Resistor value for LED

What the data sheet tells us is that the LED's absolute forward current level is 50 mA continuous, or 100 mA if fed a low-duty-cycle high-speed pulse train. More current than that will damage it.

Running an LED near to its absolute-maximum rating is, I believe, likely to shorten its lifetime significantly. I usually prefer to keep them at half-maximum or a bit less. This LED's brightness (which is considerable) is rated by the manufacturer at a forward current of

20 mA, so that's what I'd suggest using.

At that current level, the LED is going to have a forward voltage of

2.3 to 2.6 volts. Let's use 2.5 volts as a working figure (it won't matter very much if the particular LED you use is a bit higher or lower).

So... a 5-volt supply, with 2.5 volts appearing across the diode. This means that we must choose a resistor which will have (5.0 - 2.5) = 2.5 volts at a forward current level of 20 mA.

You can solve for this by using Ohm's Law, which states that E=I*R, or (by solving for R) R=E/I.

R = 2.5 / .02 (if we do it in volts and amps) or R = 2500 / 20 (if we do it millivolts and milliamps)

In either case, R works out to be 125 ohms.

Let's figure out the power dissipation. P = E*I. We know that E (across the resistor) is 2.5, and I is .02, so P = .05 watts of heat dissipated in the resistor. It's generally a good idea to "de-rate" resistors by 2:1 (that is, don't ask them to dissipate more than half of their rated capability) to ensure long life. In this particular case that's no problem, since the smallest standard leaded resistors are 1/8 watt.

So - choose a resistor value in the neighborhood of 125 ohms (whatever you have that's closest, plus or minus 10-20%), in whatever size is convenient (1/8 watt or larger), and you should be fine.

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Notice in the data sheet, that the forward current (If) listed for those maximum voltages is 20 mA. You need to drop half of your voltage across the resistor (~2.5 V), so use Ohms law.

R = V / I R = 2.5 / 0.02 R = 125 Ohm

The resistor power is calculated thus: P = V^2 / R P = 2.5^2 / 125 P = 0.05 W

so a 1/8 W resistor or larger will do just fine.

Since the LED will handle up to 50 mA continuously (although I would not recommend operating it like that) the resistor value can go as low as 50 Ohms. The LED will PROBABLY light up dimly with as little as 10 mA, so the resistor value can go as high as 250 Ohms.

Typical operation is with a current of 20 mA. I am surprised that they say

2.3 volts, because the normal forward voltage drop of a red LED is 1.8 volts, and that's what I'll assume.

You need a voltage drop of 3.2 volts (to take 5.0 down to 1.8 volts) at 20 mA = 0.020 A. Ohm's Law:

R = E / I = 3.2 / 0.020 = 160 ohms.

They don't make 160-ohm resistors, so use either 150 or 220.

Now... Will an ordinary 1/8-watt resistor handle the current? Let's see.

P = E I = 3.2 * 0.020 = 0.06 watt

Yes.

Thanks for the excellent replies!

Looks like 68 ohms would be a good choice. Use at least a quarter watt resistor.

Um, you're assuming the datasheet is *wrong*?

I just measured a 635nm red (that's less orange) of similar brightness, and it came to 2.1V @ 20mA.

Generally, forward voltages seem to be creeping up (for the same color of emitted light) as efficiencies increase.

Best regards, Spehro Pefhany

--
"it's the network..."                          "The Journey is the reward"
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I'm astonished, nay FLABBERGASTED, that none of the regulars on this newsgroup told this fellow to use a microcontroller to do the job.

{;-)

Jim

I wonder why. I thought the voltage was related to the wavelength by a basic law of physics. Is the internal resistance also increasing, I wonder?

AVR or PIC ? :)

Yeah, PIC or 555. They seem to be the solution to every problem. GG

Yeah, PIC or 555. They seem to be the solution to every problem. GG

they say

The older GaAs LEDs were lower, maybe below 2VDC. But I took a bunch of indicator grade red LEDs, maybe just a few mCd, and measured them at

20mA. They were all 2VDC or even more. Last night I took a single high brightness red LED and put 25mA thru it, and it dropped 2.3VDC. All the newer GaInAs whatever LEDs seem to be well over 2V drop.

I put this high brightness red LED into a '50s vintage 'jewel' indicator. These have a red glass multifaceted 'jewel' in a threaded brass base a half inch across, which was mounted on the equipment panel with a #47 lamp behind it. The red LED looks really good behind the jewel. Maybe a bit too bright at 20mA, tho. I looked into it and had spots in my vision for a minute afterwards. :-P

at 20

Again, you've missed the boat. The 160, 1.6k, etc., 5% resistor values are no harder to get than 1k or 2.2k. Standard resistor values can be found here. _Read_and_remember_them_.

see.

they say

1.8

Yeah, and as the number of elements increase: gallium, indium, antimony, aluminum, arsenic, rat poison...

And that doesn't even get into the various phosphors they use to get white LEDs.

reward"

I know what the standard values. Some people have to shop at Radio Shack, which has 220 but not 160 or 150 available singly. Radio Shack has assortments that have 150 but not 160. Admittedly, "they don't make" was an erroneous oversimplification.

I have actually not seen 160 (or 1.6, 16, 1600, etc.) very often at all. Not all of the 5% standard series is equally widely available. Long-established good practice, even with 5% resistors, is to prefer the values in the 20% series first, then the 10% series, then the 5% series.

The 20% series goes: 1.0, 1.5, 2.2, 3.3, 4.7, 6.8, 10.

I spent a long time writing construction project articles for magazines and developed a fairly strong aversion to hard-to-get parts. That, and I started designing things back when resistors were 10% carbon composition!

Make fun if you like, but it's way more power efficient than a resistor, even if it's not cheaper.....yet. ;-) AFAIK, even an exceptional SMPS couldn't do it any more efficiently. IME a nano-watt PIC running at

32kHz should be ~99% efficient driving the LED with at least 10mA average current.

How difficult would it be to beat that using another method?

By your definition a resistor is 100% efficient, then?

Best regards, Spehro Pefhany

--
"it's the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
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this

resistor,

SMPS

I don't know what you mean. I was thinking the dropping resistor approach was much less than 50% efficient since the resistor ends up dissipating more power than does the LED. Is that not correct?

I was thinking that by PWM'ng the LED and using a bit of inductance to limit the current surges thru the i/o pin, would be much more efficient. A nano-watt 16f88 claims to dissipate a max of 40uA at 5V running at

32kHz. That's only 200uW of power used by the PIC. The LED dissipates many times (at least 100?) as much power. What am I doing wrong?

Yes.

The increased efficiency would be because of the inductor and switch(s) or switch + diode, not the control circuitry. I didn't see either taken into account in your efficiency calculation. They'll not likely be very close to being ideal.

Best regards, Spehro Pefhany

--
"it's the network..."                          "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

The learning curve. When someone asks how to turn on a LED. There was a space constraint to the project box as I recall. Most of everyone thinks of a DC source,1 resistor and the LED. Simple enough. Someone posts the actual calculations, E=IR and W=EI . Without knowing about E=IR or W=EI, you expect to jump into Microcontrollers? Writing uC programs ? How long will it be before you research about PICs, build or buy a pic programmer, learn to program a PIC. There's a big jump between a transistor battery and a PIC project, don't you think? To run, you 1st got to learn to crawl & stand up.