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Transcript of 7.BJT Transistor Modeling.ppt
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BJT Transistor ModelingCHAPTER 5
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Topic objectives
At the end of the course you will beable to
Understand about the small signalanalysis of circuit network using remodel and hybrid equivalent model
Understand the relationship between
those two available model for smallsignal analysis
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To begin analyze of small-signal AC response of BJTamplifier the knowledge of modeling the transistor is
important.
The input signal will determine whether its 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) remodel
b) hybrid equivalent model
INTRODUCTION: TRANSISTOR
MODELING
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How does the amplification be
done? Conservation; output
power of a system
cannot be large than its
input and the efficiencycannot be greater than 1
The input dc plays the
important role for the
amplification tocontribute its level to the
ac domain where the
conversion will become
as =Po(ac)/Pi(dc)
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Disadvantages
Remodel
Fails to account the output impedancelevel 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
RC
R1
R2
VS
+
-
Vo
RS
Vi
+
-
RCR1
R2
I/p couplingcapacitors/c
Large values
Block DC and
pass AC signal Bypass
capacitors/c
Large values
DC supply
0 potential
Voltage-divider configurationunder AC analysis
Redraw the voltage-divider
configuration after removing dc
supply and insert s/c for the
capacitors
O/p coupling
capacitors/c
Large values
Block DC and
pass AC signal
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VS
RS
R2R1 R
c
Transistor small-
signal ac
equivalent cct
Vo
Zi
Ii
Zo
Io
Vi
++
- -
B
E
C
Redrawn for small-signal AC analysis
Modeli ng of
BJT begin
HERE!
VS
+
-
Vo
RS
Vi
+
-
RC
R1
R2
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1. Kill all DC sources
2. Coupling and Bypass capacitors are short circuit.
The effect of there capacitors is to set a lower cut-
off frequency for the circuit.
3. Inspect the cct (replace BJTs with its small signal
model:reor hybrid).
4. Solve for voltage and current transfer function, i/o
and o/p impedances.
AC bias analysis
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Input impedance, ZiOutput impedance, ZoVoltage gain, AvCurrent gain, Ai
Input Impedance, Zi(few ohms M)
The input impedance of an amplifier is the value as a loadwhen connecting a single source to the I/p of terminal of the
amplifier.
IMPORTANT PARAMETERS
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VS Two-port
systemVi
Rsense
Ii
Zi
+
-
Determining Zi
+
-
sense
isi
R
VVI
i
ii
I
VZ
Two port system
-determining input impedance Zi
The input impedance of transistor can beapproximately determined using dc biasingbecause it doesnt simply change when the
magnitude of applied ac signal is change.
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Demonstrating the impact of Zi
VS=10mV
Two-portsystem
Vi
Rsource
Zi
+
-
+
-
1.2 k
600
mV6.6600k2.1
)m10(k2.1
RZ
VZV
600Rimpedance,sourceWith
systemthetoapplied10mVFull
0Rsource,Ideal
sourcei
sii
source
source
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Example 6.1: For the system of Fig. Below,
determine the level of input impedance
VS=2mV Two-port
systemV
i=1.2mV
Rsense
Zi
+
-
+
-
1 k
A8.0k1
m8.0
k1
m2.1m2
R
VVI
sense
isi
:So l u t i o n
k5.18.0
m2.1
I
VZ
i
ii
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Output Impedance, Zo(few ohms2M)
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-port
system
Rsource
Vs=0V
Rsense
V
+
-
+
-
Io
Zo
Vo
Determining Zo
sense
oo
R
VVI
o
oo
I
VZ
cctopenbecomeZRZ oLo R
LZo=R
o
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 k
A16k20
m320
k20
m6801
R
VVIsense
oo
:So l u t i o n
k5.4216
m680
I
VZ
o
oo
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Example 6.3: For the system of Fig. below, determine Zoif
V=600mV, Rsense=10kand Io=10A
Two-port
system
Rsource
Vs=0V
Rsense
V
+
-
+
-
Io
Zo
Vo
mV500
k1010m600
RIVV
R
VVI
senseoo
sense
oo
:So l u t i o n
k5010
m500
I
VZ
o
o
o
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Example 6.4: Using the Zoobtained in example 6.3,
determine ILfor the configuration of Fig below if
RL=2.2 kand Iamplifier=6 mA.
RLZ
o=R
o
Iamplifier
IRo
IL
mA747.5 k2.2k50
)m6(k50
RZ)(IZI
:ruledividerCurrent
Lo
amplifieroL
:So l u t i o n
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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.
Its the ratio of circuits output to circuits input.
The small-signal AC voltage gain can be determined by:
i
ov
V
VA
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VS AvNL
Vi
Rsource
Zi
+
-
+
-
Vo
+
-
Determining the no load voltage gain
By referring the network below the analysis are:
cct)(openR
i
oLvNL
V
VA
loadno
vNLARZ
Z
V
VA
:resistancesourcewith
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=40mV
BJT amplifierA
vNL=320
Vi
Rs
Zi
+
-
+
-V
o=7.68V
+
-
1.2 k
mV24320
7.68
A
VV
V
VAa)
vNL
oi
i
ovNL
:Solution
sources
s
isi
RR
A33.13k2.1
m24m40
R
V-VIb)
k8.133.13m24
IVZc)
i
ii
192)320(k2.1k8.1
k8.1A
RZ
ZAd) vNL
si
ivs
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Current Gain, Ai
This characteristic can be determined by:
i
oiIIA
BJT
amplifierV
i
Zi
+
-
Vo
+
-
Ii
RL
Determining the loaded current gain
Io
L
ivi
RZAA
Li
io
ii
Lo
RVZV
Z/VR/V
L
oo
R
VI
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Employs a diode and controlled current source toduplicate the behavior of a transistor.
