I am new, but I wanted to create this thread to ask a question that I haven't seen an answer to anywhere (drumroll please)

Batteries can be thought of as voltage sources with a resistor in series to show the "internal resistance" that comes from conduction through metal in the battery, volts lost moving ions across the separator, and volts lost actually getting the reaction to go.

My question is: in the case of a battery made of lots of anode/cathode plates side by side having all the anodes in parallel, and all the cathodes in parallel- shouldn't there be a measurable capacitance of all these plates separated by electrolyte, which would maybe act as a dielectric?

This way, a circuit that pulls a constant current but almost instantaneously turns on would receive a capacitor-like voltage curve coming from the battery as the battery drops from the open circuit voltage to whatever the voltage of the battery stabilizes to at the constant current draw.

What do you think?

 

Batteries resemble capacitors so much that measuring the Xc is a valid factor in estimating the health of a battery. Excellent thing to do if buying a used car battery at a junque yard.

 

In theory, Yes, in practice, no.

A lead acid battery has resistive and inductive components that "compete" with the capacitance of the battery. The plates of a battery are spaced relatively far apart and represent a pretty small capacitance. The capacitive effects are only visible for about 5 to 10 microseconds during a pulsed condition (normal car starter battery). Then the inductive and resistive elements are the predominate features.

 

In some ways it is very reasonable to treat a battery as a capacitor and in other ways it is very unreasonable.

I deal with one of the more unreasonable ways in the following blog entry:

http://forum.allaboutcircuits.com/blog/a-battery-isnt-a-capacitor.588/

This is basically treating a battery as a DC capacitor (i.e., using Q = C·V) and it doesn't even come close.

But if you look at it only as an AC capacitor (i.e., using dQ = C·dV), then it is a pretty good model (or at least has a number of useful applications).

 

I see now that my answer was only about paragraph 3 of post #1, very limited and poorly defined. I think WBahn fixed that up.

 

In theory, Yes, in practice, no.

A lead acid battery has resistive and inductive components that "compete" with the capacitance of the battery. The plates of a battery are spaced relatively far apart and represent a pretty small capacitance. The capacitive effects are only visible for about 5 to 10 microseconds during a pulsed condition (normal car starter battery). Then the inductive and resistive elements are the predominate features.

I am just looking at some voltage data on a battery and am seeing capacitor-like upward curvature (about 5/100th of a second) as the battery transitions from a lower rate of discharge to the higher pulse discharge. At the end of the pulse, as the battery voltage returns back up to near its open circuit voltage, I note a similar downward curvature, again lasting about 5/100th of a second.

Could this be explained by inductive effects rather than capacitance of the circuit?

 

Lack of time labeling on the graph leaves too many uncertainties.
In addition, the voltage data seems to curve downward as the current is first turned on, not upwards. Typos?

 

In theory, Yes, in practice, no.

A lead acid battery has resistive and inductive components that "compete" with the capacitance of the battery. The plates of a battery are spaced relatively far apart and represent a pretty small capacitance. The capacitive effects are only visible for about 5 to 10 microseconds during a pulsed condition (normal car starter battery). Then the inductive and resistive elements are the predominate features.

A sulphated battery does a better impersonation of a capacitor than a of a battery - the distance (assuming any life in the electrolyte) is basically the thickness of the sulphate layer.

*PURE* distilled water is actually a non-conductor - so after all the H2SO4 has gone into crystalising the plates, the rest depends on how contaminated the water is.

 

It is much easier to use a capacitor as a substitute battery than to use a battery as a substitute capacitor.

For one thing the battery is a voltage source, whereas you charge a capacitor up to any voltage, (within reason).

 

It is much easier to use a capacitor as a substitute battery than to use a battery as a substitute capacitor.

For one thing the battery is a voltage source, whereas you charge a capacitor up to any voltage, (within reason).

Supercapacitors muddy the definition a bit - most are used as a battery for ULP things like RTC etc.

 

Lack of time labeling on the graph leaves too many uncertainties.
In addition, the voltage data seems to curve downward as the current is first turned on, not upwards. Typos?

Good point #12, here I get the axis in there.


Maybe that will clear up the curvature question.
(the bottom square one is current, the line higher up is battery voltage. You're right that the curvature is briefly downward when the current is first drawn, but I am talking about the positive , longer voltage curve from 62.04 or so to 62.07... and then also the negative longer voltage curve towards the end from 62.24 to 62.3)

See how just after the current goes from near zero to about 7 amps fairly instantly, and then it takes 5/100ths off a second to mostly level off instead of a sharp angle like the current profile shows? Equally on the other side right after the current drops to about zero, the voltage takes its time coming back up asymptotically to the initial voltage.

