Capacitor Charge Time Calculator

How long will your capacitor take to charge to a target voltage?

Enter your circuit values and find out precisely how long your capacitor needs to reach a target voltage. The result is based on the RC time constant and the exponential charging equation used in circuit design.

Updated July 2026 · How this works

Example calculation — edit any field to use your own numbers

Worth knowing
How It Works
The formula, explained simply

Imagine trying to fill a water tank through a pipe with a valve partially closed. When the tank is empty, water rushes in fast because the pressure difference is large. As the tank fills, the pressure difference between the supply and the tank shrinks, so the flow rate drops. The tank never quite fills to the supply pressure in finite time — it just gets closer and closer. A capacitor charging through a resistor behaves exactly the same way, with voltage playing the role of pressure and charge playing the role of water.

The resistor controls how much current can flow at any instant. A larger resistance slows the charging rate, stretching the curve over more time. A larger capacitance means more charge must be moved to raise the voltage by one volt, so the same current takes longer to do the job. Together these two values define the time constant, which sets the fundamental pace of the charging event regardless of the voltage levels involved.

The initial voltage on the capacitor shifts the starting point on the exponential curve without changing its shape. A partially charged capacitor has less voltage gap to close, so it reaches the target faster — but the same exponential slowdown applies from wherever it starts. This tool solves for the exact time by inverting the charging equation, giving you the answer directly rather than asking you to read it off a graph or estimate from multiples of tau.

When To Use This
Right tool, right situation

Use this calculator when you need to confirm that a capacitor will reach a specific voltage within a defined window — setting a power-on reset delay, verifying that an RC filter settles before a measurement window opens, or sizing a timing component in a relaxation oscillator. It is equally useful in the reverse direction: if you know the time budget and the voltage threshold, experiment with R and C values to find a combination that hits your target.

This calculation is appropriate when the resistance in the charging path is a known, fixed value. It works well for resistor-limited charging circuits, current-sense timing loops, and manual precharge circuits. It is not appropriate for switch-mode charging topologies, where the effective charging current is controlled by a switching regulator rather than a passive resistor, because the exponential model does not describe that process. It is also not the right tool when temperature shifts are significant enough to change capacitance materially, such as with Class II ceramic capacitors over a wide operating range.

When component tolerances matter, run the calculation twice with the worst-case high and low values of both R and C to bound your timing range. A single nominal calculation tells you where the center of the distribution is, but a design that must always charge within a deadline needs to be verified at both extremes before committing to component selection.

Common Mistakes
Why results sometimes look wrong

Using the capacitor's nominal value without accounting for tolerance. Electrolytic capacitors commonly carry a plus-or-minus 20% tolerance, meaning a 470 uF part could measure anywhere from roughly 380 uF to roughly 560 uF. If charge time is critical to your design — driving a reset circuit or a timing oscillator — verify the actual capacitance with a meter rather than trusting the body marking. The formula is exact; the component values may not be.

Ignoring equivalent series resistance (ESR) in the capacitor itself. This calculator models the charging resistance as a single external series resistor. In reality, every capacitor has internal ESR, which adds to the total resistance in the path. For low-value external resistors, the capacitor's ESR can be a significant fraction of the total, making the real time constant longer than calculated. This matters most in fast-switching power supply designs where the resistor is intentionally small.

Treating the supply voltage as fixed when it is not. Many real power supplies sag under load, especially at startup when large capacitors draw heavy inrush current. If your supply voltage drops during charging, the effective Vs is lower than the rated value, and the charge time extends beyond the formula's prediction. Bench testing with the actual load present reveals this discrepancy where a calculation alone will not.

The Math
Worked examples and deeper derivation

The voltage on a charging capacitor at time t follows: V(t) = Vs x (1 - e^(-t/tau)) + V0 x e^(-t/tau), where Vs is supply voltage, V0 is initial voltage, and tau = R x C is the time constant. To find the time needed to reach a specific target voltage Vc, rearrange by isolating t. Subtract V0 x e^(-t/tau) from both sides and factor to get V(t) - Vs = (V0 - Vs) x e^(-t/tau). Divide both sides by (V0 - Vs) to isolate the exponential term, then take the natural logarithm of both sides.

