Calculate how long a capacitor takes to charge or discharge through a resistor to a target voltage. RC time constant, milestones, and curve.
How long does a capacitor take to charge? The answer depends on three things: the capacitance (C), the series resistance (R), and the target voltage as a fraction of the supply. The time constant τ = RC sets the scale — after 1τ the capacitor reaches 63.2% of the supply voltage, and after 5τ it is 99.3% charged.
For a 100 µF capacitor charging through 10 kΩ, τ = 1 second. To reach 63.2% of the supply takes 1s; to reach 95% takes 3s; to reach a specific target voltage requires t = −τ × ln(1 − V_target/V_supply). Discharging follows the inverse exponential: after 1τ the voltage drops to 36.8% of its initial value.
This calculator computes the exact charge or discharge time to any target voltage, with support for non-zero initial conditions. It visualizes the voltage curve, shows milestone times, and provides current and power dissipation data for component selection.
RC time calculations require logarithmic functions and exponential evaluations that are tedious by hand. Designing timing circuits, backup power systems, debounce circuits, and flash capacitor chargers all require knowing the exact time to reach a specific voltage — not just the time constant.
This calculator provides exact time-to-target with non-zero initial conditions, visual charge curves, milestone tables, and current/power data for component rating — everything needed for RC circuit timing design in one tool.
Charging: V(t) = Vs(1 − e^(−t/τ)). Discharging: V(t) = V₀e^(−t/τ). Time to target: t = −τ·ln((Vs − Vt)/(Vs − V₀)). τ = RC. Current: I(t) = I₀·e^(−t/τ).
Result: 1.099 seconds
τ = 10,000 × 100×10⁻⁶ = 1 second. Time to 8V: t = −1 × ln((12−8)/(12−0)) = −ln(0.333) = 1.099 s. At this point, the charging current has dropped from 1.2 mA initial to 0.4 mA.
The time constant τ = RC is perhaps the most important parameter in analog electronics. It appears in filters, oscillators, timing circuits, debounce circuits, power supplies, and signal processing. One time constant is the time to reach 63.2% of the final value — a number that arises from 1 − 1/e ≈ 0.632.
The factor of e (2.718...) appears because the charging rate is proportional to the remaining voltage difference. This produces a first-order linear ODE: dV/dt = (Vs − V)/(RC), whose solution is the exponential V(t) = Vs(1 − e^(−t/RC)).
**555 Timer:** The most popular timing IC charges a capacitor through R₁ + R₂ to 2/3 Vs, then discharges through R₂ to 1/3 Vs. Frequency = 1.44/((R₁ + 2R₂)C).
**Debounce circuits:** A simple RC with τ ≈ 10-50 ms smooths the contact bounce of mechanical switches. The Schmitt trigger input of the logic gate provides clean switching.
**Camera flash:** A photoflash capacitor (330 µF/300V) stores 14.85 J. Charged through a boost converter in 5-10 seconds, discharged through the flash tube in about 1 millisecond.
Real circuits often start with the capacitor partially charged. The general solution is V(t) = Vs − (Vs − V₀)e^(−t/τ), where V₀ is the initial voltage. If V₀ > Vs (discharged toward a lower voltage), the exponential still applies — the capacitor voltage decreases exponentially toward Vs.
This calculator handles all initial conditions, including charging from a non-zero start and discharging to any target voltage, making it suitable for complex timing scenarios.
As the capacitor voltage approaches the supply voltage, the voltage difference (driving force) decreases, reducing the current. This produces the characteristic exponential curve — fast at first, then asymptotically approaching the final value.
Mathematically, never — the exponential approaches but never reaches the supply voltage. Practically, after 5τ the capacitor is within 0.7% of the supply voltage, which is "fully charged" for any practical purpose.
A pre-charged capacitor starts at a higher voltage, so it reaches any target above its initial voltage faster. The formula accounts for this: the effective charging range is (Vs − V₀), not the full Vs.
At t = 0, the capacitor acts like a short circuit, and the current is limited only by the resistance: I₀ = (Vs − V₀)/R. This is also the peak power dissipation point for the resistor.
Yes — the 555 timer charges to 2/3 Vs and discharges to 1/3 Vs. Set target voltage to 2/3 × V_supply for charge time or 1/3 × V_supply for discharge time to find the oscillation period.
With very low resistance, τ is very small and the initial current spike is very large. This can damage components — always include adequate series resistance to limit inrush current.