Diode Rectifiers: Half-Wave, Full-Wave and Bridge With Ripple
Mains electricity in India is 230 V AC at 50 Hz, but almost every electronic circuit needs DC. The classic linear power supply steps the voltage down with a transformer, converts AC to pulsating DC with diodes, smooths it with a capacitor and finally regulates it. This tutorial deals with the rectifier and filter stages: how each rectifier works, what voltage you actually get, and how big the smoothing capacitor must be.
The Diode in One Paragraph
A PN junction diode conducts when its anode is more positive than its cathode by about 0.6 to 0.7 V for silicon, and blocks current in the opposite direction up to its reverse voltage rating. The cathode is marked by a band. The 1N4001 to 1N4007 family are 1 A general-purpose rectifiers with reverse ratings from 50 V to 1000 V; the 1N4007 is rated 1000 V and is the usual choice for mains-transformer supplies. For higher currents, 3 A parts such as the 1N5408 or a moulded bridge module are used.
RMS to peak: Vm = Vrms x sqrt(2) = Vrms x 1.414 Silicon drop: Vd ~ 0.7 V per conducting diode
The Three Rectifier Circuits
Half-wave
A single diode passes only the positive half-cycles. The output has a 50 Hz ripple and the transformer is used only half the time, so efficiency is poor. It is acceptable only for very small loads.
Full-wave centre-tap
A transformer with a centre-tapped secondary feeds two diodes. Each half of the winding conducts on alternate half-cycles, so the output pulses twice per mains cycle at 100 Hz. Only one diode drop appears in the path, but each diode must block twice the peak voltage of half the winding.
Bridge
Four diodes arranged in a bridge give full-wave rectification from an ordinary secondary without a centre tap. Two diodes conduct on each half-cycle, so the output loses two diode drops, but each diode only blocks the peak voltage.
| Parameter | Half-wave | Centre-tap full-wave | Bridge |
|---|---|---|---|
| Diodes | 1 | 2 | 4 |
| Ripple frequency (50 Hz mains) | 50 Hz | 100 Hz | 100 Hz |
| Peak output (no filter) | Vm - 0.7 | Vm - 0.7 (Vm of half winding) | Vm - 1.4 |
| Peak inverse voltage per diode | Vm | 2Vm | Vm |
| Ripple factor (resistive load, no capacitor) | 1.21 | 0.482 | 0.482 |
| Max theoretical efficiency | 40.6% | 81.2% | 81.2% |
Smoothing Capacitor and Ripple
A capacitor across the output charges to the peak voltage on each pulse, then supplies the load between pulses, sagging until the next peak. The peak-to-peak sag is the ripple voltage. For ripple that is small compared with the output, this approximation is used:
Vripple(p-p) ~ I_load / (f_ripple x C) Rearranged: C ~ I_load / (f_ripple x Vripple) f_ripple = 50 Hz for half-wave, 100 Hz for full-wave on 50 Hz mains Average DC ~ Vpeak - Vripple / 2
Worked example: 12 V transformer, bridge, 1 A load
A 230 V to 12 V RMS transformer feeds a bridge rectifier made of 1N5408 diodes and a 2200 µF capacitor. The load draws 1 A. Find the peak DC voltage, ripple and minimum voltage.
Vm = 12 x 1.414 = 16.97 V Vpeak(out) = 16.97 - 1.4 = 15.57 V Vripple = 1 / (100 x 0.0022) = 4.55 V p-p Vmin = 15.57 - 4.55 = 11.0 V Vaverage ~ 15.57 - 2.27 = 13.3 V
If this feeds a 7805 regulator that needs about 2 V of headroom, a minimum of 11 V is fine. If it feeds a 12 V regulator, the dips to 11 V will appear at the output as ripple. Doubling the capacitor to 4700 µF reduces the ripple to 1 / (100 x 0.0047) = 2.13 V, raising the minimum to about 13.4 V.
Same supply, half-wave
With a single diode the peak is 16.97 - 0.7 = 16.27 V, but ripple frequency is only 50 Hz, so with 2200 µF the ripple doubles to 1 / (50 x 0.0022) = 9.1 V p-p. This clearly shows why half-wave supplies are avoided for anything but tiny loads.
Sizing the capacitor for a target ripple
Work backwards when the ripple limit is known. Suppose the same bridge supply must deliver 500 mA with no more than 2 V of ripple. Then C = I / (f x Vripple) = 0.5 / (100 x 2) = 0.0025 F = 2500 µF, so the next standard value, 3300 µF, would be used. Larger capacitors reduce ripple but increase the peak charging current through the diodes and transformer at switch-on, so do not oversize them without reason.
Choosing Parts
- Diode current rating: at least the DC load current, preferably with 2x margin, because capacitor charging current flows in short, high peaks.
- Diode reverse rating: above the PIV with margin; for a 12 V bridge (PIV about 17 V) a 1N4001 would technically do, but the 1N4007 or 1N5408 costs about the same and survives mains spikes.
- Capacitor voltage: above the no-load peak; 25 V for the 12 V example, since the transformer output rises at light load.
- Capacitor ripple-current rating: check the datasheet; large capacitors in heavy-load supplies get warm from ripple current.
- Fuse the transformer primary, and use a bleeder resistor across large capacitors.
Summary
The bridge rectifier is the default choice: full-wave 100 Hz output from a simple secondary at the cost of two diode drops. The peak output is about 1.414 x Vrms minus the diode drops, and the ripple is roughly I / (f C). Size the filter capacitor so that the lowest point of the ripple stays above what the following regulator needs, and choose diodes and capacitors with comfortable voltage and current margins.