Small Signal BJT Amplifi ers
3 Chapter ObjectivesX Develop h-parameter model of BJT for AC analysis.X Analyze various CE, CB and CC ampli ers
to determine input impedance, output impedance, voltage gain and current gain.
X Compare the performance of CE, CB and CC ampli ers.
X Learn method to increase input impedance such as Darlington circuit and bootstrapping technique.
3.1INTRODUCTIONThe basic construction, characteristics and DC biasing of the bipolar junction transistors were discussed in Chapter 1. The purpose of biasing is to establish a Q-point about which variations in current and voltage can occur in response to an AC input signal. The term small signal refers to the use of signals that take up a relatively small percentage of an amplifi ers operational range. The amplifi ers designed to handle these small AC signals are called small signal ampli-fi ers. The h-parameter equivalent model is commonly used in the small signal AC analysis of bipolar junction transistors. We will defi ne this model and analyze the amplifi er characteristics in terms of this model.
3.2SMALL SIGNAL BJT AMPLIFIERFigure 3.1 shows a voltage divider biased transistor amplifier. The capacitor CC1 couples the sinusoidal AC source to the base of the transistor. The capacitor CC2 couples the output to the load. These coupling capacitors block DC and thus prevents the source resistance RS and the load resistance RL from changing the DC bias voltages at the base and collector. The capacitors act as short circuits to the signal voltage. The emitter resistor RE improves the stability of the system but provides negative feedback which reduces the gain of the amplifier. This AC degeneration can be avoided by connecting the capacitor CE in parallel with resistor RE. For DC operation, capacitor CE acts as open circuit and improves the stability. For AC operation, capacitor CE acts as short circuit and bypasses the resistor RE.
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3.2Electronic Circuits I
Hence, CE is known as the emitter bypass capacitor. The input signal voltage Vi causes the base voltage to vary sinusoidally above and below its DC bias level. The resulting variation in base current produces a larger variation in collector current. As the collector current increases, the collector voltage decreases. The collector current varies above and below its Q-point value in phase with the base current. The collector to emitter voltage varies above and below its Q-point value 180o out of phase with the base voltage, as shown in Fig. 3.2.
IC
VCC
IBQ
ICQ
VCEQ
VCE
Q
0
Fig. 3.2Small Signal Waveforms
VS
RS
RE CE
R1 RC
CC1Vo
+VCC
CC2
RLR2
Fig. 3.1Common-emi er Amplifi er
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3.3h-parameter Model of BJT3.3
3.3h-PARAMETER MODEL OF BJTConsider the black box representation of the BJT as shown in Fig. 3.3.
V1
+
V2
+
I2I1
BJT
Fig. 3.3Two-port Network
The two-port network equations in terms of h-parameter are given by
V1 = h11 I1 + h12 V2I2 = h21 I1 + h22 V2
where h11, h12, h21 and h22 are h-parameters.
h-parameter(a) Case I: When output port is shorted, i.e. V2 = 0.
h11 = VI
V
1
1 02 =
It is called input impedance and is denoted by hi.
h21 = II
V
2
1 02 =
It is called forward current gain and is denoted by hf.
(b) Case II: When input port is opened, i.e. I1 = 0.
h12 = VV
I
1
2 01 =
It is called reverse voltage gain and is denoted by hr.
h22 = IV
I
2
2 01 =
It is called output admittance and is denoted by ho.
Since these four parameters represent input impedance, voltage gain, current gain and out-put admittance, these parameters are called hybrid parameters.
Let V1 = Vi, I1 = Ii, I2 = Io and V2 = Vo
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3.4Electronic Circuits I
Hence, these equations can be written as
Vi = hi Ii + hr VoIo = hf Ii + ho Vo
The fi rst equation represents Kirchhoffs voltage law to the input circuit which can be rep-resented as shown in Fig. 3.4.
Vi
+
Ii hi
hrVo
Fig. 3.4Input Circuit
The second equation represents Kirchhoffs current law to the output circuit which can be represented as shown in Fig. 3.5.
hoVo
+
Io
hf Ii 1
Fig. 3.5Output Circuit
Combining these two circuits, the h-parameter model is obtained, which is shown in Fig. 3.6.
hoVo
+
Io
hf Ii 1Vi
+
Ii hi
hrVo
Fig. 3.6h-parameter Model for BJT
The h-parameter model is used for representing transistor in CB, CE and CC confi gurations. The h-parameters, however, will change with each confi guration. A second subscript is added to the h-parameter notation. For the CB confi guration, the letter b is added. For CE and CC confi gurations, the letter e and c are added, respectively.
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3.3h-parameter Model of BJT3.5
3.3.1h-parameter Model for CB ConfigurationFigure 3.7 shows the transistor in the CB confi guration.
E
B
C
B
Ie Ic
Ve
+
Vc
+
Fig. 3.7Common-base Confi guration
h-parameter equations are written as,
Ve = hib Ie + hrb VcIc = hfb Ie + hob Vc
The h-parameter model for the CB confi guration is shown in Fig. 3.8.
hob
Ic
hf bIe1
Ie hib
hrbVc
C
B
E
B
Vc
+
Ve
+
Fig. 3.8h-parameter Model for CB Confi guration
3.3.2h-parameter Model for CE ConfigurationFigure 3.9 shows transistor in the CE confi guration.
Ic
Ib
Vb
Vc
C
EE
B
+
+
Fig. 3.9Common-emi er Confi gurationh-parameter equations are written as,
Vb = hie Ib + hre VcIc = hfe Ib + hoe Vc
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3.6Electronic Circuits I
The h-parameter model for the CE confi guration is shown in Fig. 3.10.
hoe
Ic
hf eIb1
Ib hie
hreVc
C
E
B
E
Vc
+
Vb
+
Fig. 3.10h-parameter Model for CE Confi guration
3.3.3h-parameter Model for CC ConfigurationFigure 3.11 shows transistor in the CC confi guration.
Ie
Ib
Vb
Ve
E
C
B
C
+
+
Fig. 3.11Common-collector Confi guration
h-parameter equations are written as,
Vb = hic Ib + hrc VeIe = hfc Ib + hoc Ve
The h-parameter model for the CC confi guration is shown in Fig. 3.12.
hoc
Ie
hf cIb1
Ib hic
hrcVe
B
C
E
C
Ve
+
Vb
+
Fig. 3.12h-parameter Model for CC Confi guration
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3.3h-parameter Model of BJT3.7
3.3.4h-parameter Conversion Generally h-parameter values for the CE confi guration are given in transistor data sheet. The h-parameter values for CB and CC confi gurations can be calculated from h-parameter values for the CE confi guration. Table 3.1 gives h-parameter conversion formulae from CE to CB and from CE to CC confi gurations.
Table 3.1h-parameter Conversion Formulae
Conversion from CE to CB Conversion from CE to CC
(i) hib = h
h1ie
fe+ (i) hic = hie
(ii) hob = h
h1oe
fe+ (ii) hfc = (1 + hfe)
(iii) h =
hh1fe
fe+ (iii) hoc = hoe
(iv) hrb = h h
h1oe ie
fe+ hre (iv) hrc = 1 hre
For CC confi guration hic = hie hfc = (1 + hfe) hfe hoc = hoe hrc = 1 hre 1
Hence, the h-parameter model of the CE confi guration is used for AC analysis of CC amplifi er.
3.3.5Advantages of h-parameter Model1. The h-parameter model consists of all linear elements. Hence, the analysis of the tran-
sistor circuit becomes easier by using Kirchhoffs voltage and current laws.2. h-parameters can easily be obtained from input and output characteristics of transistor.3. h-parameters are normally specifi ed by transistor manufacturer.4. It is useful at low frequency operations.
3.3.6Disadvantages of h-parameter Model1. It is not suitable at high frequency operations.2. h-parameter vary with device and temperature.
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3.8Electronic Circuits I
3.4GRAPHICAL DETERMINATION OF h-PARAMETERS The relationship between VBE and IC in terms of IB and VCE for the CE amplifi er is given by
VBE = f1 (IB, VCE)IC = f2 (IB, VCE)
The input characteristics give the relationship between the input voltage VBE and the input current IB for different values of output voltage VCE. The output characteristics give the relationship between the output voltage VCE and the output current IC for different values of input current IB. The h-parameters can be determined graphically from input and output characteristics of the amplifi er. Determination of hie: From the defi nition of hie,
hie = VIBE
B V = constantCE
= Q
1Slope of input characteristic at the -point
The parameter hie can be obtained as the change in base-emitter voltage, VBE2 VBE1 , divided by the change in base current, IB2 IB1, for a constant collector-emitter voltage VCE at the quiescent point Q. The reciprocal of the slope of input characteristic at the quiescent point Q gives the value of hie as shown in Fig. 3.13.
VCE1VCEQ
VCE2
VBE
IB
VBE
IB
0
Fig. 3.13Determination of hie
Determination of hre: From the defi nition of hre,
hre = VV
BE
CE I = constantB
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3.4Graphical Determination of h-parameters 3.9
The parameter hre can be obtained as the change in base-emitter voltage, VBE2 VBE1, divided by the change in collector-emitter voltage, VCE2 VCE1, for a constant base current IB at the quiescent point Q. A horizontal line on the input characteristics of Fig. 3.14 represents a constant base current.
VCE1VCEQ
VCE2
VBE
IBQ
VBE1 VBE2
IB
0
Fig. 3.14Determination of hre
Determination of hfe: From the defi nition of hfe,
hfe = II
C
B V = constantCE
The parameter hfe can be obtained as the change in collector current, IC2 IC1, divided by the change in base current, IB2 IB1, for a constant collector-emitter voltage VCE at the quies-cent point Q. A vertical line on the output characteristics of Fig. 3.15 represents a constant collector-emitter voltage.
