0
\$\begingroup\$

For a certain application I will be using this resistor. I might also go for this other resistor if needed.

The working conditions are:

  • Ambient temperature (max): 70 ºC
  • Power to be dissipated: 15 W
  • "On" time (max): 60 seconds

The selected power resistor is a 15 Ohm / 25 W TO-220.

The question is: is a heatsink necessary for a 60 second (max) operating time? Can I get away with a large enough copper plane? Assume that the down-time between operations is larger than the operating time.

If it was the case for continuos operation at 15W, I'm already aware of the need of a heatsink and how to calculate it, but I'm wondering how I should proceed for this "non-continuos" intermittent operation

\$\endgroup\$
2
  • 4
    \$\begingroup\$ The data sheet you linked to says, in big red letters, "THIS PRODUCT IS DESIGNED FOR USE WITH PROPER HEATSINKING." Think of a 15 W light bulb - how hot does that get after a minute? 60 s counts as a continuous rating for a TO-220 package. Unless someone else knows otherwise. What you need is a big ceramic-package resistor that isn't touching anything meltable. \$\endgroup\$
    – Andrew Morton
    Commented Oct 2, 2018 at 8:39
  • \$\begingroup\$ Hi @AndrewMorton! Thx for the answer "60s counts as a continuos rating for a TO-220". This would be a valid answer to me as I can not categorize it as non-continuos anymore \$\endgroup\$
    – Radiohead
    Commented Oct 3, 2018 at 10:36

2 Answers 2

Reset to default
1
\$\begingroup\$

Can I get away with a large enough copper plane?

Not likely. That's too much heat concentrated in too small an area.
Also the area required to dissipate than much heat would required so much area it would not be economically feasible.

Here is an example of two LEDs dissipating only 1.5 W each.

enter image description here

The question is: is a heatsink necessary for a 60 second (max) operating time? Assume that the down-time between operations is larger than the operating time.

So 60 s on and 60 s off. It will only take longer before it burns the FR-4.
What will happen is the PCB temp will increase in steps. When the 60 second off cycle ends the PCB will be warmer than when the previous on cycle began.

Each cycle the temperature is warmer than the previous. That is until the temperature differential between the copper surface and ambient air is so great that natural convection becomes effective enough to where the temperatures stops increasing.

As shown in the convection formula the heat transfer flux (q) increases as the temperature differential increases. By that point the FR-4 is burned.

enter image description here

I do not know your reasoning behind choosing a TO-220 package, but it is not a great choice. You would be better off using a through hole ceramic wire-wound power resistor where the body of the resistor is offset from the PCB.

enter image description here

\$\endgroup\$
1
\$\begingroup\$

The question is: is a heatsink necessary for a 60 second (max) operating time?

Wikipedia says: -

When a TO-220 package is used without a heatsink, the package acts as its own heatsink, and the heatsink-to-ambient thermal resistance in air for a TO-220 package is approximately 70 °C/W.

This means that without an external heatsink, the T0-220 heatsink will rise 70 °C for each watt dissipated.

Can I get away with a large enough copper plane?

I'm sure you can but you are going to have to do some thermal analysis.

\$\endgroup\$
2
  • \$\begingroup\$ I was looking for an answer in the style of how the temperature rise works when the dissipation is not continuos, but in my case limited to 60 seconds. As Andrew pointed out in the comment under my question, 60 seconds can be counted as continuos \$\endgroup\$
    – Radiohead
    Commented Oct 3, 2018 at 10:39
  • \$\begingroup\$ @Radiohead A T0-220 package has a thermal capacity (the ability to soak away fast thermal transients like a capacitor can soak up voltage transients) that is very low i.e. less than 1 second hence, the thermal resistance figure is the only relevant figure and, this figure doesn't involve time as a constraint - i.e. it can be regarded as an instantaneous event. \$\endgroup\$
    – Andy aka
    Commented Oct 3, 2018 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.