What is the capacitor discharge formula?
V(t) = V0 × e^(−t/RC). The voltage falls to 36.8 % of its starting value after one time constant (τ = RC), to 13.5 % after two and below 1 % after five. The current follows the same curve, starting at V0 / R.
Voltage after a time, or time to a voltage: this capacitor discharge calculator solves both, charging or discharging, through a resistor or at a constant current. Size a bleeder resistor that brings a bus or X capacitor below a safe voltage in time, and see what it dissipates.
Voltage after 1 s
4.415 V
Ideal capacitor: no leakage current and no ESR. Leakage speeds a discharge slightly; ESR adds to R.
Electrolytic and film capacitors recover part of their voltage after a quick discharge (dielectric absorption). Leave the discharge path connected and measure before touching the terminals.
Use t = RC × ln(V0 / V). A 100 µF capacitor discharging through 10 kΩ has τ = RC = 1 s, so going from 12 V to 1 V takes 1 s × ln 12 = 2.48 s. The voltage at any time is V0 × e^(−t/RC): after one time constant it is down to 36.8 % (4.415 V here), and after five it is under 1 %.
Through a resistor, the capacitor voltage moves exponentially from where it starts toward where the circuit is pushing it: 0 V for a discharge, the supply voltage for a charge.
The capacitor discharge equation (to 0 V) and the capacitor charging equation (from 0 V) are the two common cases:
Solved for the time to reach a target voltage V:
At a constant current I the voltage changes in a straight line, because i = C dV/dt. A bleeder resistor is the largest R that still reaches the safe voltage in time:
The energy left on the capacitor at any point is ½CV²; during a discharge through a resistor, the rest has become heat in R.
τ = 10 kΩ × 100 µF = 1 s. After 1 s the voltage is 12 × e^(−1) = 4.415 V. To reach 1 V takes t = 1 s × ln(12 ÷ 1) = 2.48 s. The discharge starts at 12 V ÷ 10 kΩ = 1.2 mA and turns 7.15 mJ into heat in the resistor on the way down to 1 V.
R ≤ 60 s ÷ (220 µF × ln(400 ÷ 50)) = 131 kΩ. The nearest E24 value below is 130 kΩ, which reaches 50 V in 59.5 s. The catch is that it dissipates 400² ÷ 130 kΩ = 1.23 W for as long as the bus is up, and it has the full 400 V across it. Two 62 kΩ resistors in series (124 kΩ) reach 50 V in 56.7 s, with 200 V and 645 mW on each. The capacitor holds 17.6 J at 400 V.
Unplugged at the crest, the capacitor starts at √2 × 230 V = 325.3 V. R ≤ 1 s ÷ (470 nF × ln(325.3 ÷ 34)) = 942 kΩ, so 910 kΩ (E24) gets there in 966 ms and dissipates 230² ÷ 910 kΩ = 58.1 mW while plugged in. That standby loss is why X capacitor discharge ICs exist: Power Integrations puts a 1 MΩ discharge resistor at 53 mW on 230 V AC (52.9 mW by the same formula). Check the test conditions in the standard: the highest rated line voltage and the component tolerances can push R lower.
Every RC charge and discharge has the same shape; only τ = RC stretches it. Multiply τ by the first column to get the time.
| Time | Discharge: voltage left | Charge: share of the step |
|---|---|---|
| 0.5 τ | 60.7 % | 39.3 % |
| 0.693 τ (half) | 50 % | 50 % |
| 1 τ | 36.8 % | 63.2 % |
| 2 τ | 13.5 % | 86.5 % |
| 3 τ | 4.98 % | 95 % |
| 4 τ | 1.83 % | 98.2 % |
| 4.61 τ | 1 % | 99 % |
| 5 τ | 0.674 % | 99.3 % |
| 7 τ | 0.0912 % | 99.9 % |
The largest resistor that meets the 50 V in 60 s limit for each capacitor, the nearest E24 value below it, and the power it burns at 400 V. Bigger capacitors need smaller resistors and more watts.
