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I have a 1.25V 2Ah battery and I'm trying to calculate a equivalent capacitance with rated voltage of 2.7V for each of those batteries. This is what I did:

Work of Battery = \$1.25V * 2A * 3600s = 9000J\$

From the capacitor work equation:

$$W = 0.5*C*V^2$$

$$9000J = 0.5*C*2.7V^2$$

$$C=2469.1358F$$

Is this correct?

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No, it is not correct. There is absolutely no way you can possibly know the capacitance value to 8 significant digits! Think about it. Even a fraction of a degree temperature change will cause more change in the stored energy of a battery than 1 part in 10**8, and of course the initial accuracy is nowhere remotely close to that. Your conclusion is simply absurd. –  Olin Lathrop Jan 29 at 18:23
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Olin is pedantically addressing the way you have used a high degree of precision in your calculation when it was not essential to do so. When he says that your answer is absurd he is essentially misleading you as he is not saying that the general principle of what you did was wrong - just the way that you stated it. Your formula for energy content of a capacitor is correct. Whether the energy is all usable is another matter. Your battery energy formula is correct for an idealised battery. –  Russell McMahon Jan 30 at 15:44
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4 Answers

What you have calculated is not an equivalent capacitance but, instead, the capacitance required to store 9kJ of energy at 2.7V.

That fact that the battery may also store that much energy does not mean that there is a capacitor equivalent to a battery.

While an ideal battery maintains the voltage across its terminals until the stored energy is exhausted, the voltage across an ideal capacitor will gradually approach zero as the stored energy is depleted.

If the attached circuit will only function properly above some minimum voltage, not all of the energy stored in the capacitor is available to the attached circuit.

Thus, one must first specify the allowed drop in voltage to determine the required capacitance.

For example, stipulate that \$9kJ\$ of energy must be supplied by the capacitor before the voltage falls to \$1V \$.

Then:

$$\frac{C(2.7V)^2}{2} - \frac{C(1.0V)^2}{2} = 9kJ $$

Solve for the required C:

$$C = \frac{2}{(2.7V)^2 - (1.0V)^2}9kJ = 2.86kF$$

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You have provided energy content formulae for idealised battery and an idealised capacitor.
This logically suggests that when you talk about an "equivalent capacitance" to a battery that you mean a capacitor that stores or can deliver the same energy as the example battery.

In theoretical terms your calculation is correct for an idealised battery (constant voltage throughout discharge, defined mAh capacity) and an idealised capacitor.

In real world situations the formulae will indicate a capacitance that is smaller than would be needed in practice. How much larger the capacitor would need to be depends on what form the load takes. As the capacitor discharges its voltage drops. To extract all the stored energy the voltage would have to drop to 0V, which is impractical.

  • If the load is eg an electronic "boost converter" which can accept the range of voltages 'offered' and convert the output to a useful voltage then the amount of energy able to be extracted in real-world situations may be over 80% + of the total stored capacitor energy. In addition to the energy that cannot be extracted for practical reasons you need to allow for the inefficiencies of the converter - in practice the best achievable will be not much over 90% efficient and in many cases around 70% to 80% is more likely.

  • If the load requires eg constant voltage and you do not use a "converter" but instead use a linear regulator then available energy will be reduced or much reduced compared to what is stored in the capacitor. The result can be calculated if the required load voltage is known.
    For a capacitor charged to V = Vmax, the energy provided to a load at some lower voltage V = Vout is given by
    Energy = 0.5 x C x (Vmax^2 - Vmax x Vout)
    [Derivation of this simple but seldom seen formula is left as an exercise for the student :-) ]
    eg for a Capacitor charged to 4V driving a 2V load via an idealised linear regulator the available energy is
    0.5 x C x (4^2-4x2) = 4C.
    The energy loss in the capacitor is 0.5 x C x (Vmax^2 - Vou^2) = 6C
    So the use of a linear regulator produces 4C/6C ~= 67% of the capacitor energy loss in this case.

