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I'm playing around with workplace illumination and have developed a 20 V –> 38 V PWM'able constant current source to drive my power LEDs (max power about 64W). So far, so good. However, I've nearly thermally killed one LED by fixing it on a significantly undersized heat sink ("luckily", the wire contacts unsoldered themselves just in time, stopping the process).

Now, I'm considering cooling options. Wanting to avoid active cooling (i.e. the humming of a fan), I was considering the "lazy" way out (dimension far from final, I don't have a heatsink candidate yet):

simple mechanical drawing

I'd like to mount the 19 x 19 mm LED directly onto an aluminum bar or profile. Now, I'm already playing around with thermal simulation software, but that seems over the top (and so far, it mostly crashes, plus I have a lot of theory to catch up on). So:

  • Is there a well-known analytic model for heat distribution when attaching a constant-power heat source to a piece of metal?
    • if not, is there a go-to simulation software? So far, I'm playing with Elmer.
  • Is simulation the way to go here, at all, or is passive cooling damned for 60W LEDs?

Data (from LED datasheet):

  • Junction-Case Thermal Resistance 0.8 K/W
  • 19x19 mm
  • max rated power 64.2 W
  • continuous power I'm planning to use: 36.6 V · 0.72 A = 26.352 W
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  • \$\begingroup\$ highlighted the fact that I don't have the metal bar yet. \$\endgroup\$ – Marcus Müller Mar 5 '17 at 14:19
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    \$\begingroup\$ If your wires start unsoldering themselves, you may want to have second thoughts about using whatever got hot again. Parts may have been partially but permanently damaged. \$\endgroup\$ – Mast Mar 5 '17 at 19:39
  • \$\begingroup\$ jep, considered that, the affected LED is now the one reserved for experimental use, but so far, for bursts, the I/V curve of the LED array is still ok \$\endgroup\$ – Marcus Müller Mar 5 '17 at 19:41
  • \$\begingroup\$ The suggested calculator seems to assume that the heat is applied evenly to the baseplate of heat sink. Your heat source is a 19x19mm spot. You should be using a copper heat spreader at least, to get near the calculated heat transfer efficiency. Also, for free convection to be efficient at reasonable delta, fin spacing should be 7-8mm, with leads to bigger sink for required area, and heat spreading is more and more important. You may want to consider omnidirectional heat sinks, this sink will have ~2C/W from 1sq.inch heater to ambient free convection, micforg.co.jp/en/c_n80e.html \$\endgroup\$ – Ale..chenski Mar 6 '17 at 4:47
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If my understanding is correct, you want to estimate the thermal resistance of a heatsink or a slab of thermally conductive material to ambient, without any airflow (= natural convection).

There is a nice online calculator for finned rectangular heatsinks that implements the natural convection model for heatsinks (a more academic, detailed explanation of the model is here).

Here's an example relevant to your design problem (55x55x55mm outer dimensions, 10x1mm fins, baseplate thickness 10mm and a rather conservative 2,000 W/m2ºC contact conductance):

Calculator snapshot

The resulting source temperature for 25ºC ambient temperature and 26.35 W of heat flowing into the heatsink is aprox 110ºC, which means that heatsink would have a 3.23 ºC/W thermal resistance in natural convection conditions.

Experiment with the calculator in order to find the outer dimensions that suits best your design.

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  • \$\begingroup\$ which browser are you using? I can never make any Source Temperature appear. Ah. Got it to work. \$\endgroup\$ – Marcus Müller Mar 5 '17 at 19:31
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    \$\begingroup\$ Safari. Be aware that the calculator is quite picky. For instance: if the sum of all your fin spaces and fin thicknesses don't add up to the total width, then the calculator will silently refuse to compute a source temperature without throwing any error code or warning. \$\endgroup\$ – Enric Blanco Mar 5 '17 at 19:34
  • \$\begingroup\$ I must admit it's very tricky. I thought the formula was total width = N_fins * fin_width * (N_fins -1) * fin_spacing, but that doesn't work \$\endgroup\$ – Marcus Müller Mar 5 '17 at 19:59
  • \$\begingroup\$ ah wait, it doesn't like if H+tp don't exactly add up, either \$\endgroup\$ – Marcus Müller Mar 5 '17 at 20:01
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    \$\begingroup\$ @Misunderstood to be fair, I just went and read the script's source code, and figured things out. Also seems that modern browsers might be supressing the mechanisms the thing uses to display warnings, so not really the tool's fault \$\endgroup\$ – Marcus Müller Mar 5 '17 at 20:47
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I've been down that road but the simulators cost way too much and have a steep learning curve. If you are not a thermal dynamics engineer you may have some problems understanding the jargon, I did. I read text books on thermal dynamics and all sorts of heat sink design papers and heatsink simulators.

