What is the formula for the energy stored in a capacitor?
The capacitor energy storage formula is E = ½ × C × V², with C in farads and V in volts giving joules. 1000 µF at 25 V stores 312.5 mJ. The charge is Q = C × V = 25 mC.
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Capacitor energy calculator
How much energy can a load take from a capacitor before the voltage falls too low? This capacitor energy calculator gives the stored energy, the charge and the usable part between two voltages. Work out the hold-up time at a constant power, or the capacitance a hold-up time needs, rounded to standard values with tolerance.
Solve for
Examples
F
V
V
Optional: where the load stops working.
Stored energy E = ½CV²
312.5 mJ
- Charge Q = CV
- 25 mC 6.94 µAh
- Energy in watt-hours
- 86.81 µWh
Enter the lowest voltage your load works at to see how much of this energy it can use.
Uses the nominal capacitance. Class 2 ceramics (X5R, X7R) lose capacitance with DC bias, so they store less than ½CV² at their rated voltage; use the manufacturer’s DC bias curve. Electrolytic tolerance (often ±20 %) scales the energy directly.
How to calculate energy stored in a capacitor
Use E = ½CV². A 1000 µF capacitor at 25 V stores ½ × 0.001 × 25² = 312.5 mJ, with a charge of Q = CV = 25 mC. A load that stops at V2 can only use ½C(V1² − V2²), and at a constant power P that energy lasts t = C(V1² − V2²) ÷ (2P).
Capacitor energy formula and hold-up time equations
The capacitor energy equation, E = ½CV², means the energy in a capacitor grows with the square of its voltage. What a load can use is the difference between the energy at the start and the energy left when it stops working.
E = ½ × C × V² Q = C × V
- E
- stored energy in J
- Q
- stored charge in C
- C
- capacitance in F
- V
- voltage across the capacitor in V
Between a starting voltage V1 and the lowest voltage the load accepts, V2:
ΔE = ½ × C × (V1² − V2²) usable share = 1 − (V2 ÷ V1)²
A switching regulator draws roughly constant power, so the hold-up time is the usable energy divided by the power. With a converter of efficiency η between the capacitor and a load of P, the capacitor supplies P ÷ η. Rearranged for the capacitance, this is the equation TI uses for the bulk capacitor in the UCC28180 datasheet.
t = η × C × (V1² − V2²) ÷ (2P) C = 2 × P × t ÷ (η × (V1² − V2²))
- t
- hold-up time in s
- P
- load power in W
- η
- converter efficiency (1 if P is drawn from the capacitor directly)
During the hold-up the voltage follows V(t) = √(V1² − 2Pt ÷ (ηC)), and the current P ÷ (ηV) rises as it falls.
Worked examples
Example 1: 470 µF on a 400 V bus
E = ½ × 470 µF × 400² = 37.6 J. If the downstream converter runs down to 300 V, it can use ½ × 470 µF × (400² − 300²) = 16.45 J, which is 43.8 % of the stored energy. The remaining 21.15 J at 300 V is still there after the supply is unplugged, which is what a bleeder resistor is for.
Example 2: the PFC bulk capacitor in TI’s UCC28180 datasheet
TI sizes the output capacitor of a 360 W PFC stage so that its 390 V output stays at or above 300 V for one line cycle at 47 Hz, t = 1 ÷ 47 Hz = 21.3 ms. C ≥ 2 × 360 W × 21.3 ms ÷ (390² − 300²) = 247 µF, the figure TI quotes. TI then recommends 10 % of margin and uses 270 µF, the next E12 value, which gives 23.3 ms at nominal capacitance.
Example 3: a 1 F supercapacitor at 100 mW
From 5 V to 3 V the usable energy is ½ × 1 F × (25 − 9) = 8 J, so a 100 mW load drawn straight from the capacitor runs for t = 8 J ÷ 0.1 W = 80 s. If the 100 mW is the output of a 90 % efficient converter, the capacitor supplies 111 mW and the time drops to 72 s. The current rises from 20 mA to 33.3 mA as the voltage falls.
