I am learning how to use capacitors (ceramic and electrolytic from 1pf - 1000uf) and am trying out various experiments using a breadboard. I am constantly adding/removing things to my layout to see what happens... but waiting for a capacitor to discharge sometimes takes a long time! The book I am currently reading (Make: Electronics) suggested to "discharge a capacitor by touching a resistor across it for a second or two". Is this safe/recommended way? Can I just hold the resistor with my fingers and touch it across both terminals?

Note: I have to admit I am a bit paranoid with capacitors after seeing pictures of exploded capacitors and resulting fallout such as melted breadboards, burnt tables and even reading about people losing fingers!

Edit: I am currently working with 1.5 - 12V though I also have some 24V stepper motors which I'd eventually like to get working.

  • \$\begingroup\$ I am currently scrambling inside a broken tv. And I have now clue what I am doing. All I know is that a few seconds before- I got shocked. And as intensity it wasn't far from an regular 220 V electric shocks (I know what I'm talking about, I had plenty of those too). Oh, and ya, it was the capacitors, as the tv is not currently plugged in. \$\endgroup\$
    – user40942
    Commented Apr 26, 2014 at 21:00

4 Answers 4


With small capacitors up to 1 mF, there is little to worry about. I suppose it's a good idea to make sure they are discharged before plugging them in where the voltage that could be on the cap could damage something, but this is something not generally worried about until you get to some real energies or high voltages.

For small electrolytic caps like what you are working with, just short them against something metal, like a bare component lead, metal chassis, or handy screw driver.

Don't waste brain cycles thinking about this for anything small enough to be a ceramic cap you can plug into a breadboard. By the time you plug it in, your fingers will have discharged it. Even if not, do the math. 1 µF at 10 V is only 50 µJ. Yes microJoules. Big deal.

  • 4
    \$\begingroup\$ Can you please define "real energies" and "high voltages"? \$\endgroup\$
    – glenneroo
    Commented Mar 25, 2012 at 2:40
  • 8
    \$\begingroup\$ At 1000 uF = 1 mF you may want to start modifying Olin's advice at above about 30V. Above there energy level starts to get noticeable and shock risk starts to matter. Even at 30V you may get "spatter" from the discharge with the very very unlikely but possible outcome of something in your eye. For 30V at 1000 uF energy = 0.5CV^2 = 0.5 x E-3 x 900 ~= 0.5 Joule . A Joule is ~ the energy dissipated in dropping a 100 gram mass 1 metre so 0.5J = 100 gram x 500 mm. Just as a drop of something like that MIGHT eject a small particle so shorting a cap with that energy JUST MIGHT do the same. \$\endgroup\$
    – Russell McMahon
    Commented Mar 25, 2012 at 7:37
  • 3
    \$\begingroup\$ @glenneroo: I was also going to say 1 mF and 30 V too before reading Russell's comment. There isn't a hard line, but at that level the energy is small and finite enough, and the voltage limited to a safe level unless you do something deliberately stupid. I remember playing with a much larger (44 mF?) and I think 15 V cap in college. I even took some pictures lit only by the sparks from shorting it with a screw driver. The sparks were cool, but a long way from hurting me even with fingers right there. The high current pulses thru the cap probably weren't the best for it though. \$\endgroup\$ Commented Mar 25, 2012 at 12:36
  • 2
    \$\begingroup\$ @Jonny: No, not really, because as you say batteries are not capacitors. Capacitors of the size we are talking about have much much less energy storage than a car battery. That's a totally different case. \$\endgroup\$ Commented Mar 25, 2012 at 21:47
  • 4
    \$\begingroup\$ You should probably change "Big deal" to "No big deal", because sarcasm isn't obvious to many people who didn't grow up in your language. \$\endgroup\$
    – Jay Bazuzi
    Commented Mar 29, 2012 at 2:41

Instead of holding the resistor with your fingers try gluing it to the end of a Popsicle stick or some other insulated material. That way your fingers will be much less likely to come in contact with the capacitor. If what you are dealing with is 20 volts or less this should be fine.

