57.7 80.1 = 33.54 k Ω
We find
      ⎛ R1 R2 Rib ⎞             ⎛ 33.54 ⎞
Vs′ = ⎜                ⎟ ⋅ Vs = ⎜             ⎟ ⋅ Vs
      ⎝ R1 R2 Rib + RS ⎠        ⎝ 33.54 + 0.5 ⎠
or
Vs′ = ( 0.985 ) Vs
Now
         ⎡ ⎛1+ β ⎞ ⎤              ⎡ ⎛ 121 ⎞            ⎤
Vs′ = Vπ ⎢1 + ⎜      ⎟ RE ⎥ = Vπ ⎢1 + ⎜      ⎟ ( 0.6 ) ⎥
         ⎢ ⎝
         ⎣      rπ ⎠ ⎥     ⎦      ⎣   ⎝ 7.46 ⎠         ⎦
or
Vπ = ( 0.0932 ) Vs′ = ( 0.0932 )( 0.985 ) Vs
So
        Vo
Av =       = − (16.08 )( 0.0932 )( 0.985 )( 5.6 )
        Vs
or
 Av = −8.27
EX6.7
RTH = R1 R2 = 15 85 = 12.75 k Ω
      ⎛ R2 ⎞                ⎛ 85 ⎞
VTH = ⎜           ⎟ ⋅ VCC = ⎜         ⎟ (12 )
      ⎝ R1 + R2 ⎠           ⎝ 15 + 85 ⎠
or
VTH = 10.2 V
Now
VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
so
12 = (101) I BQ ( 0.5 ) + 0.7 + I BQ (12.75 ) + 10.2
which yields
I BQ = 0.0174 mA
and
 I CQ = β I BQ = (100 )( 0.0174 ) = 1.74 mA
 I EQ = (101)( 0.0174 ) = 1.76 mA
VECQ = VCC − I EQ RE − I CQ RC
      = 12 − (1.76 )( 0.5 ) − (1.74 )( 4 )
or
VECQ = 4.16 V
Now
        β VT       (100 )( 0.026 )
rπ =           =                     = 1.49 k Ω
        I CQ           1.74
We have
Vo = g mVπ ( RC || RL )
           ⎛        V ⎞
Vs = −Vπ − ⎜ g mVπ + π ⎟ RE
           ⎝         rπ ⎠
Solving for Vπ and noting that β = g m rπ , we find
        Vo         − β ( RC   RL )
Av =           =
        Vs         rπ + (1 + β ) RE
                     − (100 )( 4 2 )
               =
                   1.49 + (101)( 0.5 )
or Av = −2.56
EX6.8
Dc analysis
           10 − 0.7
I BQ =                   = 0.00439 mA
       100 + (101)( 20 )
I CQ = 0.439 mA, I EQ = 0.443 mA
Now
        (100 )( 0.026 )
 rπ =                = 5.92 k Ω
           0.439
       0.439
 gm =         = 16.88 mA / V
       0.026
        100
  ro =        = 228 k Ω
       0.439
(a)
Now
Vo = − g mVπ ( ro RC )
     ⎛ RB rπ ⎞
Vπ = ⎜            ⎟ ⋅ Vs
     ⎝ RB rπ + RS ⎠
RB rπ = 100 5.92 = 5.59 k Ω
          ⎛ 5.59 ⎞
Then Vπ = ⎜            ⎟ ⋅ Vs = ( 0.918 ) Vs
          ⎝ 5.59 + 0.5 ⎠
          V
Then Av = o = − (16.88 ) ( 0.918 ) ( 228 10 )
          Vs
or Av = −148
(b)      Rin = RS + RB rπ = 0.5 + 100 5.92 = 6.09 k Ω
and
 Ro = RC ro = 10 228 = 9.58 k Ω
EX6.9
(a)
      I CQ    0.25
 gm =      =       = 9.615 mA / V
      VT     0.026
       VA    100
ro =       =     = 400 k Ω
       I CQ 0.25
 Av = − g m ( ro rc ) = − ( 9.615 )( 400 100 ) = −769
(b)
 Av = − g m ( ro rc rL ) = − ( 9.615 )( 400 100 100 )
 Av = −427
EX6.10
             5 − 0.7
  I BQ =                  = 0.00672 mA
         10 + (126 )( 5 )
 I CQ = 0.84 mA, I EQ = 0.847 mA
VCEQ = 10 − ( 0.84 )( 2.3) − ( 0.847 )( 5 )
or
VCEQ = 3.83 V
dc load line
VCE ≅ (V + − V − ) − I C ( RC + RE )
or
VCE = 10 − I C ( 7.3)
ac load line (neglecting ro)
vce = −ic ( RC RL ) = −ic ( 2.3 5 ) = −ic (1.58 )

IC (mA)


   1.37                 AC load line



                            Q-point
   0.84


                                          DC load line




                         3.83                            10 V
                                                            CE (V)


EX6.11
(a)
dc load line
VEC ≅ VCC − I C ( RC + RE ) = 12 − I C ( 4.5 )
ac load line
vec ≅ −ic ( RE + RC     RL ) or vec = −ic ( 0.5 + 4 2 ) = −ic (1.83)
Q-point values
         12 − 0.7 − 10.2
I BQ =                      = 0.0150 mA
       12.75 + (121)( 0.5 )
I CQ = 1.80 mA, I EQ = 1.82 mA
VECQ = 3.89 V
IC (mA)


      2.67            AC load line




                              Q-point
      1.80




                                               DC load line




                           3.89                               12 V (V)
                                                                  EC


(b)          ΔiC = I CQ = 1.80 mA
             ΔvEC = (1.8 )(1.83) = 3.29 V
             vEC ( min ) = 3.89 − 3.29 = 0.6 V
So maximum symmetrical swing = 2 × 3.29 = 6.58 V peak-to-peak
EX6.12
              ⎛ R2 ⎞             1
(a)     VTH = ⎜         ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
              ⎝ R1 + R2 ⎠        R1
             RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)(1)
so
                            1
RTH = 12.1 k Ω, VTH =          (12.1)(12 )
                            R1
We can write
VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
We have
                         1.6
I CQ = 1.6 mA, I BQ =        = 0.0133 mA
                         120
Then
                                1
                    12 − 0.7 −    (12.1)(12 )
                               R1
I BQ = 0.0133 =
                         12.1 + (121)(1)
which yields
R1 = 15.24 k Ω
Since RTH = R1 R2 = 12.1 k Ω, we find R2 = 58.7 k Ω
Also VECQ = 12 − (1.6 )( 4 ) − (1.61)(1) = 3.99 V
(b)          ac load line vec = −ic ( RC     RL )
Want Δic = I CQ − 0.1 = 1.6 − 0.1 = 1.5 mA
Also Δvec = 3.99 − 0.5 = 3.49 V
          Δvec 3.49
Now           =     = 2.327 k Ω = RC          RL
          Δic   1.5
So 4 RL = 2.327 k Ω which yields RL = 5.56 k Ω
EX6.13
RTH = R1 R2 = 25 50 = 16.7 k Ω
      ⎛ R2 ⎞              ⎛ 50 ⎞
VTH = ⎜         ⎟ ⋅ VCC = ⎜         ⎟ (5)
      ⎝ R1 + R2 ⎠         ⎝ 25 + 50 ⎠
or
VTH = 3.33 V
           VTH − VBE ( on )            3.33 − 0.7
I BQ =                           =
          RTH + (1 + β ) RE          16.7 + (121)(1)
or
I BQ = 0.0191 mA
Also
I CQ = (120 )( 0.0191) = 2.29 mA
Now
          I CQ        2.29
gm =             =         = 88.1 mA / V
          VT         0.026
          β VT       (120 )( 0.026 )
 rπ =            =                     = 1.36 k Ω
          I CQ            2.29
          VA    100
 ro =         =     = 43.7 k Ω
          I CQ 2.29
(a)
      ⎛ R1 R2 Rib ⎞
Vs′ = ⎜                    ⎟ ⋅ Vs
      ⎝ R1 R2 Rib + RS ⎠
Rib = rπ + (1 + β ) ( RE ro ) = 1.36 + (121)(1 43.7 )
or
Rib = 120 k Ω and R1 R2 = 16.7 k Ω
Then
R1 R2 Rib = 16.7 120 = 14.7 k Ω
Now
      ⎛ 14.7 ⎞
Vs′ = ⎜            ⎟ ⋅ Vs = ( 0.967 ) Vs
      ⎝ 14.7 + 0.5 ⎠
and
      ⎛V            ⎞                  ⎛ 1+ β       ⎞
Vo = ⎜ π + g mVπ ⎟ ( RE ro ) = Vπ ⎜                 ⎟ RE ro
      ⎝ rπ          ⎠                  ⎝ rπ         ⎠
We have
Vs′ = Vπ + Vo
then
            Vs′            ( 0.967 )Vs
Vπ =                 =
       ⎛ 1+ β ⎞           ⎛1+ β ⎞
    1+ ⎜      ⎟ RE ro 1 + ⎜      ⎟ RE ro
       ⎝  rπ ⎠            ⎝ rπ ⎠
We then obtain
                            ⎛1+ β     ⎞
                 ( 0.967 ) ⎜          ⎟ RE   ro
     V                      ⎝ rπ      ⎠
 Av = o =
     Vs                ⎛1+ β ⎞
                   1+ ⎜       ⎟ RE ro
                       ⎝ rπ ⎠
          ( 0.967 )(1 + β ) RE ro
      =
            rπ + (1 + β ) RE     ro
Now
RE ro = 1 43.7 = 0.978 k Ω
Then
       ( 0.967 )(121)( 0.978 )
Av =                           = 0.956
        1.36 + (121)( 0.978 )
(b)
 Rib = rπ + (1 + β )( RE ro )
or
Rib = 1.36 + (121)( 0.978 ) = 120 k Ω
EX6.14
For RS = 0, then
               r
 Ro = RE ro π
              1+ β
Using the parameters from Exercise 4.13, we obtain
             1.36
 Ro = 1 43.7
              121
or
 Ro = 11.1 Ω
EX6.15
(a)      For I CQ = 1.25 mA and β = 100, we find
               I EQ = 1.26 mA and I BQ = 0.0125 mA
Now VCEQ = 10 − I EQ RE
Or
4 = 10 − (1.26 ) RE
which yields
RE = 4.76 k Ω
Then
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 4.76 )
or
RTH = 48.1 k Ω
We have
      ⎛ R2 ⎞                 1
VTH = ⎜         ⎟ (10 ) − 5 = ⋅ RTH (10 ) − 5
      ⎝ R1 + R2 ⎠            R1
or
        1
VTH =      ( 481) − 5
        R1
                        VTH − 0.7 − ( −5 )
We can write I BQ =
                        RTH + (1 + β ) RE
Or
         1
            ( 481) − 5 − 0.7 + 5
         R1
0.0125 =
          48.1 + (101)( 4.76 )
which yields
 R1 = 65.8 k Ω
Since R1 R2 = 48.1 k Ω, we obtain
 R2 = 178.8 k Ω
(b)
β VT       (100 )( 0.026 )
rπ =          =                     = 2.08 k Ω
       I CQ           1.25
       VA    125
ro =       =     = 100 k Ω
       I CQ 1.25
We may note that
g mVπ = g m ( I b rπ ) = β I b
Also
Rib = rπ + (1 + β )( RE RL ro )
    = 2.08 + (101) ( 4.76 1 100 )
or
Rib = 84.9 k Ω
Now
     ⎛ RE ro ⎞
     ⎜ R r + R ⎟(
Io = ⎜              1 + β ) Ib
                 ⎟
     ⎝ E o     L ⎠

where
     ⎛ R1 R2 ⎞
Ib = ⎜              ⋅I
     ⎜R R +R ⎟ s  ⎟
     ⎝ 1 2     ib ⎠

