This tool calculates the time it takes to discharge a capacitor to a specified voltage level. The schematic consists of an RC network as shown in the title picture.
As a first step calculate the ratio between the initial (Vo) and final (V) voltage level.
As an example, if the initial voltage is 5V and the final voltage is 1V, the voltage ratio V/Vo = 1/5 = 0.2. As the capacitor is discharging, the final voltage value will always be lower than the initial value and V/Vo < 1.
The next step in calculating the time it takes to discharge a capacitor is to enter:
- Resistance (R)
- Capacitance (C)
- Voltage Ratio (V/Vo calculated above)
Formula
V = Vo*e−t/RC
t = RC*Loge(Vo/V)
The time constant time constant Ï„ = RC, where R is resistance and C is capacitance. The time t is typically specified as a multiple of the time constant.
Example Calculation
Example 1
Use values for Resistance, R = 10 Ω and Capacitance, C = 1 µF.
For an initial voltage of 10V and final voltage of 2V, we use the first calculator to find a voltage ratio calculator of 0.2. Use this value in the second calculator to find the time it takes to discharge to this level is 16.1 µs.
For the same RC values the time it takes to get to a ratio of 1/100 or 0.01 is 46.1 µs.
Example 2
Find the time to discharge a 470 µF capacitor from 240 Volt to 60 Volt with 33 kΩ discharge resistor.
Using these values in the above two calculators, the answer is 21.5 seconds.
Frequently Asked Questions
How fast does a capacitor discharge?
The speed at which a capacitor discharges depends on its capacitance and the resistor it is connected to. It depends on the RC time constant.
In general, a capacitor is considered fully charged when it reaches 99.33% of the input voltage. Conversely a cap is fully discharged when it loses the same amount of charge. The amount of charge remaining on the cap in this case is 0.67%.
The ratio Vo/V = 0.67/100 = 0.0067 can be used in the calculator above. For a 470 µF capacitor and 33 kΩ it takes 77.64 seconds. This is approximately the same as 5*RC (or five time constants).
The lower the RC time constant the quicker the discharge.
💡 This time constant is the reason that an LED light on your TV or any electronic equipment will stay on even after you turn it off. It’s also why you should be careful when opening up electronics – the capacitors can carry a significant voltage.
Why do we use five time constants?
In a related post we explained why it takes 5 time constants to charge a capacitor. The reasoning is similar for capacitor discharging. The table below shows the multiple of time constant vs. % charge.
| Number of time constants | % charge |
|---|---|
| 0 | 100.0000000000 |
| 0.5 | 60.6530659713 |
| 1 | 36.7879441171 |
| 2 | 13.5335283237 |
| 3 | 4.9787068368 |
| 4 | 1.8315638889 |
| 5 | 0.6737946999 |
| 6 | 0.2478752177 |
| 7 | 0.0911881966 |
| 8 | 0.0335462628 |
| 9 | 0.0123409804 |
| 10 | 0.0045399930 |
At time t=0, the capacitor is fully charged (100%). As time progresses, the charge decreases exponentially. At 5 time constants the amount of charge remaining is less than 1%.