I need advice on selecting a heatsink for a high-power LED driver circuit.

Can someone help me out regarding the proper heatsink to use for Q2 on step six of this instructable-


My LED (part number 475-2581-1-ND on digikey) draws 1.4A with a forward voltage of 2.5V. I am using a wall-wart that gives me about 10V. My LED runs smoothly, but the two heatsinks I've tried using with Q2 get *very* hot to the touch. I bolted both of them on using thermal compound. Both are TO-220 "U" shaped. One is about 2cm tall by 1cm wide (about the height and width of Q2). The other is about 3cm tall by 2.5cm wide. How hot is too hot for a heatsink? Would a different shape be better?

Thank you so much

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seandogue5 years ago
Simply put, too hot is when the attached device temperature approaches or exceeds its maximum temperature rating. (Voltage regulators often fall into thermal overload mode when their die temp begins to approach their max rating, first pinching down the available current, then shutting down completely or failing.)

Increasing the airflow around the heat sink may help to wick the heat, as will choosing a forced convection heat sink (ie,. with attached fan)

Measure the temperature with a PTC, thermocouple, etc and compare with the device ratings for Q2. If they're anywhere near close, you'll, need to reconsider your system cooling. In fact, if you're still in the "breadboard" stages, if the temp is within 50-60% of the max rating you may want to think harder about how the device is cooled, or replace the device with a similar component that is rated for higher delivery.

In many cases, you can expect to be burned by touching a device's heat sink, since they're operating temperatures are often far more than sufficient to burn flesh while operating normally. We're rated for 0-70, they're rated for much higher values.

(For instance, your LED has a max rating of ~150C, a temperature at which you can cook meat...the easy bake oven used a standard light bulb to make cakes.)
I looked at the circuit in the instructable you linked to. That circuit uses Q1 in a linear mode, such that it keeps the current through the LED(s) constant. When operating in a linear mode, Q2 will dissipate much more power and hence get much hotter than it would if operated as a switch.

If your application can tolerate a change in LED brightness, then you probably don't need to use a constant current drive circuit. If your input voltage will always be relatively stable, such as from a fixed power supply, you should be fine. If you are running from a battery, the voltage will of course drop somewhat as the batteries discharge and the LED brightness will decrease.

Most any FET for with for lower voltage operation will have a on resistance of only tens of milliohms, and very likely much less. At 1.4 Amps, a FET with 50 milliohms on resistance will only dissipate about 0.1 watt, and it should remain very cool without a heatsink.
That's only true if the OP's supply is very close to 2.8V
Jayefuu5 years ago
How hot is hot? If you can hold your finger on it it's not too hot. If and only if it's painful to touch then maybe you should start worrying about a better heat sink.
erikp (author)  Jayefuu5 years ago
It's way too hot to hold a finger on.
Jayefuu erikp5 years ago
Hrmmmm 60C+ then. What size heat sink do you have at the moment?

Press it against a sheet of metal, if you have drill/tin snips/file you can make yourself one as easily as buy one from a piece of aluminium.

Try pressing it against a chisel or a baking pan you'll find it won't get warmer than bath temperature then.

If you just want to mount it pointing at something in a large space you'll find bolting it to some sheet metal will suffice. If you have to cram it into a tighter space you might want something fancier.
You are dropping a lot of power in whatever else you are using - 10 W is going somewhere.....and not into light.

The device heatsinks will need to be bigger than you have . You have to think about the device temperature, which is complex to work back from the heatsink temperature, biut is ALWAYS a LOT hotter than the heatsink. Aim for no more than 50 C. Also, put the barest possible smear of heatsink compound between the device and its heatsink.

Try at least 4 X more surface area.

The LED on its own, on a perfect heatsink will get 5 deg C hotter for every watt of dissipation in its package.