Bridge Rectifier Calculator

Find the peak and DC output voltage, load current, and ripple voltage of a full-wave bridge rectifier from the AC input voltage, diode forward drop, load resistance, and filter capacitance.

Quick Facts

Diode drop
Two diodes conduct at once in a bridge
Peak output = peak input − 2 × diode forward drop (about 1.4 V total for silicon).
Ripple frequency
Twice the AC line frequency
A bridge rectifies both halves of the cycle, so the filter capacitor is recharged 2× per input cycle.

Your Results

Calculated
Peak output voltage
-
After 2 diode drops
DC (average) output voltage
-
2 × Vpeak / π
DC load current
-
Vdc / Rload
Ripple voltage (peak-to-peak)
-
Idc / (2 × f × C)

Ready

Enter the AC input, diode drop, load, and filter capacitance, then press Calculate.

About the Bridge Rectifier

A bridge rectifier converts alternating current (AC) into pulsating direct current (DC) using four diodes arranged in a diamond ("bridge") pattern. Unlike a half-wave rectifier, which throws away half of every AC cycle, a bridge rectifier uses both halves of the waveform, so it is far more efficient and produces a smoother, higher-average output for the same transformer. This calculator converts an AC RMS voltage, diode drop, load resistance, and filter capacitance into the peak output voltage, DC output voltage, load current, and ripple voltage you would measure at the load.

Understanding the formula

The AC input voltage is normally specified as an RMS value, so the peak of that waveform is Vpeak,in = Vrms × √2. In a bridge rectifier, current always passes through two diodes in series (one on each side of the bridge), so the output peak loses two forward-voltage drops: Vpeak,out = Vpeak,in − 2 × Vf. For a full-wave rectified sine feeding a resistive load, the average (DC) output voltage is Vdc = 2 × Vpeak,out / π ≈ 0.637 × Vpeak,out. Dividing that by the load resistance gives the DC load current, Idc = Vdc / Rload. If a filter (smoothing) capacitor is connected across the load, the peak-to-peak ripple voltage is approximated by Vripple = Idc / (2 × f × C), where f is the AC line frequency and the factor of 2 accounts for the bridge recharging the capacitor twice per input cycle.

Working with units

  • Enter the AC input as an RMS voltage (the value a multimeter or a transformer's rated secondary voltage normally reports), not a peak value.
  • Diode forward drop is typically about 0.7 V per silicon diode and about 0.3 V per Schottky or germanium diode; the calculator uses whatever value you enter for each of the two conducting diodes.
  • Filter capacitance is entered in microfarads (µF); the calculator converts it internally to farads for the ripple formula.

Knowing the limits

This calculator uses idealized diode and capacitor models: it assumes matched diodes with a fixed forward drop (not a full diode I-V curve), a purely resistive load, and the standard small-ripple approximation for the filter capacitor, which is most accurate when the ripple is small relative to the DC output. It does not model transformer winding resistance, diode reverse-recovery losses, or surge current at power-up. For precision power-supply design, treat these results as a solid first-pass estimate and verify with a full circuit simulation or bench measurement.

Frequently Asked Questions

Why does a bridge rectifier lose two diode drops instead of one?
In a full-wave bridge, current always flows through two diodes in series on its way from the AC source to the load and back, no matter which half of the AC cycle is active. Each silicon diode drops about 0.7 V when conducting, so the peak output voltage is the peak input voltage minus 2 times the diode drop, roughly 1.4 V total for silicon diodes.
What is the formula for the DC output voltage of a bridge rectifier?
For a full-wave rectified sine wave feeding a resistive load, the average (DC) output voltage is Vdc = 2 × Vpeak / π, where Vpeak is the peak voltage after the diode drops. This is roughly 63.7% of the peak voltage, versus about 31.8% for a half-wave rectifier.
How is ripple voltage calculated for a bridge rectifier with a filter capacitor?
The standard approximation is Vripple(peak-to-peak) = Idc / (2 × f × C), where Idc is the DC load current, f is the AC line frequency, and C is the filter capacitance. The factor of 2 appears because a full-wave bridge rectifies both halves of the AC cycle, so the capacitor is recharged twice per input cycle instead of once.
How do I reduce ripple in a bridge rectifier circuit?
Increase the filter capacitance, reduce the load current (increase load resistance), or both, since ripple voltage is directly proportional to load current and inversely proportional to capacitance and frequency. Doubling the capacitor size roughly halves the peak-to-peak ripple for the same load.

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