Can I have about 4 to 5 seconds backup by 1F 5.5V Super Capacitor?

Power supply output voltage: 5v 600mA I need 5v 0.5A for my system.

I want to use it as ups for only 4 to 5 seconds so I wasn't thinking about using batteries. should I change to any other capacitor?
If it's not possible by capacitor then how can I do it?

A Schematic would be amazing. Thankq


4 Answers 4


A Capacitor alone isn't enough

A capacitor stores energy \$E=\frac{1}{2}CV^2\$ . As soon as you start to take energy from the capacitor, the voltage drops. You can only charge to 5V, then as soon as you take any energy at all, you drop below 5V. And you say you need 5V for your circuit.

Capacitor and boost converter

For 5V, 0.5A, 4 seconds you need 10J, and a 2.2F capacitor charged to 5V stores 27J, so it has enough energy. You just need a circuit or module to convert the steadily dropping voltage from the capacitor to something your circuit can use. This would be a "boost converter". The boost converter won't be able to use all the energy from the capactor, and some will be lost in convesion, but you'll get most of it.

Making it work

A capacitor and a boost converter will get you started. You'll also need a way to automatically switch between your normal power supply and the boost converter when the normal supply fails. If you had a supply which is just a little higher voltage, then you could use diode or, but you don't so you'll need to do something more complicated.

  • 1
    \$\begingroup\$ Your analysis does not make explicit the driving constraint: how much voltage drop is permissable? Until that is made clear, you're not really answering the question. For a given (non-zero) voltage drop and a large enough capacitor, a capacitor is indeed enough. \$\endgroup\$ Jun 21, 2018 at 17:17
  • \$\begingroup\$ Without the boost converter, the permissible voltage drop is (based on my reading of the question) zero. So it can't possibly work. With the boost converter the permissible drop is down to the minimum input of the converter. So one would have to trade off buying a low-dropout converter vs buying a bigger capacitor. But that felt like it was beyond the scope of the answer. \$\endgroup\$
    – Jack B
    Jun 21, 2018 at 17:25
  • \$\begingroup\$ @WhatRoughBeast, I disagree. Do we need to provide full design? My answer had a sufficient hint. This answer is better. \$\endgroup\$ Jun 21, 2018 at 18:31

It's borderline doable with a 1F cap. I'd also suggest increasing the cap if possible, to be on the safe side.

As others have suggested, you need a boost converter to keep your output voltage, even after the cap is discharged well below 5V.

The boost converter you choose to use will dictate whether the task is doable with just 1F of capacitance.

Say you choose a garden variety of boost converter, which can't keep up with the 500 mA output current at Vin < 2V. This will be fairly typical, as output power is 2.5W (5V*0.5A), so the input current has to be at least 1.25A (2.5W/2V). This is probably too much for most small converters.

With these parameters, the Vstart is 5V, Vend is 2V, so the energy available is ((5V)²*1F/2) - ((2V)²*1F/2) = 10.5J. If you factor in converter efficiency, which will be unlikely to be >95% across the whole Vin range, we get below the 10J required for 5V*500mA*4s.

However, we have two tricks up our sleeve:

  1. Charge the cap to slightly higher than 5V. If the boost converter features a diode at the output, it can actually regulate Vout < Vin, if the Vout is no more than the diode drop lower than Vin. With this in mind, you can safely charge the cap to 5.3V. If your load can have some leeway around 5V, even 5.5V might be possible.

  2. If you select a good boost converter, it can probably cope with 2-3A input current, putting your Vend somewhere around 1V. Say it's 1.2V.

With these specs, you get ((5.3²*1/2) - (1.2²*1/2)) * η (efficiency). If we assume η >= 90% across the range, you get about 12J - so you even have some energy to spare.

A good boost converter like the LTC3124 can probably do this, and they even have a very similar application example on the last page of the datasheet:

LTC3124 supercap backup example

They don't have the diodes here, but mention them as a possibility in the datasheet.

Keep in mind that this design requires careful circuit layout, many parts (this is a dual-phase converter), expensive IC, and even then we're about borderline on doability (supercaps might be -20%..+80% spec'd, so the performance of this design isn't guaranteed).

To me, a much simpler approach would be to take any garden variety boost converter, possibly as a premade unit, and use higher capacitance. 2F should be plenty.

  • \$\begingroup\$ +1 for full design. \$\endgroup\$ Jun 21, 2018 at 20:02

No, 1 F capacitor will drop 1 V per 1 second at 1 A. If you want to keep he voltage above 5 V, your maximum allowable drop is 0.5 V, and at 0.5 A the supercap will discharge to 5V in 1 second. In short, you will have only one second out of this straightforward approach. See the cap discharge "theory" here.

You can have more than 1 second if you would employ a DC-DC upconverter, and discharge the cap to lower voltages, say, down to 2.5 V, and 5 seconds is likely doable.

CORRECTION: For initial V0=5V (5.5 would be technically stretching), the cap would hold 12.5J, while the OP needs 10J (5v*500mA*4s). With assumed 80% efficiency of DC-DC converter, the cap must be fully discharged to zero, which will be out of range of any upconverter. So the answer for given parameters (1F cap) is NO, 1F cap isn't enough under any approach. The cap needs to be bigger.

  • \$\begingroup\$ I don't think it's really 100% impossible. It's borderline, see my answer. \$\endgroup\$
    – anrieff
    Jun 21, 2018 at 19:57
  • \$\begingroup\$ @anrieff, I don't disagree, borderline it is. But it is clearly impossible at the OP's skill level. \$\endgroup\$ Jun 21, 2018 at 20:00

A 1F capacitor is designed for low power backup purposes and would struggle to deliver a few hundred milliamps let alone 600mA. You will find the voltage drop significantly under load. Perhaps a 25F or 100F capacitor would be better suited to your application.

  • \$\begingroup\$ A lot of 1F caps are high ESR (intended for memory backup applications only), but that doesn't have to be the case. E.g. see this one. \$\endgroup\$
    – anrieff
    Jun 22, 2018 at 7:52
  • \$\begingroup\$ @anrieff Nice component. But ESR aside, the voltage will drop linearly and quickly due to capacity(<10 seconds assuming the capacitor is full). I would think it would drop below the desired voltage threshold too quickly. \$\endgroup\$ Jun 22, 2018 at 10:18
  • \$\begingroup\$ This is exactly what we have calculated in the answers. The OP only needs 4 seconds, so it is borderline doable with 1F low-ESR cap. Should be definitely doable with a >2F low-ESR cap. \$\endgroup\$
    – anrieff
    Jun 22, 2018 at 10:33

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