For a sinusoidal signal: peak = RMS × √2; average DC after a half-wave rectifier = peak / π; after a full-wave rectifier = 2 × peak / π.
Result
All calculations run locally in your browser: parsing, chart, export. Disconnect the network and the page keeps working.
Facts and limits of this method
| RMS → peak | peak = RMS × √2 ≈ RMS × 1.41421 |
|---|---|
| Peak → peak-to-peak | peak-to-peak = peak × 2 |
| Half-wave rectifier | average DC = peak / π ≈ peak × 0.31831 |
| Full-wave rectifier | average DC = 2 × peak / π ≈ peak × 0.63662 |
When it misleads you
- The formulas are valid for a pure sinusoidal signal — for non-sinusoidal waveforms (pulse, triangular, etc.), the relationships between RMS, peak, and average will be different.
- The rectifier calculation is idealized: it doesn't account for voltage drop across diodes (typically 0.7 V per silicon diode), ripple from a smoothing capacitor, or other practical effects of a real power supply circuit.
- "Average DC after rectifier" here means the average of the rectified but not yet smoothed signal — after a smoothing capacitor, the real DC voltage is usually closer to the peak value.
- Rounding is applied only to the displayed result — internal calculations use the full precision of JavaScript floating-point numbers.
How it is calculated
The entered value is first converted to peak voltage: unchanged if it's already peak, multiplied by √2 if it's RMS, divided by 2 if it's peak-to-peak.
The peak voltage is then converted to the selected output quantity: to RMS by dividing by √2, to peak-to-peak by multiplying by 2, to average half-wave DC by dividing by π, to full-wave DC by multiplying by 2/π.
All formulas assume an ideal sinusoidal signal and ideal diodes in the rectifier.
The result is rounded to the chosen number of decimal places and updates instantly whenever any parameter changes.
Questions and answers
Why is household 230V mains voltage an RMS value, not the peak?
Because 230V is the RMS (root mean square) value, which is equivalent to a DC voltage in terms of power delivered; the actual peak mains voltage is about √2 times larger — roughly 325V.
What's the difference between a half-wave and full-wave rectifier?
A half-wave rectifier passes only one half of the signal cycle (average DC = peak/π), while a full-wave rectifier uses both halves via a multi-diode circuit, yielding twice the average DC (2 × peak/π).
Does it account for the voltage drop across diodes?
No, the calculation is idealized — a real circuit loses about 0.7V per silicon diode, which needs to be accounted for separately in precise power supply calculations.
Is my data sent anywhere?
No, the whole calculation runs in your browser.