I suggest you get the aluminum bar at online metals $1.23 (0.125 x 1.5 x 12) (the 6061 T6511 is the least expensive), mount the LED get it working, put the bar in the fridge. Take it out into a humid room where it condensates. Then put it in the freezer, get it frosty, take it out fire it up and watch the patterns the ice crystals make a as they melt as the bar warms up. The result is similar to the output of a simulator. Real life is astonishingly accurate too.

Besides it's not a wasted effort, if you do the simulation, you still need the bar to see how far off the simulations were.

But the problem is you will within an hour or so end up with a very hot bar of aluminum nearly as hot as the LED. But you don't need much air flow with a large surface. An aluminum bar @ $1.23 or less per foot is a damn cheap heat sink.

I do not like fans either. This one is very quiet because it moves only 13 CFM @ 12VDC, 30.3 dB, 2300 RPM but it was effective.

36V 2.4 Amp max.
Pattern shown only on one side, it was actually symmetrical.
COB LED on Heat Sink

Measuring the temperature backside.

enter image description here

Current turned way down and diffused.
enter image description here

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    \$\begingroup\$ that is an excellent idea for observation of temperature diffusion! Great! Not in the US, so that shop won't be of use to me, but I'll find a (metrically sized :D ) Alu bar locally and do this. \$\endgroup\$ – Marcus Müller Mar 5 '17 at 19:15
  • \$\begingroup\$ I went to water cooling. What I got out of that exercise I described, was the spacing between COBs. I could see how far the heat was going down the bar away from the COB. And I have already purchased an extra test COB for when (not if) this one can no longer take the abuse. \$\endgroup\$ – Misunderstood Mar 5 '17 at 19:30
  • \$\begingroup\$ Pumping 2.4 Amp the a forward voltage was 39.5V (102 Watts) the temperature of the bar directly under the LED was about 42.5° C when stabilized after 20 minutes and was the same 10 minutes later. \$\endgroup\$ – Misunderstood Mar 5 '17 at 21:04
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The good news: There is indeed a simple mathematical model that's fairly accurate.

Basically you can model most thermal problems as a simple electrical circuit:

  1. Thermal Power = Electrical Current
  2. Thermal Temperature Difference = Electrical Voltage
  3. Thermal Resistance = Electrical Resistor
  4. Thermal Mass = Electrical Capacitor

Your case is even simpler: since you don't care about time constants you don't need to worry about thermal mass.

So your model should look like this

LED Junction -> {R1} -> LED Mounting Surface -> {R2} -> Al Bar -> {R3} -> Ambient

Where

  • R1: thermal resistance from the LED junction to the LED mounting surface
  • R2: thermal resistance for the LED to Al connection
  • R3: thermal resistance from the Al to ambient Air

They are all in series so you can simply add them up. If you have R1 = 1.2K/W, R2 = 0.8K/W and R3 = 0.1 K/W your total resistance would be 2.1K/W. For 40W of dissipated heat your LED junction would be at 2.1K/W*40W = 84 Kelvin (or Celsius) above the Ambient temperature. At 25C ambient the junction would be at 109C.

The bad news: The data you need to model this is notoriously hard to predict

You will need three thermal resistances and the max allowable LED junction temperature.

  1. If you are lucky you can find R1 and the Max temp for the LED in the data sheet.
  2. R2 is very tricky, since it depends on the exact material, the exact shape, the amount of flatness, the exact surface treatments of both your mounting surface and the Al bar. Even the color and details of the anodizing process of the Aluminum do matter here.
  3. R3: IF the bar is reasonably large, that should be quite small

What to do depends on the measurement abilities that you have. In general this has a good chance of working. Make sure that the LED is firmly attached to the AL bar and put a thermal pad or some heat paste on the connection.

