RC Time Constant Calculator (Tau, Charge and Discharge)
Enter the resistor, capacitor and source voltage. Choose whether you want the voltage after a given time, or the time to reach a percentage, while charging or discharging. Tau = R × C is always shown, with 5 tau.
Updated
Report a problem
Formula
- Time constant:
tau = R × C(seconds, with R in ohms and C in farads) - Charging:
v(t) = V × (1 − e^(−t / tau)) - Discharging:
v(t) = V × e^(−t / tau) - Time to reach p % while charging:
t = −tau × ln(1 − p / 100) - Time to fall to p % while discharging:
t = −tau × ln(p / 100) - Settling rule of thumb:
5 tau(about 99.3 % charged)
The percentage is only used in the two "time to reach" modes and the time t only in the two "voltage at time t" modes. Charging to 100 % or discharging to 0 % never completes, so those targets are rejected.
Worked example
R = 10 kΩ, C = 100 µF, source 12 V: tau = 10,000 × 0.0001 = 1 s, so 5 tau = 5 s.
- Charging for 1 s:
12 × (1 − e^−1) = 7.585 V(63.2 %). - Discharging for 1 s:
12 × e^−1 = 4.415 V(36.8 %). - Time to charge to 90 %:
1 × ln(10) = 2.303 s, at which point the voltage is 10.8 V. - Time to discharge to 50 %:
1 × ln(2) = 0.693 s.
Frequently asked questions
What is the RC time constant?
Tau = R × C, in seconds when R is in ohms and C in farads. A 10 kΩ resistor with a 100 µF capacitor gives tau = 10,000 × 0.0001 = 1 s. After one tau a charging capacitor has reached 63.2 % of the source voltage.
How do you calculate the capacitor voltage while charging?
v(t) = V × (1 − e^(−t/tau)). With V = 12 V, tau = 1 s and t = 1 s: 12 × (1 − 0.3679) = 7.585 V. While discharging, v(t) = V × e^(−t/tau), which gives 4.415 V at the same moment.
Why is 5 tau used as the settling time?
After 5 tau, e^(−5) = 0.67 %, so a charging capacitor is within about 0.7 % of its final value (99.3 % charged). It is a common rule of thumb for calling the circuit settled; with tau = 1 s that is 5 s.
How long does it take to reach a given percentage?
Charging: t = −tau × ln(1 − p/100). To reach 90 % with tau = 1 s that is ln(10) = 2.303 s. Discharging to p % of the start voltage: t = −tau × ln(p/100); falling to 50 % takes ln(2) = 0.693 tau. 100 % charged is never reached exactly, so it is not allowed.
Can I use this for safety or final design sign-off?
No. This calculator is for learning and design checks, not safety certification. It assumes an ideal resistor and capacitor with no leakage, tolerance or source resistance, so check tolerances and the datasheet, and discharge large capacitors safely.