explain the use of a single diode for the half-wave rectification of an alternating current

Rectification and Smoothing – Half‑Wave Rectifier

Objective

Explain how a single diode can be used to convert an alternating current (AC) into a pulsating direct current (DC) by half‑wave rectification.

Principle of Operation

A diode conducts current only when it is forward‑biased. In an AC supply the voltage varies sinusoidally:

\$v{\text{in}}(t)=V{\!m}\sin(\omega t)\$

When the instantaneous voltage is positive (forward‑bias), the diode conducts and the output follows the input. When the voltage is negative (reverse‑bias), the diode blocks current and the output is forced to zero.

Circuit Diagram

Waveform Description

The resulting output voltage \$v_{\text{out}}(t)\$ is:

\$\$v_{\text{out}}(t)=

\begin{cases}

V_{\!m}\sin(\omega t), & \text{for } 0\le \omega t \le \pi \\

0, & \text{for } \pi < \omega t \le 2\pi

\end{cases}\$\$

This is a series of positive half‑cycles separated by intervals of zero voltage.

Key Parameters

• Peak output voltage: \$V{\!p}=V{\!m}\$ (ignoring diode forward drop).
• Average (DC) output voltage:

  \$V{\text{avg}}=\frac{1}{\pi}\int{0}^{\pi}V{\!m}\sin\theta\,d\theta=\frac{V{\!m}}{\pi}\$

• RMS output voltage:

  \$V{\text{rms}}=\sqrt{\frac{1}{2\pi}\int{0}^{\pi}V{\!m}^{2}\sin^{2}\theta\,d\theta}= \frac{V{\!m}}{2}\$

• Ripple factor (measure of AC content):

  \$r=\frac{\sqrt{V{\text{rms}}^{2}-V{\text{avg}}^{2}}}{V_{\text{avg}}}\approx 1.21\$

Advantages and Limitations

1. Simple and inexpensive – only one diode required.
2. Produces a pulsating DC that can be filtered to smoother DC with a capacitor.
3. Low efficiency: only half of the input waveform contributes to output power.
4. Large ripple voltage; not suitable for applications requiring low ripple without additional filtering.

Adding a Smoothing Capacitor

Connecting a capacitor \$C\$ across the load stores charge during the conducting half‑cycle and releases it during the non‑conducting interval, reducing ripple.

The peak‑to‑peak ripple voltage \$\Delta V\$ can be approximated by:

\$\Delta V \approx \frac{I_{\text{load}}}{f C}\$

where \$I_{\text{load}}\$ is the load current and \$f\$ is the AC supply frequency.

Summary Table

Key Take‑away

A single diode allows only the positive half of an AC waveform to pass, producing a pulsating DC. While simple, the output contains significant ripple, which can be reduced by adding a smoothing capacitor or by using more advanced rectifier circuits.