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At the moment I am working on a project which implies LEDs for street lighting. I need to dissipate around 24 W of heat. In order to do so, I calculated how much heat is dissipated into the air, and now I am considering using a heatsink and maybe a peltier plate. The problem is that I would like to use only the heatsink without the peltier plate, due to high power consumption of the plate. Is that possible?

OK. I will explain more detailed the problem that I have. We as a group, are trying to develop an LED street light powered by a solar panel. The project is build for Denmark and its weather conditions.

We did some testing on different types of LEDs (modules with different light output and energy consumption). The result of the test is that one of the LEDs used is, of course, the most efficient compared to the other ones. But in the datasheet is stated that it should use 14W, while during the test it was using 40W. This gives us a low efficiency, so I assume that the rest of 26W is transformed into heat.

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    \$\begingroup\$ Almost all heat sinks are used without Peltier element. In fact adding a Peltier element is more an exception. To know if it is possible in your case we would need more data. \$\endgroup\$ – Oldfart May 8 '18 at 9:58
  • \$\begingroup\$ Most people just add a fan, if a heatsink by itself isn't enough. \$\endgroup\$ – Simon B May 8 '18 at 10:08
  • \$\begingroup\$ ok. i will explain more detailed the problem that i have. we as a group, are trying to develop an LED street light powered up by a solar panel. the project is build for Denmark and its weather conditions. we did some testing on different types of LEDs (modules with different lumens and energy consumption). the result of the test is that one of the LEDs used is ofc most efficient compared to the other ones. but in datasheet is stated that it should use 14W, while during the test it was using 40W. this gives us a low efficiency, so i assume that the rest of 26W is transformed into heat. \$\endgroup\$ – Rsevlad May 8 '18 at 10:46
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    \$\begingroup\$ You should add that data to your question (Edit button underneath at the LHS). I also feel we have an XY problem. You should first investigate why you need 40W to run a 14W LED. Something is seriously wrong! \$\endgroup\$ – Oldfart May 8 '18 at 10:50
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    \$\begingroup\$ The data sheet almost certainly is saying that the input power should nominally be 14 watts. That you are driving 40 watts says that you need to change your driver to reduce power. The alternative is that the data sheet is saying that the LED will put out 14 watts of optical power (although this is unlikely - ordinarily output will be specified in lumens). In this case the LED efficiency would be 14/40 watts, or 35%, and this is certainly achievable. So, you need to do more research and learn what your data sheet really means. \$\endgroup\$ – WhatRoughBeast May 8 '18 at 11:24
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so i assume that the rest of 26W is transformed into heat

This is troubling. When you said 24W of heat I was assuming a 100W street lamp.

The best LED for this project, in my opinion, would be used would be the Cree XP-G3. 70% of the electricity would be converted to light and 30% heat.

XP-G3 2.82V @ 350mA produces 730 mW of light when powered with 987 mW (2.62V x 350 mA) of electricity.

enter image description here

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Note: Royal Blue is used in all white (phosphor pushed) XP-G3 LEDs.

The Samsung LH351B is a viable alternative. Half the price at an efficacy cost of 6 lumens per watt.

Since it is solar/ battery powered project a peltier cooler is out of the question. Your only option is a passive heatsink. You could add a fan, when needed, to improve efficiency. That would only be on hot days.

Denmark is a good place for LED street lights with a high temperature of 32°C.

The only place to use solar power is where no other source is available. The cost of deep discharge batteries makes solar a poor choice over most other sources of electricity.

but in datasheet is stated that it should use 14W,

Do not buy products from vendors that lie. Cree or Samsung are the only vendors for this project. A battery powered LED must be efficient. Cree currently makes the the most efficient high power LED. Samsung makes the most efficient white mid-power LED (LM301B 228 lm/W) but would require too much real estate for a street light.

A 25W street light could get you 9,000 lumens using Cree XP-G3 LEDs.

