# help with heat sink

Discussion in 'General Electronics Discussion' started by tehtehteh, May 19, 2016.

1. ### tehtehtehGuest

I thought there would be some easy learning stuff about heatsinks out there but if there is I can't find it

is it safe to assume if I am using a switching regulator that produces 5v 3a output, and is 80% efficient, that 5v * 3a = 15 watts is 80% of the power used and so total power is (15 watts / 80) * 100 = 18.75 watts, and so power wasted as heat would be 18.75 watts - 15 watts = 3.75 watts

I have 2 of these regulators at the same end of the board so is it as simple as finding a heatsink that can dissipate at least 3.75 watts * 2 = 7.5 watts and attaching them both to it?

I have been trying to work out the heatsink needed for this but I don't know what I'm doing really, the data sheet gives me a graph of temperature rise above ambient for the 3a load but this is using a heatsink and doesn't explain the thermal properties of this heatsink, so how am I supposed to apply that to a different heatsink

2. ### dorke

2,342
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Jun 20, 2015
Your assumptions of what is the power in watts that will create heating is correct.
But ,where did you get the 80% number?
The number usually varies and depends on the actual working condition of the device.

Another question you need to answer is:
What is the allowable rise in temperature of the "power device" due to this heat.
The absolute limit would be the "Max Temperature" as per the device manufacture,
that shouldn't be reached in any case.
You also need to define the ambient temperature the device would operate in ,
the rise of temperature is above the ambient.

3. ### tehtehtehGuest

http://docs-europe.electrocomponents.com/webdocs/0780/0900766b80780150.pdf

the data sheet gives a typical efficiency of 80% for 12v input to 5v output at 3a load, which is mainly how I'll be using it, don't know what the % is for lower loads but I think the waste wattage will be lower anyway if the load is lower, so I'll stick with that as a ballpark figure for now

this will be in a small enclosure in a car, so I was thinking like worst case scenario 85c for ambient temperature, but that's a bit of a guess really, maximum junction temperature is 150c but temperature range only goes up to 125c so I don't know which to account for

another thing I don't know how to account for is the tab on the to220 is connected to ground so I will have to have one of those insulating thermal pads between it and the heatsink, which will change the heat transfer properties some more

4. ### tehtehtehGuest

http://uk.rs-online.com/web/p/heatsinks/6744712/

this heatsink is PERFECT in size for what I need, so any tips on how to make sense of the values would be appreciated, I would like to learn to do it myself rather than someone just do it for me

it says 6k/w, if I assume that's kelvin per watt, would that also be the same as celcius per watt? so is it saying that it will dissipate 6c for every watt I put into it? so with approximately 7.5 watts for both regulators it will dissipate 45c of heat?

Last edited by a moderator: May 19, 2016
5. ### dorke

2,342
665
Jun 20, 2015
85c ambient in a car?
Is it in the engine compartment?
Very demanding environment.

Tj of 150C is device destruction ,you should be below 125C for opertation.

look at page 25-26 in the data sheet ,
with a "good" heat sink To-220 reis above ambient for 5V/3A is about 35C
putting the junction at 85+35=120C ,not good.
Is forced air (fan)a possibility?

6. ### dorke

2,342
665
Jun 20, 2015
Yes K and C the same per W.
Your device will be at 85+45=130 ,not good at all

7. ### tehtehtehGuest

85 ambient is just a guess, and I was trying to go worst case scenario, it's not in the engine compartment but it will be a small enclosure so I don't know much the air inside will heat up, a fan is possible but it would have to be small

thinking about it that 45c is for both regulators, so wouldn't that make it 22.5c per regulator, in which case it may be ok, or maybe it's not as simple as that!

if I've got the right idea with working out the heatsink though I'm very happy for now

8. ### tehtehtehGuest

yeah the data sheet felt a bit vague to me, I see the graph but it's with their heat sink, so I don't know how it applies to me, I guess the other info they have given (junction to case, junction to ambient) is supposed to help but I don't know what these values are or how to use them

9. ### dorke

2,342
665
Jun 20, 2015
Well,
Here is a guide for Heat Sink Design it may help you .
The number you should use from the data sheet is 2C/W for theta junction to case.

Since you need an insulator it will have about 0.5C/W.(that can be lowered to 0.1C/W with better ones).
So we have 2.5C/W ,to that we need to add the heat sink thermal resistance itself.

10. ### BobK

7,682
1,689
Jan 5, 2010
Also consider: not all of the power is lost in the transistor. Could will be some in the inductor, and possibly in resistors.

Bob

11. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
And the rectifier diodes often dissipate a fair amount.

12. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
And the rectifier diodes often dissipate a fair amount.

13. ### tehtehtehGuest

so you're saying it may not get as hot as I calculated there?

14. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
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Jan 21, 2010
Well, the total loss will be the same, but where that heat comes from is the question.

You can be pretty much guaranteed that the circuit was designed with a 25C (or thereabouts) ambient temperature in mind. Running the unit at highly elevated ambient temperatures can cause otherwise conservatively rated components to get far hotter than they probably should. For example, any capacitor rated at 85C is already at its maximum temperature. If you look at (say) capacitor life vs temperature you can quite easily get components with an operating life under 500 hours in these conditions.

A sealed enclosure is not a real problem as long as things that get hot can conduct their heat to the case and then to whatever it is mounted on. Naturally you don't want to mount it on something that gets too hot!

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