Some time ago, I built a soapbox-sized battery testing device. One of its functions was to discharge batteries at a specified current, and it handled most of the common battery sizes well, but its maximum discharge current was quite limited. Thus I didn't need to worry about heat dissipation.

I'm now designing an enhanced version, and this time I want to support draining middle-sized batteries (like the ones in cordless screwdrivers, they are around 12 Wh) in reasonable time. I'd be happy if I'm able to dissipate around 10-15 watts of heat in a device with the same form factor (think 85×60×40mm or similar). The heat-producing element is a TO-220 MOSFET. I'm thinking about either heatsinking it (and selecting a plastic enclosure with vents), or using an aluminum box, with the MOSFET dissipating directly to the enclosure. Would that work? Which approach is better (and of course, other ideas are also welcome)?

I'm quite new to this, so links, recommended reads, calculators and stuff would also be greatly appreciated :)

  • \$\begingroup\$ Is there a reason you're using a MOSFET instead of just a plain old power resistor? \$\endgroup\$ – Joe Baker Jan 19 '13 at 21:05
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    \$\begingroup\$ @JoeBaker Constant current discharge circuit with a MOSFET. \$\endgroup\$ – Nick Alexeev Jan 19 '13 at 21:12
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    \$\begingroup\$ Yes, it's an active current source with a MOSFET. I want to support 1-15V, 0-3A (within the heat dissipation limitations, of course). \$\endgroup\$ – anrieff Jan 19 '13 at 21:15

A soapbox is hardly a standard unit of measurement, but 12 W doesn't require anything too huge, even without forced air, provide you can at least let natural convection be, well, natural. Here's how you calculate the heatsink you need.

I've picked the datasheet for IRF510 as an example. It's a very common TO-220 MOSFET and should work for your application.

The first thing you will see is that the datasheet lists power dissipation as 43 W. This of course requires an ample external heatsink, but it should cover your application with a healthy margin.

The absolute maximum junction temperature \$T_J\$ is listed as \$175 ^\circ C\$, and let us assume ambient temperature is \$35 ^\circ C\$. That means the temperature can't rise more than \$175 ^\circ C - 35 ^\circ C = 140 ^\circ C\$. And to be safe, let's add a safety margin and design for no more than \$ 100 ^\circ C \$ rise.

The datasheet lists the maximum junction-to-case thermal resistance as \$ R_{\theta UC} = 3.5 ^\circ C/W \$. That is, for every watt, the junction temperature will rise \$ 3.5 ^\circ C\$ assuming the heatsink can magically remove all heat. At 15 W, that's a rise of \$ 3.5 ^\circ C/W \cdot 15 W = 52.5 ^\circ C \$.

We are hoping for no more than a \$ 100 ^\circ C \$ rise, so we will have to find a heatsink that won't raise the temperature more than another \$ 100^\circ C - 52.5 ^\circ C = 47.5 ^\circ C \$. That means our thermal resistance budget for the heatsink is \$ 47.5 ^\circ C / 15W = 3.17 ^\circ C / W \$.

This is pushing the edge of what can be done with natural convection, but it's doable. Thumbing through my Mouser catalog I can find an Ohmite heatsink FA-T220-64E with a thermal resistance of \$ 3 ^\circ C / W \$ with natural convection. It's the biggest one they sell for TO-220. It's about 1 x 1.6 x 2.5 inches and Mouser will sell you just one for $2.17 plus shipping.

Strictly speaking, I haven't taken into account the thermal resistance of the transistor case to the heatsink. The IRF510 datasheet gives a typical value of \$ 0.5 ^\circ C / W \$ for a greased surface, which at 15 W will mean another \$ 7.5 ^\circ C \$ rise in junction temperature. But remember we included a margin of \$ 40 ^\circ C \$ and assumed a rather high ambient temperature of \$ 35 ^\circ C \$. We should be safe.

Even so, something may obstruct your heatsink, so you may do well do build in some sort of thermal protection. You can implement this yourself, but there are also MOSFETs out there with thermal protection built in. If you do this, you don't need such a margin, and you may very well be able to dissipate more than 15 W if you don't mind the possibility that the thermal protection kicks in.

And, it bears mentioning that even though the transistor shouldn't fail, it will get mighty hot. A plastic box with poor ventilation is probably no good. You will have to keep fingers away for sure. If you want to keep things cooler I'm afraid you have no choice but forced air or spreading the heat over more area: big power resistors, multiple transistors, etc.

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    \$\begingroup\$ excellent post, exactly what the OP needed to learn this concept \$\endgroup\$ – justing Jan 20 '13 at 4:37
  • \$\begingroup\$ Thanks a lot, I'd probably go for an aluminum box, and settle on the max wattage limit through trial and error :) \$\endgroup\$ – anrieff Jan 21 '13 at 6:09

Heat sink would work, if you expose the heat sink to the outside air. If you add a fan, you would be able to dissipate more.

Using an aluminium enclosure as a heat would work too.

The kind of device and form factor, which you're describing can be done. In fact, there is a device with a similar form factor.

enter image description here

This battery analyzer can dissipate 100W continuously.
3.5" H x 2.8" W x 3.6" D.
Specs here.

  • \$\begingroup\$ How much "exposed" should the heatsink be? Please, elaborate. I'm thinking of a plastic box with "normal" vents, and a fan is a no-no, it should be a portable pocket device. \$\endgroup\$ – anrieff Jan 19 '13 at 22:53
  • \$\begingroup\$ @anreiff, the math to calculate this on your own is straightforward. What analysis have you done so far? \$\endgroup\$ – HikeOnPast Jan 19 '13 at 23:21
  • \$\begingroup\$ @anrieff - If the fan is a no-no then for a small device you may have to card along your supply of liquid nitrogen to cool the unit while in use. \$\endgroup\$ – Michael Karas Jan 19 '13 at 23:52
  • \$\begingroup\$ Exposed to the heat sink to the outside air as much as practical. The photo, which I've posed shows a well-exposed heatsink, although the one in the photo is probably larger than what you require. Look into online heat sink calculators (like this simple one) too. \$\endgroup\$ – Nick Alexeev Jan 20 '13 at 0:11

A soap box is larger than a couple of 7.5-watt light bulbs, and obviously those can dissipate 7.5 watts each. If you heat-sink a resistor to a soap-box-sized case and dump 15 watts into it, it will, once it heats up sufficiently, start dissipating 15 watts of heat. The $50,000 question is whether it will remain acceptably cool while doing so. And the answer to that probably depends upon what one considers "acceptably cool"; I would expect that for a temperature to be considered acceptable cool, three conditions must be met

  1. The parts of the electronics that are actually dissipating the heat must not get so hot as to be damaged or malfunction.
  2. Other parts of the electronics must not get so hot as to be damaged or malfunction.
  3. Given the nature of your product, exposed parts of the device should not get so hot as to pose a burn injury, and quasi-exposed parts should not get so hot as to pose a fire risk.

A soap-box sized block of metal with 15 watts of heat generation spread throughout sitting in open air would feel warm to the touch, but not get hot enough to pose a hazard. If someone connects your device to a battery and then throws it into a storage box and then tosses a bunch of blankets on top of it, however, the device may be incapable of dissipating 15 watts without getting quite hot. I would therefore suggest that the best approach is probably to use a heat sink which will spread heat throughout much of the surface of a metal box, along with a temperature sensor which will cause your device to reduce power or shut down if things heat up too much. I would not expect that temperature-based throttling will be necessary often, but it's relatively simple to implement and could provide valuable and protection if users of your device are insufficiently mindful of the heat it could generate.


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