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RC Time Constant Calculator

A capacitor charges through a resistor along a curve, not a straight line. The time constant τ = R × C is how long it takes to reach 63.2 % of the final voltage, and after about 5 τ it is practically full (99.3 %).

Result

Time constant τ = R × C

1 s

Charging from empty

After 1 τ (1 s): 63.2 % · 3.161 V
After 2 τ (2 s): 86.5 % · 4.323 V
After 3 τ (3 s): 95.0 % · 4.751 V
After 4 τ (4 s): 98.2 % · 4.908 V
After 5 τ (5 s): 99.3 % · 4.966 V

Your questions

Time to reach 50 %while charging from empty: 693.1 ms
Charging, after 1 s: 3.161 V (63.2 %)
Discharging, after 1 sstarting full: 1.839 V (36.8 %)

After about 5 τ a capacitor is practically full (99.3 %). It never reaches exactly 100 %.

The formula

The time constant

\htmlData{sym=tau}{\tau }= \htmlData{sym=R}{R}\,\htmlData{sym=C}{C}

Charging from empty

\htmlData{sym=U_C}{U_{\mathrm{C}}} = \htmlData{sym=U_s}{U_{\mathrm{s}}}\left(1 - \htmlData{sym=e}{\mathrm{e}}^{-\htmlData{sym=t}{t}/\htmlData{sym=tau}{\tau}}\right)

Discharging from full

\htmlData{sym=U_C}{U_{\mathrm{C}}} = \htmlData{sym=U_s}{U_{\mathrm{s}}}\,\htmlData{sym=e}{\mathrm{e}}^{-\htmlData{sym=t}{t}/\htmlData{sym=tau}{\tau}}

Time to reach x percent

\htmlData{sym=t}{t} = -\htmlData{sym=tau}{\tau}\,\ln\left(1 - \dfrac{\htmlData{sym=x:percent}{x}}{100}\right)

What each letter means

$\tau$: Time constant τ (tau) (s · second) : How long an RC or RL circuit takes to get 63 % of the way to its final value. After 5τ it is basically done.
$R$: Resistance (Ω · ohm) : How strongly something opposes current.
$C$: Capacitance (F · farad) : How much charge a capacitor stores per volt.
$U_{\mathrm{C}}$: Voltage across the capacitor (V · volt)
$U_{\mathrm{s}}$: Supply voltage (V · volt) : The source or supply voltage.
$\mathrm{e}$: Euler’s number e ≈ 2.718 (no unit) : A fixed number used for anything that grows or decays smoothly, such as a capacitor charging. Not a variable.
$t$: Time (s · second) : Time since something started.
$x$: Charge level wanted (% · percent) : How full you want the capacitor, as a percent of its final voltage. It must be below 100 %, because it only ever gets closer.

How it works

While a capacitor is nearly empty, a lot of current flows into it and it charges quickly. As it fills, the voltage across the resistor shrinks, the current drops and the charging slows down. Every time constant τ closes 63.2 % of the gap that is left, which is where the table of 63.2 %, 86.5 %, 95.0 %, 98.2 % and 99.3 % comes from: each one is 1 − e^−n for n = 1 to 5.

Discharging is the mirror image: the voltage falls to 36.8 % after 1 τ, 13.5 % after 2 τ, and so on. The supply voltage does not change τ. A bigger resistor or a bigger capacitor makes everything slower in proportion.

Worked example

R = 10 kΩ and C = 100 µF. The time constant is τ = R × C = 1 s.
After 1 τ the capacitor is 63.2 % charged, and after 5 τ it is 99.3 % charged. With a 5 V supply that is 3.161 V and 4.966 V.
To reach 50 %: t = −τ × ln(1 − 0.5) = 693.1 ms (that is τ × ln 2).
After 1 s a capacitor charging from empty is at 3.161 V. One that started full and is discharging is at 1.839 V. They add up to the 5 V supply.

Common questions

Why can a capacitor never reach 100 %?

Each time constant only closes 63.2 % of the distance that remains, so there is always a small gap left. In practice the gap becomes too small to measure after a few τ, which is why 100 % is not a valid answer in the time box.

Why do people use 5 τ?

After 5 τ the capacitor has reached 99.3 % of its final voltage. That is close enough to “full” for most circuits, such as a power-on reset delay or a button debounce, so 5 τ is the usual waiting time.

Does the supply voltage change the time constant?

No. τ depends only on R and C. A higher supply makes the capacitor end up at a higher voltage, but it still takes the same time to reach the same percentage.

Try it live

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