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Introduction to BJT Small Signal Analysis by moti

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									 Introduction to BJT
Small Signal Analysis
      CHAPTER 5




        Sem I 0809/rosdiyana
                  Introduction
•To begin analyze of small-signal AC response of BJT
 amplifier the knowledge of modeling the transistor is
 important.
•The input signal will determine whether it’s a small signal
 (AC) or large signal (DC) analysis.
•The goal when modeling small-signal behavior is to make of
 a transistor that work for small-signal enough to “keep
 things linear” (i.e.: not distort too much) [3]
•There are two models commonly used in the small signal
 analysis:
       a) re model
       b) hybrid equivalent model
                                                          2
               Introduction

Disadvantage
 Re model

    • Fails to account the output impedance
      level of device and feedback effect from
      output to input
   Hybrid equivalent model
    • Limited to specified operating condition
      in order to obtain accurate result

                                                 3
    Amplification in the AC domain
   The transistor can be employed as
    an amplifying device. That is, the
    output sinusoidal signal is greater
    than the input signal or the ac input
    power is greater than ac input
    power.
   How the ac power output can be
    greater than the input ac power?

                                            4
    Amplification in the AC domain
   Conservation; output
    power of a system
    cannot be large than its
    input and the efficiency
    cannot be greater than 1
   The input dc plays the
    important role for the
    amplification to
    contribute its level to the
    ac domain where the
    conversion will become
    as η=Po(ac)/Pi(dc)               5
    Amplification in the AC domain
   The superposition theorem is applicable for the
    analysis and design of the dc & ac components of
    a BJT network, permitting the separation of the
    analysis of the dc & ac responses of the system.
   In other words, one can make a complete dc
    analysis of a system before considering the ac
    response.
   Once the dc analysis is complete, the ac response
    can be determined using a completely ac
    analysis.



                                                    6
          BJT Transistor Model
   Use equivalent circuit
   Schematic symbol for the device can be replaced
    by this equivalent circuits.
   Basic methods of circuit analysis is applied.
   DC levels were important to determine the Q-
    point
   Once determined, the DC level can be ignored in
    the AC analysis of the network.
   Coupling capacitors & bypass capacitor were
    chosen to have a very small reactance at the
    frequency of applications.

                                                      7
         BJT Transistor Model
The AC equivalent of a network is
obtained by:
1.  Setting all DC sources to zero & replacing them
    by a short-circuit equivalent.
2.  Replacing all capacitors by a short-circuit
    equivalent.
3.  Removing all elements bypassed by short-
    circuit equivalent.
4.  Redrawing the network.




                                                  8
9
10
Example




          11
Example




          12
Example




          13
         The re transistor model

   Common Base PNP Configuration




                                    14
Common Base PNP Configuration

                Transistor is replaced by
                 a single diode between E
                 & B, and control current
                 source between B & C
                Collector current Ic is
                 controlled by the level of
                 emitter current Ie.
                For the ac response the
                 diode can be replaced by
                 its equivalent ac
                 resistance.


                                         15
     Common Base PNP Configuration

   The ac resistance
    of a diode can be
    determined by the
    equation;
           26 mV
      re 
             IE

    Where ID is the dc
    current through
    the diode at the
    Q-point.
                                     16
Common Base PNP Configuration
                  Input impedance is
                   relatively small and
                   output impedance
                   quite high.
                      Z i  reCB
                  range from a few Ω
                   to max 50 Ω

                       Zo   CB
                  Typical values are in
                   the M Ω
                                          17
The common-base
  characteristics




                    18
                           Voltage Gain
         tage : Vo   I o RL
output vol
                       ( I C ) RL
                        I e RL
input voltage :     Vi  I i Z i
                        Ie Zi
                        I e re
                        VO I e RL
voltage gain :     AV     
                        Vi   I e re
                           RL
                       
                         re
                        RL
                  AV 
                        re                19
                   Current Gain
                      I o  I C  I e
                 Ai          
                      Ii   Ie     Ie
                 Ai    1
   The fact that the polarity of the Vo as determined
    by the current IC is the same as defined by figure
    below.
   It reveals that Vo and Vi are in phase for the
    common-base configuration.




