I'm trying to determine the max heating allowable in an insulated copper inductor. This is a pulsed application where peak current in the 5kA range is delivered over <1ms. My task is to figure out how much higher we can go (theoretically) above 5kA without things blowing up / melting.

The fusing current of, say, #8 copper is 14kA over 32ms, which is an order of magnitude more time than this design's pulse length. (We are only considering a single pulse right now, i.e., assume full cooling to ambient between each pulse.)

Based on that, it seems like fusing current should not be the limiting factor. So would the limiting factor be? The melting point of the insulation around the wire? Standard insulation is ok up to ~100degC...but my intuition says that raising the temperature of my inductor say, 50degC in 300 microseconds would still cause something to go very wrong very fast. Or maybe not! I'd appreciate help thinking this through, I'm not an EE! What factors should I be thinking about when heating at this fast a ∆T, if not the insulation melting point?

At the moment, we're thinking about this without regard to cost - an exotic insulation material might be ok if that solves problems.

(as a note: this is only a back of the envelope calculation to explore the design space - if we were actually going to build this, someone who knows what they're doing would be on board!)


2 Answers 2

  • Put in 5A DC measure ‘C/W rise above ambient at 64% thermal time constant then simulate the real power with actual Z(f) V(t) for each part.

  • estimate insulated temp rise of copper from rise in DCR Resistance with the change in current,

  • Compute Rjc,then fusing margin from T rise. Done.

    The internal wire time constant from current will be 
     faster yet higher ‘C/W than the exterior using thermocouple which is cooler yet  more mass.
  • remember to shunt the current with RC snubber when opening the switch or reduce Vdc slowly.

i prefer falstad’s simulator for power measurements.

  • so the critical constant is the tempcoefficient (tempco) for copper with resistance and the thermal resistance coefficient ‘C/W due to insulation.

Next after high duty cycle is external temp.

N.B. I forgot to mention there will be ahellofalot of energy stored, so there are safer methods using low frequency alternating current limited by square wave and current sense regulation.

  • \$\begingroup\$ Thanks. I'm not familiar enough to parse all of this. Can I ignore time constants since I'm thinking about the heating due to a single pulse and assuming total cooling? I have already estimated DCR heating in the copper. I don't know what RJC and delta C/W mean and am having trouble figuring out what you mean by searching. \$\endgroup\$ Commented Aug 25, 2020 at 1:28
  • \$\begingroup\$ More generally: I am predicting something on the order of a 50C ∆T over 300us. I'd be interested to hear if an expert immediately thinks that sounds crazy or what range is ok. Is your comment suggesting that there are no other factors that start to matter, even at very fast very high ∆T? Safety/feasibility is entirely a function of staying (well) under the melting point of the insulator? \$\endgroup\$ Commented Aug 25, 2020 at 1:29
  • \$\begingroup\$ No you must do this way. Unless you know what you are computing otherwise with complex impedances. \$\endgroup\$ Commented Aug 25, 2020 at 12:21

You can get common insulating materials that are good to 200°C continuous (eg. PTFE). It's available in AWG 8, and you can get high-temperature rated conductors too if you want.

There will be some forces on the wire from pulses of magnitude so if the wire isn't fastened down you'll get some vibration and maybe eventual breakage.

5kA into an AWG 8 conductor does not actually sound all that scary. Fault current in an industrial circuit will be in that range, and nothing happens to the conductors, the CB or fuse opens in tens of ms and all is well. I think you have not accurately calculated the heat capacity of that chunk of copper if you think it will increase 50°C in 0.3ms.

Let's do a rough back-of-envelope calculation:

At 5000A a 1m length will dissipate 51.5kW since the resistance will be about 2m\$\Omega\$. In 1ms you have 51.5J. Heat capacity of copper is 0.385 J/g°C 1m of AWG 8 has a mass of 116g so the wire will only increase in temperature by about 1°C in 1ms.

For short pulses, with plenty of time between for the wire to return to room temperature, the heat capacity is what you should be looking at. Fusing current takes into account the increase in resistance with temperature and, of course, the heat loss, neither of which matter much for a short pulse.

  • \$\begingroup\$ Thanks! Yes, that's about the same numbers I got for heating at 5kA. To be clearer, 5kA is the normal operating range of the device. My question is, if I increase current 2X or 10X I can pretty easily calculate the resulting heating, but I'm concerned that I don't understand how much heating is too much. It seems like a very very fast and relatively large ∆T could cause, say, micro air bubbles trapped between the insulation to explode. Or maybe insulation behaves strangely when heated that fast. That's the type of intuition I'm asking about. \$\endgroup\$ Commented Aug 25, 2020 at 14:15

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