If it can, what kind would I get? Two potatoes if that helps. I know you have to have a fair amount of potatoes to power an LED, so outside of having eight potatoes wired up, I thought maybe two could charge a capacitor.

I thought if I could charge a capacitor with a potato or two, then I could let the capacitor charge up and light the LED, since a continuous flow from a couple of potatoes isn't enough to power the LED alone (I have to have a bunch of taters for that.)

  • \$\begingroup\$ Sure it can, but what is your goal? How will you determine the charge on the capacitor? The practical question isn't if it can charge a capacitor, but if it can charge a capacitor to a degree detectable by the method you are going to use to measure. \$\endgroup\$ – Chris Stratton Mar 23 '18 at 2:58
  • \$\begingroup\$ I edited it, but essentially I thought the capacitor could be charged by a single potato or two and that could then be used to power the LED, even if for a moment. \$\endgroup\$ – johnny Mar 23 '18 at 3:29
  • \$\begingroup\$ To power an LED you'd need to achieve a voltage in excess of its forward voltage. Without a boost converter you won't get a higher voltage on the capacitor than the voltage of the battery you start with. So if your potato battery (or several in series) can light the LED then it can probably charge the right capacitor to briefly light it as well. \$\endgroup\$ – Chris Stratton Mar 23 '18 at 3:48
  • \$\begingroup\$ @ChrisStratton More accurately: the voltage of the capacitor won't get higher than the no-load voltage of the potatoes. I have no idea of the internal resistance of a potato voltage source. Once charged, the cap would be a source with a potentially lower output resistance allowing to light up a LED that can't be light up using the potatoes alone. \$\endgroup\$ – Blup1980 Mar 23 '18 at 7:54
  • \$\begingroup\$ What kind would I get? King Edward, or Maris Piper. \$\endgroup\$ – Brian Drummond Mar 23 '18 at 12:38

Yes, that should work. Use a large value electrolytic capacitor, like 2200uF or 3300uF (voltage common value 16V, but any voltage would work.) Let it charge for half a minute.

You'll need at least two taters wired in series, and use a red LED, since those only need about 1.7V or so. (Green and blue LEDs need much higher forward voltage.)

| improve this answer | |
  • \$\begingroup\$ But as I understand the comments, is it true the voltage from the existing potatoes doesn't somehow increase or build up in the capacitor? I thought if the capacitor was "load up" charge and release it when the designer determined in the circuit, but that that exiting charge from the capacitor was somehow more or powerful or whatever and would power the light. I don't think I fully understand what the capacitor does beyond hold an amount of current or volts. \$\endgroup\$ – johnny Mar 23 '18 at 14:40
  • \$\begingroup\$ @johnny Capacitors store potential energy in the form of separated opposite charges. The potato-voltage and the capacitor voltage will be the same (when you connect them in parallel, then wait for half a minute.) But the potato alone has a large internal resistance (perhaps 5000 ohms, perhaps far more.) The internal R of the capacitor is 1000X less (perhaps 10 ohms.) When you connect the LED to the potato alone, the LED current is tiny. (View the glow in a totally dark room!) When you connect the LED to the capacitor, the current is large for a moment, and the LED will flash brightly. \$\endgroup\$ – wbeaty Mar 23 '18 at 20:32
  • \$\begingroup\$ One potato two potato three potato works. Two potato might not give enough volts, so be sure to use #14 copper wire and zinc nail (more than one of each, stabbed into the potato near each other, to form crude "plate" shapes. \$\endgroup\$ – wbeaty Mar 23 '18 at 20:46

There are two problems here.

a) The ultimate voltage you are going to get from one tater cell depends on the composition of the metal electrodes that you use in your cell.

b) The amount of current that you can generate depends on the area of the electrodes and the composition of the electrolyte (the tater).

The trick here is to use a capacitor to store up the current until the voltage across the capacitor reaches that in part a. Then you can use the capacitor to illuminate a LED. How long you can illuminate it for and how long the capacitor takes to recharge depends on the size of the capacitor, the current requirement of the LED and the current limits set by part b.

Good luck with this experiment and please let us know if any particular potato gives better results.

If you are going to use this to power microelectronic can I suggest the megachip variety, if you have sunshine available how about the solara variety.

| improve this answer | |
  • \$\begingroup\$ What size capacitor do you think I need? I have a pack of them, but don't know what to use. \$\endgroup\$ – johnny Mar 23 '18 at 18:32
  • 1
    \$\begingroup\$ A big capacitor will produce a long flash and take a long time to recharge. I cant be more precise we would need to know what current your LED takes and what current your tater battery produces. Experiment this is what it is about! \$\endgroup\$ – RoyC Mar 23 '18 at 18:42
  • \$\begingroup\$ If you use a large capacitor - say more than 100 µF or so - I would consider putting a resistor in series with the LED. Otherwise there might be enough energy stored in the capacitor to burn your LED to a crisp. \$\endgroup\$ – nekomatic Mar 6 '19 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.