Calculate Cockcroft-Walton multiplier input peak voltage, stage count, or ideal no-load output voltage from any two known values in V, kV, or mV.
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Cockcroft-Walton Multiplier Formula
The calculator uses the ideal no-load Cockcroft-Walton voltage multiplier approximation. It assumes the input voltage is the peak AC voltage, not RMS voltage.
- Vout = ideal no-load output voltage
- Vp = peak input voltage
- n = number of multiplier stages
If you enter input voltage and stages, the calculator solves for the ideal output voltage. If you enter output voltage and stages, it solves for the required peak input voltage. If you enter input voltage and output voltage, it solves for the number of stages.
If your AC source is specified as RMS voltage, convert it to peak voltage first:
Voltage Unit Conversions and Result Meaning
Use these tables to check units and understand what the ideal result means before applying it to a real circuit.
| Unit | Equivalent in volts | Example |
|---|---|---|
| 1 kV | 1000 V | 5 kV = 5000 V |
| 1 V | 1 V | 250 V = 250 V |
| 1 mV | 0.001 V | 500 mV = 0.5 V |
| Condition | Effect on actual output |
|---|---|
| No load or very light load | Closest to the ideal value from the calculator |
| Higher load current | Output voltage drops below the ideal value |
| More stages | Higher ideal voltage, but more ripple and voltage sag in real circuits |
| Low capacitor value or low frequency | More ripple and poorer voltage regulation |
Example Calculations
Example 1: Find output voltage
You have a 500 V peak input and a 4-stage Cockcroft-Walton multiplier.
The ideal no-load output voltage is 4000 V, or 4 kV.
Example 2: Find number of stages
You want an ideal output of 12 kV from a 1.5 kV peak input.
You need 4 stages for an ideal no-load output of 12 kV.
FAQ
Does the calculator use RMS or peak input voltage?
It uses peak input voltage. If your source is listed as RMS, multiply the RMS voltage by √2 before entering it. For example, 120 V RMS is about 169.7 V peak.
Why is the real output lower than the calculated output?
The formula gives the ideal no-load output. A real Cockcroft-Walton multiplier has diode voltage drops, capacitor losses, ripple, and voltage sag under load. The more current the load draws, the more the output voltage usually drops.
Can the number of stages be a decimal?
The formula can return a decimal when solving mathematically, but a physical multiplier uses whole stages. If the result is not a whole number, you normally round up to the next whole stage, then check that the parts are rated for the resulting voltages.