We can then write
      I   ⎛ RE ro ⎞                ⎛ R1 R2             ⎞
 AI = o = ⎜             ⎟ (1 + β ) ⎜                   ⎟
      I s ⎜ RE ro + RL ⎟
          ⎝             ⎠
                                   ⎜R R +R
                                   ⎝ 1 2   ib
                                                       ⎟
                                                       ⎠
We have
RE ro = 4.76 100 = 4.54 k Ω
so
      ⎛ 4.54 ⎞           ⎛    48.1 ⎞
 AI = ⎜          ⎟ (101) ⎜             ⎟
      ⎝ 4.54 + 1 ⎠       ⎝ 48.1 + 84.9 ⎠
or
AI = 29.9
                            r            2.08
(c)        Ro = RE ro π = 4.76 100
                          1+ β           101
or Ro = 20.5 Ω
EX6.16
(a)
 RTH = R1 R2 = 70 6 = 5.53 k Ω
        ⎛ R2 ⎞                  ⎛ 6 ⎞
VTH = ⎜           ⎟ (10 ) − 5 = ⎜        ⎟ (10 ) − 5
        ⎝ R1 + R2 ⎠             ⎝ 70 + 6 ⎠
or
VTH = −4.2105 V
We find
        −4.2105 − 0.7 − ( −5 )
I BQ1 =                           ⇒ 2.91 μ A
          5.53 + (126 )( 0.2 )
and
I CQ1 = β I BQ1 = (125 )( 2.91 μ A ) = 0.364 mA
I EQ1 = (1 + β ) I BQ1 = 0.368 mA
At the collector of Q1,
5 − VC1          V − 0.7 − ( −5 )
        = I CQ1 + C1
  RC1             (1 + β )( RE 2 )
or
5 − VC1          V − 0.7 − ( −5 )
        = 0.364 + C1
   5               (126 )(1.5)
which yields
VC1 = 2.99 V
also
VE1 = I EQ1 RE1 − 5 = ( 0.368 )( 0.2 ) − 5
or
VE1 = −4.93 V
Then
VCEQ1 = VC1 − VE1 = 2.99 − ( −4.93) = 7.92 V
We find
         V − 0.7 − ( −5 )
I EQ 2 = C1                  = 4.86 mA
               1.5
and
         ⎛ β ⎞              ⎛ 125 ⎞
I CQ 2 = ⎜      ⎟ ⋅ I EQ1 = ⎜     ⎟ ( 4.86 ) = 4.82 mA
         ⎝ 1+ β ⎠           ⎝ 126 ⎠
We find
VE 2 = VC1 − 0.7 = 2.99 − 0.7 = 2.29 V
and
VCEQ 2 = 5 − VE 2 = 5 − 2.29 = 2.71 V
(b)
 The small-signal transistor parameters are:
         β VT          (125 )( 0.026 )
rπ 1 =            =                      = 8.93 k Ω
         I CQ1             0.364
          I CQ1        0.364
g m1 =             =         = 14.0 mA / V
          VT           0.026
         β VT          (125 )( 0.026 )
rπ 2 =             =                     = 0.674 k Ω
         I CQ 2             4.82
          I CQ 2         4.82
gm2 =              =          = 185 mA / V
           VT           0.026
We find Rib1 = rπ 1 + (1 + β ) RE1 = 8.93 + (126 )( 0.2 )
Or
Rib1 = 34.1 k Ω
and
Rib 2 = rπ 2 + (1 + β )( RE 2 RL )
      = 0.674 + (126 )(1.5 10 ) = 165 k Ω
The small-signal equivalent circuit is:
ϩ                                 ϩ
                                  V␲1    r␲1           gm1V␲1     V␲2     r␲2          gm2V␲2
                                    Ϫ                                 Ϫ


Vs     ϩ                 R1͉͉R2                                 RC1                        Vo
       Ϫ



