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Capacitor charge time calculator

Series resistance
Ω
Capacitance
µF
Supply voltage
V
To 99% charged 2,350 ms
5 × R × C to reach 99%
One time constant 470 ms
To half voltage 325.78 ms
Voltage after 1τ 7.585 V
Voltage after 3τ 11.403 V
τ in seconds 0.47 s
63.2% at 1τ · 99% at 5τ
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A capacitor charging through a resistor reaches 63.2% of the supply in one time constant (R × C) and 99% after five. A 470 µF capacitor through 1 kΩ has τ = 470 ms, so it is effectively full after 2.35 seconds.

How to work out charge time

1 Enter the series resistance the capacitor charges through.
2 Enter the capacitance and supply voltage.
3 Read the time to 99%, which is five time constants.
4 For a target other than 99%, use the voltage at 1τ and 3τ rows as reference points.

The exponential shape is the thing to internalise: charging is fastest at the start and slows continuously, so the last few per cent take as long as the first sixty. That is why inrush limiters work — a resistor in series limits the initial surge into a large reservoir capacitor, and is then shorted out by a relay once the capacitor is mostly charged. Without it, an empty capacitor looks like a dead short at the instant of switch-on, which is what welds contacts and blows fuses on large supplies.

Questions

Five time constants to 99%. One time constant is resistance times capacitance.

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