B4 · Diodes

Bridge Rectifier and Smoothing: AC into DC

Four diodes fold even the negative half-cycles of AC upward. Then grow a capacitor and watch the bumpy pulsation get smoothed into steady DC.

Grow the capacitor and fill the valleys

The faint curve is the rectified humps and the bold curve is the output after the capacitor. Grow the capacitor and the valleys between the humps fill in, flattening the output.

Smoothing capacitor CC = 6

Output ripple ΔVr

ΔVr ≈ Vpeak

Almost no capacitor. The output follows the rectified humps all the way down to zero in the valleys.

Large ripple

The bridge gathers both half-cycles

A bridge is four diodes woven into a diamond. When the input is positive, one opposing pair turns on and sends current one way through the load; when the input is negative, the other pair turns on. Yet in both cases the direction of current through the load is the same. So the negative half-cycle is not thrown away but folded upward. This is full-wave rectification.

Full-wave: shallow, frequent valleys

Unlike half-wave rectification, which passes only one half-cycle, full-wave turns both half-cycles into humps. So the humps are packed at twice the input frequency. Because the empty valleys between them are shallow and come often, the same capacitor can smooth them far better. Since each path through the bridge passes two diodes, the hump height is two diode drops below the input peak.

The capacitor fills the valleys

Place a capacitor in parallel with the output. As a hump rises, the capacitor charges quickly to the peak; once the hump passes and the input falls, the capacitor discharges slowly into the load and holds the voltage up. It holds on until the next hump refills it. The small remaining wobble is the ripple, and the larger the capacitor, the slower the discharge and the smaller the ripple. Combine full-wave rectification with a large capacitor and you get nearly flat DC.

ObserveVout = |Vin| - 2VD

The full-wave output is the input magnitude minus two diode drops.

Choosefripple = ? fin

Being full-wave, the ripple frequency is twice the input.

Fill inΔVr ≈ Iload / (2 fin ?)

The ripple is inversely proportional to the capacitor.

On your ownΔVr ≈ ?

Large load current and a small capacitor make the ripple grow.

Back to the first screen

When the capacitor was small, the output plunged with the humps all the way to zero in the valleys. As you grew the capacitor, the charge filled on a hump rode through the valley and the output grew flatter. Two things together underlie that flatness: the bridge folding even the negative half-cycle into a hump, which makes the valleys shallow and frequent, and the capacitor filling those shallow valleys. The rectification of B1 grows to use both half-cycles, and with smoothing added, the front end of a power supply is complete.

A bridge rectifier uses four diodes to send both half-cycles of AC through the load in the same direction, folding the negative half-cycle upward (full-wave rectification). The output is positive humps at twice the input frequency. A smoothing capacitor charges on the humps and discharges through the valleys, filling them to make nearly flat DC with only a small ripple. ΔVr ≈ Iload / (2 fin C).

What comes next

Up to here the diode set, cut, or fixed a flow in one direction. The next group goes a step further, to a device that controls a much larger current with a small input current: the transistor. Its first unit shows how a BJT governs a large collector current with a small base current, starting from that principle of current control.