Introduction: Safe Capacitor Discharge Tool

Picture of Safe Capacitor Discharge Tool

Discharging capacitors is often necessary when working on troubleshooting and repair of electronic equipment. In the old days, tube radios and amplifiers found in every households contained capacitors that continued to hold dangerous levels of charge long after the appliance had been taken off the mains. Then it was CRT televisions, but now that televisions are LED flat screen it looks like everything has turned to low-voltage digital circuitry, so what is the problem?

Well, the actual devices may be low-voltage, but the associated power supplies just got from dangerous (tube era B+ supplies) to (relatively) safe (transistorized equipment with linear power supplies) to dangerous again. The culprit today is the switched mode power supply, the efficient and lightweight modern king of power conversion for electronic equipment.

A switched mode power supply is based on an input stage that rectifies the mains voltage, yielding a direct voltage of around 330 V (for 230 V mains voltage, 170 V for 120 V mains voltage) for further use by the circuit. Turns out these little nice black boxes that hang from our laptops and are built into practically everything electronic around us that is not running on battery have really some lethal voltages inside.

Filter capacitors in the power supply input stage are charged at this high DC voltage, and will maintain their charge for some time after the plug has been removed from the wall. By the way, this is the reason why you see these safety warning stickers saying: “Do not open the box”. If you thought, well, after I pull the plug of course I can open the box, guess what, you were wrong.

Disclaimer: This circuit can operate at potentially lethal voltages. Do not attempt to build this if you are not fully understand how it works and/or you are not experienced working with high voltage. In any case, do so at your own risk.

Step 1: Discharge Circuit Working Principle

Picture of Discharge Circuit Working Principle

You will read advice and see people in videos discharging a capacitor by shorting its terminals, typically using a screwdriver. Will this work? Yes, in the sense that it will definitively discharge the capacitor. Is this the right way of doing it? No. It may damage the capacitor, and yourself in the process, and that is all I am saying on this. Actually, it will even damage the screwdriver, just for you to have an idea of the energy levels that may be involved.

Key to the right way of doing this is to release the energy stored in the capacitor in a controlled fashion, and this means discharging the capacitor on a resistor. How long will the discharge process take? The mathematician will say, an infinite time (and of course this is the correct theoretical answer). But remember, if you want a usable answer from a mathematician you always have to ask the right question. What we really want to know is how long we have to wait until the voltage is low enough as not to be a concern any longer.

So, how long will it take before the voltage is reduced to, say, less than 5% of the original voltage? Some high-school math is enough to show that for a capacitor of capacitance C discharging on a resistor of resistance R, you will have to wait for a time equal to 3RC for the voltage to drop to 5% of the original voltage. Notice that 5% of the original voltage is a relative measure: That 5% could still kill you, if, say, the capacitor was initially charged at 10 KV. But for an initial voltage of, say, max 500 V, waiting 3RC would be fine as the residual voltage on the capacitor would be 25 V.

So all we need is to connect a resistor to the capacitor and wait, right? Yes, in principle, but here is a catch here. Assume we fix R by figuring out a reasonable value, still we would need to read the capacitance value from our target capacitor, calculate 3RC and use a stopwatch to see when 3RC seconds have passed. Not really practical.

It would be a lot nicer to have a visual clue telling us when the discharge process is “finished” – meaning the voltage has dropped “enough”.

A little circuit that tells us when it is done discharging the capacitor is simple enough, and can be found online (see http://www.repairfaq.org/sam/captest.htm#ctcds). Here I am focusing on explaining the working principle, with a slight modification in order to lower the number of diodes, and on showing a packaging idea.

We can use a string of 3 standard 1N4007 diodes in series (D1,D2,D3) to establish a reasonably fixed potential point where we can connect a LED with his current-limiting resistor. We need to use 3 diodes in series because we need around 1.6 V to turn the LED on. The diodes will be conducting, and the LED will be on, until the voltage at the anode of D3 drops below the combined forward voltage of the string.

