ELEC 356 Electronics II Amplifier Frequency Response
BJT Amplifier Frequency Response
Bandwidth of an Amplifier
Most amplifiers have relatively constant gain over a certain range (band) of frequencies, this is called
the bandwidth (BW) of the amplifier.
Fig. 1 Typical frequency response of an amplifier
As the frequency response curve shows, the gain of an amplifier remains relatively constant across a
band of frequencies.
When the operating frequency starts to go outside this frequency range, the gain begins to drop off.
Two frequencies of interest, fC1 and fC2, are identified as the lower and upper cutoff frequencies.
The Bandwidth is found as:
BW = fC2 – fC1
The operating frequency of an amplifier is equal to the geometric center frequency fo,
fo = √(fC1 fC2 )
Notice that the ration of fo to fC1 equals the ratio of fC2 to fo , this is:
fo / fC1 = fC2 / fo
Therefore we also have that:
fC1 = fo2 / fC2
;
fC2 = fo2 / fC1
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ELEC 356 Electronics II Amplifier Frequency Response
Fig. 2 Typical frequency response of an amplifier in dB
Bode Plots
The Bode Plot is a variation of the basic frequency response curve.
A Bode plot assumes the amplitude (transfer function) is zero until the cutoff frequency is reached.
Then the gain of the amplifier is assumed to drop at a set rate of 20 dB/decade (this is true for first
order transfer functions (or one RC time constant).
Fig. 3 Representation of a Frequency Response using Bode Plots
Bode plots are typically used because they are easier to read.
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ELEC 356 Electronics II Amplifier Frequency Response
Cutoff frequencies of an RC-coupled amplifier.
Fig. 4 The Base Circuit of a typical amplifier
The lower cutoff frequency of the Base Circuit is:
f1B = 1 / (2π RC)
where :
R = Rs + Rin ; Rin = R1 ║R2 ║hie
C = value of the Base coupling capacitor, CC1
The Collector Circuit of the BJT amplifier works as the same principle as the Base Circuit.
Refer to Fig.5. The cutoff frequency of the Collector Circuit is:
f1C = 1 / [2π (RC + RL ) C]
RC + RL = sum of the resistance in the collector circuit
C = value of the Base coupling capacitor, CC2
For the Emitter Circuit cutoff frequency, we need to refer to the following relationships derived in
previous chapters,
Rout = RE ║ ( re + R’in / hfe );
re = V T / I E
R’in = R1 ║ R2 ║ RS
In most cases RE >> ( re + R’in / hfe ) and we can approximate
Rout = ( re + R’in / hfe )
f1E = 1 / (2π Rout CE )
Refer to Fig. 5.
The lower cutoff frequency of a given common emitter amplifier will be given by the highest of the
individual transistor terminal circuits, this is,
fC1 = MAX(f1B, f1C. f1E )
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ELEC 356 Electronics II Amplifier Frequency Response
Example.
Determine the value of the lower cutoff frequency for the following amplifier. Consider the following
component values:
RS = 600 Ω, R1 = 18 kΩ, R2 = 4.7 kΩ, RC = 1.5 kΩ, RE = 1.2 kΩ, RL = 5 kΩ,
CC1 = 1 μF, CC2 = 0.22 μF, CE = 10 μF, hfe = 200, hie = 4.4 kΩ, Vcc = 10 V
Fig. 5 Typical Common Emitter Amplifier
Ans.
f1B = 61.2 Hz
f1C = 111 Hz
f1E = 650 Hz
Lower cutoff frequency of the amplifier = fC1 = max (61.2, 111, 650) = 650 Hz
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