ElectroHub

Rectifier Calculator

Rectifiers turn AC into pulsating DC; a reservoir capacitor smooths it. The numbers that matter are the average output, how much ripple is left, and the peak inverse voltage each diode must survive.

Rectifier

Average DC, RMS, ripple factor, efficiency and PIV for the three classic rectifier circuits. Add a load resistor and a smoothing capacitor to estimate the residual ripple.

Average DC V_dc
Output RMS V_rms
Ripple factor (no filter)
Rectification efficiency
Peak inverse voltage (PIV)
Load DC current I_dc
Ripple V_r (peak-peak)
Ripple factor (with cap)
Output waveform
V_dc
  1. 1.Given (Full-wave bridge): V_p = 17 V
  2. 2.FormulaV_dc = 2·V_p ÷ πSubstituteV_dc = 2 × 17 ÷ πResultV_dc = 10.8225 V
  3. 3.Ripple factor = 0.482, rectification efficiency ≈ 81.2 %, PIV = 17 V
  4. 4.FormulaLoad DC current I_dc = V_dc ÷ R_LSubstituteI_dc = 10.8225 ÷ 1000ResultI_dc = 10.8225 mA
  5. 5.FormulaCapacitor ripple V_r(pp) = I_dc ÷ (f_r·C)SubstituteV_r(pp) = 0.0108 ÷ (100 × 470e-6)ResultV_r(pp) = 230.2667 mV
Common trap: A full-wave rectifier ripples at twice the line frequency (100 Hz on a 50 Hz supply), which is why the same capacitor smooths it far better than a half-wave circuit. And watch V_p vs V_rms — the 230 V mains is 325 V peak, which sets the PIV your diodes must block.

The formula

Half-wave: V_dc = V_pk/π ; Full-wave: V_dc = 2V_pk/π ; Ripple: V_r ≈ I_dc / (f_ripple × C)

V_pk
peak of the secondary waveform, less diode drops (volts)
V_dc
average output without smoothing (volts)
V_r
peak-to-peak ripple with a reservoir capacitor (volts)
PIV
peak inverse voltage each diode must withstand

Worked example

A 12 V RMS secondary feeding a bridge rectifier.

  1. V_pk = 12 × √2 = 16.97 V
  2. A bridge puts two diodes in series: 16.97 − 1.4 = 15.57 V peak
  3. Unsmoothed average: V_dc = 2 × 16.97 / π = 10.8 V

About 15.6 V peak with a reservoir capacitor, or 10.8 V average without one.

Where you'll use it

Linear power supply design and the standard comparison question between half-wave, full-wave centre-tapped and bridge circuits. A bridge doubles the ripple frequency to 100 Hz on a 50 Hz supply, so it needs half the smoothing capacitance for the same ripple.

The laws behind it

Parts this applies to

Resistor (fixed)Potentiometer / Rheostat / TrimmerResistor Array / NetworkThermistor (NTC / PTC)Varistor / MOVSupercapacitorLED (light-emitting diode)MOSFET

Related calculators

RMS, Average and Form Factor CalculatorSine wave: V_rms = V_peak / √2Transformer Turns Ratio CalculatorN_p / N_s = V_p / V_s = I_s / I_pZener Diode Regulator CalculatorR_s = (V_in − V_z) / (I_load + I_z(min))Capacitor and Inductor Stored Energy CalculatorE = ½ × C × V² (capacitor)