Diode as a rectifier: half-wave, full-wave and the capacitor filter
A diode conducts one way, so it turns AC into pulsating DC. A half-wave rectifier uses one half cycle, a full-wave rectifier uses both; a capacitor smooths the output.
Half-wave rectifier: how it works
- 1Mains AC (230 V, 50 Hz; your NIOS lesson gives the older figure, 220 V) goes to a step-down transformer, because most devices need a lower voltage.
- 2The secondary is joined to a diode D in series with a load RL.
- 3Positive half cycle (0 to T/2): D is forward biased and conducts. Current flows through RL.
- 4Negative half cycle (T/2 to T): D is reverse biased and does not conduct. The output is zero.
- 5The output is a row of positive half cycles with gaps: pulsating DC. Its average value is Vdc = Vm/π.
- 6In the off half cycle, the diode faces a reverse voltage of Vm. Choose a diode whose PIV is more than Vm, or it will be damaged.
Full-wave rectifier: how it works
- 1A centre-tapped transformer has two equal halves in its secondary, with ends X and Z and centre point Y. X and Z are opposite in phase with respect to Y.
- 2The anode of diode D1 joins X and the anode of D2 joins Z. Their cathodes are joined, and the load RL goes from that joint back to Y.
- 3First half cycle: X is positive and Z negative. D1 conducts and D2 does not. Current flows through RL.
- 4Next half cycle: Z is positive and X negative. D2 conducts and D1 does not. Current flows through RL in the same direction.
- 5★So the output is positive in every half cycle, and Vdc = 2Vm/π, twice the half-wave value. For 50 Hz mains there are 100 pulses per second.
★ key step
Smoothing with a capacitor filter
Rectified output flows one way but rises and falls; this unwanted part is the ripple. A capacitor C joined across RL smooths it. In the first quarter cycle, C charges to nearly Vm. As the input falls, C discharges slowly through RL and keeps the current going. So the output becomes nearly steady DC. A larger C and a larger RL give smaller ripple. Better power supplies use LC or C-L-C (π) filters. With a capacitor, the reverse voltage across a half-wave diode can reach 2Vm, so it needs a higher PIV.
How to draw the circuits and waveforms
- Half-wave circuit: transformer T (two coils with lines for the core), terminals X and Y, then diode D and load RL in series. Mark the current direction in RL.
- Full-wave circuit: a transformer with secondary ends X and Z and centre tap Y. Draw D1 from X and D2 from Z, cathodes joined at a point B, and RL from B to Y.
- Waveforms: the input sine wave on top and the output below it, lined up in time. Half-wave: positive humps with gaps. Full-wave: positive humps with no gaps.
- With a filter: draw C across RL. The output is a nearly flat line just below Vm, with small saw-tooth dips (ripple).
Half-wave and full-wave rectifiers
| Feature | Half-wave | Full-wave (centre-tap) |
|---|---|---|
| Diodes | One | Two |
| Transformer | Ordinary step-down | Centre-tapped step-down |
| Half cycles used | One | Both |
| Vdc | Vm/π ≈ 0.32 Vm | 2Vm/π ≈ 0.64 Vm |
| Pulses per second (50 Hz input) | 50 | 100 |
| PIV needed (no filter) | Vm | 2Vm |
Worked example: DC output of rectifiers
A rectifier gets an AC input of peak voltage Vm = 20 V and feeds a 1 kΩ load. Find Vdc and Idc for (a) a half-wave rectifier and (b) a full-wave rectifier. Ignore the diode drop and take π = 3.14.
- (a) Vdc = Vm/π = 20 V ÷ 3.14 ≈ 6.37 V.
- Idc = Vdc/RL = 6.37 V ÷ 1000 Ω ≈ 6.37 × 10⁻³ A = 6.37 mA.
- (b) Vdc = 2Vm/π = 40 V ÷ 3.14 ≈ 12.7 V.
- Idc = 12.7 V ÷ 1000 Ω ≈ 12.7 mA.
(a) about 6.37 V and 6.37 mA; (b) about 12.7 V and 12.7 mA, twice as much
Every mobile charger, laptop adapter and home inverter has a rectifier and a filter inside. They turn 230 V, 50 Hz mains into the steady low-voltage DC that batteries and chips need.
3 marks: “Explain, with a circuit diagram and waveforms, how two diodes and a centre-tapped transformer keep the load current in one direction through both half cycles of the AC input.” Centre-tapped transformer, D1 and D2, RL from the joined cathodes to the centre tap. First half: D1 conducts; next half: D2 conducts; current through RL is in the same direction both times. Waveforms: sine input; output humps with no gaps.
5 marks: “With a circuit diagram, explain how a half-wave rectifier works. How does a capacitor filter make the output smoother? Draw the output with and without the filter.” Transformer, diode and RL in series; the diode conducts only in the positive half cycle; Vdc = Vm/π. A capacitor across RL charges to Vm and discharges through RL as the input falls; a larger C and RL give less ripple. Draw both waveforms.
2 marks: “A half-wave rectifier gets an input of peak 31.4 V and feeds a 2 kΩ load. Find Vdc and Idc. (π = 3.14)” Vdc = Vm/π = 31.4 V ÷ 3.14 = 10 V. Idc = 10 V ÷ 2000 Ω = 5 mA.
Half: one diode, one hump, Vm/π. Full: two diodes, two humps, 2Vm/π. The capacitor fills the gaps.