RC Time Constant Calculator

How to use the RC time constant calculator

1. Type R and C the way they are written on parts: 4.7k, 4k7, 100n, 10uF. τ appears at once.
2. Enter the source voltage and choose Charging (the capacitor starts empty) or Discharging (it starts full and empties through the resistor).
3. Pick a time to look at, such as 2ms or 0.5 (seconds); leave it blank for exactly one τ. The drawing marks that moment on the curve and the tiles show the capacitor's voltage and the current then.
4. Want to know when it reaches a voltage? Type the voltage and the time to it is worked out, in seconds and in time constants.
5. Milestones lists 1τ to 5τ; Calculation writes the maths out; click the resistor or the capacitor in the drawing for its own figures.

The time constant τ = R × C

τ = R × C

Ohms times farads come out in seconds: 10 kΩ × 100 nF = 1 ms. A bigger resistor lets less current in, a bigger capacitor needs more charge, and either one makes the charging slower. τ is the time the capacitor would take to fill if the starting current kept up. It does not, because the current is driven by the difference between the source and the capacitor's voltage, and that difference shrinks as the capacitor fills, so the capacitor only gets 63.2% of the way in each τ.

Your circuit τ = 10 kΩ × 100 nF = 1 ms; practically full after 5τ = 5 ms.

The charging and discharging curves

V_C = V (1 − e^−t/τ)    V_C = V e^−t/τ

The first is charging from empty, the second discharging from full; the current is I = V_R / R in both cases and follows the same e^−t/τ shape, starting at V/R. The fractions at whole time constants are always the same, which is the useful rule of thumb:

So "five time constants" is the engineer's word for "done": within 1%. Half way is reached at 0.693 τ (ln 2), which is why timer formulas are full of 0.693.

How long until it reaches a voltage

t = −τ × ln(1 − V_target / V)

Turn the charging curve round with a logarithm. For discharging, t = −τ × ln(V_target / V). This is how a power-on-reset circuit is timed, how a 555 timer sets its period, and how long a supply's capacitor holds a circuit up when the power drops.

The same RC as a filter

f_c = 1 / (2π R C)

Feed a signal into R and take the output across C and you have a low-pass filter: slow changes get through, fast ones are smoothed away. The corner (−3 dB, where the output has fallen to 70.7%) is at f_c. Swap the parts (output across R) for a high-pass with the same corner. A long τ means a low corner frequency.

Worked example: 10 kΩ and 100 nF at 5 V

τ = 10 kΩ × 100 nF = 1 ms. Charging from empty, after 1 ms the capacitor is at 5 V × 0.632 = 3.161 V; after 3 ms at 4.751 V; after 5 ms, 4.966 V, as good as 5 V. The current starts at 5 V / 10 kΩ = 500 µA and is down to 183.9 µA after one τ. To reach 2.5 V takes −1 ms × ln(0.5) = 693.1 µs. As a low-pass filter the corner is at 1 / (2π × 1 ms) = 159.2 Hz.

Questions

Does the source voltage change the time constant?

No. τ depends only on R and C. A higher voltage means more current and more charge, in the same proportion, so the curve has the same shape and timing; only its height changes.

Why does the capacitor never quite reach the source voltage?

Mathematically the curve only reaches V at infinite time. In practice the last fraction of a percent is below anything you can measure, which is why 5τ (99.3%) counts as full.

What if the capacitor is not empty to start with?

The same curve applies to the difference between where it is and where it is heading: it closes 63.2% of that gap in each τ. Enter the difference as the source voltage to get the timing.

How do I pick R and C for a delay?

Choose the time you want and the fraction you need (a reset chip might trip at 2.5 V out of 3.3 V), turn the formula round for τ, then pick a C that is large enough for leakage not to matter (100 nF to 10 µF is usual) and R = τ / C. The parallel and series capacitor calculators help when one part is not enough.