Bridge Rectifier Calculator
Estimate AC-to-DC rectifier voltage, ripple frequency, diode stress, and conduction loss for half-wave and full-wave bridge circuits.
The tool converts RMS voltage to peak voltage, accounts for conducting diode drops, and reports average unfiltered output, PIV, and ideal rectification efficiency.
Use worst-case transformer regulation, line tolerance, load current, diode temperature, and filter-capacitor charging current for final component selection.
Engineering tool
Bridge Rectifier Calculator
Estimate rectified peak and average DC voltage, ripple frequency, PIV, diode loss, and ideal conversion efficiency.
Transformer secondary or AC source RMS voltage.
Input line or source frequency.
Forward drop of one conducting diode.
Estimated average DC load current.
Peak DC output
15.570563 V
Vdc peak = Vrms × √2 - conducting diode drops
Result console
- Peak AC voltage
- 16.970563V
- Peak DC voltage
- 15.570563V
- Average DC output
- 9.912528V
- Ripple frequency
- 100Hz
- Peak inverse voltage (PIV)
- 16.970563V
- Diode power loss
- 700mW
- Total rectifier voltage drop
- 1.4V
- Rectifier efficiency (ideal)
- 81.2%
Formula reference
Rectifier Formulas
A half-wave path contains one conducting diode. A full-wave bridge contains two conducting diodes during each half-cycle. Average values describe an unfiltered resistive-load waveform.
Peak voltage: Vp = Vrms × √2Rectifier drop: Vdrop = n × VfPeak DC: Vdc_peak = max(0, Vp - Vdrop)Half-wave average: Vavg = Vdc_peak / πBridge average: Vavg = 2 × Vdc_peak / πRipple: fripple = f (half-wave) or 2f (bridge)Diode loss: Ploss = Vdrop × IloadPIV = VpVariable definitions
- n
- 1 for half-wave or 2 for bridge
- Vf
- Forward voltage of one diode
- Vrms
- AC RMS input
- Iload
- Average load current
Worked Example
12 Vrms, 50 Hz Full-wave Bridge
Vp = 12 × √2 ≈ 16.97 V
Vdrop = 2 × 0.7 = 1.4 V
Vdc_peak = 16.97 - 1.4 ≈ 15.57 V
Vavg = 2 × 15.57 / π ≈ 9.91 V
Ripple frequency = 2 × 50 = 100 Hz
Diode path loss = 1.4 × 0.5 = 0.7 W
Engineering Notes
Two diodes conduct
A bridge current path includes two forward-biased diodes during every half-cycle.
Ripple frequency doubles
Full-wave rectification produces output pulses at twice the AC input frequency.
Transformer regulation matters
Winding resistance and regulation reduce secondary voltage as load current rises.
Current creates heat
Large load and capacitor charging currents increase diode conduction loss and junction temperature.
Capacitors approach the peak
A reservoir capacitor raises DC output toward peak voltage but creates ripple and narrow current pulses.
Support reference
FAQ
How does a bridge rectifier work?
A four-diode bridge routes both AC half-cycles through the load in the same direction, producing full-wave pulsating DC without requiring a center-tapped transformer.
Why do two diodes conduct in a bridge rectifier?
Each half-cycle uses one diode to carry current from the active AC terminal to DC positive and another to return current from DC negative to the opposite AC terminal.
Why is DC output lower than the AC peak voltage?
The conducting diode path subtracts forward voltage from the AC peak. Transformer regulation, winding resistance, ripple, and load current reduce practical output further.
What is PIV in a rectifier?
Peak inverse voltage is the maximum reverse voltage a non-conducting diode must withstand. Select a diode with adequate repetitive reverse-voltage margin above the calculated value.
How do filter capacitors affect rectifier output voltage?
A reservoir capacitor charges near the rectified peak and supplies the load between peaks, raising average DC output while introducing ripple and higher charging-current pulses.
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