Lighting a bunch of LEDs-Do I need a resistor?

DosManDan

I have to light about 300 LEDs from a battery pack. If I cut the power from the battery (7.2V) down using a voltage regulator to 3.2V (low end of forward voltage for the LED) do I still need to use resistors?

The LEDs use 3.2V and 20mAh. If I can avoid the resistors, I won't be wasting valuable mA. The 7.2V batteries have 4400mAh. By my calculations, I should be able to light the LEDs for about 44 min.

4400mAh battery power
6000mAh LED current draw

(4400mAh/6000mAh) * 60 = 44 min

Thanks!
Dan

Phil Pilgrim (PhiPi)

LEDs need to be driven by a current source, not a voltage source. Moreover, if you "cut the power" to 3.2V with a linear regulator, your precious milliamps get wasted in the regulator. It would be better to wire pairs of LEDs in series and drive each pair from 7.2V through a resistor to limit current. That's assuming that twice the LEDs' top-end forward voltage doesn't exceed the battery's output voltage over its useful charge. If it does, some kind of DC-DC boost converter to drive longer strings of LEDs from a higher voltage would be appropriate.

-Phil

DosManDan

I sure hope this doesn't sound dumb, but I thought the purpose of the resistors was to drop the voltage flowing through the LEDs?

I thought that if I dropped the voltage, it would mean I wouldn't need reistors.

Using the formula: R= (Vs-Vl) / I

where Vs=Voltage supplied, Vl = voltage required by the LED and I is the mAh draw by the LED.

Using 7.2V as the supply and 3.2V as the voltage forward and 20mAh as the current needed.

R=(7.2-3.2)/0.020

R is 200

If I only supply 3.2V, the equation becomes:

(3.2-3.2)/0.020

R=0

Am I using messed up logic? I thought it made sense, but I've been wrong plenty of times

Thanks!
Dan

FireHopper

leds need current limiting..
and voltage regulators waste juice, they convert some current into heat.
its better to just drive 2 leds with a current limiting resistor. [noparse]:)[/noparse]

Phil Pilgrim (PhiPi)

The purpose of the series resistor is to limit current. Because the forward voltage of an LED at a particular current can vary from LED to LED, it's never a good idea to drive an LED with a low impedance voltage source. If you do (and don't end up destroying the LEDs outright or reducing their lifetimes), you will likely be disappointed with the variation in brightness among them, due to the wide range of currents they will be drawing.

-Phil

DosManDan

Thanks Phil and FireHopper!

I'll wire two·LEDs in parallel and calculate the resistor needed.·I just need to make up 150 pairs...yuk.

I appreciate all of your help,
Dan

StarMan

Wire them in series (with a 51 ohm resistor), not in parallel.· With your 7.2V power, you'll get slightly less than 20 mA through the circuit but you probably won't notice any loss of brightness.

·R= (7.2 - 3.2 - 3.2)/.020=40· --> next size up is 51

Chris I.


Post Edited (StarMan) : 5/28/2008 12:41:25 AM GMT

Phil Pilgrim (PhiPi)

Ditto: SERIES, not PARALLEL. LEDs wired in parallel, because their forward voltages may vary, will almost never achieve equal brightness levels. Moreover, the dominant one (i.e. the one with the lower forward voltage) could prematurely fail due to overcurrent.

BTW: The next size up from 40 (5% values) is 43. However, use the "typical" or "maximum" forward voltage for this calculation, not the "minimum". It's best to start with a high resistor value and work your way down, until you measure the current level you're looking for.

-Phil

Post Edited (Phil Pilgrim (PhiPi)) : 5/28/2008 2:10:55 AM GMT

DosManDan

Oooh thanks!

Good thing I read this, I was about to go start wiring it up. In series it is!

Thanks again, I do appreciate it.
Dan