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Chapter 10 Other BJT Amplifiers

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Chapter 10 Other BJT Amplifiers Powered By Docstoc
					          Chapter 10
      Other BJT Amplifiers
Common-Collector & Common-Base Amp.
   Pictures are redrawn (with some modifications) from
     Introductory Electronic Devices and Circuits
                            By
                     Robert T. Paynter




                                                         1
              Objectives
• Perform a complete dc analysis of an emitter
  follower.
• Describe the use of the emitter follower as a
  current amplifier and as a buffer.
• Perform a complete dc analysis of a common-
  base amplifier.
• Describe the use of the common-base amplifier
  as a voltage amplifier and a buffer.
• List the overall gain and impedance
  characteristics of all commonly used BJT
  amplifiers.


                                                  2
 Fig 10.1 The emitter follower (common-
          collector amplifier).
                    VCC 1. Convert voltage divider circuit to its
 DC                        corresponding Thevenin equivalent circuit:
 Analysis:                                    R2
                                Vth  VCC
                                            R1  R2
               R1
1V      CC1                      Rth  R1 R2
                                                      2. Find IB and IE.
                         CC2                0.95V
                                                                 Vth  0.7V
                                                      IB 
                                                             Rth   hFE  1 RE
               R2      RE      Load
                                                      I E   hFE  1 I B

              3. Check VCEQ whether the transistor is active or not.
                 VCEQ  VCC  I E RE
                                                                               3
Fig 10.2 Example 10.1.
      10V            Step 1:
                                  R2               20kΩ
                    Vth  VCC            10V               5V
                                R1  R2         20kΩ  20kΩ
                    Rth  R1 R2  20kΩ 20kΩ  10kΩ
      R1
CC1   20k                         Step 2:
             hFE = 150
                                             Vth  0.7V
                                  IB 
                 CC2                     Rth   hFE  1 RE
                                             5V  0.7V
      R2     RE                                              5.621μA
                          RL            10kΩ  151 5kΩ
      20k   5k                  I E   hFE  1 I B  151 5.621μA
                                      848.8μA

                Step 3:
               VCEQ  VCC  I E RE  10V  848.8μA  5kΩ  5.756V
                                                                         4
         Fig 10.3 Example 10.2.
 Derive and draw the dc load line for the circuit
 shown in Figure 10.2.

IC(mA)                                                              VCC 10V
                                                    I C  sat             2mA
                                                                    RE 5kΩ

 3                      VCC                         VCE (off )  VCC  10V
         I C (sat )   
 2                      RE
 1                               VCE (off )  VCC
                                    VCE(V)
         2    4       6   8 10

                                                                                    5
Fig 10.5 Emitter follower and
         its ac equivalents.
       12V



                                         ib          ic = hfeib
       R1
 CC1   25k
                        vin          R1||R2          r'e
                CC2
                                                             vout
       R2     RE      RL
                                                     rE  RE RL
       33k   2k     5k


                                     vout    r
                              Av          E
                                     vin re  rE
                                1      rE   re
                                                                    6
        Ai of CC Amp.
From CE amp.,

                    Z in rC
  Ai   h fe                     rC  RC     RL 
                 Z in(base) RL

So for CC amp.,

                   Z in rE
   Ai  h fc                      rE  RE    RL 
                Z in(base) RL
where

        h fc  h fe  1 (from Appendix C)
   Z in(base)  h fc (re  rE )  h fe (re  rE )
        Z in  R1 R2 Z in(base)                       7
                  Example 10.5.
Determine the input impedance of the amplifier
shown in Figure 10.5a. Assume that the value of
hfe for the transistor is listed on the spec sheet as
220.
             R2             33kΩ
VB  VCC            12V              6.828V
           R1  R2       25kΩ  33kΩ
VE  VB  0.7V  6.828V  0.7V  6.128V
     VE 6.128V
IE            3.064mA
     RE   2kΩ

re 
        25mV
             
                 25mV
                        8.16Ω        Z in(base)  h fc (re  rE )   h fe  1 (re  rE )
          IE   3.064mA
                                                 2211.437kΩ  317.5kΩ
rE  RE RL  2kΩ 5kΩ  1.429kΩ
                                           Z in  R1 R2 Z in(base)
         rE    1.429kΩ
Av                    0.9943                  25kΩ 33kΩ 317.5kΩ
     re  rE 1.437kΩ
                                                 13.61kΩ                                       8
                     Zout of CC Amp.
                                         RS        ib
               +8V                                        b              c

                                                  R1||R2      hie        ic = hfeib   h fc  h fe  1  h fe
         R1                                                                           hrc  1  hre  1
                                                                    e
                                                                    RE       Zout     hic  hie
 RS                                                                                   hoc  hoe
         R2                              RS        ib
                   RE                                   b                    e
  vS                        Zout
                                                 R1||R2       hic        ie = hfcib       RE
                                                                                               Zout