BJT amplifiers are referred to as current-controlled
devices.
Common-Base Configuration
Common-base BJT transistor
remodel
reequivalent circuit
reTRANSISTOR MODEL
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E
BB
C
Common-base BJT transistor - pnp
Ic
Ie
e
b b
c
ec II
IcIe
remodel for the pnp common-base
configuration
e
b b
c
ec II
IcI
e
common-base reequivalent cct
re
currentemitter
oflevelDCtheisII
26mVr E
E(dc)
e
isolation
part,
Zi=re
e
b b
c
A0Ic
Ic
Ie=
0A
Determining Zofor common-base
re
Vs=0V
Zo
Therefore, the input impedance, Zi = re
that less than 50.
For the output impedance, it will be as
follows;
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The common-base
characteristics
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e
b b
c ec II Ie
re
Defining Av
=Vo
/Vi
for the common-base configuration
BJT common-base
transistor 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
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1A
gain,Current
i
e
e
e
c
i
oi
I
I
I
I
I
IA
e
b b
c ec II Ie
re
Defining Ai=Io/Iifor the common-base configuration
BJT common-base
transistor amplifier
Vi Vo+
-
+
-
ZioZ RL
Io
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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 Zib) Calculate Avif RL=0.56kc) Find Zoand Ai
e
b b
c
ec II
IcIe
common-base reequivalent cct
re
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Solution:
5.6
m4
m26
I
26mrZa)
E
ei
43.845.6
)k56.0(98.0
r
RAb)
e
Lv
98.0I
IA
Zc)
i
oi
o
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e
b b
c
ec II
IcIe
common-base re equivalent cct
re
iI
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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) Voif RL=1.2k c) Av d)Ai e) Ib
20m5.0
m10
I
VZa)
:Solution
e
ii
88mV5
(1.2k)0.98(0.5m)
RIRIVb) LeLco
8.58m10
m588
V
VAc)
i
ov
98.0Ad) i
A10
)98.01(m5.0
)1(m5.0
I-II-IIe)
ee
ceb
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Common-emitter BJT transistorremodel
reequivalent cct.
Still remain controlled-current source (conducted
between collector and base terminal)Diode conducted between base and emitter
terminal
Input OutputBase & Emitterterminal
Collector & Emitterterminal
Common-Emitter Configuration
c
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common-emitter BJT transistor
EE
B
C
Ib
Ic
bc II
e e
b
Ic
Ib
remodel npn common-emitter configuration
bc II
c
e e
b
Ic
Ii=I
b
Determining Ziusing r
eequivalent model
re
Ie+
-
Vbe
+
-
Vi
(1)
Ii
ViZi
gives(1)intosubtitute
andIbreIereVbeVi
b
eb
b
be
I
rI
I
VZi
erZi
7k~6tohundredbetweenrangesi
Z
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The output graph
Output impedance Z
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bI
c
e
bIi=I
b
remodel for the C-E transistor configuration
re ro
e
0AbI
c
e
bIi=I
b
re
ro
e
Vs=0V
= 0A
oZ
impedance)highcct,(openZ
thethusignoredisrif
rZ
o
o
oo
Output impedance Zo
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e
b b
c
bco III Ii=Ib
re
Determining voltage and current gain for the
common-emitter amplifier
BJT common-emitter
transistor 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
ovrIRI
VVA
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 kis applied c) Ai with the 2 kload
975)125.8(120rZ
125.8m2.3
m26
I
26mra)
ei
Ee
:Solution
15.246125.8
k2
r
Rb)A
e
Lv
120I
IAc)
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
26m
ra)
ei
Ee
:Solution
bI
cbIi=I
b
remodel for the C-E transistor configuration
re
ro
e
RL
Io
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67.77
)80(
k2.1k40
k40
Rr
r
I
Rr
)I(r
A
Rr
)I(rI
I
I
I
IiAb)
(cont)Solution
Lo
o
b
Lo
bo
i
Lo
boL
b
L
i
o
6.89
13
k40k2.1
r
rRvAc)
e
oL
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Hybrid Equivalent Model
re model is sensitive to the dc level of operationthat result input resistance vary with the dcoperating point
Hybrid model parameter are defined at anoperating point that may or may not reflect theactual operating point of the amplifier
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Hybrid Equivalent Model
The hybrid parameters: hie, hre, hfe, hoeare developed andused to model the transistor. These parameters can be found
in a specification sheet for a transistor.
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Determination of parameter
0VVo
i
12
0VVi
i
11
o12i11i
o
o
V
Vh
I
Vh
VhIhV
0AIo
o22
0VVo
i21
o
o22i21O
o
o
V
Ih
IIh
,0VVSolving
VhIhI
H22is a conductance!
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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
C itt h b id
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Common emitter hybrid
equivalent circuit
C b h b id i l t
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Common base hybrid equivalent
circuit
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Simplified General h-Parameter ModelThe model can be simplified based on these approximations:
hr 0 therefore hrVo = 0 and ho (high resistance on the output)
Simplified
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Common-Emitter re vs. h-Parameter
Model
hie = re
hfe =
hoe = 1/ro
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Common-Emitter h-Parameters
acfe
eie
h
rh
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Common-Base re vs. h-Parameter
Model
hib = re
hfb = -
C
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Common-Base h-Parameters
1
fb
eib
hrh