                                                 RE1                            RE2   RL




We can write
Vo = (1 + β ) I b 2 ( RE 2 RL )
where
      ⎛ RC1 ⎞
Ib2 = ⎜             ⎟ ( − g m1Vπ 1 )
      ⎝ RC1 + Rib 2 ⎠
       V
Vπ 1 = s ⋅ rπ 1
       Rib1
Then
      V
Av = o
      Vs
                           ⎛ RC1 ⎞ ⎛ − g m1rπ 1 ⎞
     = (1 + β )( RE 2 RL ) ⎜             ⎟⎜       ⎟
                           ⎝ RC1 + Rib 2 ⎠ ⎝ Rib1 ⎠
so
                        ⎛ 5 ⎞ ⎛ 125 ⎞
 Av = − (126 )(1.5 10 ) ⎜         ⎟⎜       ⎟
                        ⎝ 5 + 165 ⎠ ⎝ 34.1 ⎠
or
 Av = −17.7
(c)
 Ri = R1 R2 Rib1 = 70 6 34.1 = 4.76 k Ω
               ⎛ r + RC1 ⎞        ⎛ 0.676 + 5 ⎞
and Ro = RE 2 ⎜ π 2       ⎟ = 1.5 ⎜           ⎟
               ⎝ 1+ β ⎠           ⎝ 126 ⎠
or
Ro = 43.7 Ω
Test Your Understanding Exercises
TYU6.1
      I CQ    0.25
 gm =      =        = 9.62 mA / V
      VT     0.026
       β VT       (120 )( 0.026 )
rπ =          =                     = 12.5 k Ω
       I CQ            0.25
       VA    150
ro =       =     = 600 k Ω
       I CQ 0.25
TYU6.2
    V          V     75
ro = A ⇒ I CQ = A =
    I CQ        ro 200 k Ω
or
ICQ = 0.375 mA
TYU6.3
                                 RC
As a first approximation, Av ≅ −
                                 RE
Resulting gain is always smaller than this value. The effect of RS is very small.
Set
 RC
    = 10
 RE
Now
5 ≅ I C ( RC + RE ) + VCEQ
or
5 = ( 0.5 )( RC + RE ) + 2.5
which yields
RC + RE = 5 k Ω
So
10 RE + RE = 5
or
RE = 0.454 k Ω and RC = 4.54 k Ω
We have
       I CQ 0.5
I BQ =     =      = 0.005 mA
        β 100
and
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.454 )
or
RTH = 4.59 k Ω
also
      ⎛ R2 ⎞             1
VTH = ⎜         ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
      ⎝ R1 + R2 ⎠        R1
or
       1                  23
VTH =     ( 4.59 )( 5 ) =
      R1                  R1
We can write
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
or
 23
    = ( 0.005 )( 4.59 ) + 0.7 + (101)( 0.005 )( 0.454 )
 R1
which yields
R1 = 24.1 k Ω
and since R1 R2 = 4.59 k Ω
we find R2 = 5.67 k Ω
TYU6.4
dc analysis
RTH = R1 R2 = 15 85 = 12.75 k Ω
       ⎛ R2 ⎞              ⎛ 85 ⎞
VTH = ⎜          ⎟ ⋅ VCC = ⎜         ⎟ (12 )
       ⎝ R1 + R2 ⎠         ⎝ 15 + 85 ⎠
or
VTH = 10.2 V
We can write
        12 − 0.7 − VTH         12 − 0.7 − 10.2
I BQ =                    =
       RTH + (1 + β ) RE 12.75 + (101)( 0.5 )
or
IBQ = 0.0174 mA
and
I CQ = β I BQ = 1.74 mA
ac analysis
Vo = h fe I b ( RC   RL )
and
               −Vs
Ib =
       hie + (1 + h fe ) RE
so
       Vo   −h fe ( RC RL )
Av =      =
       Vs hie + (1 + h fe ) RE
For ICQ = 1.74 mA, we find
h fe ( max ) = 110, h fe ( min ) = 70
 hie ( max ) = 2 k Ω, hie ( min ) = 1.1 k Ω
We obtain
                 −110 ( 4 2 )
Av ( max ) =                      = −2.59
               1.1 + (111)( 0.5 )
and
                 −70 ( 4 2 )
Av ( min ) =                      = −2.49
               2 + ( 71)( 0.5 )
TYU6.5
                                    RC
First approximation, Av ≅ −
                                    RE
                                         RC
This predicts a low value, so set           =9
                                         RE
Now
VCC ≅ I CQ ( RC + RE ) + VECQ
or
7.5 = ( 0.6 )( 9 RE + RE ) + 3.75
which yields
RE = 0.625 k Ω and RC = 5.62 k Ω
We have
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.625 )
or
RTH = 6.31 k Ω
Also
        1                1
VTH = ⋅ RTH ⋅ VCC = ( 6.31)( 7.5 )
       R1                R1
We have
       I CQ 0.6
I BQ =     =        = 0.006 mA
        β 100
The KVL equation around the E-B loop gives
VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
or
                                                           1
7.5 = (101)( 0.006 )( 0.625 ) + 0.7 + ( 0.006 )( 6.31) +      ( 6.31)( 7.5)
                                                           R1
which yields
R1 = 7.41 k Ω
Since RTH = R1 R2 = 6.31 k Ω, then R2 = 42.5 k Ω
TYU6.6
− β RC                   ⎛R ⎞
We have Av =                      = − ( 0.95 ) ⎜ C ⎟
                 rπ + (1 + β ) RE              ⎝ RE ⎠
                   ⎛ 2 ⎞
or Av = − ( 0.95 ) ⎜     ⎟ = −4.75
                   ⎝ 0.4 ⎠
Assume rπ = 1.2 k Ω from Example 4.6.
Then
      −β ( 2)
                      = −4.75
1.2 + (1 + β )( 0.4 )
or
β = 76
TYU6.7
dc analysis: By symmetry, VTH = 0
RTH = R1 R2 = 20 20 = 10 k Ω
We can write
        0 − 0.7 − ( −5 )
 I BQ =                  = 0.00672 mA
        10 + (126 )( 5 )
 I CQ = β I BQ = (125 )( 0.00672 ) = 0.84 mA
Small-signal transistor parameters:
     β VT (125 )( 0.026 )
rπ =      =                = 3.87 k Ω
     I CQ        0.84
       I CQ        0.84
gm =          =         = 32.3 mA / V
        VT        0.026
       VA    200
ro =       =     = 238 k Ω
       I CQ 0.84
(a)
We have
Vo = − g mVπ ( ro RC       RL ) and Vπ = Vs
so
       Vo
Av =          = − g m ( ro RC   RL )
       Vs
          = − ( 32.3)( 238 2.3 5 )
or
 Av = −50.5
(b)
 Ro = ro RC = 238 2.3 = 2.28 k Ω
TYU6.8
We find I CQ = 0.418 mA, then
                              ⎛ 121 ⎞
VCEQ = 5 − ( 0.418 )( 5.6 ) − ⎜     ⎟ ( 0.418 )( 0.6 )
                              ⎝ 120 ⎠
or
VCEQ = 2.41 V
So
ΔvCE = ( 2.41 − 0.5 ) × 2
or
ΔvCE = 3.82 V peak-to-peak
TYU6.9
dc load line
VCE ≅ (10 + 10 ) − I CQ ( RC + RE )
or
VCE = 20 − I C (10 + RE )
ac load line
vce = −ic RC = −ic (10 )
Now
ΔvCE = VCEQ − 0.7 = ΔiC (10 ) = I CQ (10 )
so
VCEQ − 0.7 = I CQ (10 )
We have that
VCEQ = 20 − I CQ (10 + RE )
then
 I CQ (10 ) + 0.7 = 20 − I CQ (10 + RE )
or
I CQ ( 20 + RE ) = 19.3        (1)
The base current is found from
          10 − 0.7
I BQ =
       100 + (101) RE
so
          (100 )( 9.3)
I CQ =
         100 + (101) RE
Substituting into Equation (1),
⎡ (100 )( 9.3) ⎤
⎢                ⎥ ( 20 + RE ) = 19.3
⎢100 + (101) RE ⎥
⎣                ⎦
which yields
RE = 16.35 k Ω
Then
           (100 )( 9.3)
I CQ =                      = 0.531 mA
       100 + (101)(16.35 )
and
VCEQ ≅ 20 − ( 0.531)(10 + 16.35 ) = 6.0 V
Now
ΔvCE = VCEQ − 0.7 = 6 − 0.7 = 5.3 V
or, maximum symmetrical swing = 2 × 5.3 = 10.6 V peak-to-peak
TYU6.10
We can write
        0 − 0.7 − ( −10 )
I BQ =                    ⇒ 6.60 μ A
       100 + (131)(10 )
and
I CQ = (130 )( 6.60 μ A ) = 0.857 mA
Assume nominal small-signal parameters of:
hie = 4 k Ω, h fe = 134
                             1
hre = 0, hoe = 12 μ S ⇒         = 83.3 k Ω
                            hoe
We find
                        ⎛        1 ⎞
Rib = hie + (1 + h fe ) ⎜ RE RL     ⎟
                        ⎝       hoe ⎠
    = 4 + (135 )(10 ||10 || 83.3) = 641 k Ω
To find the voltage gain:
RB Rib             100 641
Vs′ =               ⋅ Vs =              ⋅ Vs
        RB Rib + RS        100 641 + 10
or
Vs′ = ( 0.896 ) Vs
Also
Vo
   =
       (1 + h fe ) R′
Vs′ hie + (1 + h fe ) R′
where
                      1
R ′ = RE       RL        = 10 10 83.3 = 4.72 k Ω
                     hoe
Then
        Vo ( 0.896 )(135 )( 4.72 )
Av =       =                       = 0.891
        Vs    4 + (135 )( 4.72 )
To find the current gain:
           ⎛          1     ⎞
           ⎜ RE             ⎟
      Io ⎜           hoe ⎟               ⎛ RB ⎞
                            ⎟(
 Ai = =                       1 + h fe ) ⎜          ⎟
       Ii ⎜       1                      ⎝ RB + Rib ⎠
           ⎜ RE
           ⎜          + RL ⎟⎟
           ⎝     hoe        ⎠
      ⎛ 10 83.3 ⎞               ⎛ 100 ⎞
    =⎜                   (135 ) ⎜             ⎟
      ⎝ 10 83.3 + 10 ⎟ ⎠        ⎝ 100 + 641 ⎠
or
 Ai = 8.59
To find the output resistance:
            1 hie + RS RB
Ro = RE
           hoe    1 + h fe
                     4 + 10 100
      = 10 83.3                 ⇒ 96.0 Ω
                         135
TYU6.11
We find
RTH = R1 R2 = 50 50 = 25 k Ω
      ⎛ R2 ⎞              ⎛1⎞
VTH = ⎜         ⎟ ⋅ VCC = ⎜ ⎟ ( 5 ) = 2.5 V
      ⎝ R1 + R2 ⎠         ⎝ 2⎠
Now
      V − VEB ( on ) − VTH
I BQ = CC
        RTH + (1 + β ) RE
            5 − 0.7 − 2.5
       =                   = 0.00793 mA
           25 + (101)( 2 )
and
I CQ = (100 )( 0.00793) = 0.793 mA
The small-signal transistor parameters:
     I CQ 0.793
gm =     =         = 30.5 mA / V
     VT     0.026
        β VT       (100 )( 0.026 )
rπ =           =                     = 3.28 k Ω
        I CQ           0.793
        VA    125
ro =        =      = 158 k Ω
        I CQ 0.793
(a)
Define R ′ = RE RL ro = 2 0.5 158 ≅ 0.40 k Ω
Now
          (1 + β ) R′       (101)( 0.4 )
Av =                    =
       rπ + (1 + β ) R ′ 3.28 + (101)( 0.4 )
or
Av = 0.925
(b)
 Rib = rπ + (1 + β ) R ′ = 3.28 + (101)( 0.4 )
or
Rib = 43.7 k Ω
                      rπ          3.28
Also Ro = RE ro           = 2 158
                     1+ β         101
or Ro = 32.0 Ω
TYU6.12
                          VCC − VECQ       5 − 2.5
For VECQ = 2.5, I EQ =                 =           = 5 mA
                              RE             0.5
then
       ⎛ 75 ⎞
I CQ = ⎜ ⎟ ( 5 ) = 4.93 mA
       ⎝ 76 ⎠
         5
I BQ =     = 0.0658 mA
       76
Small-signal transistor parmaters:
      β VT ( 75 )( 0.026 )
rπ =       =               = 0.396 k Ω
      I CQ        4.93
       VA    75
ro =       =     = 15.2 k Ω
       I CQ 4.93
Define the small-signal base current into the base, then
g mVπ = − β I b
Now,
     ⎛ RE ro ⎞
Io = ⎜             ⎟ (1 + β ) I b
     ⎝ RE ro + RL ⎠
and
     ⎛ R1 R2 ⎞
Ib = ⎜                ⎟ ⋅ Ii
     ⎝ R1 R2 + Rib ⎠
The current gain is
      I     ⎛ RE ro ⎞                   ⎛ R1 R2 ⎞
AI = o = ⎜                   ⎟ (1 + β ) ⎜             ⎟
       I i ⎝ RE ro + RL ⎠               ⎝ R1 R2 + Rib ⎠
We have
RE = RL = 0.5 k Ω
Rib = rπ + (1 + β )( RE RL ro )
    = 0.396 + ( 76 )( 0.5 0.5 15.2 ) = 19.1 k Ω
and
RE ro = 0.5 15.2 = 0.484 k Ω
Then
           ⎛ 0.484 ⎞              ⎛ R1 R2        ⎞
 AI = 10 = ⎜             ⎟ ( 76 ) ⎜              ⎟
           ⎝ 0.484 + 0.5 ⎠        ⎝ R1 R2 + 19.1 ⎠
which yields
R1 R2 = 6.975 k Ω
Now
⎛ R2 ⎞               1
VTH = ⎜         ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC
      ⎝ R1 + R2 ⎠          R1
or
       1
VTH = ( 6.975 )( 5 )
       R1
We can write
      V − VEB ( on ) − VTH
I BQ = CC
         RTH + (1 + β ) RE
or
                   5 − 0.7 − VTH
0.0658 =
                6.975 + ( 76 )( 0.5 )
which yields
                    1
VTH = 1.34 =           ( 6.975)( 5 )
                    R1
or
 R1 = 26.0 k Ω and R2 = 9.53 k Ω
TYU6.13
(a)
dc analysis:
       V − VEB ( on ) 10 − 0.7
 I EQ = EE           =         = 0.93 mA
             RE          10
       ⎛ β ⎞           ⎛ 100 ⎞
I CQ = ⎜      ⎟ I EQ = ⎜     ⎟ ( 0.93) = 0.921 mA
       ⎝ 1+ β ⎠        ⎝ 101 ⎠
VECQ = VEE − I EQ RE − I CQ RC − ( −VCC )
         = 10 − ( 0.93)(10 ) − ( 0.921)( 5 ) − ( −10 )
or
VECQ = 6.1 V
(b)
Small-signal transistor parameters:
      β VT (100 )( 0.026 )
 rπ =      =               = 2.82 k Ω
      I CQ      0.921
         I CQ 0.921
gm =            =    = 35.42 mA / V
       VT     0.026
Small-signal current gain:
I o = g mVπ and Vπ = Vs
also
        Vs                ⎛ 1           ⎞
Ii =         + g mVπ = Vs ⎜
                          ⎜ R r + gm ⎟  ⎟
      RE rπ               ⎝ E π         ⎠
Then
      I            g mVπ           g m ( RE rπ )
 AI = o =                       =
       Ii      ⎛ 1            ⎞ 1 + g m ( RE rπ )
            Vπ ⎜         + gm ⎟
               ⎝ RE rπ        ⎠
           ( 35.42 ) (10 2.82 )
     =
         1 + ( 35.42 ) (10 2.82 )
or
 AI = 0.987
(c)
Small-signal voltage gain:
Vo = g mVπ RC = g mVs RC
or
       V
 Av = o = g m RC = ( 35.42 )( 5 )
       Vs
or
 Av = 177
TYU6.14
(a)
        V − VBE ( on )        10 − 0.7
 I BQ = EE              =
        RB + (1 + β ) RE 100 + (101)(10 )
or
I BQ = 8.38 μ A and I CQ = 0.838 mA
        β VT        (100 )( 0.026 )
 rπ =          =                       = 3.10 k Ω
        I CQ              0.838
        I CQ       0.838
gm =           =         = 32.23 mA / V
        VT         0.026
        VA     ∞
 ro =       =      =∞
        I CQ 0.838
(b)
Summing currents, we have
          V    ⎛ −V ⎞ ⎛ −V − Vs ⎞
 g mVπ + π = ⎜ π ⎟ + ⎜ π        ⎟
           rπ ⎝ RE ⎠ ⎝ RS ⎠
or
    ⎡⎛ 1 + β ⎞ 1       1 ⎤   V
Vπ ⎢⎜        ⎟+     + ⎥=− s
    ⎢⎝
    ⎣    rπ ⎠ RE RS ⎥    ⎦   RS
We can then write
         V ⎡⎛ r ⎞          ⎤
Vπ = − s ⎢⎜ π ⎟ RE RS ⎥
         RS ⎣⎝ 1 + β ⎠     ⎦
Now
Vo = − g mVπ ( RC RL )
So

