I bought 5 LEDs from OddOne. How do I add a current limiting resistot to the group to maximize battery life? Will one 330 Ohm resistor in series do the trick?

-- Customer of OddOne (@ .), October 01, 1999

Answers

You are right on target. One resistor in series will do it, AND, it must be in the circuit to prevent frying the LEDs!

-- Bob (truthseeker@seektruth.always), October 01, 1999.

what value? 330?

-- (@ .), October 01, 1999.

How did you wire the 5 LEDs? In series or parallel? What voltage source do you plan on using? 3v? 12v?

-- brett45 (brett45@bigfoot.com), October 01, 1999.

I KNOW the Odd One left a ref for the spec sheet for the LED's around here SOMEPLACE. Has anyone seen it??

Chuck

Oh, I'm SO absent minded, I'd probably lose my head if it weren't ........

C

-- Chuck, a night driver (rienzoo@en.com), October 02, 1999.

How much air do I put in my tires? You need ELECTRICAL SPECIFICATION information. I'm assuming the LED operates at a specific voltage, and as you pump more current through it, it gets brighter. Too much current and it burns out. WHAT ARE THE VOLTAGE AND CURRENT SPECIFICATIONS?

NOTE: 330 OHMS MEANS NOTHING. YOU NEED TO KNOW THE ABOVE INFORMATION AND WHAT THE VOLTAGE OF YOUR POWER SOURCE IS.

Also, if you have more than one, will they be in series or parallel? In series, you would need just one resistor. In parallel, you will need a resistor for each.
Furthermore, it would appear the terminology "pullup resistor" is used inappropriately. I would think it would be "series" or "dropping" resistor.

When you can answer the above questions, then the proper size of the resistor can be determined. R = ( E_source - E_LED ) / I
where
R is the resistance desired
E_source is your battery pack total voltage (assuming in series)
E_LED the voltage of one LED if they're in parallel, or total if in series
I
Rated current of the LEDs

Some caveats -- If in series, if one blows, then the lights go out. This would be the simplest construction, assuming your battery voltage is sufficient. In parallel, with just one resistor, one or more may exceed current rating due to slight variations in the voltage drop, then all will blow, because the one resistor will not limit the current sufficiently to the survivots. Ping, then ping,ping,ping... That's why, in parallel, you need a resistor for each. Run each resistor/LED series circuit independently off the same battery.

-- A (A@AisA.com), October 02, 1999.

For longest battery life, the ideal would be a source (battery?) with a voltage just a bit more than the (sum) of the LED voltage. This would give you a smaller dropping resistor than if the source voltage is higher. This minimizes the energy gone to waste in the dropping resistor(s).

OTH, the source and resistor have to be large enough so that tolerances in the source voltage, the LED voltage, and the resistor do not result in a current exceeding the LED specs.

-- A (A@AisA.com), October 02, 1999.

A,

I'm not sure I'm using your formula correctly. For an LED rated at 2.8 volts and a continuous current of 50ma, powered by a 6 volt battery I come up with 64 ohms, which sounds low. Here's how I plugged the numbers in:

R=6v-2.8v devided by 50ma

Was there a flaw in my math somewhere or is 64 ohms right?

-- Bokonon (bok0non@my-Deja.com), October 02, 1999.

voltage check:
battery source: 6v
circuit drain: 2.8 volts for LED + 3.2 volts for resistor.
If all in series, and the circuit current wanted is 50 ma (.05 amp)
R=E/I = 3.2/.05 = 64 ohm
or in ma and Kohm --- 3.2/50 = .064 Kohm (which is 64 ohms).
I'd be willing to hook it up and throw the switch, if the specs you gave are OK. i

-- A (A@AisA.com), October 02, 1999.

Thanks A. I'll try it out. If that's the case, then the 47 ohm resistor I'm using on two 3 volt set ups is way over what I need. It looks like a 10 ohm would be more than adequate.

-- Bokonon (bok0non@my-Deja.com), October 03, 1999.

BTW, if it helps (and it will!) the specs for the white LEDs are:

Forward voltage: 3.5-3.8 VDC, 4.0 VDC max (Their lifespan degrades at over 4 volts. Over 4.5-5, BANG!) Forward-bias current: 30 mA typical, 100 mA pulsed (10% duty cycle) Max. reverse-bias voltage: 50 VDC (over this voltage in reverse and they'll break down like a zener diode - result: BANG!)

If you're using an even multiple of 4 for the supply voltage (MAXIMUM!), wire that number of them in series and tie directly to the supply. 12 volts? Three in series. Got six? Two sets of three in series each. Just make sure the voltage doesn't exceed the multiple under any conditions. If it does, you'll need a driver or resistor. Drivers cost more and are more complex but offer more control and much greater energy efficiency, but a resistor is a cheap, one part solution for the problem if you don't need the efficiency.

If you're NOT using an even multiple, get as close as you can, figure oput how much current they will pull total, figure out how many volts would put 3.8 let's say across each, and use the Ohm's Law formula R=V/I (Resistance (R) in ohms = Voltage (V) in volts divided by Current (I) in amps) where R will be your resistor, V is the supply minus what they'll run on, and I is the current.

Example: Four LEDs, 9VDC supply. Two in series wants 7.6-8.0 volts, so let's assume that you want 7.6 to put 3.8 volts across each. 9 - 7.6 = 1.4. There's V, the amount of excess voltage you'll need to drop. Each set of two pulls 30 mA (in series their current drain is NOT cumulative. But you have two sets to tie in parallel and parallel DOES add the drain, so 30 mA per set times two sets is 60 mA. There's I. Your dropping resistor will be R = 1.4 / .06 = 23.333~ ohms. Go to the next HIGHER standard value (compensating for variations and tolerances so you don't toast anything) which is 27 ohms. Voila.

Since Watt's Law says that P=VI, (Power (P) in watts = Voltage (V) in volts times current (I) in amps) V dropped by the resistor is 1.4, and I flowing through it is 60 mA, the resistor needs to be rated for P=VI = 1.4 = .06 = .084 watt, so a 1/4-watt resistor will be plenty. Poof, four white LEDs in a two-paralleled-set-of-two-each-in-series and a 27-ohm 1/4 watt resistor will slam it on a nine volt battery. :- )

Confusing? Probably. [lol] But hey, it's a crash course in Ohm's and Watt's Laws and their application to the problem of not turning several dollars' worth of LEDs into little useless clear-epoxy-and- wire doohockeys. :-)

That crazed guy that SEEMS to think he knows just what the hell he just said, that people call...

-- OddOne (mocklamer_1999@yahoo.com), October 03, 1999.