Download 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
2
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
3
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
5
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
-
6
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
7
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.
8
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.
9
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.
10
Demonstrating the impact of Zi
Rsource
+
VS=10mV
600 Ω
Zi
1.2 k Ω
-
+
Vi
-
Two-port
system
Ideal source, Rsource  0Ω
Full 10mV applied to the system
With source impedance, Rsource  600 Ω
ZiVs
1.2k (10 m)
Vi 

 6.6mV
Zi  Rsource 1.2k  600
11
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
:
Vs  Vi 2m  1.2m 0.8m
Ii 


 0.8A
Rsense
1k
1k
Vi 1.2m
Zi 

 1.5k
Ii
0.8
12
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
For Ro  RL
IRo
Zo=Ro
RL
IL  IRo
Zo  RL  Zo become open cct
13
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  680 m 320 m
Io 


 16A
Rsense
20 k
20 k
Zo 
Vo 680 m

 42.5k
Io
16
14
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
 600 m  10  10 k
Zo 
Vo 500 m

 50 k
Io
10
 500 mV
15
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.
Iamplifier
IL
IRo
Zo=Ro
RL
Solution
:
Current divider rule :
Zo(Iamplifier )
IL 
Zo  RL
50 k (6m)

50 k  2.2k
 5.747 mA
16
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
17
By referring the network below the analysis are:
no load
Rsource
+
VS
-
Zi
+
+
Vi
AvNL
-
Vo
Vo
Av NL 
Vi
RL  Ω (open cct)
-
with source resistance :
Determining the no load voltage gain
Vo
Zi
Avs 

Av NL
Vs
Zi  Rs
18
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
-
BJT amplifier
AvNL=320
-
b) Ii 
+
a) AvNL 
Vo=7.68V
-
Vi 
Vo
Vi
Vo
7.68

 24 mV
AvNL
320
Vs - Vi 40 m  24 m

 13 .33A
Rs
1.2k
Rs  Rsource
c) Zi 
Vi
24 m

 1.8k
Ii 13 .33
d) Avs 
Zi
1.8k
AvNL 
(320 )  192
Zi  Rs
1.8k  1.2k
19
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
20
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.
21
Ic
C
Ie
E
re 
B
B
Common-base BJT transistor - pnp
26mV
 IE is t heDC level of
IE(dc)
emit t er 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  
22
The common-base
characteristics
23
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
Voltage gain,
RL
RL
Av 

re
re
24
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
Current gain,
Ai    1
25
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
26
Solution:
26m 26 m
a) Zi  re 

 6.5
IE
4m
RL 0.98 (0.56 k )
b) Av 

 84 .43
re
6.5
c) Zo  Ω
Io
Ai      0.98
Ii
27
Ii  Ie
e
re
Ic
c
Ic  α Ie
b
b
common-base re equivalent cct
28
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 10 m
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 588 m

 58 .8
Vi
10 m
29
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
30
c
C
Ic
Ic
B
b
Ib
E
E
Ic   Ib
Ib
e
e
common-emitter BJT transistor
Zi 
Vi
Ii
re model npn common-emitter configuration
(1)
c
Vi  Vbe  I ere  I bre and
Ic
s ubt it ut eint o( 1)gives
Zi 
Vbe Ibre

Ib
Ib
Zi  re
Z i ranges bet w eenhundred t o6 ~ 7k
b
+
Vi
e
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
re
 Ib
ro
e
e
re model for the C-E transistor configuration
b
Vs=0V
Ii=Ib = 0A
c
re
e
Ib  0A
Zo
ro
e
Zo  ro
if ro is ignored thus the
Zo  Ω (open cct, high impedance)
33
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
Voltage gain,
Vo
IbRL
Av 

Vi
Ibre
RL
Av  
re
Current gain,
Io Ic Ib
Ai 


Ii Ib
Ib
Ai  
34
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 26 m
a) re 

 8.125 
IE
3.2m
Zi  re  120 (8.125 )  975 
RL
2k
b)Av  

  246 .15
re
8.125
c) Ai 
Io
   120
Ii
35
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 26 m
a) re 

 13
IE
2m
Zi  re  80 (13)  1.04 k
re model for the C-E transistor configuration
36
Solution (cont)
Io IL

Ii Ib
ro( Ib)
IL 
ro  RL
ro( Ib)
ro
40 k
Ai  ro  RL 
 
(80 )
Ib
ro  RL
40 k  1.2k
 77 .67
b)Ai 
c)Av  
RL ro
re

1.2k 40 k
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
Solving Vo  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]
48