Av =
       Vo
           = gm
                 ( RC RL ) ⎡⎛ rπ ⎞ R R ⎤
                           ⎢⎜       ⎟ E S⎥
       Vs            RS    ⎣⎝ 1 + β ⎠    ⎦
       ( 32.23) (10 1) ⎡ 3.10       ⎤
     =                  ⎢ 101 10 1⎥
             (1)        ⎣           ⎦
or
 Av = 0.870
Now
     ⎛ RE            ⎞        ⎛ 10 ⎞
Ie = ⎜               ⎟ ⋅ Ii = ⎜           ⎟ ⋅ I i = ( 0.763) I i
     ⎝ RE + rπ       ⎠        ⎝ 10 + 3.10 ⎠
     ⎛ β           ⎞        ⎛ 100 ⎞
Ic = ⎜             ⎟ ⋅ Ie = ⎜     ⎟ ⋅ Ie
     ⎝ 1+ β        ⎠        ⎝ 101 ⎠
     ⎛ RC ⎞             ⎛ 10 ⎞
Io = ⎜         ⎟ ⋅ Ic = ⎜         ⎟ ⋅ I c = ( 0.909 ) I c
     ⎝ RC + RL ⎠        ⎝ 10 + 1 ⎠
So we have
      I             ⎛ 100 ⎞
AI = o = ( 0.909 ) ⎜       ⎟ ( 0.763)
      Ii            ⎝ 101 ⎠
or
 AI = 0.687
(c)
We have
           r        3.10
 Ri = RE π = 10
          1+ β       101
or
 Ri = 30.6 Ω
Also
 Ro = RC = 10 k Ω
TYU6.15
dc analysis:
We can write 5 = I BQ RB + VBE ( on ) + I EQ RE
or
            5 − 0.7         4.3
I BQ =                =
         RB + (101) RE RB + (101) RE
          (100 )( 4.3)
I CQ =
         RB + (101) RE
Also
5 = I CQ RC + VCEQ + I EQ RE − 5
or
                 ⎡      ⎛ 101 ⎞ ⎤
VCEQ = 10 − I CQ ⎢ RC + ⎜     ⎟ RE ⎥
                 ⎣      ⎝ 100 ⎠ ⎦
ac analysis:
Vo = − g mVπ ( RC RL )
and
             Vπ                 ⎛ RB ⎞
Vs = −Vπ −      ⋅ RB = −Vπ      ⎜1 +    ⎟
             rπ                 ⎝    rπ ⎠
or
       ⎛ r       ⎞
Vπ = − ⎜ π ⎟ ⋅ Vs
       ⎝ rπ + RB ⎠
Then
     V         β
Av = o =
     Vs rπ + RB
                   ( RC RL )
where β = g m rπ
                              β VT       (100 )( 0.026 )
For I CQ = 1 mA, rπ =                =                     = 2.6 k Ω
                              I CQ             1
Now
             (100 ) ( 2 2 )
Av = 20 =
             2.6 + RB
which yields
RB = 2.4 k Ω
Also
            (100 )( 4.3)
I CQ = 1 =
           2.4 + (101) RE
which yields
 RE = 4.23 k Ω
TYU6.16
(a)
dc analysis:
For I EQ 2 = 1 mA,
         ⎛ 100 ⎞
I CQ 2 = ⎜      ⎟ (1) = 0.990 mA
         ⎝ 101 ⎠
         I EQ 2     1
I EQ1 =         =       = 0.0099 mA
        1 + β 101
and
          I EQ1 0.0099
I BQ1 =         =           = 0.000098 mA
        1+ β         101
and
I CQ1 = (100 )( 0.000098 ) = 0.0098 mA
and
VB1 = − I BQ1 RB = − ( 0.000098 )(10 )
or
VB1 = −0.00098 ≅ 0
so
VE1 = −0.7 V
and
VE 2 = −1.4 V
Now
 I1 = I CQ1 + I CQ 2 = 0.0098 + 0.990 ≅ 1 mA
then
VO = 5 − (1)( 4 ) = 1 V
and
VCEQ 2 = 1 − ( −1.4 ) = 2.4 V
VCEQ1 = 1 − ( −0.7 ) = 1.7 V
(b)
Small-signal transistor parameters:
        β VT (100 )( 0.026 )
 rπ 1 =       =              = 265 k Ω
        I CQ1   0.0098
         I CQ1         0.0098
g m1 =             =          = 0.377 mA
          VT           0.026
         β VT          (100 )( 0.026 )
rπ 2 =             =                     = 2.63 k Ω
         I CQ 2            0.990
          I CQ 2        0.990
gm2 =              =          = 38.1 mA / V
           VT           0.026
 ro1 = ro 2 = ∞
(c)
Small-signal voltage gain: From Figure 4.73 (b)
Vo = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC
Vs = Vπ 1 + Vπ 2
and
        ⎛V               ⎞         ⎛1+ β ⎞
Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ⋅ rπ 2 = ⎜      ⎟ ⋅ Vπ 1rπ 2
        ⎝ rπ 1           ⎠         ⎝ rπ 1 ⎠
Then
       ⎡                  ⎛1+ β          ⎞            ⎤
Vo = − ⎢ g m1Vπ 1 + g m 2 ⎜              ⎟ ⋅ rπ 2Vπ 1 ⎥ ⋅ RC
       ⎣                  ⎝ rπ 1         ⎠            ⎦
Also
⎛1+ β      ⎞
Vs = Vπ 1 + ⎜          ⎟ ⋅ rπ 2Vπ 1
            ⎝ rπ 1     ⎠
            ⎡             ⎛ r ⎞⎤
     = Vπ 1 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥
            ⎣             ⎝ rπ 1 ⎠ ⎦
or
                 Vs
Vπ 1 =
                      ⎛r ⎞
         1 + (1 + β ) ⎜ π 2 ⎟
                      ⎝ rπ 1 ⎠
Now
                ⎡                       ⎛ rπ 2 ⎞ ⎤
                ⎢ g m1 + g m 2 (1 + β ) ⎜ ⎟ ⎥ ⋅ RC
                                        ⎝ rπ 1 ⎠ ⎦
 Av = o = − ⎣
      V
      Vs                              ⎛r ⎞
                         1 + (1 + β ) ⎜ π 2 ⎟
                                      ⎝ rπ 1 ⎠
so
          ⎡                         ⎛ 2.63 ⎞ ⎤
          ⎢ 0.377 + ( 38.1)(101) ⎜ 265 ⎟ ⎥ ( 4 )
                                    ⎝         ⎠⎦
 Av = − ⎣
                                ⎛ 2.63 ⎞
                    1 + (101) ⎜         ⎟
                                ⎝ 265 ⎠
or
 Av = −77.0
(d)
 Ri = rπ 1 + (1 + β ) rπ 2 = 265 + (101)( 2.63)
or
 Ri = 531 k Ω
TYU6.17
(a)
dc analysis:
For R1 + R2 + R3 = 100 k Ω
           VCC       12
I1 =               =    = 0.12 mA
       R1 + R2 + R3 100
Now
VE1 = I CQ 2 RE = ( 0.5 )( 0.5 ) = 0.25 V
VC1 = VE1 + VCEQ1 = 0.25 + 4 = 4.25 V

and
VC 2 = VC1 + VCEQ 2 = 4.25 + 4 = 8.25 V
             VCC − VC 2 12 − 8.25
So RC =                =          = 7.5 k Ω
                I CQ       0.5
Also
VB1 = VE1 + VBE ( on ) = 0.25 + 0.7 = 0.95 V
And
       ⎛      R3       ⎞                 R3
VB1 = ⎜                ⎟ ⋅ VCC ⇒ 0.95 =     (12 )
       ⎝ R1 + R2 + R3 ⎠                 100
which yields
R3 = 7.92 k Ω
We have
VB 2 = VC1 + VBE ( on ) = 4.25 + 0.7 = 4.95 V
and
⎛ R2 + R3 ⎞
VB 2 = ⎜                 ⎟ ⋅ VCC
         ⎝ R1 + R2 + R3 ⎠
or
           ⎛ R + 7.92 ⎞
 4.95 = ⎜ 2           ⎟ (12 )
           ⎝ 100 ⎠
which yields
 R2 = 33.3 k Ω
Then
 R1 = 100 − 33.3 − 7.92 = 58.8 k Ω
(b)
Small-signal transistor parameters:
               βV     (100 )( 0.026 )
 rπ 1 = rπ 2 = T =
               I CQ           0.5
or
 rπ 1 = rπ 2 = 5.2 k Ω
and
                 I CQ    0.5
 g m1 = g m 2 =       =
                 VT     0.026
or
 g m1 = g m 2 = 19.23 mA/V
and
 ro1 = ro 2 = ∞
(c)
Small-signal voltage gain:
We have
Vo = − g m 2Vπ 2 ( RC RL )
Also
 Vπ 2
      + g m 2Vπ 2 = g m1Vπ 1
 rπ 2
or
                 ⎛ r ⎞
Vπ 2 = g m1Vπ 1 ⎜ π 2 ⎟ and Vπ 1 = Vs
                 ⎝ 1+ β ⎠
We find
       V                              ⎛ r ⎞
  Av = o = − g m 2 ( RC RL ) g m1 ⎜ π 2 ⎟
       Vs                             ⎝1+ β ⎠
                                  ⎛ β ⎞
              = − g m 2 ( RC RL ) ⎜      ⎟
                                  ⎝ 1+ β ⎠
We obtain
                             ⎛ 100 ⎞
 Av = − (19.23) ( 7.5 2 ) ⎜        ⎟
                             ⎝ 101 ⎠
or
 Av = −30.1
TYU6.18
(a)
dc analysis: with vs = 0
 RTH = R1 R2 = 125 30 = 24.2 k Ω
      ⎛ R2 ⎞              ⎛ 30 ⎞
VTH = ⎜         ⎟ ⋅ VCC = ⎜          ⎟ (12 )
      ⎝ R1 + R2 ⎠         ⎝ 125 + 30 ⎠
or
VTH = 2.32 V
Now
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
or
              2.32 − 0.7
I BQ =                         = 0.0250 mA
           24.2 + ( 81)( 0.5 )
and
I CQ = ( 80 )( 0.025 ) = 2.00 mA
Also
                    ⎡     ⎛1+ β ⎞ ⎤
VCEQ = VCC − I CQ ⎢ RC + ⎜        ⎟ RE ⎥
                    ⎣     ⎝ β ⎠ ⎦
                  ⎡ ⎛ 81 ⎞          ⎤
     = 12 − ( 2 ) ⎢ 2 + ⎜ ⎟ ( 0.5 ) ⎥
                  ⎣ ⎝ 80 ⎠          ⎦
or
VCEQ = 6.99 V
Power dissipated in RC:
PRC = I CQ RC = ( 2.0 ) ( 2 ) = 8.0 mW
        2                      2


Power dissipated in RL:
I LQ = 0 ⇒ PRL = 0
Power dissipated in transistor:
 PQ = I BQVBEQ + I CQVCEQ
    = ( 0.025 )( 0.7 ) + ( 2.0 )( 6.99 ) = 14.0 mW
(b)
With vs = 18cos ω t ( mV )
        β VT       (80 )( 0.026 )
rπ =           =                    = 1.04 k Ω
        I CQ             2.0
We can write
           β
vce =
    rπ
       ( RC RL )VP cos ω t
Power dissipated in RL:
           vce ( rms )
                         2                               2
                              1 1 ⎡β              ⎤
pRL =                        = ⋅ ⋅ ⎢ ( RC RL ) VP ⎥
               RL             2 RL ⎣ rπ           ⎦
                                                     2
                          ⎡ 80
                                ( 2 2 ) ( 0.018)⎤
           1    1
       =     ⋅           ⋅⎢                     ⎥
           2 2 × 103      ⎣1.04                 ⎦
or
 pRL = 0.479 mW
Power dissipated in RC:
Since RC = RL = 2 k Ω, we have pRC = 8.0 + 0.479 = 8.48 mW
Power dissipated in transistor: From the text, we have
                               2
                ⎛β ⎞ ⎛V ⎞
                                      2

pQ ≅ I CQVCEQ − ⎜ ⎟ ⎜ P ⎟ ( RC RL )
                ⎝ rπ ⎠ ⎝ 2 ⎠
                                                 2           2
                               ⎛            ⎞ ⎛ 0.018 ⎞
     = ( 2 × 10−3 ) ( 6.99 ) − ⎜                      ⎟ ( 2 × 10 2 ×       )
                                    80                          3      3
                                          3 ⎟ ⎜
                               ⎝ 1.04 × 10 ⎠ ⎝ 2 ⎠
or
 pQ = 13.0 mW
TYU6.19
(a)
dc analysis:
RTH = R1 R2 = 53.8 10 = 8.43 k Ω
        ⎛ R2 ⎞              ⎛ 10 ⎞
VTH = ⎜           ⎟ ⋅ VCC = ⎜           ⎟ (5)
        ⎝ R1 + R2 ⎠         ⎝ 53.8 + 10 ⎠
or
VTH = 0.7837 V
Now
       0.7837 − 0.7
I BQ =                 = 0.00993 mA
            8.43
and
I CQ = (100 )( 0.00993) = 0.993 mA
We have
VCEQ = VCC − I CQ RC
or
2.5 = 5 − ( 0.993) RC
which yields
 RC = 2.52 k Ω
(b)
Power dissipated in RC:
PRC = I CQ RC = ( 0.993) ( 2.52 )
        2                 2


or
PRC = 2.48 mW
Power dissipated in transistor:
PQ ≅ I CQVCEQ = ( 0.993)( 2.5 )
or
PQ = 2.48 mW
(c)
ac analysis:
Maximum ac collector current:
ic = ( 0.993) cos ω t ( mA )
average ac power dissipated in RC:
       1              1
 pRC = ( 0.993) RC = ( 0.993) ( 2.52 )
                 2              2

       2               2
or
 pRC = 1.24 mW
Now
              pRC        1.24
Fraction =          =              = 0.25
            PRC + PQ 2.48 + 2.48

