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Thread ID: 90270	2008-05-28 11:50:00	Powering IrLEDs and the Minority Report UI	Chilling_Silence (9)	PC World Chat

Post ID	Timestamp	Content	User
673438	2008-05-29 00:32:00	The link must have been fixed recently then, it was dead last night	Speedy Gonzales (78)
673439	2008-05-29 13:23:00	Hi Think I have found your problem. I have made the following assumptions based on your replies. 1. You are using a series circuit. Forget it use parallel. See # 2. 2. Source voltage is 3v, using 2x 1. 5 AA. To drive two leds at 20ma in a series circuit you require at least 4V. Not surprised to hear your current circuit just aren't workin'!! 3. Resistor calculations done by yourself and others are for 20ma forward current. The Specification Sheet clearly shows this is 100ma. Therefore based on the above assumptions, heres what to do: 1. Get two 18 ohm resistors and wire/solder one to the Anode of each Ir Led. 2. Wire/solder the two Cathodes together. This is your negative (-) connection. Wire a length of Black wire here too. 3. Now you should have a end of the two resistors left. Wire/solder together with a length of Red wire. This is the positive (+) connection. 4. Connect your battery and test. 5. You are now in business. Just a word of warning. I would ideally raise the voltage level, this would require a resistor recalculation, however. Batteries are not going to last long so think of powering from a power pack. Hope this helps. BURNZEE	Burnzee (6950)
673440	2008-05-30 00:27:00	That would make sense if he was using 3v. But if you read the first post again you'll notice it's actually 6v	Agent_24 (57)
673441	2008-05-30 03:39:00	From memory, LEDs (and diodes in general) use a constant forward voltage drop (the IR LEDs in the spec have a drop of 1.6V). If you connect enough of them in series, the total voltage drop will be more than the supplied voltage, hence nothing will go. As Burnzee said, you'd need to connect them in parallel so that the voltage drop is below the voltage supplied. If you're using 6V, you could connect groups of three series LEDs (for a total drop of 4.8V each) in parallel, and use a current-limiting resistor in series with the whole lot to absorb the remaining 1.2V. So to get a continuous current of 100ma through each LED, it'll have to be at most 12 ohms (for one series group). Or you can add a 12-ohm series resistor to each series group of LEDs, so that with 20 groups (60 LEDs) you won't need a theoretical 0.6 ohm resistor. The power pack option is looking more attractive...	D. McG (3023)
673442	2008-05-30 05:05:00	Thanks Guys Yes, 6v is the better option, I missed this on first read . My mistake . Trouble with a series circuit is the forward voltage drop across each Led is constant but the current isn't . After all, you may design it ok but slowly dying battery voltage guarantees this circuit is doomed to fail . Better with a power pack . Common mistake for computer geeks without electronic knowledge to make . In this case, a circuit would comprise of a 33 ohm resistor and 2x Ir Leds connected in series wired to a 6v power source . You are quite correct D . Mc G with 3 x Ir Leds a 12 ohm resistor would be required . Hope this helps BURNZEE	Burnzee (6950)
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