ELECTRONICS LABORATORY
EXPERIMENT 6
WIDE-BAND AMPLIFIERS
6.1 Objective
In this experiment, the frequency response of wide band amplifiers will be studied, and the
cutoff frequencies will be determined.
6.2 Introduction
The voltage gain of the amplifiers that are used to amplify audio and video signals depends on
frequency. The frequency response of such a “wide-band” amplifier (Figure 6.1) has three
ranges: mid frequency, low frequency and high frequency. The gain in low and high frequency
ranges is lower than that of mid frequency range. The internal capacitances of the active devices
and the parasitic inductances lower the gain at high frequencies. The coupling capacitors and
bypass capacitors cause the gain decrease as the frequency decreases.
The mid-frequency range is determined by the 3-dB drop in the gain. At the low cut-off
frequency, the voltage gain is 3-dB lower than the mid-band gain. Likewise, high cut-off
frequency marks the 3-dB drop in the mid-band gain at the high frequency side.
|Gain|
3 dB
———————————————
frequency
low cut-off frequency
high cut-off frequency
Figure 6.1 Wide-band amplifier frequency response
To determine the high cut-off frequency, the high frequency small signal model of the
active device, BJT or FET (Figure 6.2), is substituted in the high frequency equivalent circuit of
the amplifier. The high cut-off frequency is determined analytically through the analysis of the
circuit.
(a)
(b)
Figure 6.2 High frequency small signal models: a) BJT b) FET
— 6.1 —
The low cut-off frequency of the amplifier, too, may be determined through the analysis
of the amplifier equivalent circuit: The low frequency model of the active devices is substituted
in the amplifier circuit, and the external capacitors are kept.
The frequency domain analysis of the amplifier circuit may be complicated to determine
the cut-off frequencies. The so-called dominant pole technique may be used to find the pole
frequencies due to capacitors. With this technique, only one capacitor is taken at a time. The
other capacitor are assumed to be short circuit or open circuit at the pole frequency of the
capacitor under consideration. Then the Thevénin resistance seen by this capacitor is calculated.
The pole frequency of the capacitor is given by the RC time constant. The pole frequencies of
all capacitors are calculated similarly.
In the low frequency equivalent circuit, the capacitor with the largest pole frequency is
called the dominant capacitor. In this circuit, while the pole frequency of a capacitor is
calculated, it is assumed to be dominant, and the other capacitors are taken as short circuits
because, at the dominant pole frequency, the impedances of the non-dominant capacitors are
negligibly small. The largest of all low-frequency poles will be the true dominant pole
frequency; the non-dominant pole frequencies may have to be recalculated because the
dominant capacitor will not be a short circuit at the non-dominant pole frequencies. The low 3dB cut-off frequency of the amplifier can be estimated with the superposition of poles.
For the high frequency circuit, the dominant capacitor will have the lowest pole
frequency. While the Thevénin resistance seen by each capacitor is calculated, the other
capacitors are taken as open circuit, because they will be assumed to be non-dominant
capacitors. Once the actual dominant pole is determined, the non-dominant pole frequencies
may have to be recalculated with the dominant capacitor short circuit. In the end, the parallel
superposition of high frequency poles is used to estimate the high 3-dB cut-off frequency.
6.3 Common Emitter Amplifier
A common emitter amplifier and its mid, low and high frequency equivalent circuits are given in
Figure 6.3. The voltage gain vout/vs for mid frequencies can be found easily:
R p & (rx + rπ )
vout
v
(6.1)
= out
vs
vin Rs + R p & (rx + rπ )
where,
vout
rπ
= – gm(ro||RC||RL)
(6.2)
vin
rx + rπ
RP = R1||R2
(6.3)
Let’s use Figure 6.3c to find the low cut-off frequency of the amplifier. For the
simplicity of the calculations, we can ignore ro compared to RC and RL. The Thevénin resistance
RTH1 which is seen by C1 is not affected by C2. However, CE affects RTH1.
When CE is open circuit:
RTH1 = Rs + RP||[(rx + rπ +( βo + 1) RE]
(6.4)
When CE short circuit:
RTH1 = Rs + RP||(rx + rπ )
— 6.2 —
(6.5)
(a)
(b)
(c)
(d)
Figure 6.3 a) Common emitter amplifier circuit, b) Mid-frequency equivalent circuit, c) Lowfrequency equivalent circuit, d) High-frequency equivalent circuit
— 6.3 —
Similarly, RTHE, the Thevénin resistance seen by CE is not affected by C2. When C1 is
open circuit
RTHE = RE ||
R p + rx + rπ
(6.6)
βo + 1
When C1 is short circuit:
RTHE = RE ||
Rs & R p + rx + rπ
(6.7)
βo + 1
The Thevénin resistance seen by C2 is found to be
RTH2 = RC + RL
(6.8)
The pole frequency of each capacitor is found and compared with the other pole frequencies.
Calculations may be repeated after the dominant capacitor is determined. The sum of all lowfrequency poles will approximate the low 3-dB cut-off frequency.
