Capacitor Energy Calculator

Introduction to the energy stored in a capacitor

A capacitor does not consume energy, it borrows it. Move charge onto one plate and pull the same amount off the other and the separated charge sets up an electric field in the dielectric between them. That field is where the energy actually resides, and an ideal capacitor hands every joule of it back when a load is connected across the terminals. This page turns two numbers you can read straight off a datasheet or a bench supply, the capacitance $C$ and the terminal voltage $V$, into the joules held in that field, plus the stored charge, the watt-hour equivalent, the number of electrons that had to be moved, and the energy the source must spend to put them there.

The SI unit of capacitance is the farad. NIST Special Publication 330 lists it as the coherent derived unit for capacitance, equal to the coulomb per volt, $F=\frac{C}{V}$, and in base units $F={\mathrm{kg}}^{-1}\cdot m^{-2}\cdot s^4\cdot A^2$. One farad is an enormous capacitance, so real parts live far below it: a decoupling ceramic is 100 nF, a snubber film capacitor 1 µF, a supply bulk electrolytic a few hundred µF, and a hold-up supercapacitor several farads. The calculator therefore takes capacitance in pF, nF, µF, mF or F and voltage in mV, V or kV, so you never have to hand-convert 4.7 µF into 4.7e-6 and hope you counted the zeros correctly.

Two features of the physics dominate every design decision that follows. The first is that stored energy goes as the square of the voltage, so derating a part to 60 % of its rated voltage does not cost 40 % of the energy, it costs 64 % of it. The second is that when a constant-voltage source charges a capacitor through any series resistance, exactly half the energy the source delivers is dissipated before the capacitor is full, and the fraction does not depend on the resistance. Both results are derived in full below, and both are reported by the calculator alongside the stored energy so that the numbers you carry into a design already account for them.

How to use the unit selectors, the derating table and the share link

1. Enter the capacitance and pick its unit. Use the value you expect the part to have in circuit, not necessarily the nameplate number. For Class II ceramics that means the capacitance measured under your DC bias, which can be substantially lower than the marking (see the limitations section).
2. Enter the working voltage and pick its unit. This is the voltage the capacitor actually sits at, not its rated voltage. If you want the nameplate energy, enter the rated voltage and read the derating table, which shows what several standard derating factors leave you.
3. Optionally enter a cut-off voltage. Real hold-up circuits never discharge a capacitor to zero, because the downstream regulator drops out. Enter the voltage at which the load stops working and the calculator reports the usable energy $\Delta E=\frac{1}{2}C(V_0^2-V_1^2)$ rather than the full stored energy, which is the number that actually matters.
4. Read the breakdown. Energy is given in an automatically chosen SI prefix and again in joules and watt-hours; charge, electron count, source energy and series-resistance loss appear beneath it.
5. Check the curve and the derating table. The chart plots $E(v)=\frac{1}{2}C{v}^2$ from zero up past your working point so the quadratic shape is visible rather than merely asserted. The table applies published voltage derating factors and shows how little energy survives them.
6. Share or export. The share button rewrites the address bar with your inputs so the link reproduces the exact calculation, and the CSV button downloads the full breakdown for a design review.

Every input is validated in JavaScript rather than by the browser, so an empty box, a negative capacitance, a cut-off voltage above the starting voltage or a value large enough to overflow a double all produce a specific message instead of a blank panel or a stale number.

Formula derivation: where the factor of one half comes from

Capacitance is defined by the charge a device holds per volt across it, $C=\frac{Q}{V}$, which rearranges to $Q=CV$. That linear relationship is the whole reason the energy formula carries a one half, and it is the single most common source of confusion on this topic. If you multiply the final charge by the final voltage you get $QV=C{V}^2$, which is twice the answer. The mistake is assuming every electron was pushed across the full voltage.

