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LED help - What resistor do I need to make 3.7 LED's work on 12v's?

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  • LED help - What resistor do I need to make 3.7 LED's work on 12v's?

    i have 2 led's.. rated at 3.7 volts..
    since cars are 12 volts... what resistor would i need.. and i need to know in ohm... since they don't say in volts.. and i can't remember from school.. so anyone?

    please
    u should have died in that wreck. natural selection just isnt what it used to be...
    http://www.cardomain.com/id/sunsetprobe


    shoot down a ufo.. they have a sweet drivetrain

  • #2
    a 470ohm will do 12:5 or maybe less...i am not sure, use the little chart at radioshack.

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    • #3
      It will depend on the LED itself. Remember from school V=IR. You need to know how much current that certin LED your interested in will draw.

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      • #4
        Originally posted by WSU_EE_PhD
        It will depend on the LED itself. Remember from school V=IR. You need to know how much current that certin LED your interested in will draw.
        If I remember right, isn't it:

        V = I^2 * R?
        Jason
        Back in /X\IPOC Land!

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        • #5
          Here is the correct formula, just plug in the numbers:

          R=(12volts -LED voltage required)/(current in milliamps)*1000

          You need to get the MA to complete your formula. In my case it was a 2v led, that was rated 20ma, so it came to 500...

          Joe

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          • #6
            Ok guys, assuming you want 25mA of loop current (you can modify the 25mA depending on what you're using for an LED), the circuit is as follows:

            Sum of voltages in a loop = 0 or Sum of voltage drops in circuit = Battery Voltage

            +12V = Vdrop(LED) + I(loop)*R(resistor) = 3.7V + 0.025mA * R

            or

            R(min) = (V(car) - V(resistordrop))/I(LEDmax) = (12-3.7)/0.025 = 330ohm MINIMUM

            and, since P(resistor)=I(resistor)^2*R(resistor)=0.025^2*330 = 0.20W<1/4 Watt, so 1/4 Watt rating is just fine.

            Now, LEDs vary in their brightness, so you may want to play around with some values till you get the right one. But resistors are cheap, so you can try it and see. Mind you, you will see a bit of variance if your supplied car voltage goes higher or lower, so you may want to set the resistance a little higher.
            Mike's White '94 Probe GT - K&N Air Box Filter | NGK 8mm Wires | NGK V-groove Platinum Plugs | Aftermarket Power Moonroof | Sony 10 CD Changer | Spring Lowering Kit | DEAD TRANNY AND AXLE BY MEINEKE | more to come...
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            • #7
              I'll make it simple

              1...4 kOhm. Lower resistance = brighter LED, but don't use bellow 1k cuz the current will be too high and the LED will burn out.

              Voltage in the car can reach up to 14.something Volts.
              '95 DSM Eclipse GST
              '96 Ford Probe GT - R.I.P.

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              • #8
                ok... im still lost. that and my grandma just had a stroke.. and i cna't thik

                but anyways... the led is 3.7 to 4 or 4 somthing..
                so what would i use then?

                thanks
                u should have died in that wreck. natural selection just isnt what it used to be...
                http://www.cardomain.com/id/sunsetprobe


                shoot down a ufo.. they have a sweet drivetrain

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                • #9
                  Originally posted by jbr12


                  If I remember right, isn't it:

                  V = I^2 * R?
                  I'm pretty sure it is V=IR. I am in ALOT of trouble if it isnt

                  V in volts
                  I in amperes
                  R in ohms

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                  • #10
                    It is also important the u hook the resistor up in SERIES and not parrallel. Meaning hook up the wire to one end of the resistor, then hook the other end of the resistor to the led.
                    Probe is gone.....

                    2000 monte carlo SS

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                    • #11
                      You need a resistor 3 times your LEDs resistance.
                      3 x (your Leds resistance) = what you need.

                      BY KVL the other resistor will draw 3/4 of the 12 volts will go to the larger resistor and your diode will get 3 volts.
                      KVL simply states that in a series circuit the voltage applied to that circuit must drop over the resistance of the curcuit.

                      Ohms law can be used to tell what the current draw will be
                      I = V / Rt

                      or in your case, which i doubt will matter.

                      I = 12 / (your LED's resistance + the other resistor)

                      this will tell you how many amps will pass through your LED and resistor so that you know what wattage rating of resistor to buy and if your LED can take it.
                      Ohms law helps here again:
                      P = I x V
                      P (wattage rating your LED and resitor need) = I x 12v

                      Good luck
                      Front and rear sway bars, Tokico Shocks, Racing beat eghuast, racing beat 17" wheels, Dunlop 9000 tires

                      -Friends don't let friends buy SONY!
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                      • #12
                        Just to throw a bone

                        You might consider a Zener Diode to maintain your voltage level....that'll help w/ the altering brightness of the LED.

                        bet you didnt think you'd get this much support heh

                        and yes to support the doctor: V = IR
                        Stock 96 PGT
                        Pioneer, Polk Audio, Rockford, Eclipse

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                        • #13
                          thanks alot guys.

                          finally someppl who help instead of telling me to search..
                          u should have died in that wreck. natural selection just isnt what it used to be...
                          http://www.cardomain.com/id/sunsetprobe


                          shoot down a ufo.. they have a sweet drivetrain

                          Comment

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