LED info - 12V battery wired to four 3V LEDs in series...

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Two resistors are connected as shown in the following diagram: The output voltage V out is related to V in as follows: As a simple example, if R 1 = R 2 then Any ratio between 0 and 1 is possible. Luminous flux lumen cd * sr = lm R = (V1 - V2) / I R = (4.5V - 1.7V) / .02 A = 140 ohms I = LED current (when wiring in parallel, multiply the current of one LED by the total number of LEDs) 20mA x 2 = 40 mA = .04A (we had been using 20 mA in our other calculations but since wiring LEDs in parallel draws more current I had to multiply the current that one LED draws by the total number of LEDs I was using. 20 mA x 2 = 40 mA, or .04A. Whenever using a resistor on an LED it should get placed before the LED on the positive electrode. If you wire a whole bunch of LEDs in series, rather than dividing the power supplied to them between them, they all share it. So, a
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Unformatted text preview: 12V battery wired to four 3V LEDs in series would distribute 3V to each of the LEDs. But that same 12V battery wired to four 3V LEDs in parallel would deliver the full 12V to each LED - enough to burn out the LEDs for sure! Wiring LEDs in parallel allows many LEDs to share just one low voltage power supply. Wiring in parallel only works if all the LEDs you are using have exactly the same power specifications. Do NOT mix and match different types/colors of LEDs when wiring in parallel. http://www.instructables.com/id/E9NFEFO2ZGEV2Z9QKZ/?ALLSTEPS The package didn't come with a 140 ohm resistor but it did come with a 150 ohm one. Its always better to use the next closest value resistor greater than what you calculated. Using a lower value could burn out your LED....
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