93Views8Replies

Author Options:

modify an LED circuit? Answered

Picture of


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
Please Help :)(

9 Replies

user
steveastroukBest Answer (author)2013-12-14

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.

Select as Best AnswerUndo Best Answer

user
MYawar (author)steveastrouk2013-12-20

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).

Select as Best AnswerUndo Best Answer

user
steveastrouk (author)2013-12-20

Measure the current. Report back !

Select as Best AnswerUndo Best Answer

user
MYawar (author)steveastrouk2013-12-22

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)

Select as Best AnswerUndo Best Answer

user
steveastrouk (author)MYawar2013-12-22

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

Select as Best AnswerUndo Best Answer

user
MYawar (author)steveastrouk2013-12-23

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

Select as Best AnswerUndo Best Answer

user
MYawar (author)steveastrouk2013-12-23

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.

Select as Best AnswerUndo Best Answer

user
MYawar (author)2013-12-20

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

Select as Best AnswerUndo Best Answer

user
iceng (author)2013-12-14

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.

Select as Best AnswerUndo Best Answer