Is the formula for capacitance of a supercapacitor C = epsilon(A/d) ?
Since a supercapacitor does not have a dielectric, then will the permitivity be the permitivity of free space ?
Electrical – Calculate the capacitance of a supercapacitor
supercapacitor
Related Solutions
Overview
In general capacitor lifetime (including supercapacitors) is dependent on three things:
- Electrolyte Life
- Voltage Derating
- Temperature / Power Dissipation
If you want the capacitor to last a long time, limit the applied voltage, keep it cool, and limit the output current. All this should be in the datasheet of your capacitor.
Electrolyte Life
If the capacitor is, say, ceramic or tantalum, the electrolyte is a solid and the cap will basically never go bad. If it's an electrolytic, then it contains fluid which will evaporate and eventually cause the cap to fail. In an electrolytic double-layer super-capacitor, the electrolyte is a combination of a fluid and of activated carbon, so it is mildly vulnerable to evaporation.
Voltage Derating
More important, though, is the voltage derating of the capacitor. If the voltage burns the cap up on the first use, you won't need to worry about evaporation. A capacitor is a carefully constructed device which separates two conductive films with a thin, thin layer of insulating material over a wide area (folded or rolled into a package). Decrease the separation, and you've got higher capacitance with the same area. This thin separation is vulnerable to high voltages; that's why capacitors have specific voltage maximums, often printed on the case. In a supercap, this barrier is often just nanometers thick, and the dielectric will not insulate high voltages across this short distance.
Using a cap at close to its maximum voltage will cause it to fail more quickly than using it at a lower voltage, this tradeoff is known as a derating curve. It should be available from your capacitor manufacturer.
Temperature / Power Dissipation
A capacitor is negatively affected by heat. It causes the electrolyte to evaporate more quickly, causes the dielectric to be weakened, and it can damage the thin conducting elements in the capacitor. Both environmental heat and self-heating effects should be considered. If the capacitor is discharged very rapidly, the small resistance of the foil and leads will be inconsequential compared to the square of the current.
Start by reading about how to use the capacitor charge, current and energy formulae. Q=CV, Ic=CdV/dt, E= ½CV²
Then realize that Batteries have more capacitance per volume than super caps, but only above the final Vf cell voltage, so stored energy is E= ½C(Vi²-Vf²).
Current is limited by the ESR of the charging circuit and Cap/battery. Parts with lower ESR are generally rated for higher currents but this value changes with voltage and aging.
example photo not to same scale
--
>3kJ supercap vs 
>50kJ
This Supercap with a cost of $100 for 1000F with 10mΩ ESR has a T=RC=1second with a much faster response time than a smaller LiPo battery which has greater energy storage capacity.
This LiPo has much more Capacitance value but also much higher ESR.
E= ½C(Vi²-Vf²) for C=1000F and Vi=2.5V and Vf=0.0 (complete discharge) E= 3125 Joules
A LiPo pack with a cost of $20 is rated at 3.7V 4.4Ah for 1.0C and 0.2C to 3.0V
E= ½(Vi+Vf) *Ah 3600s/h = ½(3.7+3.0) 4.4Ah *3600 = 53,064 Joules [Watt-seconds]
You may compare the energy , cost and size with life expectancy then answer your own questions.
Best Answer
You can also solve the capacitance of a capacitor using the capacitor charge equation.
\$V = V_p{e^{-t/RC}}\$
Based on what you can see in the equation, you have to wire the resistor and the capacitor in parallel, and then put a voltmeter in the capacitor. Then, all you have to do is to have a supply voltage (Vp), the resistor value (R), find the voltage of the capacitor (V) at time (t), and then you'll get the Capacitance of the capacitor. Assuming that you have a very accurate resistor reading, an accurate voltage source that does not sag when a load is connected to it, and an accurate voltmeter, given time (t), you will get an accurate enough value of the capacitor.
Here's an example:
Let's say we have an unknown capacitor just like in the schematic. I grab my stopwatch, and start it the moment the switch is closed. I will then check the Voltmeter and take note of the time that passed to reach the capacitor voltage to, let's say 5 Volts. The capacitor reached this voltage in roughly 14 minutes and 30 seconds. Converting that to seconds and we will get 870 seconds. So we now have, V = 5V, Vp = 12V, t=870s, and R which is 20 Ohms. Plugging that into the equation will give me 49.69 F, rounding it up will give me 50 F.
Since you're dealing with a super capacitor here, charging times will be large enough to get a good reading as what I have demonstrated in the example above. However, if you need to measure a small capacitor, you can use a larger resistor that will charge the capacitor slower to prevent sub-second stopwatch readings, since we humans are limited to our ~20ms response time :)