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Note in Figs. 24.29 and 24.30 that the waveform for v C approaches an
average value of V/2.
EXAMPLE 24.10 The 1000-Hz square wave of Fig. 24.31 is applied
to the R-C circuit of the same figure.
a. Compare the pulse width of the square wave to the time constant of
the circuit.
b. Sketch v C.
c. Sketch i C.
V = 10 mV
v i
f = 1000 Hz
0 t
T T
2
+
v i
–
FIG. 24.31
Example 24.10.
R
5 k�
Solutions:
a. T � � � 1 ms
tp � � 0.5 ms
t � RC � (5 � 10 3 �)(0.01 � 10 �6 F) � 0.05 ms
� � 10 and
tp � 10t � � T
2 �
1 1
� �
f 1000
T
�
2
tp 0.5 ms
� �
t 0.05 ms
C
R-C RESPONSE TO SQUARE-WAVE INPUTS ⏐⏐⏐ 1107
iC 0.01 mF
The result reveals that vC will charge to its final value in half the
pulse width.
b. For the charging phase, Vi � 0 V and Vf � 10 mV, and
vC � Vf � (Vi � Vf)e �t/RC
� 10 mV � (0 � 10 mV)e �t/t
and vC � 10 mV(1 � e �t/t )
For the discharge phase, Vi � 10 mV and Vf � 0 V, and
vC � Vf � (Vi � Vf)e �t/t
� 0 V � (10 mV � 0 V)e �t/t
and v C � 10 mVe �t/t
The waveform for vC appears in Fig. 24.32.
c. For the charging phase at t � 0 s, VR � V and IRmax � V/R �
10 mV/5 k� �2 mA, and
iC � Imaxe �t/t � 2 mAe �t/t
For the discharge phase, the current will have the same mathematical
formulation but the opposite direction, as shown in Fig. 24.33.
+
vC –
10 mV
v C
0
5t
tp = 10t
T
2
T
t
FIG. 24.32
v C for the R-C network of Fig. 24.31.
i C
2 mA
0 T T
2T t
5t 2
–2 mA
FIG. 24.33
i C for the R-C network of Fig. 24.31.