Seems capacitor-esque to me. As if the capacitance of the wires and plates within the battery is providing high voltage for longer than the battery would under the higher draw, and then slowing the battery from coming back up immediately when the draw is lowered.

 

In some ways it is very reasonable to treat a battery as a capacitor and in other ways it is very unreasonable.

I deal with one of the more unreasonable ways in the following blog entry:

http://forum.allaboutcircuits.com/blog/a-battery-isnt-a-capacitor.588/

This is basically treating a battery as a DC capacitor (i.e., using Q = C·V) and it doesn't even come close.

But if you look at it only as an AC capacitor (i.e., using dQ = C·dV), then it is a pretty good model (or at least has a number of useful applications).

Ha! I liked that post. It IS a common misconception.
I am thinking more about transient, short lived behavior on the DC battery circuit that (I am thinking) act like AC. (See the better version of the graph I posted in response to #12)

 

From your graph:
I = 7 amps
delta V = .2volts
delta T = .2 seconds

Using those numbers in, delta V/ delta T = I/C, .2V/.2sec = 7/C, C (capacitance) would equal 7 farads. (give or take a few hundred thousand microfarads)

Not possible

 

Of course another difference is that batteries obtain electricity by making and breaking chemical bonds, and of course some can do this as a primary source.
Capacitors store electric energy in an electric field and at most deform (polarise) chemical bonds.

Ian, I had thought you could charge super capacitors to any required voltage as with any other.

Mark it has already been mentioned that battery impedence is complex and may inlcude a capacitive component. but that does not mean you could use one to block DC between stages in an amplifier.

 

From your graph:
I = 7 amps
delta V = .2volts
delta T = .2 seconds

Using those numbers in, delta V/ delta T = I/C, .2V/.2sec = 7/C, C (capacitance) would equal 7 farads. (give or take a few hundred thousand microfarads)

Not possible

Right, I think mainly what the graph is showing is the Diffusion ion drift rate as the batteries chemical rodox reaction kicks in to get the cell back to equilibria. You see a small current tip from electrode surface charge and a quick voltage Exponential type drop as that is used up then a rebuild of voltage from redox spinning up as energy flow builds over time.

 

Right, I think mainly what the graph is showing is the Diffusion ion drift rate as the batteries chemical rodox reaction kicks in to get the cell back to equilibria. You see a small current tip from electrode surface charge and a quick voltage Exponential type drop as that is used up then a rebuild of voltage from redox spinning up as energy flow builds over time.

I would assign the chemical changes to that slow slope in the middle. Caused by the equilibrating concentrations and resulting changes in activity at each electrode creating electrode voltages effects.

When you say "spinning up" what do you mean?

 

From your graph:
I = 7 amps
delta V = .2volts
delta T = .2 seconds

Using those numbers in, delta V/ delta T = I/C, .2V/.2sec = 7/C, C (capacitance) would equal 7 farads. (give or take a few hundred thousand microfarads)

Not possible

I would assign the chemical changes to that slow slope in the middle. Caused by the equilibrating concentrations and resulting changes in activity at each electrode creating electrode voltages effects.

those exponential looking curves though, using your equation I just calculated what the capacitance must be at two points on each curve and I got
for the first curve: 0.356369 Farads and 0.195261 Farads
for the second curve: 0.5779224 Farads and 1.7530752 Farads

Which would mean that the electrolyte in the middle at those four points would need to have a permittivities
0.0071808 F/m
0.0039345 F/m
0.0116451 F/m
0.0353242 F/m

Am I right in saying that even looking just at these more curvy sections, the values are beyond what is possible?

I was just reading that the permittivity of free space is 8.854× 10^−12 F/m, and that the best dielectrics would be maybe a hundred thousand times that. Nowhere near what I am calculating. Then again, most dielectrics don't have dissolved ions in a liquid solution.

 

"spinning up"

The reaction is exothermic and the reaction rate increases with temperature so they build on each other. In extreme conditions this can cause runways and a explosion.

 

I was just reading that the permittivity of free space is 8.854× 10^−12 F/m, and that the best dielectrics would be maybe a hundred thousand times that. Nowhere near what I am calculating. Then again, most dielectrics don't have dissolved ions in a liquid solution.

Those curvy sections are not classical capacitance discharge/charge curves. The topmost reaction surface charge on the battery plate is still mainly electrochemical not electrostatic so the equivalent capacitance would be very high.