7/30/2019 6 Chptr 1 - Freq Response of BJT Amplifier(Part I)
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ANALOGUEELECTRONICS
Frequency Response of BJT Amplifiers (Part 1)
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The Decibel (dB)
A logarithmic measurement of the ration of power or voltage
Power gain is expressed in dB by the formula:
where a p is the actual power gain, Pout/Pin
Voltage gain is expressed by:
If av is greater than 1, the dB is +ve, and if av is less than 1, thedB gain is –ve value & usually called attenuation
vdBV a A log20)(
P P a A log10
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Amplifier gain vs frequency
Midband range
Gain falls of due to theeffects of CC and CE
Gain falls of due to the effectsof stray capacitance and
transistor capacitance effects
L H BW f f f
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Definition Frequency response of an amplifier is the graph of its gain versus
the frequency .
Cutoff frequencies : the frequencies at which the voltage gain
equals 0.707 of its maximum value.
Midband : the band of frequencies between 10f L
and 0.1f H
where
the voltage gain is maximum. The region where coupling &
bypass capacitors act as short circuits and the stray capacitance
and transistor capacitance effects act as open circuits.
Bandwidth : the band between upper and lower cutoff frequencies
Outside the midband, the voltage gain can be determined by
these equations:
21/1 f f
A A mid
22/1 f f
A A mid
Below midband Above midband
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Lower & Upper Critical
frequency
Critical frequency a.k.a the cutoff frequency
The frequency at which output power drops by 3
dB. [in real number, 0.5 of it’s midrange value.
An output voltage drop of 3dB represents about a0.707 drop from the midrange value in real
number.
Power is often measured in units of dBm. This is
decibels with reference to 1mW of power. [0 dBm= 1mW], where;
.dBm0mW1
mW1log10
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Gain & frequencies
Gain-bandwidth product : constant value
of the product of the voltage gain and
the bandwidth.
Unity-gain frequency : the frequency at
which the amplifier’s gain is 1
BW A f mid T
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LOW FREQUENCY
At low frequency range, the gain falloff due to
coupling capacitors and bypass capacitors.
As signal frequency , the reactance of thecoupling capacitor, XC - no longer behave
as short circuits.
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Short-circuit time-constant method
(SCTC)
To determine the lower-cutoff frequency having n
coupling and bypass capacitors:
n
i iiS
LC R1
1
RiS = resistance at the terminals of the ith capacitor C i with all
the other capacitors replaced by short circuits.
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Common-emitter Amplifier
30 k
V CC = 12V
10 k
RS C 1
2 F
C 2
C 3
10 F
0.1 F
1 k
1.3 k
4.3 k
R1
RC
R E R2 vS
vO
R L
100 k
Given :
Q-point values :1.73 mA, 2.32 V
= 100, VA = 75 V
Therefore,
r = 1.45 k,
r o =44.7 k
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Common-emitter Amplifier - Low-frequency ac equivalent circuit
RS
R B
RC R L
R E
C 1
C 2
C 3
v s
vo
In the above circuit, there are 3 capacitors (coupling plus bypass
capacitors). Hence we need to find 3 resistances at the terminals of the
3 capacitors in order to find the lower cut-off frequency of the amplifier
circuit.
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Circuit for finding R1S
22201450750010001 r R R R R R R BS inCE BS S
7500 where 21 R R R B
srad
F k C R S
/22500.222.2
11
11
RS
R B
R1S
RC R L
RinCE
Replacing C2 and C3 by
short circuits
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Circuit for finding R2S
k k k k r R R R R R R oC LoutCE C LS 1047.443.41002
srad
F k C R S
/1.96100.0104
11
22
R L
RS R B
R2S
RC
RoutCE
Replacing C1 and C3 by
short circuits
k r 7.44 where 0
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Circuit for finding R3S
7.22101
8821450
130013
TH
E outCC E S
Rr
R R R R
srad F C R
S
/4410107.22
11
33
Replacing C1 and C2 by
short circuits
RS R B
R3S R E
RC ||R L
RoutCC
RTH
882 BS TH R R R
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Estimation of L
srad C Ri iiS
L /473044101.9622513
1
Hz f L L 753
2
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Common-base Amplifier
Given :
Q-point values : 0.1
mA, 5 V
= 100, VA = 70 V
Therefore,
gm = 3.85 mS, r o = 700 k
r = 26
R E
RS
RC R L
C 1 C 2
vO
vS 22 k
100
43 k 75 k
1 F 4.7 F
+V CC -V EE
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Common-base Amplifier - Low-frequency ac equivalent circuit
R E
RS
RC R L
C 1C 2
vo
v s
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Circuit for finding R1S
Replacing C2 by shortcircuit
10026.04300100
11
r R R R R R R E S inCB E S S
srad
F C R S
/1013.27.4100
11 3
11
RC || R L
R1S
R E
RS
RinCB
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Circuit for finding R2S
k k k R R R R R R C LoutCBC LS 9722752
srad
F k C R S
/309.10197
11
22
Replacing C1 by short circuit
RS || R E
R2S
RC R L
RoutCB
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Estimation of L
srad C Ri iiS
L /309.10309.101013.2
1 32
1
Hz f L L 64.1
2
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Common-collector Amplifier
R B
RS
R E
R L
C 1
C 2
vS vO
-V EE
+V CC
100 k
1 k
3 k 47 k
100 F
0.1 F
Given :
Q-point values : 1 mA, 5 V
= 100, VA = 70 V
Therefore,
r = 2.6 k, r o =70 k
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Common-collector Amplifier - Low-frequency ac equivalent circuit
R B
RS
R E R L
C 2
C 1
vov s
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Circuit for finding R1S
Replacing C2 by short circuit
L E o BS inCC BS S R Rr r R R R R R R 11
srad
F k C R S
/18.1361.043.74
11
11
k 43.74
R B
RS
R E || R L
R1S RinCC
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Circuit for finding R2S
o
TH E LoutCC E LS r
r R R R R R R R
12
srad
F k C R S
/213.0100038.47
11
22
Replacing C1 by short circuit
k 038.47
R E R L
RTH = RS || R B
R2S
RoutCC
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Estimation of L
srad
C Ri iiS
L /393.136213.018.13612
1
Hz f L L 7.21
2
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Example
v S
62 k
V CC = 10V
22 k
RS C 1
0.1 F
C 2
C 3 10 F
0.1 F
600
1.0 k
2.2 k
R1 RC
R E R2
vO
R L
10 k
Given :
Q-point values : 1.6
mA, 4.86 V
= 100, VA = 70 V
Therefore,
r = 1.62 k, r o = 43.75 k, gm
= 61.54 mS
Determine the total low-
frequency response of the
amplifier.
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Low frequency due to C1 and
C2 C3
k k k r R R R BS S 07.262.124.166001
k R R R B 24.1621
Hz Hz
F k C R f
S
C 76986.7681.007.22
1
2
1
111
k k k k r R R R oC LS 09.1275.432.2102
Hz Hz
F k C R f
S
C 13264.1311.009.122
1
2
1
222
Low frequency due to C1
Low frequency due to C2
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32.21101
58.062.1
113
k k
k
Rr
R R
TH
E S
Hz Hz F C R f S
C 7475.7461032.212
1
2
1
333
k R R R BS TH 58.0
Low frequency due to C3
Low frequency due to C3