IC2 IB2
IB1
IBQ
IC1
0
IC
VCE VCEQ
Fig. 3.15Determination of hfe
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3.10Electronic Circuits I
Determination of hoe: From the defi nition of hoe,
hoe = I
VC
CE I = constantB
= Slope of output characteristic at the Q-pointThe parameter hoe can be obtained as the change in collector current, IC2 IC1 , divided by
the change in collector-emitter voltage VCE2 VCE1 for a constant base current IB at the quiescent point Q. The slope of output characteristic at the quiescent point Q gives the value of hoe as shown in Fig. 3.16.
VCE IC
IB2
IB1
IBQ
0
IC
VCE
Fig. 3.16 Determination of hoeSimilarly, common-base h-parameters can be determined graphically from input and output
characteristics of the CB confi guration as shown in Figs. 3.173.20.Determination of hib
hib = VIBE
E V = constantCB
= Q
1Slope of input characteristic at the -point
VCB2 VCBQVCB1
VBE
IE
0
IE
VBE
Fig. 3.17Determination of hib
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3.4Graphical Determination of h-parameters 3.11
Determination of hrb
hrb = VV
BE
CB I = constantE
IEQ
VCB1
VBE1VBE2
VCBQVCB2
VBE
IE
0
Fig. 3.18Determination of hrb
Determination of hfb
hfb = II
C
E V = constantCB
IC1
IC
IE1
IE2
IEQ
VCBQVCB0
IC2
Fig. 3.19Determination of h
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3.12Electronic Circuits I
Determination of hob
hob = I
VC
CB I = constantE
= Slope of output characteristic at the Q-point
VCB IC
IC
IE1
IE2
IEQ
VCB0
Fig. 3.20Determination of hob
3.5EXACT ANALYSIS OF TRANSISTOR AMPLIFIER Figure 3.21 shows the general h-parameter equivalent circuit with load RL connected to the output.
ho RL
Io
hf Ii1
RS
VS
Ii hi
hrVo
Zi Zo
Vo
+
Vi
+
Fig. 3.21General h-parameter Equivalent Circuit
Current gain (Ai)
It is the ratio of output current Io to the input current Ii.
Ai = II
o
i
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3.5Exact Analysis of Transistor Amplifi er 3.13
Applying Kirchhoffs current law to the output circuit,
Io = hf Ii + hoVoBut Vo = IoRL Io = hf Ii ho IoRL (1 + ho RL) Io = hf Ii
Io = h Ih R1f i
o L+
Ai = II
o
i
= hh R1
f
o L+
Voltage gain (Av)
It is the ratio of output voltage Vo to the input voltage Vi.
Av = VV
o
i
Applying Kirchhoffs voltage law to the input circuit,
Vi hi Ii hr Vo = 0 Vi = hi Ii + hr Vo
We know that, II
o
i
= hh R1
f
o L+
Ii = I h R
h(1 )o o L
f
+
But Io = VR
o
L
Ii = VR
h Rh
(1 )oL
o L
f
+
Hence, Vi = hV h R
h R(1 )i o o L
f L
++ hr Vo
= h h R h h h R
h Rf r L i i o L
f L
Vo
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3.14Electronic Circuits I
VV
o
i
= h R
h h h R h h Rf L
i i o L f r L +
Av = VV
o
i
= h R
h h h R h h Rf L
i i o L f r L
+
= h R
h h h h h R( )f L
i i o f r L
+
= h R
h h Rf L
i L+ , where h = hi ho hf hr
Input impedance (Zi)
It is the ratio of input voltage Vi to the input current Ii.
Zi = VI
i
i
Applying Kirchhoffs voltage law to the input circuit,
Vi hi Ii hrVo = 0 Vi = hi Ii + hr Vo = hi Ii hr Io RL
= hi Ii hrh Ih R1f i
o L+
RL
= hh h R
h R1if r L
o L
+
Ii
Zi = VI
i
i
= hi h h R
h R1f r L
o L+
Output impedance (Zo)
It is the ratio of output voltage Vo to the output current Io when input voltage VS = 0.
Zo = VI
o
o V = 0s
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3.5Exact Analysis of Transistor Amplifi er 3.15
Applying Kirchhoffs voltage law to the input circuit,
VS hi Ii RS Ii hr Vo = 0
Ii = V h V
R hS r o
S i
+
When VS = 0,
Ii = h V
R hr o
S i+
Applying Kirchhoffs current law to the output circuit,
Io = hf Ii + ho Vo
= hf h V
R hr o
S i++ ho Vo
= hh h
R hof r
S i
+
Vo
Zo = VI
o
o
= h
h hR h
1
of r
S i
+
Voltage gain (Avs)
It is the ratio of output voltage Vo to the input voltage VS.
Avs = VV
o
S
Considering source resistance RS as shown in Fig. 3.22,RS
VS Zi Vi
Fig. 3.22Thevenin's Equivalent for the Input Circuit
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3.16Electronic Circuits I
VV
i
S
= Z
Z Ri
i S+
Now, Avs = VV
o
S
= VV
o
i
VV
i
S
= Av VV
i
S
= Av Z
Z Ri
i S+
Hence, voltage gain Avs is always less than Av.
Current gain (Ais)It is the ratio of output current Io to the input current IS.
Ais = II
o
S
Considering source resistance as shown in Fig. 3.23,
VS
RSZi
Ii
Fig. 3.23Input Circuit
By source transformation (Fig. 3.24),
ZiIS RS
Ii
Fig. 3.24Norton's Equivalent for the Input Circuit
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3.5Exact Analysis of Transistor Amplifi er 3.17
II
i
s
= R
R ZS
S i+
Ais = II
L
s
= II
L
i
II
i
s
= Ai II
i
s
= Ai R
R ZS
S i+
Hence, current gain Ais is always less than Ai.
Example 3.1: Calculate the values of Ai, Av, Zi and Zo for the circuit shown in Fig. 3.25 if the CE amplifi er uses transistor with the specifi cations given in the fi gure and a resistive load of 1 k.
hoe RL
Io
hf eIi1VS
Ii hie
hreVo
Zi Zo
Vo
+
Vi
+
hfe = 220hie = 2.7 k
hre = 1.5 104
hoe = 18
Fig. 3.25Example 3.1
Solution:
RL = 1 k RS = 0
(i) Current gain Ai = hh R1
fe
oe L+
= 2201 18 10 1 106 3+
= 216.11
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3.18Electronic Circuits I
(ii) Voltage gain Av = h R
h h h h h R( )fe L
ie ie oe fe re L
+
= 220 1 10
2.7 10 (2.7 10 18 10 220 1.5 10 )(1 10 )
3
3 3 6 4 3
+
= 81.48The negative sign indicates a phase shift of 180 between input and output voltages.
(iii) Input impedance Zi = hie h h R
h R1fe re L
oe L+
= 2.7 103 220 1.5 10 1 101 18 10 1 10
4 3
6 3
+
= 2.66 k
(iv) Output impedance Zo = h
h hR h
1
oefe re
S ie
+
= 1
18 10 220 1.5 100 2.7 10
64
3
+
= 173.07 k
3.6EXACT ANALYSIS OF COMMON-EMITTER AMPLIFIERFigure 3.26 shows a common-emitter amplifi er.
VS
RS
RE CE
R1 RC
CC1Vo
+VCC
CC2
RLR2
Fig. 3.26Common-emi er Amplifi er
To analyze the AC signal operation of an amplifi er, an AC equivalent circuit is developed. The capacitors CC1, CC2 and CE are replaced by short circuit because the capacitive reactance is approximately zero at signal frequency. The DC source is replaced by ground. The AC equivalent circuit is shown in Fig. 3.27.
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3.6Exact Analysis of Common-emi er Amplifi er3.19
VS
RS
Vo
RLRCR1 R2
Fig. 3.27AC Equivalent Circuit
The BJT is replaced by the h-parameter model as shown in Fig. 3.28.
hoeRC RL
Ic
hf eIb
Io
1RS
RB
VS
Ii Ib hie
hreVo
Zi Z iZin Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.28h-parameter Equivalent Circuit
RB = R1 || R2 RL = RC || RL
Input impedance
Zi = hie h h R
h R1fe re L
oe L
+
Zi = Zi || RB Zin = Zi + RS
Output impedance
Zo = h
h hR R h
1
( || )oefe re
B S ie
+
Zo = Zo ||RC ||RL
Voltage gain
Av = h R
h h h h h R( )fe L
ie ie oe fe re L
+
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3.20Electronic Circuits I
The negative sign indicates a phase shift of 180 between input and output voltages.Considering voltage gain with source resistance RS,
Avs = Av Z
Z Ri
i S+
Hence, voltage gain Avs is always less than Av.
Current gain
For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Io = VR
o
L
Ii = VZ
i
i
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Considering current gain with source resistance,
Ais = Ai R
R ZS
S i+
Hence, current gain Ais is always less than Ai.
Example 3.2: Find Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.29.
