How much ripple is acceptable?

For most linear regulators the input ripple should be less than the dropout voltage. As a general rule 5 percent ripple is acceptable for most applications. Sensitive analog circuits may need less than 1 percent.

Why does full-wave have less ripple than half-wave?

Full-wave rectification charges the capacitor twice per mains cycle instead of once. The capacitor has half the time to discharge between charging pulses so the voltage drops less. Ripple is halved for the same capacitance.

How do I reduce ripple?

Increase the filter capacitance, use full-wave instead of half-wave rectification, reduce the load current, or add a voltage regulator after the filter capacitor. A linear regulator rejects ripple by its power supply rejection ratio.

What size filter capacitor do I need?

Use the formula C equals I divided by n times f times delta V. For 1A load at 50Hz full-wave with 1V ripple: C equals 1 divided by 2 times 50 times 1 equals 10000 microfarads. Switch to Find Capacitance mode in the calculator.

Does ESR affect ripple?

Yes. The capacitor ESR adds a resistive ripple component on top of the capacitive ripple. At high frequencies ESR dominates. Use low-ESR capacitors for switching power supplies. For ESR analysis use the ESR Capacitor Calculator.

Capacitor Ripple Voltage Calculator – Filter Cap Sizing & Ripple

ΔV = I / (n × f × C)

Calculate ripple voltage or find minimum capacitance for a target ripple.

Calculate Ripple Find Capacitance

I Load Current

C Filter Capacitance

f Mains Frequency

n Rectifier Type

V_dc DC Output Voltage (optional — for % ripple)

Ripple Analysis

Ripple Voltage
--
peak-to-peak

Ripple %

Ripple Frequency

Ripple Voltage on a Filter Capacitor

After rectification, the capacitor charges to the peak voltage and discharges between peaks under load. The ripple voltage is the sawtooth variation — the difference between the peak and the minimum. Larger capacitance or lower load current reduces ripple.

ΔV — Peak-to-peak ripple voltage. The sawtooth variation on the DC output.
I_load — DC load current. More current drains the capacitor faster between peaks.
C — Filter capacitance. Larger C stores more charge and reduces ripple.
n × f — Ripple frequency. Full-wave rectifier doubles the mains frequency (100 Hz at 50 Hz mains).

Capacitor Ripple Voltage Calculator

After a rectifier converts AC to pulsating DC, a filter capacitor smooths the output. But the smoothing is never perfect — under load, the capacitor discharges between rectification peaks, producing a sawtooth variation called ripple. Too much ripple causes audible hum in audio circuits, unstable operation in digital systems, and noise in sensor readings. This calculator finds the ripple voltage for a given capacitor, or finds the minimum capacitance for a target ripple.

What Is Ripple Voltage?

The peak-to-peak variation in the DC output caused by the capacitor charging and discharging each half-cycle (full-wave) or full cycle (half-wave). It is the difference between the peak voltage (when the capacitor charges) and the minimum voltage (just before the next charging pulse). Lower ripple means a cleaner, more stable DC output.

The Formula

ΔV = I_load / (n × f × C)

I_load = DC load current (A)
n = 1 (half-wave) or 2 (full-wave)
f = mains frequency (Hz) — 50 Hz UK/EU, 60 Hz US
C = filter capacitance (F)

Ripple % = (ΔV / V_dc) × 100%

This is the linear approximation, valid when the ripple is small compared to the DC voltage (which is the normal operating condition for a well-designed supply). For the exact exponential discharge between peaks, use the Capacitor Discharge Calculator.

Half-Wave vs Full-Wave

Half-wave (n = 1) — One diode. Capacitor charges once per mains cycle. Ripple frequency equals mains frequency (50 Hz). More ripple for the same capacitance. Used in simple, low-current supplies.

Full-wave bridge (n = 2) — Four diodes. Capacitor charges twice per mains cycle. Ripple frequency is double the mains (100 Hz). Half the ripple for the same capacitance. Standard for nearly all power supplies.

