I am trying to power a string of 24 LEDs which are connected in series (re-used from an LED bulb) from 120 V mains AC. Each LED has a Vf of 3.2 V, and a 150 mA maximum rating.
To limit the LED current to 100 mA, I was planning on using the LC5202D. It's a constant-current driver IC that comes in a DIP package (which I prefer over SMD.)
My plan was to use a buck topology as shown in the datasheet:
This leaves a problem. If my AC input is 170 V0 after rectification (120 * 1.414), but my LEDs are only consuming 76.8 V (24 * 3.2), what am I going to do with the rest of the voltage?
Considering the IC is only rated at 1.78 W power dissipation, it won't be able to drop 5 W of power (50 V * 100 mA).
Can I simply put a Zener diode after the rectifier to drop the 50 V, assuming I can find one with the proper power dissipation?
Another solution might be to use a flyback circuit with a transformer, but that's going to involve more parts and complexity than I want.
Maybe this is much easier and I'm just missing something.