VS
1 k
1 k CE
5 k
CC1
Vo
+VCC
CC2470 k
Zin Zo
hfe = 220hie = 2.7 k
hre = 1.5 104
hoe = 18
Fig. 3.29Example 3.2
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3.6Exact Analysis of Common-emi er Amplifi er3.21
Solution:
AC equivalent circuit (Fig. 3.30)
VS
RS
Vo
RCRB
Fig. 3.30AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.31)
hoeRC
Io = Ic
hf eIb1
RSRB
VS
Ii Ib hie
hreVo
Zi Z iZin Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.31h-parameter Equivalent Circuit
(i) Input impedance Zi = hie h h R
h R1fe re C
oe C+
= 2.7 103 220 1.5 10 5 10
1 18 10 5 10
4 3
6 3
+
= 2.54 k
Zi = Zi || RB = (2.54 103) || (470 103) = 2.53 k
Zin = Zi + RS = 2.53 103 + 1 103 = 3.53 k
(ii) Output impedance Zo = h
h hR R h
1
( || )oefe re
B S ie
+
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3.22Electronic Circuits I
= 1
18 10 220 1.5 10[(470 10 ) || (1 10 )] 2.7 10
64
3 3 3
+
= 110.12 k Zo = Zo || RC = (110.12 10
3) || (5 103) = 4.78 k
(iii) Voltage gain Av = h R
h h h h h R( )fe C
ie ie oe fe re C+
=
= 395.97
Avs = Av Z
Z Ri
i S+
= 395.97 2.53 10
2.53 10 1 10
3
3 3
+ = 283.79
(iv) Current gain Ai = Av ZR
i
C
= (395.97) 2.53 10
5 10
3
3
= 200.35
Ais = Ai R
R ZS
S i+
= 200.35 1 10
1 10 2.53 10
3
3 3
+ = 56.75
Example 3.3: Find Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.32.
VS
1 k
1 k
5 k
CE
5 k
CC1
Vo
+VCC
CC2470 k
Zin Zo
hfe = 220hie = 2.7 k
hre = 1.5 104
hoe = 18
Fig. 3.32Example 3.3
220 5 102.7 10 (2.7 10 18 10 220 1.5 10 )(5 10 )
3
3 3 6 4 3
+
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3.6Exact Analysis of Common-emi er Amplifi er3.23
Solution:
AC equivalent circuit (Fig. 3.33)
VS
RS
Vo
RCRB
RL
Fig. 3.33AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.34)
hoeRC RL
Ic
hf eIb
Io
1RS
RB
VS
Ii Ib hie
hreVo
Zi Z iZin Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.34h-parameter Equivalent Circuit
RL = RC || RL = (5 103) || (5 103) = 2.5 k
(i) Input impedance Zi = hie h h R
h R1fe re L
oe L
+
= 2.7 103 220 1.5 10 2.5 101 18 10 2.5 10
4 3
6 3
+
= 2.62 k Zi = Zi || RB = (2.62 10
3) || (470 103) = 2.61 k Zin = Z i + RS = 2.61 10
3 + 1 103 = 3.61 k
(ii) Output impedance Zo = h
h hR R h
1
( || )oefe re
B S ie
+
= 1
18 10 220 1.5 10[(470 10 ) || (1 10 )] 2.7 10
64
3 3 3
+
= 110.119 k Zo = Zo || RC || RL = (110.119 10
3) || (5 103) || (5 103) = 2.44 k
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3.24Electronic Circuits I
(iii) Voltage gain Av = h R
h h h h h R( )fe L
ie ie oe fe re L
+
=
= 200.8
Avs = Av Z
Z Ri
i S+
= 200.8 2.61 102.61 10 1 10
3
3 3
+
= 145.17
(iv) Current gain Ai = AvZR
i
L
= (200.8) 2.61 105 10
3
3
= 104.8
Ais = Ai R
R ZS
S i+
= 104.8 1 101 10 2.61 10
3
3 3
+
= 29.03
Example 3.4: Determine Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.35.
VS
10 k
10 kRE
3 k
CE
5 k
CC1
Vo
+VCC
CC2100 k
Zin Zo
hfe = 50hie = 1.1 k
hre = 2.5 104
hoe = 24
Fig. 3.35Example 3.4
220 2.5 102.7 10 (2.7 10 18 10 220 1.5 10 )(2.5 10 )
3
3 3 6 4 3
+
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3.6Exact Analysis of Common-emi er Amplifi er3.25
Solution:
AC equivalent circuit (Fig. 3.36)
RS RCR2R1
RL
VS+
Fig. 3.36AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.37)
hoeRC RL
Ic
hf eIb
Io
1RS
RB
VS
Ii Ib hie
hreVo
Zi Z iZin Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.37h-parameter Equivalent Circuit
RB = R1 || R2 = (100 103) || (10 103) = 9.09 k
RL = RC || RL = (5 103) || (3 103) = 1.875 k
(i) Input impedance Zi = hie h h R
h R1fe re L
oe L
+
= 1.1 103 50 2.5 10 1.875 101 24 10 1.875 10
4 3
6 3
+
= 1.077 k
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3.26Electronic Circuits I
Zi = Zi || RB = (1.077 103) || (9.09 103) = 0.963 k
Zin = Zi + RS = 0.963 103 + 10 103 = 10.963 k
(ii) Output impedance Zo = h
h hR R h
1
( || )oefe re
B S ie
+
= 1
24 10 50 2.5 10[(9.09 10 ) || (10 10 )] 1.1 10
64
3 3 3
+
= 45.73 k
Zo = Zo || RC || RL = (45.73 103) || (5 103) || (3 103) = 1.8 k
(iii) Voltage gain Av = h R
h h h h h R( )fe L
ie ie oe fe re L
+
=
= 83.25
Avs = AvZ
Z Ri
i S+
= 83.25 0.963 100.963 10 10 10
3
3 3
+
= 7.313
(iv) Current gain Ai = AvZR
i
L
= (83.25) 0.963 103 10
3
3
= 26.72
Ais = Ai R
R ZS
S i+
= 26.72 10 1010 10 0.963 10
3
3 3
+ = 24.75
50 1.875 101.1 10 (1.1 10 24 10 50 2.5 10 )(1.875 10 )
3
3 3 6 4 3
+
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3.7Exact Analysis of Common-base Amplifi er3.27
3.7EXACT ANALYSIS OF COMMON-BASE AMPLIFIERFigure 3.38 shows a common-base amplifi er.
RERS
R1 RC
CC1Vo
+VCC
CC3
CC2RL
R2
VS
Fig. 3.38Common-base Amplifi er
The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replac-ing DC source by ground as shown in Fig. 3.39. Two resistors R1 and R2 are shorted to ground.
RERS
Vo
RC RL
VS
Fig. 3.39AC Equivalent Circuit
Figure 3.39 can be redrawn by rotating the BJT as shown in Fig. 3.40.
RS
Vo
RCRE RL
VS
Fig. 3.40AC Equivalent Circuit
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3.28Electronic Circuits I
The h-parameter equivalent circuit is obtained by replacing the BJT by its h-parameter model as shown in Fig. 3.41.
hob
Ic
hf bIe
Io
1ViRS
+
Vo
+
IeIi hib
hrbVoRE RC RL
E C
B
ZiZin Zi Zo Zo
VS
Fig. 3.41h-parameter Equivalent Circuit
RL = RC || RLInput impedance
Zi = hib h h R
h R1fb rb L
ob L
+
Zi = Zi || RE Zin = Zi + RS
Output impedance
Zo = h
h hR R h
1
( || )obfb rb
E S ib
+
Zo = Zo || RC ||RL
Voltage gain
Av = h R
h h h h h R( )fb L
ib ib ob fb rb L
+
Considering source resistance,
Avs = Av Z
Z Ri
i S+Current gain
For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
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3.7Exact Analysis of Common-base Amplifi er3.29
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Considering source resistance,
Ais = Ai R
R ZS
S i+
Example 3.5: Calculate Zi, Zo, Av and Ai for the circuit shown in Fig. 3.42.
3 k1 k
50 k 2.2 k
CC1Vo
+VCC
CC2
CC3
50 k
Zi Zo
VS
hrb = 2 104hib = 14.41
hfb = 0.991hob = 0.18 106
Fig. 3.42Example 3.5
Solution:AC equivalent circuit (Fig. 3.43)
RERS
Vo
RC
VS
Fig. 3.43AC Equivalent Circuit
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3.30Electronic Circuits I
Figure 3.43 can be redrawn by rotating the BJT as shown in Fig. 3.44.Vo
RCRE
RS
VS
Fig. 3.44AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.45)
hob
Ic
hf bIe1Vi
RS
+
+
Vo
+
IeIi hib
hrbVoRE RC
E C
B
Zi Zi Zo Zo
VS
Fig. 3.45h-parameter Equivalent Circuit
(i) Input impedance Zi = hib h h R
h R1f b rb C
ob C+
= 14.41 ( 0.991)(2 10 )(2.2 10 )1 0.18 10 2.2 10
4 3
6 3
+ = 14.84
Zi = Zi || RE = 14.84 || (3 103) = 14.76
(ii) Output impedance Zo = h
h hR R h
1
( || )obf b rb
E S ib
+
= 1
0.18 10 ( 0.991)(2 10 )[(3 10 ) || (1 10 )] 14.41
64
3 3
+
= 2.276 M Zo = Zo || RC = (2.276 10
6) || (2.2 103) = 2.197 k
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3.7Exact Analysis of Common-base Amplifi er3.31
(iii) Voltage gain Av = +h R
h h h h h R( )fb C
ib ib ob fb rb C
= ( 0.991) 2.2 10
14.41 [14.41 0.18 10 2 10 ( 0.991)](2.2 10 )
3
6 4 3
+
= 146.7
(iv) Current gain Ai = Av ZR
i
C
= 146.7 14.762.2 103
= 0.9847
Example 3.6: Calculate Zi, Zo, Avs and Ais for the circuit shown in Fig. 3.46.
