Bridge Rectifier

The bridge rectifier converts both halves of an AC waveform into DC, doubling the output compared to a half-wave circuit. Every phone charger, laptop adapter, and DC power supply uses this four-diode arrangement as its first stage.

Core Concept

Four diodes arranged in a diamond allow current to flow in the same direction through the load during both halves of the AC cycle. During the positive half-cycle, D1 and D2 conduct. During the negative half-cycle, D3 and D4 conduct. The load always sees current flowing in the same direction.

The peak inverse voltage (PIV) across each non-conducting diode in a bridge rectifier equals the peak secondary voltage Vm. This is half the PIV required in a center-tap full-wave rectifier, which is why the bridge circuit is preferred even though it uses twice as many diodes. Four 1N4007 diodes (PIV = 1000 V each) handle mains-derived supplies comfortably.

The ripple frequency at the output is twice the input frequency. For a 50 Hz AC supply, the ripple is at 100 Hz. This higher frequency makes the capacitor filter much more effective, which is why bridge rectifiers produce smoother DC with smaller filter capacitors than half-wave circuits of the same power rating.

Key Equations

Peak output voltage: V_m = Vm_secondary - 2 * V_D — two diodes conduct at once, each dropping 0.7 V.

Average (DC) output voltage: V_dc = (2 * Vm) / pi = 0.636 * Vm

RMS output voltage: V_rms = Vm / sqrt(2) = 0.707 * Vm

Ripple factor: r = 0.482 (for resistive load, no filter).

Form factor: FF = V_rms / V_dc = 1.11

PIV per diode: PIV = Vm (peak secondary voltage, not 2*Vm as in center-tap).

Efficiency: eta = 81.2% (theoretical maximum for full-wave rectifier).

Given:
Transformer secondary: 230V AC RMS, 50 Hz
Diodes: 1N4007 (V_D = 0.7 V each)
Load: RL = 500 Ω

Why this formula:
Bridge rectifier conducts two diodes at a time, so deduct 2 x 0.7 V from peak.

Formula:
Vm_secondary = 230 * sqrt(2) = 325.3 V (peak)
V_m_out = Vm_secondary - 2 * 0.7 = 325.3 - 1.4 = 323.9 V

DC output voltage:
V_dc = 0.636 * V_m_out
V_dc = 0.636 * 323.9 = 205.9 V

Load current:
I_dc = V_dc / RL = 205.9 / 500 = 0.412 A = 412 mA

PIV per diode:
PIV = Vm_secondary = 325.3 V
1N4007 rated 1000 V, so safe.

Ripple factor:
r = 0.482 (standard result, no capacitor)

Final Answer:
V_dc = 205.9 V, I_dc = 412 mA, PIV = 325.3 V per diode

Exam Tip: In a bridge rectifier, two diodes conduct simultaneously, so the voltage drop is 2 * 0.7 = 1.4 V, not 0.7 V. Also, PIV per diode = Vm (not 2*Vm). Students often confuse this with the center-tap full-wave rectifier, where PIV = 2*Vm. The ripple frequency is 2f (100 Hz for 50 Hz input). These three numbers are the most frequently tested: 1.4 V drop, PIV = Vm, ripple = 2f.

Key Properties

• Two diodes conduct per half-cycle. Total diode forward voltage drop = 2 * 0.7 = 1.4 V for silicon diodes like 1N4007.
• PIV per diode = Vm (peak secondary voltage). This is lower than the center-tap circuit, so lower-rated diodes can be used.
• Ripple frequency = 2 * f_supply = 100 Hz for 50 Hz mains. Easier to filter than 50 Hz ripple.
• DC output: V_dc = 0.636 * Vm (without filter). Efficiency: 81.2% theoretical maximum.
• Ripple factor without filter: 0.482. This reduces to under 0.05 with a properly sized capacitor filter.
• No center tap is needed on the transformer secondary, unlike the center-tap full-wave rectifier. This simplifies the transformer design.
• Bridge rectifier modules like the W04G (400 V, 1.5 A) integrate all four diodes in a single package for compact designs.

Quick Revision

• Bridge rectifier uses 4 diodes in diamond configuration.
• Both half-cycles appear at output. Ripple frequency = 2 * f_in.
• V_dc = 0.636 * Vm. V_rms = 0.707 * Vm.
• Two diodes conduct at a time: subtract 2 * 0.7 = 1.4 V from Vm.
• PIV per diode = Vm. PIV in center-tap = 2*Vm.
• Efficiency = 81.2%. Ripple factor = 0.482 (no filter).
• No center-tap transformer needed. Uses more diodes but simpler transformer.
• Exam trap: Subtracting only 0.7 V (one diode drop) instead of 1.4 V gives a wrong peak output voltage. Always subtract two diode drops in a bridge circuit.