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 ?
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12\$\begingroup\$ "Since a supercapacitor does not have a dielectric" Say what? \$\endgroup\$– WhatRoughBeastCommented Sep 10, 2017 at 12:32
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\$\begingroup\$ "In a supercapacitor, there is no dielectric as such" explainthatstuff.com/how-supercapacitors-work.html \$\endgroup\$– DeadshotCommented Sep 10, 2017 at 12:49
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6\$\begingroup\$ Misleading content: The thin isolating layer is the dialectric and has a much larger permitivity than free space. \$\endgroup\$– Turbo JCommented Sep 10, 2017 at 14:27
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2\$\begingroup\$ Actually, the "isolating layer" is NOT the dielectric -- that's really just there to keep the two plates and their porous carbon coating from physically touching. The electrolyte permeates the isolation layer and allows current to flow through it. The actual dielectric is a thin layer that forms through chemical reaction between the electrolyte and the porous carbon, and is formed directly on the surface of the carbon itself, which is why it has such a huge surface area. There is such a coating on both plates, which is why they are called double-layer capacitors. \$\endgroup\$– Dave TweedCommented Sep 10, 2017 at 16:10
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\$\begingroup\$ I don't get this question, how would you know the area? \$\endgroup\$– pipeCommented Nov 22, 2018 at 10:42
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2 Answers
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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 :)
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Measuring capacitance of a supercapacitor is pretty easy if you have a power supply capable of constant current. Apply a constant current for a constant amount of time and measure the difference in voltage. The capacitance is then F = As/V.
For example, lets say I apply 100 mA for a minute and the capacitor voltage raises by 500mV. F = (100 mA)*(60 s)/(.5 V) = 12 farads.
This method also works for the usual sizes of capacitors, you just need a quick hand and a very accurate stopwatch. :)
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\$\begingroup\$ If you try to measure say a 5F supercap this way, its very likely that you won't get 5F. Forget off hand, but I believe its IEC62391 that specifies the test conditions to get 5F. It's measured on discharge, and the capacitance you calculate will also vary with discharge current. \$\endgroup\$ Commented Feb 3, 2018 at 0:23
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\$\begingroup\$ @MattYoung Constant current discharge equipment is less common for amateurs. Constant current power supplies are more common and it's good enough for most use cases. \$\endgroup\$– DaffyCommented Feb 3, 2018 at 0:29
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\$\begingroup\$ All true, I'm just saying it won't be accurate. :) \$\endgroup\$ Commented Feb 3, 2018 at 0:30
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1\$\begingroup\$ @MattYoung well technically it will be accurate, it will just be an accurate measurement of a slightly different parameter than the one specified in the datasheet \$\endgroup\$– BeB00Commented Aug 15, 2018 at 0:31
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\$\begingroup\$ Sounds like a wonderful way to measure your reaction time, assuming you know the capacitance. \$\endgroup\$ Commented Sep 11, 2019 at 1:31