Simple heat sink

Not really a question; just a surprising observation that has some benefit to all of us electron shepherds.

I set up a circuit involving 24 neopixels, run through a 5 v. 7805. The 7805 is only rated to 0.5 amp, and it runs very hot (over 100 C.) at that level. So I dare not go to full intensity on the LED's.

So I chose an intensity level to try, and didn't get the brightness I want, but the 7805 was already at about 90 C. Since this was already in a circuit, I couldn't rejigger components to add an official heat sink. But I decided that any heat sink is better than no heat sink, and I had room to squeeze about 3/4 sq. in. of aluminum in the space.

So I cut this little basically flat Al heat sink, added it, and found the temperature dropped way down to 35 C. I know nothing about heat transfer, but I was quite surprised.

Compare the exposed aluminum back of the regulator to the area of your hestsink and tell us what is the ratio of exposed metal on your heatsink to exposed metal of the tab on the 7805 ?

Probably about 5 to 1. So it doesn't take much extra metal to do a lot better than that 3/8 x 1/4 metal tab.

Are you counting both sides of the part of the heatsink that is exposed on both sides ?

@jrdoner Just a query for myself regarding that heatsink: did you use any thermal bonding or just clamp the aluminium plate onto the 7805?

The reason for my rather general question is that I found heatsink compound (whether paste or epoxy or double-sided thermal bonding pad) has always worked so much better than just plain metal so your, well, massive drop in temperature is of interest to me.

So, was it with or without thermal bonding of some sort please? I could save myself some money in the future!

Hi,
Paint the aluminium black if it already isn't, it should radiate more heat energy, run cooler.

Tom... :slight_smile:

it should radiate more heat energy, run cooler.

Ok. Now I'm confused. I always thought that black ABSORBS heat, which is why you don't wear black t-shirts on a hot day.

Unless the black paint (Which is mainly plastic, after all) insulates the 7805 from the heatsink...I believe you want a black anodized heatsink.

The main point is that air is bad. When you put two pieces of metal together that aren't perfectly flat there are a lot of gaps between them. The gaps are filled with air. Silicon heatsink compound - used sparingly - will fill those gaps, but still allow as much metal-to-metal contact as possible. This is the best case solution.

raschemmel:
Ok. Now I'm confused. I always thought that black ABSORBS heat, which is why you don't wear black t-shirts on a hot day.

That is with an external heat source. In this case, the heatsink is the heatsource. But I don't think it matters since we are really looking at convection here, not radiation.

http://www.aavid.com/product-group/extrusions-na/anodize

At work we are told to ALWAYS use thermal paste. If someone installs a TO220 device WITHOUT the thermal paste, they are told to disassemble the unit and redo it with thermal paste.

Yes, paint (any color) is very bad, it insulates. An anodized finish is good, but on such a small heat sink, it won't do much, if anything.

A THIN layer of thermal compound is good.

FIRMLY attaching the 7805 to the heat sink is important. You want a strong mechanical connection for the best thermal transfer.

Our rule of thumb was to put a dab on and spread it with a wooden Q-Tip until you could see through it.

Hi,

raschemmel:
Ok. Now I'm confused. I always thought that black ABSORBS heat, which is why you don't wear black t-shirts on a hot day.

google black body radiation
It depends on what its hotter.

KeithRB:
Our rule of thumb was to put a dab on and spread it with a wooden Q-Tip until you could see through it.

Exactly, I use the blade of a Stanley Knife blade.
Tom.... :slight_smile:

Our rule of thumb was to put a dab on and spread it with a wooden Q-Tip until you could see through it.

I sometimes use an old credit card to scrape off the excess.

jrdoner:
Not really a question; just a surprising observation that has some benefit to all of us electron shepherds.

I set up a circuit involving 24 neopixels, run through a 5 v. 7805. The 7805 is only rated to 0.5 amp, and it runs very hot (over 100 C.) at that level. So I dare not go to full intensity on the LED's.

So I chose an intensity level to try, and didn't get the brightness I want, but the 7805 was already at about 90 C. Since this was already in a circuit, I couldn't rejigger components to add an official heat sink. But I decided that any heat sink is better than no heat sink, and I had room to squeeze about 3/4 sq. in. of aluminum in the space.

So I cut this little basically flat Al heat sink, added it, and found the temperature dropped way down to 35 C. I know nothing about heat transfer, but I was quite surprised.

