SparkFun Forums

[Chupa]

Im using a AP1117 LDO in a SOT89-3L package. They have a current limit of 1.1A. On my proto type board I dident do anything as far as thermal management for the device, and it enters thermal shutdown when I have .5A going through it. I obviously want to fix this for the next proto. What are the options for such a task? Can I just put large planes on the output tab on both sides of the board and chuck many vias in it to act like a heat sink, or should i do just a plane on top with out solder mask so i can solder a flat HS to it or something. I have never had to compensate for thermals before in a design so this is something new to me. Thanks

[inventore123]

I think that you need to calculate the power dissipation [ (Vin-Vout)*I ] and decide if using the pcb as heatsink is possible, or if a real heatsink is required.

[Chupa]

12-5*1 would be 7W in a worst possible senario.

I would say 3.5W average though. The end design will be in an enclosure without a lot (if any) air flow.

[Philba]

the data sheet you linked to shows 182 C/W for what looks like the case you describe (no sink, no airflow). That means each W raises the temperature by 182C. Needless to say, even .14A (1 W) would push the device into shutdown.

Two things you can do - figure out how to heatsink and reduce Vin. If you reduced Vin to 6V, you could get up to 500 mA without dropout. That would cut your heat to 500 mW. The regulator would run around 121C (91+30 ambient). That would be below thermal cutoff (150C) but still way hot. Adding a heatsink (large area of copper on PCB) would probably help get a fair amount lower but I don't know how to calculate that.

I would also figure out, with some accuracy, the current your circuit will draw as that will have a bearing on what you do.

[Chupa]

Ill probably will be lowing the input as low as I can as you suggested to around 6. I dont think I will be drawing more than 500mA. But i need to finish before I know for sure.
Thanks

[bigglez]

Greetings Chupa,

Can I just put large planes on the output tab on both sides of the board and chuck many vias in it to act like a heat sink, or should i do just a plane on top with out solder mask so i can solder a flat HS to it or something. I have never had to compensate for thermals before in a design so this is something new to me.

Two options come to mind. Use a switching regulator
with better efficiency (you are currently at 5/12 = 41%).

Alternatively, by-pass the regulator with one or more
resistors. These will still dissipate the same power and
efficiency, but the heat will be stread over two or more
components. For example, if the load is 5V at 1A and
two resistors each carried 425mA the regulator power
would be reduced to 7*0.15 = 1.05W, and the resistors
approximately 3W each (USe 5W resistors mounted clear
of the PCB to allow air movement). Resistors can
be in series with the regulator input, as a third option.

A practical limit for direct to PCB thermal dissipation
in still air is one watt. (A rule of thumb that I use).

Comments Welcome!

[Philba]

here's a calculator that may help. I didn't try it out though..
http://www.daycounter.com/Calculators/H ... ator.phtml

By the way, any chance of using one of the other packages for that part? They have a lot better thermal resistance. The one you picked looks bad in comparison.

[Chupa]

Yea I could go to the SOT223-3L package. I probably will do that. Im gona play with that calculator

[Chupa]

Im looking at these surface mount heat sinks here and i cant quite figure out how it works.
part number 218-40CT3
http://www.wakefield.com/pdf/Low%20Powe ... 20Spec.pdf
Do you solder it to a large copper pad above the IC? Which part of this heat sink is in contact with the copper pad on the PCB?