I am using a low-current red led here (a Kingbright WP710A10LID) that has a typical 1.7 V forward voltage and turns on already with a 0,5 mA forward current, which is nice because it allows us using just 3 diodes. Corresponding to the low current flowing through the LED, the value of the current limiting resistor is relatively high at 2700 Ohm 1/4 W.

The capacitor discharge resistor is a 2200 Ohm 3W resistor, which is sized for a max input voltage of 400V. This is adequate to work on standard switching mode power supplies. Notice that if you look at the datasheet for the 1N4007 diode you will see a rated forward voltage of 1 V, so one may think that two diodes will be enough to turn on the LED. Not quite so, as the 1 V forward voltage for the 1N4007 is rated at 1 A forward current, a value we will never reach (hopefully), as it would mean we have applied a voltage of 2200 V at the input of our circuit. The forward voltage in our operating current range is in the order of 500-600 mV, so we need three diodes.

The lesson here is, always take note of at the conditions under which a rating is indicated in a datasheet. Do they apply in your circuit? It may be a good idea not to stop at the front page, but to look at the characteristic curves as well!

Step 2: Proper Discharge Circuit

Picture of Proper Discharge Circuit

One final note, the circuit described above is useful to illustrate the working principle, but is should not be built and used in practice – actually, it is dangerous. This is because you have to connect the capacitor with the right polarity to the discharge circuit (Vcc terminal must be positive with respect to GND terminal), otherwise no current will flow through the D1-D2-D3 diode string! So if you accidentally connect the capacitor the wrong way, no current will flow, and the full input voltage will appear at the LED1 terminals as reverse voltage. If the applied reverse voltage is higher than a few volts, LED1 will burn out, and stay off. This may lead you to believe the capacitor is not charged, while it still is…

To make this circuit safe we need to provide a symmetric current path to discharge the capacitor when Vcc - GND is negative. This can be easily done by adding D4-D5-D6 and LED2 as in the schematic. When Vcc - GND is positive, current will flow only via D1-D2-D3 and LED1. When Vcc-GND is negative, current will flow only via D4-D5-D6 and LED2. In this way, regardless of the applied polarity, we will always know if the capacitor is initially charged and when the voltage has dropped to a safe level.

Step 3: The Case

Picture of The Case

Now that we worked out what to put in the package, it’s time to think about packaging. The whole thing could be put together either as a probe-like tool or as a little box you keep on the workbench and connect to the capacitor using test leads. I opted for the second option, so I don’t have to hold the probe in proper contact with the capacitor terminals the whole time while the capacitor discharges.

I turned a little circular box in two halves from a bar of plastic material. With the turning process it is easy to obtain a tight interference fit between the two halves, so no screws are needed.

The hole in the top shell is to fit the aluminum “button” that helps cooling the discharge resistor. The button was turned from an aluminum bar, and then a slot was milled on the top (slot not shown in the picture) to hold the resistor in place and secure good thermal transfer. Also there is a small hole that can be used to secure an additional external heat sink.

The fit between button and case is also an interference fit. As you will see in the next step, the button also helps holding all components in place. And I like the look and feel of this little case, which is just ¾” by 2”.

Step 4: Putting It All Together

Picture of Putting It All Together

This is just a little bit of precision work, where you need to pay attention to the isolation – remember, there can be a couple of hundreds volts in here! Couple of points:

  • Notice the heat shrinking tube on the aluminum button, which is the only conductive path to the outside of the box. The aluminum button must be isolated from the circuit. It is good idea to use some silicon based sealant or epoxy to fix components in the enclosure once you have tested your build (not shown in the picture).
  • The copper mesh around the resistor helps to securely hold it in place in the slot and reduces thermal resistance to the button.
  • Use 600V rated wires, don’t just grab some signal wire that you may have lying around, solder it, discover it turns out to be rated for 30 V (how do I know?)

Happy and foremost safer discharging!

