Rectifier circuits

YOU WILL BE ABLE TO

Explain half-wave and full-wave rectification.
Compute output amplitude accounting for diode drops.
Describe the role of the smoothing capacitor and estimate ripple.
Identify the bridge rectifier and its four-diode operation.

Here is the diode's original job, and still its most common: turning AC into DC. Ontario's outlets deliver a sinusoid that spends half its time negative; almost every electronic device needs a steady positive rail. The simplest converter is one diode in series with the load — a half-wave rectifier. During positive half-cycles the diode conducts and the load sees the input minus one diode drop; during negative half-cycles the diode blocks and the output sits at zero.

The half-wave rectifier: one diode passes the positive half-cycles to the load.

Input and half-wave output: the negative half-cycles are simply gone.

Half-wave rectification throws away half the input and produces a lumpy output. A full-wave rectifier does better by inverting the negative half-cycles instead of discarding them, and the standard implementation is the bridge: four diodes arranged so that whichever way the input swings, two of them conduct and steer current through the load in the SAME direction. The cost is two diode drops in the path rather than one.

v_{o,\text{peak}} = v_{s,\text{peak}} - 0.7 \quad\text{(half-wave)}

v_{o,\text{peak}} = v_{s,\text{peak}} - 1.4 \quad\text{(bridge)}

peak output, constant-drop model

The bridge: four diodes, all conducting toward the + rail. Whichever way v_s swings, two of them steer load current the same direction.

Even full-wave output still pulses from zero to peak twice per cycle. The cure is CA-5's mathematics: a capacitor across the load, charged to the peak on each pulse and discharging slowly through the load between them. The capacitor holds the output up; the remaining sag is the ripple voltage, and it follows directly from $i = C\,dv/dt$ with the load current assumed roughly constant.

V_{\text{ripple}} \approx \frac{I_{\text{load}}\, T}{C}

peak-to-peak ripple, T = time between charging pulses

WORKED EXAMPLE

A full design pass

A transformer supplies $12$ V peak into a bridge, load current $100$ mA, on the $60$ Hz mains.
Peak output: $12 - 1.4 = 10.6$ V.
Bridge charging interval: $T = 1/(2 \times 60) \approx 8.3$ ms.
For $0.5$ V of ripple: $C = I T / V_{\text{ripple}} = (0.1 \times 0.0083)/0.5 \approx 1.7$ mF — so a $2200$ µF electrolytic, the next stock size up.
Result: roughly $10.1$ – $10.6$ V of DC, ready for a regulator to trim to a clean rail.

C \approx 2200

µF for about 0.5 V of ripple

PRACTICE

A half-wave rectifier is fed a $9$ V peak sinusoid. What is the peak output voltage across the load, using the constant-drop model?

The same $9$ V peak into a bridge rectifier. What is the peak output, and why is it lower than the half-wave answer?

A bridge rectifier on 60 Hz mains feeds a $50$ mA load through a $1000$ µF capacitor. Estimate the peak-to-peak ripple.

Why does the same capacitor produce twice the ripple in a half-wave circuit as in a bridge, and what would happen to ripple if the supply frequency were 400 Hz (as on aircraft power)?