Estimate peak-to-peak rectifier ripple, nominal smoothing capacitance or load current at a ripple limit. Select half-wave or single-phase full-wave recharge frequency, and enter an optional total capacitance reduction from nominal. These are preliminary capacitor-discharge estimates, not component ratings or a power-supply approval.

Constant-load, small-ripple discharge approximation. Full-wave uses two recharge pulses per line cycle; half-wave uses one. This model cannot establish regulator headroom or circuit safety.

60 Hz is a default. Enter the incoming line frequency, not an already doubled ripple frequency.

Loss, DC diagnostic and display

0% assumes the full nominal capacitance. Enter one total reduction relative to nominal; it is not an ESR allowance.

Blank skips DC-relative percentages and the triangular-model low point. This input does not set regulator headroom.

Ripple Voltage Formula

Ripple voltage is the residual AC variation on a DC output. This calculator uses the capacitor-input approximation Vpp ≈ I / (fr × Ceff), where fr is the recharge frequency and Ceff is effective operating capacitance. For single-phase full-wave rectification fr = 2 × line frequency, giving the equation below; half-wave uses fr = line frequency. It assumes nearly constant load between charging peaks.

Vₚₚ = (I) / (2 f C)
  • Vpp = peak-to-peak ripple voltage
  • I = load current in amperes
  • f = AC line frequency in hertz
  • C = effective operating filter capacitance in farads

This calculator uses AC line frequency as the input. For a full-wave rectifier, the capacitor is recharged twice during each AC cycle, so the ripple frequency is double the line frequency.

fᵣᵢₚₚₗₑ = 2 fₗᵢₙₑ

Rearranged Design Equations

For the full-wave case, the rearrangements below estimate effective capacitance or load current at an entered ripple limit. A total fractional reduction L from nominal is applied as Ceff = Cnom × (1 − L), so required nominal capacitance is Ceff ÷ (1 − L). The single allowance includes any reductions you choose to model; do not count the same tolerance or loss twice.

C = (I) / (2 f Vₚₚ)
I = 2 f C Vₚₚ
f = (I) / (2 C Vₚₚ)

How to Use the Calculator

  1. Enter the load current drawn by the DC side of the circuit.
  2. Enter the AC line frequency, typically 50 Hz or 60 Hz.
  3. Enter the total filter capacitance connected across the rectifier output.
  4. Select the solve task and rectifier type, then calculate the estimate. Average DC voltage is optional and only enables diagnostic ratios and an ideal waveform trough; it is not needed by the primary ripple equation.

Input Guide

Input What It Represents Effect on Ripple
Current Average current demanded by the load from the filtered DC output Higher current increases ripple
Frequency The incoming AC line frequency feeding the rectifier Higher frequency reduces ripple
Capacitance Total smoothing capacitance across the output Higher capacitance reduces ripple

Example Calculation

Suppose a full-wave power supply delivers 1.5 A from a 60 Hz source and uses a 4700 µF filter capacitor.

Vₚₚ = (1.5) / (2 × 60 × 4700 × 10⁻⁶)
Vₚₚ ≈ 2.66 V

This means the DC output rises near each charging peak and then falls by about 2.66 V before the next recharge pulse.

How Each Input Changes Ripple

  • Increase current: ripple rises almost proportionally because the capacitor discharges faster between peaks.
  • Increase capacitance: ripple falls because the capacitor stores more charge and holds the output voltage up longer.
  • Increase line frequency: ripple falls because the capacitor is recharged more often.
  • Reduce ripple: common fixes include a larger capacitor, a lighter load, a higher-frequency source, or an added filtering/regulation stage.

Why Ripple Voltage Matters

  • Too much ripple can introduce hum and noise into audio circuits.
  • Large ripple reduces the minimum instantaneous DC voltage available to the load.
  • Voltage regulators may drop out if ripple pulls the input below the regulator’s required headroom.
  • Sensitive digital, analog, and communication circuits generally perform better with lower ripple.

Common Unit Conversions

1 mF = 10⁻³ F
1 μF = 10⁻⁶ F
1 nF = 10⁻⁹ F

If multiple capacitors are connected in parallel at the output, their capacitances add together, which lowers ripple.

Full-Wave vs. Half-Wave Rectification

Both half-wave and single-phase full-wave or bridge rectifiers are available. Half-wave recharges once per source cycle and full-wave twice. At the same load and effective capacitance, this approximation therefore gives twice the ripple for half-wave. It does not model multiphase supplies, switching converters or arbitrary pulsed loads.

Vpp, half - wave ≈ (I) / (f C)

Practical Limits of the Estimate

The formula is a preliminary constant-load approximation. Actual ripple and minimum voltage depend on the discharge waveform, ESR, diode conduction angle, transformer and source impedance, recharge pulses and changing loads. Approximate AC RMS ripple is Vpp / √12 only for the linear-ramp waveform assumption; Vpp / average DC is a different ratio from the RMS ripple factor. The optional average-DC midpoint-minus-half-ripple value is not a measured minimum and cannot establish regulator headroom. Check actual waveforms and capacitor voltage, RMS-current and temperature ratings separately.