Electronic – safely place a 60w between a 50w LED and a large heat sink that can dissipate 100 watts

heatheatsinkhigh-powerledpeltier

Today I have learned that Peltier devices can be used to generate electricity.

I have a set up with a 50 watt LED which is glued to a very large heat sink that cools it without any fans (that can dissipate up to 100w).

The idea of being able to sandwich the Peltier between the LED (heat source), and the large heat sink (cool surface) to generate current sounds awesome.

Before I order a 60w Peltier device off eBay, I'd like to ask:

Can I safely place this Peltier between my LED and heat sink to:

Generate electricity through the difference in surface temperature (LED bay and heat sink)?
Will my set up still work without trouble (LED works fine and heat sink still cools it down)?

Best Answer

Your 50-watt LED requires a heatsink, because otherwise it would overheat and destroy itself. The temperature of the LED is directly related to the thermal resistance between the LED chip inside the package, through the heatsink, to the ambient air.

Your Peltier device will add a considerable amount of thermal resistance to this path, which means that your LED will run much hotter with it than without it. This is probably not a good idea.

Furthermore, Peltier devices are not very efficient at converting heat flow to electricity. If you need electricity for some reason, you'd be better off tapping what you need from the LED power supply itself.

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Electronic – Large difference between specced values and real values for LED strips

The 'specs' (pasted below) indicate 12V, not 12.0V. I'd guess that this product is made to be compatible with automotive 12V, which is usually in the 13.8 to 14.2V range. Your measurements somewhat confirm this.

Specs from the reseller's page:

Item Type: Strip
Place of Origin: Guangdong China (Mainland)
is_customized: Yes
Brand Name: G-Lighting
Average Life (hrs): 50000
LED Chip Brand: Edison
Warranty: 2
Wattage: 75 
Certification: CCC,CE,UL
Waterproof: Yes
Voltage: 12V 
Power Source: DC
LEDs Number/M: 60 pcs/m
Model Number: GL60-5630C
Power: 75W Per Pcs
waterproof: yes
Length: 5 m
Beam Angle(°): 120

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How to power a 12V6A peltier device with a 14.4V battery

Q1.1: The smaller the resistance, the higher the current passing through it. The resistor heated up so much, because the current was around I=6/10 = 0.6A P = VI = 6*0.6 = 3.6W which is probably much higher than the (assumed 1/4 or 1/2W) rating of the resistor. This resulted in the overheating and burning. (The calculation is estimate. I'm not sure what is the voltage drop on the peltier if there is any. The calculation assumes it is a short circuit, but probably it has some resistance.)

Q1.2: It doesn't make any difference. In a circuit where all components are in series, the passing current in all components will be the same. Basically all current goes into the circuit will come out at the end. The current will not be consumed. (Kirchhoff's current law).

Q2: Assuming that the peltier has its internal resistance, it should be generally no problem to connect it directly. However, if too much current flows through it, the peltier's hot side will heat up quickly faster than the heat sink can cool it and will heat up eventually the cool side as well. Should that happen, you will either need to control the current, reducing it or improve the cooling method e.g. increase the heat sink, put a fan on it or use water cooling.

To control the current you may want to either introduce a big resistor, but given the relatively high current, it should be a big power resistor which are not too cheap, or you should get a power supply which allows you to control the output current.

Best Answer

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Electronic – Large difference between specced values and real values for LED strips

How to power a 12V6A peltier device with a 14.4V battery

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