Download - Chap3 Utem (BJT Transistor Modeling)
Chapter 7: BJT Transistor ModelingChapter 7: BJT Transistor Modeling
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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 modelb) 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…
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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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VS
VCC
C1
C2
C3
+
-
Vo
RS
Vi
+
-RE
RCR1
R2
VS
+
-
Vo
RS
Vi
+
-
RCR1
R2
•I/p coupling capacitor s/c• Large values• Block DC and pass AC signal • Bypass
capacitor s/c•Large values
DC supply “0” potential
Voltage-divider configuration under AC analysis
Redraw the voltage-divider configuration after removing dc
supply and insert s/c for the capacitors
• O/p coupling capacitor s/c• Large values• Block DC and pass AC signal
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VS
RSR2 R1
Rc
Transistor small-signal ac
equivalent cct
Vo
Zi
Ii
Zo
Io
Vi
+ +
- -
B
E
C
Redrawn for small-signal AC analysis
Modeling of BJT begin
HERE!
VS
+
-
Vo
RS
Vi
+
-
RCR1
R2
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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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VS Two-portsystem
Vi
Rsense
IiZi
+
-
Determining Zi
+
-
sense
isi
R
VVI
i
ii
I
VZ
Two port system-determining input impedance 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
VS=10mVTwo-portsystem
Vi
Rsource
Zi
+
-
+
-1.2 kΩ
600Ω
mV6.6600k2.1
)m10(k2.1
RZ
VZV
Ω600R impedance, sourceWith
system the toapplied 10mV Full
0ΩR source, Ideal
sourcei
sii
source
source
12
Example 6.1: For the system of Fig. Below, determine the level of input impedance
VS=2mV Two-portsystem
Vi=1.2mV
RsenseZi
+
-
+
-
1 k Ω
A8.0k1
m8.0
k1
m2.1m2
R
VVI
sense
isi
:Solution
k5.18.0
m2.1
I
VZ
i
ii
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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.
Two-portsystem
Rsource
Vs=0V
Rsense
V
+
-
+
-
Io
Zo
Vo
Determining Zo
sense
oo
R
VVI
o
oo
I
VZ
cctopen become ZRZ oLo RLZo=Ro
Iamplifier
IRo
IL
RoL
Lo
II
RRFor
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Example 6.2: For the system of Fig. below, determine the level of output impedance
Two-portsystem
Vs=0V
Rsense
V=1 V
+
-
+
-
Zo
Vo=680mV
20 kΩ
A16k20
m320
k20
m6801
R
VVI
sense
oo
:Solution
k5.4216
m680
I
VZ
o
oo
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Example 6.3: For the system of Fig. below, determine Zo if V=600mV, Rsense=10k and Io=10A
Two-portsystem
Rsource
Vs=0V
Rsense
V
+
-
+
-
Io
Zo
Vo
mV500
k1010m600
RIVVR
VVI
senseoo
sense
oo
:Solution
k5010
m500
I
VZ
o
oo
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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.
RLZo=Ro
Iamplifier
IRo
IL
mA747.5k2.2k50
)m6(k50RZ
)(IZI
:ruledivider Current
Lo
amplifieroL
:Solution
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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:
i
ov
V
VA
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VS AvNLVi
Rsource
Zi
+
-
+
-Vo
+
-
Determining the no load voltage gain
By referring the network below the analysis are:
cct)(open ΩRi
oLvNL
V
VA
load no
vNLARZ
Z
V
VA
:resistance sourcewith
si
i
s
ovs
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Example 6.5: For the BJT amplifier of fig. below, determine: a)Vi b) Ii c) Zi d) Avs
VS=40mVBJT amplifier
AvNL=320Vi
Rs
Zi
+
-
+
-Vo=7.68V
+
-
1.2 kΩ
mV24320
7.68
A
VV
V
VA a)
vNL
oi
i
ovNL
:Solution
sources
s
isi
RR
A33.13k2.1
m24m40
R
V-VI b)
k8.133.13
m24
I
VZ c)
i
ii
192)320(k2.1k8.1
k8.1A
RZ
ZA d) vNL
si
ivs
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Current Gain, Ai
• This characteristic can be determined by:
i
oi
I
IA
BJTamplifier
Vi
Zi
+
-
Vo
+
-
Ii
RL
Determining the loaded current gain
Io
L
ivi
R
ZAA
Li
io
ii
Lo
RV
ZV
Z/V
R/V
L
oo
RV
I
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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 transistorre modelre equivalent cct.
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E
BB
C
Common-base BJT transistor - pnp
Ic Ie
e
b b
c
ec I αI
IcIe
re model for the pnp common-baseconfiguration
e
b b
c
ec I αI
IcIe
common-base re equivalent cct
re
current emitter
of level DC the isII
26mVr E
E(dc)
e
isolation part,Zi=re
e
b b
c
A0Ic
IcIe=0A
Determining Zo for common-base
reVs=0V
Zo
Therefore, the input impedance, Zi = re
that less than 50Ω.