  • One less familiar example of a load that can accept a wide range of capacitor voltages without use of a boost converter or similar is a PWM driven DC load that can accept energy at a low continuous voltage AND also accept energy in short high current pulses. A heating element might be an example of this. Such an arrangement allows the capacitor to be driven by low duty cycle PWM when Vcap~= Vmax and for duty cycle to be increased a Vcap falls. In this case energy is used AT the capacitor voltage, there is no need for energy conversion and efficiency is limited mainly by the PWM switch losses. Using a modern low Rdson MOSFET as a switch can allow efficiencies of 98 - 99% in practical situations. [I am presently investigating such an arrangement to allow a PV panel charged capacitor to power a heating element over a wide range of solar insolation].
    An alternative which achieves much the same result is to use a switched load where a number of resistors are switched in or out of circuit as required. Using binary weighted resistor values a load able to accept a wide range of voltages, at APPROXIMATELY constant power, can be constructed.

As can be seen, a battery holds an immense amount of energy for its size and cost, compared even to the most energy dense "super" capacitors.


Notes:

The reason that in real-world cases you usually need more capacitance than calculated is because, to extract all the energy from the capacitor you have to drain it to zero volts. No real world process is overly happy at starting at say 2.7V and finishing at 0.1V or 0.05V or 0.001V etc. So you need to measure the energy change when discharging from Vmax to Vlowest_usable.

Fortunately, because capacitor energy content is proportional to V^2, most of the energy has been extracted before it gets to very low voltages, so you do not reduce effective energy capacity vastly. At V = 50% x Vmax energy remaining is (50%/100%)^2 = 25% and energy taken taken is 100-25= 75%. At 20% of Vmax remaining energy = (20/100)^2 = 4%.

If the capacitor drives a boost converter and starts at 2.7V then 20% = 2.7 x .2 = 0.54V. This is 'on the low side' but a number of boost converters will operate at 0.5V even though they need say 0.8V to 1.0V to start.

Energy taken when discharged across a range =

= 0.5*C*Vmax^2 - 0.5*C*Vmin^2

= 0.5*C*(Vmax^2 - Vmin^2)

So to establish the required capacitance for a given battery use.
C = 2 x mAh x Vbat_mean /(Vmax^2 - Vmin^2)

In this case, discharge to 0.54V would increase capacitance needed only by about 5%.

For an endpoint voltage of 1V you have remaining energy of 1V^2 / 2.7V^2 =~ 14% energy remaining.
So you need to increase capacitance by about 100/(100-14) =~ 16%

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Appealing to the fact that stored energy is proportional to \$V^2\$ is only relevant if the capacitor is driving a non-linear load, like a boost converter. If it's instead driving a linear load, like a linear voltage regulator, then the load requires effectively constant current, not constant power, and the advantage of \$V^2\$ will be squandered simply by heating the regulator more when the capacitor voltage is higher. –  Phil Frost Jan 29 at 19:53
    
@PhilFrost You seem to be revisiting what I have already said, in slightly more detail. eg " ... No real world process is overly happy at starting at say 2.7V and finishing at 0.1V or 0.05V or 0.001V etc. ... " & " ... If the capacitor drives a boost converter". -> The OP seems tome to have a better grasp of the core issues than any are giving him credit for. –  Russell McMahon Jan 30 at 15:38
    
My point is that if you want to solve the problem of the load not being happy with the voltage decrease, and the load is linear, then you are going to need to add a lot more than 10% to 20% capacitance. The voltage isn't going to go down slowly at first because the higher voltage current has more energy. Rather, The voltage is going to go down quickly at first. Think of the exponential discharge curve you get with a simple RC circuit and an initially charged capacitor. That's very different than the constant energy sink you describe with your math. It really depends on the particular load. –  Phil Frost Jan 30 at 15:42
    
I meant to write constant power sink, not constant energy sink. A resistor load isn't constant anything: the power in the resistor is \$P=V^2/R\$. The \$V^2\$ here negates that "capacitor energy content is proportional to \$V^2\$". –  Phil Frost Jan 30 at 15:51
    
My example answer specifically noted the use of a boost converter and was correct for an idealised capacitor. The calculations correctly demonstrated the point that I was making. –  Russell McMahon Jan 30 at 15:57
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One problem with your calculations is that you assume the battery voltage will remain constant at 1.25V until it is completely discharged. However, the capacitor equation uses a change in voltage so it assumes that the capacitor voltage falls to 0.0V when all of the energy is removed from the capacitor. This is an important difference if you are actually planning to replace a battery with a capacitor.

