Series resistor

Discussion in 'General Electronics Chat' started by knoblame, May 24, 2005.

  1. knoblame

    Thread Starter New Member

    May 24, 2005
    i am trying to figure out how ohms law will change in regards to voltage in a series, does it affect the resistance. for example. if i have a source voltage of X and the voltage drops at a (LED) to Y. I have a current value of Z amps. How would I figure out the value of resistor at LED To achieve my Z amps. Do I have to take in affect that there was a change in voltage at the LED.


    R= V (would i need to figure out the difference of X and Y by subtracting or just use Y value if that is where i need to find the resistance) divided by I (Z amps)
  2. pebe

    AAC Fanatic!

    Oct 11, 2004
    Ohms law will never change!

    Your resistor will have across it a voltage equal to the source voltage X minus the LED voltage Y. The current through both the resistor and LED is Z amps.

    So from Ohms law R = V/I then R = (X - Y)/I
  3. Brandon

    Senior Member

    Dec 14, 2004
    Say you have a typical red led at with a on voltage of 0.7v and an ideal current of 20 mA to acheive the brightness you want and you hve a 5v source to use. Since the led can not control the current only a resistor can, we need to generate the current limiting resistor.

    You need to set up the KVL loop equasion to represent the circuit first.

    5v (+ terminal) - 0.7v (led Vdrop) - Vr (Vdrop over res) = 0v (- terminal)


    Since we want 20mA and V=IR we set up the equasion.equasion.

    5v - 0.7v - Vr = 0
    5v - 0.7v - Ir*Rr = 0
    5v - 0.7v - 0.020A * Rr=0

    Rr= - (0.7v - 5v) / 0.020A

    Anytime you need to solve for a led, you follow a method like this.
  4. n9xv

    Senior Member

    Jan 18, 2005
  5. Brandon

    Senior Member

    Dec 14, 2004
    oops.. thats right.

    Concepts still the same, lol.