Vo
10 k 47 k7.5 k
6 V +12 V
100
CC1 CC2
Zi Zo
hfb = 0.99hob = 2.47 107
hib = 12
hrb = 3 105VS
Fig. 3.46Example 3.6
Solution:AC equivalent circuit (Fig. 3.47)
Vo
RC RLRE
RS
VS
Fig. 3.47AC Equivalent Circuit
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3.32Electronic Circuits I
h-parameter equivalent circuit (Fig. 3.48)
hob
Ic
hf bIe
Io
1ViRS
+
Vo
+
IeIi hib
hrbVoRE RC RL
E C
B
Zi Zi Zo Zo
VS
Fig. 3.48h-parameter Equivalent Circuit
RL = RC || RL = (10 103) || (47 103) = 8.25 k
(i) Input impedance Zi = hib
+
h h R
h R1fb rb L
ob L
= 12 +
( 0.99)(3 10 )(8.25 10 )1 2.47 10 8.25 10
5 3
7 3
= 12.24 Zi = Zi || RE = 12.24 || (7.5 10
3) = 12.22
(ii) Output impedance Zo = h
h hR R h
1
( || )obfb rb
E S ib
+
= 1
2.47 10 ( 0.99)(3 10 )[(7.5 10 ) ||100] 12
75
3
+
= 1.94 M
Zo = Zo || RC || RL = (1.94 106) || (10 103) || (47 103) = 8.209 k
(iii) Voltage gain Av =
+
h R
h h h h h R( )fb L
ib ib ob fb rb L
= ( 0.99)(8.25 10 )
12 [12 2.47 10 3 10 ( 0.99)](8.25 10 )
3
7 5 3
+
= 665.64
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3.8Exact Analysis of Common-collector Amplifi er3.33
Avs = Av Z
Z Ri
i S+
= 665.64 +
12.2212.22 100
= 72.6
(iv) Current gain Ai = Av ZR
i
L
= 665.64 12.2247 103
= 0.173
Ais = Ai R
R ZS
S i+
= 0.173 100100 12.22+
= 0.154
3.8EXACT ANALYSIS OF COMMON-COLLECTOR AMPLIFIERFigure 3.49 shows a common-collector amplifi er.
RE
R1
CC1
+VCC
VoCC2
RLR2
RS
VS
Fig. 3.49Common-collector Amplifi er
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3.34Electronic Circuits I
The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.50.
RE
Vo
RL
R2R1
RS
VS
Fig. 3.50AC Equivalent Circuit
The h-parameter equivalent circuit is obtained by replacing the BJT by its h-parameter model as shown in Fig. 3.51.
hoc
Ie
hf cIb
Io
1ViRS
+
Vo
+
IbIi hic
hrcVoRB RE RL
B E
C
VS
Fig. 3.51h-parameter Equivalent Circuit
RB = R1 || R2RL = RE || RL
Input impedance
Zi = hic
+
h h R
h R1fc rc L
oc L
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3.8Exact Analysis of Common-collector Amplifi er3.35
Zi = Zi || RB
Zin = Zi + RS
Output impedance
Zo = h
h hR R h
1
( || )ocfc rc
B S ic
+
= Zo || RE || RL
Voltage gain
Av = h R
h h h h h R( )fc L
ic ic oc fc rc L
+
Considering source resistance,
Avs = AvZ
Z Ri
i S+
Current gain
For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Considering source resistance,
Ais = AiR
R ZS
S i+
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3.36Electronic Circuits I
Example 3.7: Calculate Zi, Zo, Av and Ai, for the circuit shown in Fig. 3.52.
5 k
10 k
1 k
CC1
+VCC
VoCC2
20 k10 k
Zi Zo
hic = 1.2 khrc = 1hfc = 101
hoc = 25 106
VS
Fig. 3.52Example 3.7
Solution:AC equivalent circuit (Fig. 3.53)
RE
Vo
RL
R2R1
RS
VS
Fig. 3.53AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.54)
hoc
Ie
hf cIb
Io
1ViRS
+
Vo
+
IbIi hic
hrcVoRB RE RL
B E
C
VS
Fig. 3.54h-parameter Equivalent Circuit
RB = R1 || R2 = (10 103) || (10 103) = 5 k
RL = RE || RL = (5 103) || (20 103) = 4 k
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3.8Exact Analysis of Common-collector Amplifi er3.37
(i) Input impedance Zi = hic h h R
h R1fc rc L
oc L
+
= 1.2 103 ( 101)(1)(4 10 )
1 25 10 4 10
3
6 3
+
= 368.47 kZi = Zi || RB = (368.47 10
3) || (5 103) = 4.93 k
(ii) Output impedance Zo = h
h hR R h
1
( || )ocfc rc
B S ic
+
= 1
25 10 ( 101)(1)[(5 10 ) || (1 10 )] 1.2 10
63 3 3
+
= 20.12 Zo = Zo || RE || RL = 20.12 || (5 10
3) || (20 103) = 20.02
(iii) Voltage gain Av = h R
h h h h h R( )fc L
ic ic oc fc rc L
+
= ( 101)(4 10 )
1.2 10 [1.2 10 25 10 ( 101)(1)](4 10 )
3
3 3 6 3
+
= 0.996
Avs = Av Z
Z Ri
i S+
= 0.996 4.93 104.93 10 1 10
3
3 3
+ = 0.828
(iv) Current gain Ai = AvZR
i
L
= 0.996 4.93 1020 10
3
3
= 0.246
Ais = Ai R
R ZS
S i+
= 0.246 1 101 10 4.93 10
3
3 3
+ = 0.041
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3.38Electronic Circuits I
3.9APPROXIMATE ANALYSIS OF TRANSISTOR AMPLIFIERFor the CE and CB confi gurations, the magnitude of hr and ho is often such that the parameters Zi, Zo, Av and Ai are only slightly affected if hr and ho are not included in the model. Since the value of hr is very small, controlled source hrVo becomes very small. It is approximated by hrVo = 0 and is replaced by a short circuit. Similarly, the value of ho is very small, and the impedance
h1
o
becomes very large and can be ignored in comparison to a parallel load. It is approxi-
mated by h1
o
= and is replaced by an open circuit. Thus, the h-parameter model is simpli-
fi ed by short circuiting controlled source hrVo and open circuiting impedance h1
o
. Generally, if
hoeRL < 0.1, the approximate model can be used. The simplifi ed h-parameter model is shown in the Fig. 3.55.
RL
Io
hi
VS
RS
Ii
hfIi Vo
+
Vi
+
Fig. 3.55Simplifi ed h-parameter Model
Input impedance Zi = hi
Output impedance Zo = h1
o
Voltage gain Av = h R
hf L
i
Current gain Ai = hf
3.10APPROXIMATE ANALYSIS OF COMMON-EMITTER AMPLIFIER (FIXED-BIAS CONFIGURATION)
Fig. 3.56 shows a fi xed-bias common-emitter amplifi er.
Case I: When RE is bypassed
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3.10Approximate Analysis of Common-emi er Amplifi er3.39
RE
RS
RB RC
CC1Vo
+VCC
CC2
RL
CEVS
Fig. 3.56Common-emi er Amplifi er with Bypassed RE
The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.57.
Vo
RC RLRB
RS
VS
Fig. 3.57AC Equivalent Circuit
The BJT is replaced by the approximate h-parameter model as shown in Fig. 3.58.
Ic
hf eIb
Io
ViRS
+
Vo
+
IbIi
hieRB RC RL
B C
E
ZiZin Zi Zo Zo
VS
Fig. 3.58h-parameter Equivalent Circuit
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3.40Electronic Circuits I
Input impedance Zi = hieZi = Zi || RBZin = Zi + RS
Output impedance The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.
Zo =
Zo = Zo || RC || RL = RC || RL
Voltage gain
RL = RC || RLVo = hfe Ib RLVi = Ib hie
Av = VV
o
i
= h I R
I hfe b L
b ie
= h R
hfe L
ie
The negative sign indicates a phase shift of 180 between input and output voltages.Considering voltage gain with source resistance RS,
Avs = Av Z
Z Ri
i S+
Current gainFor transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Considering current gain with source resistance,
Ais = Ai R
R ZS
S i+
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Case II: When RE is not bypassed Figure 3.59 shows a CE amplifi er with unbypassed RE.
RE
R1 RC
CC1
+VCC
VoCC2
RLR2
RS
VS
Fig. 3.59Common-emi er Amplifi er with Unbypassed RE
The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.60.
RE RC RL
Vo
R2R1
RS
VS
Fig. 3.60AC Equivalent Circuit
The BJT is replaced by the approximate h-parameter model as shown in Fig. 3.61.
Input impedance Vi = Ibhie + IeRE
= Ibhie + (1 + hfe) Ib RE = [hie + (1 + hfe) RE] Ib
Zi = VI
i
b
3.10Approximate Analysis of Common-emi er Amplifi er3.41
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3.42Electronic Circuits I
= h h R I
I[ (1 ) ]ie fe E b
b
+ +
= hie + (1 + hfe) RE
Due to unbypassed RE, input impedance Zi increases by a value (1 + hfe) RE. When RE is transferred to base side, it gets multiplied by 1 + hfe. Hence, input impedance looking from the base side, i.e. Zi is the sum of hie and (1 + hfe) RE.
Zi = Zi || RB= [hie + (1 + hfe) RE] || RB
Zin = Zi + RSOutput impedance The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.
