Supercapacitor backup calculator

How much capacitance a power backup needs to carry a constant-power load for a given time, or how long a given capacitor lasts, between the voltage it is charged to and the lowest voltage the load accepts. The ESR step at the first instant, the tolerance and the end-of-life loss are included, in the order Abracon's application note works through them, and the stored energy is split into what the load gets and what it never can.

Fig 1 — 20.0 W constant power from 1348 F: a 171 mV ESR step from 4.20 V to 4.03 V, then V = √(V₀² − 2Pt/C) down to 3.30 V after 3.00 min. Below V_min the curve is energy the backup never uses.

Solve for

Load power P (W)

Hold-up time t (s)

Start voltage V_start (V)

Minimum voltage V_min (V)

ESR (mΩ)

Capacitance tolerance (%)

Capacitance lost at end of life (%)

Capacitance needed at end of life, worst case (Abracon step 5): 1348 F
Nominal capacitance to specify, −20 % tolerance and −30 % at end of life: 2408 F
With the ESR drop tracked through the discharge: 1442 F worst case · 2575 F nominal
Energy the load needs, P × t · ideal capacitor, 2E/V_start²: 3.60 kJ · 408 F
Current at the start, P/V_start · ESR step V_IR: 4.76 A · 171 mV, so the discharge starts at 4.03 V
Current at V_min, P/V_min · ESR drop there: 6.06 A · 218 mV; must be within the rated discharge current
Stored at V_start · usable down to V_min, at 1348 F: 11.9 kJ · 3.60 kJ
Stored energy: used · lost to the ESR step · left below V_min: 30.3 % · 8.0 % · 61.7 %

The single ESR step, taken at the starting current, overstates the hold-up by 7 %: the current rises to 6.06 A by V_min and the drop with it. Size from the tracked figure when the two differ by this much.

How this is calculated

Standard: Abracon — Designing with Supercapacitors (Power Backup, April 2024)

$E = P\,t, \qquad E = \tfrac{1}{2}\,C\,V^2, \qquad C = \frac{2E}{V^2}$: The work the backup must do, and the capacitance that stores it if every joule down to 0 V could be used. Abracon's 20 W for 180 s is 3,600 J, and 2 × 3600 / 4.2² ≈ 408 F: "In an ideal world, a capacitor of at least 408 farads would be sufficient to supply backup power of 20W for 3 minutes. In reality, it needs to be much larger."
$T_{discharge} = \frac{1}{2P}\,C\,\left(V_{dis,start}^2 - V_{min}^2\right)$: Abracon's constant power equation. Only the energy between the two voltages is available: "the total usable capacity is the amount of charge between the starting and ending voltages". The 408 F part from 4.2 V to 3.3 V at 20 W lasts ≈ 68 s. Solved for V, the same energy balance gives the curve in the figure, V(t) = √(V₀² − 2Pt/C).
$V_{min} = \frac{P}{I_{max}}$: A constant-power load draws more current as the voltage falls; the note's step 2 is to "calculate the minimum capacitor voltage allowed without exceeding the rated current". Its 6 A part at 20 W gives 3.3 V.
$V_{IR} = ESR \cdot \frac{P}{V_{start}}, \qquad V_{dis,start} = V_{start} - V_{IR}$: "The voltage drop is calculated as Vdrop = ESR * Idischarge", at the starting current. Step 4: "The ESR used in this calculation should be 2-3x of the datasheet value to compensate for the aging effects on supercapacitors." The note's 2 × 0.018 Ω at 4.76 A drops ≈ 0.17 V, leaving 4.03 V. Abracon writes the post-drop voltage as V_start in its step 5; it is V_dis,start here so that V_start stays the charge voltage.
$C = \frac{2\,P\,t}{V_{dis,start}^2 - V_{min}^2}, \qquad C_{nominal} = \frac{C}{(1 - tol)(1 - fade)}$: Step 5, then steps 6 and 7: the capacitance required at the worst case, and the nominal part whose low-tolerance, end-of-life value is still that much. The note's final figures divide — "~3,500 farads" at end of life becomes "~6250 farads at the beginning of service", which is 3500 / (0.8 × 0.7). "As an industry standard, a supercapacitor’s lifetime is considered ‘expired’ when the capacity has decreased by 30%."
$\begin{aligned} t &= \frac{C}{2P}\Big[H(V_{c0}) - H(V_{c,end})\Big], \qquad a^2 = 4PR \\ H(x) &= \tfrac{1}{2}\Big[x^2 + x\sqrt{x^2 - a^2} - a^2 \ln\!\left(x + \sqrt{x^2 - a^2}\right)\Big] \end{aligned}$: Not in the note: the same discharge with the ESR R in the loop the whole way, derived from P = (V_c − IR)·I and C·dV_c/dt = −I. It ends where the terminal voltage reaches V_min, at a cell voltage V_c,end = V_min + R·P/V_min. With R = 0, H(x) = x² and it is Abracon's equation; with R > 0 it is shorter, because the drop grows as the current does. The calculator shows both.

