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Full-Wave Bridge Rectifier Calculator

Calculate output voltage, ripple factor, and peak inverse voltage for bridge rectifier circuits. Optimize your power supply design.

⚡ Calculate Bridge Rectifier

RMS voltage from transformer secondary
50Hz (Europe) or 60Hz (US)
×2 for bridge (two diodes conduct)
Expected DC load current
Smoothing capacitor value

📊 Bridge Rectifier Output

Peak Voltage
15.6 V
DC Output
14.6 V
Ripple Frequency
100 Hz
Ripple Voltage
1.0 V
Min Capacitor
100 µF
Efficiency
81.2%

📝 Detailed Calculation Steps

1 Peak Voltage (Vpeak)
Vpeak = √2 × Vrms − 2×Vf
Note: 2×Vf because two diodes conduct at any time in a bridge configuration.
2 Ripple Frequency (fripple)
fripple = 2 × finput
Full-wave rectification produces ripple at twice the input frequency.
Without Capacitor Filter
3a DC Output Voltage (No Filter)
VDC = 2×Vpeak / π ≈ 0.636 × Vpeak
Without a filter capacitor, the DC output is the average of full-wave pulses.
4a Ripple Voltage (No Filter)
Vripple = γ × VDC (γ = 0.48 for full-wave)
Ripple factor (γ) = 0.48 for unfiltered full-wave rectification.
With Capacitor Filter
3b Ripple Voltage (With Filter)
Vripple = Iload / (fripple × C)
4b DC Output Voltage (With Filter)
VDC = Vpeak − (Vripple / 2)
5 Minimum Capacitor for 10% Ripple
Cmin = Iload / (fripple × 0.1×Vpeak)
6 Maximum Efficiency (η)
ηmax = 81.2% (theoretical maximum for full-wave)
The bridge rectifier utilizes both halves of the AC cycle, resulting in higher efficiency compared to half-wave rectifiers.

Bridge Rectifier Circuit

Vin D3 D1 D2 D4 Vout RL

The bridge rectifier uses four diodes in a diamond configuration. During each half-cycle of the AC input, two diodes conduct, providing full-wave rectification without needing a center-tapped transformer.

📈 Input & Output Waveforms

t V AC Input DC Output

Notice how the output signal (green) uses both halves of the input AC waveform, doubling the ripple frequency and improving efficiency compared to half-wave rectification.

📐 Bridge Rectifier Formulas

VDC = 2×Vpeak π ≈ 0.636×Vpeak
Average DC output voltage (without filter)

Peak Voltage

Vp = √2 × Vrms − 2Vf

Ripple Frequency

fr = 2 × fin

Ripple Voltage

Vr = I ÷ (2 × f × C)

Filter Capacitor

C = I ÷ (2 × f × Vr)

Advantages of Bridge Rectifier

  • Higher Efficiency: Maximum theoretical efficiency of 81.2% - double that of half-wave
  • Lower Ripple: Ripple frequency is 2× input frequency, easier to filter
  • Full AC Utilization: Uses both halves of the AC cycle
  • No Center Tap: Works with any transformer, not just center-tapped
  • Smaller Filter: Needs smaller filter capacitor for same ripple

PIV (Peak Inverse Voltage)

Each diode in a bridge rectifier must withstand a PIV equal to the peak voltage:

PIV = Vpeak

⚖️ Rectifier Comparison

Parameter Half-Wave Bridge Center Tap
Diodes Required 1 4 2
Max Efficiency 40.6% 81.2% 81.2%
Ripple Frequency fin 2×fin 2×fin
Transformer Standard Standard Center-tapped
Voltage Drop 1×Vf 2×Vf 1×Vf
PIV per Diode Vpeak Vpeak 2×Vpeak

🔗 Related Calculators

Diode Bias Calculator Half-Wave Rectifier Center Tap Rectifier