# Capacitor bank as a constant current source?

Can a capacitor bank of high power density, be integrated to a circuit such that it acts as a constant current source to a load?

Even for a short duration of time.

• What kind of load do you need to drive? – Vladimir Cravero Jan 13 at 21:58

Yes. Simply treat the capacitor bank as if it were a battery. That means that you need to add a constant-current circuit to the power source (capacitor bank) in order to get a constant current output.

Do note that the actual capacity of the cap bank is much, much less than either a primary or secondary battery of similar physical size.

• Completely wrong. A capacitor is essentially a voltage source with variable current depending on load. – Jack Creasey Jan 13 at 22:31
• @Jack Creasey: Oh, Kay. How would you describe a battery using your terms? – Dwayne Reid Jan 13 at 22:35
• @DwayneRied. A battery is a constant voltage source (over a short period of time) and will supply as much current as it can limited only by the chemistry and architecture. If you remove the 'Yes' from your answer it actually reads almost correct …..but a capacitor or battery are NOT Constant current devices. – Jack Creasey Jan 13 at 22:37
• @Jack Creasey: Exactly the same as a capacitor bank. – Dwayne Reid Jan 13 at 22:38
• @DwaneReid. I don't disagree, but the question was "Capacitor bank as a constant current source?" …..to which you answered 'Yes" …...care to modify that? – Jack Creasey Jan 13 at 22:43

A constant current source, as its name says, delivers the same current over a (wide) range of loads.

The opposite is a constant voltage source, which delivers the same voltage over a (wide) range of loads.

Over a short period a loaded capacitor behaves as a constant voltage source.

So No, a loaded capacitor can't be used as a constant current source. Unless you add a constant current circuit, but then every voltage source can be used.

A switched load of value R from a charged C bank of voltage Vc can deliver a constant current only for the duration of the tolerance of what is considered "constant"

Since I is hoped to be constant and C and Vi or initial voltage are constant, the result is ;

V(t) = Vi-dV/dt where $$\Ic(t)=\dfrac{C~dV(t)}{dt}\$$ and $$\Ic(t)=\dfrac{Vc(t)}{R }\$$ thus $$\\dfrac{C ~dV_C(t)}{dt}=\dfrac{V_C(t)}{R} \$$ or $$\\dfrac{dV_C(t)}{V_C(t)}=RC\$$

So if the tolerance for CC was 10% current droop = voltage droop then $$\dt~ = 0.1 RC ~\$$max

This assumes R is much greater than the ESR of the C bank to be able to neglect the additional $$ΔV = I*ESR$$ drop.

A capacitor will not be the greatest choice for that particular usage, i recommand a battery instead. But you can use a capacitor in this case with a correct circuit.

Some IC convert the output of capacitor to constant voltage, you just have to change it to constant current. Look at TI they have some ic that can suit your need.

Also if you say by high power density, super-capacitor. You have some application online that already use super-capacitor for long time usage (exemple: a tramway in Germany that only use super-capacitor between station). So yes your idea is ok.

A capacitor will have a "Punch" current but not for a long time (low energy density): (From https://www.global.tdk.com/techmag/electronics_primer/vol1.htm)