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Resistor values for a series circuit

  1. Jul 21, 2011 #1
    Suppose we have a bulb or a forward-biased LED (or any other component) which needs no more than say 3 V across it to operate. And we want to connect it in series with a voltage source of say 12 V, and a resistor. So without this resistor, the LED would quickly burn out and be destroyed.

    I've edited a diagram to illustrate the situation:

    [PLAIN]http://img225.imageshack.us/img225/6515/circt2.jpg [Broken]

    How can we calculate the value of this resistor in series with the LED and the battery, that limits the current flow through the LED to a safe value, and reduce the 12 V to 3 V? I know that there many websites and softwares that can calculate the value of the resistor, but I just want to know how engineers would do the calculation without using them. Is there a quick method or a formula?

    And what if there are two (or more) LEDs instead of just one?

    I couldn't find any methods for this in my physics textbook so any guidance is greatly appreciated.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 21, 2011 #2
    For an LED, take the forward voltage drop rating (Vf) from the datasheet.
    Add all of those forward voltage drops together for multiple LEDs. If you don't have a datasheet, you can make assumptions (~2V for red, etc.)
    Subtract that number from your source voltage. This give the voltage drop across the resistor.

    Then it becomes a simple Ohm's law calculation (V=IR) - if you know the current you want to use, then you need to solve for the resistance - Take the voltage across the resistor and divide by the desired current to get the resistor value.
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