Zo = Zo = Zo || RC || RL = RC || RL
Voltage gain
R R R||L C L =
Vo = hfe Ib RLVi = [hie + (1 + hfe) RE] Ib
Av = VV
o
i
RC
RE
RL
Ic
hfeIb
Io
RSRB
VS
Ii Ib
Ie
hie
Zi Z iZin Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.61h-parameter Equivalent Circuit
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=
+ +
h I R
h h R I[ (1 ) ]fe b L
ie fe E b
= h R
h h R(1 )fe L
ie fe E
+ +
The negative sign indicates a phase shift of 180 between input and output voltages. Due to unbypassed RE, voltage gain decreases.
Considering voltage gain with source resistance RS,
Avs = Av Z
Z Ri
i S+Current gain For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Considering current gain with source resistance,
Ais = Ai R
R ZS
S i+
Example 3.8: Calculate Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.62.
VS
1 k
RE
10 k
CE
4.7 k
CC1
Vo
+VCC
CC2470 k
Zin Zo
hfe = 110hie = 1.6 k
Fig. 3.62Example 3.8
Solution: AC equivalent circuit (Fig. 3.63)
3.10Approximate Analysis of Common-emi er Amplifi er3.43
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3.44Electronic Circuits I
VS
RS
Vo
RCRB
RL
Fig. 3.63AC Equivalent Circuith-parameter equivalent circuit (Fig. 3.64)
RC RL
Ic
hfeIb
Io
RSRB
VS
Ii Ib
hie
Zi Z iZin Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.64h-parameter Equivalent Circuit
RL = RC || RL = (4.7 103) || (10 103) = 3.197 k
(i) Input impedance Zi = hie = 1.6 k Zi = Zi || RB = (1.6 10
3) || (470 103) = 1.595 k Zin = Zi + RS = 1.595 10
3 + 1 103 = 2.595 k(ii) Output impedance Zo =
Zo = RC || RL = (4.7 103) || (10 103) = 3.197 k
(iii) Voltage gain Av = h R
hfe L
ie
=
110 3.197 101.6 10
3
3
= 219.8
Avs = Av Z
Z Ri
i S+
= 219.8 +
1.595 101.595 10 1 10
3
3 3
= 134.94
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(iv) Current gain Ai = Av ZR
i
L
= (219.8)
1.595 1010 10
3
3
= 35.05
Ais = Ai R
R ZS
S i+
= 35.05 +
1 101 10 1.595 10
3
3 3
= 13.53
Example 3.9: Calculate Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.65.
VS
1 k
4.7 k
1.2 k
10 k
4.7 k
CC1
Vo
+VCC
CC2470 k
Zin Zo
hfe = 110hie = 1.6 k
Fig. 3.65Example 3.9Solution:AC equivalent circuit (Fig. 3.66)
RE RC RL
Vo
RB
RS
VS
Fig. 3.66AC Equivalent Circuit
3.10Approximate Analysis of Common-emi er Amplifi er3.45
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3.46Electronic Circuits I
h-parameter equivalent circuit (Fig. 3.67)
RC
RE
RL
Ic
hfeIb
Io
RSRB
VS
Ii Ib
Ie
hie
Zi Z i Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.67h-parameter Equivalent Circuit
RL = RC || RL = (4.7 103) || (10 103) = 3.197 k
(i) Input impedance Zi = hie + (1 + hfe) RE= 1.6 103 + (1 + 110) (1.2 103)= 134.8 k
Zi = Zi || R B = (134.8 103) || (470 103) = 104.76 k
(ii) Output impedance Zo = Zo = RC = 4.7 k
(iii) Voltage gain Av = h R
h h R(1 ) fe L
ie fe E
+ +
= 110 3.197 10
1.6 10 (1 110)(1.2 10 )
3
3 3
+ + = 2.61
Avs = Av Z
Z Ri
i S+
= 2.61 +
104.76 10104.76 10 1 10
3
3 3
= 2.56
(iv) Current gain Ai = Av ZR
i
L
= (2.61)
104.76 1010 10
3
3
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= 27.34
Ais = Ai R
R ZS
S i+
= 27.34 +
1 101 10 104.76 10
3
3 3
= 0.26
3.11 APPROXIMATE ANALYSIS OF COMMON-EMITTER AMPLIFIER (VOLTAGE DIVIDER BIAS CONFIGURATION)
Figure 3.68 shows a common-emitter amplifi er with voltage divider bias circuit.
Case I: When RE is bypassed
RE
RS
R1 RC
CC1Vo
+VCC
CC2
R2
RL
CEVS
Fig. 3.68Common-emi er Amplifi er with Bypassed REThe AC equivalent circuit is obtained by replacing all capacitors by short circuits and
replacing DC source by ground as shown in Fig. 3.69.
RC RL
Vo
R2R1
RS
VS
Fig. 3.69AC Equivalent Circuit
3.11Approximate Analysis of Common-emi er Amplifi er3.47
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3.48Electronic Circuits I
The BJT is replaced by the approximate h-parameter model as shown in Fig. 3.70.
Ic
hf eIb
Io
ViRS
+
Vo
+
IbIi
hieRB RC RL
B C
E
ZiZin Zi Zo Zo
VS
Fig. 3.70h-parameter Equivalent Circuit
RB = R1 || R2Input impedance
Zi = hieZi = Zi || RB
Zin = Zi + RS
Output impedance
The output impedance Zo of the circuit is the impedance looking from collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.
Zo =
Zo = Zo || RC || RL = RC || RL
Voltage gain
RL = RC || RLVo = hfe Ib RLVi = Ib hie
Av = VV
o
i
= h I R
I hfe b L
b ie
= h R
hfe L
ie
The negative sign indicates a phase shift of 180 between input and output voltages.Considering voltage gain with source resistance RS,
Avs = Av Z
Z Ri
i S+
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Current gain
For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Considering current gain with source resistance,
Ais = Ai R
R ZS
S i+
Case II: When RE is not bypassed Figure 3.71 shows a CE amplifi er with unbypassed RE.
RE
R1 RC
CC1
+VCC
VoCC2
RLR2
RS
VS
Fig. 3.71Common-emi er Amplifi er with Unbypassed RE
The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.72.
3.11Approximate Analysis of Common-emi er Amplifi er3.49
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3.50Electronic Circuits I
RCRE RL
Vo
R2R1
RS
VS
Fig. 3.72AC Equivalent Circuit
The BJT is replaced by the approximate h-parameter model as shown in Fig. 3.73.
RC
RE
RL
Ic
hfeIb
Io
RSRB
VS
Ii Ib
Ie
hie
ZiZin Z i Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.73h-parameter Equivalent Circuit
RB = R1 || R2Input impedance
Vi = Ibhie + IeRE = Ibhie + (1 + hfe) Ib RE = [hie + (1 + hfe) RE] Ib
Zi = VI
i
b
= + +h h R I
I[ (1 ) ]ie fe E b
b
= hie + (1 + hfe) REDue to unbypassed RE, input impedance Zi increases by (1 + hfe) RE.
Zi = Zi || RB= [hie + (1 + hfe) RE] || RB
Zin = Zi + RS
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Output impedance The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.
Zo =
Zo = Zo || RC || RL = RC || RLVoltage gain
R R R||L C L =
Vo = hfe Ib RLVi = [hie + (1 + hfe) RE] Ib
Av = VV
o
i
=
+ +
h I R
h h R I[ (1 ) ]fe b L
ie fe E b
=
+ +
h Rh h R(1 )
fe L
ie fe E
The negative sign indicates a phase shift of 180 between input and output voltages. Due to unbypassed RE, voltage gain decreases.
Considering voltage gain with source resistance RS,
Avs = Av Z
Z Ri
i S+Current gain For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Ai = II
o
i
=
VRVZ
o
L
i
i
= VV
o
i
ZR
i
L
= AvZR
i
L
Considering current gain with source resistance,
Ais = Ai R
R ZS
S i+
Example 3.10: Calculate Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.74.Solution:AC equivalent circuit (Fig. 3.75)
3.11Approximate Analysis of Common-emi er Amplifi er3.51
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3.52Electronic Circuits I
RE
500
120 k 3.3 k
CC1Vo
+VCC
CC2
39 k
10 k
CE
Zin Zo
hie = 2 khfe = 120
VS
Fig. 3.74Example 3.10
RC RL
Vo
R2R1
RS
VS
Fig. 3.75AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.76)
RC RL
Ic
hfeIb
Io
RSRB
VS
Ii Ib
hie
ZiZin Z i Z o Zo
Vo
+
Vi
+
B
E
C
Fig. 3.76h-parameter Equivalent Circuit
RB = R1 || R2 = (120 103) || (39 103) = 29.43 k
RL = RC || RL = (3.3 103) ||(10 103) = 2.48 k
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(i) Input impedance Zi = hie = 2 kZi = Zi || RB = (2 10
3) || (29.43 103) = 1.87 kZin = Zi + RS = 1.87 10
3 + 500 = 2.37 k
(ii) Output impedance Zo = Zo = RC || RL = (3.3 10
3) || (10 103) = 2.48 k
(iii) Voltage gain Av = h R
hfe L
ie
=
120 2.48 102 10
3
3
= 148.8
Avs = Av Z
Z Ri
i S+
= 148.8 1.87 101.87 10 500
3
3
+
= 117.4
(iv) Current gain Ai = Av ZR
i
L
= ( 148.8)
1.87 1010 10
3
3
= 27.83
Ais = Ai R
R ZS
S i+
= 27.83 +
500500 1.87 103
= 5.87
Example 3.11: Calculate Zi, Zo, Av and Ai for the circuit shown in Fig. 3.77.
