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I'm considering using a Vishay FVT series resistor for precharging some capacitors.

  • Capacitance: 3300uF
  • Voltage: 730V
  • Resistance: 500 ohms
  • Resistor continuous wattage: 25W

This series, like many wirewound resistors I've seen, have a spec'd overload rating of 10x for five seconds. So my 25W resistor can handle 250W for five seconds, or 1250 joules.

In my precharge configuration, the peak power dissipation in the resistor will be 1080 watts, 43x the continuous rating. Obviously the resistor can't survive that indefinitely, but the power dissipation will decay exponentially. My reasoning is that if the total energy dissipated in the resistor is under 1250 joules, it should be safe.

The total energy dissipated in the resistor is the same as the total energy stored in the caps, giving me ~880 joules, less than my 1250-joule limit. Is my analysis valid? Should my resistor be safe? Or will the high instantaneous peak power break something even though I'm within my computed joule limit?

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  • \$\begingroup\$ depending how long it takes for the discharge event from the capacitors takes, it may pop. You say that the dissipation decays exponentially, but how fast is that in reality? Can you simulate this in PSpice or another simulator? That will be the real test, and allow you to integrate/sum the power for the time period to see if it's within spec long enough \$\endgroup\$
    – KyranF
    Oct 9, 2014 at 14:08

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The 5-second overload rating tells you a couple of things. One, as you note, is an implied "total Joule" rating, which says something about the thermal characteristics of the resistor assembly overall.

The other is that the manufacturer expects that the resistance wire can handle about 700 mA for at least that amount of time. You're asking it to briefly carry twice that current, which may or may not push the resistance wire into a nonlinear region, which could lead to localized failures. With a time constant of 1.65 seconds, it takes about 1.2 seconds for the current to drop from 1.46 A to 700 mA.

I would expect that you'll be fine, but it would be worthwhile to do some tests of your own, running short pulses of current in the range of 1A - 2A through the resistor and monitoring for anomalous resistance changes (i.e., excessive voltage drop).

Or you could just ask the manufacturer directly ...

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    \$\begingroup\$ Knowing almost noting about high power stuff, I would guess it should be OK too, the 1.65 seconds is not too different from 5 seconds. Which I would guess is enough time for the most of the ceramic to get hot. For a very short time, I'd worry that the heat would have no chance to get out of the wire and into the ceramic. (Stephen, If you contact vishay will you let us know what they say?) \$\endgroup\$
    – George Herold
    Oct 9, 2014 at 15:00
  • \$\begingroup\$ "Push the resistance wire into a nonlinear region". What does that mean? That the current will be high enough to alter the resistance of the wire? \$\endgroup\$
    – Stephen Collings
    Oct 9, 2014 at 15:41
  • \$\begingroup\$ @StephenCollings: yes, exactly. Most materials have a positive temperature coefficient, and this can lead to excessive heating. That is, after all, how you "blow" a fuse. \$\endgroup\$
    – Dave Tweed
    Oct 9, 2014 at 15:43

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