Why rectifier ripple comes out smaller than (Vp)/(2fRC)
Using full-wave rectifier simulation results to see when the textbook estimate breaks down.
Full-wave rectifier ripple is usually estimated with this formula:
With a small smoothing capacitor, though, the ripple you measure or simulate comes out well below that. Here is why, using PSpice results for a center-tap full-wave rectifier (5 V secondary peak, 60 Hz, 1 kΩ load, D1N4004 diodes).
In this guide 5
1. The results
| C | Output peak (PSpice) | Ripple (PSpice) | Ripple (discharge model) | (Vp)/(2fRC) | Estimate error |
|---|---|---|---|---|---|
| C10 µF | Output peak (PSpice)4.35 V | Ripple (PSpice)1.85 V | Ripple (discharge model)1.88 V | (Vp)/(2fRC)3.63 V | Estimate error+96% |
| C22 µF | Output peak (PSpice)4.35 V | Ripple (PSpice)1.09 V | Ripple (discharge model)1.10 V | (Vp)/(2fRC)1.65 V | Estimate error+51% |
| C56 µF | Output peak (PSpice)4.44 V | Ripple (PSpice)0.53 V | Ripple (discharge model)0.52 V | (Vp)/(2fRC)647 mV | Estimate error+22% |
The estimate is almost double at 10 µF and still +22% at 56 µF. The discharge model explained below lands within 3% of PSpice in all three cases. You can run the same numbers in the rectifier ripple calculator.
2. What the estimate assumes
The estimate assumes the capacitor discharges for the whole half-cycle (8.33 ms for full-wave at 60 Hz) at a constant current. Both are nearly true when the ripple is tiny compared with the output.
3. A small capacitor breaks both
- The discharge is exponential: 22 µF × 1 kΩ is a 22 ms time constant, close to the half-cycle. As the voltage falls the load current falls too, so even a full half-cycle of discharge only takes it down to 68% of the peak, less than the straight-line estimate.
- Charging starts early: when the capacitor has sagged a lot, the next half-cycle catches up with it before its peak and starts recharging it. The real discharge time is shorter than a half-cycle.
At 22 µF the capacitor falls from 4.35 V to 3.25 V and then recharges, a ripple of 1.10 V. The estimate of 1.65 V ignores both effects.
Try it: change the capacitor and the load
The same rectifier (5 V secondary peak, 60 Hz, 0.65 V diode drop) through the discharge model. It starts at the 22 µF, 1 kΩ case above.
- Ripple (discharge model)
- 1.1 V
- Estimate
- 1.65 V
- Estimate error
- +50%
- Highest
- 4.35 V
- Lowest
- 3.25 V
- RC ÷ discharge time
- 2.64
Things to try: a larger C shrinks the ripple next to the output and the estimate error with it. Half-wave doubles the discharge time and the ripple; a smaller load resistance makes the estimate drift off again at the same C.
4. When the estimate is fine
Under the same conditions the estimate improves as the capacitor grows, and its error falls within 10% from about 250 µF, where the ripple is 3.0% of the output peak. In other words, the estimate is only trustworthy when the capacitor is big enough to keep the ripple to a few percent. With the tens of microfarads typical in a lab, compare against the discharge model or a simulation.
5. Example discussion for a report
"The smaller the capacitor, the further the measured ripple fell below (Vp)/(2fRC). The formula assumes a constant-current discharge over a full half-cycle, but here the RC time constant is close to the period, so the discharge is exponential and recharging starts early."
Check the average output and the capacitor you need in the rectifier ripple calculator, and the RC time constant in the RC filter calculator.