# BJT Modeling (AC Analysis)

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```					  Analogue Electronics I
EAB 2014

Jan 2009

Lecture 5 – BJT AC Analysis
Small Signal Transistor Modeling

1
Topics To Cover
• Small-signal vs Large-signal techniques
• Reviewing the models used in small signal ac
analysis
re, hybrid equivalent and hybrid π model

2
INTRODUCTION
• Transistor equivalent models that represents the
operation of a transistor during ac mode.

• AC analysis techniques are determined by the
magnitude of the input signal
→ Small Signal (low power, low I & V)
→ Large Signal (high power, high I & V)

• Small-signal transistor equivalent models:
→ re equivalent model.
→ hybrid equivalent model.
→ hybrid π model.

3
BJT TRANSISTOR MODELING
What is a BJT modeling?

A combination of circuit elements that best
approximates the actual behavior of BJT under
specific operating conditions

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BJT TRANSISTOR MODELING
Small signal modeling steps:
• Obtain all the Q-Points through DC-Biasing (ICQ,
VCEQ, IBQ, ……etc).
• For a Common-Emitter configuration,
VCC

R1         RC
C2

C1

RS
VO
VS   Vi        R2         RE        C3

5
BJT TRANSISTOR MODELING
1. Set all DC supplies potential to zero (Ground) and assume
all capacitors are short circuited (due to small reactance value).

R1        RC

RS
VO
VS       Vi   R2

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BJT TRANSISTOR MODELING
2. Rearrange the elements then choose the appropriate small
signal equivalent circuit for the analysis.
Ii                         IO
B   smallsignal   C
cct.
RS
R1 || R2
E                RC   VO
VS       Vi
Zi                         ZO

What are the important quantities to be determined?
Input impedance, output impedance, voltage gain, current gain

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TWO-PORT NETWORK REVISITED
Ii                                  Io

Two port
Vi      Zi                                   Zo    VO
sy stem

Input Impedance, Zi :
Ii
V − Vi                           Rsense
Ii = s
Rsense                                             Two port
Vs             Zi       Vi
sy stem
Vi RsenseVi
Zi = =
I i Vs − Vi

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TWO-PORT NETWORK REVISITED
•The importance of the input impedance.
Ii      Rsource

Vs           Zi        Vi
Two port
sy stem

Ω           Ω
Given: Rsource = 600Ω, Zi = 1.2kΩ, Vs = 10 mV
Using voltage divider to the input circuit,
Z iVs
Vi =                    = 6.67mV
Z i + Rsource
Ω
When Zi = 1.2kΩ, Vi = 6.67 mV → 66% Vs available at Vi.
When Zi = 600 Ω, Vi = 5 mV → 50% Vs available at Vi.
Ω
When Zi = 8.2kΩ, Vi = 9.31mV → 93.2% Vs available at Vi.
→ The higher the i/p impedance, the better it is !!!

9
TWO-PORT NETWORK REVISITED
Output Impedance, Zo :
• The o/p impedance is determined at the o/p terminals
looking back into the system with the applied signal set to
zero.
• Vs is applied to the o/p terminal.
Rso urce                      R sense    o
I

Vs=0                   Two port    Vo         Zo       Vs
sy stem

Vo RsenseVo
Zo =    =                        Vs − Vo
I o Vs − Vo            Io =
Rsense

10
TWO-PORT NETWORK REVISITED
Voltage Gain (with the load connected) , Av

Vo
Defined by,               Av =
Vi

Ii                        Io

Two port              RL
Zi        Vi                    Vo
system

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TWO-PORT NETWORK REVISITED

Voltage Gain with no load , AVNL :
Defined by,               Vo
AVNL   =
Vi   RL =∞Ω

Rsource

Vs                       Two port
Zi      Vi                     Vo
sy stem

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TWO-PORT NETWORK REVISITED

Voltage Gain with source connected , AVS :
Rsource
Z iVs
Vi =
Z i + Rsource   Vs                   Two port
Zi      Vi                       Vo
sy stem
Vi      Zi
=
Vs Z i + Rsource
      Zi         
Vo Vi Vo                AVS =                   AVNL
AVS =   =   ⋅                        Z i + Rsource   
VS VS Vi