Touch the bar: it should be noticeably warmer very close to the LED. If not, that means you are not getting any heat transferred into the bar and the thermal connection is no good. If the entire bar feels warm or even hot, you are not getting enough thermal coupling to the environment. Consider more surface area for the bar.

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    \$\begingroup\$ So, yeah, modelling R3 is exactly the problem! R1 is (of course) given in the datasheet; R2 is an interesting aspect, though I planned to keep that below 5K/W by pressure-mount and heat paste. However, as said in my question, I don't yet have a bar yet to measure on, so while I wished this was an answer to my question, it isn't. \$\endgroup\$ – Marcus Müller Mar 5 '17 at 14:17
  • \$\begingroup\$ I like that touch the bar. Works so much better than the formulas. I would keep increasing the current, wait an hour for things to stabilize, measure the temp on the back side, then up the current, repeat. You got my vote. \$\endgroup\$ – Misunderstood Mar 5 '17 at 20:52
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One 60W LED is a thermal challenge because the heat source is small and very powerful. Therefore, you will need thick metal to spread the heat laterally into a large enough heatsink.

This is similar to a desktop PC cpu: small surface area, lots of power. Many desktop PC heat sinks use heat pipes to solve the heat spreading problem. A fanless PC heatsink should work.

However, this doesn't solve your other problem, which is that one 60W LED is a very bright point source, and it is not ideal for workplace illumination. It will be blindingly bright and will cast harsh shadows.

You can solve both problems by using LED strips like this:

http://www.leds.de/en/LED-strips-modules-oxid-oxid-oxid-oxid-oxid/High-power-LED-strips/

I used these in a project:

http://www.leds.de/en/LED-strips-modules-oxid-oxid-oxid-oxid-oxid/High-power-LED-strips/PowerBar-LED-Strip-12-Nichia-LEDs-CRI-90.html

They come on a metal PCB, and the strip can be cut into individual LEDs. I then glued them onto aluminium L-profiles using thermal conductive epoxy (one LED every 10cm).

Spreading the heat-generating LEDs over a length of aluminium profile allows much easier cooling, and generates a more pleasant light.

EDIT

OK, lets go with the 60W LED.

I suppose it is pointing down. You want the heat sink fins to be vertical for optimum convection. This points towards this kind of form factor:

Link Link

If you use a flat heat sink, you'll need to mount the LED on a thick aluminium square, then mount this on a heat sink.

Since your problem is spreading heat generated by a small source, you could also use flat heat pipes:

Link Link

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  • \$\begingroup\$ Though I value the estimation, I do have diffusors in place, plus a healthy distance between light source and desk top :) \$\endgroup\$ – Marcus Müller Mar 5 '17 at 14:14
  • \$\begingroup\$ Your edit still does not address my question, sorry. \$\endgroup\$ – Marcus Müller Mar 5 '17 at 15:16
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There's Lisa, a finite element analysis tool that is free at least for models that have max. about 1000 nodes.

The simulation is difficult, needs a deep understanding and it's based on assumptions about the boundary conditions. Real tests, if safe and possible are better. If you already have the led and the heatsink candidate, you can well try it. Run it at a known, but safe power level, let it reach the equilibrium (=no more measurable temperature rise) and store that final temperature. You must have proper equipment for the measurements. The temperature difference between the led and the ambient is directly propostional to the dissipated power. Of course you can't go inside the led until you use itself as your sensor. The manufacturer can possibly give some useful data of the relation between the forward voltage, current and temperature.

But yo can also measure at the border between the led and the heatsink. There's surely available the thermal resistance between that point and the semiconductor or the allowed temperature limits are directly told as temperatures at the heatsink border.

If your temperature rise at 10W is say 1/3 of the allowed rise, you can maximally have the dissipation =30W.

Note that in a cabinet the ambient temperature also rises and that must be taken into the account. An adjacent other heating device also must be taken into the account. It warms up the ambience and also radiates heat. You see now and probably already have known that thermal design is an area full of challenges and traps.

ADDENDUM: The problem is interesting. I had took it granted that mounting onto an aluminium plate solves the heat problem with leds. Some quick calculations showed that no thin plate will nail it. The dissipation is quite the same as in a 100W audio amplifier per one of the 2 output transistors, so the samelike heatsinks are needed. Their performance suffer drastically if the dust clogs them. Remember to demad the regular cleaning as a condition for the warranty or make heavily oversized heatsinks.

