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 \cdot 2A \cdot 3600s = 9000J\$
From the capacitor work equation:
$$W = 0.5 \cdot C \cdot V^2$$
$$9000J = 0.5 \cdot C \cdot 2.7V^2$$
$$C=2469.1358F$$
Is this correct?
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:
$$C = \frac{2}{(2.7V)^2 - (1.0V)^2}9kJ = 2.86kF$$
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%
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:
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.
I´ve actually been looking at something similar - which is how I came across this thread. A friend found some videos of a guy who is using Boost/supercaps to start his car (there are several videos on YT).
This made me wonder about the relationship between the car battery and a capacitor. All the above is interesting (and accurate), but maybe could be simplified:
A 2Ah battery has an equivelent charge flow of 2*3600 = 7200 coulombs
So equivalent C = 7200/1.25 = 5760F
Which is quite a big capacitor!
C=Q/V = (2AH)*(1coulomb/amp-sec)(7200sec)/1.25v = 5670 Farads. This is the correct answer to your question.
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.
Using Phil Frost's battery, its voltage falls from 1.5V to 1.2V within 1.6 hours at a constant rate of 0.1 A (suppose that the horiz axis is in hours, not in AH). The capacitor that does the same thing is:
$$C = \frac{i}{dv/dt} = \frac{0.1}{\frac{0.3}{1.6 \cdot 3600}} = 1920 \;farads$$
Now compare the cost of C with an equivalent rechargeable battery.
