The Common-Base Amplifier Basic Circuit DC Solution by lanyuehua

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									The Common-Base Amplifier
Basic Circuit
Fig. 1 shows the circuit diagram of a single stage common-base amplifier. The object is to solve
for the small-signal voltage gain, input resistance, and output resistance.




                                Figure 1: Common-base amplifier.



DC Solution
(a) Replace the capacitors with open circuits. Look out of the 3 BJT terminals and make Thévenin
equivalent circuits as shown in Fig. 2.

                                    V + R2 + V − R1
                            VBB =                          RBB = R1 kR2
                                        R1 + R2

                    VEE = V −       REE = RE            VCC = V +   RCC = RC
   (b) Make an “educated guess” for VBE . Write the loop equation between the VBB and the VEE
nodes. To solve for IC , this equation is
                                                             IC             IC
               VBB − VEE = IB RBB + VBE + IE REE =              RBB + VBE +    REE
                                                              β              α

   (c) Solve the loop equation for the currents.

                                                       VBB − VEE − VBE
                             IC = αIE = βIB =
                                                       RBB /β + REE /α

   (d) Verify that VCB > 0 for the active mode.

   VCB = VC − VB = (VCC − IC RCC ) − (VBB − IB RBB ) = VCC − VBB − IC (RCC − RBB /β)

                                                   1
                                       Figure 2: DC bias circuit.

Small-Signal or AC Solutions
(a) Redraw the circuit with V + = V − = 0 and all capacitors replaced with short circuits as shown
in Fig. 3.




                                        Figure 3: Signal circuit.

   (b) Calculate gm , rπ , re , and r0 from the DC solution..
                              IC             VT               VT              VA + VCE
                      gm =            rπ =             re =            r0 =
                              VT             IB               IE                 IC
    (c) Replace the circuits looking out of the base and emitter with Thévenin equivalent circuits
as shown in Fig. 4.
                                                               RE
                    vtb = 0        Rtb = 0        vte = vs                Rte = Rs kRE
                                                             Rs + RE


                                                       2
                        Figure 4: Signal circuit with Thévenin emitter circuit.

Exact Solution
(a) Replace the BJT in Fig. 4 with the Thévenin emitter circuit and the Norton collector circuit
as shown in Fig. 5.




                            Figure 5: Emitter and collector equivalent circuits.

   (b) Solve for ic(sc) .
                                                                      RE
                                    ic(sc) = −Gme vte = −Gme vs
                                                                   Rs + RE
                                          1       αr0 + re0              rx
                                                                   0
                             Gme =          0           0
                                                                  re =      + re
                                     Rte + re kr0 r0 + re              1+β
   (c) Solve for vo .
                                                                  RE
                            vo = −ic(sc) ric kRC kRL = Gme vs           ric kRC kRL
                                                                Rs + RE
                                                            0
                                                     r0 + re kRte
                                         ric =                0
                                                 1 − αRte / (re + Rte )

                                                        3
   (d) Solve for the voltage gain.

                                           vo     RE
                                    Av =      =         Gme ric kRC kRL
                                           vs   Rs + RE
   (e) Solve for rin .

                                                          0          r0 + Rtc
                            rin = R1 kR2 krie      rie = re    0
                                                              re + r0 + Rtc / (1 + β)

   (f) Solve for rout .
                                                 rout = ric kRC

Example 1 For the CB amplifier in Fig. 1, it is given that Rs = 100 Ω, R1 = 120 kΩ, R2 = 100 kΩ,
RC = 4.3 kΩ, RE = 5.6 kΩ, R3 = 100 Ω, RL = 20 kΩ, V + = 15 V, V − = −15 V, VBE = 0.65 V,
β = 99, α = 0.99, rx = 20 Ω, VA = 100 V and VT = 0.025 V. Solve for Av , rin , and rout .

    Solution. Because the dc bias circuit is the same as for the common-emitter amplifier example,
the dc bias values, re , gm , rπ , and r0 are the same.
    In the signal circuit, the Thévenin voltage and resistance seen looking out of the emitter are
given by
                                  RE
                      vte =             vs = 0.9825vs   Rte = Rs kRE = 98.25 Ω
                              Rs + RE
The Thévenin resistances seen looking out of the base and the collector are

                                   Rtb = 0       Rtc = RC kRL = 3.539 kΩ
                       0
   Next, we calculate re , Gme , ric , and rie .

                       Rtb + rx                                      1      αr0 + re 0     1
                 0
                re =            + re = 12.03 Ω       Gme =             0           0
                                                                                       =       S
                        1+β                                     Rte + re kr0 r0 + re     111.4
                           0
                    r0 + re kRte                                                r0 + Rtc
                                                                   0
        ric =                0
                                       = 442.3 kΩ           rie = re    0
                                                                                                 = 12.83 Ω
                1 − αRte / (re + Rte )                                 re   + r0 + Rtc / (1 + β)
   The output voltage is given by
                                                                      RE
                       vo = Gme (ric kRtc ) vte = Gme (ric kRtc )           vs = 30.97vs
                                                                    Rs + RE
Thus the voltage gain is
                                                   Av = 30.97
The input and output resistances are

                          rin = R1 kR2 krib = 12.81 Ω        rout = ric kRC = 4.259 kΩ

Approximate Solutions
These solutions assume that r0 = ∞ except in calculating ric . In this case, ic(sc) = i0 = αi0 = βib .
                                                                                       c     e




