To the other people trying to help, please read the AAC book chapter I recommended before trying to help. You are confusing the issue without researching the background.

I think you need to revise your section. It is wasteful to use a reverse parallel diode to shunt voltage around the LED. In this case, we are dealing with 354Vpk AC. The ballast resistor is going to have to dissipate that negative shunt IR loss ...despicable.

The back-to-back 25 LED idea is more reasonable, but runs into reverse polarity protection problems (5Vreverse x 25) + (3V fwd x 25) = 225V which is not enough to protect against 354VACpk. Or, if they are a 25 series-parallel arrangement then each leg would still need a series protection diode.

Consider that 50 LEDS in series is 50 x 3V = 150V (forward voltage) and that 250VAC x 1.414 = 354VACpk. That means the diodes will conduct during the (positive) 150VAC to 354VAC range -- that's ~200V. Who's saying that's not enough or it won't work?

By using a series diode (1N4007 or such) to protect against reverse conduction, which manifests as a half-wave rectifier, the ballast resistor only needs to dissipate the remaining IR drop (the negative half is eliminated). Note that creating a smoothed DC with a filter cap is NOT necessary.

Although the PRF of the half-wave is 50 Hz, the effect is similar to 25Hz or so because the LEDS are only on during the +150V to +354V region. That's a lot of off time, but in this case it shouldn't matter.

However, I don't think messing directly with 250VAC is the best, or safest, way to do this. Using a step-down transformer and connecting the LED's in parallel or series-parallel would be a better approach.

Regardless, see attached:

 

If you really read the chapter you'd notice they are using a capacitor for the ballast reactance. The power isn't wasted, no more than any electronics does at least. 240VAC X .02A = 4.8 Watts, not bad for a light bulb. More LEDs would increase the brightness of course, but would also be correspondingly more expensive.

Whether you agree or not, it's in the book. Try reading it sometime, it's a good long read. Your illustration makes it obvious you haven't.

I haven't read the entire thing either of course, but I do read the appropriate sections when someone is referencing it to make a point.

A transformer might indeed be safer, but ultimately you have to connect SOMETHING to AC, unless you're using batteries. The capacitor as ballast does offer a much smaller and lighter package.

 

240VAC X .02A = 4.8 Watts, not bad for a light bulb.

And if ~1/2W is being expended by the resistor, how much is the capacitor radiating? In the example, I don't see a calculation for power in the capacitor. After all, that 4.8W has to go somewhere, and only a small amount is in the LED's (20mA x ~3V).

Per the back-to back diode example given, do you realize capacitor dissipation is on the order of 3W RMS? I would much rather have this dissipated by a properly sized resistor, than torture an innocent capacitor by broiling it from the inside out.

Try reading it sometime, it's a good long read. Your illustration makes it obvious you haven't.

No, I just like to look at the pictures. Don't care for the articles.

 

Do you even understand reactance? The capacitor doesn't dissipate anything, it resists AC current flow through an entirely different mechanism, no wattage involved, the same principle as a ballast coil uses. This is what makes fluorescent lights so efficient.

You really do need to do some reading. This is pretty basic AC theory.

The calculated 4.8 watts is deceptive. The power used is only what the LEDs absorb (at 3V each,150V X .02A, 3 Watts total), the diodes (at .7V each, 35V X .02A, .7 Watts), and the resistor (1.2KΩ, ½W). All at 20ma, which the cap limits through reactance, which involves no wattage at all.

All told, this arrangement dissipates around 4 watts throughout the whole circuit.