Capacitor Energy and Backup Time Calculator
Capacitor Energy and Backup Time Calculator
Energy stored in a capacitor in joules, watt-hours and milliamp-hours — and the question people actually arrive with: how long it holds a load up between Vₘₐₓ and Vₘᵢₙ at constant current, constant power or constant resistance, or how much capacitance a required hold-up needs.
Capacitor energy and hold-up
a 1 F supercapacitor at 5 V holding up a 100 mA constant-current load down to 3.3 V, 50 mΩ ESR
Stored energy, and three different hold-up times
- V₁
- the starting voltage, less the instant IR drop across the ESR
- V₂
- the lowest voltage the load still works at
- E
- joules; divide by 3600 for watt-hours
- C(V₁−V₂)
- the charge delivered, in coulombs; divide by 3600 for amp-hours
Worked example
a 1 F supercapacitor at 5 V holding up a 100 mA constant-current load down to 3.3 V, 50 mΩ ESR
E = ½CV² = 0.5 × 1 × 5² = 12.5 J, which is 3.472 mWh
Only the part above 3.3 V is usable: ½ × 1 × (5² − 3.3²) = 7.055 J
The ESR drops 100 mA × 50 mΩ = 5 mV the instant the load appears, so the discharge really starts at 4.995 V
At constant current, t = C·ΔV ÷ I = 1 × (4.995 − 3.3) ÷ 0.100 = 16.95 s; the charge moved is 471 µAh
The same capacitor and the same window would give 14.06 s into a 500 mW constant-POWER load and 20.75 s into a 50 Ω resistor — three answers, one stored energy
The three load laws from the same capacitor
| Load | Hold-up equation | This example |
|---|---|---|
| Constant current (a current sink, an LED driver) | t = C(V₁ − V₂) ÷ I | 16.95 s at 100 mA |
| Constant power (any switching regulator) | t = C(V₁² − V₂²) ÷ 2P | 14.06 s at 500 mW |
| Constant resistance (a plain resistive load) | t = RC · ln(V₁ ÷ V₂) | 20.75 s into 50 Ω |
Energy, and how long it lasts
A capacitor charged to V holds ½CV² joules. The square is the whole story: a supercapacitor at 5 V holds more than twice what the same part holds at 3.3 V. Watt-hours are joules ÷ 3600, and the milliamp-hour figure people ask for is the charge, C×ΔV ÷ 3.6 — a 1 F capacitor swinging 1.7 V moves about 0.47 mAh, which is why a supercapacitor backs up a clock chip and not a phone.
You never get all of it. Only the energy between the voltage you start at and the lowest voltage the load still works at is usable, and that is a fraction: from 5 V down to 3.3 V you can take out 1 − (3.3/5)² = 56% of what is stored. Widening the window is far more effective than adding capacitance, which is why hold-up circuits put a buck-boost converter after the capacitor and run it down to a volt or two.
Three loads, three answers. A constant-current load takes the voltage down in a straight line, so the time is just C·ΔV ÷ I. A resistor is an ordinary exponential decay: t = RC·ln(V₁/V₂), the same curve the RC time constant calculator draws in full, so it is not repeated here. A switching regulator is neither: it holds its output power, so as its input falls it draws MORE current, and the only honest way to get the time is from the energy. Set P·t equal to the usable energy ½C(V₁²−V₂²) and t = C(V₁²−V₂²) ÷ 2P. Treating a switching regulator as a fixed current is the commonest mistake in hold-up design and it always flatters the answer.
ESR, and what supercapacitors really do. Equivalent series resistance costs you voltage the instant the load appears — i×ESR, before any energy has been taken out — and that step comes straight off the top of your window. With a hundred milliohms and an amp, half a volt disappears at t = 0. Supercapacitors bring three more problems worth knowing before you design one in. They leak: a float current in the microamps to tens of microamps that never stops, which matters on a coin-cell-backed design. They cannot be stacked in series without balancing — cell voltages drift apart, one cell goes over its rating and dies — so a series stack needs bleed resistors or an active balancer. And their life is set by voltage and temperature together: Eaton quotes up to twenty years at 25 °C with a 10–15% derate on the float voltage, and far less hot and at full rating. Use two to three times the datasheet ESR in this page to represent an aged part. For the battery alternative see the battery life calculator, and for reading a capacitor’s printed value the capacitor code calculator.
Frequently asked questions
How do I calculate the energy stored in a capacitor?
E = ½CV², in joules with C in farads and V in volts. A 1 F supercapacitor at 5 V holds 12.5 J, which is 3.472 mWh.
How long will a capacitor power my circuit?
It depends on the load law. At a constant current I, t = C(V₁−V₂) ÷ I. At constant power P, t = C(V₁²−V₂²) ÷ 2P. Into a resistor, t = RC·ln(V₁÷V₂). V₂ is the lowest voltage your circuit still works at.
What size capacitor do I need for a given hold-up time?
Turn the same equation round: at constant power, C = 2Pt ÷ (V₁²−V₂²). Then round up to a standard part, because capacitance tolerances are wide and capacitance falls with age.
How does ESR affect hold-up time?
It costs you an immediate i×ESR volts the moment the load appears, which comes off the top of your usable window before any charge has moved. On a high-current load from a supercapacitor that step often decides the answer.
Can I put supercapacitors in series?
Only with balancing. Leakage currents differ between cells, so the voltages drift apart and one cell ends up over its rating. Use bleed resistors across each cell or an active balancing IC, and derate the stack voltage.
Related calculators
References
- Eaton. Supercapacitor modules: frequently asked questions. Leakage and float current depend on the balancing scheme; life is set by voltage and temperature together, with a 10–15% float-voltage derate at 25 °C giving up to twenty years.
- Richardson RFPD. How to Calculate Supercapacitor Requirements, white paper, July 2023. C = 2E ÷ (V₁² − V₂²) for a constant-power discharge; Vₜ = V₀ − It/C − I·ESR; take end-of-life ESR as 200% of the datasheet figure.
- Abracon. Designing with Supercapacitors, April 2024. Discharge time t = C(Vₛₜₐᵣₜ² − Vₘᵢₙ²) ÷ 2P for a constant-power load, and 2–3× the datasheet ESR to allow for ageing.
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. Chapter 1: capacitors, stored energy and RC discharge.