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TWO-PORT NETWORK REVISITED
Current Gain, Ai :
Io
Defined by,    Ai =
Ii
Ii                      Io

Vi                              Two port
Zi     Vi                      RL Vo
Ii =                                  sy stem
Zi
Vo
Io = −                       Io    Vo / RL         Zi
RL               Ai =    =−          = − AV
Ii    Vi / Z i        RL

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Example 5.1

Ω
Given VS = 40mV, Rsource = 1.2kΩ, AVNL = 320, Vo = 7.68 V
Find : Vi, Ii, Zi, AVS

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re TRANSISTOR MODEL

• Derived directly from the operating conditions of the
transistor.
• Can also be derived directly from the hybrid parameters.
• Employs a diode and controlled current source to describe
the transistor behavior.
• re model for the CB, CE and CC BJT transistor will be
introduced

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re TRANSISTOR MODEL
Common Base Configuration (Zi, Zo, AV, Ai)
Ic
IE                   IC                         Ie
E                             C                 e                             c

Ic = αIe

B                             B                 b                             b

Ii=Ie                 Io= − Ic
Ii                    Io               e                                           c

Vi Z
CB                                            re          Zo
Zo Vo
i
Amplifier                        Vi Zi                              Vo
Ic = αIe
b                                           b

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re TRANSISTOR MODEL
V
Input Impedance, (Zi)       Z i = i = re  Since, Vi=Iere=Iire
Ii
Output Impedance, (Zo)
•Set input source to zero (short circuit).
Ii                           Ic         Io
Ie

re       Ic = αIe
Vi                            Vo        Vs

Vo                                    Since, Ii=Ie=0A
Zo =    = ∞Ω                                          α
IC=αIE=0A
Io
→Typically in MΩ range.
Ω                                       Io=-IC=0A

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re TRANSISTOR MODEL
Ii                           Ic         Io
Ie

Voltage Gain, (AV)
re
Vi   Zi                           Zo   Vo    RL
Ic = αIe

Vo
AV =        Vo = − I o RL = −( − I C ) RL = −( −α I e ) RL = α I e RL
Vi
Vi = I e Z i = I e re
Vo β I e RL α RL RL
AV =      =           =       ≈
Vi        I e re     re     re
Io    I     αI
Current Gain, (Ai)        Ai =      = − c = − e = −α ≈ −1
Ii    Ie     Ie

19
re TRANSISTOR MODEL
Common Emitter Configuration (Zi, Zo, AV, Ai)

Ic                                              Ic
C                                               C
Ic = βIb
Ib                                      Ib
B                                       B

re

E                          E            E                                  E

Ii                                        Io

Vi                CE
Zi                              Zo    Vo
Amplif ier

20
re TRANSISTOR MODEL
Io=Ic
Input Impedance, (Zi)
Ic = βIb
Ii=Ib
Vo

Vi             Vbe   re

Vi Vbe
Zi = =            Vi = Vbe = I e re = ( β + 1) I b re
Ii Ib

( β + 1) I b re
=
Zi =                  = ( β + 1) re ≈ β re
Ib

21
re TRANSISTOR MODEL
Output Impedance, (Zo)
• Applied signal is set to zero, Ic=0A.
Io =Ic
Ic = β Ib

Ii =Ib
ro
Vo      Vs

Vi             V be   re

Z o = ro
•If ro is ignored, then                      Z o = ∞Ω

22
re TRANSISTOR MODEL
Io = Ic = β Ib
Ic = β Ib
Voltage Gain, (AV)
Ii =Ib            Common
Emitter    Zo = ∞          RL Vo

V i Z i = β re
Amplifier
Vbe  re

Vo
Av =                Vo = − I o RL = − I C RL = − β I b RL
Vi
Vi = I i Z i = I b β re

− β I b RL    RL
AV =            =−
I b β re    re
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re TRANSISTOR MODEL

