# Getting the most from an LED

Hi guys. I have recently written the walkthorugh on how to transmit serial data with an IR led and it works fine. Now to the next step: i want to achieve the highest range possible with my LED, without using external aids such as lenses. (that is: only with electronics).

I have a TSAL 6100 from Vishay ( http://www.farnell.com/datasheets/93358.pdf ). If I am right, to achieve the most out of it, i have to get a current similar to the "peak current" value from the manual in the LED. But as for the voltage i guess i just need to give it at least a value equal to its forward voltage.

Problem is, i am not sure about the things i have just written... :) Is it really the right way to do it?

Suppose it is, i cannot give enough current to the LED with an output pin, so i think i'll be using either a mosfet or a standard NPN. But again, how do i exactly know what current i will get with it? As far as i know i have to measure the "amplification value"( hFe?) with my multimeter and multiply that to the current entering the base. Am I right?

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Yep, I was thinking of the LED power, not the resistor power dissipation. Both pass the same current, but not the same watts, so a 1 watt resistor would be best. THe only reason a half watt meight be ok, is that you are only passing that current part of the time, and estimating a data stream to be half 0s and half ones would give you an estimated half the power dissipated. But being safe would spec a 1 watt there.

Here is a web page listing what resistors are available according to what toerance they are :

http://www.elexp.com/t_eia.htm

You can see from it that 10% tolerance resistors only are mannufactured in certain values, 5% in more values, 15 even more. So if you want a cheaper resistor (10%) you have to get one close to the value calculated.

How do you get 1.35V from a 5V supply?

Don't you use resistors in parallel? Wait no, that increases current? I dunno, I'm so confused:X

haha my newbie book is sayin somethin about using resistors to divide voltage.

(Vin) X R2/(R1 + R2)

I don't get why, but everytime I've used a resistor with my learning lab, its only been a 1k resistor, nothin less, even when dealing with higher voltages, I don't get it :/

A voltage divider (two or more resistors) will only work accurately when each resistor has the same current flowing through it. If you divert some current in the middle (the divider point?), the top R will experience a higher current than the bottom one. And your careful calculation fails.

So use V-dividers only when you are not drawing a lot of current (relatively speaking) from it. If you must draw more current (as in your problem here), you should use clever voltage regulators that maintain the right voltage, regardless of the current drawn.

Maybe BoA is hinting at several diodes in series?

8ik

Yes, but then I realised that 6 is too many and 5 isn't enough.

Maybe you could use an appropriately biased transistor to drop the voltge...

But now you've brought you potential dividers, can you see a way of (ab)using the desired forward current as part of the equation? Is this what's known as Thevenin's Theorem? See we're into analog electronics and that's where I step aside and leave it to you.

I am not an electronics expert. Analog or otherwise. I have no idea how to bias a transistor. I just learn about this stuff on websites (mostly this one). I realized the misconception is common when I saw the poster make the same mistake as I made only a few days ago.

Thevenin Who?  Black boxes? Predicting their behaviour? Wow. Don't let Frits get wind of this!

Huh?

As detailled elsewhere (can't quite pinpoint it) LEDs cannot possibly exist as they defy the laws of physics. One of the physics-defying features is that they have no internal resistance and will therefore "suck" as much current as you supply (for a brief period until they burn out and produce lots of really smelly Zauberhafte Rauch).

Yes, I think it's a good idea to think of volts as being pushed by the source and current as being pulled by the device. You might "give" a device a certain voltage, but you "allow" it a certain current.

So, to prevent them burning out, you use a series resistor. It is this resistor which dictates the current allowed to flow through the LED by the formula V=IR (V=Voltage, I=Current, R=Resistance).

If you have a 1.35V source and your Imax is 100mA: V=IR becomes R=V/I = 1.35/0.1 = 13.5, so you need a 13.5 ohm resistor.

Now ask me how you get 1.35V from a 5V supply.

Don't forget the voltage drop across the led (most diodes have a drop of 0.7V but I'm not sure about LED's)  So I think it would actually be R = (1.35V-Vdiode)/0.1A.

Gabe

"lots of really smelly Zauberhafte Rauch)."

:P

/Nick