Assumptions

The load draws constant power, as in Abracon's example. A resistive load draws less as the voltage falls and lasts longer; the capacitor charge time calculator covers that case.
The load power is what the capacitor delivers. Behind a DC/DC converter that is the load divided by the converter efficiency, which the note warns "is not constant and may decrease as the input voltage to the boost circuit decreases".
The ESR is one fixed resistance. Abracon's model subtracts a single step at the starting current; the tracked result keeps it in the loop to V_min. Neither includes the capacitor's frequency-dependent resistance or its leakage current.
Tolerance and end-of-life loss both reduce the capacitance, and both are applied at once as the worst case. Temperature and voltage derating for life are the designer's choice of V_start; the note's example drops from 4.2 V to about 3.8 V for two years at 60 °C.
One capacitor, or a stack entered as one: its total capacitance, its total ESR and its total voltage. Balancing a series stack is outside the note and outside this page.

What sets a supercapacitor's backup time

A supercapacitor holding up a load after the supply fails is an energy budget. The load needs a certain amount of work done in a certain time, E = P·t, and the capacitor stores E = ½CV². Abracon's application note starts there: a 20 W load held for 180 s is 3,600 J, and 3,600 J at 4.2 V is about 408 F. The note's own verdict on that number follows immediately: "In an ideal world, a capacitor of at least 408 farads would be sufficient to supply backup power of 20W for 3 minutes. In reality, it needs to be much larger."

The first reason is that the load cannot use the energy down to zero. Most backup loads draw constant power, either directly or through a DC/DC converter that regulates their rail, and at constant power the current is P/V. As the capacitor discharges the voltage falls and the current rises to keep the product fixed. The note puts it bluntly: "The current must increase proportionately to maintain 20 watts as the voltage decreases. In fact, as the capacitor’s voltage approaches zero, the current approaches infinity!" Something has to stop the discharge before that, and in the note it is the capacitor's rated discharge current. A part rated for 6 A feeding 20 W reaches its limit at 20 W / 6 A ≈ 3.3 V, and that is the minimum voltage, V_min. Below it the capacitor still holds charge, but the load cannot have it.

So the usable energy is only the slice between two voltages, ½C(V_start² − V_min²), and the time it lasts at constant power is that slice divided by P. That is the note's constant power equation, T = (1/2P)·C·(V_start² − V_min²). Because energy goes as the square of voltage, the slice is smaller than the voltage ratio suggests. Discharging from 4.2 V to 3.3 V gives up just over a fifth of the voltage but uses only 38 % of the stored energy; the other 62 % stays on the capacitor. The note says so without hedging: "in reality, using the full capacity of the supercapacitor can never be achieved."

The second reason is the capacitor's internal resistance. The moment the load switches over, its current flows through the ESR and the terminal voltage steps down by V_IR = ESR × I. The note computes the step at the starting current, P/V_start, and subtracts it from the starting voltage, so the discharge really begins at V_dis,start = V_start − V_IR. The ESR to use is not the datasheet's: Abracon's step 4 says "The ESR used in this calculation should be 2-3x of the datasheet value to compensate for the aging effects on supercapacitors." Every millivolt of that step comes off the top of the window, where each volt is worth the most energy.

Then the part itself. The datasheet capacitance is typical, not guaranteed, and it falls with age. The note sizes for the worst of both: the tolerance, ±20 % for its example part, and the end-of-life loss, for which "a supercapacitor’s lifetime is considered ‘expired’ when the capacity has decreased by 30%." The capacitance the equation gives is what the part must still have at the end; the part to buy is larger by 1/((1 − 0.2)(1 − 0.3)), a factor of 1.79. The calculator carries both, and draws the discharge in the figure with the worst-case value, since that is the capacitor the backup will eventually have.

Supercapacitor chart: how the voltage window sets the capacitance

The same kilojoule, delivered across different voltage windows, computed by the calculator with the ESR and the margins left out so that only the window changes. Scale the capacitance column by the energy the load needs: a 20 W load for 180 s is 3.6 kJ, so multiply by 3.6. The last column is how much larger the current is at V_minthan at the start, and it is the column the capacitor's current rating has to cover.

