shissqbob

iTrader: (14)

Alright, so i have 5 leds in a row right now on my pcb board and i have resistors in my hand that are supposedly for a car battery. Now, the question is, after i solder all the positive leads together in that row, and then solder the negatives, you have a straight current running through those leads. i'd like to make the board 25 leds total...5 rows X 5 columns. For the electrical savvy guys, how many resistors do i need per row? led? or board? I'd really appreciate your help.


Thanks

MOHFpro90

iTrader: (11)

Wll essentially, only one would be needed per board, if wired correctly. Think about it: if you ahve 25 LEDs, the power source is 12V at 5 amps, and the resistor brings it down to 4 amps(hypothetically), with one resistor installed on the power into the board, all fo the power into the board will be at 4 amps.

Now another issue is, by the time you get to the 25th LED, the resistance and loss of current may be an issue. I am not a pro at this, but I see it possible that it may not light up. Solution: make each row have a resistor. Thus, the 4 amps is only going though 4 other LEDs, not 24.

Hope that helps.

the_3d_man

iTrader: (1)

I think you're a little confused...

each LED has a forward voltage drop (they should tell you on the package). usually about 3-3.5 volts for white or blue LEDs, which means that at the recommended current (and oddly enough at almost any current it changes little) the voltage drop across the LED will be about 3 volts. so if you are driving the LED with 14 volts (use 14 for car because alternator puts out about 14 volts) then subtract the voltage drop of that LED from the 14 volts and then divide by the rated current for that LED (if the LED is rated at 20milliamps then divide by .020 amps) to determine the needed resistance.

You cannot just put a bunch of LEDs in series and expect to be able to drive them all unless you have a large supply voltage. Take the forward voltage of the LED and see how many times if will go into 14 volts (12.5 if you want those things to light up when the engine is not running). if the LED has a forward drop of 3.2 volts, then you may want to only put 3 in series, although you can always wire it up with 4 and test it out to see how well that will work.

Another way to do this is to just use one big resistor and just wire all the LEDs in parallel and put the one big resistor in series with the bundle of LEDs (ie. all the positive leads of the LEDs soldered together, all the Negative Leads soldered together and then a big resistor in line with those.) you will need to calculate the total current needed to go through all those LEDs, then subtract the voltage drop of one of those LEDs from the supply voltage and then divide that by the total current you calculated. The resistance value will be low, but the power dissipated will be high so you will need a high power resistor. I don't like doing it that way, because it is inefficient, and because sometimes the LEDs have a mind of their own. Sometimes an LED that was manufactured a little off and has a slightly higher forward voltage will light up dim when you do that. But these days is not too common.

JSMax

iTrader: (13)

yeah, what he said...

markd248

iTrader: (8)

talk to kevlo he might be able to help out with this since hes been doin his own LED boards.

besides that im no help.

MOHFpro90

iTrader: (11)

The_3d_man: That is what I was trying to say, that after each LED, the current drops, thus by the time you get to the end, you have no more current(or not enough).

Sorry...hope that helps BTW, thank you for clearing it up 3D_man.

the_3d_man

iTrader: (1)

Your just confusing current and voltage. The voltage drops across each LED, not the current. The current is constant along the entire circuit.

shissqbob

iTrader: (14)

alright, thanks for the replies guys, it sparked ideas. as a result, i build a square of led's 5 X 5... and soldered all pos. and neg. My intention is to use one resistor per row, but just to see if the power to the lights diminished, i took a 9volt battery and tested three rows of 5 with a resistor. To my disbelief, i didn't notice a difference in intensity, so i bet you could get away with one resistor per 5 rows. I'm not quite sure as to what i'm going to do but thanks for the advice