About this calculator
A capacitor charging through a resistor does not ramp linearly — it rushes at first and then creeps, following an exponential curve that never quite arrives. Knowing where on that curve you are at a given moment is the basis of timing circuits, power-up delays, sample-and-hold stages and soft-start networks.
This calculator plots that curve and answers both questions you can ask about it: what voltage the capacitor has reached after a given time, and how long it takes to reach a given voltage.
How it works
When charging, the voltage across the capacitor approaches the supply along V(t) = Vs × (1 − e^(−t/τ)), where τ is R × C. The rate of change is proportional to how far it still has to go, which is exactly what produces an exponential.
Discharging is the mirror image: V(t) = V₀ × e^(−t/τ), decaying toward zero from wherever it started.
Rearranging either expression gives the time to hit a particular voltage. Because the logarithm is involved, the time depends on the fraction of the supply you are targeting, not the absolute voltage — reaching half of 5 V takes exactly as long as reaching half of 12 V in the same network.
The current tells the other half of the story. At the instant charging begins the capacitor looks like a short circuit, so current peaks at Vs / R and then decays exponentially as the capacitor voltage rises to oppose the supply.
Worked example
A 100 µF capacitor charging through 10 kΩ from a 5 V supply. How long until it crosses the 2.5 V logic threshold?
- τ = 10000 × 100e-6 = 1 s
- Target fraction: 2.5 / 5 = 0.5
- t = −1 × ln(1 − 0.5) = −ln(0.5) = 0.693 s
- Peak current at t = 0: 5 / 10000 = 0.5 mA
- Energy at full charge: 0.5 × 100e-6 × 25 = 1.25 mJ
It crosses 2.5 V after 693 ms — the ln(2) result that shows up throughout timer design. Full settling to 99.3% takes 5 seconds.
Practical notes
- The 63.2% and 36.8% figures are not arbitrary — they are
1 − 1/eand1/e. One time constant is defined as the time to traverse that fraction. - A large capacitor charging through a low resistance draws a substantial initial surge. Check that your supply and any series switch can survive
Vs / Ramps. - A charged capacitor holds its energy after power is removed. In mains-powered gear with big reservoir capacitors, this is genuinely dangerous — bleed resistors exist for a reason.
- Real capacitors leak. Electrolytics in particular self-discharge over minutes to hours, so very long RC delays built with them are unreliable.
- Class 2 ceramics lose significant capacitance under DC bias, so a nominal 10 µF MLCC at its rated voltage may behave as 5 µF and time correspondingly short.
Frequently asked questions
How do I calculate capacitor voltage over time?
For charging, V(t) = Vs × (1 − e^(−t/RC)). For discharging, V(t) = V₀ × e^(−t/RC). Both need time and RC in the same units — seconds.
How long does it take a capacitor to discharge?
It falls to 36.8% after one time constant, 5% after three, and 0.7% after five. Five time constants is the usual rule for "fully discharged".
How much energy does a capacitor store?
E = ½ C V², in joules. Note the square — doubling the voltage quadruples the stored energy, which is why capacitor voltage ratings matter so much.
Why does the charging current start high and fall?
An uncharged capacitor has no voltage across it, so initially the full supply appears across the resistor and current is at its maximum. As the capacitor charges it opposes the supply, leaving less across the resistor and so less current.
Does the time to reach a voltage depend on the supply?
Only through the fraction. Reaching 50% of the supply always takes 0.693τ, whatever that supply is. Reaching a fixed 2.5 V takes longer from a 3 V supply than from a 12 V one, because 2.5 V is a much larger fraction of 3 V.