It was not. Charging is incremental. When the capacitor already holds charge $q$, its terminal voltage is $v=\frac{q}{C}$, and moving one more increment $dq$ against that potential costs work

Formula: d W = v d q = q / C d q

$$dW=vdq=\frac{q}{C}dq$$

The very first increment crosses zero volts and costs nothing at all. The last increment crosses the full voltage $V$. Because the voltage climbs linearly with the charge already delivered, the average potential the charge is pushed through is $V/2$, not $V$. Integrating from an empty capacitor to a final charge $Q$ makes that exact:

Formula: E = ∫_0^Q q / C d q = Q^2 / (2 C)

$$E={\int }_0^Q\frac{q}{C}dq=\frac{Q^2}{2C}$$

Substituting $Q=CV$ into that result gives the three equivalent forms the calculator uses, all of which describe the same joules:

Formula: E = 1 / 2 C V^2 = 1 / 2 Q V = Q^2 / (2 C)

$$E=\frac{1}{2}C{V}^2=\frac{1}{2}QV=\frac{Q^2}{2C}$$

The units work out because one joule is one coulomb-volt: NIST Special Publication 330 gives the joule as $J=N\cdot m=\mathrm{kg}\cdot m^2\cdot s^{-2}$ and the volt as the watt per ampere, so farads times volts squared is indeed joules with no stray constant.

The same energy can be counted a second way, by looking at the field rather than the plates. In a linear dielectric of permittivity $\epsilon$ carrying field $ℰ$, the energy density is $u=\frac{1}{2}\epsilon {ℰ}^2$. Integrating that density over the volume between the plates reproduces $\frac{1}{2}C{V}^2$ exactly. This is the view Purcell and Morin take in Electricity and Magnetism, and it is the reason it is wrong to say a capacitor consumes energy: the joules sit in the field in the dielectric and come back out when the field collapses.

Because the calculator also reports how many electrons were displaced, it uses the 2019 SI value of the elementary charge, which is exact by definition: $e=1.602176634\times {10}^{-19}C$, so the electron count is simply $N=Q/e$.

The charging loss that no resistor value can avoid

Now connect an ideal source of fixed voltage $V$ to an uncharged capacitor through a resistance $R$. The source pushes out a total charge $Q=CV$ while its own terminals remain at $V$ the whole time, so the energy it gives up is

Formula: E_src = V Q = C V^2

$$E_{\mathrm{src}}=VQ=C{V}^2$$

The capacitor keeps only $\frac{1}{2}C{V}^2$, so the difference must have gone into the resistance:

Formula: E_R = C V^2 − 1 / 2 C V^2 = 1 / 2 C V^2

$$E_R=C{V}^2-\frac{1}{2}C{V}^2=\frac{1}{2}C{V}^2$$

There is no $R$ in that answer. The direct calculation confirms it. The current is $i(t)=\frac{V}{R}{e}^{-t/RC}$ with time constant $\tau =RC$, and the heat dissipated is

Formula: E_R = ∫_0^∞ i^2 R d t = V^2 / R ⋅ (R C) / 2 = 1 / 2 C V^2

$$E_R={\int }_0^{\infty }{i}^{2}Rdt=\frac{V^2}{R}\cdot \frac{RC}{2}=\frac{1}{2}C{V}^2$$

The resistance cancels. Halving $R$ doubles the peak current and halves the time constant, and the same joules end up as heat, just faster. The charging efficiency from an empty start is therefore exactly $\eta =50\%$. MIT OpenCourseWare 6.622 Power Electronics states the general form of this result, $\Delta E_{\mathrm{loss}}=\frac{1}{2}C{(V_s-v_{\mathrm{ci}})}^2$, and notes that it does not depend on the resistance value or even on the resistor being linear. The practical corollary is the one that matters: efficiency improves only if the voltage step is made small, which is why switched-capacitor converters transfer charge in many small steps, and why inductive converters exist at all.

Worked example: sizing a supercapacitor hold-up bank

A logging device runs from a 2.7 V rail and draws 250 mW. When mains power fails it must stay alive long enough to close its files, and its regulator drops out below 1.6 V. A single 10 F, 2.7 V supercapacitor is proposed. How long does it actually last?

The total stored energy at 2.7 V is

Formula: E = 1 / 2(10 F) (2.7V)^2 = 36.45 J

$$E=\frac{1}{2}(10F){(2.7V)}^2=36.45J$$

That looks like 146 seconds of runtime at 250 mW, and that number is wrong, because the last 1.6 V of charge is unusable. The energy available between 2.7 V and the 1.6 V dropout is

Formula: Δ E = 1 / 2(10)(2.7^2 − 1.6^2) = 23.65 J

$$\Delta E=\frac{1}{2}(10)({2.7}^2-{1.6}^2)=23.65J$$

which is 94.6 seconds at 250 mW, or 65 % of the headline figure. Enter 10 F, 2.7 V and a 1.6 V cut-off into the calculator above and it reports exactly these numbers, along with a stored charge of $Q=(10)(2.7)=27C$, an electron count of about $1.685\times {10}^{20}$, and 36.45 J of loss in the charging path if the bank is charged straight from a 2.7 V rail through a resistor.