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Ch06p

  • 1. Chapter 6 Exercise Problems EX6.1 VBB − VBE ( on ) 2 − 0.7 (a) I BQ = = ⇒ 2 μA RB 650 Then I CQ = β I BQ = (100 )( 2 μ A ) = 0.20 mA VCEQ = VCC − I CQ RC = 5 − ( 0.2 )(15 ) = 2 V (b) Now I CQ 0.20 gm = = = 7.69 mA / V VT 0.026 and β VT (100 )( 0.026 ) rπ = = = 13 k Ω I CQ 0.20 (c) We find ⎛ r ⎞ Av = − g m RC ⎜ π ⎟ ⎝ rπ + RB ⎠ ⎛ 13 ⎞ = − ( 7.69 )(15 ) ⎜ ⎟ = −2.26 ⎝ 13 + 650 ⎠ EX6.2 V − VBE ( on ) 0.92 − 0.7 I BQ = BB = RB 100 or I BQ = 0.0022 mA and I CQ = β I BQ = (150 )( 0.0022 ) = 0.33 mA (a) I CQ 0.33 gm = = = 12.7 mA / V VT 0.026 β VT (150 )( 0.026 ) rπ = = = 11.8 k Ω I CQ 0.33 VA 200 ro = = = 606 k Ω I CQ 0.33 (b) ⎛ rπ ⎞ vo = − g m vπ ( ro RC ) and vπ = ⎜ ⎟ ⋅ vs ⎝ rπ + RB ⎠ so vo ⎛ rπ ⎞ Av = = − gm ⎜ ⎟ ( ro RC ) vπ ⎝ rπ + RB ⎠ ⎛ 11.8 ⎞ = − (12.7 ) ⎜ ⎟ ( 606 15 ) ⎝ 11.8 + 100 ⎠ which yields Av = −19.6 EX6.3 (a) V − VEB ( on ) 1.145 − 0.7 I BQ = BB = RB 50 or I BQ = 0.0089 mA Then I CQ = β I BQ = ( 90 )( 0.0089 ) = 0.801 mA
  • 2. Now I CQ 0.801 gm = = = 30.8 mA / V VT 0.026 β VT ( 90 )( 0.026 ) rπ = = = 2.92 k Ω I CQ 0.801 VA 120 ro = = = 150 k Ω I CQ 0.801 (b) We have Vo = g mVπ ( ro RC ) ⎛ r ⎞ and Vπ = − ⎜ π ⎟ Vs ⎝ rπ + RB ⎠ so V ⎛ rπ ⎞ Av = o = − g m ⎜ ⎟ ( ro RC ) Vs ⎝ rπ + RB ⎠ ⎛ 2.92 ⎞ = − ( 30.8 ) ⎜ ⎟ (150 2.5 ) ⎝ 2.92 + 50 ⎠ which yields Av = −4.18 EX6.4 Using Figure 6.23 (a) For I CQ = 0.2 mA, 7.8 < hie < 15 k Ω, 60 < h fe < 125, 6.2 × 10−4 < hre < 50 × 10−4 , 5 < hoe < 13 μ mhos (b) For I CQ = 5 mA, 0.7 < hie < 1.1 k Ω, 140 < h fe < 210, 1.05 × 10−4 < hre < 1.6 ×10−4 , 22 < hoe < 35 μ mhos EX6.5 RTH = R1 R2 = 35.2 || 5.83 = 5 k Ω ⎛ R2 ⎞ ⎛ 5.83 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (5) ⎝ R1 + R2 ⎠ ⎝ 5.83 + 35.2 ⎠ or VTH = 0.7105 V Then V − VBE ( on ) 0.7105 − 0.7 I BQ = TH = RTH 5 or I BQ = 2.1 μ A and I CQ = β I BQ = (100 )( 2.1 μ A ) = 0.21 mA VCEQ = VCC − I CQ RC = 5 − ( 0.21)(10 ) and VCEQ = 2.9 V Now I CQ 0.21 gm = = = 8.08 mA VT 0.026 VA 100 ro = = = 476 k Ω I CQ 0.21 And Av = − g m ( ro RC ) = − ( 8.08 ) ( 476 10 ) so Av = −79.1 EX6.6
  • 3. RTH = R1 R2 = 250 75 = 57.7 k Ω ⎛ R2 ⎞ ⎛ 75 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (5) ⎝ R1 + R2 ⎠ ⎝ 75 + 250 ⎠ or VTH = 1.154 V VTH − VBE ( on ) 1.154 − 0.7 I BQ = = RTH + (1 + β ) RE 57.7 + (121)( 0.6 ) or I BQ = 3.48 μ A I CQ = β I BQ = (120 )( 3.38 μ A ) = 0.418 mA (a) Now I CQ 0.418 gm = = = 16.08 mA / V VT 0.026 βVT (120 )( 0.026 ) rπ = = = 7.46 k Ω I CQ 0.418 We have Vo = − g mVπ RC We find Rib = rπ + (1 + β ) RE = 7.46 + (121)( 0.6 ) or Rib = 80.1 k Ω Also R1 R2 = 250 75 = 57.7 k Ω R1 R2 Rib = 57.7 80.1 = 33.54 k Ω We find ⎛ R1 R2 Rib ⎞ ⎛ 33.54 ⎞ Vs′ = ⎜ ⎟ ⋅ Vs = ⎜ ⎟ ⋅ Vs ⎝ R1 R2 Rib + RS ⎠ ⎝ 33.54 + 0.5 ⎠ or Vs′ = ( 0.985 ) Vs Now ⎡ ⎛1+ β ⎞ ⎤ ⎡ ⎛ 121 ⎞ ⎤ Vs′ = Vπ ⎢1 + ⎜ ⎟ RE ⎥ = Vπ ⎢1 + ⎜ ⎟ ( 0.6 ) ⎥ ⎢ ⎝ ⎣ rπ ⎠ ⎥ ⎦ ⎣ ⎝ 7.46 ⎠ ⎦ or Vπ = ( 0.0932 ) Vs′ = ( 0.0932 )( 0.985 ) Vs So Vo Av = = − (16.08 )( 0.0932 )( 0.985 )( 5.6 ) Vs or Av = −8.27 EX6.7 RTH = R1 R2 = 15 85 = 12.75 k Ω ⎛ R2 ⎞ ⎛ 85 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (12 ) ⎝ R1 + R2 ⎠ ⎝ 15 + 85 ⎠ or VTH = 10.2 V Now VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
  • 4. so 12 = (101) I BQ ( 0.5 ) + 0.7 + I BQ (12.75 ) + 10.2 which yields I BQ = 0.0174 mA and I CQ = β I BQ = (100 )( 0.0174 ) = 1.74 mA I EQ = (101)( 0.0174 ) = 1.76 mA VECQ = VCC − I EQ RE − I CQ RC = 12 − (1.76 )( 0.5 ) − (1.74 )( 4 ) or VECQ = 4.16 V Now β VT (100 )( 0.026 ) rπ = = = 1.49 k Ω I CQ 1.74 We have Vo = g mVπ ( RC || RL ) ⎛ V ⎞ Vs = −Vπ − ⎜ g mVπ + π ⎟ RE ⎝ rπ ⎠ Solving for Vπ and noting that β = g m rπ , we find Vo − β ( RC RL ) Av = = Vs rπ + (1 + β ) RE − (100 )( 4 2 ) = 1.49 + (101)( 0.5 ) or Av = −2.56 EX6.8 Dc analysis 10 − 0.7 I BQ = = 0.00439 mA 100 + (101)( 20 ) I CQ = 0.439 mA, I EQ = 0.443 mA Now (100 )( 0.026 ) rπ = = 5.92 k Ω 0.439 0.439 gm = = 16.88 mA / V 0.026 100 ro = = 228 k Ω 0.439 (a) Now Vo = − g mVπ ( ro RC ) ⎛ RB rπ ⎞ Vπ = ⎜ ⎟ ⋅ Vs ⎝ RB rπ + RS ⎠ RB rπ = 100 5.92 = 5.59 k Ω ⎛ 5.59 ⎞ Then Vπ = ⎜ ⎟ ⋅ Vs = ( 0.918 ) Vs ⎝ 5.59 + 0.5 ⎠ V Then Av = o = − (16.88 ) ( 0.918 ) ( 228 10 ) Vs or Av = −148 (b) Rin = RS + RB rπ = 0.5 + 100 5.92 = 6.09 k Ω
  • 5. and Ro = RC ro = 10 228 = 9.58 k Ω EX6.9 (a) I CQ 0.25 gm = = = 9.615 mA / V VT 0.026 VA 100 ro = = = 400 k Ω I CQ 0.25 Av = − g m ( ro rc ) = − ( 9.615 )( 400 100 ) = −769 (b) Av = − g m ( ro rc rL ) = − ( 9.615 )( 400 100 100 ) Av = −427 EX6.10 5 − 0.7 I BQ = = 0.00672 mA 10 + (126 )( 5 ) I CQ = 0.84 mA, I EQ = 0.847 mA VCEQ = 10 − ( 0.84 )( 2.3) − ( 0.847 )( 5 ) or VCEQ = 3.83 V dc load line VCE ≅ (V + − V − ) − I C ( RC + RE ) or VCE = 10 − I C ( 7.3) ac load line (neglecting ro) vce = −ic ( RC RL ) = −ic ( 2.3 5 ) = −ic (1.58 ) IC (mA) 1.37 AC load line Q-point 0.84 DC load line 3.83 10 V CE (V) EX6.11 (a) dc load line VEC ≅ VCC − I C ( RC + RE ) = 12 − I C ( 4.5 ) ac load line vec ≅ −ic ( RE + RC RL ) or vec = −ic ( 0.5 + 4 2 ) = −ic (1.83) Q-point values 12 − 0.7 − 10.2 I BQ = = 0.0150 mA 12.75 + (121)( 0.5 ) I CQ = 1.80 mA, I EQ = 1.82 mA VECQ = 3.89 V
  • 6. IC (mA) 2.67 AC load line Q-point 1.80 DC load line 3.89 12 V (V) EC (b) ΔiC = I CQ = 1.80 mA ΔvEC = (1.8 )(1.83) = 3.29 V vEC ( min ) = 3.89 − 3.29 = 0.6 V So maximum symmetrical swing = 2 × 3.29 = 6.58 V peak-to-peak EX6.12 ⎛ R2 ⎞ 1 (a) VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC ⎝ R1 + R2 ⎠ R1 RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)(1) so 1 RTH = 12.1 k Ω, VTH = (12.1)(12 ) R1 We can write VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH We have 1.6 I CQ = 1.6 mA, I BQ = = 0.0133 mA 120 Then 1 12 − 0.7 − (12.1)(12 ) R1 I BQ = 0.0133 = 12.1 + (121)(1) which yields R1 = 15.24 k Ω Since RTH = R1 R2 = 12.1 k Ω, we find R2 = 58.7 k Ω Also VECQ = 12 − (1.6 )( 4 ) − (1.61)(1) = 3.99 V (b) ac load line vec = −ic ( RC RL ) Want Δic = I CQ − 0.1 = 1.6 − 0.1 = 1.5 mA Also Δvec = 3.99 − 0.5 = 3.49 V Δvec 3.49 Now = = 2.327 k Ω = RC RL Δic 1.5 So 4 RL = 2.327 k Ω which yields RL = 5.56 k Ω EX6.13
