# Calculating power dissipation for short periods of time.

#### FrancescoC

Joined Nov 22, 2014
31
HI,
If you have a resistor, say 10 Ohms, and you apply a voltage across it, say 10V
the resistor will dissipate V*R=10W. So you would need a 10W resistor for the job.

But what about if you draw current for a short period of time, say 0.1 sec?
Could you then use a lower rating resistor?
Is there any way that you can calculate, and make a chart, the time-on vs power dissipation?

Regards
Francesco C

#### AnalogKid

Joined Aug 1, 2013
8,824
Some power resistors have such a data point. I used one that had a 10 W continuous rating, but also a 7 A peak current rating for one second max. Most small resistors do not have that level of detail specified, but the manufacturer can help.

ak

#### FrancescoC

Joined Nov 22, 2014
31
Hi Sabouras,
The final formula was: E=VxIxt. Energy =Voltage x Current x time.
But i think that relates to to power been delivered rather to power been dissipated.

I will continue my search!

Regards

Francesco C

#### AnalogKid

Joined Aug 1, 2013
8,824
Continuous energy is watts. Total energy for any particular period of time is watt-seconds or watt-hours. Battery people relate the total energy in batteries of the same voltage in amp-hours because the battery loads frequently are quantified in amps rather than watts, but it's the same idea. Transients are described in volt-seconds; a voltage spike caused by a lightning strike a mile away might be 6 kV for 10 ms (milliseconds). So, the question to you is what is the transient voltage or current you are dealing with?

It comes down to the testing the manufacturer does. For example, some zener diodes also are sold as transient suppressors. They are rated for 5 W continuous power dissipation, or 1500 W for a specified transient period. Or something like that; working from memory.

ak

#### crutschow

Joined Mar 14, 2008
26,057
HI,
If you have a resistor, say 10 Ohms, and you apply a voltage across it, say 10V
the resistor will dissipate V*R=10W. ..................
Both your formula and math are incorrect. Interestingly the two cancel out to give the correct answer for these particular values of voltage and resistance.
Power is V²/R = 10W.
V*R would be 100, which is not correct.

#### FrancescoC

Joined Nov 22, 2014
31
Hi,
thank you for the info.
The reason i am investigating this is because i am experimenting with loads when are not On continuously.
I would like to found out, what happens is substitute a 10W resistor, which was designed to be permanently conducting,
with a one of 1W which will be on for only 100ms. Then Off for 100ms then On again........

It must be an issue for designer engineers. If you can calculate this you can use a small resistor and save money and space.
Am I been too naive in thinking this way?
(and yes my formula was wrong)
Regards

Francesco c

#### dl324

Joined Mar 30, 2015
12,263
It must be an issue for designer engineers. If you can calculate this you can use a small resistor and save money and space.
Am I been too naive in thinking this way?
It depends on the application. Most engineers are conservative in their power dissipation estimates; often derating parts by 50%.

If I was contemplating doing this, I'd contact the manufacturer for data. If they didn't have any, I'd look for a supplier that did. If none had data, I'd take the conservative approach and use the high wattage resistor.

If it's a design for your personal use, you can do whatever you want. I have a Stanley X10 remote that has become discolored in one spot on the case due to a component (likely a resistor) getting hot. Clearly this is a commercial product that has a design flaw; I keep reminding myself I need to unplug it before it starts a fire...

#### WBahn

Joined Mar 31, 2012
26,398
HI,
If you have a resistor, say 10 Ohms, and you apply a voltage across it, say 10V
the resistor will dissipate V*R=10W. So you would need a 10W resistor for the job.
If you would have tracked your units, you would have seen that

V*R = 10 V * 10 Ω= 100 V⋅Ω

Since V⋅Ω is not a unit of power, you would have seen that your equation was fundamentally wrong.
But what about if you draw current for a short period of time, say 0.1 sec?
Could you then use a lower rating resistor?
Is there any way that you can calculate, and make a chart, the time-on vs power dissipation?
Yes and no. You can easily determine the average power dissipation and if the peak power is small enough then you can use a resistor that is closer to the average power. But this depends on a lot of factors. As you are applying power the resistor will be heating up and the max temperature it gets to before you remove the power and let it start cooling is the big determiner. That depends on the size and makeup of the specific resistor in question.

For a duration as short as 0.1 s you can pretty safely assume that a resistor that is sized for the average power is sufficient unless there are other factors at play, such as maximum voltage or current ratings imposed for other reasons.

Track down the data sheet for the resistor family you want to use and see what it says. It may not say much so you might then look at another manufacture's data sheet.[/QUOTE]

#### crutschow

Joined Mar 14, 2008
26,057
How much pulse energy a device can tolerate depends upon its thermal time-constant, and that info would be need to be obtained from the manufacturer.

#### ian field

Joined Oct 27, 2012
6,539
Both your formula and math are incorrect. Interestingly the two cancel out to give the correct answer for these particular values of voltage and resistance.
Power is V²/R = 10W.
V*R would be 100, which is not correct.
The example equations in appnotes for pulsed operation of LEDs might give some useful guidance on the topic.

Sharp did a good opto applications handbook, but I think they may have got out of that line. Probably worth searching for an archived copy.

#### AnalogKid

Joined Aug 1, 2013
8,824
I would like to found out, what happens is substitute a 10W resistor, which was designed to be permanently conducting, with a one of 1W which will be on for only 100ms. Then Off for 100ms then On again...
The resistor will burn up.

On and off with a 50% duty cycle is very different from your question in post #1. In this case the average power in the resistor will be 5 W, and a part rated for 1 W will fail quickly because it does not have enough off time between pulses to cool down. For a 1 W part to work reliably its average power should be below 0.5 W. For a 10 W pulse that means 0.1 seconds on and 2 seconds off.

ak

#### WBahn

Joined Mar 31, 2012
26,398
The resistor will burn up.

On and off with a 50% duty cycle is very different from your question in post #1. In this case the average power in the resistor will be 5 W, and a part rated for 1 W will fail quickly because it does not have enough off time between pulses to cool down. For a 1 W part to work reliably its average power should be below 0.5 W. For a 10 W pulse that means 0.1 seconds on and 2 seconds off.

ak
But also keeping in mind that the pulses have to be short compared to the thermal time constant of the part. For instance, a 10 W pulse of 1 minute followed by 20 minutes off would yield the same 0.5 W average power, but the part will almost certainly burn out during that 1 minute.

#### AnalogKid

Joined Aug 1, 2013
8,824
True, but the ON pulse period already is a part of this thread, so I didn't expand to the genral cases.

ak