Rectifier Calculator: Average Output, Ripple Factor, Efficiency and PIV for Half-Wave, Full-Wave, Bridge and Three-Phase

Get the output voltage, ripple, efficiency and diode ratings of a single-phase or three-phase rectifier with a resistive load.

Calculated from standard textbook formulas, with every step shown. The formulas are fixed, so the working does not need checking: what matters is the numbers you type in. Use these calculators only if you are competent to choose the inputs (loads, material strength, dimensions, supports) and check your units and assumptions. For learning and checking your own working: design and sign-off are for a qualified engineer. Drafting Central does not store the numbers you type.

How this calculator works

Choose the rectifier, enter the rms input voltage, the load resistance and, if you want it, the diode voltage drop. For the controlled types enter the firing angle.

The calculator gives the peak voltage, the average (dc) and rms output voltage, the load current and power, the form factor, the ripple factor and the rectification efficiency, the diode peak inverse voltage and average current, and the ripple frequency.

Enter a smoothing capacitor to estimate the peak-to-peak ripple. A firing angle above 90 degrees on a fully controlled converter gives a negative dc voltage: the converter works as an inverter.

Formulas used

  • V_{dc} = \frac{V_m}{\pi}\ \ (\text{half-wave})
  • V_{dc} = \frac{2V_m}{\pi}\ \ (\text{full-wave})
  • V_{dc} = \frac{2V_m\cos\alpha}{\pi}\ \ (\text{full controlled})
  • \mathrm{RF} = \sqrt{\left(\frac{V_{rms}}{V_{dc}}\right)^2 - 1}
  • \Delta V = \frac{I_{dc}}{f_r C}

Worked example

With the default numbers in the calculator above:

  1. Peak input voltage: 325.27 V
  2. Average (dc) and rms output voltage: Vdc = 207.07 V; Vrms = 230.00 V
  3. Average load current and dc power: 4.14 A and 857.58 W
  4. Form factor and ripple factor: Form factor 1.11, ripple factor 0.48
  5. Rectification efficiency (dc power over the total output power): 81.06 %
  6. Peak inverse voltage across each diode: 325.27 V (allow a margin of at least 1.5 times)
  7. Average current in each diode: 2.07 A
  8. Ripple frequency in the output: 100.00 Hz

Change the numbers in the calculator to see the full working, every step, for your own case.

A second worked example

A different case, with these numbers: Rectifier Single-phase half-wave.

  1. Peak input voltage: 325.27 V
  2. Average (dc) and rms output voltage: Vdc = 103.54 V; Vrms = 162.63 V
  3. Average load current and dc power: 2.07 A and 214.40 W
  4. Form factor and ripple factor: Form factor 1.57, ripple factor 1.21
  5. Rectification efficiency (dc power over the total output power): 40.53 %
  6. Peak inverse voltage across each diode: 325.27 V (allow a margin of at least 1.5 times)
  7. Average current in each diode: 2.07 A
  8. Ripple frequency in the output: 50.00 Hz

Common questions

What is the ripple factor?

A measure of how much ac remains in the dc output: the rms of the ac part over the dc value. A half-wave rectifier is 1.21, a full-wave 0.482 and a three-phase bridge about 0.04.

Why is a bridge better than a centre-tap?

The bridge needs no centre-tap, so the diodes only see the peak voltage Vm, where the centre-tap needs diodes rated for twice the peak.

What is the efficiency of a half-wave rectifier?

Ideally 40.6 percent with a resistive load. A full-wave rectifier reaches 81.2 percent.

What does the firing angle do?

It delays the turn-on of the thyristors, which lowers the average output voltage, so the dc voltage can be controlled.

Related calculators

Got a question about this? Ask an engineering question and get a plain-language answer with references.

Further reading: Rashid, Power Electronics: Circuits, Devices and Applications; Mohan, Undeland and Robbins, Power Electronics.

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