  • 7. RTH = R1 R2 = 25 50 = 16.7 k Ω ⎛ R2 ⎞ ⎛ 50 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (5) ⎝ R1 + R2 ⎠ ⎝ 25 + 50 ⎠ or VTH = 3.33 V VTH − VBE ( on ) 3.33 − 0.7 I BQ = = RTH + (1 + β ) RE 16.7 + (121)(1) or I BQ = 0.0191 mA Also I CQ = (120 )( 0.0191) = 2.29 mA Now I CQ 2.29 gm = = = 88.1 mA / V VT 0.026 β VT (120 )( 0.026 ) rπ = = = 1.36 k Ω I CQ 2.29 VA 100 ro = = = 43.7 k Ω I CQ 2.29 (a) ⎛ R1 R2 Rib ⎞ Vs′ = ⎜ ⎟ ⋅ Vs ⎝ R1 R2 Rib + RS ⎠ Rib = rπ + (1 + β ) ( RE ro ) = 1.36 + (121)(1 43.7 ) or Rib = 120 k Ω and R1 R2 = 16.7 k Ω Then R1 R2 Rib = 16.7 120 = 14.7 k Ω Now ⎛ 14.7 ⎞ Vs′ = ⎜ ⎟ ⋅ Vs = ( 0.967 ) Vs ⎝ 14.7 + 0.5 ⎠ and ⎛V ⎞ ⎛ 1+ β ⎞ Vo = ⎜ π + g mVπ ⎟ ( RE ro ) = Vπ ⎜ ⎟ RE ro ⎝ rπ ⎠ ⎝ rπ ⎠ We have Vs′ = Vπ + Vo then Vs′ ( 0.967 )Vs Vπ = = ⎛ 1+ β ⎞ ⎛1+ β ⎞ 1+ ⎜ ⎟ RE ro 1 + ⎜ ⎟ RE ro ⎝ rπ ⎠ ⎝ rπ ⎠ We then obtain ⎛1+ β ⎞ ( 0.967 ) ⎜ ⎟ RE ro V ⎝ rπ ⎠ Av = o = Vs ⎛1+ β ⎞ 1+ ⎜ ⎟ RE ro ⎝ rπ ⎠ ( 0.967 )(1 + β ) RE ro = rπ + (1 + β ) RE ro Now RE ro = 1 43.7 = 0.978 k Ω
  • 8. Then ( 0.967 )(121)( 0.978 ) Av = = 0.956 1.36 + (121)( 0.978 ) (b) Rib = rπ + (1 + β )( RE ro ) or Rib = 1.36 + (121)( 0.978 ) = 120 k Ω EX6.14 For RS = 0, then r Ro = RE ro π 1+ β Using the parameters from Exercise 4.13, we obtain 1.36 Ro = 1 43.7 121 or Ro = 11.1 Ω EX6.15 (a) For I CQ = 1.25 mA and β = 100, we find I EQ = 1.26 mA and I BQ = 0.0125 mA Now VCEQ = 10 − I EQ RE Or 4 = 10 − (1.26 ) RE which yields RE = 4.76 k Ω Then RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 4.76 ) or RTH = 48.1 k Ω We have ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10 ) − 5 = ⋅ RTH (10 ) − 5 ⎝ R1 + R2 ⎠ R1 or 1 VTH = ( 481) − 5 R1 VTH − 0.7 − ( −5 ) We can write I BQ = RTH + (1 + β ) RE Or 1 ( 481) − 5 − 0.7 + 5 R1 0.0125 = 48.1 + (101)( 4.76 ) which yields R1 = 65.8 k Ω Since R1 R2 = 48.1 k Ω, we obtain R2 = 178.8 k Ω (b)
  • 9. β VT (100 )( 0.026 ) rπ = = = 2.08 k Ω I CQ 1.25 VA 125 ro = = = 100 k Ω I CQ 1.25 We may note that g mVπ = g m ( I b rπ ) = β I b Also Rib = rπ + (1 + β )( RE RL ro ) = 2.08 + (101) ( 4.76 1 100 ) or Rib = 84.9 k Ω Now ⎛ RE ro ⎞ ⎜ R r + R ⎟( Io = ⎜ 1 + β ) Ib ⎟ ⎝ E o L ⎠ where ⎛ R1 R2 ⎞ Ib = ⎜ ⋅I ⎜R R +R ⎟ s ⎟ ⎝ 1 2 ib ⎠ We can then write I ⎛ RE ro ⎞ ⎛ R1 R2 ⎞ AI = o = ⎜ ⎟ (1 + β ) ⎜ ⎟ I s ⎜ RE ro + RL ⎟ ⎝ ⎠ ⎜R R +R ⎝ 1 2 ib ⎟ ⎠ We have RE ro = 4.76 100 = 4.54 k Ω so ⎛ 4.54 ⎞ ⎛ 48.1 ⎞ AI = ⎜ ⎟ (101) ⎜ ⎟ ⎝ 4.54 + 1 ⎠ ⎝ 48.1 + 84.9 ⎠ or AI = 29.9 r 2.08 (c) Ro = RE ro π = 4.76 100 1+ β 101 or Ro = 20.5 Ω EX6.16 (a) RTH = R1 R2 = 70 6 = 5.53 k Ω ⎛ R2 ⎞ ⎛ 6 ⎞ VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎝ 70 + 6 ⎠ or VTH = −4.2105 V We find −4.2105 − 0.7 − ( −5 ) I BQ1 = ⇒ 2.91 μ A 5.53 + (126 )( 0.2 ) and I CQ1 = β I BQ1 = (125 )( 2.91 μ A ) = 0.364 mA I EQ1 = (1 + β ) I BQ1 = 0.368 mA At the collector of Q1, 5 − VC1 V − 0.7 − ( −5 ) = I CQ1 + C1 RC1 (1 + β )( RE 2 ) or
  • 10. 5 − VC1 V − 0.7 − ( −5 ) = 0.364 + C1 5 (126 )(1.5) which yields VC1 = 2.99 V also VE1 = I EQ1 RE1 − 5 = ( 0.368 )( 0.2 ) − 5 or VE1 = −4.93 V Then VCEQ1 = VC1 − VE1 = 2.99 − ( −4.93) = 7.92 V We find V − 0.7 − ( −5 ) I EQ 2 = C1 = 4.86 mA 1.5 and ⎛ β ⎞ ⎛ 125 ⎞ I CQ 2 = ⎜ ⎟ ⋅ I EQ1 = ⎜ ⎟ ( 4.86 ) = 4.82 mA ⎝ 1+ β ⎠ ⎝ 126 ⎠ We find VE 2 = VC1 − 0.7 = 2.99 − 0.7 = 2.29 V and VCEQ 2 = 5 − VE 2 = 5 − 2.29 = 2.71 V (b) The small-signal transistor parameters are: β VT (125 )( 0.026 ) rπ 1 = = = 8.93 k Ω I CQ1 0.364 I CQ1 0.364 g m1 = = = 14.0 mA / V VT 0.026 β VT (125 )( 0.026 ) rπ 2 = = = 0.674 k Ω I CQ 2 4.82 I CQ 2 4.82 gm2 = = = 185 mA / V VT 0.026 We find Rib1 = rπ 1 + (1 + β ) RE1 = 8.93 + (126 )( 0.2 ) Or Rib1 = 34.1 k Ω and Rib 2 = rπ 2 + (1 + β )( RE 2 RL ) = 0.674 + (126 )(1.5 10 ) = 165 k Ω The small-signal equivalent circuit is:
  • 11. ϩ ϩ V␲1 r␲1 gm1V␲1 V␲2 r␲2 gm2V␲2 Ϫ Ϫ Vs ϩ R1͉͉R2 RC1 Vo Ϫ RE1 RE2 RL We can write Vo = (1 + β ) I b 2 ( RE 2 RL ) where ⎛ RC1 ⎞ Ib2 = ⎜ ⎟ ( − g m1Vπ 1 ) ⎝ RC1 + Rib 2 ⎠ V Vπ 1 = s ⋅ rπ 1 Rib1 Then V Av = o Vs ⎛ RC1 ⎞ ⎛ − g m1rπ 1 ⎞ = (1 + β )( RE 2 RL ) ⎜ ⎟⎜ ⎟ ⎝ RC1 + Rib 2 ⎠ ⎝ Rib1 ⎠ so ⎛ 5 ⎞ ⎛ 125 ⎞ Av = − (126 )(1.5 10 ) ⎜ ⎟⎜ ⎟ ⎝ 5 + 165 ⎠ ⎝ 34.1 ⎠ or Av = −17.7 (c) Ri = R1 R2 Rib1 = 70 6 34.1 = 4.76 k Ω ⎛ r + RC1 ⎞ ⎛ 0.676 + 5 ⎞ and Ro = RE 2 ⎜ π 2 ⎟ = 1.5 ⎜ ⎟ ⎝ 1+ β ⎠ ⎝ 126 ⎠ or Ro = 43.7 Ω Test Your Understanding Exercises TYU6.1 I CQ 0.25 gm = = = 9.62 mA / V VT 0.026 β VT (120 )( 0.026 ) rπ = = = 12.5 k Ω I CQ 0.25 VA 150 ro = = = 600 k Ω I CQ 0.25 TYU6.2 V V 75 ro = A ⇒ I CQ = A = I CQ ro 200 k Ω or ICQ = 0.375 mA
  • 12. TYU6.3 RC As a first approximation, Av ≅ − RE Resulting gain is always smaller than this value. The effect of RS is very small. Set RC = 10 RE Now 5 ≅ I C ( RC + RE ) + VCEQ or 5 = ( 0.5 )( RC + RE ) + 2.5 which yields RC + RE = 5 k Ω So 10 RE + RE = 5 or RE = 0.454 k Ω and RC = 4.54 k Ω We have I CQ 0.5 I BQ = = = 0.005 mA β 100 and RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.454 ) or RTH = 4.59 k Ω also ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC ⎝ R1 + R2 ⎠ R1 or 1 23 VTH = ( 4.59 )( 5 ) = R1 R1 We can write VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE or 23 = ( 0.005 )( 4.59 ) + 0.7 + (101)( 0.005 )( 0.454 ) R1 which yields R1 = 24.1 k Ω and since R1 R2 = 4.59 k Ω we find R2 = 5.67 k Ω TYU6.4 dc analysis RTH = R1 R2 = 15 85 = 12.75 k Ω ⎛ R2 ⎞ ⎛ 85 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (12 ) ⎝ R1 + R2 ⎠ ⎝ 15 + 85 ⎠ or VTH = 10.2 V We can write 12 − 0.7 − VTH 12 − 0.7 − 10.2 I BQ = = RTH + (1 + β ) RE 12.75 + (101)( 0.5 ) or IBQ = 0.0174 mA
  • 13. and I CQ = β I BQ = 1.74 mA ac analysis Vo = h fe I b ( RC RL ) and −Vs Ib = hie + (1 + h fe ) RE so Vo −h fe ( RC RL ) Av = = Vs hie + (1 + h fe ) RE For ICQ = 1.74 mA, we find h fe ( max ) = 110, h fe ( min ) = 70 hie ( max ) = 2 k Ω, hie ( min ) = 1.1 k Ω We obtain −110 ( 4 2 ) Av ( max ) = = −2.59 1.1 + (111)( 0.5 ) and −70 ( 4 2 ) Av ( min ) = = −2.49 2 + ( 71)( 0.5 ) TYU6.5 RC First approximation, Av ≅ − RE RC This predicts a low value, so set =9 RE Now VCC ≅ I CQ ( RC + RE ) + VECQ or 7.5 = ( 0.6 )( 9 RE + RE ) + 3.75 which yields RE = 0.625 k Ω and RC = 5.62 k Ω We have RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.625 ) or RTH = 6.31 k Ω Also 1 1 VTH = ⋅ RTH ⋅ VCC = ( 6.31)( 7.5 ) R1 R1 We have I CQ 0.6 I BQ = = = 0.006 mA β 100 The KVL equation around the E-B loop gives VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH or 1 7.5 = (101)( 0.006 )( 0.625 ) + 0.7 + ( 0.006 )( 6.31) + ( 6.31)( 7.5) R1 which yields R1 = 7.41 k Ω Since RTH = R1 R2 = 6.31 k Ω, then R2 = 42.5 k Ω TYU6.6