Street lights have been 100-400W HPS.

Below is a table from a 2011 US Dept. of Energy street light study. Lumen output ranges from 6,000 to 32,000 lumens. 9,000 lumens is not that bright.

enter image description here

In December Denmark has 12.25 hours of night plus 2 hours for Astronomical Dusk and Dawn. That means a minimum of 14 hours battery charge. At 25W you will need 2A from a 12V lead-acid battery which equates to 30 Ah AGM battery. You will need at least a 100W solar panel. You will need more capacity because it takes a long time (14-16 hours) to charge lead acid. and on a sunny day you will have less than 8 hours of sunlight sufficient for charging. For cloudy days you should have batteries with about 3 days worth of charge.

Your biggest issue is keeping the batteries warm in winter and cool in summer. The capacity of batteries are calculated at 25°C. Heat is very bad for batteries. Keep in mind in summer the capacity of a lead acid battery drops 6.25% for every degree over 25°C. A battery damaged by heat cannot be repaired.

Sounds to me you guys are in over your head.

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  • \$\begingroup\$ You sure about those efficiency numbers? im \$\endgroup\$ – Spehro Pefhany May 8 '18 at 23:48
  • \$\begingroup\$ Yes. I fact check what I write. XP-G3 is ≈185 lm/w. I do horticulture research fixtures so I look at photons per watt. I will look at the deep (royal) blue performance becasue they are spec'd in radiometric watts. All white use the deep blue with yellow phosphor. Xp-G3 at 350mA, 730 mW of light when powered with 987 milli-watts of electricity. \$\endgroup\$ – Misunderstood May 9 '18 at 0:41
  • \$\begingroup\$ Okay, seems shockingly high compared to the last time I looked at it. \$\endgroup\$ – Spehro Pefhany May 9 '18 at 1:15
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So the answer to your question is "YES", but the consensus of a lot of experts seems to be that you should not go further down the cooling/heat-sink track for now.

Instead try to find why your circuit is using 40W. If you get stuck you are welcome to open a new question for us to help you with that. I advice you to be well prepared:

  • Show a schematic of as much as possible. This includes the energy source and how you provide the LED module with power.
  • Give a link to the LED module datasheet.
  • Also perform some measurements e.g. the voltage over the LED module and the current through it.

If it turns out that it is unavoidable that your circuit is burning 40W you can open a question about cooling again. In my opinion and reading Andy aka's answer you might need to start with the housing of the whole system to get rid of the heat.

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24 watts of heat can easily raise a small device to an absurdly high temperature. A heat sink turns a small device (with small surface area and low thermal mass) into a large surface area with large thermal mass. But you still need to get rid of that heat into the atmosphere and heatsinks are generally used because they have a parameter called "thermal resistance".

A typical figure might be 5 °C/watt and this, when inputted with 24 watts of heat would produce a contact area temperature rise of 120 °C. A heatsink with 1 °C/watt will only cause a temperature rise of 24 °C with 24 watts being "inputted".

So you need to work out how much your LEDs can rise in temperature above the maximum ambient temperature and choose your heatsink with care. You also need some natural air flow around the heatsink because you still need to remove the 24 watts of heat. The bigger the heatsink the less air flow you need BUT you still need air flow. You also need to consider that the heat you are trying to remove may indeed add the the local ambient temperature around you heatsink.

If there are space constraints you need to consider forced air flow and possibly ventilation.

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  • \$\begingroup\$ The temperature that the LED module is reaching is around 70degC. And the average temp during summer in Denmark is around 20degC. Thanks for the explanation. \$\endgroup\$ – Rsevlad May 8 '18 at 10:51
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The size of heat sink can be estimated by comparing to pc cpu coolers, a cpu usually put out 65 -95 watt of heat, 20-40 watt cooling need probably won´t even require a fan if ambient airflow is nominal. If heat source and exterior is not very close consider using heat pipes to move the thermal load to where you can remove it best.

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