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Supercapacitor specific capacitance, energy density and battery C-rate calculator

Q: How do you calculate specific capacitance from a charge–discharge curve?

Use C = I · Δt / (m · ΔV): discharge current times discharge time, divided by the active mass and the voltage window. Take the IR drop off the window first.

Q: How is energy density calculated for a supercapacitor?

E = 1/2 · C · ΔV², with C the cell capacitance per total mass and ΔV the voltage window. In F/g and V this gives J/g; divide by 3.6 for Wh/kg. Power density is E divided by the discharge time.

Specific capacitance from a galvanostatic discharge is C = I·Δt / (m·ΔV), with I the current, Δt the discharge time, m the active mass and ΔV the voltage window without the IR drop; from a cyclic voltammogram it is C = ∮I dV / (2·m·v·ΔV) at scan rate v. For a two-electrode cell the energy density is E = ½·C·ΔV² and the power density P = E / Δt; for a battery, the current at n C is n × the capacity and a full charge or discharge takes 1/n hours.

What it does: Calculates specific capacitance from charge–discharge or CV curves, the energy and power density of a two-electrode cell, and battery C-rate, current and time.
Formula: $\begin{aligned} \text{GCD: } C &= I \cdot \Delta t \div (m \cdot \Delta V) \\ \text{CV: } C &= \textstyle\oint I\,dV \div (2 \cdot m \cdot v \cdot \Delta V) \\ E &= \tfrac{1}{2} \cdot C \cdot \Delta V^2 \\ P &= E \div \Delta t \\ \text{battery current} &= \text{C-rate} \times \text{capacity} \end{aligned}$
Example: A symmetric cell (2 mg total) discharged at 1 mA for 100 s over 0.95 V after the IR drop: C = 0.001 × 100 ÷ (0.002 × 0.95) = 52.6 F/g, E = 6.60 Wh/kg.

Last updated 8 November 2026

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At a glance

Supercapacitor & battery calculator at a glance
GCD	$C = I \cdot \Delta t / (m \cdot \Delta V)$ , $\Delta V$ without the IR drop
CV	$C = \oint I\,dV$ ${}/ (2 \cdot m \cdot v \cdot \Delta V)$ , whole loop
Cell vs electrode	Symmetric cell: $\text{per-electrode value}$ ${}= 4 \times \text{cell value}$ per total mass
Energy, power	$E = \tfrac{1}{2} \cdot C \cdot \Delta V^2$ (Wh/kg), $P = E / \Delta t$ (W/kg); two-electrode cells only
Battery	$\text{current} = \text{C-rate} \times \text{capacity}$ ; $\text{time} = 1 / \text{C-rate}$ hours
Output	F/g, mF/cm², Wh/kg, W/kg, µWh/cm²; CSV, curve (PNG, SVG)
Runs in	Your web browser; nothing is uploaded
Price	Free, no account needed

What it does

How to use the supercapacitor & battery calculator

Pick the measurementChoose charge–discharge (GCD), cyclic voltammetry (CV) or battery C-rate.
Say what was measuredTwo-electrode cell (enter the mass of both electrodes) or three-electrode single electrode (enter the working electrode’s mass). Add the electrode area for areal values.
Enter the curveGCD: discharge current, discharge time, voltage window and IR drop, or paste the discharge curve. CV: scan rate and the loop area and window, or paste the whole loop.
Read the resultsGet the capacitance in F/g (and per electrode for a symmetric cell), energy and power density for a cell, and copy or download them.

How it works

$\begin{aligned} C &= \frac{I \cdot \Delta t}{m \cdot \Delta V} && \text{(GCD)} \\ C &= \frac{\oint I\,dV}{2 \cdot m \cdot v \cdot \Delta V} && \text{(CV)} \\ E &= \tfrac{1}{2} \cdot C \cdot \Delta V^2 \\ P &= E / \Delta t \end{aligned}$

Galvanostatic charge–discharge. The gravimetric specific capacitance is $C = I \cdot \Delta t / (m \cdot \Delta V)$ , with $I$ the discharge current, $\Delta t$ the discharge time, $m$ the mass of active material and $\Delta V$ the potential window during discharge, excluding the initial IR drop (Idris, Mamba & Xolile 2026, RSC Adv). GCD is the method closest to how a device is used.