The result is t = -RC x ln((Vs - Vc) / (Vs - V0)). For the example with supply voltage 12 V, target 7.6 V, 470 uF, 10 kohm, and zero initial voltage, the time constant tau = 4.7 s. The logarithm argument becomes (12 - 7.6) / (12 - 0), which simplifies to a ratio less than one. Taking the natural log of this ratio and multiplying by -tau gives the charge time 4.716 s. The negative sign in the formula corrects for the fact that the log of a fraction less than one is negative — the double negative produces a positive time.

The formula breaks down if Vc equals Vs because the logarithm of zero is undefined. Physically this confirms what the water-tank analogy shows: the capacitor asymptotically approaches but never quite reaches the supply voltage. For practical circuit design, 99.3 of supply voltage after five time constants is the accepted definition of full charge in most applications.

Power supply bulk capacitor in a 12 V audio amplifier
470 uF capacitor, 10 kohm series resistance, 12 V supply, charging to 7.6 V from zero
The RC time constant is 4.7, meaning the capacitor charges through roughly 4.7 per exponential step. To reach 7.6 V on a 12 V rail from zero, the formula gives a charge time of 3.25. At one time constant the capacitor would reach 7.585 V V. This 7.6 V threshold represents the midpoint commonly used in power-on reset circuits, where the designer needs the rail stable enough to release a microcontroller from reset without waiting for full charge.
Ceramic capacitor timing circuit at 5 V logic level
100 nF capacitor, 1 kohm series resistance, 5 V supply, charging to 3.16 V from zero
With 100 nF and 1 kohm, the time constant 100 is very short. The target of 3.16 V is close to the classic RC threshold of one time constant into charging, yielding a charge time of 69.3. This sub-microsecond timing is typical for oscillator RC networks in timer circuits, where the threshold voltage of the comparator sets the oscillation frequency. A small change in resistance here has an outsized effect on timing because the time constant is so short.
Partially charged capacitor in a motor soft-start circuit
1000 uF capacitor, 5 kohm series resistance, 9 V supply, charging from 2 V to 6 V
Starting from 2 V rather than zero shrinks the effective voltage gap the capacitor must cross. The time constant 5 is the natural charging pace. The partial initial charge means the logarithm argument shifts, reducing total charge time to 1.95. In motor soft-start designs the capacitor often retains residual charge between cycles, so the actual time to reach the gate-drive threshold is shorter than a cold-start calculation would predict. Ignoring initial voltage in this scenario overstates inrush delay and can lead to conservative — and unnecessary — component sizing.
Expert Unlock
The thing most explanations skip

The formula assumes the charging source is an ideal voltage supply with zero output impedance. In real circuits the source impedance adds to the series resistance, making the effective time constant longer than the declared resistor value alone would suggest. Battery-powered circuits are especially susceptible: a battery's internal resistance rises as it discharges, shifting the time constant mid-cycle. For precision timing applications, a low-dropout regulator with well-characterized output impedance is a better charging source than a battery or an unregulated rail.

The other hidden assumption is that the capacitor's capacitance is constant over the voltage range. X5R and X7R ceramic capacitors lose a significant fraction of their capacitance near their rated voltage due to DC bias derating, meaning the effective C when the capacitor is nearly full is lower than the rated value. The charging curve departs from the ideal exponential in this regime, and measured charge times near the supply voltage will be shorter than the formula predicts. Electrolytic and film capacitors are far more stable over voltage and are the better match for this model.

Why does my capacitor charge slower than the formula predicts?

What is the RC time constant and what does it tell me?
The RC time constant is the product of resistance and capacitance and represents how quickly the charging process moves. One time constant is the time after which the capacitor has reached about 63.2 of its remaining voltage gap from zero initial charge. It is not the total charge time but a pace indicator: a large time constant means slow charging, a small one means fast. Knowing tau lets you estimate behavior at any point in the curve without recalculating from scratch.
How many time constants does it take to fully charge a capacitor?
In practice, engineers treat a capacitor as fully charged after five time constants, at which point it has reached 99.3 of the supply voltage. The curve never reaches exactly the supply voltage in finite time because each step only closes a fraction of the remaining gap. For most circuits, the difference between five tau and true full charge is negligible, but high-precision analog circuits may require more time to settle within their error budget.
Why does charging slow down as the capacitor fills up?
As the capacitor voltage rises toward the supply, the voltage difference driving current into it shrinks. Current through the resistor equals the voltage difference divided by the resistance, so less voltage difference means less current, which means slower charging. This is the exponential decay at the heart of the RC charging equation. The rate of change is always proportional to how much headroom remains, which is why the curve appears to flatten near the supply voltage even though it is still technically climbing.

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