I want to make a capacitor discharge tool

Question

I want to make a capacitor discharge tool for use in an industrial application where the voltage may be as high as 400 volts and I have no idea what the maximum capacitance may be. I have been weighing the pros and cons of different resistor values. Then I thought, Why not ask some of the best engineers on the planet? If you have some insights for me it would be appreciated.

John

mcb1 · Answer

Sounds like it's for a switch mode power supply.

Most of them are 470-1000uF, and to prevent damage you'd want to restrict the current to 1A.

R= E/I so 400/1 = 400 ohms.

Discharge from here https://www.electronics-tutorials.ws/rc/rc_2.html

t= R (ohms) x C (in Farads) t= 400 x .oo1 (1000uF) = 0.4 secs.

5T is considered fully discharged so 5 x 0.4 = 2 secs.

I'm sure you can either add it or make it from a couple of probes with the resistor in series.

Mark

jw0752 · Answer

Last Post: I put the unit together this afternoon using the advice and wisdom from my Forum friends. The series resistor was lowered to 330 ohms and wattage increased to 5 Watts. I discharged the 2000 uF 200 volt capacitor several times and I also discharged a 33,000 uF 60 Volt capacitor. Both capacitors were brought to safe voltage in less than 20 seconds. The components were all salvage parts. The bridge diode's specification were increased to 2.5 A 1000V. One problem that I see is that the indicator light doesn't go out until the voltage of the capacitor drops below 3 volts. (90+ seconds on the 33,000 uF cap). Here is what it looks like: Raw Circuit Board In the Box Thanks John

jw0752 · Answer

Hi Jon, As promised I have run some experiments after returning home. I have taken a 2000 uF 200 Volt capacitor and charged it up. The formula for energy in a capacitor is 1/2 CV^2 which works out to about 40 Joules for this capacitor. I built the circuit in my schematic above using a 1/2 Watt 470 ohm series discharge resistor. As you can see from the color of the resistor it sustained heat damage after three discharges. Here is a video of the action:

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A second test using a 1 Watt 470 Ohm resistor was able to handle the discharge without any problem. The resistor did heat up but was able to be touched after the experiment. I would estimate it got to about 45 degrees C.

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We were originally discussing a 100 uF 400V capacitor like one we might find in a small switching power supply. The energy storage for 100UF at 400 V is E=1/2CV^2 or 8 Joules. I have no easy way to put 400 volts on a 100 uF capacitor so I decided to simulate this by putting only 8 Joules of charge into the Capacitor in the above experiment. This works out using the formula V = SQRT(2E/C) or about 126 Volts. I will run this experiment using a fresh 1/2 watt resistor. Keep in mind this is an imperfect match as 400 volts from 100 uF would discharge the 8 Joules of energy much more quickly than our 126 volts from the 1000 uf capacitor. Here is a video of the test:

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My conclusion is that the design in the schematic works well and while the 1/2 watt resistor would probably work for the 100 uF 400 V situation it would not be adequate for the 2000 uF 200 Volt capacitor. Jon's suggestion to use a higher wattage resistor is confirmed as the best way to go. John

jw0752 · Answer

Hi Doug,

I will take a look at the UJN1205K. I would like to keep the build simple and I would like to have some sort of indicator of the voltage. Here is an idea that came to mind. Ideally I will be able to find a sensitive LED that will indicate at low current levels. I think this will indicate voltage down to less than 10 volts though I am away from the shop so I can't actually build and test. Voltage above approximately 2 volts will cause current to flow through the shunt diodes D1 - D3. The bridge will make the polarity of the discharge probes irrelevant. R1 will limit discharge current to less than 1 amp at 400 volts. Watching the LED go from lit to off will encourage the operator to leave the discharge shunt in place long enough for voltages to drop to safe levels. I was already thinking of adding the non-slip tips into the build. The larger 1.2 mm size tips would be ideal for ensuring that the probe stay in place and make good contact with the capacitor leads. Thanks for you suggestions.

John

ciorga · Answer

Hi John, One more detail: after you short the capacitor terminals and "fully" discharge it then you remove the short and leave the capacitor not connected to anything you may notice that the capacitor "charges back" - yes, probing with a voltmeter the voltage on the capacitor terminals starts to increase even though there is nothing connected to those terminals. This effect is caused by "dielectric absorption". I have published a paper on this topic - here is a link for who has access to IEEE journals library (there is also some material available online that describes this effect): http://ieeexplore.ieee.org/document/841807/ Best Wishes, Cosmin

shabaz · Answer

Hi John,

Maybe a voltage alert could also be added, although a voltmeter could be used too.

I was thinking maybe some comparator circuit that would tell you if the voltage is above (say) a few volts. It would have to work both-ways, so something like this window comparator maybe:

http://www.ti.com/lit/ug/tidub01/tidub01.pdf

It could be run from a dual supply, so that the window straddles 0V. For a suitable comparator, maybe the one used in this circuit: The Engineer’s Multi-Tool - Project Hydra , TLV1701 could be handy, because it could be powered from two 1.5V cells.

