# Will a resistor be damaged if power rating is surpassed momentarily?

I am using a power resistor to discharge an 82uF capacitor from 500V to 40V in 5 seconds.

Using this online calculator, the peak power on the 24k resistor will be 10W. Because the capacitor is discharging, this large power dissipation is only momentary, down to 6.33W after 500ms and down to 3.75W after 1000ms.

Because 10W resistors are massive, can I use a resistor with a lower power rating?

• What is the voltage rating of the resistor? That will be important. Add the info into your question. – Transistor Mar 31 at 19:25
• I was going to use any with a rating over 500V. How would the voltage rating affect the power dissipation of the resistor? – A.S. Mar 31 at 19:27
• No one can tell you for certain what will happen if you exceed the specified maximum ratings of any device. It might fail instantly, its characteristics might change a bit, or there may be no significant effect at all. If you go beyond what the manufacturer specifies then you are on your own. How much of a risk are you willing to take? – Elliot Alderson Mar 31 at 19:31
• Voltage rating doesn't affect the power rating. It affects whether or not its internal insulation breaks down and causes a fire. – Transistor Mar 31 at 19:31
• Wirewound resistors are best at peak pulse current handling and you probably will find specifications for pulse capability for them. There's a number of phrases to look for (different terms are used for the same/similar thing, unfortunately.) For example, "action integral" and "fuse action" are two ways. There are other phases. Essentially, this is $\int I_t^2\:\text{d}t$ and is usually specified by the manufacturer by assuming there isn't time for the heat to move from where it is being generated by the current. Use devices that specify pulse capability. – jonk Mar 31 at 19:53

Maybe. Some resistors have surge current ratings right on the data sheet (usually curves), and it would be best to use one of those within ratings.

You may see significant changes in resistance from the surges, even within the maximum surge rating.

You also need to observe voltage rating, and I would suggest keeping it within operating rather than surge voltage rating. In this case, maybe two 12K 2W 2512 resistors in series would fill the bill with a fair bit of margin. Smaller ones have less voltage rating but in series, you might consider them adequate.

Check out the single pulse surge rating:

You should also consider what might happen if the bleeder resistor was somehow switched across the power supply with it still energized. You probably want to ensure it does not cause a fire. There are resistors that are designed to be "fusible" and fail in a controlled manner to handle such situations (though maybe not in SMT form), generally they are wirewound or metal oxide film types.

Resistors have multiple ratings, and you must stay within all of them. However, not all resistors publish all of their ratings.

'Small' resistors only tend to publish maximum voltage and maximum continuous power.

'Power' resistors also often publish 'single pulse energy', which is a measure of how much they can heat up during a single event. Calculate the energy that the capacitor will dump into the resistor when discharging. Choose a power resistor with an energy rating greater than this.

Power resistors tend to have several power ratings as well. A continuous power rating, and several higher single pulse power ratings, depending on the length of the pulse.

Decades ago my employer hired some new-graduate technicians. Their "leader" (who already was a good tech) came to me, and asked what happens when resistors are overloaded.

So we placed some 10 ohm 1/4 watt resistors across 5 volt supply.

Within about 2 seconds, the resistor was on fire. This was at 10X overload, for a couple seconds.

Why was this? If the resistor was rated for 1/4 watt at room-temperature-case, derated to 1/8watt at 125C with 125C case temperature??? (assume that), then at 10X overload (2.5 watts) the overload factor is 2.5 watts / 1/8watt = 20X.

Might we expect the temperature to rise by 20x * 125C, or to about 2,500 degree?