Transcript Chapter 6a: BJT Transistor Modeling

Chapter 7:
BJT Transistor Modeling
Topic objectives
• At the end of the course you will be able to
– Understand about the small signal analysis of circuit
network using re model and hybrid equivalent model
– Understand the relationship between those two
available model for small signal analysis
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INTRODUCTION:TRANSISTOR MODELING
• 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
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How does the amplification be
done?
• 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)
• Simply speaking…
4
Disadvantages
• 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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VCC
DC supply 
“0” potential
•I/p coupling
capacitor  s/c
• Large values
• Block DC and
pass AC signal
R1
RC
C1
C2
RS
+
Vi
Vo
R2
RE
C3
-
VS
+
-
Voltage-divider configuration
under AC analysis
R1
+
Vi
VS
• Bypass
capacitor  s/c
•Large values
RC
+
RS
• O/p coupling
capacitor  s/c
• Large values
• Block DC and
pass AC signal
Vo
R2
-
Redraw the voltage-divider
configuration after removing dc
supply and insert s/c for the
capacitors
-
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Modeling of
BJT begin
HERE!
Ii
B
R1
+
RC
Zi
+
RS
+
Vi
VS
Vo
R2
-
RS
VS
Vi
R1 R2
Transistor smallsignal ac
equivalent cct
C
Io
+
E
Rc
Zo
-
Vo
-
-
Redrawn for small-signal AC analysis
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AC bias analysis :
1) Kill all DC sources
2) Coupling and Bypass capacitors are short cct.
The effect of there capacitors is to set a lower cut-off
frequency for the cct.
3) Inspect the cct (replace BJTs with its small signal
model:re or hybrid).
4) Solve for voltage and current transfer function,
i/o and o/p impedances.
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IMPORTANT PARAMETERS
• Input impedance, Zi
• Output impedance, Zo
• Voltage gain, Av
• Current gain, Ai
Input Impedance, Zi(few ohms  M)
The input impedance of an amplifier is the value as a
load when connecting a single source to the I/p of
terminal of the amplifier.
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Two port system
-determining input impedance Zi
Rsense
+
VS
+
Ii
Zi
-
Vi
-
Two-port
system
Vi
Zi 
Ii
Vs  Vi
Ii 
Rsense
Determining Zi
• The input impedance of transistor can be
approximately determined using dc biasing because it
doesn’t simply change when the magnitude of applied
ac signal is change.
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Demonstrating the impact of Zi
Rsource
+
VS=10mV
600 Ω
Zi
1.2 k Ω
-
+
Vi
-
Two-port
system
Ideal source,Rsource  0Ω
Full10mV appliedto the system
Withsourceimpedance,Rsource  600Ω
ZiVs
1.2k (10m)
Vi 

 6.6mV
Zi  Rsource 1.2k  600
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Example 6.1: For the system of Fig. Below, determine
the level of input impedance
1k Ω
+
VS=2mV
-
Rsense
+
Zi
Vi=1.2mV
Two-port
system
-
Solution
:
Ii 
Vs  Vi 2m  1.2m 0.8m


 0.8A
Rsense
1k
1k
Vi 1.2m
Zi 

 1.5k
Ii
0.8
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Output Impedance, Zo (few ohms  2M)
The output impedance of an amplifier is determined at
the output terminals looking back into the system with
the applied signal set to zero.
Rsense
Rsource
+
Two-port
system
Vs=0V
Vo
Io
Zo
-
Determining Zo
Iamplifier
+
V
V  Vo
Io 
Rsense
-
Vo
Zo 
Io
IL
IRo
Zo=Ro
RL
For Ro  RL
IL  IRo
Zo  RL  Zo become open cct
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Example 6.2: For the system of Fig. below, determine the
level of output impedance
Rsense
Two-port
system
Vs=0V
+
20 k Ω
Zo
Vo=680mV
-
+
V=1 V
-
Solution
:
V  Vo 1  680m 320m
Io 


 16A
Rsense
20k
20k
Zo 
Vo 680m

 42.5k
Io
16
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Example 6.3: For the system of Fig. below, determine Zo
if V=600mV, Rsense=10k and Io=10A
Rsense
Rsource
+
Two-port
system
Vs=0V
Io
Vo
Zo
-
+
V
-
Solution
:
V  Vo
Io 
Rsense
Vo  V  IoRsense 
 600m  10 10k
Zo 
Vo 500m

