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I am going to be working with many capacitors that can easily kill me (kV range), and I was wondering if there was any kind of industry recommended method/tool for discharging a capacitor to make sure it is safe to handle. I'd prefer a general tool that gives a decently high load across the terminals. The only things I could find were seemingly application-specific large value resistors.

The perfect tool in my mind would be a rod of some high resistance, high specific heat material with a thick rubber handle. A resistor with dinky little leads can be dangerous in itself to use, and a longer rod seems just like a better idea.

Suggestions?

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    \$\begingroup\$ Any sane designs that use high voltage capacitors have bleed resistors across them. So as a precaution let the device under test sit without power for a while before probing, then proceed with your "tool". I just use a long insulted screwdriver to short the cap :p. \$\endgroup\$
    – Adil Malik
    Jan 22 '17 at 16:36
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    \$\begingroup\$ @AdilMalik My concern is that a small resistance will overload the capacitor and damage it as well \$\endgroup\$ Jan 22 '17 at 16:39
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    \$\begingroup\$ Perfect, as this is your design, be sure to include bleed resistors across the capacitors! \$\endgroup\$
    – Adil Malik
    Jan 22 '17 at 16:59
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    \$\begingroup\$ Recently, discharging 20,000 microfarads at 100V, hardly a remarkably large capacitance, by an accidental short with a screwdriver gave me a rather spectacular bang. I don't believe it's good for the capacitors, terminals or whatever is shorting them. \$\endgroup\$
    – Ian Bland
    Jan 22 '17 at 17:13
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    \$\begingroup\$ Not just "a disorienting bang." I once amputated the screwdriver blade when doing that by accident! \$\endgroup\$
    – alephzero
    Jan 22 '17 at 21:32
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Is there an ideal part? Not really. depends on budget.

  • Energy in the cap is \$E=\frac{1}{2} CV^2\$ may be greater than the Pd of the resistor perhaps Pd>=10%Pk to prevent fusing open in 1 second. (component specs need to be verified)
  • Voltage withstanding rating of the resistors should be greater than the stored voltage.
  • decay time constant can be determined by the RC=T product with 40% voltage remaining and after 5T about 1% of the energy is remaining in the cap.
  • This creates 3 equations and 3 unknowns for choosing the ideal R for any given cap that may be ok for a limited range of C,V calues.
    • the 3 unknowns are; Vmax, Pmax and R value for the resistor(s).
    • other unknowns are the time limit for discharge and the thermal resistance time constant, so let's use a 1 second time limit for 5T. The peak power in the resistor decays exponentially. Power transfer in the capacitor rises with current then peaks about 0.6T followed by exponential decay, thus Pd of R can be 10% of Ppk, where \$Ppk=V^2/R\$

Examples

  1. 1kV 2uF , E= 2J = 2 watt-sec, R=0.2s/C = 10 kΩ Ppk =100W
    • so choose 10% of Ppk,
    • Pd = 10W, 10 kΩ , >1 kV rating
  2. 10kV 0.2uF, E= 10J , R=0.2s/C= 1 MΩ, Ppk=100W,
    • Pd=10W, 1 MΩ >10 kV rating
  3. 10kV 2uF , E= 100J, R=0.2s/C= 10 kΩ, Ppk=1kW,

    • Pd=100W 10 kΩ >10 kV rating

    added

  4. 100V, 100 mF E= 500J , Screwdriver ESR and cap bank with 5% D.F.est. ESR=1mΩ, Imax=100kA (ignoring inductance), T=RC=0.1s thus if 5kW*0.1s=500J the 5kW pulse ought to have vaporized the screwdriver tip. (Ian can confirm)

This should give you a rough idea of what is ideal for safe discharging a high voltage, high value capacitor. Pd rating may need to be increased for safety margin to prevent fusing open depending on Imax pulse rating of part.

If a long time constant than 1 second is chosen then R can be limited by Watt-seconds/ seconds or E/(RC)= \$Pd=\frac{1}{2}CV^2/RC=\frac{V^2}{2R}\ \$ for a pulse discharge and \$Pd=\frac{V^2}{R}\$ for continuous load to stay within power rating of R.

I recommend Vitreous enamel axial long wire-wound R's. V rating tends to be limited by value of R that affects continuous Pd rating and high V rated ones are expensive, but unnecessary for this.

You can string a "bunch" of >=500V rated 1/4W axial parts in series for a cheap and dirty solution to accommodate Pd, Vr and R values needed if equal. Then mount with silicone onto a nylon or dry wood stick with a ground alligator clip.

e,g, $30 High V rated example

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Any sane designs that use high voltage capacitors have bleed resistors across them. So as a precaution let the device under test sit without power for a while before probing, then proceed with your "tool".

