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Controlling LED mA's

Discussion in 'LEDs and Optoelectronics' started by jeeep, Feb 5, 2010.

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  1. jeeep


    Feb 5, 2010
    Hello All,
    I have a quick question concerning the mA's being used by a series of LED's that are equal to the total input voltage.

    For example, 12vdc input, 6 LED's @ 2vdc can output 20mA or 40mA, etc.

    How do you control the output of the LEDs in this situation?

    The reason I ask is that I bought several modules configured in this fashion and can make no sense of it.

    Thank you!

    Reference: http_//
    12 = Source Voltage
    2 = Diode forward voltage
    20 or 40 = Diode forward current (mA)
    6 = Number of LEDs in array
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    OK, you need to ensure the combined voltage of the LEDs is less than the DC voltage.

    In your case I'd connect the 6 LEDs in 2 strings of three each with the appropriate resistor that would drop 6 volts at 20 (or 40) mA.

    Using ohms law, R = V/I = 6/0.02 = 300 ohms.

    so a 300 ohm resistor in series with 3 LEDs having a 2V drop would draw 20 mA.

    Note that the voltage you use to determine the size of the resistor is the supply voltage (12V) minus the voltage across the LEDs (3 LEDs x 2V = 6V) which is 6V. (In this case both voltages are the same, it may not always be!)

    If you wanted 4 x 2V LEDs in series using 40 mA from a 12V source, then R = V/I = 4/0.04 = 100 ohms.

    for a string of 3 LEDs at 40 mA, R = 6/0.04 = 150 ohms.

    so you would need (in the last example 2 x 150 ohm resistors (1/2 watt)). Place 1 resistor in series with three LEDs. Make 2 of these and connect each of them across the 12V supply.

    The total current will be 40mA or 80mA because each of the strings requires 20 or 40 mA

    EDIT: OK, I checked your link. It gives VERY wrong answers when the sum of the LED voltages equal the battery voltage. try setting the LED voltage to 2.001 volts. Also the "wizard" gives the nearest E12 resistor value, not the theoretical value. The difference is marginal, but will explain why my calculations yeild slightly different answers.
    Last edited: Feb 5, 2010
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