Something that might be interesting to you is that heat transfer is often modeled thermal circuits -> 16.4 Thermal Resistance Circuits

I looks like you have enough data to draw your before/after thermal circuits, and calculate the thermal resistances of your interfaces.

Now I'm confused. I always thought that black ABSORBS heat, which is why you don't wear black t-shirts on a hot day.

Anything that is a good absorber of heat is also a good radiator of heat. It has to that way otherwise it would not obey the second law of thermodynamics and it could never obtain thermal equilibrium.

What you are doing here is reducing the thermal resistance of the case to ambient by adding a heat sink. Radiation is only one source of heat loss, there is also convection. A good passive heat sink design will promote a natural air flow rather like a chimney. A trick that Apple has used more than once in its computer design.

You will also find if mounting things in a sealed box then just a few holes at each end reduce the internal temperature dramatically.

I looks like you have enough data to draw your before/after thermal circuits, and calculate the thermal resistances of your interfaces.

YIKES !

I think my brain is going to explode...(I'm having flashbacks from college physics)

Grumpy_Mike:
Anything that is a good absorber of heat is also a good radiator of heat.

Sure, but some are better absorber-conductor than other. And other are better radiator-sink than other.
As example copper is a better conductor-absorber than aluminum, and aluminum is a better radiator than cooper. As result: copper transfers heat faster, retains heat longer. Aluminum transfers heat more slowly, but retains it for a much shorter period of time.
You can see it on intel based processor heat sink. Centre core is copper, and body aluminum.
Just to say: copper is good absorber-conductor, but bad radiator, and aluminum is a bad conductor-absorber but good radiator.

raschemmel:
YIKES !
I think my brain is going to explode…(I’m having flashbacks from college physics)

Lol…from what I’ve seen your brain is strong enough to take it.

His first circuit is only 2 resistors and they’re both specified in the datasheet on page 2: https://www.sparkfun.com/datasheets/Components/LM7805.pdf

heat_transfer_2r.png

Relating the datasheet values to the diagrams in the MIT link, R1 = 3°C/W and R2 = 16°C/W. I’m assuming the ambient temperature is 25°C and the OP told us the case temperature was 90°C. That enables us to calculate his Q (aka the losses in his regulator) as (90°C - 25°C)/(16°C/W) = 4W. That’s pretty high for the 7805 but possible.

Adding the heat sink replaces the datasheet’s resistance value to air with two resistors: one from the IC case to the heat sink and one from the heat sink to the air. In this picture T1 would be the internal temperature of the IC, T2 would be the case temperature, T3 would be the heat sink temperature, and T4 would be the ambient temperature.

heat_transfer_3r.png

The OP told us that adding the heatsink brought his case down to 35°C. Unfortunately we don’t have enough info to calculate the value of both R2 and R3, but we do have enough to calculate their total: (R2 + R3) = (35°C - 25°C)/4W = 2.5°C/W. Honestly, that sounds really low. I have a feeling that the circuit was still heating up and hadn’t reached steady state yet when this data was taken.

What you can now do, is see how hot you’ll get for different amounts of power. Assuming our resistance calculations were correct, the regulator could burn (150°C-25°C)/(3°C/W+2.5°C/W) = 22.7W. That’s really high, so again I think there’s a problem with the data used to calculate 2.5°C/W.

If the OP measured both the case and the heat sink temperature, we could figure out the values of R2 and R3. At that point we could determine if it’s worth making R2 lower with heat sink compound or not.

I think this is the point where we ask the OP what device he used to measure the temperature. I have some DS18B20s but for something like a heatsink I would use a thermocouple with my Adafruit MAX31855 thermocouple breakout board.

PS- Got it. "Package Thermal Data" and "Note-1"
What's wrong with P = I*V | where I = 1.5A (max current)
V = 5V (output voltage)

P = 1.5 A*5 V = 7.5W

Delta T of home made heatsink = 90 deg - 35 deg = 60 deg

Let's go back to the OP's statement:

I set up a circuit involving 24 neopixels, run through a 5 v. 7805. The 7805 is only rated to 0.5 amp, and it runs very hot (over 100 C.) at that level. So I dare not go to full intensity on the LED's.

I see a problem with the OP stating a 7805 with a datasheet rating of 1.5A is only rated for 500 mA
Second, I don't see any current measurement performed on the OP's load. Do you ?
Can you tell me definitively how much current his leds are drawing ?