Comments

Waste Of Space (author)2017-05-04

Thanks for this circuit. I am currently building it but I just need to get a 3 watt 2k2 . The rest is done. I have decided that I will use some heat transfer cement from a computer CPU to transfer the heat to a few 20c coins that will be siliconed in place.

I didn't have enough space to fit in 20c coins, I had to use 5c coins. I feel so cheap.

PS I had to order in the 3 watt resistor from

http://au.rs-online.com/web/p/through-hole-fixed-resistors/2142803/

SPenrod (author)2017-04-18

Thank you Gentlemen! My early electronics experience is very well buried in my head, but things do come back when you see the relationships again! I am sourcing parts to put one together for the shop.

SPenrod (author)2017-04-16

I work on outboard boat motors. These have a 50000 uF @63 volts capacitor in the 55 volt circuit which powers fuel injectors, coils and a number of other things. I assume it stabilizes and filters the power for these components. What changes would be needed in this circuit to safely discharge this big cap? The screw driver method can produce fireworks and arc welding! Several have gone bad, leaking fluid and corroding terminals.

mecanicafina (author)SPenrod2017-04-17

Thanks for your question. The good news is, you can use the circuit as is, but there is a catch. I suggest you optimize it for your application, here is why and how. For a 50 mF capacitor charged at 63 V the circuit as it stands will work. In fact, the max input voltage is 63 V, the max input current is 29 mA, and the max instantaneous power dissipated on the resistor is 1,80 W, all well within the component specifications. However the discharge time constant RC is 110 sec, so for the voltage to drop to 5% of 63V you will have to wait 3RC = 330 sec. That is way too long if you do this on a regular basis. To drop RC the only option we have is to lower the value of R, which shortens the safe discharge time, but also increases the power dissipation in the resistor. Sizing the resistor is a bit tricky because the dissipated power decreases exponentially from its initial maximum value. If we were to size the resistor as if the power dissipated was constant and equal to the initial value, we would oversize it quite a bit. Here is what I mean.

The instantaneous power dissipated in the resistor is given by W(t)=Wo exp (-2t/RC), where Wo=Vo^2/R is the instantaneous power dissipation when the discharge process is started. Let’s say we want to drop the safe discharge time to 30 sec. That means RC has to be about 10 sec, this gives a resistance R of 200 Ohm. Now Wo is almost 20 W, that is a bit of a resistor!

Math comes to help, because if we plot the curve W(t) for 0<t<30 we see that the instantaneous power dissipated in the resistor drops under 10 W already after about 3 sec. and under 5 W after about 7 sec. This suggests that we can go quite lower than 20 W.

A jellybean 200 Ohm 10W resistor mounted on a heat sink will happily handle this (you will need to use an enclosure big enough for the circuit to accommodate the resistor and heat sink). I am thinking here about the jellybean power resistors with aluminum body, and they need a heat sink. You could probably go even lower in terms of wattage but then you would really need to be picky about the construction of the power resistor you choose, for example try a vetroceramic power resistor. They can glow red for a few seconds and live happily even after. I would not go this route as those components are quite specialized and may not be easy to source.

In summary, swap the 2200 Ohm 3W resistor for a 200 Ohm 10W with an adequate heat sink and you’ll be fine.

SPenrod (author)mecanicafina2017-04-17

You wrote 50 mF and made calculations based on that. Does that match my 50000 uF? It seems off to me but it has been over 30 years since I took basic electronics in college, and i haven't used it much lately... The cap in question is about 2" in diameter and nearly 5" long, polarized. Huge! I don't mind waiting a couple of minutes for it to discharge. If I split the difference and go with 1k ohms at 5 watt resistor, I think the discharge time could be around 2 minutes... assuming 50 mF = 50000 uF otherwise i think we're off somewhere.

mecanicafina (author)SPenrod2017-04-18

As Fezder already pointed out, 1 mF = 1 milliFarad = 1000 uF. With a 1k resistor the 3RC discharge time is two and half minutes, if you are happy with that 5 watt is plenty as Wo is just under 4W.

