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# if I used a buck/boost voltage regulator and a 2.7 volt 200F supercapacitor, how long could I output 1.5 volts for?

It would also be nice if someone could explain exactly how a capacitor discharges over time in terms of voltage and remaining capacity. Is the rate of decay constant, and does it vary according to the capacitor?

thx! =)

## Comments

11 years ago

From those figures, you have 1450 Joules stored, or a charge of 540 Coulomb If you have a perfect converter AND there are no leakage losses AND you are driving a 1A load, it will run for 540 seconds. At half the load, you have twice the run time. However, all real PSUs have a minimum operating voltage, depending on the topology and components used, so you will hit a point when the PSU can't work.

11 years ago

I found the graph here, so not mine to grant permission. Looks like a useful site for basic info.

11 years ago

Like Sean says, Can't say without a load, and a super cap LEAKS electricity too, so you need to consider its not inconsiderable leakage current. Steve

11 years ago

The voltage across a discharging capacitor looks like THIS. If you increase the capacitor (C), the discharge time increases, but the graph will always be that shape so not at all suitable for driving anything which needs a constant voltage.