I just use a long insulated screwdriver to short the capacitor. Proceed at your own risk ofcourse.

Just remember, shorting a large capacitor unleashes a lot of power and give off sparks and huge bang.

PS: i was reluctant to present this as a answer as opposed to a comment because i dont want a quick browser to think this is the way to go. Again, be very careful!

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    \$\begingroup\$ "a long insulted screwdriver" - lmao - think you mean insulated. \$\endgroup\$ Jan 22 '17 at 16:54
  • \$\begingroup\$ Haha, i can imagine someone getting the looks for asking for an insulted screwdriver at a hardware store!! Thankyou for pointing that out! \$\endgroup\$
    – Adil Malik
    Jan 22 '17 at 16:57
  • \$\begingroup\$ Yes, can just imagine it - + 1 for the laugh. \$\endgroup\$ Jan 22 '17 at 17:00
  • \$\begingroup\$ There ARE insulated screwdrivers meant for use around high voltage. I had a set of them back when I worked as a radio technician. \$\endgroup\$
    – JRE
    Jan 22 '17 at 18:30
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The theory is covered pretty well in another post already, but here are some practical considerations and a circuit which I use when discharging the HV anode on a CRT in any of my arcade cabinets (50kV, minimal capacitance).

Disclaimer: while the circuit presented here is taken from a reputable source (see attribution below) and has worked just fine for me, don't build it if you don't understand what it does or how it works!

  • Always allow the existing bleed-off resistors time to do their job; this tool exists as a way of making sure that a circuit is not still storing HV unexpectedly due to those having failed.
  • Always discharge to a true earth ground.
  • Make certain that all of your parts are rated well above what you expect.
  • Ensure that whatever you are using (this tool or anything else) is designed to fail safe across any permutation of part failures — you do not want the possibility of a "cascade failure".
  • Put as much distance between yourself and the contact point as you reasonably can, given the circumstances. Electricity may not be prone to flying around the room, but molten bits of plastic and metal sure can be.
  • The main issue with using 1/4w style resistors is that the dielectric voltage of air is low enough that the voltage can "jump" the resistor entirely, given how short it is. While using several in a row helps to mitigate this, it is still better to get a single resistor that is large enough to handle the load, if you can. Whether this is a point of concern depends on whether you're talking "a couple" of kV or "several tens" of them, which the original question did not address.
  • Put the resistor as close to the probing tip as you reasonably can. This minimizes the portion of your discharge circuit which is carrying a high voltage relative to ground.
  • Use an appropriate gauge of wire.
  • Remember that various failure modes can result in more of the circuit being exposed to HV than are intended, and insulate / contain all components accordingly.
  • While it may seem self-evident: don't be tempted to put a switch into the circuit

With all that said, the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Notes on the circuit:

  • D1-D8 can be any 1N400x for X=1-7, or an equivalent; they exist to provide a voltage clamp across the indicator LEDs
  • LED1-2 can be pretty much any LED without an internal resistor
  • R1 should have different values depending on what you're discharging; this circuit reflects my personal "rig". Recommended values from the original source are:
    • 2kΩ 25W ("low" voltage (400V), high capacitance (1000µF): power supplies, etc.)
    • 100kΩ 25W (high voltage (5kV), low capacitance (2µF): microwaves)
    • 1MΩ 10W (extremely high voltage (50kV), extremely low capacitance(2nF): CRT anodes)
  • R2 limits the current to LED1 and LED2, and thus can be substituted with anything that has an appropriate value for the ratings of the LEDs.

Attribution: one of my main sources when researching this years ago, and the source of the circuit, is the sci.electronics.repair FAQ, in particular the section on safe capacitor discharge.

My actual implementation of the circuit is:

  • A probe built around:
    • 2-3 feet of PVC (I want to say 1/4" but it isn't particularly crucial)
    • A sacrificial voltmeter probe (CRT anodes are under a rubber "cap" so you need something narrow enough to get in under that)
    • A 2MΩ 20W resistor (about the size of my pinky finger, and the hardest part to find by far)
    • 12 gauge solid/bare copper wire (the ground wire from some old 12/3 I had laying around)
  • More of the 12/3 conductor, and a "blank" plug from the home supply store, to build a "ground only" plug

I am deliberately omitting a description of my "indicator circuit" portion, because the original design was unsafe and I have stopped using it due to both that and the lack of any useful indication of discharge (when dealing with "EHV" values, the discharge may be too short to notice — an issue that was raised in the original FAQ).

Edit 1: corrected schematic to use an earth ground and orient it properly

Edit 2: spaced out implementation detail list so that it formats correctly

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