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> let's talk about another common failure mode of outdoor LED lighting: flashing. LED lights use a constant current power supply (often called a driver in this
by thayne 3y ago
> let's talk about another common failure mode of outdoor LED lighting: flashing. LED lights use a constant current power supply (often called a driver in this context) that regulates the voltage applied to the LEDs to achieve their rated current. Unfortunately, several failure modes can cause the driver to continuously cycle.
This is a problem I've frequently encountered with household LED lighting.
What I don't understand is why modern houses aren't set up with a central, high quality "driver" with adequate cooling to convert AC to a steady DC current, then wire that to light outlets so you only need the LED itself there. That seems much more efficient, and cost effective than putting small, cheap drivers in every single bulb that fail after a few months or a few years.
- praptak 3y agoA single current source for multiple receivers means the receivers have to be connected in series. This implies that the light switches would have to work as bypasses. Also all LEDs would need to be rated for the same current.
- Terr_ 3y ago> What I don't understand is why modern houses aren't set up with a central, high quality "driver" with adequate cooling to convert AC to a steady DC current Perhaps to avoid a new extra set of nonstandard electrical wiring to much if not all of the home?
- thayne 3y agoWell it could, and probably should, be standaedized.
- dreamcompiler 3y ago12 gauge wire can easily handle the current needs of household LED lighting. Most modern houses in the US have 14 gauge. That can still be made to work. The real reason -- as another commenter here said -- is that the lights would have to be in series and all must require the same current.
- fdsafdsafdol 3y ago>What I don't understand is why modern houses aren't set up with a central, high quality "driver" with adequate cooling to convert AC to a steady DC current This comes up fairly often here (because, after all, it isn't a hardware design forum), and the answer every time is - I^2 x R is a bitch. Losses increase with the square of current, instead of directly (as in the case of voltage). This is why it's significnatly more efficient to run power lines at high voltage, not high current. Running lighting circuits at LED voltage would require surprisingly thick cables to avoid losing most of the power in the cable runs. On the other hand, it might make sense to create a 48v LVDC circuit, and regulate down from that at the point of load. This would avoid all of the issues with designing every single little LED bulb to handle 110v AC (or here, 240v AC), and the inevitable failure of the highly-stressed components.