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IR LED overheating

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giftiger_wunsch

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Hi guys,

Simple question: I have recently been experimenting with a couple of simple circuits which use 9V batteries to transmit or receive an IR signal, but I have noted that the IR transmitter overheats very rapidly if on for more than about 2 seconds. I have also noted that most remote controls appear to solve this problem by flicking the transmitter on and off rapidly.

My initial thought about how to do this would be to use alternating current instead of direct current provided by the battery: since it's an LED, the forward current would power the transmitter, while the reverse current would be resisted and have no effect, causing the transmitter to flicker on an off at the same frequency as the current. Is this a valid solution or is there a hole in my logic?

Thanks in advance.
 
In most circuits, IR led's are turned on and off very quickly (38-40 KHz), but pulsed DC is used, not AC. An led of any sort doesn't usually like to be reverse biased, which is what would happen during the negative portion of an AC drive waveform. Any more than a few volts of reverse bias will probably destroy them.

You should post a schematic of the circuit you are using. It would help us to help you find out why your led's are overheating.

Jeff
 
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Hi poisonous_desire (literally translated from giftiger_wunsch) :)

You missed to tell us what forward current you apply to the IR-LED.

Most of them can stand a continuous forward current of 100mA (surge current 3A)

If you use an IR-receiver chip you should modulate the IR-TX-LED with the appropriate frequency. If a continuous IR-signal is received the built in amplifier reduces sensitivity automatically. (which you might not want because of reduced range)

Boncuk
 
You should post a schematic of the circuit you are using. It would help us to help you find out why your led's are overheating.

Jeff

Drawing a schematic will probably be more hassle than it's worth (especially as I haven't drawn one in years): it's a very simple circuit; 9V battery (in a case with a switch), IR LED, and 10-ohm resistor connected in series.

Boncuk, I think that should also answer your question. By the way, you're the first person to translate my handle rather than asking me what it means :p I think the avatar sums it up quite well :D


There's no chip involved, I got a very cheap (£2) little kit to demonstrate how infra red technology works.
 
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***** N00B DISCLAIMER *****

10 Ω resistor?

I don't know about IR LEDs, but for a normal cheesy LED there's something like a 2.2v forward voltage. So with 9v and 10 Ohms...

(9v-2.2v) = I(10)
6.8 = 10I
.68A = I

I don't now how many amps a 9V battery can produce but it's going nearly full-bore into your LED.

With a 250Ω resistor you see 6.8 = 250I -> 27mA of current which is approaching the limit of normal LEDs.
 
You are lucky your LED isn't kaput.
 
No IR-LED likes a constant forward current of 770mA!

The forward voltage drop is 1.3V. Using a (fresh) 9V battery the current is calculated: If(A)=(UB (9V) -Uf (forward voltage))/R (Ω).

In your case it is If = (9V-1.3)V/10=0.77A! Allowed forward current If=0.1A (100mA)!

As Crutschow already said: You're lucky the LED is still alife.

Boncuk

P.S. Your avatar shows a modified "Aladins lamp" :D
 
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I assembled the transmitter according to the instructions provided with the kit, which also warned that it may overheat if switched on for more than a couple of seconds. I also read on another thread that 1A pulsed-DC is used in TV remotes etc. to send a signal, so I figured my setup was fine except that the current is constant rather than pulsed...

As for the comment on the limit of normal LEDs, this also seems to be reflected by the receiver circuit; again I made this circuit to the specification of the guide: it's the phototransistor, a 100Ω resistor, and a standard LED (simply to indicate that the signal is being received) in series. It was meant to be powered by a 9V battery as well, but it's currently being powered by a wall wart with 6VDC output since I didn't have a spare battery.

Could someone confirm that there is too much current reaching my IR LED? Should I be using a higher resistor?


P.S. Your avatar shows a modified "Aladins lamp" :D

It does! With poison coming out of it. Hence Giftiger Wunsch, poisonous wish :D
 
Pulsed current is normal for IR LEDs, it's only on 30% of the time so the average current won't do it any damage.
Why not modulate it? What's it for?
 
Sorry, I'm an electronics noob. What do you mean by modulate it?

And what sort of component would I use to produce pulsed DC current?
 
Hi,

if you are unable to calculate for the correct current limiting resistor here is the value for 6V supply: Use 47Ω.

At a supply voltage of 6V and a resistor value of 10Ω you're still miles away from the required value having reduced the forward current to 470mA versus 100mA max.

P.S. Aladin never had bad wishes. :)
 
Hi,

if you are unable to calculate for the correct current limiting resistor here is the value for 6V supply: Use 47Ω.

At a supply voltage of 6V and a resistor value of 10Ω you're still miles away from the required value having reduced the forward current to 470mA versus 100mA max.

Could you explain how you calculated these values? I'm somewhat confused about how you arrived at the 47Ω and 470mA values.

Boncuk said:
P.S. Aladin never had bad wishes. :)

His lamp never had poison oozing out of the spout :D



Thanks to both Tony Ennis and blueroomelectronics for the information about 555 chips. The page I looked at sold them for 47 cents each but with a minimum order of 1,000 of them; hopefully I'll be able to find somewhere which sells them in considerably lower numbers :D
 
Thanks, will do.

EDIT: Hmm, the site won't process my order unless it's $20 or over >________<
 
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You probably need other supplies too :D

Maybe some proto boards, resistor and capacitor assortment kits, wire kits, a few 555s, some counter chips, etc.

$8 shipping up to $50, too.
 
Of course any cheap TV remote will make for a hand IR transmitter. Of course since the OP hasn't mentioned the purpose of the transmitter it's difficult to offer useful advice.
 
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