Fezder (author)SPenrod2017-04-18

50mF=50*10^-3 and 50000uF=50000*10^-6, they are the same value

KT Gadget (author)2017-04-17

In Step 2, this sentence doesn't make quite sense:

"When Vcc - GND is positive, current will flow only via D1-D2-D3 and LED1. When Vcc-GND is positive, current will flow only via D4-D5-D6 and LED2."

Shouldn't the second sentence be "negative" or written to show if it was connected in reverse polarity? Just a thought haha.

Wonder how this would work with some of the larger Farad capacitors when discharging them.

mecanicafina (author)KT Gadget2017-04-17

Well spotted and corrected, thank you.

andrewty (author)2017-04-17

I use a pair of croc clips on leads with a 2k2 resistor soldered in circuit.

The 2k2 limits the current of a 63V fully charged capacitor to <=29mApk

A 2W resistor will initially get very warm if trying to discharge a 15mF cap from 63Vdc but it does cool down quite quickly as the charge dissipates.

The 2k2 on a 15mF will be down to <5% of starting voltage in ~160seconds.

mecanicafina (author)andrewty2017-04-17

Funny, same resistance I picked up for generic usage. I am
suggesting a lower resistance value in my answer to SPenrod
below because his caps are 50 mF,
so I feel the discharge time will be too long. Actually, in your case the discharge time to 5% of the initial 63V should not
be 160 sec, but about 100 sec.

Cliffystones (author)2017-04-16

I build tube amps. I'm old. I'm lazy. I always find myself looking around for a resistor on my bench (one I'm using during prototyping). If I don't find it fast enough, then the laziness re-asserts itself. Then I feel the wrath of that 500uF, 800VDC cap I was too lazy to properly discharge. it's really amazing the new ways your mind can use old, and invent new swear words when a hand muscle gets that amount of power through it for that instant!

I want to thank you, mecanicafina for inspiring me into making your circuit. But I'll probably use an old prescription medicine bottle as I have quite a few of those in my old age :).

You are welcome, thanks for your feedback

Ardutronico (author)2017-04-16

BTW, if the full input voltage appeared at the LED, it could explode, and it would be even more dangerous.

mecanicafina (author)Ardutronico2017-04-16

The full voltage could appear at the LED terminals if one of the 1N4007 diodes would fail with an open junction. In that case one of the LED’s would see the full input voltage applied as direct voltage, the other one as reverse voltage. For an input voltage of 300 V one of the LED (hard to say which one) would blow up, however I am not sure it would explode.One could add some sort of overvoltage protection to the circuit of course, but maybe the simplest course of action is to just modify the case using LED mounting holes that are not pass through. The material is “transparent” enough that you would still see the LED lighting up.

e5frog (author)2017-04-16

I like the idea, looked fine until you just stuffed the components into the case. What about isolation distance, have you checked?
It seems like a bad idea having something conductive touchable from the outside in case something goes wrong.
Maybe some heat transfer compound would be good, perhaps even the two component epoxy type, looks like you just rolled some desoldering braid around that resistor and stuffed it in the slot.

mecanicafina (author)e5frog2017-04-16

Good point. The resistor fits quite snugly in that milled slot, and I added the braid (from a piece of coax) to improve heat transfer while padding the fit a bit (the resistor body is not a cylinder). I have added some silicone like sealant afterwards anyway to fix everything in place even better. Unfortunately I had already taken the pictures, but I will add a note in the text.

jimdkc (author)e5frog2017-04-16

Agree. I would encapsulate all the exposed wiring in a blob of silicone sealant.

rocrem (author)2017-04-16

Great tutorial !

Thank you very much.

Yes . . . the isolation could be better, but it's a good start and description of how it works, and what to care about.

Eric Brouwer (author)2017-04-12

Very interesting Instructble. Thank you for sharing this.

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