                                                     20
   Common Base PNP Configuration




Approximate model for a common-base npn transistor configuration



                                                                   21
Example 1: For a common-base configuration in figure
below with IE=4mA, =0.98 and AC signal of 2mV is
applied between the base and emitter terminal:
a) Determine the Zi b) Calculate Av if RL=0.56k
c) Find Zo and Ai



                   Ie                Ic
             e                                    c

                     re               Ic  α Ie
             b                                    b

                 common-base re equivalent cct



                                                       22
Solution:
                         26m 26m
            a) Zi  re           6.5
                          IE   4m

                    RL 0.98(0.56k)
            b) Av                  84.43
                     re     6.5

            c) Zo  Ω
                Io
            Ai      0.98
                Ii



                                              23
Example 2: For a common-base configuration in previous
example with Ie=0.5mA, =0.98 and AC signal of 10mV is
applied, determine:
a) Zi b) Vo if RL=1.2k c) Av d)Ai e) Ib

    Solution:
                                d) Ai    0.98
            Vi 10m
    a) Zi            20
            Ie   0.5m
                                e) Ib  Ie - Ic
                                 Ie - Ie
    b) Vo  IcRL  IeRL
     0.98(0.5m) (1.2k)          0.5m(1   )
     588mV                      0.5m(1  0.98)
                                 10A

              Vo 588m
    c) Av             58.8
              Vi   10m


                                                         24
     Common Emitter NPN Configuration

   Base and emitter
    are input
    terminal
   Collector and
    emitter are
    output terminals



                                        25
    Common Emitter NPN Configuration

   Substitute re
    equivalent circuit

          I c  I b
   Current through
    diode
     I e  I c  I b  I b  I b
     I e  (  1) I b  I b
                                       26
    Input impedance

                      Vi Vbe
input impedance: Z i  
                      Ii Ib
input voltage :      Vi  I e re
                        (   1) I b re
                         (   1) I b re
 so that            Zi 
                              Ib
                   Z i  (   1)re
 usually greater than 1 ;     Z i  re

                                           27
The output graph




                   28
      Output impedance Zo
         Ii=Ib
  b                                                   c

                                      Ib
                      re                         ro       Zo
  e                                                   e
 re model for the C-E transistor configuration

             Ii=Ib = 0A
         b                                        c


 Vs=0V                          Ib  0A
                           re               ro

         e                                        e


Zo  ro
if ro is ignored thus the
Zo  Ω (open cct, high impedance)                             29
     Voltage Gain                   Current Gain

output voltage : Vo   I o RL
                  Vo   I c RL          I o I C I b
                                     Ai       
                        I b RL        Ii Ib    Ib
input voltage : Vi  I i Z i         Ai  
                       I b re
                Vo  I b RL
so that    AV     
                Vi   I b re
                  RL
           AV 
                  re
                                                        30
re model for common-emitter




                              31
Example 3: Given =120 and IE(dc)=3.2mA for a common-
emitter configuration with ro=  , determine:

a) Zi b)Av if a load of 2 k is applied c) Ai with the 2 k load
                 Solution :
                        26m 26m
                a) re              8.125
                         IE    3.2m
                Zi  re  120(8.125)  975

                         RL     2k
                b)Av               246.15
                         re    8.125

                          Io
                c) Ai          120
                          Ii

                                                              32
Example 4: Using the npn common-emitter configuration,
determine the following if =80, IE(dc)=2 mA and ro=40 k

a) Zi            b) Ai if RL =1.2k         c) Av if RL=1.2k 
        Ii=Ib
  b                       c                          Solution :
                                                Io
                                                            26m 26m
                                Ib                  a) re               13
                  re                   ro   RL                IE    2m
                                                     Zi  re  80(13)  1.04k
 e
re model for the C-E transistor configuration




                                                                             33
Solution (cont)
         Io IL
b) Ai       
         Ii Ib
     ro( Ib)
IL 
     ro  RL
      ro( Ib)
Ai   ro  RL  ro       40k
                                    (80)
          Ib   ro  RL   40k  1.2k
 77.67

           RL ro        1.2k 40k
c)Av                             89.6
            re            13




                                              34
    Common Collector Configuration
   For the CC configuration, the model
    defined for the common-emitter
    configuration is normally applied
    rather than defining a model for the
    common-collector configuration.




                                           35

								
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