Library / Electrical / ELE-13
Electrical
Capacitor discharge half time
Time for a discharging capacitor’s voltage to fall to half of its starting value.
- Ref
- ELE-13
- Rev
- v1
- Field
- Electrical
- Result
- t½ [ms]
0.693 is ln 2. t½ in seconds, R in ohms, C in farads. With R in kΩ and C in µF the answer is in milliseconds. The calculator uses ln 2 exactly (0.6931…).
A discharging capacitor loses voltage by the same fraction in every equal interval, so the time to halve is the same whether you start at 10 V or at 1 V. It equals 0.693 time constants.
Calculator
- Time constant τ = RC1,000 ms
- Half time in seconds0.6931 s
- Time to fall to 1% of the start4.605 s
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Half time depends only on RC, not on the starting voltage. Each half time halves the voltage again: after 3 half times it is 12.5% of its start, and after 10 half times about 0.1%.
- t½ = 0.693 τ, so the time constant is about 1.44 times the half time.
- Any process that loses a constant fraction per unit time, such as radioactive decay, has the same ln 2 relation between half-life and time constant.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (ms) | Local step | Change in result |
|---|---|---|---|---|---|---|
| C Capacitance | 0.1 to 10,000 µF | 0.6931 | 69,310 | 69,310 | +10% | +10 % |
| R Resistance | 0.1 to 1,000 kΩ | 6.931 | 69,310 | 69,310 | +10% | +10 % |
Sweep of Resistance
| Resistance (kΩ) | Half time (ms) |
|---|---|
| 0.1 | 6.931 |
| 0.3162 | 21.92 |
| 1 | 69.31 |
| 3.162 | 219.2 |
| 10 | 693.1 |
| 31.62 | 2,192 |
| 100 | 6,931 |
| 316.2 | 21,920 |
| 1,000 | 69,310 |
Sweep one variable
Variables and how each one contributes
Half time
What the equation gives you. Everything else feeds it.
Resistance
Resistance the capacitor discharges through. Directly proportional: doubling R doubles the half time.
Capacitance
Capacitance of the capacitor. Directly proportional: doubling C doubles the half time.
Worked example with the values above
- Time constant: τ = RC = 10 × 100 = 1,000 ms.
- Half time: t½ = ln 2 × τ = 0.6931 × 1,000 = 693.1 ms.
Assumptions and limits
- Discharge through a fixed resistor towards 0 V, so the decay is exponential.
- Ideal capacitor without leakage or dielectric absorption.
- The 0.693 factor is ln 2 and applies only to exponential decay.
Common mistakes
- Expecting the capacitor to be empty after two half times. It still holds 25% of its starting voltage.
- Mixing up t½ and τ. They differ by the factor 0.693.
- Reading kΩ × µF as seconds. The product is in milliseconds.
Statistics and uncertainty
Half time is easy to read off an oscilloscope trace, which makes it a practical way to estimate an unknown capacitance or leakage resistance, at the price of reading error. For independent errors, δt½/t½ ≈ √[(δR/R)² + (δC/C)²], the same as for τ. A measured t½ gives C = t½/(0.693 R).
Where it is used
- Estimating how long stored charge takes to fall to a safe level.
- Measuring an unknown capacitance or resistance from a discharge curve.
- Setting timing and delay intervals.
Maths you need
- Natural logarithm: 0.693 = ln 2 comes from solving e−x = ½
- Exponential decay: a constant fraction is lost in each equal interval