Rectifier Ripple Calculator (Half-Wave, Full-Wave, Bridge)
Rectify AC with diodes, smooth it with a capacitor, and get the output voltage, ripple and the capacitor you need, worked out cycle by cycle.
For a center-tap rectifier, enter one half of the secondary (measured from the center tap). Use 0 for C if there is no capacitor.
Output waveform
How it is calculated
The textbook Vr ≈ Vp / (2fRC) (or Vp / (fRC) for half-wave) only holds when the ripple is much smaller than the output. This calculator follows the capacitor as it discharges exponentially into the load until the rectified wave rises to meet it and recharges it, so it stays within a few percent of PSpice even with a small capacitor and large ripple. Real circuits peak a little lower because of transformer winding resistance and diode current.
Worked example: full-wave with 22 µF
- The output peak is the input peak minus one diode drop: 5 − 0.65 = 4.35 V
- Estimate: Vr ≈ Vp / (2fRC) = 4.35 / (2 × 60 × 1 kΩ × 22 µF) = 1.65 V
- In reality the capacitor only falls to 3.252 V before it recharges, so the ripple is 1.1 V. The estimate is 50% too high.
Ripple by capacitor size
| C | Ripple (this calculator) | Vp/(2fRC) | Estimate error | Average output |
|---|---|---|---|---|
| 10 µF | 1.88 V | 3.63 V | +93% | 3.47 V |
| 22 µF | 1.1 V | 1.65 V | +50% | 3.83 V |
| 56 µF | 515 mV | 647 mV | +26% | 4.1 V |
| 100 µF | 309 mV | 363 mV | +17% | 4.2 V |
| 220 µF | 149 mV | 165 mV | +11% | 4.28 V |
| 470 µF | 72.2 mV | 77.1 mV | +7% | 4.31 V |
| 1 mF | 34.7 mV | 36.3 mV | +4% | 4.33 V |
Center-tap full-wave, 5 V secondary peak, 0.65 V diode, 60 Hz, RL = 1 kΩ. A PSpice simulation of the same circuit (D1N4004 diodes) gave 1.85, 1.09 and 0.53 V of ripple at 10, 22 and 56 µF.
Frequently asked questions
Why is my measured ripple smaller than Vp/(2fRC)?
That formula assumes the capacitor discharges at a constant current for a whole half-cycle. When the ripple is large, the capacitor voltage drops exponentially and the rectified wave catches it early, so the real ripple is smaller. With 22 µF, 1 kΩ and 60 Hz the estimate is about 50% too high.
Center-tap or bridge?
A center-tap rectifier uses two diodes and loses one diode drop, but needs a transformer with a center tap and each diode sees about twice the peak voltage. A bridge works with any transformer winding and loses two diode drops.