How this instrument works
A capacitive dropper does not step voltage down the way a transformer does — it sits in the current's path and lets its reactance, not its resistance, decide how much gets through. Capacitive reactance Xc = 1/(2πfC) falls as capacitance rises, so a larger capacitor passes more current at a given mains voltage and frequency. Rearranging that relationship for C gives the sizing formula this page solves: C = I/(2πfV).
The shape of the formula follows straight from Ohm's law applied to a reactive element: current equals voltage divided by reactance, I = V/Xc, and substituting Xc = 1/(2πfC) then solving for C produces exactly what sits above. Because an ideal capacitor stores and releases energy every half-cycle rather than converting it to heat, a dropper capacitor barely warms up even while limiting tens of milliamps — the excess voltage is deflected by phase shift, not burned off the way a series resistor would burn it.
The formula is an approximation that holds only when the downstream load's own voltage is small next to the mains — typically a rectifier bridge feeding a zener-regulated rail well under 24 V from 230 V or 120 V mains. It also assumes a purely reactive path, when real designs always add a little resistance on purpose: a series resistor to blunt the inrush surge at switch-on, and a bleeder resistor across the capacitor so stored charge cannot outlive the plug being pulled. Skip either and the capacitor becomes a genuine hazard rather than a convenience.
- Enter the supply voltage in Mains voltage — 230 V across most of the world, 120 V in North America.
- Set Mains frequency to 50 Hz or 60 Hz, matching the AC supply the circuit will actually run from.
- Enter the current the downstream circuit needs in Desired output current — the load after the rectifier and regulator, not a guess at the capacitor's own rating.
- Read Required dropper capacitance, then switch its unit between nF and µF to match what's printed on real capacitor bodies.
- Round up to the nearest standard X2-rated value at or above the figure shown — undersizing starves the load of current.
Worked example — sizing a 30 mA dropper on 230 V mains
A small mains-powered gadget needs a steady 30 mA to feed its rectifier and regulator from 230 V, 50 Hz mains. Plugging straight into the formula: C = 0.03 ⁄ (2π × 50 × 230) = 0.03 ⁄ 72256.63 = 4.15186808066 × 10⁻⁷ F, which is 0.415186808066 µF — call it 0.415 µF. No off-the-shelf part is sold to eight decimal places, so the nearest standard X2 value at or above that figure, 0.47 µF, is what actually goes on the parts list.
That rounding direction matters: dropping to a smaller stocked value like 0.33 µF would under-deliver current and starve the regulator, while rounding up slightly is harmless because the downstream zener simply clamps the extra. The arithmetic scales cleanly, too — doubling the target to 60 mA on the same 230 V, 50 Hz mains doubles the capacitance to roughly 0.83 µF, and moving the original 30 mA target to 120 V, 60 Hz mains raises the requirement to about 0.663 µF, since a lower mains voltage delivers less current per farad and needs more farads to make up the difference.
Questions
Why use a capacitor instead of a resistor to drop mains voltage?
Because a capacitor limits current through reactance, which does not dissipate real power the way a resistor's I squared R drop does. A resistor sized to drop, say, 200 V while passing 30 mA would burn about 6 W as heat — a real problem in a small sealed enclosure. A capacitor doing the same job barely warms up, since the current it passes leads the voltage by 90 degrees and an ideal capacitor converts none of that to heat.
Does this kind of supply provide any isolation from the mains?
No — none at all. Every part of the low-voltage side sits at mains potential relative to earth until proven otherwise, which is why capacitive dropper circuits need double insulation, an X2-rated capacitor built to fail open rather than short, and a bleeder resistor that discharges the capacitor within a couple of seconds after unplugging. A transformer's magnetic coupling is what actually isolates; a capacitor's electric field does not.
Why doesn't the formula include the load's own resistance?
Because the approximation assumes the capacitor's reactance dominates the circuit's impedance and the regulated output voltage is small next to the mains voltage — commonly true when a few volts DC comes from 230 V or 120 V AC. Under that assumption nearly the full mains voltage appears across the capacitor, so its reactance alone sets the current, and the load's own resistance barely moves the answer.
What happens if the dropper capacitor is undersized?
The circuit gets less current than it needs. Since C = I/(2πfV), a smaller capacitance passes proportionally less current at the same voltage and frequency — halve the capacitance and the available current halves too, which can starve a regulator or make an LED flicker under load. Sizing up to the next standard value rather than down avoids the problem entirely.
Why does mains frequency change the required capacitance?
Because frequency sits in the denominator of C = I/(2πfV) — reactance falls as frequency rises, so a smaller capacitor delivers the same current more readily. Holding voltage and target current fixed, a 60 Hz supply needs about 17 percent less capacitance than an otherwise identical 50 Hz design, which matters when one product ships to both North American and European mains without a redesign.
Is the current this formula gives an RMS or a peak value?
RMS. The formula is built from V and I as root-mean-square quantities, matching how mains voltage and load current are normally specified and measured. The instantaneous current through the capacitor is a sinusoid whose peak is the square root of two times the RMS figure — about 42 mA peaking for the 30 mA RMS example above — and that peak is what the rectifier diodes and reservoir capacitor downstream actually have to handle.