Electronic – High voltage Capacitor Discharge Tool

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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. Id 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?

Best Answer

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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