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Infra red LED connection

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RUDRA

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Hi everyone,
I am building an infra red remote control I have finished building the decoder and the encoder. To transmit the control signal I am using 3 high output infra red LED. Now the question is what is the best way to connect the infra red LEDs?
Currently I am driving the 3 LEDs in parallel configuration with a single 5 ohm resistor and the power supply is 5volt. Why do I want to drive more than one led? because I want more power output.
Thanks in advance for help.
 

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LEDs each have a slightly different voltage. If they are in parallel then the one with the lowest voltage hogs most of the current and burns out, quickly followed by the remaining ones. You are very lucky if they each have the same voltage because they were made at exactly the same time and from the same wafer. A current-limiting resistor should be used in series with each LED.

LEDs in series all share the same amount of current and work fine. They need only a single current-limiting resistor.
 
Yes, you don't want to put LED's in parallel - either use seperate resistors, or put them in series. Check my PIC tutorial for an example, where I use two LED's in series (you can't do three in series off 5V).

I don't know what you're using to encode it?, but presuming it's a micro-controller (as it should be really), you can generate the carrier directly, again check my PIC tutorials.
 
Thanks for the replies,
now for the questions
# how did I come up with 5 ohm? well the LEDs that I am using can handle about 1amp for very short duration.My carrier frequency is 40KHz which is modulated by pulses that are 200 or 500 micro second long. So in short I guessed that since the LEDs are on only for very short duration its ok to use 5ohm.
# 5volt is the dc supply voltage since the circuit contains some ttl ics.
# I am not using any microcontroller everything is hardwared.

Anyway I will put a resistance for each of the LEDs as everyone has said.

I think I am extremely lucky because just before reading all of your posts I connected 6 LEDs in parallel.So far all 6 seems to be ok.
thanks again
 
actually, you need to subtract the Vf of the LED before applying ohms law so your LEDs were seeing 600 mA (assuming your LED's Vf is 2V).

I would not drive your LEDs at 1A, that's just asking for trouble. can you point us to a data sheet for the LED?
 
Hi Philba
I understand what you are saying yes I substracted the LED voltage from 5 volts and divided it by 5ohms to get the LED current (I didnt simply divide 5volt by 5ohm). I used 5ohm simply because this is the smallest resistance I have.

What I meant by driving the LED with 1 amps is these LEDs can handle upto 1amp in short bursts.That doesnot mean that I am driving them with 1 amp.Anyway here is a link to the LED and a related article.
http://www.rentron.com/Fire-Stick-III.htm
**broken link removed**
 
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I would probably use the peak forward current as an upper limit. They don't say in the spec but that 100 uS pulse is for a very low duty cycle. The peak current is with a 50% duty cycle.

It may be worth playing around with higher current as the the LED at 1A has >7X the "brightness" as at 100 mA, They don't say it explicitly but I think figure 3 shows that surge current of 1A is .5% duty cycle. Note also that as the current goes up, so does Vf. At 1A, Vf is 2.6 while at 100 mA, Vf is 1.35. You should figure out what your duty cycle will be and pick a current that falls within the curves listed in fig 3.
 
the other thing to consider is the myth of pulse driving leds yields higher brightness.

google returns several results on this subject; from amature experimenters to seasoned researchers. the summary is, older generation leds would benefit from it, new generation leds don't. so all you might gain from your additional current is a shorter life on the led and wasted current from your supply.

oh, the firestick on Rentron is using a 9v battery to power the led, not 5v ... they use the 5.1 ohm resistor (with the 9v supply)... you'll need to reduce that down if your supply for the led is only 5v.
 
The circuit designer is crazy to operate the IR LRDs with such a high current. But he does say to use the circuit only for a moment then let the LEDs cool. He also says to have spare IR LEDs handy to replace the ones that burn out.
 
justDIY said:
the other thing to consider is the myth of pulse driving leds yields higher brightness.

true, but the datasheet very clearly shows significant brightness at the high current. It's not proportional - 10X current gets you 7X gain. That point in the datasheet is very misleading for the inexperienced though that probably wasn't the target audience...
 
With visible light, it is definitely much better to use DC drive, but when building sensor "systems", pulsing the LED gives you a definite S/N benefit.

https://www.sbprojects.com/knowledge/ir/ir.htm
In any case, it looks like most IR protocols end up having 25% duty cycles or less - the 38KHz modulation gives you 50%, and the bi-phase gives you another 50% multiplier. Pulse durations are 1/38KHz = 2.6E-2 msec. This means that the part is specified for ~400mA or so. Obviously recheck this if you plan on doing some other protocol.

Incidentally, about putting the diodes in parallel, there seems to be a ~.4V spread on the forward voltages. At a current of 400mA, the forward voltage is ~2V, and if you look at the V/I curve, there looks like a knee in the plot. Looking at the slope of the curve at that point, a .5V difference can cause a .4A change. This .5V/.4A=1.3 Ohm "dynamic resistance" will balance out the currents a little bit - but putting in little more resistance is a good idea. A +-.2V difference roughly translates into a (.2V / "1.3Ohm") = +-150mA difference in currents.

And finally, power dissipation: 400mA * 2.4V (worst case) * 25% (duty cycle) = ~250mW which is actually above the maximum power dissipation of the device. In normal use, there are delays between pulses, so this is probably okay, but at these settings, the thermal resistance value predicts a pretty insane (.24W*350K/W) = 84Deg. C temperature rise over ambient - i.e it's going to smoke if it continously transmits. To keep the junction temperature under 100C, we should limit the temperature rise to ~60C or so - so we should somehow scale the power back by (84-60/84) = ~30%, so one quick fix is to take the 400mA number and push it down to ~300mA. Obviously this is conservative - LED forward voltage also goes down a couple tenths and you can heatsink the LED a bit as well - but it's guaranteed to work.

James
 
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good write up. A couple of points.

Some IR protocols go for several seconds and a lot of units repeat the whole thing 3 times. so the duty cycle computations are pretty important.

I'd target the IR temperature to less than 60C. especially if this is a hand held unit or it can be otherwise touched.
 
just for grins, check out some of the high-power IR leds Roithner has to offer!

LED920-66-16100, IR, 920 nm, 650 mW at 2.0 A, Uf: 5.5 V, +/-60°, TO-66 ... that's an 11 watt LED with 650mw of radiant energy output! ow my eye!

http://www.roithner-laser.com/LED_HP_multi_chip.html
 
well, to be fair, those are IR Lasers and the duty cycle for their max current is .5% still , pretty impressive piece of thechnology
 
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