Half Wave and Full Wave Rectifiers

By the end you should be able to: Describe the operation of half wave, full wave centre-tapped and bridge rectifiers, and calculate DC output, ripple factor and rectification efficiency, including the effect of a capacitor filter.

Mains supply is AC; almost every circuit needs DC. A rectifier is the first stage of that conversion, and it exploits the one property established in Module 3 — a diode conducts one way only.

Half wave rectifier

One diode in series with the load.

Positive half cycle: the diode is forward biased, conducts, and the load sees the input less the diode drop.
Negative half cycle: reverse biased, blocks. Output is zero.

The output is a train of positive humps with gaps where the negative half cycles were. Its frequency equals the supply frequency, 50 Hz.

V_{dc} = \frac{V_m}{\pi} = 0.318\,V_m, \qquad V_{rms} = \frac{V_m}{2} = 0.5\,V_m

$V_{dc}$ is the average over a whole cycle — the conducting half averages $2V_m/\pi$ , halved by the idle half.

PIV, the peak inverse voltage the diode must withstand, is $V_m$ .

Full wave rectifier

Both half cycles used. Two constructions.

Centre-tapped, two diodes: a transformer with a centre-tapped secondary. On each half cycle one diode conducts, and both deliver current the same way through the load. Each diode sees only half the secondary winding, so each handles $V_m$ , but the PIV is $2V_m$ because the non-conducting diode has the whole winding across it.

Bridge, four diodes: no centre tap. Diagonally opposite pairs conduct alternately. PIV is $V_m$ , and the transformer is used more efficiently — which is why the bridge dominates in practice, despite two extra diodes. Its cost is two diode drops in the conduction path instead of one, about 1.4 V, which matters in low-voltage supplies.

For either full wave circuit:

V_{dc} = \frac{2V_m}{\pi} = 0.636\,V_m, \qquad V_{rms} = \frac{V_m}{\sqrt{2}} = 0.707\,V_m

The output frequency is twice the supply frequency, 100 Hz from a 50 Hz mains. That doubling matters for filtering: ripple at 100 Hz is easier to smooth than at 50 Hz.

Ripple factor

How much AC survives in the output:

r = \frac{\text{rms value of the AC component}}{\text{DC value}} = \sqrt{\left(\frac{V_{rms}}{V_{dc}}\right)^{2} - 1}

Half wave:

r = \sqrt{\left(\frac{0.5}{0.318}\right)^{2}-1} = \sqrt{2.467 - 1} = 1.21

Full wave:

r = \sqrt{\left(\frac{0.707}{0.636}\right)^{2}-1} = \sqrt{1.234 - 1} = 0.48

A half wave output contains more AC than DC — $r > 1$ . It is barely DC at all. The full wave figure of 0.48 is much better and still far too poor to run anything without filtering.

Rectification efficiency

\eta = \frac{P_{dc}}{P_{ac}} \times 100\%

Ignoring diode and winding resistance:

Half wave: 40.6%
Full wave: 81.2% — exactly twice, as you would expect from using both half cycles.

These are maxima. Real efficiency is lower once forward drops and transformer resistance are counted.

Capacitor filter

A capacitor across the load charges to near $V_m$ on each peak and discharges into the load between peaks. The output becomes a near-constant voltage with a small sawtooth ripple.

r = \frac{1}{4\sqrt{3}\,f\,R_L\,C}

Read the formula for what it tells you to do: ripple falls with larger $C$ , larger load resistance (that is, smaller load current), and higher $f$ . The frequency term is why full wave rectification helps twice over — better ripple before filtering, and 100 Hz instead of 50 Hz to filter.

The trade-off is current. The capacitor recharges only near the peaks, so conduction happens in short, tall spikes rather than smooth humps. Peak diode current is therefore far above the average load current, and the diode must be rated for it.