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Not so simple LED series resistor value?

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?

-Bill

Reply to
Bill Bowden
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Bill Bowden wrote in news:1187134439.948558.140070 @q4g2000prc.googlegroups.com:

I got about a 4k7 ohm resistor....

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Reply to
me

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.
Reply to
Nobody

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.

Reply to
Tolstoy

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?

-Bill

Reply to
Bill Bowden

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

Reply to
Nobody

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