Bridge rectifiers

The idea

A diode is a one-way valve for current. In the symbol, current can only flow the way the triangle points, and getting through costs a small, roughly fixed voltage, the forward voltage $V_F\approx 1$ V. Backwards, it blocks, up to its limit $V_{RRM}$.

One diode on its own (a half-wave rectifier) lets the positive half of the AC through and throws the negative half away. A full bridge uses four diodes arranged in a diamond so that both halves come out positive.

How the bridge works

Name the four corners: A and B connect to the AC, + and − connect to the load.

Diode	Allows current
D1	A → +
D2	B → +
D3	− → A
D4	− → B

Two arrows point into +, two point out of −. So whatever the AC does, current can only arrive at + and leave from −. It always goes through the load from + to −.

Positive half (A pushes): A → D1 → + → load → − → D4 → B. D2 and D3 would need current to flow against their arrows, so they block.

Negative half (B pushes): B → D2 → + → load → − → D3 → A. Same direction through the load.

The two things to remember

1. Current always passes through exactly two diodes. So the output peak is $V_{pk}-2V_F$, and two diodes are always making heat.
2. The output is a row of positive humps, two per AC cycle: 100 per second at 50 Hz.

A three-phase bridge (6 diodes) on a three-phase supply gives six overlapping humps per cycle, which is nearly flat DC. That’s what big industrial kit uses.

Smoothing with a capacitor

The humps still drop to 0 V between peaks. A capacitor across the output works like a water tank: it fills at each peak, and the load drains it a little until the next hump refills it. The dip is the ripple, $\Delta V$.

Deriving the ripple:

1. Refills happen 100 times a second, so the gap is $\Delta t=\frac{1}{2F_{grid}}=10$ ms.
2. In that gap the load takes charge $\Delta Q=I\Delta t$ (current is charge per second).
3. For a capacitor $Q=CV$, so the voltage drops by $\Delta V=\Delta Q/C$.

$\boxed{\Delta V=\frac{I}{2F_{grid}C}}$

The capacitor charges to the top of each hump, minus the two diode drops:

$V_{C,max}=V_{pk}-2V_F$

Heat in the bridge

1. Power = voltage × current. A conducting diode has a fixed $V_F$ across it, so $P=V_F\times i$ at every instant.
2. Averaged over time, $V_F$ is constant, so $P=V_F\times I_{avg}$. Diodes use average current, unlike resistors and MOSFETs, which use $I^{2}R$ with RMS.
3. Two diodes are always on, so $P=2V_F{I}_{avg}$.
4. You’re usually given RMS, and for these humps $I_{avg}=0.9I_{RMS}$ (see AC values).

$\boxed{P_{D(FB)}=2\times 0.9\times V_F{I}_{D,RMS}=1.8V_F{I}_{D,RMS}}$

Choosing a bridge from its datasheet

Check	Rule	Why
$I_{F(AV)}$	Above your current, with margin (~50%)	It sets how much average current it can carry without overheating
$V_{RRM}$	Above the peak, 325 V on UK mains, with margin	When a diode is off, it has to block the full peak
$V_F$	Lower is better	Less heat, slightly higher output

Worked example

A robot charger uses a GBPC12 bridge ($V_F=1.1$ V, $I_{F(AV)}=12$ A, $V_{RRM}=800$ V) on 230 V mains. It draws 2 A RMS and has a 4700 µF smoothing capacitor. Find the bridge’s heat, the capacitor’s peak voltage and the ripple. Is the bridge suitable?

Answer

$P_{D(FB)}=1.8\times 1.1\times 2=3.96$ W $V_{C,max}=325.3-2(1.1)=323.1$ V $\Delta V=\frac{2}{2\times 50\times 4700\times 10^{-6}}=\frac{2}{0.47}=4.26$ V Suitable: 12 A ≫ 2 A, and 800 V > 325 V.

Common mistakes

• Forgetting the two diode drops: it’s $V_{pk}-2V_F$, not $-V_F$.
• Forgetting µF is $\times 10^{-6}$. 4700 µF = 0.0047 F.
• Using 50 instead of $2\times 50$ in the ripple formula. A full bridge refills twice per cycle.
• Checking $V_{RRM}$ against 230 V instead of the 325 V peak.

Check yourself

1. In the positive half, why doesn't current go A → D3?

D3 only allows − → A. Going A → − would be against its arrow, so it blocks.

2. You double C. What happens to the ripple?

It halves, because $\Delta V\propto 1/C$.

3. A bridge with $V_F=0.9$ V carries 3 A RMS. How much heat does it make?

$1.8\times 0.9\times 3=4.86$ W.