| Capacitor | Largest R | E24 value | Time to 50 V | Dissipation at 400 V |
|---|---|---|---|---|
| 47 µF | 614 kΩ | 560 kΩ | 54.7 s | 286 mW |
| 100 µF | 289 kΩ | 270 kΩ | 56.1 s | 593 mW |
| 220 µF | 131 kΩ | 130 kΩ | 59.5 s | 1.23 W |
| 470 µF | 61.4 kΩ | 56 kΩ | 54.7 s | 2.86 W |
| 1 mF | 28.9 kΩ | 27 kΩ | 56.1 s | 5.93 W |
The calculator treats the capacitor as ideal. Leakage current adds a parallel path, so a real discharge is a little faster; ESR and wiring add to R, so a low-value discharge resistor is slower than the formula and its peak current is lower. Aluminum electrolytics have a wide capacitance tolerance (check the datasheet), and a bleeder has to meet its limit with the capacitor at the top of its tolerance and the line at its highest voltage.
Size the resistor for three things, not one: the continuous power V²/R while the circuit is on, the working voltage across it (several resistors in series share both), and, for a fast discharge, the pulse energy ½CV² it absorbs at once. Electrolytic and film capacitors also show dielectric absorption: shorted briefly, they recover part of their voltage, so leave the bleeder connected and measure before touching the terminals. A constant-current sink stops regulating when its own headroom runs out, so its straight line bends toward zero near the end.
V(t) = V0 × e^(−t/RC). The voltage falls to 36.8 % of its starting value after one time constant (τ = RC), to 13.5 % after two and below 1 % after five. The current follows the same curve, starting at V0 / R.
V(t) = Vs × (1 − e^(−t/RC)) when charging from 0 V through a resistor. The capacitor reaches 63.2 % of the supply after one time constant (τ = RC) and 99.3 % after five.
Through a resistor, rearrange the capacitor charge formula to t = RC × ln((V0 − Vf) ÷ (V − Vf)), where V0 is the starting voltage, V the target and Vf the supply voltage. Charging 1 mF from 0 V toward 24 V through 100 Ω reaches 90 % in 230 ms.
t = RC × ln(V0 / V). For 100 µF and 10 kΩ (τ = 1 s), going from 12 V to 1 V takes 1 s × ln 12 = 2.48 s.
Through a resistor, never exactly: the voltage approaches zero exponentially. Engineers use 5 τ as "discharged" (0.67 % left) or work out the time to a specific safe voltage with t = RC ln(V0 / V).
Use R = t ÷ (C × ln(V0 / Vsafe)) and round down to a standard value. A 220 µF capacitor on a 400 V bus needs at most 131 kΩ to reach 50 V within 60 s; 130 kΩ does it in 59.5 s but dissipates 1.23 W the whole time the bus is up.
It depends on the rule that applies to your product. Electrical Construction & Maintenance (2014) quotes US National Electrical Code section 460.6(A) as 50 V, nominal, or less within 1 minute of disconnection, for capacitors rated not over 600 V installed on electrical circuits, and California’s Low-Voltage Electrical Safety Orders (Title 8 §2534.6) set the same limit; check the code edition adopted where you are, and note that a capacitor built into other equipment may fall under that equipment’s standard instead. For X capacitors in appliances, Power Integrations quotes IEC 60335 as under 34 V within 1 s after unplugging.
Through a resistor the current falls with the voltage, so the curve is exponential. A constant-current load (an electronic load or a current-regulated LED driver) pulls the voltage down in a straight line at I / C volts per second, and t = C × ΔV / I. Charging 100 µF to 5 V at 10 mA takes 50 ms.
Charging from 0 V, the resistor turns as much energy into heat as the capacitor stores, whatever its value: half of what the supply delivers. Charging 1 mF to 24 V through 100 Ω heats the resistor by 288 mJ, the same as the 288 mJ stored.
See all free electronics tools →
ProtoFlow is free AI PCB design software: describe a circuit and get a schematic with real parts, PCB layout with an autorouter, SPICE simulation and Gerber files. Native KiCad, Altium and EAGLE import and export. Free on macOS, Windows and Linux.
Free to start. No card required.