I am trying to re use this cob led which came out of a rechargeable battery powered LED light.
It powers up fine from the original Lithium battery pack.
The battery specs are 3.7 volt and 800ma. I have checked these with a multimeter and this seems accurate.
What I cannot work out is how the LED's are wired. If they are in series this would make the current through each one 28.5 m.a. I believe. This seems very low to me although the lamp is very bright.
Also I cannot work out how to get to the stated output of the lamp which is 10Watts. The numbers do not seem to add up.
My last question is what are the small components on the LED board? Are they resistors?
Are they a kind of current limiting circuit?
The battery specs are 3.7 volt and 800ma. I have checked these with a multimeter and this seems accurate.
Batteries are rated in milliamp hours, which is a measure of battery life. You have to dig a little deeper into the specs to get the current rating.
Did you measure the current?
The current in series components is the same (think of water flow/current in pipes). The voltages across series component's add-up so there's no way they are in series.
With just 3.7V available, they have to be in parallel. The resistors are probably in a series/parallel with multiple resistors used simply to handle the power.
Some "cheap" flashlights rely on the internal resistance of battery but I don't know if that's done with li-po batteries, but it's possible that your circuit is using both. In those cases, you can get into trouble if you try to use a power supply or a different battery.
Yes the plate on the lamp says that the battery is 3.7volt 8000mAh. I have measured the current with a multimeter as 800mA.
In my first post I mistakenly said series when I meant that the parallel, current through each LED would be 28.5mA.
So if I can assume they are in parallel what I would like to work out is how to arrive at the 10Watt stated output.
I am hoping to replace the batteries with a more permanent power supply fed through a buck converter current limiting board. Will this will be ok? The Led works fine when tested from an external power supply.
My best guess is I see what looks like 14 LEDs on the top with 14 resistors and the same on the lower half. So 28 total LEDs and 28 current limiting resistors. The resistors are hard to read but from the image I see what looks like 6R2 which lenads me to believe 6.2 Ohm resistors? The 3.7 volts is a typical LiPo battery. What I can't get to add up is 10 watts and your numbers. Using 3.7 volts applied and 800 mA Itotal I see about 2.96 Watts or roughly 3 watts.
With it powered by 3.7 volts you could measure the voltage drop across a few of the resistors, they should all be about the same. That assumes 28 LEDs with each a series resistor in parallel. Knowing the resistor voltage drop the remainder subtracted from 3.7 should be the Vfwd of the LED. Measure from - of supply to each side of a few LEDs and you should see what V difference was. Vresistor + Vled should add up to Vsupply of 3.7 volts.
That's my best guess and I assume the traces are not visible on the other side of the board.
I still need to find a suitable way to power this Cob LED.
I have tried using a Buck Converter LED controller board which will give the required 800mA.
However this board requires a minimum 9volts input and I need 3.7volts to mimic the lithium battery.
Can I use a buck voltage converter after the current control board to drop the voltage?
I would ideally like to power the whole setup from 12 volts as I have lots of these power supplies
and many devices which use these.
I just need the simplest solution to achieve 3.7volts at 800 ma.
If anyone could suggest a way to do this I would be very grateful.
If its a 10W array then each LED uses 100mA or so, for a total current 2.8A.
The 6.2 ohm resistors would then be dropping about 0.6V, seems right for 3.7LiPo
driving white LEDs (about 3.0--3.2V forward voltage normally)
The fact you measured 800mA total probably just means there was too much stray
resistance in your meter leads, or your power source wasn't upto 3.7V at significant
current drain. Even 0.05 ohms in each meter lead could explain most of this.
With low voltage and high current circuits the wiring resistance is usually a factor
to consider.
Can I use a buck voltage converter after the current control board to drop the voltage?
No, but you could use a buck converter to power it directly. You never said even a word about the application - whether it just has to shine, or be controlled by the Arduino in some way? Does the brightness have to vary? Any other special needs?
aarg:
No, but you could use a buck converter to power it directly. You never said even a word about the application - whether it just has to shine, or be controlled by the Arduino in some way? Does the brightness have to vary? Any other special needs?
Hello, Yes it just needs to shine, doesn"t need to be controlled.
It is to be installed in an artwork.
Would it be a long term solution to power it from the buck converter at 3.7 volts?
Would the current take care of itself?
Its in its original housing with a heatsink and lots of space around.
Thanks
k
