Question 24.10: The 1000 Hz square wave in Fig. 24.31 is applied to the R-C ...
The 1000 Hz square wave in 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}.
\text { a. } \quad T=\frac{1}{f}=\frac{1}{1000}=1 ms.
t_{p}=\frac{T}{2}=0.5 ms.
\tau=R C=\left(5 \times 10^{3} \Omega\right)\left(0.01 \times 10^{-6} F \right)=0.05 ms.
\frac{t_{p}}{\tau}=\frac{0.5 ms }{0.05 ms }=10 \text { and }t_{p}=10 \tau=\frac{T}{2}.
\text { The result reveals that } v_{C} \text { charges to its final value in half the pulse width.}\text { b. For the charging phase, } V_{i}=0 V \text { and } V_{f}=10 mV \text {, and }
v_{C}=V_{f}+\left(V_{i}-V_{f}\right) e^{-t / R C}.
=10 mV +(0-10 mV ) e^{-t / \tau}.
and v_{C}=10 mV \left(1-e^{-t / \tau}\right).
\text { For the discharge phase, } V_{i}=10 mV \text { and } V_{f}=0 V , \text { and }v_{C}=V_{f}+\left(V_{i}-V_{f}\right) e^{-t / \tau}.
=0 V +(10 mV -0 V ) e^{-t / \tau}.
and v_{C}=10 mVe ^{-t / \tau}
\text { The waveform for } v_{C} \text { appears in Fig. 24.32. }\text { c. For the charging phase at } t=0 s , V_{R}=V \text { and } I_{R_{\max }}=V / R=10 mV / 5 k \Omega=2 \mu A \text { and }
i_{C}=I_{\max } e^{-t / \tau}=2 \mu A e^{-t / \tau}.
For the discharge phase, the current will have the same mathematical formulation but the opposite direction, as shown in Fig. 24.33.
