Copyright 20 13 Dorling Kindersley (India) Pvt. Ltd

11.5.5 
parallel 
RC
 circuit for signal Bypassing
The need for making a resistor offer its full resistance to DC current flow and near-zero resistance 
to signal (i.e., alternating component) current flow arises often in Electronic Circuit applications. 
Constructing a Parallel RC Circuit by adding a capacitor in parallel to the concerned resistor is the 
standard solution to this ‘signal bypassing’ problem. The capacitor is sized suitably such that it offers 
a phasor impedance with very small magnitude compared to the resistance value of the resistor to be 
bypassed even for the lowest frequency sinusoidal component that is present in the signal current. 
Parallel
RC
Circuitisusedasalow-passfilterwithitsbandwidthmuchlowerthanthe
lowestfrequencypresentinthesignalin
signal bypassing
application.
The DC component of total current flowing into the parallel combination goes through the resistor 
under steady-state. But all the AC components flow almost entirely through the capacitor since it offers 
much lower impedance than the parallel resistor at the frequencies concerned. The voltage across the 
combination remains nearly DC. The next example illustrates this application of a Parallel RC Circuit.
Example: 11.5-2
Consider the ‘signal bypassing’ context in Fig. 11.5-
10. 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). The coupling capacitor C
1
may be assumed to be so large that it is effectively a 
short-circuit at all frequencies except at DC. It is desired 
that more than 95% of applied signal voltage appear 
across the 1 k
W
resistor for the entire frequency range 
without affecting the DC content of current through 
that resistor. (i) Calculate the value of C needed for this 
purpose. (ii) Assume v
S
(t) is a 10 mV amplitude 20 Hz 
sine wave and plot the waveforms of total current through 1k
W
, 99k
W
and the capacitor with the value 
of C calculated above.
Solution
The equivalent circuits needed for calculating the steady-state response contributions from the DC 
source and signal source are shown in of Fig. 11.5-11 (a) and (b), respectively. 
(a)
A
10 k
Ω
1 k
Ω
99 k
Ω
2 k
Ω
12 V +
–
A
1 k
Ω
2 k
Ω
//10 k
Ω
V
S
(
t
)
(b)
1
99 k
Ω
j
ω
C
2
+
–
Fig. 11.5-11 
Steady-stateequivalentcircuitsinExample:11.5-2
Fig. 11.5-10 

Circuitforsignal
bypassingproblemin
Example:11.5-2
+
+
–
–
A
10 k
Ω
1 k
Ω
99 k
Ω
2 k
Ω
V
S
(
t
)
C
1
C
2
12 V +
–

---

FrequencyResponseofFirstOrder
RC
Circuits

11.27
Solving the circuit in Fig. 11.5-11 (a), we get the DC potential at the point A as 1.97 V, the DC 
current through 1 k
W
and 99 k
W
resistors as 0.0197 mA and the DC potential across 99 k
W
resistor 
(and hence across the capacitor C
2
) as 1.95 V.
Consider the circuit in of Fig. 11.5-11 (b). We desire the amplitude of signal voltage across 
1 k
W
to be 
≥
95% of amplitude of applied signal voltage even when the signal is a 20 Hz 
sinusoid. 
Ratio of phasor voltage across 1k to applied phasor volta
Ω
gge
=
+
1000
1000 99000
1
40
2
/ /
j
C
p
We want the magnitude of this ratio to be 0.95. Solving for C
2
we get C
2
=
24.5 
m
F.
Substituting this value of C
2
in the phasor ratio above, we get the ratio as 0 95 0 32
.
.
∠
rad.
Applied signal voltage
V
Signal voltage across
=
(
)
∴
0 01
40
. sin
p
t
1k
0.0095 
V
Signal current through 1k
0.0
Ω
Ω
=
+
(
)
∴
=
sin
.
40
0 32
p
t
0095 
mA
Total current through 1k
0.0197 0.0
sin
.
40
0 32
p
t
+
(
)
∴
=
+
Ω
0095 
mA
sin
.
40
0 32
p
t
+
(
)
Phasor impedance of 24.5 F capacitor at 20 Hz
Pa
m
= −
j
 
325
Ω
rrallel Combination of 24.5 F capacitor at 20 Hz and 99k
m
Ω =
3325
1.567rad 
∠ −
Ω
∴
=
×
Signal voltage across parallel combination 0.0095 0.325 sinn
.
.
sin
.
40
0 32 1 567
40
1 247
p
p
t
t
+
−
(
)
=
−
(
)
∴
V 
0.00309 
V
Signal curreent through 99k
0.00003
mA
Total current th
Ω =
−
(
)
∴
sin
.
40
1 247
p
t
rrough 99k
0.0197 0.00003
mA
Signal current 
Ω =
+
−
(
)
sin
.
40
1 247
p
t
tthrough 24.5 F capacitor
0.00309 V
0.325 k
m
p
=
−
Ω
sin
.
40
1 247
t
++
(
)
=
+
(
)
p
p
/
sin
.
2
40
0 323
0.0095
mA
Total voltage across paralle
t
ll combination
0.00309 
V
=
+
−
(
)
1 95
40
1 247
.
sin
.
p
t
Almost the entire signal current flowing through 1 k
W
resistor goes into the capacitor. Thus the 
99 k
W
resistor has been bypassed very effectively by a 24.5 
m
F even at 20 Hz. The relevant waveforms 
are shown in Fig. 11.5-12. Voltages shown are with respect to the negative terminal of the DC
source.

---

11.28


First-Order
RC
Circuits
Current
(ma)
0.02
0.01
0.05
(a)
(c)
(d)
(e)
Time (s)
Time (s)
1.925
1.975
2
Volts
1.95
0.05
0.15
0.1
0.1
0.15
–0.01
Fig. 11.5-12 

WaveformsforExample:11.5-2(a)Totalcurrentin1k
w
resistor(b)Total
currentinthecapacitor(c)Totalcurrentin99k
W
resistor(d)Totalvoltage
atA(e)Totalvoltageacrosscapacitor