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

Rectification and Smoothing – Half‑Wave Rectifier (Single Diode)

Learning Objectives

• Describe how a single diode converts an alternating current (AC) into a pulsating direct current (DC) – the principle of half‑wave rectification.
• Derive the expressions for peak, average and RMS output voltages, efficiency and ripple factor.
• Compare half‑wave rectification with full‑wave rectification (centre‑tap and bridge) and explain the effect of diode‑drops.
• Explain the role of a smoothing capacitor and calculate the expected ripple voltage.
• Identify practical limitations – diode forward drop, peak inverse voltage (PIV) rating, efficiency and ripple.
• Plan and carry out a simple laboratory investigation, recording the key quantities V_p, V_avg, V_rms and ripple.

1. Principle of Half‑Wave Rectification

A diode conducts only when it is forward‑biased (anode voltage higher than cathode by at least the forward‑bias voltage \(V_f\)). For a silicon diode \(V_f\approx0.7\;{\rm V}\).

For an ideal sinusoidal supply

\[ v_{\text{in}}(t)=V_m\sin(\omega t) \]

the diode behaves as a unidirectional switch:

• Positive half‑cycle (\(0\le\omega t\le\pi\)): \(v_{\text{in}}>V_f\) → diode forward‑biased → conducts → the load sees the input voltage (minus \(V_f\)).
• Negative half‑cycle (\(\pi<\omega t\le2\pi\)): diode reverse‑biased → blocks current → load voltage is forced to zero.

2. Circuit Diagram

3. Output Waveform and Basic Equations

\[ v_{\text{out}}(t)= \begin{cases} V_m\sin(\omega t)-V_f, & 0\le\omega t\le\pi\\[4pt] 0, & \pi<\omega t\le2\pi \end{cases} \]

• Peak output voltage (ideal): \(V_p = V_m - V_f\)
• Average (DC) output voltage: \[ V_{\text{avg}}=\frac{1}{\pi}\int_{0}^{\pi}\!\bigl(V_m\sin\theta - V_f\bigr)\,d\theta =\frac{V_m}{\pi}-\frac{V_f}{\pi}\approx\frac{V_m}{\pi}\quad(V_f\ll V_m) \]
• RMS output voltage: \[ V_{\text{rms}}=\sqrt{\frac{1}{2\pi}\int_{0}^{\pi}\!\bigl(V_m\sin\theta - V_f\bigr)^{2}\,d\theta} \approx\frac{V_m}{2}\qquad(V_f\ll V_m) \]

4. Derivation of Efficiency and Ripple Factor

Efficiency (\(\eta\)) is the ratio of DC power delivered to the load to the apparent AC power supplied by the source.

\[ \eta=\frac{P_{\text{DC}}}{P_{\text{AC}}} =\frac{V_{\text{avg}}I_{\text{avg}}}{\frac{V_m I_{\text{avg}}}{2}} =\frac{2V_{\text{avg}}}{V_m} =\frac{2}{\pi}\approx0.40\;(40\%) \]

Thus a half‑wave rectifier can at most convert 40 % of the AC power into useful DC power.

Ripple factor (\(r\)) measures the remaining AC component in the rectified output:

\[ r=\frac{\sqrt{V_{\text{rms}}^{2}-V_{\text{avg}}^{2}}}{V_{\text{avg}}} =\sqrt{\left(\frac{V_{\text{rms}}}{V_{\text{avg}}}\right)^{2}-1} =\sqrt{\left(\frac{V_m/2}{V_m/\pi}\right)^{2}-1} =\sqrt{\left(\frac{\pi}{2}\right)^{2}-1}\approx1.21 \]

A ripple factor of 1.21 means the AC component is larger than the DC component – a clear indication that smoothing is required for most practical circuits.

5. Comparison with Full‑Wave Rectification

Full‑wave circuits use either a centre‑tap transformer with two diodes or a bridge of four diodes. Both conduct during **both** half‑cycles, doubling the utilisation of the AC supply.

5.1. Centre‑Tap Full‑Wave Rectifier

• Two diodes, each conducts for one half‑cycle.
• Peak output voltage (ideal): \(V_p = V_m - V_f\) (only one diode drop per half‑cycle).
• Average output voltage: \[ V_{\text{avg}}=\frac{2}{\pi}V_m \]
• RMS output voltage: \[ V_{\text{rms}}=\frac{V_m}{\sqrt{2}} \]
• Efficiency (ideal): \(\displaystyle \eta=\frac{2V_{\text{avg}}}{V_m}= \frac{4}{\pi}\approx0.81\;(81\%)\).

