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11.5.4
capacitor as a signal coupling Element
Another application context involving the Series RC Circuit is the ‘signal coupling problem’ in
electronic amplifiers. We abstract the problem as follows.
At a certain input point in the electronic amplifier a certain value of DC voltage has to be established
using a DC source and resistors. This DC voltage is needed to fix the operating point of transistors in
the amplifier at suitable levels. Yet, we want to connect an AC signal source to that input point without
upsetting the DC potential there. Interposing a suitably sized capacitor between the signal source and
input point achieves this objective. The capacitor used for this purpose is called coupling capacitor
and amplifiers employing this form of signal coupling are called RC-Coupled Amplifiers in the study
of Electronic Circuits. Notice that the RC circuit is used as a high-pass filter (or average absorber) in
this application.
This application is explained further in the following example.
11.24
First-Order
RC
Circuits
Example: 11.5-1
Consider the signal-coupling problem presented in Fig. 11.5-7. The signal v
S
(t) is a mixture of
sinusoids and may contain sinusoids of frequency from 20 Hz to 20 kHz (the standard audio range).
It is desired that 95% of signal amplitude appear at the point A for the entire frequency range without
affecting the DC content of voltage at that point. (i) Calculate the value of C needed for this purpose.
(ii) Assume v
S
(t) is a 500 mV amplitude 20 Hz sine wave and plot the potential at A and the voltage
across the coupling capacitor with the value of C calculated.
A
(a)
10 k
Ω
100 k
Ω
2 k
Ω
12 V
+
+
–
–
A
(b)
10 k
Ω
100 k
Ω
2 k
Ω
V
S
(
t
)
12 V
+
–
+
+
–
–
A
(c)
10 k
Ω
100 k
Ω
2 k
Ω
V
S
(
t
)
V
C
(
t
)
12 V
+
–
Fig. 11.5-7
Circuitsfor
RC
signal-couplingexample
Solution
Consider the circuit in Fig. 11.5-7 (a). The 100k
W
resistor produces only negligible loading at the
point A and hence the current drawn by it may be ignored. Then the DC voltage at that point is 12 V
×
(2/10
+
2)
=
2 V with respect to the negative of DC source.
Now if the AC signal source is connected straight at point A as in Fig. 11.5-7 (b) the DC voltage at
that point gets affected. In fact it goes to zero if v
S
(t) has no DC content. This is a two-source problem
and can be solved by applying superposition principle. When the DC source alone is considered, the AC
source has to be shorted (assuming it is an ideal independent voltage source). This short will reduce the
DC potential at point A to zero once the signal source is connected. Hence direct coupling will not work.
The circuit Fig. 11.5-7 (c) is also excited by two sources. We know that the steady-state responses
due to many sources obey superposition principle in a linear time-invariant circuit. The two equivalent
circuits needed for calculating the steady-state response are shown in Fig. 11.5-8.
+
–
A
(a)
10 k
Ω
100 k
Ω
2 k
Ω
C
V
C
(
t
)
12 V
+
–
+
+
–
–
100 k
Ω
C
A
(b)
10 k
Ω
2 k
Ω
V
S
(
t
)
V
C
(
t
)
Fig. 11.5-8
CircuitsforapplyingsuperpositioninExample:11.5-1
The DC component of a steady-state response at point A is 2 V (neglecting the effect of 100k
W
resistor) since a capacitor behaves as an open-circuit under DC steady-state, and the DC content in
v
C
(t) is also 2 V.
FrequencyResponseofFirstOrder
RC
Circuits
11.25
As far as steady-state component of AC signal is concerned, we want 95% of input amplitude to
appear at point A even at 20 Hz. The parallel combination of the three resistors is 1.64 k
W
. The signal
voltage gets divided between the phasor impedance of capacitor, 1/(j2
p
×
20C), and 1.64 k
W
.
Percentage of signal amplitude reaching point A at 20Hz
=
16440
100
1640
1
j40 C
+
×
p
%
We desire this to be
≥
95%.
∴
=
×
×
(
)
+
⇒ =
0.95
C
C
2
1640 40
1640 40
1
14 8
p
p
m
C
F
.
Therefore the capacitance value required is 14.8
m
F.
The ratio between voltage at A to voltage at input at 20 Hz has a gain value of 0.95.
Its phase will be angle of
Substituting C
1640
1640
1
40
+
=
j
C
p
.
114.8 F we get the phase angle as 18.15
rad
m
°
( .
)
0 317
Therefore with 500 mV amplitude sine wave at input there will be a steady-state component of
(500
×
0.95 mV) sin (40
p
t
+
0.317) at point A.
Therefore total steady-state voltage at A
=
2
+
0.475 sin (40
p
t
+
0.317) V.
V (j
V j
j
C
j
C
j
j
C
S
w
w
p
p
)
(
)
.
.
.
= −
+
= −
−
−
= −
1
40
1640
1
40
537 7
1640
537 7
0 312
∠
∠ −
∴
1 254
.
rad
The ac component of steady-state voltage acrosss capacitor
= −
×
−
(
)
= −
−
0 312 0 5
40
1 254
0 156
40
1
.
.
t
.
.
t
sin
sin
p
p
..
254
(
)
Therefore total steady-state voltage across capacitor, v
C
(t)
=
2
-
0.156 sin (40
p
t – 1.254) V.
Input signal
Time (s)
Total steady-state voltage across C
Total steady-state voltage at A
Volts
2.5
2.0
1.5
1
0.5
0.025 0.05
0.075
0.125
0.15 0.175
0.2
0.1
–0.5
Fig. 11.5-9
WaveformplotsforExample:11.5-1
11.26
First-Order
RC
Circuits
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