  • 14. − β RC ⎛R ⎞ We have Av = = − ( 0.95 ) ⎜ C ⎟ rπ + (1 + β ) RE ⎝ RE ⎠ ⎛ 2 ⎞ or Av = − ( 0.95 ) ⎜ ⎟ = −4.75 ⎝ 0.4 ⎠ Assume rπ = 1.2 k Ω from Example 4.6. Then −β ( 2) = −4.75 1.2 + (1 + β )( 0.4 ) or β = 76 TYU6.7 dc analysis: By symmetry, VTH = 0 RTH = R1 R2 = 20 20 = 10 k Ω We can write 0 − 0.7 − ( −5 ) I BQ = = 0.00672 mA 10 + (126 )( 5 ) I CQ = β I BQ = (125 )( 0.00672 ) = 0.84 mA Small-signal transistor parameters: β VT (125 )( 0.026 ) rπ = = = 3.87 k Ω I CQ 0.84 I CQ 0.84 gm = = = 32.3 mA / V VT 0.026 VA 200 ro = = = 238 k Ω I CQ 0.84 (a) We have Vo = − g mVπ ( ro RC RL ) and Vπ = Vs so Vo Av = = − g m ( ro RC RL ) Vs = − ( 32.3)( 238 2.3 5 ) or Av = −50.5 (b) Ro = ro RC = 238 2.3 = 2.28 k Ω TYU6.8 We find I CQ = 0.418 mA, then ⎛ 121 ⎞ VCEQ = 5 − ( 0.418 )( 5.6 ) − ⎜ ⎟ ( 0.418 )( 0.6 ) ⎝ 120 ⎠ or VCEQ = 2.41 V So ΔvCE = ( 2.41 − 0.5 ) × 2 or ΔvCE = 3.82 V peak-to-peak TYU6.9 dc load line VCE ≅ (10 + 10 ) − I CQ ( RC + RE ) or
  • 15. VCE = 20 − I C (10 + RE ) ac load line vce = −ic RC = −ic (10 ) Now ΔvCE = VCEQ − 0.7 = ΔiC (10 ) = I CQ (10 ) so VCEQ − 0.7 = I CQ (10 ) We have that VCEQ = 20 − I CQ (10 + RE ) then I CQ (10 ) + 0.7 = 20 − I CQ (10 + RE ) or I CQ ( 20 + RE ) = 19.3 (1) The base current is found from 10 − 0.7 I BQ = 100 + (101) RE so (100 )( 9.3) I CQ = 100 + (101) RE Substituting into Equation (1), ⎡ (100 )( 9.3) ⎤ ⎢ ⎥ ( 20 + RE ) = 19.3 ⎢100 + (101) RE ⎥ ⎣ ⎦ which yields RE = 16.35 k Ω Then (100 )( 9.3) I CQ = = 0.531 mA 100 + (101)(16.35 ) and VCEQ ≅ 20 − ( 0.531)(10 + 16.35 ) = 6.0 V Now ΔvCE = VCEQ − 0.7 = 6 − 0.7 = 5.3 V or, maximum symmetrical swing = 2 × 5.3 = 10.6 V peak-to-peak TYU6.10 We can write 0 − 0.7 − ( −10 ) I BQ = ⇒ 6.60 μ A 100 + (131)(10 ) and I CQ = (130 )( 6.60 μ A ) = 0.857 mA Assume nominal small-signal parameters of: hie = 4 k Ω, h fe = 134 1 hre = 0, hoe = 12 μ S ⇒ = 83.3 k Ω hoe We find ⎛ 1 ⎞ Rib = hie + (1 + h fe ) ⎜ RE RL ⎟ ⎝ hoe ⎠ = 4 + (135 )(10 ||10 || 83.3) = 641 k Ω To find the voltage gain:
  • 16. RB Rib 100 641 Vs′ = ⋅ Vs = ⋅ Vs RB Rib + RS 100 641 + 10 or Vs′ = ( 0.896 ) Vs Also Vo = (1 + h fe ) R′ Vs′ hie + (1 + h fe ) R′ where 1 R ′ = RE RL = 10 10 83.3 = 4.72 k Ω hoe Then Vo ( 0.896 )(135 )( 4.72 ) Av = = = 0.891 Vs 4 + (135 )( 4.72 ) To find the current gain: ⎛ 1 ⎞ ⎜ RE ⎟ Io ⎜ hoe ⎟ ⎛ RB ⎞ ⎟( Ai = = 1 + h fe ) ⎜ ⎟ Ii ⎜ 1 ⎝ RB + Rib ⎠ ⎜ RE ⎜ + RL ⎟⎟ ⎝ hoe ⎠ ⎛ 10 83.3 ⎞ ⎛ 100 ⎞ =⎜ (135 ) ⎜ ⎟ ⎝ 10 83.3 + 10 ⎟ ⎠ ⎝ 100 + 641 ⎠ or Ai = 8.59 To find the output resistance: 1 hie + RS RB Ro = RE hoe 1 + h fe 4 + 10 100 = 10 83.3 ⇒ 96.0 Ω 135 TYU6.11 We find RTH = R1 R2 = 50 50 = 25 k Ω ⎛ R2 ⎞ ⎛1⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ ( 5 ) = 2.5 V ⎝ R1 + R2 ⎠ ⎝ 2⎠ Now V − VEB ( on ) − VTH I BQ = CC RTH + (1 + β ) RE 5 − 0.7 − 2.5 = = 0.00793 mA 25 + (101)( 2 ) and I CQ = (100 )( 0.00793) = 0.793 mA The small-signal transistor parameters: I CQ 0.793 gm = = = 30.5 mA / V VT 0.026 β VT (100 )( 0.026 ) rπ = = = 3.28 k Ω I CQ 0.793 VA 125 ro = = = 158 k Ω I CQ 0.793 (a)
  • 17. Define R ′ = RE RL ro = 2 0.5 158 ≅ 0.40 k Ω Now (1 + β ) R′ (101)( 0.4 ) Av = = rπ + (1 + β ) R ′ 3.28 + (101)( 0.4 ) or Av = 0.925 (b) Rib = rπ + (1 + β ) R ′ = 3.28 + (101)( 0.4 ) or Rib = 43.7 k Ω rπ 3.28 Also Ro = RE ro = 2 158 1+ β 101 or Ro = 32.0 Ω TYU6.12 VCC − VECQ 5 − 2.5 For VECQ = 2.5, I EQ = = = 5 mA RE 0.5 then ⎛ 75 ⎞ I CQ = ⎜ ⎟ ( 5 ) = 4.93 mA ⎝ 76 ⎠ 5 I BQ = = 0.0658 mA 76 Small-signal transistor parmaters: β VT ( 75 )( 0.026 ) rπ = = = 0.396 k Ω I CQ 4.93 VA 75 ro = = = 15.2 k Ω I CQ 4.93 Define the small-signal base current into the base, then g mVπ = − β I b Now, ⎛ RE ro ⎞ Io = ⎜ ⎟ (1 + β ) I b ⎝ RE ro + RL ⎠ and ⎛ R1 R2 ⎞ Ib = ⎜ ⎟ ⋅ Ii ⎝ R1 R2 + Rib ⎠ The current gain is I ⎛ RE ro ⎞ ⎛ R1 R2 ⎞ AI = o = ⎜ ⎟ (1 + β ) ⎜ ⎟ I i ⎝ RE ro + RL ⎠ ⎝ R1 R2 + Rib ⎠ We have RE = RL = 0.5 k Ω Rib = rπ + (1 + β )( RE RL ro ) = 0.396 + ( 76 )( 0.5 0.5 15.2 ) = 19.1 k Ω and RE ro = 0.5 15.2 = 0.484 k Ω Then ⎛ 0.484 ⎞ ⎛ R1 R2 ⎞ AI = 10 = ⎜ ⎟ ( 76 ) ⎜ ⎟ ⎝ 0.484 + 0.5 ⎠ ⎝ R1 R2 + 19.1 ⎠ which yields R1 R2 = 6.975 k Ω Now
  • 18. ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC ⎝ R1 + R2 ⎠ R1 or 1 VTH = ( 6.975 )( 5 ) R1 We can write V − VEB ( on ) − VTH I BQ = CC RTH + (1 + β ) RE or 5 − 0.7 − VTH 0.0658 = 6.975 + ( 76 )( 0.5 ) which yields 1 VTH = 1.34 = ( 6.975)( 5 ) R1 or R1 = 26.0 k Ω and R2 = 9.53 k Ω TYU6.13 (a) dc analysis: V − VEB ( on ) 10 − 0.7 I EQ = EE = = 0.93 mA RE 10 ⎛ β ⎞ ⎛ 100 ⎞ I CQ = ⎜ ⎟ I EQ = ⎜ ⎟ ( 0.93) = 0.921 mA ⎝ 1+ β ⎠ ⎝ 101 ⎠ VECQ = VEE − I EQ RE − I CQ RC − ( −VCC ) = 10 − ( 0.93)(10 ) − ( 0.921)( 5 ) − ( −10 ) or VECQ = 6.1 V (b) Small-signal transistor parameters: β VT (100 )( 0.026 ) rπ = = = 2.82 k Ω I CQ 0.921 I CQ 0.921 gm = = = 35.42 mA / V VT 0.026 Small-signal current gain: I o = g mVπ and Vπ = Vs also Vs ⎛ 1 ⎞ Ii = + g mVπ = Vs ⎜ ⎜ R r + gm ⎟ ⎟ RE rπ ⎝ E π ⎠ Then I g mVπ g m ( RE rπ ) AI = o = = Ii ⎛ 1 ⎞ 1 + g m ( RE rπ ) Vπ ⎜ + gm ⎟ ⎝ RE rπ ⎠ ( 35.42 ) (10 2.82 ) = 1 + ( 35.42 ) (10 2.82 ) or AI = 0.987 (c) Small-signal voltage gain:
  • 19. Vo = g mVπ RC = g mVs RC or V Av = o = g m RC = ( 35.42 )( 5 ) Vs or Av = 177 TYU6.14 (a) V − VBE ( on ) 10 − 0.7 I BQ = EE = RB + (1 + β ) RE 100 + (101)(10 ) or I BQ = 8.38 μ A and I CQ = 0.838 mA β VT (100 )( 0.026 ) rπ = = = 3.10 k Ω I CQ 0.838 I CQ 0.838 gm = = = 32.23 mA / V VT 0.026 VA ∞ ro = = =∞ I CQ 0.838 (b) Summing currents, we have V ⎛ −V ⎞ ⎛ −V − Vs ⎞ g mVπ + π = ⎜ π ⎟ + ⎜ π ⎟ rπ ⎝ RE ⎠ ⎝ RS ⎠ or ⎡⎛ 1 + β ⎞ 1 1 ⎤ V Vπ ⎢⎜ ⎟+ + ⎥=− s ⎢⎝ ⎣ rπ ⎠ RE RS ⎥ ⎦ RS We can then write V ⎡⎛ r ⎞ ⎤ Vπ = − s ⎢⎜ π ⎟ RE RS ⎥ RS ⎣⎝ 1 + β ⎠ ⎦ Now Vo = − g mVπ ( RC RL ) So Av = Vo = gm ( RC RL ) ⎡⎛ rπ ⎞ R R ⎤ ⎢⎜ ⎟ E S⎥ Vs RS ⎣⎝ 1 + β ⎠ ⎦ ( 32.23) (10 1) ⎡ 3.10 ⎤ = ⎢ 101 10 1⎥ (1) ⎣ ⎦ or Av = 0.870 Now ⎛ RE ⎞ ⎛ 10 ⎞ Ie = ⎜ ⎟ ⋅ Ii = ⎜ ⎟ ⋅ I i = ( 0.763) I i ⎝ RE + rπ ⎠ ⎝ 10 + 3.10 ⎠ ⎛ β ⎞ ⎛ 100 ⎞ Ic = ⎜ ⎟ ⋅ Ie = ⎜ ⎟ ⋅ Ie ⎝ 1+ β ⎠ ⎝ 101 ⎠ ⎛ RC ⎞ ⎛ 10 ⎞ Io = ⎜ ⎟ ⋅ Ic = ⎜ ⎟ ⋅ I c = ( 0.909 ) I c ⎝ RC + RL ⎠ ⎝ 10 + 1 ⎠ So we have I ⎛ 100 ⎞ AI = o = ( 0.909 ) ⎜ ⎟ ( 0.763) Ii ⎝ 101 ⎠
