How does a supercapacitor differ from a battery?

A battery stores charge through chemical reactions; an electric double-layer capacitor stores it by lining up ions on the electrode surface. With no reaction, charge and discharge are fast and it can cycle hundreds of thousands of times. But stored energy is small: 3.04 Wh in a 3000 F cell. Voltage falls in proportion to charge, so using it down to half voltage gives 75% of the energy.

Why does capacitance fall at higher frequency?

Electrodes are carbon riddled with fine pores, and ions must travel through electrolyte to reach surfaces deep inside. That travel has resistance, so at fast rates only surface near the pore mouths is used. A transmission-line model for a 3000 F cell gives 99% usable capacitance at 0.1 Hz but 65% at 1 Hz and 19% at 10 Hz.

What should I watch for in series strings?

Capacitance spread between cells. In series the same charge flows, so lower-capacitance cells reach higher voltage. With 18 cells at 48.6 V, a cell 10% low in capacitance reaches 2.98 V, above the 2.7 V limit. Overvoltage shortens life, so balancing circuits are needed to equalise voltages.

Supercapacitor Modeling

Category: Coupled Analysis – Electrochemistry | Revised 2026-10-01

Theory: storing charge by lining up ions

Overview

🙋

Where are supercapacitors used instead of batteries?

🎓

Wherever large power must flow in or out over short times. An electric double-layer capacitor immerses porous carbon electrodes in electrolyte and stores charge by lining up ions on the electrode surface. With no chemical reaction, it charges or discharges in seconds and repeats hundreds of thousands of times. That suits buffering energy recovered by braking cranes or buses, or bridging the moment of a power cut. But stored energy is far below that of a battery of the same weight. Modelling looks at stored energy, losses and heat from internal resistance, whether pore depths can be used, and voltage spread in series strings.

Energy, power and time constant

$$ E = \tfrac12 C V^2,\qquad P_{max} = \frac{V^2}{4R_{ESR}},\qquad \tau = R_{ESR}\,C $$

R_ESR is the cell's internal (equivalent series) resistance. Here C = 3000 F, V = 2.7 V, R_ESR = 0.29 mΩ. P_max is maximum power with load resistance equal to internal resistance, at which efficiency is 50%.

Transmission-line pore model

$$ Z(\omega) = R_s + \sqrt{\frac{R_p}{j\omega C}}\,\coth\sqrt{j\omega R_p C},\qquad C_{eff} = -\frac{1}{\omega\,\mathrm{Im}\,Z} $$

R_s is electrolyte and current-collector resistance (0.19 mΩ) and R_p ionic resistance within pores (0.30 mΩ); at DC, R_s + R_p/3 equals the 0.29 mΩ internal resistance. The pore is a ladder of resistors and capacitors, the form de Levie derived for porous electrodes.

Coffee Break Trivia

From Helmholtz's layer

Helmholtz proposed in 1853 that charge lines up in two layers at the electrode–electrolyte boundary. Becker of General Electric turned it into a storage component, patenting a porous-carbon capacitor in 1957. High-surface carbon offers over 1,000 m² per gram, giving capacitance orders of magnitude beyond ordinary capacitors. Starting as memory backup supplies, they are now used in vehicles and power systems.

Worked examples

Example 1: a 3000 F cell

Item	Value
Stored energy	10,935 J（3.04 Wh）
Energy used down to half voltage	75%
Peak power	6,284 W
Time constant	0.87 s
1 kW down to half voltage	7.53 s (8.20 s if lossless)
Resistive loss in that discharge	670 J

At constant power, current rises as voltage falls, reaching 924 A at the end. Loss scales with current squared, so it concentrates at the end of discharge.

Example 2: frequency and usable capacitance

Frequency	Usable capacitance	Fraction	Resistance
0.01 Hz	3,000 F	100%	0.290 mΩ
0.1 Hz	2,979 F	99%	0.290 mΩ
1 Hz	1,939 F	65%	0.275 mΩ
10 Hz	564 F	19%	0.218 mΩ

Example 3: 18 cells in series with capacitance spread

Item	Value
Module voltage	48.6 V
Module capacitance	167 F
Module energy	54.7 Wh
Normal cell voltage	2.683 V
Voltage of a cell 10% low in capacitance	2.982 V

🙋

If only one cell hits 2.98 V and the rest sit at 2.68 V with margin, isn't that fine?

🎓

Not so. The voltage limit of a double-layer capacitor is set by where the electrolyte starts to decompose; beyond it gas forms, internal pressure rises and ageing accelerates sharply. Ageing lowers capacitance, so that cell's voltage climbs even higher on the next charge. The weakest cell degrades fastest and grows weaker still — a vicious circle. Internal resistance also rises, increasing heat, and higher temperature speeds ageing further. Series modules therefore include balancing circuits that measure each cell and bleed the high ones. Design calculations include cell-to-cell capacitance and resistance spread and module temperature distribution to find which cell reaches the limit first.

Modelling workflow

Fit internal and pore resistance to impedance measurements.
Simulate discharge with the actual current waveform to confirm usable capacitance.
Derive heat from resistive loss and solve module temperature distribution.
Include cell spread to obtain voltage distribution in the string.
Estimate life from temperature and voltage and size balancing and cooling.

Coffee Break Trivia

“A regeneration unit that emptied faster than the datasheet said”

On one crane's regeneration unit, energy stored in the capacitor module ran out sooner than calculated, hitting the lower voltage limit. The calculation used the datasheet capacitance (measured at DC), but the real load swung strongly over about a second. Recomputing with the pore model and the actual current waveform showed about 20% less usable capacitance, matching measurements. Adding cells in series to gain voltage margin solved it.

Common mistakes

Mistakes and fixes

Mistake	Effect	Fix
Using DC capacitance	Energy overestimated for fast loads	Pore model
Assuming use down to 0 V	Converter cannot operate	Use the minimum voltage
Ignoring spread in series	Some cells overvoltage	Balancing circuit
Heat from average current	Heat underestimated	Use RMS current
Constant temperature	Life overestimated	Couple with heat

🙋

I'd like to learn related topics.

🎓

Rate this article

Thank you for your feedback!

Helpful

More details

Report error