                                                        4
                                  Figure 6: Simplified T model circuit.

Simplified T Model Solution
(a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the
BJT with the simplified T model as shown in Fig. 6.
    (b) Solve for i0 and ric .
                   c

                                 ¡ 0     ¢ i0 ¡ 0     ¢               α
                    0 − vte = i0 re + Rte = c re + Rte =⇒ i0 = −vte 0
                               e                           c
                                            α                      re + Rte
                                                           0
                                                    r0 + re kRte
                                        ric =                0
                                                1 − αRte / (re + Rte )
   (c) Solve for vo .
                                           α                        RE       α
          vo = −i0 ric kRC kRL = vte
                 c                       0
                                                 ric kRC kRL = vs          0
                                                                                   ric kRC kRL
                                        re + Rte                  Rs + RE re + Rte

   (d) Solve for the voltage gain.
                                        vo      Rs      α
                                 Av =      =          0 +R
                                                             ric kRC kRL
                                        vs   Rs + RE re   te

   (e) Solve for rie and rin .
                                                                    ve
                                                     0
                                        0 − ve = i0 re =⇒ i0 = −
                                                  e        e         0
                                                                    re
                                                        ve     0
                                                rie =       = re
                                                        −i0
                                                          e
                                                        0
                                                 rin = re kRE
   (f) Solve for rout .
                                                rout = ric kRC

Example 2 For Example 1, use the simplified T-model solutions to calculate the values of Av , rin ,
and rout .

                                                        5
                                       ¡            ¢ ¡           ¢
                          Av = 0.9825 × 8.978 × 10−3 × 3.511 × 103 = 30.97
                                      rin = 12 Ω      rout = 4.259 kΩ

π Model Solution
(a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the
BJT with the π model as shown in Fig. 7.




                                    Figure 7: Hybrid-π model circuit.

   (b) Solve for i0 and ric .
                  c

                                               i0
                                                c      i0 i0                  −vte
             0 − vte = ib rx + vπ + i0 Rte =
                                     e            rx + c + c Rte =⇒ i0 =
                                                                     c
                                               β      gm   α             rx    1   Rte
                                                                            +    +
                                                                         β    gm    α
                                                          0
                                                   r0 + re kRte
                                       ric =                0
                                               1 − αRte / (re + Rte )
   (c) Solve for vo .

                                      vte                     RE           1
   vo = −i0 ric kRC kRL =                      r kR kR = vs                       r kR kR
          c                     rx    1     Rte ic C L      Rs + RE rx    1    Rte ic C L
                                   +      +                            +     +
                                β    gm      α                      β    gm     α

   (d) Solve for the voltage gain.

                                   vo     RE           1
                            Av =      =                       r kR kR
                                   vs   Rs + RE rx    1    Rte ic C L
                                                   +     +
                                                β    gm     α

   (e) Solve for rout .
                                               rout = ric kRC


                                                      6
   (f) Solve for rie and rin .
                                                    i0
                                                     e                         1+β
                        0 − ve = ib (rx + rπ ) =       (rx + rπ ) =⇒ i0 = −ve
                                                                      e
                                                   1+β                        rx + rπ
                                                      ve    rx + rπ
                                            rie =       0
                                                          =
                                                      −ie    1+β
                                                   rin = rie kRE

Example 3 For Example 1, use the π-model solutions to calculate the values of Av , rin , and rout .

                                       ¡            ¢ ¡           ¢
                          Av = 0.9825 × 8.978 × 10−3 × 3.539 × 103 = 30.97
                                       rin = 12 Ω         rout = 4.259 kΩ

T Model Solution
(a) After making the Thévenin equivalent circuits looking out of the base and emitter, replace the
BJT with the T model as shown in Fig.??.




                                         Figure 8: T model circuit.

   (b) Solve for i0 and ric .
                  c

                                                    i0
                                                     c     i0                        −vte
            0 − vte = ib rx + i0 (re + Rte ) =
                               e                       rx + c (re + Rte ) =⇒ i0 =
                                                                              c
                                                    β       α                     rx re + Rte
                                                                                    +
                                                                                  β      α
                                                              0
                                                       r0 + re kRte
                                         ric =                  0
                                                   1 − αRte / (re + Rte )
   (c) Solve for vo .
                                     vte                   RE        1
     vo = −i0 ric kRC kRL =                 r kR kR = vs                    r kR kR
            c                    rx re + Rte ic C L      Rs + RE rx re + Rte ic C L
                                   +                               +
                                 β       α                       β     α

                                                          7
   (d) Solve for the voltage gain.

                                       vo     RE        1
                              Av =        =                    r kR kR
                                       vs   Rs + RE rx re + Rte ic C L
                                                      +
                                                    β     α

   (e) Solve for rie and rin .
                                                              µ              ¶
                                         i0
                                          e                        rx                        −ve
           0 − ve =   ib rx + i0 re
                               e      =     rx + i0 re = i0
                                                  e       e           + re       =⇒ i0 =
                                                                                     e      rx
                                        1+β                       1+β                          + re
                                                                                           1+β
                                                    ve     rx
                                            rie =       =     + re
                                                    −i0
                                                      e   1+β
                                                    rin = RE krie
   (f) Solve for rout .
                                                 rout = ric kRC

Example 4 For Example 1, use the T-model solutions to calculate the values of Av , rin , and rout .

                                       ¡            ¢ ¡           ¢
                          Av = 0.9825 × 8.978 × 10−3 × 3.539 × 103 = 30.97
                                         rin = 12 Ω       rout = 4.259 kΩ




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