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True, but not highly relevant. His battery description is an idealised one. If he'd said eg Vmean it would have been more useful. BUT his purpose was clearly comparing order of energy contents. . –  Russell McMahon Jan 30 at 16:04
    
@RussellMcMahon I disagree. The OP did not say he wanted to compare energy storage, he said he wanted an equivalence and he tried to evaluate that equivalence by comparing the total energy storage. There's also no indication that the OP intended for the battery to be "idealised". You're reading intent into the question that is not there, in my opinion. –  Joe Hass Jan 30 at 21:08
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A battery and a capacitor are hardly equivalent.

A battery has a voltage that's a function of the chemistries of the materials inside it. This voltage is constant. As the stored energy in the battery is exhausted, the voltage decreases some. Some of this is due to an increase in internal resistance as the reactants inside the battery become exhausted. Even so, the voltage does not decrease linearly as the battery is discharged: it follows a more or less shallow decline, then falls off a cliff at the end.

For an example, see these discharge curves for some AA batteries. These are from a test on powerstream.com:

battery discharge curves

Also notable, battery voltage can recover if the load is removed in the middle of the test. See also: Do batteries lose voltage as they're used up?

On the other hand, capacitors aren't like this at all. If you were to draw a similar discharge curve as above for a capacitor, it would be a straight line. It would start at the left at whatever voltage you charge the capacitor to, decreasing linearly to 0V when all the stored energy has been removed.

Furthermore, your question suggests that maybe you believe "capacitance" is some measure of how much "capacity" a capacitor has. It's not. Capacitance is just a ratio of electric charge (the integral of current) to voltage:

$$ C = \frac{Q}{V} $$

The SI unit of capacitance, the Farad, is a coulomb per volt:

$$ \mathrm{F} = \frac{\mathrm{C}}{\mathrm{V}} $$

(note here the C is coulomb, where above it was capacitance)

This says nothing about how much energy the capacitor can hold. In fact, an ideal capacitor of any capacitance can hold infinite energy. Real capacitors break at some maximum voltage, and this is what limits their energy storage capacity.

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Would someone like to explain the downvote? –  Phil Frost Jan 30 at 0:23
    
If you get annoyed by the following rather than deciding if each point is factual then we are both wasting our time. Just about everything you said here is either correct but not relevant, or wrong. | He was asking about energy storage. He did not address or ask about 'equivalence' | Many battery types over a range of chemistries have discharge curves vaguely similar to what you show but the differences with load and discharge rate are so varies as to make the example graph more misleading than useful. Almost all batteries "dip in output as they discharge, but the "cliff" you show is not ... –  Russell McMahon Jan 30 at 15:50
    
... present in some cases and much reduced in others. | Your statement that " ... your question suggests that maybe you believe "capacitance" is some measure of how much "capacity" a capacitor has ... " is untrue. He has not used enough words but he has used correct enough equations to compare battery energy storage (mAh x Vmean) with capacitor energy storage (1/2 C V^2). He is not doing any of what you say when he does this. | Your closing comments about ideal capacitors are essentially correct but irrelevant to him. He clearly has a cap that he is charging to its rated voltage of 2.7V ... –  Russell McMahon Jan 30 at 15:58
    
@RussellMcMahon calm down. Each of our answers interpret the question in different ways. It's written in a human language and is inherently ambiguous. We differ in that you interpreted the question to be about equivalent energy storage, while I interpreted the question to be about functional equivalence, and thought the OP might not understand how there's more to that beyond equivalent energy storage at a particular voltage. Your answer is not wrong, it's just a different approach, based on a different interpretation of the question. –  Phil Frost Jan 30 at 16:01
    
... etc. | He essentially stated an idealised comparison and correctly identified the relevant formulae for energy content. One will always be able to think up examples that make different assumptions and produce different answers. If he'd used more words his assumptions could have been more easily seen, but they are clear from his question. –  Russell McMahon Jan 30 at 16:02
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