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enter image description hereAfter 100 mSec, switch (SW) is going to bypass the resistor permanently. what should be the wattage of resistor? In image voltage source is of 700V, R=7ohm, L=330micro Henry, C=1.36mF.

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closed as off-topic by Olin Lathrop, Brian Carlton, RoyC, winny, Dave Tweed Feb 14 '18 at 20:03

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  • \$\begingroup\$ What do you calculate to be the peak power dissipated by the resistor? \$\endgroup\$ – Andy aka Feb 8 '18 at 11:21
  • \$\begingroup\$ I won't do the calculations for you, but you should either do the math or spice it, and obtain both the peak power dissipated in the resistor, and the total amount of energy it will have to dissipate. Considering the cap is 1360 µF this will be quite large. I'm going to bet you will end up with a halogen lightbulb as a resistor, or some form of coil of wire, otherwise it'll be pretty huge. \$\endgroup\$ – peufeu Feb 8 '18 at 11:40
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    \$\begingroup\$ We are not here to do your homework for you. You haven't shown any effort, nor demonstrated any understanding of resistor power limitations. \$\endgroup\$ – Olin Lathrop Feb 8 '18 at 12:01
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    \$\begingroup\$ Also, if you are building this thing for real you need to consider the voltage rating of the resistor, the capacitor, the diode, and the inductor. \$\endgroup\$ – JRE Feb 8 '18 at 12:09
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    \$\begingroup\$ @jre and the switch. \$\endgroup\$ – Arsenal Feb 8 '18 at 12:40
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Although most resistors just seem to have a power rating, that's not the whole story. It's a continuous rating, where the power input is balanced by its heat loss to ambient.

When you start using resistors for other than continuous use, you need two more ratings. There's the pulse power rating, which is a measure of what it takes to damage the resistive element. There's the pulse energy rating, which is a measure of the thermal mass of resistor forming the element, or in intimate contact with it, as heat has no time to be dissipated to ambient but must be stored in the body of the resistor. The pulse power will usually be orders of magnitude larger than the continuous power, for a robust resistor technology like wirewound. It may only be a bit larger for fragile technologies like thin film.

To get the pulse ratings, you usually need to dig into the manufacturer's data. It's often not obtainable for general purpose resistors. You can buy pulse-rated resistors. An alternative is to make resistors out of wire, or water. You know the construction, so you know heat is going to be uniformly deposited throughout the mass of the resistor, and you can work out the heat capacity from physics.

A couple of simplifications. Your circuit has an LC frequency of about 250Hz. This is far quicker than your 100mS gate time, so given the R damping you can pretty much ignore the inductor. The RC time constant is about 10mS, so in 100mS the capacitor will be fully charged. The energy lost in the resistor will be equal to the energy in the fully charged capacitor (it just is, no need for calculus).

The capacitor energy at 700v is \$\frac{1}{2}CV^2\$ = 330J. As that energy is delivered in much less than 100mS, most of it in the first 20mS, the instantaneous power will be north of 10kW.

Look at this data sheet for easily obtainable wirewound resistors. It says that for pulses between 100mS and 1 second, the larger sizes can be overloaded by a factor of 10x their continuous power. Below 100mS, you only have to observe the total energy, which for 100W and 200W resistors of 1 to 10 ohms is in the 200J to 900J range.

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  • \$\begingroup\$ Thank you, actually, I am checking datasheets of various manufacturers since morning but I am not coming to the solution. But now you gave me proper direction by your answer. thank you again \$\endgroup\$ – Vicky Siraswar Feb 8 '18 at 12:09
  • \$\begingroup\$ Are you sure that the 10x overload rating ceases to apply below 100ms? Also, a reminder to Vicky: with 7 ohms resistor, this will draw 100 amps from the 700V source when the switch is closed. Might be a problem. Might also exceed the capacitor's current rating. I'd go with a higher resistor value. \$\endgroup\$ – peufeu Feb 8 '18 at 13:02
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    \$\begingroup\$ The peak power will be 67 kW! \$\endgroup\$ – Andy aka Feb 8 '18 at 13:09
  • \$\begingroup\$ @peufeu Ignore 10x rated limit below 100mS? It looks like it. It's encouraging that the energy rating is dependent on the resistance, which implies it's the mass of the wire only that's soaking up the heat, not relying on conduction to the adjacent fill. If the wire is uniform, so no necking down where it's attached to the terminals for instance, then you might hope that arbitrary powers can be handled tamely, at least up to risetimes where RF effects modify the distribution of heat across the resistance wire. Might be worth sacrificing one to see how it's built on that score. \$\endgroup\$ – Neil_UK Feb 8 '18 at 19:49
  • \$\begingroup\$ Hm, yes I think you're right. I'm still worried about the huge current vs whatever source powers this doomsday device... \$\endgroup\$ – peufeu Feb 8 '18 at 20:17

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