Energy stored for common capacitors and voltages
E = ½CV² for standard values. Doubling the voltage quadruples the energy; doubling the capacitance only doubles it.
| C \ V | 3.3 V | 5 V | 12 V | 24 V | 48 V |
|---|---|---|---|---|---|
| 10 µF | 54.5 µJ | 125 µJ | 720 µJ | 2.88 mJ | 11.5 mJ |
| 100 µF | 545 µJ | 1.25 mJ | 7.2 mJ | 28.8 mJ | 115 mJ |
| 470 µF | 2.56 mJ | 5.88 mJ | 33.8 mJ | 135 mJ | 541 mJ |
| 1 mF | 5.44 mJ | 12.5 mJ | 72 mJ | 288 mJ | 1.15 J |
| 10 mF | 54.4 mJ | 125 mJ | 720 mJ | 2.88 J | 11.5 J |
| 1 F | 5.44 J | 12.5 J | 72 J | 288 J | 1.15 kJ |
Hold-up time per 100 µF at 1 W
Hold-up time scales with C and inversely with P: multiply by your capacitance in units of 100 µF and divide by your power in watts. The voltage windows are examples; 390 V to 300 V is the UCC28180 case.
| V1 to V2 | Usable share | Usable energy per 100 µF | Hold-up per 100 µF at 1 W |
|---|---|---|---|
| 3.3 V to 3 V | 17.4 % | 94.5 µJ | 94.5 µs |
| 5 V to 4.5 V | 19 % | 238 µJ | 238 µs |
| 12 V to 9 V | 43.8 % | 3.15 mJ | 3.15 ms |
| 24 V to 18 V | 43.8 % | 12.6 mJ | 12.6 ms |
| 48 V to 36 V | 43.8 % | 50.4 mJ | 50.4 ms |
| 390 V to 300 V | 40.8 % | 3.1 J | 3.1 s |
Why real capacitors hold less
The formula uses the nominal capacitance, and real parts fall short of it. Aluminum electrolytics are commonly specified at ±20 %, and their endurance rating allows a further change over life (Nichicon’s UPW series allows ±20 % of the initial value), so size for the low end. Class 2 ceramic capacitors (X5R, X7R) lose a large part of their capacitance under DC bias, so a ceramic at its rated voltage stores much less than ½CV²; read the manufacturer’s DC bias curve. At high discharge currents the ESR drop takes a slice out of V1 − V2, and the converter’s own input range and undervoltage lockout set V2. A linear regulator draws a constant current rather than a constant power, so its hold-up is C × (V1 − V2) ÷ I instead.
Glossary
- Hold-up time
- How long a capacitor keeps a load running after its input power drops out.
- Usable energy
- The energy released between the starting voltage and the lowest voltage the load accepts, ½C(V1² − V2²).
- Undervoltage lockout (UVLO)
- The input voltage below which a regulator shuts off; it sets V2.
- ESR
- Equivalent series resistance: it drops voltage at high current and shrinks the usable range.
- Bulk capacitor
- The large capacitor on a supply rail that carries the load through ripple and short dropouts.
- Supercapacitor
- An electric double-layer capacitor with farads of capacitance at a few volts, used for backup power.
Related guides
- Capacitor charge and discharge calculator: voltage over time through a resistor or at constant current.
- Supercapacitor backup time calculator: hold-up including the ESR drop.
Frequently asked questions
How much energy does a capacitor store?
E = ½CV², so voltage matters most. 1000 µF at 25 V stores 312.5 mJ, 470 µF at 400 V stores 37.6 J, and a 1 F supercapacitor at 5 V stores 12.5 J.
How do you convert farads to joules?
You also need the voltage, since a farad measures capacitance and a joule measures energy: E = ½ × C × V². 1 F at 5 V stores 12.5 J, and the same 1 F at 3 V stores 4.5 J.
Why is capacitor energy ½CV² and not CV²?
The voltage rises from zero as the charge goes in, so on average each coulomb is pushed in against half the final voltage. Charging from a fixed supply through a resistor, the supply delivers QV and the other half is lost as heat in the resistor, whatever its value.
How do I calculate hold-up time from a capacitor?
At a constant power P, t = C × (V1² − V2²) ÷ (2P), where V1 is the starting voltage and V2 the lowest voltage the load works at. A 1 F supercapacitor going from 5 V to 3 V at 100 mW lasts 80 s.
How do I size a hold-up capacitor?
C = 2 × P × t ÷ (V1² − V2²), then add margin for tolerance and aging. TI’s UCC28180 datasheet sizes a 360 W PFC stage to stay above 300 V from 390 V for one 47 Hz cycle: 247 µF minimum, built with 270 µF.
Why can a load not use all the energy in a capacitor?
The load stops working at some minimum voltage V2, and the energy left below it, ½CV2², stays in the capacitor. The usable share is 1 − (V2 / V1)²: 75 % when the load runs down to half the starting voltage, only 19 % when it stops at 90 %.
Does doubling the voltage double the stored energy?
It quadruples it, since energy goes with V². This is why hold-up capacitors sit on the highest-voltage rail available, such as the 390 V PFC output in the UCC28180 example, rather than on a supply’s low-voltage output.
How do I convert capacitor energy to watt-hours or mAh?
1 Wh = 3600 J, and 1 mAh = 3.6 C of charge. 1000 µF at 25 V is 86.8 µWh, which shows why capacitors are for short hold-up, not for runtime.
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