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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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Give it a shot, just whatever gets you good solid comms.
thanks for the schematic, really helpful. One thing: where did you get the "in every 500mS period" from? That could be a problem for me, and not only for me i guess, since you sometimes need to send something like 3 or 4 signals per second.
I interperated 0.5 as half a second eg. 500mS but it is in fact the ratio/dutycycle.
The Absolute Maximums section of your datasheet has in Peak Forward Current parameter has some conditions. It shows tp/T = 0.5, which refers to a duty cycle of 50%, on half the time, off half the time. It also gives tp of being 100 us, which means the T would be 200 us for a frequency of 5000 Hz.  I was making an estimate that half your bits in a serial data stream would be on, and half off to allow you to drive at the 200 mA level. I finally went back and read your IR-serial tutorial, and realized you are modulating the LEDs with PWM so that the recievers would respond to them. So as long as you have a PWM signal above 5000 Hz (which you are probably running around 40 kHz or so) you should be fine driving the LEDs close to the 200 mA rating.

That was all very complex and a little confusing. I've tried to sum it all up with a schematic.


Note: Because of the dutycycle a 1/4W resistor will be fine as the 1/4W rating is a continuous rating. There is also a small voltage drop across the collector/emmiter of the transistor at saturation (usually 0.4-0.8V) but I usually leave this out and call it a safety margin.

Just read the datasheet, to achieve max range this LED can be pulsed at 200mA for 100mS in every 500mS period but unless you get the timing right the magic smoke will be released from the LED.

In the Absolute Maximum values section, the If forward current is a max continuous value, what to shoot for if you have the device constantly on as an illuminator. Note the Ifm value of 200 mA, combined with a tp/T duty cycle of 50% and a tp time of peak of 100us for a pulse. So if you are pulsing the LED (like you are in using it for IR communications) then you can drive it a little harder. And still stay under the Power Dissipation of 210 mW (200 mA If x estimated 1.6v Vf = 320 mW, but only half the time, so 160  mW)

So to get that if you have a 5 volt source, minus the 1.35 Vf forward voltage drop of the LED, equals 3.65 volts. This 3.65 is what would be used to calculate the limiting resistor, so 3.65v  / 200mA = 18.25 so use n 18 ohm standard value. BUt only if you are going to pulse the LED in communucaitons. And note that an MCU pin will not deliver this 200 mA current, a transistor or darlington driveris needed, which adds it's on voltage drop into the equation. 

You know I didn't even realise forward voltage was a voltage drop. I often wondered why LED resistor values were so high (compared to what I expected tehm to be)!!

Ok i get it, so it's gonna be: output goes to the base of the NPN, collector gets the pwm signal at 50% and then there will be a 18 Ohm resistor in series with the emitter and at last the LED.

Thank you for your anwsers :0) 

oh wait one thing... 3.65V * 0.2A =  0.73W ...does that mean i have to use a 1W resistor? (ps: what did you mean by standard value?)

Yep, probably need a 1/2 watt or at least a quarter watt resistor (LED only takes 210 mW, which was calculated above).

Standard values for 10% resistors are multiples of 10 12 15 18 22 27 33 39 47 56 68 & 82 . 

Not sure what you meant by the "collector gets the PWM signal at 50%" but all that is needed is the serial signal driving the base of the transistor. Might need an inverter somewhere, as adding the NPN transistor inverts the serial signal. Maybe : serial signal feeds a PNP, which drives the base of the NPN which drives the LED. The PNP inverts the signal, which is re-inverted by the NPN to go out as it was fed in.Or just NPN to NPN if that is what is in your component box.

wait wait...i don't get why a quarter resistor is ok...isn't the power passing through the resistor 0.73 W(as i wrote before)? How come a 1/4 W is ok then? 

Sorry for my ignorance :( but i still don't get what you mean by 10% resistors and their values (unofrtunately i don't study these things at school). The only thing i can think of is... tolerance (if that is how you call it)