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  • \$\begingroup\$ nope, no heatsink candidate yet! Point is that I'd like to know before I order meters of alu. Thermal resistance is known, indeed, but the linearity assumptions (x times power in -> x times higher deltaT) seems to break down for distribution of heat in a volume – or does it not? I always thought the thermal resistance model just applied within close boundaries. \$\endgroup\$ – Marcus Müller Mar 5 '17 at 13:39
  • \$\begingroup\$ This is mostly right, but passive convection due to heating of the heatsink is non-linear. Fortunately this works to your advantage. Do everything as if LED temperture rise above ambient scales linearly with temperature, and the extra convection at high temperatures gives you a little margin. \$\endgroup\$ – Olin Lathrop Mar 5 '17 at 13:40
  • \$\begingroup\$ @MarcusMüller the commentator Olin means that heat makes the air to flow. This at low power levels can be quite laminar and the airflow decreases the total thermal resistance. Finally at high levels it will be so turbulent that its efficiency will become virtually unpredictable. But he's right. \$\endgroup\$ – user287001 Mar 5 '17 at 13:49
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    \$\begingroup\$ @MarcusMüller there's no contradiction between the nonuniform temperature distribution and the law x times power => x times delta T. The formula applies still for every point separately. \$\endgroup\$ – user287001 Mar 5 '17 at 13:54
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    \$\begingroup\$ @MarcusMüller You had the idea not to buy a pile of aluminium until you know it performs ok. Then get a smaller piece - the one led part and do the test run. \$\endgroup\$ – user287001 Mar 5 '17 at 19:21
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To give you an idea of what you are up against with a passive heatsink. Cree made a reference design as a replacement for a 1000W HPS Lamp.

The fixture is made up of four "engines". Each 130 Watt engine is 11.25" x 7.25" x 2.5". Which is basically the size of the heatsink.

 LED fixture with massive heat sinks

The heatsink used is an Aavid Black Anodized P/N 62625

Estimated Price (for Heatsink Only) $450

That's $3.46 per Watt.

For your 64 Watts that would be $222.

$450 cost is based on an Aavid Black Anodized P/N 627252 (2.28" x 9.75" x 55")

And a Aavid 701652 1.78" x 12" x 48" was $431.



Each Engine is made up of 48 LEDs pushing 130 Watts.

You would need a heat sink only half this size. This heatsink is 11.25" x 7.25" x 2.28"

LED Engine

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Check out the blog post 'How to design a flat plate heat sink' http://www.heatsinkcalculator.com/blog/how-to-design-a-flat-plate-heat-sink/. It provides a detailed explanation of how to calculate the thermal resistance of a metal plate used as a heat sink. I believe you can also get a spreadsheet that does the calculations if you give them your email address.

Essentially you need to determine the radiation and natural convection resistance from the external surfaces then determine the conduction thermal resistance. Add the three together based on the thermal circuit shown below:

enter image description here

where:

Rconv is the external convection resistance

Rrad is the external radiation resistance

Rsp is the spreading resistance

Rint/Rcont is the contact or interface resistance

Rth-jc is the case to junction resistance of the LED

Ts is the heat sink surface temperature

Tj is the LED junction temperature

The equations for Rconv and Rrad are quite involved and are explained in detail in the blog post.

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A simple spice simulator will do this: it is just like a capacitor being discharged.

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  • \$\begingroup\$ sorry, I don't see it. What I'm asking is basically for a way to derive model parameters (e.g. the thermal resistance) from the geometry and material properties of my heat sink. You're saying "sure, a linear electrical network analyzer will do that". I'm afraid that's not the case. I'm looking for the values to plug into an "equivalent circuit", not for the circuit. \$\endgroup\$ – Marcus Müller Mar 5 '17 at 15:15
  • \$\begingroup\$ @MarcusMüller I at first blamed this answer to junkmail class in my mind, but it has an idea. The temperature distribution in a long rod would be a diminishig exponential function. The variable is not the time but the distance from the led. The time constant should be replaced by the thermal diffusion length constant. Unfortunately this quite true fact dos not be a big help in this phase. \$\endgroup\$ – user287001 Mar 5 '17 at 20:12

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