LED Circuit Board and Resistors Please Help Me? :(

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Quintin Gellar

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Hello guys! alright my knowledge is not that great on these things. but i do know a little about it.

i have a light i bought from the store that i took out of the light case it was in. it runs on 220v but i wan't to run it on my portable 12volt 4.5amp Battery.

i know there must be a way! i just don't know what to change on the little Circuit board. here is some pictures of it, any help would be amazing!









 
The 220VAC uses a capacitor to limit the power. The AC is rectified and filtered into DC. The 5 LEDs are connected in parallel.

To work from 12VDC the entire circuit needs to be completely re-designed and some new parts will get hot. Re-design will need the LEDs to be tested for their voltage drop. We do not have the manufacturer's datasheet for the LEDs to see their maximum allowed current so they might quickly burn out.
 


thank you so much for the reply! now i also learned a few things.

so there is no way possible i can put direct voltage on it, without the board ?
 
audioguru said:
The 5 LEDs are connected in parallel.
Click to expand...
I think the LEDs are in series. It takes about 15 volts to make the LEDs light.
To work on 12 volts you need a circuit that takes the 12V and makes 15V while limiting the current. You do need a different little control board.
 
ronsimpson said:
I think the LEDs are in series.
Click to expand...
The pcb shows one side of each LED connected together (the hub) and the other side of each LED are also connected together (the outside ring). Then they are connected in parallel.

They can be powered from 12VDC through a series power resistor. Who will guess a safe amount of current?
 


i just tried to put 15V DC on it and nothing happened! not even a little life
 
Scratch away the paint on the pcb and connect 2 of the leds to your 12v battery via a 100 ohm resistor, that should light 2 of them, try the wires both ways on the leds they only work one way, you can put the resistor either way round in either + or - wire.

You can then isolate the other 3 leds and connect these in series to the 12v supply through a 22ohm resistor, you need to break the track completely bewteen the 2 led circuit and the 3 led circuit.
 

NOW THAT IS SOME INFORMATION! THANK YOU!!!

i will go buy the resistor and do exactly what u said and report back on what happened with a picture also if it worked
 
If my maths is correct and the dropper cap has a value of 0.56uF then the circuit as presently configured draws ~ 40mA from the mains. That implies ~ 8mA per LED since they are in parallel.
Scratch away the paint on the pcb and connect 2 of the leds to your 12v battery via a 100 ohm resistor
Click to expand...
That alone won't quite work . You would have ~9V across the 100 Ohm hence a current of ~90mA. That would be shared (equally-ish) between 2 LEDs, hence 45mA per LED which would likely kill them.
You need to separate all the LEDs from each other then reconnect 2 in series to form a first chain and connect the remaining 3 in series to form a second chain.
The first chain with a series 100 Ohm would draw ~ 50-60mA from the 12V supply, which might be ok depending on the LED spec, but I doubt it. Personally, I'd opt for 220 Ohm to be safe.
I'd use 100 Ohms in series with the second chain.
 
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i cant find it at the mo but i have seen a datasheet on a light like these and the last 2 lot of digits on the board give you the current and number of led's and what configuration, so in this pic it agrees with alec-T (-P40 -5P) oh and the other number is the manufactures size code for the Leds
 
If the total current is 40mA and there are 5 LEDs in parallel then each LED has a current of only 8mA.
That would make a very dim mains light bulb.

Maybe the current in each LED is 40mA then 5 in parallel would use about 3.5V x 40mA x 5= 0.7W which is brighter but is still fairly dim when compared with a REAL mains light bulb.
 
Series LEDs!
1) What you see as copper is actually "anti copper". (not copper) See the red lines drawn in my picture.
2) LEDs should have heat sink copper in large areas.
3) It makes no sense to have 8mA/LED. It does work to have 80mA/LED.
4) A capacitor power supply likes a 15V load much more than a 3V load.
 
The LEDs are in parallel like all Chinese LED products.
Does anybody know the size of these tiny LEDs are so we can guess their maximum allowed current?
 
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Series like it should be designed.
What you see as a pin wheel is copper voids not copper.
This is not a normal PCB. It is made of copper on aluminum. (big heat sink)
There is writing on the LEDs. We need to know the part number.
 
Quintin Gellar said:
i just tried to put 15V DC on it and nothing happened! not even a little life
Click to expand...
Don't put 15VDC into where the 220VAC goes. It will do nothing!
There are wires from the LED board to the power board. Put DC voltage across these wires. CURRENT LIMITED! If the LEDs are in series they will light at 3V. If I am right they will light at about 15V. Either way the current should not get above 100mA. Do you have a variable power supply with current limit. If you don't have current limit start out with a small voltage and work your way up.

I have been assuming 3V LEDs. I found some Seoul Semi. LEDs that look right. They are 6V 100mA parts so there is a chance that the voltages are 2x what I thought. If you are brave, plug the LED bulb in and measure the voltage across the LED board. That is where the white wires hit the LED board. Labeled (+) and (-). The voltage will be close to 3, 6, 15 or 30 volts. That will give us much information.
 
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Having seen Ron's helpful explanation I concede the LEDs are in series. From the calculated impedance of the 560nF cap we know the current through the chain when mains-driven is ~40mA. That makes a lot more sense.
So using a 12V supply, a 2-LED chain (total Vf ~ 6V, say) could be used with a series resistor of 180 Ohms (allowing a bit of margin), and a 3-LED chain (total Vf ~ 9V, say) with a 100 Ohm resistor.
 
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