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Not so simple LED series resistor value?

Discussion in 'Electronic Basics' started by Bill Bowden, Aug 15, 2007.

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  1. Bill Bowden

    Bill Bowden Guest

    Can sombody check my math to see if this LED resistor value is right?

    I have a series chain of 25 white LEDs at 3 volts each. I want to
    connect them through a resistor to a bridge rectifier connected to the
    120VAC line. There is a 50uF capacitor across the chain of LEDs. What
    is the resistor value for a RMS current of 20mA?

    The total LED voltage will be 25*3 or 75 volts so the conduction angle
    will begin at about (asin) 75/170 =.44 = 26 degrees. So, out of a 180
    degree half cycle, the resistor will conduct for 180-52 = 128 degrees
    or about 71 percent. The RMS voltage across the resistor is the peak
    input minus the LED voltage times 0.707 or (170-75)*.707 = 67 volts
    RMS. But since the duty cycle is 71 percent, the RMS voltage should be
    adjusted to 67*.71 = 48 volts RMS. Therefore, the resistor value for
    20mA of current should be 48/.02 =2400 ohms.

    Does this make sense, or did I miss something?

  2. me

    me Guest

    I got about a 4k7 ohm resistor....
  3. Nobody

    Nobody Guest

    Okay so far.
    This shouldn't have any significant impact on steady-state operation. The
    capacitor will charge to ~75V then stay there. It will just cause the LEDs
    to turn on more slowly when power is first applied.
    What is the significance of RMS for something other than a resistive
    load? Wouldn't it be more useful to work out e.g. the equivalent DC
    current for the same mean power in the LEDs?

    For a resistive load, RMS current and equivalent-power current are the
    same thing. For your LED chain, they're quite different.
    0.707 is for a sine wave.

    In your case, the mean resistor voltage is:

    / t = 2.684
    1 |
    --- * | 170*sin(t)-75 dt
    pi |
    / t = 0.457

    = ~43.94V

    The mean current will be that value divided by the resistor; a 2k2
    resistor will give ~20mA, resulting in a mean power of 20mA*75V = 1.5W in
    the LEDs.
    1. The RMS:peak ratio of 1:sqrt(2) only applies to sine waves. For square
    waves, it's 1:1. In general, you need to compute the definite integral of
    v(t)^2 over one cycle and divide that by the period. But ...

    2. RMS itself is only meaningful for resistive loads, where current and
    voltage are proportional (i.e. W = I^2*R = V^2/R). For the LED case,
    voltage is constant during conduction and irrelevant (because the current
    is zero) otherwise.
  4. Tolstoy

    Tolstoy Guest

    Just wondering why you want to do it that way. Does that
    configuration give you better efficiency than the more conventional
    way of putting the filter cap right at the bridge? Or maybe you want
    to reduce the surges through the rectifiers.
    I get about 4k7 ohms in the conventional circuit with the filter cap
    at the bridge. I'm pretty sure it's right because the arithmetic is
    simple. I din't check your math, but I doubt just putting the cap at
    the other end of the resistor would cause it to change by almost a
    factor of two. Anything's possible, though.
  5. Bill Bowden

    Bill Bowden Guest

    Thanks, I'm sure your numbers are right, but I don't follow integral
    math. Where do the numbers 2.684 and 0.457 come from?

  6. Nobody

    Nobody Guest

    arcsin(75/170) = 0.457 radians (26 degrees) or 2.684 radians (154 degrees).
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