The high cut-off frequency of the amplifier may be found with the help of the equivalent
circuit in Figure 6.3d. In this circuit, the capacitor Cµ is eliminated with Miller’s theorem. The
new equivalent circuit is given in Figure 6.4. The capacitors that are connected parallel are
taken as one capacitor. The Thevénin resistances,
RTHA = rπ||(rx + Rs||RP)
(6.9)
RTHB = ro||RC||RL
(6.10)
and the high cut-off frequency
fH ≈
1
2π [(C A + Cπ ) RTHA + CB RTHB ]
CA ≈ Cµ [1 + gm(ro||RC||RL)];
CB ≈ Cµ [1 +
(6.11)
1
] ≈ Cµ
g m (ro & RC & RL )
Figure 6.4 The application of the Miller’s Theorem to eliminate Cµ in the common emitter
amplifier high frequency equivalent circuit.
— 6.4 —
6.4 Common Base Amplifier
A common base amplifier and its equivalent circuits at mid, low and high frequency ranges are
given in Figure 6.5. Again, for the sake of simplicity, rx and ro are ignored compared to rπ and
RC||RL , respectively. The mid frequency voltage gain,
vout
v
Rin
= out
vs
vin Rs + Rin
(6.12)
vout
= gm(RC||RL)
vin
(6.13)
where
and
Rin = RE ||
rπ
βo + 1
(6.14)
(a)
(b)
(c)
(d)
Figure 6.5 a) A common base amplifier circuit, and it equivalent circuits at b) mid frequencies,
c) low frequencies, and d) high frequencies.
Let’s use Figure 6.5c to find the low cut-off frequency. The Thevénin resistance seen by
C1 is not affected by C2:
— 6.5 —
When CE is open circuit:
RTH1 = RP||[ rπ +(βo + 1) RE]
(6.15)
RTH1 = RP||[ rπ +( βo + 1) RE||Rs ]
(6.16)
When CE is short circuit:
The Thevénin resistance seen by CE is not affected by C2 . Hence, when C1 is open
circuit:
RTHE = Rs + RE ||
R p + rπ
βo + 1
(6.17)
When C1 is short circuit:
RTHE = Rs + RE ||
rπ
βo + 1
(6.18)
Thevénin resistance seen by C2 ,
RTH2 = RC + RL
(6.19)
The pole frequency of each capacitor is calculated and compared with other pole frequencies.
Once the dominant pole is determined, the non-dominant pole frequencies may be recalculated.
The low cut-off frequency is estimated by adding all pole frequencies.
The high cut-off frequency of the amplifier is found from the equivalent circuit in Figure
6.5d. The application of the Miller’s theorem to this circuit brings lengthy calculations. Instead,
the Thevénin resistances seen by Cµ and Cπ are calculated directly:
RTHπ = RE||Rs||
rπ
βo + 1
RTHµ = RC||RL
(6.20)
(6.21)
Since the base of the transistor is connected to the ground, the Thevénin resistance seen by each
capacitor is not affected by the other capacitor. The high cut-off frequency of the amplifier can
be estimated to be
fH ≈
1
2π (Cπ RTH π + Cμ RTH μ )
— 6.6 —
(6.22)
6.5 Preliminary Work
1) Design a wide-band amplifier with a gain of 15±2 dB. Use external coupling and bypass
capacitors in the range 0.1 – 1.0 µF. For the configuration, use the sum of the digits in the
student ID numbers in the group. If the sum is an even number, use common emitter
configuration, otherwise, use common base configuration. (Example: Let the student id
numbers
be
151219962015
and
151219972059.
Sum
=
1+5+1+2+1+9+9+6+2+0+1+5+1+5+1+2+ ... +5+9= 93. Since 93 is an odd number, the group
will use the common base configuration.)
2) Calculate the low and high 3-dB cut-off frequencies for the circuit you designed.
3) Obtain the frequency response of the amplifier using PSPICE or another simulation
program..
4) Reduce the load resistance into its half and determine the high cut-off frequency.
5) Double the values of the capacitors used in the circuit, and determine the low cut-off
frequency of the amplifier.
6.6 Experimental Work
1) Set up the circuit designed in the preliminary work and plot its frequency response. Keep
the amplitude of the input signal constant at a value that does not cause any distortion in the
output signal. Plot the gain versus frequency curve on a semilogarithmic paper. You may use
frequency values such as 1, 2, 5, 10, 50, 100, .... Find the low and high cut-off frequencies at
which the amplitude of the gain is 70.7% of its mid frequency value. Compare them with the
simulation results.
2) Reduce the load resistance by connecting an identical resistor in parallel. Plot the gain
versus frequency curve on a semilogarithmic paper. Find the low and high cut-off frequencies.
Compare them with the simulation results. Increase the load resistor to its original value.
3) Increase the values of the external capacitors by connecting identical capacitors in parallel.
Plot the gain versus frequency curve on a semilogarithmic paper. Find the low and high cut-off
frequencies. Compare them with the simulation results.
4) Explain possible sources of disagreements among your design values, PSPICE analysis
results and the experimental results if there are any.
— 6.7 —