I completely disagree. Where exactly are you going to put a power supply in a room? Make a special electrical box for it? Won't it be unsightly in many rooms, or need some huge special panel that looks like a breaker panel? The comments I see seem to be advocating a whole-house solution, where a power supply is mounted in the breaker panel to supply LVDC to the whole unit. But this makes no sense for several reasons, especially the voltage drop.

>I don't know if you meant it, but the sentence about "any sophomore level electrical engineering student can solve this" can easily come across as dismissive. I also think it gives too much credit to sophomore students. :)

It's supposed to be dismissive, because this whole discussion is a bunch of software people trying to make up solutions for a perceived problem when they obviously don't know one of the most basic things about electrical theory, which makes all of their solutions unworkable. It's like a bunch of people trying to make a new kind of personal vehicle to replace cars when they don't even understand Newton's Laws. It's really annoying, because I see this kind of discussion pop up every so often, over many many years.

I have another comment here I don't feel like copy-and-pasting, but basically this whole discussion is silly because people are trying to make a solution using a very expensive power supply to fix a problem they see because they're buying cheap $2 light bulbs that burn out quickly, instead of just buying light fixtures that were properly engineered in the first place. With modern SMPSs, you're not going to get any kind of benefit by centralizing the power supply to drive individual LEDs, you're only going to get problems. LEDs need a driver circuit to provide constant current, and that means the power supply needs to be matched to the emitters and kept very close to it.

It's not about DC vs AC, it's high-voltage vs low-voltage. The power dissipation by wire resistance scales with the square of the current ($P=RI^2$), and low line voltage means that you need large currents to transmit the same amount of power.