# COMMUNITY : FORUMS : TECH

## capacitor resistance

why is it that as super capacitors capacitance goes up the internal resistance goes down? I'm looking for a tiny capacitor that can supply 0.3 volts20mA (I'm going to be powering flexinol, not an LED) for 1 second. I don't want the capacitor to be huge because I'm making a beam bot where size is a factor for charging time and because of it's physical size.

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guyfrom7up (author) 10 years ago
10 years ago
I believe so. They were discussed some on the Yahoo BEAM Robotic group.
guyfrom7up (author) 10 years ago
does anybody have an idea for the capatence? I need to power the flexinol at .35 volts, .02 amps for 1 second. Does anybody know what size capacitor I should get?
10 years ago
In theory, you should be able to figure out the capacitance from the amount of energy you need to store to heat up the wire:

0.35V x 20mA x 1s = 7mJ

energy storage for a capacitor = C x V2 / 2

=> C = 7mJ x 2 / .35V2 = 114mF, or 0.114F

That's quite a hefty capacitor! Note that you could store this amount of energy in a much smaller capacitor, if you use a higher voltage. For example, let's see what happens when we store this at 5V:

C = 7mJ x 2 / 5V2 = 0.56mF, or 560uF

Note that the 1 second, 20mA contraction current is not necessarily the physical limit of the wire. The Dynalloy site mentions "On small diameter wires (</= 0.006" diameter) currents which heat the wire in 1 second can typically be left on without over-heating it. Both heating and cooling can be dramatically changed (see section 3 of the technical characteristics in the standard literature for more information.)"

I couldn't find the "standard literature" they were referring to, but I would guess that (within limits), you should be able to switch that wire by pumping those 9mJ of heat into it in a much shorter amount of time. In fact, the FAQ over on robotstore.com states "Muscle Wires contract as fast as they are heated - in one thousandth of a second or less".

So - I would suggest you pick whatever voltage works best, given the output of your solar cells, figure out what capacitance you need to store 6mJ of energy, then discharge all of that through your Flexinol wire in a fraction of a second, no limiting resistor. If that proves robust enough - great, it doesn't get any simpler than that! If you burn through your wire - well, you only wasted one centimeter of wire...
guyfrom7up (author)  Patrik10 years ago
thanks! I think I'll just experiment like you said, I mean what else to do with 100 cm of it. I"m going to probably use 1000uF, but I'll just guess and check.
Patrik10 years ago
If you put two resistors in parallel, the total resitance is halved. If you put two capacitors in parallel, the total capacitance is doubled.

Now, if you model a real capacitor as a series circuit of an internal resistance R, and a capacitance C, then putting two of these circuits in parallel results in an internal resistance R/2, and a capacitance C*2.

In practice, larger capacitance aren't made by putting multiple smaller ones in parallel. Rather, they just make a capacitor with a larger surface area between the two electrodes, but the same principles apply.

Note that you can always increase the the discharge time of your capacitor, by putting an extra resistor in series. For most typical capacitors in your parts bin, you can ignore the internal resistance. For supercaps, that doesn't seem to be the case though.
guyfrom7up (author)  Patrik10 years ago
I think if I just use a super cap and don't let it charge all the way and discharge at the voltage I want, I can control the voltage using an external voltage divider. Could I use 1 external resistor like with an led?