5.2. Bridge Full‑Wave Rectifier

• Four diodes, two in series during each half‑cycle.
• Peak output voltage (ideal): \(V_p = V_m - 2V_f\) (two forward drops).
• Average and RMS voltages are the same as the centre‑tap case (the extra drop only reduces the peak value).
• Efficiency remains ≈ 81 % (ideal), but the extra diode drop slightly lowers the DC level.

5.3. Summary Table

Feature	Half‑Wave (1 diode)	Full‑Wave Centre‑Tap (2 diodes)	Full‑Wave Bridge (4 diodes)
Diodes required	1	2	4
Conducted portion of AC cycle	50 %	100 %	100 %
Peak output (ideal)	\(V_m-V_f\)	\(V_m-V_f\)	\(V_m-2V_f\)
Average output	\(V_m/\pi\)	\(2V_m/\pi\)	\(2V_m/\pi\)
RMS output	\(V_m/2\)	\(V_m/\sqrt{2}\)	\(V_m/\sqrt{2}\)
Efficiency (ideal)	≈ 40 %	≈ 81 %	≈ 81 %

6. Peak Inverse Voltage (PIV) Requirement

The diode must survive the maximum reverse voltage it experiences when it is off. For a half‑wave rectifier:

\[ \text{PIV}_{\text{required}} \ge V_m \]

For a bridge full‑wave rectifier each diode sees a reverse voltage of up to \(V_m\) (the peak of the opposite half‑cycle). In a centre‑tap arrangement the PIV requirement is also \(V_m\). Always select a diode with a rating comfortably above this value (e.g., 1.5 × \(V_m\)).

7. Adding a Smoothing Capacitor

Connecting a capacitor \(C\) across the load stores charge during the conducting half‑cycle and releases it when the diode is reverse‑biased, thereby reducing the ripple.

For a half‑wave rectifier the approximate peak‑to‑peak ripple voltage is

\[ \Delta V \approx \frac{I_{\text{load}}}{f\,C} \]

where

• \(I_{\text{load}}\) = average load current (A)
• \(f\) = supply frequency (50 Hz or 60 Hz for half‑wave; note that the ripple frequency equals the supply frequency)
• \(C\) = smoothing capacitance (F)

Numerical Example

• Supply: 230 V rms, 50 Hz → \(V_m=\sqrt{2}\times230\approx325\;{\rm V}\)
• Load resistor: 1 kΩ → \(I_{\text{load}}\approx V_{\text{avg}}/R\approx 0.103\;{\rm A}\)
• Capacitance: 100 µF
• Ripple: \[ \Delta V \approx \frac{0.103}{50\times100\times10^{-6}} \approx 20.6\;{\rm V} \]
• Increasing the capacitance to 470 µF reduces the ripple to ≈ 4.4 V, illustrating the inverse relationship \(\Delta V\propto 1/C\).

8. Laboratory Investigation (AO3)

9. Summary Table (Typical Values for 230 V rms, 50 Hz)

Parameter	Expression	Numerical Value
Peak input voltage \(V_m\)	\(\sqrt{2}\,V_{\text{rms}}\)	≈ 325 V
Peak output voltage \(V_p\)	\(V_m - V_f\)	≈ 324 V (silicon diode)
Average output voltage \(V_{\text{avg}}\)	\(V_m/\pi\)	≈ 103 V
RMS output voltage \(V_{\text{rms}}\)	\(V_m/2\)	≈ 162 V
Ripple factor \(r\)	\(\sqrt{(\pi/2)^2-1}\)	≈ 1.21
Efficiency \(\eta\)	\(2/\pi\)	≈ 40 %
Required PIV	\(\ge V_m\)	≥ 325 V (choose a diode rated ≥ 600 V for safety)

Key Take‑away

A single diode acts as a one‑way switch, allowing only the positive half of an AC waveform to reach the load. The resulting half‑wave rectifier delivers a pulsating DC with a relatively low average voltage, high ripple (factor ≈ 1.21) and an efficiency of about 40 %. Adding a smoothing capacitor reduces the ripple according to \(\Delta V \approx I_{\text{load}}/(fC)\), but the circuit still suffers from the diode’s forward drop and the need for a diode with a PIV rating at least equal to the peak input voltage. Mastery of these concepts satisfies the Cambridge 9702 syllabus requirements for “Rectification and smoothing” and provides the foundation for studying more efficient full‑wave and bridge rectifier circuits.