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Build LED Lights

Started by branx86, April 11, 2017, 11:51:46 AM

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branx86

 (Voltage of LEDs x How many LEDs = Voltage Drop) White LEDS  3.6x3=10.8 then (Power Supply Voltage - Voltage Drop =  additional Voltage drop) 12v-10.8=1.2
  (additional Voltage drop / milliamp = ohms)   1.2/.025ma =48 ohms

As an example, for a 12 volt light, you can run a maximum of 3 white LEDs in series at full power (3.6 x 3 = 10.8 volts drop).  Subtract this from your supply voltage of 12 volts to get the additional voltage that must be dropped (in this case, 12 - 10.8 = 1.2 volts of additional drop needed).  In this case, 1.2 volts of additional drop / .025 amps (25 ma) = 48 ohms.  Use the next highest value of resistor available, 50 ohms.  You must also be sure the resistor can handle enough current.  Volts x Amps = Watts; resistors are rated in watts.  So in this case, 1.2 volts x .025 amps = 0.03 watts.  A 1/4 watt resistor will work fine, but if you run a second string of 3 LEDs in parallel, each string would need its own 50 ohm resistor. It's important that each string has its own resistor....putting them in parallel with a single resistor is bad practice.

http://www.otherpower.com/otherpower_lighting_leds.html