For the output impedance, it will be as follows;
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The common-base characteristics
24
e
b b
c ec I αI Ie
re
Defining Av=Vo/Vi for the common-base configuration
BJT common-basetransistor amplifier
Vi Vo
+
-
+
-
Zi
oZ RL
Io
LeLcLoo RIRIRIV
e
L
e
Lv
r
R
r
RA
gain, Voltage
eeiei rIZIV
ee
Lev
rI
RI
Vi
VoA
25
1A
gain,Current
i
e
e
e
c
i
oi
I
I
I
I
I
IA
e
b b
c ec I αI Ie
re
Defining Ai=Io/Ii for the common-base configuration
BJT common-basetransistor amplifier
Vi Vo
+
-
+
-
ZioZ RL
Io
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Example 6.6: For a common-base configuration in figurebelow with IE=4mA, =0.98 and AC signal of 2mV isapplied between the base and emitter terminal:a) Determine the Zi b) Calculate Av if RL=0.56kc) Find Zo and Ai
e
b b
c
ec I αI
IcIe
common-base re equivalent cct
re
27
Solution:
5.6m4
m26
I
26mr Za)
Eei
43.845.6
)k56.0(98.0
r
RA b)
e
Lv
98.0I
IA
Ω Zc)
i
oi
o
28
e
b b
c
ec I αI
IcIe
common-base re equivalent cct
re
iI
29
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
20m5.0
m10
I
V Za)
:Solution
e
ii
88mV5(1.2k)0.98(0.5m)
RIRIV b) LeLco
8.58m10
m588
V
VA c)
i
ov
98.0A d) i
A10
)98.01(m5.0
)1(m5.0
I-I
I-II e)
ee
ceb
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Common-Emitter Configuration
Common-emitter BJT transistorre modelre equivalent cct.Still remain controlled-current source (conducted between collector and base terminal)Diode conducted between base and emitter terminal
Input Output
Base & Emitter terminal Collector & Emitter terminal
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common-emitter BJT transistor
EE
B
C
Ib
Icbc I I
c
e e
b
Ic
Ib
re model npn common-emitter configuration
bc I I
c
e e
b
Ic
Ii=Ib
Determining Zi using re equivalent model
re
Ie+
-
Vbe
+
-
Vi
(1) Ii
ViZi
gives (1) into subtitute
and IbreIereVbeVi
b
eb
b
be
I
rI
I
VZi
erZi 7k~6 to hundred between ranges iZ
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The output graph
33
bI
c
e
bIi=Ib
re model for the C-E transistor configuration
rero
e
0AbI
c
e
bIi=Ib
rero
e
Vs=0V
= 0A
oZ
impedance)high cct,(open ΩZ
the thusignored is r if
rZ
o
o
oo
Output impedance Zo
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e
b b
cbco I II Ii=Ib
re
Determining voltage and current gain for the common-emitter amplifier
BJT common-emittertransistor amplifier
Vi Vo
+
-
+
-
oZ RL
Io
ei rZ
e
Lv
r
RA
Ib
Ib
Ib
Ic
Ii
IoA
gain,Current
i
LbLcLoo RIRIRIV
ebiii rIZIV
eb
Lb
i
ov
rI
RI
V
VA
gain, Voltage
iA
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Example 6.8: 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
975)125.8(120rZ
125.8m2.3
m26
I
26mr a)
ei
Ee
:Solution
15.246125.8
k2
r
Rb)A
e
Lv
120I
IA c)
i
oi
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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
a) Zi b) Ai if RL =1.2k c) Av if RL=1.2k
k04.1)13(80rZ
13m2
m26
I
26mr a)
ei
Ee
:Solution
bI
cbIi=Ib
re model for the C-E transistor configuration
rero
e
RL
Io
37
67.77
)80(k2.1k40
k40
Rr
r
IRr
)I(r
A
Rr
)I(rI
I
I
I
IiAb)
(cont)Solution
Lo
o
b
Lo
bo
i
Lo
boL
b
L
i
o
6.8913
k40k2.1
r
rRvAc)
e
oL
38
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
39
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.
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Determination of parameter
0VVo
i12
0VVi
i11
o12i11i
o
o
VV
h
IV
h
VhIhV
0AIo
o22
0VVo
i21
o
o22i21O
o
o
VI
h
II
h
, 0VV Solving
VhIhI
H22 is a conductance!
41
General h-Parameters for any Transistor Configuration
hi = input resistancehr = reverse transfer voltage ratio (Vi/Vo)hf = forward transfer current ratio (Io/Ii)ho = output conductance
42
Common emitter hybrid equivalent circuit
43
Common base hybrid equivalent circuit
44
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
45
Common-Emitter re vs. h-Parameter Model
hie = rehfe = hoe = 1/ro
46
Common-Emitter h-Parameters
[Formula 7.28]
[Formula 7.29]
acfe
eie
h
rh
47
Common-Base re vs. h-Parameter Model
hib = rehfb = -
48
Common-Base h-Parameters
[Formula 7.30]
[Formula 7.31]
1
fb
eib
h
rh