voltage drop equation

Discussion in 'Homework Help' started by jonisonvespa, Mar 30, 2012.

  1. jonisonvespa

    Thread Starter Member

    Mar 25, 2012
    49
    0
    hi
    im currently trying to make a project using a ardunio uno, the outputs are 5v and the leds i want to use are surface mount 2v max 30ma current, what equation can i use to drop the voltage to 2v, 30ma?
    thank you
     
  2. MrChips

    Moderator

    Oct 2, 2009
    12,414
    3,353
    Ohm's Law is I = V/R

    hence,

    R = V/I
     
  3. jonisonvespa

    Thread Starter Member

    Mar 25, 2012
    49
    0
    hi mr chips im confused with the v value, is that the supply or the max voltage of the led? (supply)
    ok thanks
     
    Last edited: Mar 30, 2012
  4. Jess_88

    Member

    Apr 29, 2011
    174
    1
    The way I would look at it is;
    current in a series circuit is the same, while a voltage drop can be achieved across one resistor and the led in parallel with the second (same voltage in parallel).

    IR1 + IR2 = 5V
    IR2 = 5v-3v

    ...whats how I understand it anyway
     
  5. Jess_88

    Member

    Apr 29, 2011
    174
    1
    wait... sorry, that can be right. current would change in the LED

    3v = 30mA x R

    with just a sires circuit
     
  6. MrChips

    Moderator

    Oct 2, 2009
    12,414
    3,353
    Vr = 5V - 2V = 3V
    R = 3V/30mA = 100 ohm

    30mA for LED is too much. I would reduce it to 10mA
    R = 3V/10mA = 300 ohm

    Even 3mA may be bright enough
    R = 3V/3mA = 1K ohm
     
  7. jonisonvespa

    Thread Starter Member

    Mar 25, 2012
    49
    0
    thanks mr chips will try some different values when i get the leds
     
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