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newbie heatsink question
newbie heatsink question
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Old 24th July 2012, 09:32 PM   #1
magnetman is offline magnetman  Australia
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Location: Hobart Australia
Default newbie heatsink question

Hi all,

I have a PQ5EV5 voltage regulator requiring a heat sink. Typically this would be easy but it is mounted on the underside of the board thus the heat sink contact side of the chip is facing toward the middle of the board.

Take a look at the photo, that's it with the long bolt through it. most people have a large heat sink attached to this. I have tried the maths but am getting stuck finding values to stick in the equations. Obviously i have limited space about 18mm in the direction of the bolt.

Here is the datasheet PQ5EV5 datasheet pdf datenblatt - Sharp Electrionic Components - Large Output Current Type Low Power-Loss Voltage Regulator ::: ALLDATASHEET :::

please help

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Old 28th July 2012, 07:50 AM   #2
jitter is offline jitter  Netherlands
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Unfortunately the datasheet doesn't give any info on the thermal characteristics other than those in the absolute maximum ratings, so I'm going to make some assumptions. The housing is said to be thermally equivalent to a TO-220. The regulator has overheat protection, so destroying it by using an undersized heatsink should not be possible, though.

PDmax = (Tjmax - Tamb)/(Өsa+Өjc+Өcs)
Өsa = (max.) thermal resistance of the heatsink in C/W;
Өjc = thermal resistance of junction to case in C/W;
Өcs = thermal resistance of case to heatsink in C/W;
Tjmax = max. junction temperature in C;
Tamb = ambient temperature in C;
PDmax = maximum power dissipation in W.

According to this wiki, the typical Өjc of a TO-220 = 1.5 C/W and 0.1 for the elastomer pad (Өjc), but I'd rather use 0.2 C/W as people tend to use too much themal grease which increases thermal resistance.

So, now you know most variables:
Tjmax is 150 C;
for Tamb I would use the typical 25 C (unless your board is going to be in a hot environment, in that case use a higher Tamb);
Өjc = 1.5 C/W;
Өcs = 0.2 C/W;
Now, all you need to find out is what PDmax in your circuit is expected to be.
If you don't have specifications, you must measure the voltage across the regulator (Vin - Vout) and the current going through it to calculate the power dissipation in the state in which your circuit draws maximum power from the regulator.
Then you have all variables needed to calculate the remaining Өsa.

Some examples:
If you need to go all the way up to the max of 45 W, you would need a heatsink of
PDmax = (Tjmax - Tamb)/(Өsa+Өjc+Өcs)
Өsa = [(Tjmax - Tamb)/PDmax]-Өjc-Өcs
Өsa = [(150-25)/45]-1.5-0.2 = 1.08 C/W

This is a rather large heatsink. The one in your picture seems to be very similar to the FK 222 which has 20 degr. C/W heat resistance.

With a 20 C/W heatsink you can go up to:
PDmax = (Tjmax - Tamb)/(Өsa+Өjc+Өcs)
PDmax = (150-25)/(20+1.5+0.2) = 5.76 W.
With a Tamb of 70 C, that would drop to 3.69 W max.

Edit: if you don't need to electrically isolate the IC from the heatsink, you can use Өcs = 0.1 C/W in calculations.

Last edited by jitter; 28th July 2012 at 07:59 AM.
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Old 28th July 2012, 11:13 PM   #3
magnetman is offline magnetman  Australia
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Join Date: Jun 2011
Location: Hobart Australia
Hey Thank you, I understand now. Much appreciated!

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