John Fields said:

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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

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.