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# What capacitance does a battery "look" like to connected electronics?

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#### cupoftea

Hi,
C=Q/V
So is that what capacitance the battery looks like?
Presumably not, because it can "produce" charge, but its not all available at any instant.
I mean, if a battery has a C of 1Ahr..then it can supply 1A for 3600 seconds.

However, can you then say thats its Q in the C=Q/V equation is 1x 3600 = 3600 Coulombs?

That would mean a 1AHr, 8V battery has a capacitance of 3600/8 = 450 Farads

if i put it in series with an L of 1H, then would it give a resonant frequency of w = 1/[SQRT{LC)] ?

Q=C*V describes a linear relation between accumulated charge and voltage. It doesn't apply to batteries. A linear approximation can be made by calculating dQ/dV for a specific state of charge.

Since a battery can deliver charge without a significant change in voltage, it does not look like a capacitor.

So if i put say a 1 Farad super cap stack outside a 11.1v, 2Ahr battery, then most of the ripple will come from the battery.....i will be drawing 10A pulse of current on a duty cycle of 1/6......and the ON time is 100ms.

I would say all that pulse would end up coming from the battery, would you agree?

In case of a chemical battery (which would cover all batteries typically associated to that word), chemical reactions take place that consumes electrons at the cathode and deposit it at the anode. This continues until an equilibrium is reached, where the reverse reaction happens at the same rate. If you remember from chemistry class, equilibrium occurs because the reagents for the forward and reverse reaction are in such proportions that they occur at the same rate. For the reaction happening inside a battery, this always includes the electron. In electrical terms, the relative abundance (or lack) of electrons is referred to as voltage*. E.g. the reaction inside an AA battery reaches equilibrium when there is a 1.5V potential across the terminals.

Hence, the reason shorting both ends of an AA alkaline battery to ground produces no noteworthy result is because in such a case, the supply of electrons become limited by the rate of the chemical reaction, even when the chemical reaction has all the electrons it needs thanks to the shorted circuit. Whereas shorting a car battery will create sparks, because the reaction happens at a MUCH faster rate.

So what you really have is a chemical system that tries to maintain a constant voltage as charge is drawn from it, which is very different from a capacitor. While a battery will still lose voltage, the process is fundamentally different from C=Q/V.

So generally speaking, one cannot expect it to behave like a capacitor. Although, maybe at a specific frequency, the battery behaves closely enough to a capacitor that it CAN resonate.

Despite of what's said above, battery impedance for a specific state of charge can be modeled as lossy capacitor, represented by a RC ladder circuit.

Thanks, so would you agree, the percentage of the current that would be delivered by the battery, and by the supercapacitor, would be dependent on the resistance of the battery, and the resistance of the supercapacitors?
If we can put enough supercaps in parallel, and get a low overall series internal resistance of the supercaps.....then all , or most, of our 100ms, 10A current pulse will come from the super caps?

The average current obviously has to come from the battery.

How is supercap related to this thread? Hasn't been yet mentioned.

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