# How to choose a properly rated power resistor for pulsed current?

Datasheets for power resistors typically provide a thermal junction coefficient or graph showing steady-state temperature increase for a given power dissipation, such as below for these ohmite power resistors: We'd like to test some high discharge RC LiPo batteries by pulsing them at ~150-200A for very brief periods, about a quarter to a half-second or so. Supposing that the resistor is .25 ohm, we know that our instantaneous power is

$$\begin{equation}\label{eq:er1}P=I^2R\end{equation}$$ $$P = 150A^2(.25)\Omega = 5.625kW$$

This is an enormous amount of power that would cause the above resistors to eventually (quickly) fail. We obviously don't want to spec out (huge, expensive) 6kW resistors, so our question is, what's "eventually"?

I want to say that:

$$I_{avg} = I_{pulse}\frac{t_{on}}{t_{on}+t_{off}}$$

At first I tried supposing a duty cycle of 10% e.g,

$$I_{avg} = 150\frac{1}{10} = 15A$$

so

$$P = 15A^2(.25\Omega) = 56.25W$$

...which would be satisfied with a 75, 100W (etc) resistor. But this falls apart for arbitrarily long cycles, so there must be some constraint.

Given that $Q = mc_{heat}\Delta T$, and assuming the resistor weighs 200g, can increase 125 degrees, has a coefficient of copper (.385), does not lose heat to the surroundings, and $Q=E$, then it takes

$$Q = (200g)(.385J/g^{\circ}C)(125^{\circ}C) = 9.6kJ$$

to cause the resistor to overheat and fail. This happens in 1.7 seconds at 150 amps $\big(t = \frac{E}{P_{avg}})$, so using 1.7 seconds as $T_{period}$,

$$I_{avg} = I_{pulse} \frac{t_{on}}{t_{on}+t_{off}}$$

$$I_{avg} = 150A\frac{.250s}{1.7s} = 22A$$

$$P = 22A^2(.25\Omega) = 121W$$

So I'd need a resistor rated for at least this value -- like the HS150/200.

Am I correct?

P.S: Supposing I choose a fan with some arbitrary air flow rating, this will cool the resistor by some value in watts. Would the effect on the resistor be like moving left on the graph, i.e. subtracting power dissipated?

• Some resistors have pulse ratings. I suggest you buy one that has well-defined pulse ratings. Another option is to put a strip of some resistive foil, for example, stainless steel tape (available in 0.002" thickness adhesive tape) in an oil bath. Or you can use nichrome wire in an oil bath. You can use water, too, if the voltage is not high enough to cause electrolysis problems. The stainless steel will not fail until it reaches very high temperatures, and the oil bath will make sure it can't get that hot. – mkeith Dec 4 '16 at 1:50
• You may want to contact Ohmite. The general subject you want to discuss is "pulse load" (electronics) and "thermal shock resistance in solids" (physics.) You will want cylindrical MELF-style designs, I think, too. Ohmite has a PDF that may help: ohmite.com/techdata/res_select.pdf and this one at Vishay: vishay.com/docs/28870/pulseloadsmdlimit.pdf – jonk Dec 4 '16 at 2:04
• On the last question of how forced air would affect the heat rise curves, it would flatten the curves. Look up some heat sinks, you should be able to find some data and examples of how forced air can drastically reduce heat rise per watt. For example, the HS150 would still be limited to 150W, but you can use a heat sink with smaller than the 995cm^2 surface area listed as "Standard heatsink" in the datasheet. – rioraxe Dec 4 '16 at 3:58
• The calculations you show use squared average current to workout power. Unfortunately that's wrong, you should have RMS current instead. – carloc Dec 4 '16 at 14:30