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 Post subject: Wiring LEDs to a Battery
PostPosted: 15.07.2002 18:10 
Hey. I'm thinking about making that Mouse Pad w/ the LEDs. I'm wondering, can I use a battery instead of the power supply from my computer? If so, what kind of battery would I use for 2 LEDs(blue LEDs but I would also like to know the same information for the red LEDs if it is different). Could I use a 9v or something? I plan on using the Superbright LEDs if that makes a difference. Thanks!

-Code


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 Post subject:
PostPosted: 15.07.2002 18:18 
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Just connect those two blue leds in series and then add a current limiting resistor.

http://www.metku.net/index.html?sect=vi ... _eng#kirc2

You just have to know the voltage threshold for the leds and the current they need. Usually for blue leds that is 3.5-4 volts and 20-30mA and super bright red leds 2.5-3 volts and 20-30mA.


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 Post subject:
PostPosted: 15.07.2002 18:35 
Ok, lets say I plan on using a small 9 volt battery with 2 Superbright blue LEDs. What resistor would I need? Sorry, but I'm new to wiring LEDs.

-Code


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 Post subject:
PostPosted: 15.07.2002 19:27 
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9 V - 2*4 V = 1 V

R=U/I

1 V / 20 mA = 50 ohms.


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 Post subject:
PostPosted: 15.07.2002 20:37 
Ok, I'm getting this now. Sorry, I should have read your basic wiring better before I asked this question. Now, when wiring 2 LEDS what happens? Lets say the LED is 1.6 w/ 20 amps. I want to wire 2 of them. What happens then? Remember I'm wiring to a 9 volt. So would the amps double and the voltage stay the same or what?

-Code


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 Post subject:
PostPosted: 15.07.2002 20:46 
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You still need to read that tutorial even better. ;) If you connect those leds in series as I just calculated it for you, the current will be that 20mA for every component on the circuit. Voltages will vary. If you connect components in parallel the voltage will be the same but the current will vary.

http://www.metku.net/index.html?sect=vi ... _eng#kirc1
http://www.metku.net/index.html?sect=vi ... _eng#kirc2


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 Post subject:
PostPosted: 15.07.2002 21:26 
"The voltage threshold for the LEDs are 3.8 V and the current 20 mA. Because the LEDs are in parallel, they are affected by the same voltage, but the current is doubled to 40 mA. The voltage affecting the resistor is therefore 5 V - 3.8 V = 1.2 V. The value for the resistor should therefore, according to Ohm's law, be R= U/I= 1.2 V / 0.04 A = 30 ohm. Because the E-series doesn't include a resistor of this value, we have to choose the closest match, which is 33 ohms."

Are you saying that 1 LED is 3.8 V or that both LEDs equal 3.8 V?

-Code


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 Post subject:
PostPosted: 15.07.2002 21:34 
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Because the leds are connected in parallel, there is same voltage affecting them both. This time 3.8 volts.


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 Post subject:
PostPosted: 15.07.2002 21:36 
Ok, I got it. Thanks for your help and sorry for bugging you on this. Wednesday I'll be getting all the items to built the GlowPad. Thanks again!

-Code


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 Post subject:
PostPosted: 15.07.2002 21:40 
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hehe, np. 8)


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