V_start → V_min	Stored energy used	Capacitance per 1 kJ	End current / start current
4.2 V → 3.3 V	38 %	296 F	1.27 ×
4.2 V → 2.8 V	56 %	204 F	1.50 ×
4.2 V → 2.1 V	75 %	151 F	2.00 ×
4.2 V → 1.4 V	89 %	128 F	3.00 ×
3.8 V → 3.3 V	25 %	563 F	1.15 ×
3.6 V → 3.3 V	16 %	966 F	1.09 ×

Two things are worth reading off it. Letting the voltage fall to half the start uses 75 % of the stored energy instead of 38 %, and cuts the capacitance to 51 % of it, at the price of doubling the current the capacitor and the converter behind it must handle at the end. And the bottom two rows are Abracon's derated case: charging only to 3.8 V for lifetime, then losing 0.2 V to the ESR, leaves a 3.6 V to 3.3 V window that uses 16 % of what is stored at 3.6 V and needs 3.3 times the capacitance of the 4.2 V to 3.3 V window. Derating for life is the most expensive decision in the whole calculation.

Supercapacitor discharge time for common capacitances and loads

Hold-up time at constant power from 4.2 V to 3.3 V, the note's window, for a new part at its typical capacitance and with no ESR. Time is proportional to the capacitance and inversely proportional to the load, so any cell of the table scales to its neighbours. The table is the optimistic bound: take off the tolerance and end-of-life loss (×0.56 at the note's 20 % and 30 %) and the ESR step, which for a small cell at a high load can be the larger effect, before relying on a figure.

Capacitance	100 mW	1 W	5 W	20 W
1 F	33.8 s	3.38 s	675 ms	169 ms
10 F	5.63 min	33.8 s	6.75 s	1.69 s
100 F	56.3 min	5.63 min	67.5 s	16.9 s
400 F	3.75 h	22.5 min	4.50 min	67.5 s
1000 F	9.38 h	56.3 min	11.3 min	2.81 min
3000 F	28.1 h	2.81 h	33.8 min	8.44 min

The 400 F row at 20 W is the note's first attempt, just over a minute against a three-minute requirement. The same part holds a 100 mW load for 3.75 h. That spread is why the note's first question is how long the backup must run, and its second how much work must be done in that time.

Worked example: Abracon's 20 W, 180 s backup

The note's example is a backup energy module modelled on JEDEC's JESD315, holding a constant 20 W for 180 s from a cell family rated 4.2 V with a 6 A discharge limit. The left column is the calculator's arithmetic; the right is what the note prints.

energy       E = 20 W × 180 s                             = 3600 J    note: 3,600 J
ideal        C = 2 × 3600 / 4.2²                          = 408 F     note: ≈ 408 F
hold-up      of that part, 4.2 V → 3.3 V
             T = 408 × (4.2² − 3.3²) / (2 × 20)           = 68.9 s    note: ≈ 68 s
ESR step     I = 20 / 4.2                                 = 4.76 A    note: 4.76 A
             V_IR = ESR × I = 0.036 × 4.76                = 0.171 V   note: ≈ 0.17 V
             V_dis,start = 4.2 − 0.171                    = 4.029 V   note: 4.03 V
sized        C = 2 × 20 × 180 / (4.029² − 3.3²)           = 1348 F    note: ≈ 1340 F
tolerance    C / 0.8                                      = 1686 F    note: 1,675 F (from 1,340)
derated      to 3.8 V, less 0.2 V of ESR drop: 3.6 V → 3.3 V
             C = 2 × 20 × 180 / (3.6² − 3.3²)             = 3478 F    note: ~3,500 F
new part     3500 / (0.8 × 0.7)                           = 6250 F    note: ~6250 F
check        ½ × 3500 × 3.6² − ½ × 3500 × 3.3²
             22680 − 19058                                = 3623 J    note: ≈ 3623 J

Every figure agrees with the note to within its own rounding. The first attempt is the lesson: the 408 F capacitor that holds exactly 3,600 J at 4.2 V delivers only about 1,380 J of it to a load that stops at 3.3 V, which is about 68 seconds, not 180. The note gives "about 1,360 joules" for this step, which is 20 W times its rounded 68 s; the difference is rounding, not a different model.

