Series resistor for LED?

--
Since the diodes are going to be conducting even if they haven\'t lit
up, I think if you use the RMS value of the AC source it\'ll come out
right.  That is:

           Vrms - Vled     24V - 15V
     Rs = ------------- = ----------- = 450 ohms
              Iled           0.02A 


Closest safe 5% is 470, so the RMS current will go down to:
 

             Vrms - Vled     24V - 15V
     Iled = ------------- = ----------- ~ 0.019A
                  Rs           470R

        
And the power dissipation in the resistor will be:


     P = Iled * Vrms - Vled = 0.019A * 9V ~ 0.172 watts

Yes, that's probably close, but the diodes only conduct about 600 nanoamps at 2 volts each, and probably nothing below that, so there is about 17 degrees out of 90 with no current.

So, I'm figuring the RMS value might be 20% lower, but I don't have a true RMS meter to measure it.

-Bill

Hmm... This is an interesting question... I hope people were paying attention in math class.

So uhh... What is the justification for your above formula? When I use my method I calculate a resistance of 440 Ohms, which is amazingly close to yours, but I am having trouble seeing how your method could be accurate. Suppose you hooked up a set of LEDs with V(f) of say 24V. In that case Vrms is still 24V, but then your formula would result in a resistance of 0 Ohms. That won't do. But even from 24Vrms AC, an LED string with V(f) of 24V should still be drivable.

The way I calculated 440 ohms as appropriate took more than one step.

Step 1: find out when (in seconds) the instantaneous AC voltage exceeds 15V, and also when in the cycle it also drops below 15V.

Suppose we look at one half wave of the sinusoid which at 60Hz would last

8.33ms. The instantaneous voltage is V(inst.)=34sin(377t). Solve for V(inst.)=15V. I get 1.212ms and 7.121ms.

Step 2: apply formula: I(dc average in LED) = {integral[(34sin(377t) - 15)/R]dt}/0.00833 seconds with upper limit at 0.007121 seconds, and lower limit at 0.001212 seconds. When I evaluate that only mildly ugly integral and solve for R I get:

R= 0.07323/[I(dc average in LED)*0.00833]

For 20mA this is 440 Ohms. For 470 ohms, DC average current in LED is

18.7mA.

If we integrate the instantaneous voltage formula with respect to time (during that interval when V(inst.) is above 15V) divided by the resistance R, when we should get the total coulombs of charge that flows through the LEDs during that half cycle. To get the equivalent DC average current through the LEDs, we simply divide that result by the duration of that total half cycle (8.33ms). After all, the definition of current is charge per unit time, or in other words, one ampere = one coulomb per second.

When I use the above expression and plug in the desired I(dc average in LED) equal to 20mA and solve for R, I get 440 ohms. Hopefully I pressed the right buttons on my calculator, but that number seems somewhat believable.

Note that the DC average current through the LEDs will be 20mA, so assuming the efficiency of the LEDs doesn't decrease at higher peak currents, then the LEDs should be equally bright as if powered from pure 20mA DC. In practice the forward voltage of the LEDs does increase somewhat at higher current densities, so the result won't be perfectly accurate, and the dissipation in the LEDs will be slightly higher than what it would be from

20mA pure DC. One should probably derate the forward current slightly for this kind of use.

To find the power dissipation in the resistor, one would have to integrate the instantaneous power in the resistor (during the interval any current actually flows through the resistor) divided by the total time of the cycle.

Or in other words:

Average power dissipation in resistor = {integral([(34sin(377t)-15)^2]/R)dt}/0.00833 where the limits of integration are also 1.212ms to 7.121ms. Evaluating I get:

Average power dissipation in resistor R = (1.1/R)/0.00833

For 440 ohms, this would be approximately 300mW. For 470 ohms this would be approximately 280mW.