# RC Response to Square Wave Inputs   Whatsapp  The square wave of Fig. $1$ is a particular form of pulse waveform. It has a duty cycle of $50 \%$ and an average value of zero volts, as calculated below:
$$\text { Duty cycle }=\frac{t_{p}}{T} \times 100 \%=\frac{T / 2}{T} \times 100 \%=\mathbf{5 0} \%$$
$$V_{\mathrm{av}}=\frac{\left(V_{1}\right)(T / 2)+\left(-V_{1}\right)(T / 2)}{T}=\frac{0}{T}=\mathbf{0} \mathbf{V}$$ Fig. 1: Periodic square wave.
The application of a dc voltage $V_{1}$ in series with the square wave of Fig. $1$ can raise the base-line voltage from $-V_{1}$ to zero volts and the average value to $V_{1}$ volts. Fig. 2: Raising the base-line voltage of a square wave to zero volts.
If a square wave such as developed in Fig. $2$ is applied to an R-C circuit as shown in Fig. 3, the period of the square wave can have a pronounced effect on the resulting waveform for $v_{C}$.
For the analysis to follow, we will assume that steady-state conditions will be established after a period of five time constants has passed. The types of waveforms developed across the capacitor can then be separated into three fundamental types: $$T / 2>5 t$$ $$T / 2=5 t$$ and $$T / 2<5 t$$ Fig. 3: Applying a periodic square-wave pulse train to an R-C network.

#### $T / 2>5 \tau$

The condition $T / 2>5 \tau$, or $T>10 \tau$, establishes a situation where the capacitor can charge to its steady-state value in advance of $t=T / 2$. The resulting waveforms for $V_{C}$ and $i_{C}$ will appear as shown in Fig. 4. Note how closely the voltage $v_{C}$ shadows the applied waveform and how $i_{C}$ is nothing more than a series of very sharp spikes. Also that the change of $V_{i}$ from $V$ to zero volts during the trailing edge simply results in a rapid discharge of $V_{C}$ to zero volts.
In essence, when $V_{i}=$ 0 , the capacitor and resistor are in parallel and the capacitor simply discharges through $R$ with a time constant equal to that encountered during the charging phase but with a direction of charge flow (current) opposite to that established during the charging phase.  Fig. 4: $v_{C}$ and $i_{C}$ for $T / 2>5 \tau$.

#### $T / 2=5 \tau$

If the frequency of the square wave is chosen such that $T / 2=5 \tau$ or $T=10 \tau$, the voltage $V_{C}$ will reach its final value just before beginning its discharge phase, as shown in Fig. 5. The voltage $V_{C}$ no longer resembles the square-wave input and, in fact, has some of the characteristics of a triangular waveform. The increased time constant has resulted in a more rounded $V_{C}$, and $i_{C}$ has increased substantially in width to reveal the longer charging period.  Fig. 5: $v_{C}$ and $i_{C}$ for $T / 2=5 \tau$.

#### $T / 2<5 \tau$

If $T / 2<5 \tau$ or $T<10 \tau$, the voltage $V_{C}$ will not reach its final value during the first pulse (Fig. 6), and the discharge cycle will not return to zero volts. In fact, the initial value for each succeeding pulse will change until steady-state conditions are reached. In most instances, it is a good approximation to assume that steady-state conditions have been established in five cycles of the applied waveform.  Fig. 6: $v_{C}$ and $i_{C}$ for $T / 2<5 \tau$.
As the frequency increases and the period decreases, there will be a flattening of the response for $V_{C}$ until a pattern like that in Fig. $7$ results. Figure $7$ begins to reveal an important conclusion regarding the response curve for $v_{C}$ :Under steady-state conditions, the average value of ${v}_{C}$ will equal the average value of the applied square wave. Fig. 7: $V_{C}$ for $T / 2 \ll 5 \tau$ or $T \ll 10 \tau$.
Note in Figs. 6 and 7 that the waveform for $v_C$ approaches an average value of $V/2$.
Example 1: The 1000-Hz square wave of Fig. $8$ 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 $i_{C}$. Fig. 8: $V_{C}$ for $T / 2 \ll 5 \tau$ or $T \ll 10 \tau$.
Solution:
a. $T=\frac{1}{f}=\frac{1}{1000}=1 \mathrm{~ms}$
b. For the charging phase, $V_{i}=0 \mathrm{~V}$ and $V_{f}=10 \mathrm{mV}$, and
\begin{aligned} V_{C}&=V_{f}+\left(V_{i}-V_{f}\right) e^{-t / R C}\\ &=10 \mathrm{mV}+(0 \mathrm{mV}-10 \mathrm{~V}) e^{-t / \tau}\\ V_{C}&=\mathbf{1 0} \mathbf{~ m V}\left(\mathbf{1}-e^{-t / \tau}\right) \end{aligned}
For the discharge phase, $V_{i}=10 \mathrm{mV}$ and $V_{f}=0 \mathrm{~V}$, and
\begin{aligned} V_{C}&=V_{f}+\left(V_{i}-V_{f}\right) e^{-t / \tau} \\ &=0 \mathrm{~V}+(10 \mathrm{mV}-0 \mathrm{~V}) e^{-t / \tau}\\ V_{C}&=\mathbf{1 0} \mathbf{~ m V} e^{-t / \tau}\\ \end{aligned}
The waveform for $v_C$ appears in Fig. 9. Fig. 9: $v_{C}$ for the R-C network of Fig. $8$. Fig. 10: $i_{C}$ for the R-C network of Fig. $8$.
c. For the charging phase at $t = 0 s$, $V_R = V$ and $I_{R_{max}}=V/R = 10 mV/5 kΩ = 2 \mu A$, 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. 10.
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