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DPRG: IR woes...

Subject: DPRG: IR woes...
From: Ferialb at www.dci.co.ir Ferialb at www.dci.co.ir
Date: Sun Mar 14 05:59:20 CST 1999

 Sorry, I have deleted your first message and can't remember you
explanations, but here are what I know:
 LED current must be about 10 to 50 mA (depending on its size), they
have a max. pulse current of about 100 mA. LED voltage is usually about
1.5 V, thus using a 5V supply, you need a resistor about 168 ohm. If I
remember right, longer leg of LED is positive input (Anode) and shorter
one is negative input (Cathode). On reception of IR, receiver will only
change its leakage current, thus receiver must be reversed biased using
a resistor (about 1 Meg). Connect receiver's Cathode  to +5V using 1M
resistor, and connect its Anode to ground. HTH

Best Wishes,
 Hamid Reza Badili
  mailto:Ferialb at www.dci.co.ir

DP>I assume I'm doing something drastically wrong (yeah, like putting way
DP>too much current through the LED perhaps?) Will some kind soul please
DP>explain how this should be attached?

DP>I'm just glad these are LED's and not microcontrollers.

DP>Running out of CD tracks to play and components to burn,


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