Cyclic voltammetry. The capacitance comes from integrating the current over the CV loop with the scan rate $v$ , the mass and the window (Idris et al. 2026). The tool uses the whole closed loop: each sweep passes a charge $Q = \int I\,dt = \int I\,dV / v$ , the loop contains both the charging and the discharging sweep, and $C = Q / \Delta V$ , so $C = \oint I\,dV / (2 \cdot m \cdot v \cdot \Delta V)$ . CV and GCD values for the same material often differ, because of scan rate, current density and resistance.

Two-electrode or three-electrode. A three-electrode measurement gives one electrode against a reference. In a two-electrode cell the two electrodes act as capacitors in series, so a symmetric cell’s capacitance per total mass is about one quarter of a single electrode’s (Idris et al. 2026). The tool reports the cell value and, for a symmetric cell, the per-electrode value $4 \times C_\mathrm{cell}$ . It always says which one it reports.

Energy and power. Energy density is half the capacitance times the square of the voltage, $E = \tfrac{1}{2} \cdot C \cdot \Delta V^2$ , and the average power density is the energy divided by the discharge time (Idris et al. 2026). The tool computes them only for a two-electrode cell, from the cell capacitance per total mass and the same $\Delta V$ (after the IR drop), and converts 1 J/g to 1/3.6 Wh/kg. For CV, $\Delta t$ is one sweep, $\Delta V / v$ .

Battery C-rate. At a rate of $n$ C the current is $n \times$ the capacity and a full charge or discharge takes $1/n$ hours: 1 C of a 1.9 mAh cm⁻² electrode is 1.9 mA cm⁻² and 6 C about 11.4 mA cm⁻² (Adv Sci 2024), and 10 C takes 6 minutes (Research 2026).

Worked example: a symmetric two-electrode cell

A symmetric cell with 2 mg of active material in total is discharged at 1 mA from 1.0 V; the IR drop is 0.05 V and the discharge takes 100 s.

$\Delta V = 1.0 - 0.05 = 0.95\ \mathrm{V}$ .
$C = 0.001\ \mathrm{A} \times 100\ \mathrm{s} / 0.95\ \mathrm{V}$ $= 0.1053\ \mathrm{F}$ , so $C_\mathrm{cell} = 0.1053\ \mathrm{F} / 0.002\ \mathrm{g} = \mathbf{52.6\ F/g}$ .
Per electrode: $4 \times 52.6 = 210.5\ \mathrm{F/g}$ .
$E = \tfrac{1}{2} \times 52.6 \times 0.95^2\ \mathrm{J/g} = 23.75\ \mathrm{J/g}$ $= \mathbf{6.60\ Wh/kg}$ .
$P = 6.60\ \mathrm{Wh/kg} \times 3600 / 100\ \mathrm{s}$ $= \mathbf{237.5\ W/kg}$ .

Battery: a 2,500 mAh cell at 0.5 C needs 1,250 mA and takes 2 h to charge or discharge fully.

Frequently asked questions

How do you calculate specific capacitance from a charge–discharge curve?

Use $C = I \cdot \Delta t / (m \cdot \Delta V)$ : discharge current times discharge time, divided by the active mass and the voltage window. Take the IR drop off the window first.

How do you calculate specific capacitance from a CV curve?

Integrate the current over the voltage for the whole loop and divide by twice the mass, the scan rate and the window: $C = \oint I\,dV / (2 \cdot m \cdot v \cdot \Delta V)$ . The factor 2 is there because the loop contains both the charging and the discharging sweep.

Why is the capacitance lower in a two-electrode cell than in a three-electrode test?

In a cell the two electrodes are in series, so for a symmetric cell the capacitance per total mass is about a quarter of a single electrode’s. Higher mass loading, resistance and ion transport in a real cell lower it further.

Should I subtract the IR drop?

Yes. The sudden voltage drop at the start of a discharge comes from internal resistance, not stored charge, so the voltage window used for capacitance excludes it.

How is energy density calculated for a supercapacitor?

$E = \tfrac{1}{2} \cdot C \cdot \Delta V^2$ , with $C$ the cell capacitance per total mass and $\Delta V$ the voltage window. In F/g and V this gives J/g; divide by 3.6 for Wh/kg. Power density is $E$ divided by the discharge time.

What is C-rate?

The charge or discharge current relative to the capacity: at 1 C a full charge or discharge takes one hour, at 2 C 30 minutes and at C/10 ten hours. The current is the C-rate times the capacity.

What is the difference between F/g and mAh/g?

F/g (specific capacitance) suits capacitive, double-layer storage; mAh/g (specific capacity) suits battery-type, faradaic materials. They are related by $Q = C \cdot \Delta V$ , so a conversion depends on the voltage window.

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