led
I recently took apart a cheap ($7) LED light to use in a project. It was attached to a heat-sink that I removed and discarded, thinking nothing of it.
Later when I turned on the light, I noticed that the circular piece of metal that had about 11 tiny LEDs on it was getting extremely hot!
If LEDs are suppose to be super energy-saving, why are they giving off so much heat?
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
Your LCD's LED typically accepts 4.2~4.6V. It should be self regulating, so no further resistor needed. You have 11V or 5V available. Below are the three options available. First is use a Adjustable regulator from 11V. The Second is use a fixed 5V regulator and then a regular silicon Diode (1n4001 or similar) to bring it down another ~0.7V. In either mode, the regulator wastes 1 Watt of heat, a heatsink might be recommended (attach the regulator to the module's metal casing is one method).
Or as mention, tap off the already existing 5V regulator, and just throw on the diode. 0.160A is not exactly a large load, it should be fine.
simulate this circuit – Schematic created using CircuitLab
Best Answer
Russell commented in a recent answer that efficiency(*) up to 20 % is possible. That's 4 or 5 times the efficiency of an incandescent bulb. So we're doing fine here. But of course, a 20 W LED lamp will still produce 16 W in heat (that bulb would have 19 W in heat). And if it comes with a heat sink, better leave it as it is; it's there for a reason.
(*) I'm using the word "efficiency" here because that's the word used for it in everyday language. Szymon correctly points out that "efficacy" would be better, expressed in lumen per watt. Actual efficiency as Szymon uses the word is always 100 %, because that's the ratio energy out/energy in, due to the laws of thermodynamics.