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 Vp, Vavg, Vrms 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 \(Vf\)). For a silicon diode \(Vf\approx0.7\;{\rm V}\).
For an ideal sinusoidal supply
\[
v{\text{in}}(t)=Vm\sin(\omega t)
\]
the diode behaves as a unidirectional switch:
- Positive half‑cycle (\(0\le\omega t\le\pi\)): \(v{\text{in}}>Vf\) → 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
\[
v_{\text{out}}(t)=
\begin{cases}
Vm\sin(\omega t)-Vf, & 0\le\omega t\le\pi\\[4pt]
0, & \pi<\omega t\le2\pi
\end{cases}
\]
- Peak output voltage (ideal): \(Vp = Vm - V_f\)
- Average (DC) output voltage:
\[
V{\text{avg}}=\frac{1}{\pi}\int{0}^{\pi}\!\bigl(Vm\sin\theta - Vf\bigr)\,d\theta
=\frac{Vm}{\pi}-\frac{Vf}{\pi}\approx\frac{Vm}{\pi}\quad(Vf\ll V_m)
\]
- RMS output voltage:
\[
V{\text{rms}}=\sqrt{\frac{1}{2\pi}\int{0}^{\pi}\!\bigl(Vm\sin\theta - Vf\bigr)^{2}\,d\theta}
\approx\frac{Vm}{2}\qquad(Vf\ll V_m)
\]
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.
\[
\eta=\frac{P{\text{DC}}}{P{\text{AC}}}
=\frac{V{\text{avg}}I{\text{avg}}}{\frac{Vm I{\text{avg}}}{2}}
=\frac{2V{\text{avg}}}{Vm}
=\frac{2}{\pi}\approx0.40\;(40\%)
\]
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:
\[
r=\frac{\sqrt{V{\text{rms}}^{2}-V{\text{avg}}^{2}}}{V_{\text{avg}}}
=\sqrt{\left(\frac{V{\text{rms}}}{V{\text{avg}}}\right)^{2}-1}
=\sqrt{\left(\frac{Vm/2}{Vm/\pi}\right)^{2}-1}
=\sqrt{\left(\frac{\pi}{2}\right)^{2}-1}\approx1.21
\]
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): \(Vp = Vm - V_f\) (only one diode drop per half‑cycle).
- Average output voltage:
\[
V{\text{avg}}=\frac{2}{\pi}Vm
\]
- RMS output voltage:
\[
V{\text{rms}}=\frac{Vm}{\sqrt{2}}
\]
- Efficiency (ideal): \(\displaystyle \eta=\frac{2V{\text{avg}}}{Vm}= \frac{4}{\pi}\approx0.81\;(81\%)\).
5.2. Bridge Full‑Wave Rectifier
- Four diodes, two in series during each half‑cycle.
- Peak output voltage (ideal): \(Vp = Vm - 2V_f\) (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) | \(Vm-Vf\) | \(Vm-Vf\) | \(Vm-2Vf\) |
| 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:
\[
\text{PIV}{\text{required}} \ge Vm
\]
For a bridge full‑wave rectifier each diode sees a reverse voltage of up to \(Vm\) (the peak of the opposite half‑cycle). In a centre‑tap arrangement the PIV requirement is also \(Vm\). 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)
Objective: Build a half‑wave rectifier, measure its key electrical parameters, and investigate the effect of a smoothing capacitor.
Equipment: 12 V rms transformer, silicon diode (1N4007), load resistor (1 kΩ and 100 Ω for variation), electrolytic capacitors (100 µF, 470 µF, 25 V), digital multimeter, oscilloscope or PC‑based data logger.
Procedure:
- Assemble the circuit shown in Figure 1 (no capacitor). Verify diode polarity.
- Display both the input sinusoid and the rectified output on the oscilloscope. Measure and record:
- Peak output voltage \(V_p\) (oscilloscope cursors)
- Average DC voltage \(V_{\text{avg}}\) (multimeter DC‑V)
- RMS voltage \(V_{\text{rms}}\) (multimeter AC‑V or calculate from the trace)
- Load current \(I{\text{load}} = V{\text{avg}}/R_L\)
- Replace the 1 kΩ resistor with 100 Ω to increase load current and repeat the measurements.
- Insert the 100 µF capacitor in parallel with the load. Observe the waveform and record the new peak‑to‑peak ripple \(\Delta V\).
- Repeat step 4 with the 470 µF capacitor.
- Calculate the ripple factor from the measured RMS and average values and compare with the theoretical value 1.21.
- Discuss any discrepancy – consider diode forward drop, meter tolerances and non‑ideal capacitor ESR.
Data Table (to be completed)
| Condition | RL (Ω) | C (µF) | Vp (V) | Vavg (V) | Vrms (V) | ΔV (V) | Ripple factor r |
|---|---|---|---|---|---|---|---|
| No capacitor | 1 000 | – | |||||
| 100 µF | 1 000 | 100 | |||||
| 470 µF | 1 000 | 470 | |||||
| No capacitor, R=100 Ω | 100 | – |
Analysis Questions (AO2)
- Compare the measured average voltage with the theoretical value \(V_m/\pi\). Explain any differences.
- Why does the ripple voltage decrease when the capacitance is increased? Relate your answer to the expression \(\Delta V = I_{\text{load}}/(fC)\).
- Discuss how the diode’s forward voltage drop and its PIV rating influence the peak output and the reliability of the circuit.
- Calculate the efficiency of your half‑wave rectifier using the measured values and comment on how it compares with the ideal 40 %.
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 \(V_p\) | \(Vm - Vf\) | ≈ 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.