That last figure is the sting in the tail. Charging the bank from empty through a plain series resistor costs 72.9 J from the source to store 36.45 J. If the device recharges after every outage, the charging path, not the load, dominates the energy budget, and the resistor has to be sized for the heat rather than only for inrush current.

The comparison table below runs the same calculation across the families of capacitor you are likely to be choosing between, at a capacitance and voltage typical for each. It shows why supercapacitors are used for hold-up and ceramics are not, and it applies the voltage derating factors published in NASA EEE-INST-002 Table 4 so the energy column reflects what a flight-grade design would actually be allowed to use.

The derating column is where the quadratic law bites. A factor of 0.60 on voltage leaves ${0.60}^2=36\%$ of the energy, and the 0.50 factor NASA applies to solid tantalum leaves just 25 %. In general, derating to a fraction $k$ of rated voltage gives

Formula: E_k = 1 / 2 C (kV)^2 = k^2 ⋅ 1 / 2 C V^2

$$E_k=\frac{1}{2}C{(kV)}^2=k^2\cdot \frac{1}{2}C{V}^2$$

Reading the result: what those joules will and will not buy

A joule is a watt-second, so the fastest sanity check on any capacitor energy figure is to divide it by the load power. 36.45 J at 250 mW is 146 s; 147 mJ at 5 W is 29 ms; 125 µJ at 100 mW is 1.25 ms. That immediately sorts capacitors into their real roles. Millijoule-class parts ride out switching transients and hold rails steady between converter cycles. Joule-class banks run orderly shutdowns and fire photoflash tubes. Nothing in the table comes close to a coin cell, which stores on the order of 3000 J, and that gap is the reason capacitors complement batteries rather than replace them.

The watt-hour figure is there for the same reason, because energy storage is quoted in Wh outside electronics. The conversion is exact: $1\mathrm{Wh}=3600J$, so 36.45 J is 10.1 mWh. Seeing a farad-scale supercapacitor land in the milliwatt-hour range next to an AA cell at roughly 3 Wh puts the energy density difference in perspective faster than any argument.

The charge and electron figures are diagnostic rather than decorative. If the reported charge is a large fraction of what your charging source can deliver in the time available, the capacitor will not reach the voltage you assumed and the energy figure is optimistic. The source-energy and resistive-loss rows tell you what the charging path has to survive thermally, which is a routine cause of scorched inrush resistors in supplies that cycle frequently.

Limitations of the ideal model and the assumptions behind every number

Every figure on this page comes from the ideal linear capacitor, and the following assumptions are baked into it. Stating them plainly is more useful than a disclaimer, because each one has a known size and a known direction of error.

Capacitance is assumed constant. It very often is not. Class II ceramic dielectrics, the X5R, X6S, X7R and Y5V materials, are ferroelectric barium titanate, and their permittivity falls once a DC bias polarises the material. Murata's published example puts a 100 µF, 6.3 V high dielectric constant part at roughly 10 % capacitance loss under 1.8 V of bias with X5R material, and roughly 40 % loss with Y5V. The same source notes that Class I temperature-compensating materials such as C0G, film capacitors and polymer electrolytics show essentially none of this effect. If you enter a Class II nameplate value, the energy this page reports is an upper bound and can be optimistic by a factor of two or more. Measure the capacitance under your actual bias, or use the manufacturer's bias curve, and enter that.

The voltage is assumed to be within ratings and steady. Exceeding the rated voltage does not simply store more energy, it initiates dielectric breakdown, and for an aluminium electrolytic that can mean venting. Voltage derating is standard practice for exactly this reason: NASA EEE-INST-002 Table 4 requires the sum of DC and peak AC ripple to be multiplied by 0.60 for ceramic and plastic film, 0.50 for glass and foil tantalum, 0.60 for wet-slug tantalum at 70 °C falling to 0.40 at 110 °C, and 0.50 for solid tantalum at 70 °C falling to 0.30 at 110 °C. Aluminium electrolytics are not in that table at all, since they are not a flight-qualified family; manufacturers publish their own guidance, commonly in the region of 80 % of rated voltage, and that figure is a vendor recommendation rather than a standards requirement. The calculator does not enforce any rating, so it will happily compute the energy of a part you are about to destroy.