I am assuming we are talking about relatively small capacitors and voltages here. If you start talking about high voltages that could be fatal then what you want a professionally manufactured device and extra precautions.

Here is an article where someone made a nice discharge probe from a Bic pen. He also goes into the math if you are curious. Once again - safety first! If you are dealing with lethal voltages your best alternative is to use a professionally manufactured, tested and certified probe.

enter image description here

Now having said all that, I agree with Olin that this will be overkill for the small capacitors you are currently dealing with. This information may prove helpful as you advance and perhaps start dealing with bigger capacitors and higher voltages.

  • 4
    \$\begingroup\$ I performed a HIPOT test of the pen body at 1,500 volts and there was no measureable current. Use this pen or any other at your own risk. I can't guarantee that next months lot of Bic pens will not have some impurities or chemical changes that will degrade its dielectric strength. <- This line is important! BIC pens and popsicle sticks are not known for their insulating properties. There are plenty of safer alternatives. I'd suggest repurposing a multimeter lead for this task. \$\endgroup\$ Commented Mar 25, 2012 at 2:34
  • 2
    \$\begingroup\$ @Kevin - Thanks for the safety tip. I modified my answer to emphasize safety. Particularly for people without the necessary lab to do a HIPOT test they are well advised to buy a commercial product rather than jury rigging something that could get them killed. \$\endgroup\$
    – JonnyBoats
    Commented Mar 25, 2012 at 3:11
  • \$\begingroup\$ Related: electronics.stackexchange.com/questions/52289/… \$\endgroup\$
    – zebonaut
    Commented Dec 29, 2012 at 9:19

The product of the resistance (in Ohms) and capacitance (in Farads) is the scale time for the discharge (to go to 1/e of the original charge): t = RC. With V = Q/C and I = V/R = Q/t, you can also solve for the minimum resistance to keep the discharge current to a safe value. (Keeping the current below 1 mA is a rough guideline: https://www.asc.ohio-state.edu/physics/p616/safety/fatal_current.html That's for discharge through humans, but what constitutes "safe" varies by what's connected to your discharge circuit. If you're running the current through flexible copper wiring, it can probably take a few amps.) Also note the energy stored in the capacitor, which will be deposited in the resistor which shorts it: U = 0.5 C V^2.

As long as you're dealing with the sorts of capacitors typically used with bread boards, you can probably short it with copper wire, as others have mentioned: 1 uF * 1mOhm = 1 ns discharge time. If it only has 42V on it, these formulas say it will have a high current for a few nanoseconds, but the nanoHenry-scale parasitic inductances will limit the current and slow the discharge. That 42 V at a 1uF is less than 1 mJ, which could damage sensitive electronic components - so don't short out the capacitor with that high-end CPU. Anything else should be fine.

If you get into voltages and currents where discharge takes a second or more, or where your discharge currents will be in excess of that 1 mA for more than 1 ms, or where the energy stored exceeds a few Joules, then you should be careful: Check the current and power ratings of the components in the discharge circuit, estimate the inductance, and maybe run a simple sim of the discharge process. In general, discharge-before-use won't be a significant issue unless your capacitor is comparable to a full Farad or the voltages are a few kV.


Not sure how relevant this post is, but here is something I've been developing for a relay timer circuit.

Q1 is a PNP transistor that will turn off when the voltage is released, discharging the capacitor to ground.

  • \$\begingroup\$ I don't think this will completely discharge the capacitor, at least not through the transistor. You need the voltage across R6 to be about 0.6V if the transistor is going to be active. That means the voltage across R5 will be about 10V, and the capacitor voltage will be also roughly 10V. You might as well leave out Q1 and R8. \$\endgroup\$ Commented Mar 6, 2020 at 13:06
  • \$\begingroup\$ ... and it's polarity sensitive. \$\endgroup\$
    – Transistor
    Commented Mar 6, 2020 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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