 50k
Io
10
 500mV
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Example 6.4: Using the Zo obtained in example 6.3,
determine IL for the configuration of Fig below if
RL=2.2 k and Iamplifier=6 mA.
Solution
:
Iamplifier
IL
IRo
Zo=Ro
RL
Current dividerrule:
Zo(Iamplifier)
IL 
Zo  RL
50k (6m)

50k  2.2k
 5.747mA
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Voltage Gain, AV
• DC biasing operate the transistor as an amplifier.
Amplifier is a system that having the gain behavior.
• The amplifier can amplify current, voltage and power.
• It’s the ratio of circuit’s output to circuit’s input.
• The small-signal AC voltage gain can be determined
by:
Vo
Av 
Vi
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By referring the network below the analysis are:
no load
Rsource
+
VS
-
Zi
+
+
Vi
AvNL
-
Vo
AvNL
Vo

Vi
RL  Ω (open cct)
-
withsourceresistance:
Determining the no load voltage gain
Vo
Zi
Avs 

AvNL
Vs
Zi  Rs
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Example 6.5: For the BJT amplifier of fig. below,
determine: a)Vi b) Ii c) Zi d) Avs
Solution:
Rs
+
VS=40mV
1.2 kΩ
+
Zi
Vi
-
-
b) Ii 
BJT amplifier
AvNL=320
+
a) AvNL 
Vo=7.68V
-
Vi 
Vo
Vi
Vo
7.68

 24mV
AvNL
320
Vs - Vi 40m  24m

 13.33A
Rs
1.2k
Rs  Rsource
Vi
24m
c) Zi 

 1.8k
Ii 13.33
d) Avs 
Zi
1.8k
AvNL 
(320)  192
Zi  Rs
1.8k  1.2k
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Current Gain, Ai
• This characteristic can be determined by:
Io
Ii
+
Vi
+
Zi
BJT
amplifier
RL
Vo
-
-
Vo
Io  
RL
Determining the loaded current gain
Io
Ai 
Ii
Vo / RL
VoZi


Vi / Zi
ViRL
Zi
Ai  Av
RL
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re TRANSISTOR MODEL
• employs a diode and controlled current source to
duplicate the behavior of a transistor.
• BJT amplifiers are referred to as current-controlled
devices.
Common-Base Configuration
Common-base BJT transistor
re model
re equivalent cct.
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Ic
C
Ie
E
re 
B
B
Common-base BJT transistor - pnp
26mV
 IE is the DC lev elof
IE(dc)
emitter current
Therefore, the input impedance, Zi = re
e
that less than 50Ω.
Ic
Ie
c
Ic  α Ie
b
For the output impedance, it will be as
follows;
Ie=0A
b
re model for the pnp common-base
configuration
Ic
e
Vs=0V
c
re
Ic  0A
b
e
Ie
re
b
Ic
c
Ic  α Ie
b
common-base re equivalent cct
isolation
part,
b Zi=re
Determining Zo for common-base
Zo  
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The common-base
characteristics
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BJT common-base
transistor amplifier
Ie
e
+
Vi
-
c
Zo  
re
Zi
Ic  α Ie
b
Io
RL
b
+
Vo
-
Defining Av=Vo/Vi for the common-base configuration
Vo  IoRL   Ic RL  IeRL
Vi  IeZi  Iere
Vo IeRL
Av 

Vi
Iere
Voltagegain,
 RL
RL
Av 

re
re
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BJT common-base
transistor amplifier
Ie
e
+
Vi
-
c
Zo  
re
Zi
Ic  α Ie
b
Io
RL
b
+
Vo
-
Defining Ai=Io/Ii for the common-base configuration
Io Ic
Ie
Ai  

Ii
Ie
Ie
Currentgain,
Ai    1
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Example 6.6: 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
e
Ie
re
Ic
c
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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Ii  Ie
e
re
Ic
c
Ic  α Ie
b
b
common-base re equivalent cct
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Example 6.7: 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:
Vi 10m
a) Zi 

