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 V)/(2 × 60 Hz × 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 |
|---|---|---|---|---|
| C10 µF | Ripple (this calculator)1.88 V | (Vp)/(2fRC)3.63 V | Estimate error+93% | Average output3.47 V |
| C22 µF | Ripple (this calculator)1.1 V | (Vp)/(2fRC)1.65 V | Estimate error+50% | Average output3.83 V |
| C56 µF | Ripple (this calculator)515 mV | (Vp)/(2fRC)647 mV | Estimate error+26% | Average output4.1 V |
| C100 µF | Ripple (this calculator)309 mV | (Vp)/(2fRC)363 mV | Estimate error+17% | Average output4.2 V |
| C220 µF | Ripple (this calculator)149 mV | (Vp)/(2fRC)165 mV | Estimate error+11% | Average output4.28 V |
| C470 µF | Ripple (this calculator)72.2 mV | (Vp)/(2fRC)77.1 mV | Estimate error+7% | Average output4.31 V |
| C1 mF | Ripple (this calculator)34.7 mV | (Vp)/(2fRC)36.3 mV | Estimate error+4% | Average output4.33 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.
Checked on our bench: the diode drop isn't fixed at 0.65 V
The numbers below are computed from values the site owner recorded in lab classes. Measured values are left exactly as recorded.
This calculator treats the diode as a fixed drop Vd. Here it is against a half-wave rectifier (no capacitor) driven with a sine in the lab, with the input and output peaks measured.
| Input | Input peak (measured) | Calculator (Vd 0.65 V) | Output peak (measured) | Actual drop |
|---|---|---|---|---|
| Input5 V sine | Input peak (measured)5.065 V | Calculator (Vd 0.65 V)4.415 V | Output peak (measured)4.240 V | Actual drop0.825 V |
| Input0.5 V sine | Input peak (measured)0.488 V | Calculator (Vd 0.65 V)0 V | Output peak (measured)0.040 V | Actual drop0.448 V |
- 5 V input: the real drop was 0.825 V, 0.18 V more than the 0.65 V default. At the peak the diode carries its largest current, so it drops more. Typing about 0.8 V into Vd matches this measurement.
- 0.5 V input: a fixed-drop model passes nothing below 0.65 V, yet 40 mV came out. A diode follows an exponential curve rather than switching, so a little current flows even at low voltage (see the cut-in voltage guide).
How to use that: if the output voltage matters, try Vd at 0.75 to 0.85 V as well as 0.65 V and design for both. The ripple and capacitor size hardly change. The PSpice comparison is in the rectifier ripple guide.
Building a rectifier
From transformer to smoothed DC
- Choose half-wave, centre-tapped full-wave or bridge, and enter the transformer's AC output voltage and the mains frequency (50 or 60 Hz).
- Enter the load current and the smoothing capacitor to get the DC output and the peak-to-peak ripple.
- Or enter the ripple you can accept and read the capacitor you need.
Where rectifier estimates go wrong
- Using the RMS voltage as the DC output. The capacitor charges to the peak, about 1.41 times the RMS voltage, minus the diode drops. A 12 V transformer gives roughly 15 V DC after a bridge at light load.
- Forgetting the second diode drop. Current passes through two diodes in a bridge, so about 1.4 V is lost; a centre-tapped full-wave rectifier loses only one drop.
- A capacitor with no voltage margin. Transformers put out more than their rated voltage at light load and when the mains is high. Choose a capacitor rated at least 25 to 50% above the peak.
Ripple and inrush in practice
The ripple formula assumes the capacitor discharges at a constant current for the whole cycle, so it slightly overestimates ripple; our rectifier ripple guide compares it with measurements. Large capacitors draw a big inrush current at switch-on and high ripple current in operation, so check the capacitor's ripple current rating. If the load needs a steady voltage, follow the capacitor with a regulator and make sure the lowest point of the ripple stays above the regulator's dropout.
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.