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prisonier #1 21/10/2010 - 15h39

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Hi all,

I would like to know if those schéma are the same. In fact, i would like to use only one resistance.

Posted Image

Thanks for your answers.

Bye
JB
dewey1 #2 22/10/2010 - 16h01

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Short answer: No!

Need more information. What LEDs and what is the power source?
It will be a 3 LED series wired if only one resistor is wanted.

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prisonier #3 22/10/2010 - 16h37

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thanks a lot. I won't use a series system because i want to use 1.5V battery with 1.5V LED. Do i need a resistance ?

I want to use 70mA with 100mA nominative current. Does it works ?

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Interuptor K is in the case battery 1 is empty and i'm flying. I don't want to change battery when i am flying....
dewey1 #4 22/10/2010 - 17h40

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First thing is your battery symbols in your diagram are wrong. Use the schematic that I show for parallel LEDs.

You need to have a battery supply of greater than 1.5 volts.
When your battery drops to 1.4 or less the 1.5 volt LED will not function.
Typical alkaline 1.5 volt batteries are considered "dead" at .8 to .9 volts.

Use a max current of 75 mA, not 100 mA. You should not operate at maximum ratings.

So if you use 2 AA for 3 volt supply, each LED resistor at 30 ohms will give 50 mA per LED when the batteries are new. That is a total of 150 mA. As the batteries run down to about 1 volt each, your LED currents will be about 17 mA, which may be to low for operation.
prisonier #5 22/10/2010 - 18h57

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Sorry for the symbol, in France we are used to use this one.
So, i will use:

2 AA batteries => 3.0 V
3 IR LED 1.5V => 1.5 V

What is the minimum current at the LED works ?

If it's 25-30 mA, i can use 22 Ohms. The minimal tension will be 1.025V for one battery no ?

I x R = U (total)

0.025 x 22 = 0.55

U (total) = U(battery) - U (LED)
U (total) + U (LED) = U(battery)

0.55+1.5= 2.05
2.05 / 2 = 1.025

Am i right ?

Wich lifetime can i expect for this system ?


Thanks a lot for your answers dewey1 you are helpfull.
Edited by prisonier on 22/10/2010 at 18h58.
dewey1 #6 22/10/2010 - 22h01

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Here is a spec sheet.

http://www1.duracell.com/oem/Pdf/others/ATB-5.pdf

With 150 mA total you may get about 13 to 15 hours of operation. Educated guess.

3-1.5=1.5
1.5/22=68 mA
68x3=204 mA total load.

The best method and cheapest is with a 9 VDC power adapter and 50 mA constant current supply circuit like this;

http://forum.free-track.net/index.php?showtopic=2666

Batteries get expensive after awhile!
What IR LEDs are you using? Hopefully the SFH485P.

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