Parasitics are ignored. A real capacitor is the ideal element in series with an equivalent series resistance and inductance, in parallel with a leakage resistance, plus a dielectric absorption network. ESR converts a fraction of every charge and discharge into heat and limits how fast the stored energy can be extracted; on a supercapacitor with hundreds of milliohms it can dominate a fast discharge entirely. Leakage bleeds the charge away over hours to days, which matters greatly for hold-up applications where the bank sits charged and idle. Dielectric absorption returns a small voltage after a short circuit is removed, which is a genuine safety issue on high-voltage banks that have been discharged once already. IEC 60384-1 defines the standard terms and test methods for these quantities, and the datasheet numbers you should use come from tests specified there.

Temperature and ageing are ignored. Dielectric permittivity, ESR and leakage are all temperature dependent, electrolytics lose capacitance as the electrolyte dries, and Class II ceramics also age, losing capacitance roughly logarithmically with time since the last thermal excursion above the Curie point. None of that appears in $\frac{1}{2}C{V}^2$.

Tolerance is ignored. The capacitance you buy is a preferred value with a tolerance band. IEC 60063 defines the E series and pairs each with a tolerance: E6 (10, 15, 22, 33, 47, 68) with plus or minus 20 %, E12 with plus or minus 10 %, and E24 with plus or minus 5 %. A part at the bottom of a plus or minus 20 % band delivers 20 % less energy than nominal, and capacitors frequently carry asymmetric tolerances such as -20 %/+80 %, so a worst-case energy budget should use the minimum capacitance, not the marking.

Ideal circuit behaviour is assumed elsewhere. The 50 % charging loss result assumes an ideal constant-voltage source, an initially empty capacitor and a purely resistive path; it is a hard lower bound on loss for that topology, not for all topologies. Series and parallel combinations are also outside the scope of the single-part calculation: capacitances add in parallel, $C_p=\sum _i{C}_i$, while in series they combine as $\frac{1}{C_s}=\sum _i\frac{1}{C_i}$. Compute the equivalent capacitance and the total bank voltage first, then enter those. Series strings also need balancing resistors or active balancing, or the string voltage will not divide the way the arithmetic assumes.

Safety. A joule-class bank at hundreds of volts is genuinely dangerous, and this page is an arithmetic aid, not a safety authority. Treat any bank above roughly 1 J and 50 V as capable of injury, discharge it deliberately through a resistor rated for the energy, verify with a meter, and expect dielectric absorption to put some voltage back.

Sources and further reading

Unit definitions and the elementary charge: National Institute of Standards and Technology, NIST Special Publication 330 (2019), The International System of Units (SI), which gives the farad as the coherent derived unit for capacitance equal to C/V and to kg^-1 m^-2 s⁴ A², the joule as N⋅m, and the exact elementary charge e = 1.602176634 × 10^-19 C (nvlpubs.nist.gov). Charging loss independent of series resistance: Massachusetts Institute of Technology OpenCourseWare, 6.622 Power Electronics, Prof. David Perreault, Lecture 32, Spring 2023 (ocw.mit.edu). Capacitor terms, tests and specification: International Electrotechnical Commission, IEC 60384-1:2021, Fixed capacitors for use in electronic equipment - Part 1: Generic specification, edition 6.0 (webstore.iec.ch). Preferred capacitance values and their tolerances: IEC, IEC 60063:2015, Preferred number series for resistors and capacitors, edition 3.0, Tables 1 and 3. Voltage derating factors: National Aeronautics and Space Administration, Goddard Space Flight Center, EEE-INST-002: Instructions for EEE Parts Selection, Screening, Qualification, and Derating, NASA/TP-2003-212242, Section C1 Table 4 (nepp.nasa.gov). DC bias behaviour of Class II ceramics: Murata Manufacturing Co., Ltd., The voltage characteristics of electrostatic capacitance and the ceramic capacitor FAQ on capacitance change under applied DC voltage (article.murata.com). Derivations follow the standard treatments in E. M. Purcell and D. J. Morin, Electricity and Magnetism, 3rd edition, Cambridge University Press, 2013, and P. Horowitz and W. Hill, The Art of Electronics, 3rd edition, Cambridge University Press, 2015.

Questions engineers ask about stored capacitor energy