  • 20. or AI = 0.687 (c) We have r 3.10 Ri = RE π = 10 1+ β 101 or Ri = 30.6 Ω Also Ro = RC = 10 k Ω TYU6.15 dc analysis: We can write 5 = I BQ RB + VBE ( on ) + I EQ RE or 5 − 0.7 4.3 I BQ = = RB + (101) RE RB + (101) RE (100 )( 4.3) I CQ = RB + (101) RE Also 5 = I CQ RC + VCEQ + I EQ RE − 5 or ⎡ ⎛ 101 ⎞ ⎤ VCEQ = 10 − I CQ ⎢ RC + ⎜ ⎟ RE ⎥ ⎣ ⎝ 100 ⎠ ⎦ ac analysis: Vo = − g mVπ ( RC RL ) and Vπ ⎛ RB ⎞ Vs = −Vπ − ⋅ RB = −Vπ ⎜1 + ⎟ rπ ⎝ rπ ⎠ or ⎛ r ⎞ Vπ = − ⎜ π ⎟ ⋅ Vs ⎝ rπ + RB ⎠ Then V β Av = o = Vs rπ + RB ( RC RL ) where β = g m rπ β VT (100 )( 0.026 ) For I CQ = 1 mA, rπ = = = 2.6 k Ω I CQ 1 Now (100 ) ( 2 2 ) Av = 20 = 2.6 + RB which yields RB = 2.4 k Ω Also (100 )( 4.3) I CQ = 1 = 2.4 + (101) RE which yields RE = 4.23 k Ω TYU6.16 (a)
  • 21. dc analysis: For I EQ 2 = 1 mA, ⎛ 100 ⎞ I CQ 2 = ⎜ ⎟ (1) = 0.990 mA ⎝ 101 ⎠ I EQ 2 1 I EQ1 = = = 0.0099 mA 1 + β 101 and I EQ1 0.0099 I BQ1 = = = 0.000098 mA 1+ β 101 and I CQ1 = (100 )( 0.000098 ) = 0.0098 mA and VB1 = − I BQ1 RB = − ( 0.000098 )(10 ) or VB1 = −0.00098 ≅ 0 so VE1 = −0.7 V and VE 2 = −1.4 V Now I1 = I CQ1 + I CQ 2 = 0.0098 + 0.990 ≅ 1 mA then VO = 5 − (1)( 4 ) = 1 V and VCEQ 2 = 1 − ( −1.4 ) = 2.4 V VCEQ1 = 1 − ( −0.7 ) = 1.7 V (b) Small-signal transistor parameters: β VT (100 )( 0.026 ) rπ 1 = = = 265 k Ω I CQ1 0.0098 I CQ1 0.0098 g m1 = = = 0.377 mA VT 0.026 β VT (100 )( 0.026 ) rπ 2 = = = 2.63 k Ω I CQ 2 0.990 I CQ 2 0.990 gm2 = = = 38.1 mA / V VT 0.026 ro1 = ro 2 = ∞ (c) Small-signal voltage gain: From Figure 4.73 (b) Vo = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC Vs = Vπ 1 + Vπ 2 and ⎛V ⎞ ⎛1+ β ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ⋅ rπ 2 = ⎜ ⎟ ⋅ Vπ 1rπ 2 ⎝ rπ 1 ⎠ ⎝ rπ 1 ⎠ Then ⎡ ⎛1+ β ⎞ ⎤ Vo = − ⎢ g m1Vπ 1 + g m 2 ⎜ ⎟ ⋅ rπ 2Vπ 1 ⎥ ⋅ RC ⎣ ⎝ rπ 1 ⎠ ⎦ Also
  • 22. ⎛1+ β ⎞ Vs = Vπ 1 + ⎜ ⎟ ⋅ rπ 2Vπ 1 ⎝ rπ 1 ⎠ ⎡ ⎛ r ⎞⎤ = Vπ 1 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥ ⎣ ⎝ rπ 1 ⎠ ⎦ or Vs Vπ 1 = ⎛r ⎞ 1 + (1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ Now ⎡ ⎛ rπ 2 ⎞ ⎤ ⎢ g m1 + g m 2 (1 + β ) ⎜ ⎟ ⎥ ⋅ RC ⎝ rπ 1 ⎠ ⎦ Av = o = − ⎣ V Vs ⎛r ⎞ 1 + (1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ so ⎡ ⎛ 2.63 ⎞ ⎤ ⎢ 0.377 + ( 38.1)(101) ⎜ 265 ⎟ ⎥ ( 4 ) ⎝ ⎠⎦ Av = − ⎣ ⎛ 2.63 ⎞ 1 + (101) ⎜ ⎟ ⎝ 265 ⎠ or Av = −77.0 (d) Ri = rπ 1 + (1 + β ) rπ 2 = 265 + (101)( 2.63) or Ri = 531 k Ω TYU6.17 (a) dc analysis: For R1 + R2 + R3 = 100 k Ω VCC 12 I1 = = = 0.12 mA R1 + R2 + R3 100 Now VE1 = I CQ 2 RE = ( 0.5 )( 0.5 ) = 0.25 V VC1 = VE1 + VCEQ1 = 0.25 + 4 = 4.25 V and VC 2 = VC1 + VCEQ 2 = 4.25 + 4 = 8.25 V VCC − VC 2 12 − 8.25 So RC = = = 7.5 k Ω I CQ 0.5 Also VB1 = VE1 + VBE ( on ) = 0.25 + 0.7 = 0.95 V And ⎛ R3 ⎞ R3 VB1 = ⎜ ⎟ ⋅ VCC ⇒ 0.95 = (12 ) ⎝ R1 + R2 + R3 ⎠ 100 which yields R3 = 7.92 k Ω We have VB 2 = VC1 + VBE ( on ) = 4.25 + 0.7 = 4.95 V and
  • 23. ⎛ R2 + R3 ⎞ VB 2 = ⎜ ⎟ ⋅ VCC ⎝ R1 + R2 + R3 ⎠ or ⎛ R + 7.92 ⎞ 4.95 = ⎜ 2 ⎟ (12 ) ⎝ 100 ⎠ which yields R2 = 33.3 k Ω Then R1 = 100 − 33.3 − 7.92 = 58.8 k Ω (b) Small-signal transistor parameters: βV (100 )( 0.026 ) rπ 1 = rπ 2 = T = I CQ 0.5 or rπ 1 = rπ 2 = 5.2 k Ω and I CQ 0.5 g m1 = g m 2 = = VT 0.026 or g m1 = g m 2 = 19.23 mA/V and ro1 = ro 2 = ∞ (c) Small-signal voltage gain: We have Vo = − g m 2Vπ 2 ( RC RL ) Also Vπ 2 + g m 2Vπ 2 = g m1Vπ 1 rπ 2 or ⎛ r ⎞ Vπ 2 = g m1Vπ 1 ⎜ π 2 ⎟ and Vπ 1 = Vs ⎝ 1+ β ⎠ We find V ⎛ r ⎞ Av = o = − g m 2 ( RC RL ) g m1 ⎜ π 2 ⎟ Vs ⎝1+ β ⎠ ⎛ β ⎞ = − g m 2 ( RC RL ) ⎜ ⎟ ⎝ 1+ β ⎠ We obtain ⎛ 100 ⎞ Av = − (19.23) ( 7.5 2 ) ⎜ ⎟ ⎝ 101 ⎠ or Av = −30.1 TYU6.18 (a) dc analysis: with vs = 0 RTH = R1 R2 = 125 30 = 24.2 k Ω ⎛ R2 ⎞ ⎛ 30 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (12 ) ⎝ R1 + R2 ⎠ ⎝ 125 + 30 ⎠ or VTH = 2.32 V
  • 24. Now VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE or 2.32 − 0.7 I BQ = = 0.0250 mA 24.2 + ( 81)( 0.5 ) and I CQ = ( 80 )( 0.025 ) = 2.00 mA Also ⎡ ⎛1+ β ⎞ ⎤ VCEQ = VCC − I CQ ⎢ RC + ⎜ ⎟ RE ⎥ ⎣ ⎝ β ⎠ ⎦ ⎡ ⎛ 81 ⎞ ⎤ = 12 − ( 2 ) ⎢ 2 + ⎜ ⎟ ( 0.5 ) ⎥ ⎣ ⎝ 80 ⎠ ⎦ or VCEQ = 6.99 V Power dissipated in RC: PRC = I CQ RC = ( 2.0 ) ( 2 ) = 8.0 mW 2 2 Power dissipated in RL: I LQ = 0 ⇒ PRL = 0 Power dissipated in transistor: PQ = I BQVBEQ + I CQVCEQ = ( 0.025 )( 0.7 ) + ( 2.0 )( 6.99 ) = 14.0 mW (b) With vs = 18cos ω t ( mV ) β VT (80 )( 0.026 ) rπ = = = 1.04 k Ω I CQ 2.0 We can write β vce = rπ ( RC RL )VP cos ω t Power dissipated in RL: vce ( rms ) 2 2 1 1 ⎡β ⎤ pRL = = ⋅ ⋅ ⎢ ( RC RL ) VP ⎥ RL 2 RL ⎣ rπ ⎦ 2 ⎡ 80 ( 2 2 ) ( 0.018)⎤ 1 1 = ⋅ ⋅⎢ ⎥ 2 2 × 103 ⎣1.04 ⎦ or pRL = 0.479 mW Power dissipated in RC: Since RC = RL = 2 k Ω, we have pRC = 8.0 + 0.479 = 8.48 mW Power dissipated in transistor: From the text, we have 2 ⎛β ⎞ ⎛V ⎞ 2 pQ ≅ I CQVCEQ − ⎜ ⎟ ⎜ P ⎟ ( RC RL ) ⎝ rπ ⎠ ⎝ 2 ⎠ 2 2 ⎛ ⎞ ⎛ 0.018 ⎞ = ( 2 × 10−3 ) ( 6.99 ) − ⎜ ⎟ ( 2 × 10 2 × ) 80 3 3 3 ⎟ ⎜ ⎝ 1.04 × 10 ⎠ ⎝ 2 ⎠ or pQ = 13.0 mW TYU6.19 (a) dc analysis:
  • 25. RTH = R1 R2 = 53.8 10 = 8.43 k Ω ⎛ R2 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (5) ⎝ R1 + R2 ⎠ ⎝ 53.8 + 10 ⎠ or VTH = 0.7837 V Now 0.7837 − 0.7 I BQ = = 0.00993 mA 8.43 and I CQ = (100 )( 0.00993) = 0.993 mA We have VCEQ = VCC − I CQ RC or 2.5 = 5 − ( 0.993) RC which yields RC = 2.52 k Ω (b) Power dissipated in RC: PRC = I CQ RC = ( 0.993) ( 2.52 ) 2 2 or PRC = 2.48 mW Power dissipated in transistor: PQ ≅ I CQVCEQ = ( 0.993)( 2.5 ) or PQ = 2.48 mW (c) ac analysis: Maximum ac collector current: ic = ( 0.993) cos ω t ( mA ) average ac power dissipated in RC: 1 1 pRC = ( 0.993) RC = ( 0.993) ( 2.52 ) 2 2 2 2 or pRC = 1.24 mW Now pRC 1.24 Fraction = = = 0.25 PRC + PQ 2.48 + 2.48