Io Ic β Ib
Current Gain, (Ai)            Ai =   =   =    =β
Ii Ib   Ib
Finally after considering Zi, Zo, AV, and Ai,the small
signal equivalent cct. for CE can be summarized as
Ii                          Ic      Io
Ib

β re
Vi                          ro     Vo
Ic = βIb

24
re TRANSISTOR MODEL
re equivalent circuit
Ic
Ie
e                                 c
re
Common Base                                     Ic = αIe

b                                 b

Ii                                     Ic    Io
Ib

Common Emitter                   β re
Vi                                     ro   Vo
Ic = βIb

25
HYBRID TRANSISTOR MODEL
• re model fails to account for the output impedance and the
feedback effect.
• The most popular method used.
• Normally provided at certain operating conditions such as
at IC=1mA, VCE=10V, f=1kHZ.
• Using the same two-port system,
Ii                       Io
1                                           2

Vi                                 Vo

1'                                        2'

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--- (b)
HYBRID TRANSISTOR MODEL
Vi = h11 I i + h12Vo --- (a) I o = h21 I i + h22Vo

•The parameters relating the four variables are called
h-parameters.
Setting Vo=0V (short circuit the o/p terminals)
solving for eqn (a).
Vi             ohms
h11 =
Ii   V0 = 0V

h11 - short-cct. input-impedance parameter

27
HYBRID TRANSISTOR MODEL
Setting Ii=0A (opening the input leads)
solving for eqn (a).
Vi
h12 =                  unitless
Vo    Ii = 0 A

h12 - open cct. reverse transfer voltage ratio parameter
Setting Vo=0V (short circuit the o/p terminals)
solving for eqn (b).          I
h21 = o         unitless
Ii    V0 = 0V

h21 - short-cct. Forward transfer current ratio parameter

28
HYBRID TRANSISTOR MODEL

Setting Ii=0A (opening the input leads)
solving for eqn (b).
Io
h22 =                   siemens
Vo   Ii = 0 A

h22 - open cct. output admittance parameter

29
HYBRID TRANSISTOR MODEL

By applying KVL and KCL to find the circuit that fits all the
equations obtained previously, we’ll get :
h11
Ii                                              Io

Vi                h12Vo      h21Ii            h22          Vo

Hybrid input equiv. circuit   Hybrid output equiv. circuit

30
HYBRID TRANSISTOR MODEL
hi
Ii                                        Io

Vi              hrVo     hf Ii        ho            Vo

h11 → hi
h12 → hr             Complete Hybrid equivalent circuit
h21 → hf
h22 → ho

31
HYBRID TRANSISTOR MODEL
• Since hr is normally small, therefore we can assume hr = 0
and hrVo=0 (short circuit). 1/ho=resistance is often large
enough and can be approximated by an open-circuit.

Ii                                         Io

Vi                hi      hfIi                       Vo

32
HYBRID TRANSISTOR MODEL
•To distinguish parameters to be used for different configurations,
Common Base                Common Emitter
hi → hib                     hi → hie
hr → hrb                     hr → hre
hf → hfb                     hf → hfe
ho → hob                     ho → hoe
Vo → Vcb                     Vo → Vce

Ii→ Ie                       Ii→ Ib

33
HYBRID TRANSISTOR MODEL
Common-Emitter h equivalent circuit:
Ic                                         Ic    Io
Ii       Ib

Ie

Vce             Vi
hi e    hfeIb               Vo
Vbe

Ib

Common-Base h equivalent circuit:
Ic    Io
Ii    Ie
Ie                             Ic

Vi
hi b    hfbIe               Vo
Veb            Vcb

Ib

34
HYBRID TRANSISTOR MODEL
Comparison between re and h parameters:

Common-Emitter                  Common-Base

hie = β re                       hib = re
h fe = β ac                 h fb = −α ≅ −1

35
Example 5.2

Given IE=2.5mA, hfe=140, hoe=20us, and hob=0.5us,
determine:
a) The common-emitter hybrid equivalent circuit.
b) The common-base re model.

36

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Description: This article will cover how different of small-signal and large-signal techniques to Bipolar Junction Transistor (BJT). Thenm we look at hybrid equivalent and hybrid pi model.