On the input there would be some potential divider to reduce the 400V (or higher) down to a far lower value (and some diodes to act as a limiter).

genebren · Answer

Hi John,

Isn't this what they make screwdrivers for? Just kidding, please everyone be safe out there. I would agree that discharging quickly is a good idea. shabaz idea of using a voltmeter is a good idea. First verify the voltage, then reconfigure into current mode, with a series resistor. This would allow you to verify when the current diminishes (near zero) and the capacitor is discharged.

Good luck in finding an appropriate solution.

Gene

P.S. Are you working on vacation/roadtrip???

dougw · Answer

I like the simple resistor approach, but if you want to have some fun you could get a sample of a USCi UJN1205K and play around with it. These devices can act as current limiters up to 1200 V - without power supplies. I'd like to see what jc2048 would do with it. Maybe devise some zener circuit to get Vgs to be 4 V.

By the way - good application for your special probes.

three-phase · Answer

When shorting a capacitor you have to remember that the initial energy is quite high and your components will need to be specified to add reliability onto the system, especially if this for an industrial application in use by others. You will also need to consider the safety implications of the physical design. When we manually discharge high voltage apparatus, our discharge sticks are built with the discharge element in the actual probe to limit where the high voltage appears in the circuit. Whilst a lower value discharge resistor gives a faster discharge, it will put higher voltage around the rest of the circuit. The second element of the discharge stick is used to put a complete short across the circuit after we have discharged with the resistive element, then we know for definite that there is no voltage present. In some respects some of the low voltage detectors that we use can do this as they have a switched load resistor built into them, so the voltage can initially be checked with the probe and then the load resistor switched in to increase the discharge rate. There are some designs available on stack exchange that may be useful to you; Capacitor discharge tool There are also some commercially available items, that you could take a look at, seems this particular one is limited to 350V though. Ebay capacitor discharge tool Kind regards

jw0752 · Answer

Hi Jon, I put some numbers on paper last night to see if I could make a better guess at the wattage needed for the shunt resistor. I will not get into a lot of detail but there were a couple of interesting things pop up. In most cases as the voltage of the circuit goes up the need for high capacitance goes down. Many common switching supplies use a 400 Volt 100 uF capacitor in the primary section of the supply. A 400 V 100 uF capacitor, fully charged, holds about 8 Joules of energy. If this capacitor is discharged through a 470 Ohm shunt we drop the voltage by 63.2% in the first time constant of 47 milliseconds. What surprised me was to find that the energy in the capacitor drops by 87.5% in this same time period. This puts the energy discharge predominately in the first period of 47 milliseconds. This 7 Joules of energy is manifest as heat in the core of the resistor. Since it occurs very quickly we can't expect there to be much dissipation by conduction. The best remedy to potential heat failure of the resistor is to increase the amount of resistive material that will be heated. This tells me that Jon's suggestion that a 1/2 Watt resistor is probably not adequate for this situation is correct. A real test will have to wait until I get back to the shop where a number of resistors will be treated to some real world tests to see how they react to this type off discharge. Obviously as the capacitance goes up at the higher voltage levels the challenge to discharge also increases accordingly. John

jc2048 · Answer

I'm not too good with FETs - when I started out they were a bit exotic and I've never had to use them for anything.

A constant-current configuration would be like this, wouldn't it?

The FET shuts down when the gate gets to below about -6V (below the source), so that would regulate the voltage across the resistor to about 6V and keep the current at around an amp.

It would need a reasonable heatsink - initial dissipation is 400W (on 400V), although it comes down quite rapidly. The 6 ohm resistor would need to be 10W or so.

That SiC part is very expensive though. You could build a whole constant-current load with a MOSFET for that (including a processor to give you constant-wattage rather than constant-current and other nice things).

michaelkellett · Answer

You can buy pulse rated resistors. https://www.vishay.com/resistors-fixed/high-pulse-load/ There are other suppliers. MK

jw0752 · Answer

Hi Michael, Vishay has so much depth in their product lines. I did not know that there were pulse rated product available. I always just guessed and used an over rated resistor to address this concern. Of course all the things that I have designed are relatively simple and usually only for my own use so critical engineering isn't a factor. John

three-phase · Answer

Dielectric absorption also occurs when we carry out high voltage DC tests on motor and generator windings, as when tested to earth, these just act as big capacitors. Our rule is that once discharged, the winding should be directly shorted and earthed for at least twice the duration of the test time, to ensure that the voltage does not creep back up when the winding is reconnected. In practice, we will usually try to carry out all the tests one day and leave the windings shorted to earth overnight and then reconnect the following day. Kind regards

jc2048 · Answer

Since we're talking about Vishay parts, an alternative to a resistor that you might experiment with would be a PTC thermistor. http://www.vishay.com/docs/29165/ptcel.pdf

I want to make a capacitor discharge tool

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