Solution:AC equivalent circuit (Fig. 3.78)
3.11Approximate Analysis of Common-emi er Amplifi er3.53
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3.54Electronic Circuits I
1.2 k
540 k5.6 k
CC1
+VCC
VoCC2
540 k
Zi Zo
hie = 555 hfe = 120
VS
Fig. 3.77Example 3.11
RE RC
Vo
R2R1
RS
VS
Fig. 3.78AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.79)
RE
RChfeIb
Io = Ic
RBVS
Ii Ib
Ie
hie
Zi Z i Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.79h-parameter Equivalent Circuit
R B = R1 || R2 = (540 103) || (540 103) = 270 k
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(i) Input impedance Zi = hie + (1 + hfe) RE
= 555 + (1 + 120) (1.2 103)
= 145.76 k
Zi = Zi || R B = (145.76 103) || (270 103) = 94.66 k
(ii) Output impedance Zo =
Zo = RC = 5.6 k
(iii) Voltage gain Av = + +h R
h h R(1 )fe C
ie fe E
= 120 5.6 10
555 (1 120)(1.2 10 )
3
3
+ +
= 4.61
(iv) Current gain Ai = Av ZR
i
C
= ( 4.61)
94.66 105.6 10
3
3
= 77.93
Example 3.12: Calculate Zi, Zo, Av and Ai for the circuit shown in Fig. 3.80.
0.9 k
0.1 k
33 k 3.3 k
CC1Vo
+VCC
CC2
10 k
CE
Zi Zo
hie = 2 khfe = 100VS
Fig. 3.80Example 3.12
3.11Approximate Analysis of Common-emi er Amplifi er3.55
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3.56Electronic Circuits I
Solution:AC equivalent circuit (Fig. 3.81)
RE1 RC
Vo
R2R1
RS
VS
Fig. 3.81AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.82)
RC
RE1
hfeIbRS
RB
VS
Ii Ib
Ie
hie
Zi Z i Z o Zo
Vo
+
Vi
+
B
E
Io = Ic C
Fig. 3.82h-parameter Equivalent Circuit
R B = R1 || R2 = (33 103) || (10 103) = 7.6 k
(i) Input impedance Zi = hie + (1 + hfe) RE= 2 103 + (1 + 100)(0.1 103)= 12.1 k
Zi = Zi || RB = (12.1 103) || (7.67 103) = 4.694 k
(ii) Output impedance Zo =
Zo = RC = 3.3 k
(iii) Voltage gain Av = + +h R
h h R(1 )fe C
ie fe E
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= 100 3.3 102 10 (1 100)(0.1 10 )
3
3 3
+ + = 27.272
(iv) Current gain Ai = Av ZR
i
C
= ( 27.272)
4.694 103.3 10
3
3
= 38.79
3.12 APPROXIMATE ANALYSIS OF COMMON-EMITTER AMPLIFIER (COLLECTOR TO BASE BIAS CONFIGURATION)
Figure 3.83 shows a common-emitter amplifi er with the collector to base bias circuit. The collector feedback circuit uses a feedback path from collector to base to increase the stability of the system. The resistor RB is connected between input and output.
RE
RS
RB
RC
CC1
Vo
+VCC
CC2
C
EB
RL
CEVS
Fig. 3.83Common-emi er Amplifi er
This circuit provides negative feedback which reduces the gain of the amplifi er. In the CE amplifi er, output is 180o out of phase with input. If output is fed back to input, the net gain decreases. This process is known as AC degeneration. If AC signal voltage increases, base current increases and collector current IC increases (Fig. 3.84). Hence, the net base current through the transistor decreases as If opposes Ib. Due to this the gain of the amplifi er decreases.
AC degeneration can be avoided by splitting RB into two parts and connecting a capacitor at the centre of both resistors (Fig. 3.85). For DC operation, there is no effect of the capacitor
3.12Approximate Analysis of Common-emi er Amplifi er3.57
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3.58Electronic Circuits I
(XC = ). For AC operation, as Ib increases, Ic increases. Hence, the component of base current through RB is bypassed by C. Hence, there is no change in Ib and there is no degeneration of input signal.
RS
RB
RC
CC1
Vo
+VCC
CC2
RL
If Ic
Ib
VS
Fig. 3.84AC Degeneration
RS
RC
CC1
+VCC
CC2RB2RB1
RL
C
VS
Fig. 3.85Method to Avoid AC Degeneration
The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.86.
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RB2 RC
Vo
RLRB1
RS
VS
Fig. 3.86AC Equivalent Circuit
The h-parameter equivalent circuit is obtained by replacing BJT by its h-parameter model as shown in Fig. 3.87.
Ic
hf eIb
Io
ViRS
+
Vo
+
IbIi
hieRB1 RB2 RC RL
B C
E
ZiZin Zi Zo Zo
VS
Fig. 3.87h-parameter Equivalent Circuit
Input impedance
Zi = hieZi = Zi || RB1Zin = Zi + RS
Output impedance
The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfe Ib = 0. The current source hfe Ib is replaced by an open circuit.
Zo = Zo = Zo || RB2 || RC || RL = RB2 || RC || RL
3.12Approximate Analysis of Common-emi er Amplifi er3.59
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3.60Electronic Circuits I
Voltage gain
RL = RC || RL
Vo = hfe Ib RLVi = Ib hie
Av = VV
o
i
= h I R
I hfe b L
b ie
= h R
hfe L
ie
The negative sign indicates a phase shift of 180 between input and output voltages.Considering voltage gain with source resistance RS,
Avs = Av Z
Z Ri
i S+
Current gain
For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Considering current gain with source resistance,
Ais = Ai R
R ZS
S i+
If a capacitor is not connected at the centre of the feedback circuit, Millers theorem is used for AC analysis. Resistor RB is splitted into two equivalent resistors at the input and output using Millers theorem.
Millers theorem
If an impedance Z is connected between the input and output of a circuit having voltage gain
Av = VV
2
1
, it can be replaced with two impedances Z1 and Z2 at the input and output terminals,
respectively as shown in Fig. 3.88.
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AmplifierAV
ZI1
I1 I2
I2
V1
+
V2
+
AmplifierAV
V1
+
V2
+
Z1 Z2
Fig. 3.88Millers Theorem
IV V
Z
VVV
Z
1
11 2
12
1=
=
=
V AZ
VZ
A
VZ
(1 )
(1 )
v
v
1 1 1
1
=
=
where ZZ
A1 v1 =
IV V
Z
VVV
Z
1
22 1
21
2=
=
=
VA
ZV
AA
Z
VZ
1 1
1
v
v
v
22 2
2
=
=
where ZA
AZ
1.v
v2 =
The resistor RB, which is connected between input and output, is splitted into resistors RB1 and RB2 at the input and output using Millers theorem.
The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.89.
RB2 RC
Vo
RLRB1
RS
VS
Fig. 3.89AC Equivalent Circuit
3.12Approximate Analysis of Common-emi er Amplifi er3.61
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3.62Electronic Circuits I
The h-parameter equivalent circuit is obtained by replacing BJT by its h-parameter model as shown in Fig. 3.90.
Ic
hf eIb
Io
ViRS
+
Vo
+
IbIi
hieRB1 RB2 RCRC RL
B C
E
ZiZin Zi Zo Zo
VS
Fig. 3.90h-parameter Equivalent Circuit
Voltage gain
RA
AR R
1B
v
vB B2
=
RL = RB2 || RC || RL = RB || RC || RLVo = hfe Ib RLVi = Ib hie
Av = VV
o
i
= h I R
I hfe b L
b ie
= h R
hfe L
ie
The negative sign indicates a phase shift of 180 between input and output voltages.Considering voltage gain with source resistance RS,
Avs = Av Z
Z Ri
i S+
Input impedance
RB1 = R
AB
v1
Zi = hieZi = Zi || RB1
Zin = Zi + RS
Output impedance
The output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0.
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When VS = 0, Ib = 0, hfeIb = 0. The current source hfeIb is replaced by an open circuit.
Zo =
Zo = Zo || RB2 || RC || RL= RB2 || RC || RL
Current gain
For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Considering current gain with source resistance,
Ais = Ai R
R ZS
S i+
Example 3.13: Calculate Zi, Zo, Av and Ai for the circuit shown in Fig. 3.91.
VS
3 k
CC1
Vo
CC2120 k 68 k
Zi Zo
+VCC
hfe = 140hie = 1.4 k
Fig. 3.91Example 3.13
Solution: AC equivalent circuit (Fig. 3.92)
3.12Approximate Analysis of Common-emi er Amplifi er3.63
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3.64Electronic Circuits I
VS
Vo
RB2RB1RC
Fig. 3.92AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.93)Ic
hf eIb
Io
Vi
+
Vo
+
IbIi
hieRB1 RB2 RCRC
B C
E
Zi Z i Zo Zo
VS
Fig. 3.93h-parameter Equivalent Circuit
R R R|| (68 10 ) || (3 10 ) 2.87 kL B C3 3
2 = = =
(i) Input impedance Zi = hie = 1.4 k
Zi = Zi || RB1= (1.4 103) || (120 103) = 1.38 k
(ii) Output impedance Zo =
Zo = RB2 || RC = (68 103) || (3 103) = 2.87 k
(iii) Voltage gain Av = h R
hfe L
ie
=
140 2.87 101.4 10
3
3
= 287
(iv) Current gain Ai = Av ZR
i
C
= ( 287)
1.38 103 10
3
3
= 132.02
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Example 3.14: Calculate Av, Zi, Zo and Ai for the circuit shown in Fig. 3.94.
