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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

 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

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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.

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9
10
Example

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Example

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Example

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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
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The common-base
characteristics

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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.

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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

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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

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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

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Common Emitter NPN Configuration

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

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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
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   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

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The output graph

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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
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re model for common-emitter

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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

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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

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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

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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.

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