                                                               hrcvec
                    R                      Rin 
                                                  
Z out  RE     re  in      RE    re           
                    h fc                  h fe  1              c
                                                   
where          
              Rin  RS R1 R2
                                                                                                               9
  Fig 10.6 Example 10.6.
Determine the output impedance (Zout) of the
amplifier shown in Figure 10.6.
                 +8V                      hie    hic 4kΩ
                               re                     20Ω
                                       h fe  1 h fc 200

         R1                     
                               Rin  R1 R2 RS
                   hic = 4k
         3k       hfc = 200        3kΩ 4.7kΩ 600Ω  451.9Ω
                   hFC = 180
  RS                                               Rin 
                                                      
                               Z out    RE  re       
  600   R2                                        h fc 
         4.7k     RE                                   
                   390                                   451.9Ω 
                                        390Ω  20Ω              
                                                            200 
                                        21.06Ω                       10
Applications for CC Amp.
• Current amplifier (Av  1)
• Buffer – a circuit used to compensate
  for an impedance mismatch between
  a source and its load




                                          11
Fig 10.12 Using a buffer to overcome an
          impedance mismatch.
                          Zout
                         10k


                                                 ZL
                   12V
                                                 1k



                     Source                  Load


            Zout              Av = 0.98
           10k


                              Zin         Zout          ZL
     12V
                              180k       20           1k



       Source                    Buffer                Load   12
Fig 10.14 Darlington emitter-
          follower. (1)
          VCC
                                   DC Analysis:
                         IC1+IC2            VB1  2VBE
          IC1                       IE2   
   R1                                           RE
        IB1                                   I E1         I
                  Q1     IC2        I B1          , IB2  E 2
                                             hFC1         hFC 2
   R2
                         Q2         Rin 2  hFC 2  re2  RE   hFC 2 RE
              IE1=IB2               Rin1  hFC1 (re1  Rin 2 )
                   IE2                     hFC1 Rin 2  hFC1hFC 2 RE
                   RE         RL



                                                                             13
Fig 10.14 Darlington emitter-
          follower. (2)
                    VCC
                                          AC Analysis:
                                IC1+IC2
                   IC1                     Zin(base)  hic1  h fc1  hic 2  h fc 2rE 
      R1
              IB1                          Z in  R1 R2 Z in(base)
                          Q1    IC2
                                                                   re1   Rin / h fc1  
                                                                               
      R2                                   Zout  RE        re2                         
                  R’in                                                     h fc 2         
                                Q2                                                        
                     IE1=IB2
                                                                 Z in rE
  
 Rin  R1 R2 RS           IE2              Ai  h fc1h fc 2
                                                              Z in(base) RL
                          RE         RL
            rE  RE RL


                                                                                               14
Fig 10.16 Typical common-
          base amplifier.
                   +VCC


                      RC          vin                                 vout
                                        RE                RC
                                                    R1
vin                        vout                                +VCC
      RE
                                             CB     R2

       -VEE

           Emitter bias                  Voltage-divider bias



                                                                             15
Fig 10.17 The common-base ac
          equivalent circuit.
                     iin    ie                                      iout
                                                                           vout

        vin                RE         r'e         ic = hfeib   RC           RL
                 Zin                                                Zout
                                            ib
                                                          rC  RC RL

Av 
       vout ic rC rC
                               iout    vout / RL     rC  RE re
                            Ai                     
       vin   ie re re          iin vin /  RE re re    RL
Zin  re RE  re                    RC RL 1 RE re 1                   RC RE
                                                                
                                    RC  RL  re  RE  re RL  RC  RL  RE  re
               1
Z out  RC         RC               RC
              hob                
                                   RC  RL                                                 16
        Fig 10.18 Example 10.10.
  Determine the gain and impedance values for the
  circuit shown in Figure 10.18.
                                   +10V
                                                      Zin  re  75.58Ω
              RS
                                      RC              Zout  RC  10kΩ
                                      10k
             75
                                                      rC  RC RL  10kΩ 5.1kΩ
                                             RL
100V                RE
                                             5.1k        3.377kΩ
                   13k
                                                           r    3.377kΩ
                                                      Av  C            44.69
                      -5V                                  re 75.58Ω
        0.7V   VEE        0.7V  5V
IE                                       330.8μA
              RE                 13kΩ
        25mV    25mV
re                   75.58Ω
          IE   330.8μA                                                            17
Common-Base Applications

• Voltage amplification without current
  gain
• High-frequency applications (more
  common)




                                          18
    BJT Amplifiers: A Summary

          CE              CC*         Darlington CC      CB

Av    Midrange            <1                <1         Midrange

Ai    Midrange         Midrange       Extremely high     <1

Ap       High          Midrange             High       Midrange

Zin   Midrange           High         Extremely high     Low
Zou   Midrange            Low               Low          High
t

*Another name for CC is emitter follower.



                                                                  19
           Summary
• The emitter-follower (common-
  collector amplifier.)
• Emitter-follower ac analysis.
• Emitter-follower applications.
• Darlington emitter-follower.
• The common-base amplifier and their
  applications.

                                        20

				
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