 20
Ie
0.5m
b) Vo  IcRL  IeRL
 0.98(0.5m)(1.2k)
 588mV
c) Av 
d) Ai    0.98
e) Ib  Ie - Ic
 Ie - Ie
 0.5m(1  )
 0.5m(1  0.98)
 10A
Vo 588m

 58.8
Vi
10m
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Common-Emitter Configuration
Common-emitter BJT transistor
re model
re equivalent cct.
Still remain controlled-current source (conducted
between collector and base terminal)
Diode conducted between base and emitter terminal
Input
Base & Emitter terminal
Output
Collector & Emitter terminal
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c
C
Ic
Ic
B
b
Ib
E
E
Ic   Ib
Ib
e
e
common-emitter BJT transistor
Vi
Zi 
Ii
re model npn common-emitter configuration
(1)
c
Vi  Vbe  Iere   Ibre and
Ic
subtitute into (1) gives
b
+
Vbe Ibre
Zi 

Ib
Ib
Vi
Zi   re
e
Z ranges between hundred to 6 ~ 7k
i
Ic   Ib
Ii=Ib
+
Vbe
-
Ie
re
e
Determining Zi using re equivalent model
31
The output graph
32
Output impedance Zo
b
Ii=Ib
c
 Ib
re
ro
e
e
re model for the C-E transistor configuration
b
Vs=0V
Ii=Ib = 0A
c
re
Ib  0A
e
Zo
ro
e
Zo  ro
if ro is ignoredthus the
Zo  Ω (open cct, highimpedance)
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Ii=Ib
e
+
Vi
BJT common-emitter
transistor amplifier
c
Zo  
re
Zi  re
b
-
Io  Ic   Ib
b
RL
Io
+
Vo
-
Determining voltage and current gain for the
common-emitter amplifier
Vo  IoRL  Ic RL  IbRL
Vi  IiZi  Ibre
Voltagegain,
Vo
IbRL
Av 

Vi
Ibre
RL
Av  
re
Current gain,
Io Ic  Ib
Ai 


Ii Ib
Ib
Ai  
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Example 6.8: Given =120 and IE(dc)=3.2mA for a commonemitter 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 6.9: Using the npn common-emitter configuration,
determine the following if =80, IE(dc)=2 mA and ro=40 k
b) Ai if RL =1.2k 
a) Zi
b
Ii=Ib
c) Av if RL=1.2k 
c
Io
re
 Ib
ro
RL
e
Solution:
26m 26m
a) re 

 13
IE
2m
Zi  re  80(13)  1.04k
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)
ro
40k
Ai  ro  RL 
 
(80)
Ib
ro  RL
40k  1.2k
 77.67
c)Av  
RL ro
re

1.2k 40k
13
  89.6
37
Hybrid Equivalent Model
• re model is sensitive to the dc level of operation
that result input resistance vary with the dc
operating point
• Hybrid model parameter are defined at an
operating point that may or may not reflect the
actual operating point of the amplifier
38
Hybrid Equivalent Model
The hybrid parameters: hie, hre, hfe, hoe are developed and used to model the transistor.
These parameters can be found in a specification sheet for a transistor.
39
Determination of parameter
Vi  h11Ii  h12 Vo
h11 
h12 
Vi
Ii
Vo  0V
Vi
Vo
Vo  0V
IO  h21Ii  h22 Vo
SolvingVo  0V ,
h21 
h22 
Ii
Io
Vo  0V
Io
Vo
Io  0A
H22 is a conductance!
40
General h-Parameters for any
Transistor Configuration
hi = input resistance
hr = reverse transfer voltage ratio (Vi/Vo)
hf = forward transfer current ratio (Io/Ii)
ho = output conductance
41
Common emitter hybrid
equivalent circuit
42
Common base hybrid equivalent
circuit
43
Simplified General h-Parameter Model
The model can be simplified based on these approximations:
hr  0 therefore hrVo = 0 and ho   (high resistance on the output)
Simplified
44
Common-Emitter re vs. h-Parameter Model
hie = re
hfe = 
hoe = 1/ro
45
Common-Emitter h-Parameters
hie  re
h fe   ac
[Formula 7.28]
[Formula 7.29]
46
Common-Base re vs. h-Parameter Model
hib = re
hfb = -
47
Common-Base h-Parameters
hib  re
h fb    1
[Formula 7.30]
[Formula 7.31]
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