Bridge Rectifier Calculator
Enter your AC input voltage, line frequency, load resistance, filter capacitor, and diode forward voltage drop to instantly calculate every key parameter of a full-wave bridge rectifier: DC output voltage, ripple voltage, peak inverse voltage (PIV), load current, power dissipation, rectifier efficiency, and the recommended filter capacitor size. The show-your-work panel walks through each formula step by step.
How a bridge rectifier works
A full-wave bridge rectifier uses four diodes arranged in a bridge configuration to convert both halves of an AC sine wave into pulsating DC. During the positive half-cycle, two diodes conduct and current flows through the load in one direction. During the negative half-cycle, the other two diodes conduct but current still flows through the load in the same direction. The result is a pulsating DC signal at twice the input frequency (100 Hz for 50 Hz mains, 120 Hz for 60 Hz). Adding a filter capacitor across the output smooths this pulsation to a nearly steady DC voltage. The capacitor charges to the peak voltage during each pulse and discharges slowly into the load between pulses, so larger capacitance means less ripple.
Key formulas explained
The peak voltage is Vpeak = Vrms x 1.414 (sqrt of 2). Since two diodes conduct simultaneously in a bridge, the effective peak output is Vpeak_out = Vpeak - 2Vf, where Vf is the forward voltage of one diode. For an unfiltered bridge, the average DC output is Vdc = (2 / pi) x Vpeak_out, about 0.637 times the peak output. With a filter capacitor C at load current Idc and frequency f, the ripple voltage is approximately Vripple = Idc / (2 x f x C). The peak inverse voltage each diode must block equals Vpeak (not 2 x Vpeak as in half-wave or full-wave centre-tap circuits), which is one advantage of the bridge topology. Rectifier efficiency is the ratio of DC output power to total input power: efficiency = Pload / (Pload + 2 x Vf x Idc).
Choosing the filter capacitor
A larger filter capacitor reduces ripple but increases peak diode current, which can stress the diodes and transformer. The minimum capacitance needed to keep ripple voltage below a fraction r of Vdc is C = Idc / (2 x f x r x Vdc) = 1 / (2 x f x r x RL). For 1% ripple (r = 0.01) at 60 Hz with a 100 ohm load, you need about 833 uF. For 50 Hz the same load needs 1000 uF. This calculator shows the exact value for your inputs. Electrolytic capacitors should be rated for at least 1.5 times the peak output voltage. Place the capacitor as close to the load as possible and add a 0.1 uF ceramic capacitor in parallel to handle high-frequency transients.
Transformer and diode sizing guidelines
Because the capacitor draws current only during the peaks of the rectified waveform, the transformer must supply short high-current pulses rather than a steady current. This means the transformer RMS current is roughly 1.5 to 2 times the DC load current, and the transformer VA rating should be approximately 1.8 times the DC output power. For diodes, always derate the peak reverse voltage (PRV) rating by at least 1.5 times the calculated PIV to allow for mains voltage spikes and startup transients. For load currents above 1 A, attach the diodes (or the bridge module) to a heatsink: each diode dissipates Vf x Iavg_diode watts, where the average per diode is Idc / 2 in a bridge circuit.
Diode selection guide
| Diode type | Vf (V) | Max current | Best for |
|---|---|---|---|
| 1N4001-1N4007 (silicon) | 0.7 | 1 A | Low-cost general-purpose up to 50-1000 V PIV |
| 1N5400-1N5408 (silicon) | 0.7-1.0 | 3 A | Higher-current supplies |
| Schottky (e.g. 1N5819) | 0.3-0.45 | 1-40 A | Low-voltage, low-loss, high-frequency supplies |
| KBPC bridge module | 1.0-1.1 | 4-50 A | All-in-one bridge module for mains supplies |
| SiC Schottky | 1.2-1.5 | 1-60 A | High-efficiency switch-mode power supplies |
Typical forward voltage drops and use cases for common rectifier diode families.
Frequently asked questions
What is the output voltage of a bridge rectifier?
For an unfiltered bridge rectifier, the average DC output is (2 / pi) x (Vpeak - 2Vf), where Vpeak = Vrms x sqrt(2) and Vf is the diode forward voltage. That works out to about 0.637 x Vpeak_out. With a filter capacitor the output rises to nearly Vpeak_out, minus the ripple voltage. For a 12 V RMS transformer with 0.7 V silicon diodes, the unfiltered DC is about (2/pi) x (16.97 - 1.4) = 9.9 V, while a well-filtered output approaches 15.6 V.
What is the peak inverse voltage (PIV) of a bridge rectifier?
In a bridge rectifier, the PIV each diode must withstand equals the peak input voltage Vpeak, which is Vrms x sqrt(2). This is half the PIV required in a full-wave centre-tap configuration (which needs 2 x Vpeak). For a 12 V RMS source the PIV is about 17 V, so diodes rated at 25-50 V are typically used after applying a 1.5x safety derating.
How do I reduce ripple in a bridge rectifier?
The most effective method is to increase the filter capacitor value. Ripple voltage is approximately Vripple = Idc / (2 x f x C), so doubling C halves the ripple. Other options include increasing the supply frequency (switched-mode supplies use 50-200 kHz instead of 50-60 Hz, giving much smaller ripple for the same capacitor), using a pi filter (two capacitors with an inductor between them), or adding a linear voltage regulator such as the LM78xx series downstream.
Why does a bridge rectifier use four diodes instead of two?
A two-diode full-wave rectifier requires a centre-tapped transformer, and each diode sees twice the peak voltage, meaning more expensive higher-rated diodes. The bridge uses a standard (non-centre-tapped) transformer but needs four diodes. The two extra diodes are a small cost. The bridge is also more efficient because only one half-turn of the transformer winding is used at a time in the centre-tap design, while the bridge uses the full winding every half-cycle.
What is rectifier efficiency and what is the theoretical maximum?
Rectifier efficiency is the ratio of DC output power to total AC input power: n = Pdc / Pac = (Vdc x Idc) / ((Vdc + 2Vf) x Idc) = Vdc / (Vdc + 2Vf). For an ideal diode (Vf = 0) the efficiency of a full-wave bridge is 81.2%, set by the waveform shape alone. Real silicon diodes with Vf = 0.7 V reduce efficiency somewhat more. This is the efficiency of the rectification process only, not the overall power supply efficiency.
How do I size the transformer for a bridge rectifier power supply?
First determine the required DC output voltage and load current. The required transformer RMS voltage is approximately Vac = (Vdc + 2Vf) / (sqrt(2) x (1 - ripple/2)), but in practice you can estimate Vrms = (Vdc + 2Vf + Vripple/2) / sqrt(2). The transformer must supply not just the average load current but the high peak charging pulses to the capacitor. Use a transformer VA rating of approximately 1.8 times the DC output power (watts) as a starting point, and verify with the transformer datasheet.