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Sem I 0809/rosdiyanaSem I 0809/rosdiyana
Introduction to BJT Introduction to BJT Small Signal AnalysisSmall Signal Analysis
CHAPTER 5CHAPTER 5
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IntroductionIntroduction•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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DisadvantageDisadvantage RRee model model
• Fails to account the output impedance Fails to account the output impedance level of device and feedback effect from level of device and feedback effect from output to inputoutput to input
Hybrid equivalent modelHybrid equivalent model• Limited to specified operating condition Limited to specified operating condition
in order to obtain accurate resultin order to obtain accurate result
IntroductionIntroduction
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The transistor can be employed as an The transistor can be employed as an amplifying device. That is, the output amplifying device. That is, the output sinusoidal signal is greater than the sinusoidal signal is greater than the input signal or the ac input power is input signal or the ac input power is greater than ac input power.greater than ac input power.
How the ac power output can be How the ac power output can be greater than the input ac power?greater than the input ac power?
Amplification in the AC domainAmplification in the AC domain
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Amplification in the AC domainAmplification in the AC domain
Conservation; output Conservation; output power of a system power of a system cannot be large than its cannot be large than its input and the efficiency input and the efficiency cannot be greater than 1cannot be greater than 1
The input dc plays the The input dc plays the important role for the important role for the amplification to amplification to contribute its level to the contribute its level to the ac domain where the ac domain where the conversion will become conversion will become as as ηη=P=Po(ac)o(ac)/P/Pi(dc)i(dc)
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The superposition theorem is applicable for the The superposition theorem is applicable for the analysis and design of the dc & ac components of analysis and design of the dc & ac components of a BJT network, permitting the separation of the a BJT network, permitting the separation of the analysis of the dc & ac responses of the system.analysis of the dc & ac responses of the system.
In other words, one can make a complete dc In other words, one can make a complete dc analysis of a system before considering the ac analysis of a system before considering the ac response.response.
Once the dc analysis is complete, the ac response Once the dc analysis is complete, the ac response can be determined using a completely ac can be determined using a completely ac analysis.analysis.
Amplification in the AC domainAmplification in the AC domain
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BJT Transistor ModelBJT Transistor Model Use equivalent circuitUse equivalent circuit Schematic symbol for the device can be replaced Schematic symbol for the device can be replaced
by this equivalent circuits.by this equivalent circuits. Basic methods of circuit analysis is applied.Basic methods of circuit analysis is applied. DC levels were important to determine the Q-pointDC levels were important to determine the Q-point Once determined, the DC level can be ignored in Once determined, the DC level can be ignored in
the AC analysis of the network.the AC analysis of the network. Coupling capacitors & bypass capacitor were Coupling capacitors & bypass capacitor were
chosen to have a very small reactance at the chosen to have a very small reactance at the frequency of applications.frequency of applications.
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The AC equivalent of a network isThe AC equivalent of a network isobtained by:obtained by:1.1. Setting all DC sources to zero & replacing them Setting all DC sources to zero & replacing them
by a short-circuit equivalent.by a short-circuit equivalent.2.2. Replacing all capacitors by a short-circuit Replacing all capacitors by a short-circuit
equivalent.equivalent.3.3. Removing all elements bypassed by short-Removing all elements bypassed by short-
circuit equivalent.circuit equivalent.4.4. Redrawing the network. Redrawing the network.
BJT Transistor ModelBJT Transistor Model
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1010
1111
ExampleExample
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ExampleExample
1313
ExampleExample
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The re transistor modelThe re transistor model
Common Base PNP ConfigurationCommon Base PNP Configuration
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Common Base PNP ConfigurationCommon Base PNP Configuration
Transistor is replaced by Transistor is replaced by a single diode between E a single diode between E & B, and control current & B, and control current source between B & Csource between B & C
Collector current Ic is Collector current Ic is controlled by the level of controlled by the level of emitter current Ie.emitter current Ie.
For the ac response the For the ac response the diode can be replaced by diode can be replaced by its equivalent ac its equivalent ac resistance.resistance.
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Common Base PNP ConfigurationCommon Base PNP Configuration
The ac resistance The ac resistance of a diode can be of a diode can be determined by the determined by the equation;equation;
Where IWhere IDD is the dc is the dc current through current through the diode at the Q-the diode at the Q-point. point.
EI
mVre
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Input impedance is Input impedance is relatively small and relatively small and output impedance output impedance quite high.quite high.
range from a few range from a few ΩΩ to max 50 to max 50 ΩΩ
Typical values are in Typical values are in the M the M ΩΩ
CBi reZ
CBZo
Common Base PNP ConfigurationCommon Base PNP Configuration
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The common-base The common-base characteristicscharacteristics
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Voltage GainVoltage Gain
re
RA
re
R
rI
RI
V
VA
rI
ZI
ZIV
RI
RI
RIV
LV
L
ee
Le
i
OV
ee
ie
iii
Le
LC
Loo
:gain voltage
: ageinput volt
)(
: tageoutput vol
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Current GainCurrent Gain
1
i
e
e
e
C
i
oi
A
I
I
I
I
I
IA
The fact that the polarity of the VThe fact that the polarity of the Voo as determined as determined by the current Iby the current ICC is the same as defined by figure is the same as defined by figure below.below.
It reveals that VIt reveals that Voo and V and Vii are in phase for the are in phase for the common-base configuration.common-base configuration.
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Common Base PNP ConfigurationCommon Base PNP Configuration
Approximate model for a common-base npn transistor configuration
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Example 1: 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
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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
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Example 2: 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
Common Emitter NPN ConfigurationCommon Emitter NPN Configuration
Base and emitter Base and emitter are input are input terminalterminal
Collector and Collector and emitter are emitter are output terminalsoutput terminals
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Common Emitter NPN ConfigurationCommon Emitter NPN Configuration
Substitute rSubstitute re e
equivalent circuitequivalent circuit
Current through Current through diodediode
2626
bc II
bbe
bbbce
III
IIIII
)1(
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Input impedanceInput impedance
ei
ei
b
ebi
eb
eei
b
be
i
ii
rZ
rZ
I
rIZ
rI
rIV
I
V
I
VZ
; 1an greater thusually
)1(
)1( that so
)1(
:ageinput volt
:impedanceinput
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The output graphThe output graph
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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
3030e
LV
eb
Lb
i
oV
eb
iii
Lb
Lco
Loo
r
RA
rI
RI
V
VA
rI
ZIV
RI
RIV
RIV
that so
:ageinput volt
: tageoutput vol
i
b
b
b
C
i
oi
A
I
I
I
I
I
IA
Voltage GainVoltage Gain Current GainCurrent Gain
rree model for common-emitter model for common-emitter
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Example 3: 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 4: 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
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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
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Common Collector ConfigurationCommon Collector Configuration
For the CC configuration, the model For the CC configuration, the model defined for the common-emitter defined for the common-emitter configuration is normally applied configuration is normally applied rather than defining a model for the rather than defining a model for the common-collector configuration.common-collector configuration.