The sized capacitance of about 1,340 F then grows twice. Tolerance takes it to 1,675 F, which is 1,340 / 0.8: dividing, so that a part 20 % low still has 1,340 F. The note's next figure, 2,297 F for the 30 % lifetime loss, is 1,340 × 1.2 / 0.7 and so mixes the two conventions; its final answer divides by both, 3,500 / (0.8 × 0.7) = 6,250 F, and that is the rule the calculator applies throughout. Dividing is the safe direction: a part bought at 1.2 × 1,340 = 1,608 F and delivered 20 % low has only 1,286 F.

The large step is the derating. Two years at a 60 °C ambient, in a part whose rated temperature is 60 °C, leaves only the voltage to derate, and the note derates it to about 3.8 V. With the ESR drop that leaves a 3.6 V to 3.3 V window, and the capacitance at end of life becomes about 3,500 F; with the margins, about 6,250 F at the start of service. The note's summary: "Basic calculations yielded a 408F supercapacitor, but after all the analysis, the best solution for 20W of continuous for 180s, lasting 2 years in the field, requires a 6250F supercapacitor, 15X larger in capacity."

The calculator's defaults reproduce the example before derating: 20 W, 180 s, 4.2 V to 3.3 V, 36 mΩ, 20 % tolerance and 30 % lifetime loss. Set V_start to 3.8 V and the ESR to 38 mΩ, which gives the note's 0.2 V step at 5.26 A, and the result is the derated case: 3478 F at the worst case and 6211 F nominal, within 1 % of the note's rounded 3,500 F and 6,250 F.

Where the constant-power model stops being valid

The ESR step is taken once, at the start. That is how the note computes it, and it is the right first estimate. But the current at V_min is larger than at the start, by the ratio V_start/V_min, and so is the drop across the ESR: in the note's example 0.17 V at the start and 0.22 V at the end. The calculator also solves the discharge with the ESR in the loop the whole way, a closed form derived from P = (V_c − I·R)·I that is not in the note, and shows the result beside the note's. On the derated example the difference is large: 3875 F instead of 3478 F at end of life, 11 % more, because a 0.2 V drop is two thirds of a 0.3 V window. When the two figures differ by more than a few per cent, the tracked one is the one to design to.

The load can ask for more than the capacitor can give.A source with internal resistance R can deliver at most V_c²/4R, when the terminal voltage has fallen to half the cell voltage. Below a cell voltage of 2√(P·R) the constant-power load has no stable operating point and the terminal voltage collapses. With the note's milliohms and volts that is far away, but a high-ESR cell feeding a heavy pulse load can reach it. The calculator reports the case, and a V_min that the ESR step alone already passes.

The converter is not free. The load power here is what leaves the capacitor. Behind a boost or buck-boost that is the load divided by the converter efficiency, and the note warns that the efficiency "is not constant and may decrease as the input voltage to the boost circuit decreases (supercapacitor voltage)." Use the efficiency at V_min, the worst point, and add the converter's own quiescent draw to the load.

Life is a voltage and temperature choice. The note quotes the industry convention of rating a supercapacitor's life as "1,000 hours operating at its rated temperature and voltage", a little over a month. Reaching years means charging below the rated voltage, running cooler, or both, and V_start here is whatever that choice leaves. The note's lifetime detail lives in another Abracon application note that this page does not cite.

No leakage, no charging. The model covers the discharge only. It has no term for the capacitor's leakage current, which matters when the hold-up is long and the load small, and says nothing about how long the capacitor takes to charge back up; the capacitor charge time calculator covers a resistor-limited charge.

Stacks. Enter a series stack as one capacitor: n identical cells in series have C/n, n times the ESR and n times the voltage. Keeping the cell voltages equal in a stack is its own design problem, and it is outside the note and this page.

Common supercapacitor sizing mistakes

Sizing from E = ½CV² at the full voltage. It counts energy the load can never take, and in the note's example it gives 408 F for a job that needs 6,250 F.
Choosing V_min from the load alone. A constant-power load draws its largest current at the bottom of the window, and that current has to be within the capacitor's rating and the converter's input limit. The note chooses V_min from the current, not from the load's undervoltage threshold.
Using the datasheet ESR. It is the value for a new part; the note asks for two to three times it, and the step it produces comes off the most valuable part of the window.
Adding margins by multiplication. 1,340 F plus 20 % is 1,608 F, and a part 20 % low from that is 1,286 F, short of the requirement. Divide by (1 − tolerance)(1 − loss), as the note's final figures do.
Charging to the rated voltage for a multi-year product. At the rated voltage and temperature the convention is 1,000 hours. The note's example derates to 3.8 V, and pays for it with a narrower window and a much larger part.
Forgetting the converter efficiency, and that it falls at low input voltage. The capacitor supplies the losses as well as the load.