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# modify an LED circuit? Answered

Re: I am trying to modify the circuit shown in the image to adapt to a new power source
The LED's are 10 mm  and 8 in number
but I am unaware about the forward current and sorry about the previous details it was my
error that i typed i there drop voltage although i have the pictures
showing brightness and believe me it outputs about 50 lumen I think so!
and the LED's don't get hot or don't show any signs of burning
I am quite new to this part of electronics

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## Discussions

You'll struggle to get them to light properly on 3.7V, when their forward volt drop is 3.3 - that only leaves you with 0.4V across the dropper resistor. I never recommend running LEDs in parallel.

Thanks but there was an error in the write-up , I don't know the drop voltage but I am getting quite decent light as in the picture above where the circuit is same but the battery is 3.7V 800mAh phone battery(see the picture above).

Measure the current. Report back !

Sorry i don't know about the current because i live a damn extreme part of globe,
I tried a basic setup showing about 210mA current was drawn by the circuit while operation relative the brightness depicted in the pic above (whee the lights operating)

I suspect you won't need a resistor, but you won't get full brightness. Measure the voltage across the existing resistor.

Also I removed the resistor and now it works great.
Thanks again

Thats it :)
Thanks for your help, I measured the voltage drop to be 2V whih was the problem,The light works good now and am thinking to turn it to my first instructable.

Sorry everyone but i was busywith some school stuff and thus didn't attend to your comments
and also I noticed I had written about the drop voltages actually I don't know about them
you should guess everything according to the picture showing brightness of the circuit

Like Steve says paralleling LEDs is a poor design.

Here is the math.

3 AA batteries  start at  4.5v
if we subtract the LED -3.3v
The  Resistor voltage _1.2v
...That lets us calculate the
current I = V / R = 1.2 / 10 = o.12A =>120ma

Now reverse the math to solve for a new R

The new 800 mahr battery of 3.7v
& subtracting the same LED -3.3v
The low Resistance voltage  0.4v
...This time we solve for
Resistance R = V / I = o.4 / o.12 = 1/3 = o.33 ohms
...We should also try to determine the
..resistor Power P = V × I = o.4 × o.12 = o.o48 Watt >> OK

Now to get a fractional 1/3 ohm resistor
you need to connect three one ohm resistors in parallel..

I estimate you will get you somewhere under six hours
LED light in this new configuration.