5V / 1A Full-Wave PSU (2200µF, 50 Hz)

ΔV = 1 / (2 × 50 × 0.0022) = 1 / 0.22 = 4.55 V
Ripple % = 4.55 / 5 × 100 = 91%

91% ripple is far too high — the "DC" output swings from 5V down to 0.45V each cycle. 2200 µF is nowhere near enough for 1A at 5V. You need at least 10000 µF to get the ripple below 1V (ΔV = 1/(2 × 50 × 0.010) = 1.0V = 20%). In practice, a linear regulator after the capacitor absorbs the remaining ripple. For the regulator efficiency impact, see the Electrical Efficiency Calculator.

12V / 2A Full-Wave PSU (4700µF, 50 Hz)

ΔV = 2 / (2 × 50 × 0.0047) = 2 / 0.47 = 4.26 V
Ripple % = 4.26 / 12 × 100 = 35.5%

35.5% ripple. Still high for a raw unregulated supply, but a 12V linear regulator (e.g. 7812) with a 15V transformer can tolerate this because the minimum voltage (12 − 4.26 = 7.74V... wait, the DC voltage is higher than 12V from the transformer). The real design approach: choose a transformer with enough headroom for the ripple plus dropout voltage. The capacitor just needs to keep the minimum above the regulator's dropout.

Finding Capacitor Size (500 mA, 0.5V Target Ripple)

C = I / (n × f × ΔV)
C = 0.5 / (2 × 50 × 0.5) = 0.5 / 50 = 10000 µF

10000 µF for 0.5V ripple at 500 mA — a large electrolytic capacitor. This is why most power supplies use a regulator rather than trying to achieve low ripple with capacitance alone. The capacitor only needs to keep the voltage above the regulator's minimum input.

How to Reduce Ripple

Increase capacitance — Most direct method. Doubling C halves ripple. But large capacitors are expensive and take up board space.

Use full-wave rectification — Halves ripple compared to half-wave for free (just 4 diodes instead of 1).

Add a voltage regulator — A linear regulator rejects ripple by its PSRR (power supply rejection ratio). A 7805 has ~60 dB PSRR at 120 Hz, reducing 1V ripple to ~1 mV at the output.

Use a switching regulator — Switch-mode PSUs regulate output tightly regardless of input ripple, as long as the input stays within the operating range.

ESR and Real-World Ripple

The formula assumes an ideal capacitor. Real capacitors have equivalent series resistance (ESR) that adds a resistive ripple component: V_ESR = I_peak × ESR. At low frequencies (50–100 Hz rectifier ripple), the capacitive component dominates. At high frequencies (switching power supply ripple at 100 kHz+), ESR dominates. Use low-ESR capacitors for switching supplies. For ESR analysis, see the ESR Capacitor Calculator.

Frequently Asked Questions

What is acceptable ripple voltage?

Depends on the application. For a linear regulator input, the ripple just needs to stay within the dropout margin (typically 2–3V). For direct use without regulation, below 5% is good, below 1% is excellent. Audio circuits need very low ripple to avoid hum.

Why is full-wave better than half-wave?

Full-wave charges the capacitor twice per mains cycle instead of once. The capacitor discharges for half the time between charges, so it droops less. Ripple is halved for the same capacitance. Full-wave also puts less stress on the capacitor and diodes.

Can I just use a bigger capacitor instead of a regulator?

Technically yes, but impractical. To get 1% ripple at 1A and 12V with no regulator requires C = 1/(2 × 50 × 0.12) = 83000 µF. A regulator with a modest 4700 µF capacitor achieves millivolt-level output ripple and costs less.

Does ripple damage components?

High ripple means the capacitor charges and discharges more deeply each cycle, increasing the RMS ripple current through it. This heats the capacitor (via ESR losses) and shortens its life. Electrolytic capacitors are rated for maximum ripple current — exceeding it causes premature failure.

What is the ripple frequency?

For half-wave: same as mains (50 or 60 Hz). For full-wave: double the mains (100 or 120 Hz). For switching power supplies: the switching frequency (typically 50 kHz to 1 MHz), where ESR dominates capacitive reactance.

How does this relate to the Capacitor Charge Calculator?

The Charge Calculator finds Q = CV for a given voltage. The Ripple Calculator uses the same relationship in reverse — the ripple is caused by the load current draining charge (Q = I × t) from the capacitor between charging pulses. ΔV = ΔQ / C = I × t / C.

Last updated: March 2026