What's the formula to calculate how many seconds a supercapacitor can provide power when employing a buck/boost converter?

Also, how different would that calculation be when using a pair of supercaps in serie (eg. 2x 2.7V @ 1F)

Example data:

Supercap: 5.5V, 1F; Panasonic EEC-S5R5V105 http://www.mouser.com/ds/2/315/ABC0000C22-947554.pdf

Buck/Boost (5V out): XL6009, 94% efficiency; are there other relevant specs? https://www.pollin.de/productdownloads/D351434D.PDF

Load: 5V, 250mA (Raspberry Pi)

(intended application: to provide a few seconds to save settings at power loss for an embedded RasPi)

  • 2
    \$\begingroup\$ Google: "energy in capacitor calculator", fill in the values and profit. Why would the calculation change with 2 caps in series? Two 1 F caps in series behave as a total capacitor of 0.5 F but since the voltage doubled the total stored energy remains...? \$\endgroup\$ – Bimpelrekkie Feb 9 '18 at 14:47
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    \$\begingroup\$ Your coin cell type cap has up to 30 Ohms internal resistance and will thus not provide enough output current for the RPi. \$\endgroup\$ – Turbo J Feb 9 '18 at 15:00
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    \$\begingroup\$ @Bimpelrekkie to be fair, I don't think OP knew that in series the capacitance halves, so that was what they were looking for. \$\endgroup\$ – BeB00 Feb 9 '18 at 15:02
  • \$\begingroup\$ You question lets me suspect that you want to maintain the 5V. BUT on a discharging capacitor the voltage drops. It is NOT a battery! \$\endgroup\$ – Oldfart Feb 9 '18 at 15:26
  • \$\begingroup\$ @oldfart don't you think that's what the buck/boost is for? \$\endgroup\$ – Arsenal Feb 9 '18 at 15:35

Hold up time is

T= \$\frac{C(V_s - V_f)}{I}\$

where I is the current, C is the capacitance, Vs is initial voltage on the capacitor, Vf is final voltage on the capacitor (perhaps the minimum voltage at which the system will work).

That's for an ideal capacitor. If the capacitor has significant internal resistance the voltage will drop an additional amount I*R, so the hold up time will be reduced. For a non-ideal capacitor, also adjust I to add the internal leakage current.

If you're trying to hold up a RPi long enough for an orderly shutdown I think you're going to require a very large supercapacitor with low internal resistance or a battery.

  • \$\begingroup\$ How to take into account the buck/boost efficiency? just add it to the load? Also, it's embedded Linux which doesn't really need an orderly shutdown, but I do like to save a few settings on power off, which really only takes 1-2 sec. \$\endgroup\$ – svenema Feb 9 '18 at 16:36
  • \$\begingroup\$ The current will increase as the regulator input voltage decreases. You can estimate it from the average input current or look at the energy C(Vi^2-Vf^2)/2 of the capacitor using power at one point (I*V) and assuming constant efficiency (which isn't quite correct but will probably get you within spitting distance). \$\endgroup\$ – Spehro Pefhany Feb 9 '18 at 17:22
  • \$\begingroup\$ Where does ESR fit into this formula? \$\endgroup\$ – ubiquibacon Aug 22 '18 at 1:31
  • \$\begingroup\$ @ubiquibacon Hopefully it's low enough to be irrelevant. It will provide an immediate constant voltage drop of ESR * I. That number should be checked when choosing a capacitor. Good point. \$\endgroup\$ – Spehro Pefhany Aug 22 '18 at 10:10

If your buck mode is most efficient you should go from higher voltage to lower. If boost is most efficient just the opposite.

Check the voltage inside the PI. I think internally it only needs 3V3. At least 90% runs of 3V3 if I remember correctly. Thus you are having 5V capacitors which discharge. You boost to 5V which then inside the PI gets converted back to 3V3: not ideal!

For Linux shutdown you can ignore everything which is 5V anyway as only the CPU and DRAM need to keep working and that is all 3V3. Oh! and the SDCARD, also 3V3.

Post edit.
Assuming the CPU etc. need 3.3V, add internal Pi regulator voltage drop ~200mV. Use your 2 seconds run time: Then you external voltage can drop from 5V to 3.5V in two seconds. Using @ Spehro Pefhany formula gives you ~0.33F without need for a buck/boost converter. I would take one a bit bigger as we used a number of estimated values.

Be aware that when you switch the 5V supply on, those capacitors will need to charge and look almost like a short circuit for a while. Your 5V supply might not like that. You can work around that by adding an R plus parallel diode in series with the cap, but that gives an additional voltage drop which you have to compensate for.


I found an site calculating super capacitor discharge time that might be helpful: http://www.circuits.dk/calculator_capacitor_discharge.htm


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