VS
10 k
CC1
Vo
CC2200 k
Zi Zo
+VCC
hfe = 50hie = 1.1 k
Fig. 3.94Example 3.14
AC equivalent circuit (Fig. 3.95)
Ic
hf eIb
Io
Vi
+
Vo
+
IbIi
hieRB1 RB2 RCRC
B C
E
Zi Z i Zo Zo
VS
Fig. 3.95AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.96)
Ic
hf eIb
Io
Vi
+
Vo
+
IbIi
hieRB1 RB2 RCRC
B C
E
Zi Z i Zo Zo
VS
Fig. 3.96h-parameter Equivalent Circuit
RA
AR R
1B
v
vB B2
=
RL = RB2 || RC = (200 103) || (10 103) = 9.52 k
3.12Approximate Analysis of Common-emi er Amplifi er3.65
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3.66Electronic Circuits I
(i) Voltage gain Av = h R
hfe L
ie
= 50 9.52 101.1 10
3
3
= 432.73
(ii) Input impedance RB1 = R
A1B
v
= 200 101 ( 432.73)
3
= 461.12 Zi = hie = 1.1 k
Zi = Zi || RB1 = (1.1 103) || (461.12) = 324.92
(iii) Output impedance Zo = Zo = RB2 || RC = (200 10
3) || (10 103) = 9.52 k
(iv) Current gain Ai = Av ZR
i
C
= (432.73) 324.9210 103
= 14.06
Example 3.15: Calculate Av, Zi, Zo and Ai for the circuit shown in Fig. 3.97.
VS
10 k
CC1
Vo
CC2100 k
Zi Zo
+VCC
hfe = 100hie = 1 k
50
10 k
Fig. 3.97Example 3.15
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AC equivalent circuit (Fig. 3.98)
VS
Vo
RB2RB1 RE
RC RL
Fig. 3.98AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.99)
RB2
RE
RC
Ic
hfeIb
Io
RLRB1VS
Ii Ib
hie
Zi Z i Z o Zo
Vo
+
Vi
+
B C
E
Fig. 3.99h-parameter Equivalent Circuit
RA
AR R
1B
v
vB B2
=
RL = RB2 || RC || RL = (100 103) || (10 103) || (10 103) = 4.76 k
(i) Voltage gain Av =
+ +
h Rh h R(1 )
fe L
ie fe E
= 100 4.76 10
1 10 (1 100)(50)
3
3
+ +
= 78.68
(ii) Input impedance RB1 = R
A1B
v
= 100 10
1 ( 78.68)
3
= 1.26 k
3.12Approximate Analysis of Common-emi er Amplifi er3.67
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3.68Electronic Circuits I
Zi = hie + (1 + hfe) RE
= 1 103 + (1 + 100) (50)
= 6.05 k
Zi = Zi || RB1 = (6.05 103) || (1.26 103) = 1.04 k
(iii) Output impedance Zo =
Zo = RB2 || RC || RL = (100 103) || (10 103) || (10 103) = 4.76 k
(iv) Current gain Ai = Av ZR
i
L
= (78.68) 1.04 1010 10
3
3
= 8.18
3.13 APPROXIMATE ANALYSIS OF COMMON-BASE AMPLIFIERFigure 3.100 shows a common-base amplifi er.
RERS
VS
R1 RC
CC1Vo
+VCC
CC3
CC2RL
R2
Fig. 3.100Common-base Amplifi er
The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.101. Two resistors R1 and R2 are shorted to ground.
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3.13Approximate Analysis of Common-base Amplifi er3.69
RE VS
Vo
RC RL
Fig. 3.101AC Equivalent Circuit
Figure 3.102 can be redrawn by rotating BJT as shown in Fig. 3.101.
RS RCRE RL
VS
Vo
Fig. 3.102AC Equivalent Circuit
The h-parameter equivalent circuit is obtained by replacing the BJT by its h-parameter model (Fig. 3.103).
Ic
hf bIe
Io
ViRS
+
Vo
+
IeIi
hibRE RCRC RL
E C
B
ZiZin Zi Zo Zo
VS
Fig. 3.103h-parameter Equivalent Circuit
Input impedance
Zi = hibZi = Zi || RE
Zin = Zi + RSOutput impedanceThe output impedance Zo of the circuit is the impedance looking from the collector side when input voltage VS = 0. When VS = 0, Ib = 0, hfb Ie = 0. The current source hfb Ie is replaced by an open circuit.
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3.70Electronic Circuits I
Zo = Zo = Zo || RC || RL = RC || RL
Voltage gainRL = RC || RL Vo = hfb Ie RLVi = Ie hib
Av = VV
o
i
= h I R
I hfb e L
e ib
= h R
hfb L
ibConsidering source resistance,
Avs = Av Z
Z Ri
i S+Current gain For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Considering source resistance,
Ais = Ai R
R ZS
S i+
Example 3.16: Calculate Zi, Zo, Av and Ai for the circuit shown in Fig. 3.104.
Solution:
1 k
33 k 3.3 k
CC1Vo
+VCC
CC2
CC3
10 k
Zi Zo
hfe = 99hie = 2 k
VS
Fig. 3.104Example 3.16
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3.13Approximate Analysis of Common-base Amplifi er3.71
AC equivalent circuit (Fig. 3.105)
RE
Vo
RC
VS
Fig. 3.105AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.106)
hf bIe
Io = Ic
Vi
+
Vo
+
IeIi
hibRE RC
E C
B
Zi Z i Zo Zo
VS
Fig. 3.106h-parameter Equivalent Circuit
hib = h
h1ie
fe+ = 2 10
1 99
3
+= 0.02 k
hfb = +h
h1fe
fe
= 991 99+
= 0.99
(i) Input impedance Zi = hib = 0.02 k Zi = Zi || RE = (0.02 10
3) || (1 103) = 19.6
(ii) Output impedance Zo = Zo = RC = 3.3 k
(iii) Voltage gain Av = h R
hfb C
ib
= ( 0.99)(3.3 10 )0.02 10
3
3
= 163.35
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3.72Electronic Circuits I
(iv) Current gain Ai = Av ZR
i
C
= 163.35 19.63.3 103
= 0.97
Example 3.17: Calculate Zin, Zo, Avs and Ais for the circuit shown in Fig. 3.107.
2.7 k
100 k 3.9 k
1 k
CC1Vo
+VCC
CC2
CC3
33 k10 k
Zi Zo
hfe = 99hie = 2 k
VS
Fig. 3.107Example 3.17
Solution:AC equivalent circuit (Fig. 3.108)
RE
RS
Vo
RC RL
VS
Fig. 3.108AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.109)
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3.13Approximate Analysis of Common-base Amplifi er3.73
Ic
hf bIe
Io
ViRS
+
Vo
+
IeIi
hibRE RCRC RL
E C
B
ZiZin Zi Zo Zo
VS
Fig. 3.109h-parameter Equivalent Circuit
hib = h
h1ie
fe+= 2 10
99 1
3
+= 0.02 k
hfb = h
h1fe
fe
+
991 99+
= 0.99
R R R|| (3.9 10 ) || (10 10 ) 2.81 kL C L3 3
= = =
(i) Input impedance Zi = hib = 0.02 k
Zi = Zi || RE = (0.02 103) || (2.7 103) = 19.85
Zin = Zi + RS = 19.85 + 1 103 = 1.02 k
(ii) Output impedance Zo =
Zo = RC || RL = (10 103) || (3.9 103) = 2.81 k
(iii) Voltage gain Av = h R
hfb L
ib
= ( 0.99) 2.81 100.02 10
3
3
= 139.1
Avs = Av Z
Z Ri
i S+
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3.74Electronic Circuits I
= 139.1 19.8519.85 1 103+
= 2.71
(iv) Current gain Ai = Av ZR
i
L
= 139.1 19.8510 103
= 0.28
Ais = Ai R
R ZS
S i+
= 0.28 1 101 10 19.85
3
3
+= 0.27
3.14 APPROXIMATE ANALYSIS OF COMMON-COLLECTOR AMPLIFIER
Figure 3.110 shows a common-collector amplifi er.
RE RL
RS
R1
CC1
+VCC
Vo
CC2
R2
VS
Fig. 3.110Common-collector Amplifi er
The AC equivalent circuit is obtained by replacing all capacitors by short circuits and replacing DC source by ground as shown in Fig. 3.111.
hic = hiehfc = (1 + hfe) hfe
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3.14Approximate Analysis of Common-collector Amplifi er3.75
RE RL
RSVo
R1 R2VS
Fig. 3.111AC equivalent Circuit
Hence, the h-parameter model of the CE confi guration is used for the AC analysis of the CC amplifi er (Fig. 3.112).
RE RL
Ic
hfeIb
Io
RS
RB
VS
Ii Ib
Ie
hie
Zi Z iZin Z o Zo
Vo
Vi
+
B C
E
Fig. 3.112h-parameter Equivalent Circuit
RB = R1 || R2 RL = RE || RLInput impedance
Vi = Ib hie + Ie RL = Ib hie + (1 + hfe) Ib RL = [hie + (1 + hfe) RL]Ib
Zi = VI
i
b
= hie + (1 + hfe) RL Zi = Zi || RB = [hie + (1 + hfe) RL] || RB Zin = Zi + RS
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3.76Electronic Circuits I
Output impedance The base circuit can be converted into Thevenins equivalent circuit (Fig. 3.113).
VTh = V = R
R RB
S B+VS
RTh = (RB || RS) + hie
Io
Ib
Ie
Z o
Vo
V+
B E(RB||RS) + hie
Fig. 3.113Thevenin's Equivalent Circuit
The output impedance Zo of the circuit is the impedance looking from the emitter side when input voltage V = 0.
