What Cable Thickness for Capacitor Discharges?

[nsled]

Is there a general rule of thumb for what thickness of cable is required for capacitor discharges? My specific need is 5kV and 40uF.

6mm2/9AWG is very affordable, and seems the largest available short of specialty distributors who aren't likely to sell it by the metre?

6mm2 cross section has resistance of 3.4mOhm per metre, so it'll draw up to 1.5 million amps at 5kV. Is this likely to impede the discharge?

Does anyone have examples for fried cables from not being thick enough?

[Twospoons]

Inductance will be your limiting factor.  A mere 1uH would limit your current to around 30,000 A peak. 
If you ever did achieve 1.5million amps the force pushing your cables apart would be on the order of 50 million newtons per metre.

[klugesmith]

Twospoons made the important point that circuit inductance matters.   Also R and L within the capacitor, and R and L of the load.  If load resistance were zero, load energy from the capacitor discharge would be zero.

There's an easier way to know that the wire is big enough, at least for isolated shots.
Nsled's simplistic estimate of 1.5 MA (peak) is consistent with the resistance of 1 meter of 6mm2 copper wire.
That would weigh about 55 grams, and have heat capacity of about 21 joules per degree. If it absorbed 100% of the energy from capacitor, temp rise would be about 24 degrees.

[MRMILSTAR]

One thing to consider is whether it is DC or AC. If the shot induces oscillation (a damped sinusoid) then the vast majority of current will flow in the outer mm or so of the conductor. For this case a solid round conductor will mostly be wasted copper. The frequency of oscillation will determine the skin depth. The skin depth will then determine the amount of copper to carry the current. For this case round tubing, flat tubing, or thin solid flat conductors would be best.

[Benbmw]

As pointed out by Twospoons and klugesmith, the total circuit inductance will be the main current limiter in most pulsed power situations. It would be helpful to know what capcitor(s) you are using to get 40uF at 5kV. Also what does your discharge circuit look like? Is it two meter-long wires attached to chicken sticks (not recommended). Or are you wiring up a relatively low inductance discharge path by minimizing loop area etc. ?

• Unless you are being very conscious of inductance in your design, it is unlikely that you are below 500nH. A big contribution to your ESR will be from your capacitor (all depending on its type), so lets say you aren't getting away with anything less than 10-20mOhm total. This puts your peak current about 40kA and quite underdamped, ringing at ~35kHz for several cycles. Note, f = 1/(2pi*sqrt(LC))

As MRMILSTAR eludes to, the entire cross section of your conductor is not all used. Commonly this is referred to as the skin effect, which for AC signals can be reduced to:

• d = sqrt(1/(pi*f*sigma*u_r*u₀))
  
  Where d is the depth in meters at which the current density has reduced by 63% (ie. most of the current is within this depth). sigma is the material conductivity, which for copper is about 5.8*10⁷ and u_r is 1 for copper. So this can condense to:
  
  d = 0.066/sqrt(f)
  
  Which is 0.35mm at 35kHz. In your case you will likely be utilizing less than a 0.5mm layer on the outside of your conductor, almost certainly less than a 1mm layer. This effect is not limited to true "AC" signals. Any time changing pulse has the same effect present. This is due to the magnetic field taking time do diffuse into conductors (details are out of the scope of this discussion).

Getting back to your question, klugesmith's approach of assuming an adiabatic temperature rise is excellent for getting an upper bound on the current capacity of a conductor. In fact there is a famous paper from 1944 by Onderdonk addressing the problem this way (including temperature varying resistivity etc.) a more modern review of this is attached here and produces the following equation for the fusing of copper conductors:

• (i/A)²t = 9.6*10⁴
  
  Or, rearranged:
  
  i_fuse = A*sqrt( 9.6*10⁴ / t )
  
  Where i is the current, A is the conductor cross section in mm², and t is time. Per this equation, the fusing action of 6mm² wire is ~3.5*10⁶ A²s. Your discharge will probably have an action about 100x less than this. However, going back to the skin depth consideration, you will want to scale this fusing equation by the factor:
  
  (2d/r - d²/r²)²
  
  Where d is the skin depth from the previous equation and r is the wire radius, both in meters. Which still puts you well in the clear by this approach.

In reality you will not want the copper to get anywhere near its melting point of 1083 C so maybe a better thing to do is say we don't want its temperature to rise more than X degrees, for practical reasons like not starting a fire or melting its insulation (maybe 100-200 C is a safe limit). Per my second attachment (the TRM white paper) we get an equation that calculates the temperature rise of the wire:

• T_rise = 253(10^{8.9*10-6(i/A)2t} - 1)
  
  OR rearranged to give a maximum current for some pulse width t and an allowable T_rise:
  i_max = A*sqrt( log₁₀(T_rise/253 + 1) / (8.9*10^-6 t) )
  Again where T_rise is in degrees C and A is in mm²
  

There will be other considerations in specific situations such as the forces involved when conductors are in close proximity to each other (or themselves as in a coil of wire). In these situations the wire can fail mechanically well below its fusing current. All of this to say, your 6mm² wire will probably be totally fine for discharging a capacitor bank into random loads such as popping aluminum foil etc. but if you are trying to achieve another goal you may be better served with larger wire or even some form of bus bar like MRMILSTAR suggests. Let us know what your setup and use case are and we can advise further!

[klugesmith]

Thanks, Ben. I've long been familiar with the fancy formula for temperature rise from a given i2t integral, and the complication from skin effect.

You mentioned: "In these situations the wire can fail mechanically well below its fusing current."
Reminds me of when Hackaday staff did a coin shrinking exercise featuring high speed videography.
The exuberantly said "that glow is work coil being vaporized".
No, the work coil was broken into many small bits of copper wire.  Entire capacitor energy was not enough to even melt more than a tiny fraction of the metal.

[Benbmw]

I don't have much experience with the formula, but dug up those articles on it while answering this to refresh my memory and learn a bit more. Since most of my experience is via my career (ie. employer is footing the bill!) I've never had to skimp on buswork enough to worry about adiabatic temperature rise much.

Reminds me of when Hackaday staff did a coin shrinking exercise featuring high speed videography.

I was picturing my coin shrinking setup as I was writing that bit! Great example