Zo = VI
oo V 0=
Applying Kirchhoffs voltage law to the base-emitter circuit,
V Ib [(RB || RS)+ hie] Vo = 0
When V = 0, Vo = Ib [(RB || RS) + hie] Io = Ie = (1 + hfe) Ib
VI
o
o
= R R h
h( || )
(1 )B S ie
fe
+
+
Zo = R R h
h( || )
(1 )B S ie
fe
+
+
When base resistance [(RB || RS) + hie] is transferred to the emitter side, it gets divided by
(1 + hfe). Hence, output impedance looking from the emitter side, i.e. Zo is R R h
h( || )
1B S ie
fe
+
+
.
Zo = Zo || RL
= R R h
h( || )
1B S ie
fe
+
+
|| RL
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3.14Approximate Analysis of Common-collector Amplifi er3.77
When RS is not present, i.e. RS = 0,
Zo = h
h1ie
fe+
|| RL
Generally output impedance of the common-collector confi guration is very low. Hence, it is used as buffer for impedance matching.
Voltage gain
Vo = Ie RL
= (1 + hfe) Ib RL Vi = [hie + (1 + hfe) RL ] Ib
Av = VV
o
i
= h I R
h R h I(1 )
[ (1 )]fe b L
ie L fe b
+
+ +
= h R
h h R(1 )
(1 )fe L
ie fe L
+
+ +
If (1 + hfe) RL >> hieAv 1
Thus, Vo Vi and output at the emitter follows the input at the base. Hence, the common-collector amplifi er is called as emitter follower.
Current gain
For transistor, the voltage gain is the most important gain. Hence, the current gain can be determined directly from the voltage gain, the defi ned load and the input impedance.
Ai = II
o
i
=
VR
VZ
o
L
i
i
= VV
o
i
ZR
i
L
= Av ZR
i
L
Example 3.18: Determine Zi, Zo, Av and Ai for the emi er follower circuit shown in Fig. 3.114.
Solution:AC equivalent circuit (Fig. 3.115)
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3.78Electronic Circuits I
3.3 k
220 k
CC1
Vo
Vi
+VCC
CC2
ZoZi
hfe = 98hie = 1.275 k
Fig. 3.114Example 3.18
RBRE
Vo
Vi
Fig. 3.115AC Equivalent Circuith-parameter equivalent circuit (Fig. 3.116)
RE
Ic
hfeIb
IoRB
Ii Ib
Ie
hie
Zi Z i Zo
Vi
+
Vo
+
B C
E
Fig. 3.116h-parameter Equivalent Circuit
(i) Input impedance Zi = hie + (1 + hfe) RE= 1.275 103 + (1 + 98) (3.3 103)= 327.98 k
Zi = Zi || RB = (327.98 103) || (220 103) = 131.67 k
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3.14Approximate Analysis of Common-collector Amplifi er3.79
(ii) Output impedance Zo = h
h1ie
fe+
|| RE
= 1.275 101 98
3
+
|| (3.3 10
3)
= 12.83
(iii) Voltage gain Av = h R
h h R(1 )
(1 )fe E
ie fe E
+
+ +
= (1 98)(3.3 10 )
1.275 10 (1 98)(3.3 10 )
3
3 3
+
+ +
= 0.996
(iv) Current gain Ai = Av ZR
i
E
= 0.996 131.67 103.3 10
3
3
= 39.74
Example 3.19: Find Zin, Zo, Av and Ai for the circuit shown in Fig. 3.117.
20 k
91 k
10 k
100 k
CC1
Vo
+VCC
ZoZin
hfe = 50hie = 1.1 k
VS
Fig. 3.117Example 3.19
Solution: AC equivalent circuit (Fig. 3.118)
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3.80Electronic Circuits I
R2RE
Vo
R1
RS
VS
Fig. 3.118AC Equivalent Circuit
h-parameter equivalent circuit (Fig. 3.119)
RE
Ic
hfeIb
Io
RS
RB
VS
Ii Ib
Ie
hie
Zi Z iZin Zo
Vi
+
B C
E
Vo
+
Fig. 3.119h-parameter Equivalent Circuit
RB = R1 || R2 = (91 103) || (100 103) = 47.64 k
(i) Input impedance Zi = hie + (1 + hfe) RE= 1.1 103 + (1 + 50) (20 103)= 1021.1 k
Zi = Zi || RB = (1021.1 103) || (47.64 103) = 45.52 k
Zin = Zi + RS = 45.52 103 + 10 103 = 55.52 k
(ii) Output impedance Zo = R R h
h( || )
1B S ie
fe
+
+
|| RE
= {(47.64 10 ) || (10 10 )} 1.1 10
1 50
3 3 3 +
+
|| (20 10
3)
= 181.954
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3.14Approximate Analysis of Common-collector Amplifi er3.81
(iii) Voltage gain Av = h R
h h R(1 )
(1 )fe E
ie fe E
+
+ +
= (1 50)(20 10 )
1.1 10 (1 50)(20 10 )
3
3 3
+
+ +
= 0.998
(iv) Current gain Ai = Av ZR
i
E
= 0.998 45.52 1020 10
3
3
= 2.27
Example 3.20: Find Zi, Zo and Av for the circuit shown in Fig. 3.120.
5 k 2 k
1 k
50 k
CC1
+VCC
Vo
CC2
25 k
Zi Zo
hfe = 100hie = 2 k
VS
Fig. 3.120Example 3.20
Solution:AC equivalent circuit (Fig. 3.121)
R2RE
Vo
RL
R1
RS
VS
Fig. 3.121AC Equivalent Circuit
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3.82Electronic Circuits I
h-parameter equivalent circuit (Fig. 3.122)
RE RL
Ic
hfeIb
Io
RS
RB
VS
Ii Ib
Ie
hie
Zi Z i Zo
Vo
Vi
+
B C
E
Fig. 3.122h-parameter Equivalent Circuit
RB = R1 || R2 = (50 103) || (25 103) = 16.67 k
RL = RE || RL = (5 103) || (2 103) = 1.43 k
(i) Input impedance Zi = hie + (1 + hfe) RL
= 2 103 + (1 + 100) (1.43 103)
= 146.28 k
Zi = Zi || RB = (146.28 103) || (16.67 103) = 14.96 k
(ii) Output impedance Zo = R R h
h( || )
1B S ie
fe
+
+
|| RE
= {(16.67 10 ) || (1 10 )} 2 10
1 100
3 3 3 +
+
|| (5 103)
= 28.97
(iii) Voltage gain Av = h R
h h R(1 )
(1 )fe L
ie fe L
+
+ +
= (1 100)(1.43 10 )
2 10 (1 100)(1.43 10 )
3
3 3
+
+ +
= 0.9863
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3.15Comparison of CE, CB and CC Amplifi ers3.83
3.15COMPARISON OF CE, CB AND CC AMPLIFIERS
Parameter CE CB CC
(1) Voltage gain (Av) Very high Very high Less than 1
(2) Current gain (Ai) Very high Less than 1 High
(3) Input impedance (Zi) Medium Low High
(4) Output impedance (Zo) High High Very low
(5) Phase of output signal Out of phase with input In phase with input In phase with input
(6) Application Voltage amplifi er Non-inverting amplifi er Bu er between high impedance source and low impedance load
SUMMARY OF BJT AMPLIFIERS
Confi guration Circuit Equation
CE amplifi er with bypassed RE
VS
RS
R1 RC
CC1Vo
+VCC
CC2
RL
RE CE
R2
Zi Z i Z o Zo
Zi = hieZi = Zi R1 R2
Zin = Zi + RSZo =
Zo = Zo RC RL
Av = h R R
h( || )fe C L
ie
Avs = Av Z
Z Ri
i S+
Ai = Av ZR
i
L
Ais = Ai R
R ZS
S i+
(Continued)
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3.84Electronic Circuits I
Confi guration Circuit Equation
CE amplifi er with unbypassed RE
VS
RS
R1 RC
CC1Vo
+VCC
CC2
RL
RE
R2
Zi Z i Z o Zo
Zi = hie + (1 + hfe) RE
Zi = Zi R1 R2
Zin = Zi + RSZo =
Zo = Zo RC RL
Av = h R R
h h R( || )(1 )
fe C L
ie fe E+ +
Avs = Av Z
Z Ri
i S+
Ai = Av ZR
i
L
Ais = Ai R
R ZS
S i+
CB amplifi er
VS
R1 RC
CC1Vo
+VCC
CC3
CC2RL
RERS
R2
Zi Z i Z o Zo
Zi = hib
Zi = Zi REZin = Zi + RS
Zo =
Zo = Zo RC RL
Av = h R R
h( || )fb C L
ib
Avs = Av Z
Z Ri
i S+
Ai = Av ZR
i
L
Ais = Ai R
R ZS
S i+
SUMMARY OF BJT AMPLIFIERS (Continued)
(Continued)
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Confi guration Circuit Equation
CC amplifi er
VS
RS
R1
CC1
Vo
+VCC
CC2
RLRE
R2
Zi Z i Z o
Zi = hie + (1 + hfe) RE
Zi = Zi R1 R2
Zin = Zi + RS
Zo = RE R R h
h[( || ) ]
1S B ie
fe
+
+
Zo = Zo RL
Av = h R R
h h R R(1 )( || )
(1 )( || )fe E L
ie fe E L
+
+ +
Avs = Av Z
Z Ri
i S+
Ai = Av ZR
i
L
Ais = Ai R
R ZS
S i+
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