I always cringe a bit when folks approach hardware like they do software
Not sure what you mean - was there some terrible mistake in what the author was doing?
Not sure what you mean - was there some terrible mistake in what the author was doing?
So I don't doubt this works for the author's setup, but it is by no means the correct solution to distributing power across his lights. There are three problems with this, first the PC power supply (being unspecified) may, or may not, actually regulate the 5V line when the 12V line is unloaded. So someone reading this might spend hours trying to figure out why their supply isn't producing 5V when the answer in the general case is to load the 12V supply with a 15W load (a 12V lightbulb works well for this).
Next up he adds two 5V supplies that can provide 15A in "parallel" but the "grounds are isolated". So he is assuming there might be ground cross talk, but what he really should be concerned about is whether or not the power supplies are designed to run in parallel. If supply A produces 5.05V (well within spec) and supply B provides 4.95V, then supply A will try to drive all the LEDS and back drive supply B. Of course in so doing its output will come down, and eventually supply B will get a chance to contribute some power, the supply with the slightly higher output however will run at 100% of its rated output, and can then heat up. This can sometimes cause it to go into thermal shutdown, and suddenly supply B will have all the load. Now it over heats and supply A is called upon again to carry the load. Power supplies that are designed to be run in parallel will have a current mirror which lets each supply see what the other is producing and split the load between them, they are essentially current controlled supplies at that point.
Third, there are lots of places that sell 60A 5V supplies if you want one, although a more typical arrangement might be to bus a higher power and then run independent supplies with a common return at each group of LEDs. So perhaps a 48V bus, with 5V step down supplies every 5M.
The simplest solution is of course to just use a bus bar. Provide a suitable gauge wire (8M of 8Ga wire will have a resistance of about 10 mOhms. Less than .5V drop over the length. Have the bus wire go 8M along two walls, and then out the other direction 8M along the other two walls. That solves his power supply issue caveat he needs a 300W 5V power supply. Since high power GPU boards are so common, if he wanted to use a PC power supply he could run 12V on one half of the room and 12V on the other half split into two 150W feeds, with 12V->5V supplies to down convert at the LED strips.
However since he missed all that (and I think its just missing some of the EE basics) and may (or may not) get a fire in his dual 15A supplies, I cringed.
Dmitry has designed at least two of the consumer electronics products you have at your place.
He takes pretty brilliant shortcuts with his personal projects (see his others). While you're cringing, I'm grinning. A true 'hack' should do exactly that.
Or are you grinning because Dmitry has so far gotten away with a risky approach? You, Dmitry, and Chuck probably all know more about electrical engineering that I do, but I've done just enough with low voltage power supplies to share Chuck's concern. The result is cool, but there might be a better way to get the same result.
Alternatively, perhaps Dmitry meant "parallel" in the sense of "separate", rather than electrically parallel. That interpretation might make more sense, since the "just hook them up in parallel" approach seems so obviously poor.
Yeah, well, none of his "calculations" made sense to me. Actually, there is no calculation, just random figures dropped randomly.
There is ABSOLUTELY NO WAY he gets 4V at the end of a 4m strip of 240 LEDs when they are at full intensity if the strip has a resistance of 2 Ohm.
Without even doing the full calculation, let's just calculate the drop caused by the single last diode group at the end of the strip: dV = R * I = 2 Ohm * 60 mA = 0,12 V
The others diodes groups will face less and less resistance as they get closer to the power source, but still, with that one single group, we ate 0,12V out of a 0,50V budget. Just imagine adding the voltage drops caused by the... 239 remaining diode groups.
(The total voltage drop supposing all groups pull 60mA is actually over 14V, so it just does not work. It only _seems_ to work because LEDs emit light with less than 4V and much less than 20mA each.)
He also writes: "Lessons learned and improvements to make: Really, what I should do is construct better power supplies, with sense lines running down from the top of the wires, since even now there is a significant voltage drop over the wires that run the height of the room. With sense lines, a supply could at least compensate for that drop."
Yeah, sure, and how do you compensate the voltage drop at the end of the line? By rising the source voltage. Which means you fry the LEDs which are closest to the source...
No, just add a pair of wires behind the strip and connect it to the power bus of the strip from place to place. Its resistance will be much lower than the 2 Ohms of the strip (supposing he did not screw this measurement too).
> Dmitry has designed at least two of the consumer electronics products you have at your place.
One must be the new satellite receiver that takes 2 minutes to boot and crashes every other day. As for the second one, I wonder.
Low voltage systems are a lot safer, but not immune to trouble. Even USB ports occasionally catch on fire if there's a damaged cable or crossed pins.
If there was the potential for danger, the hardware should have had a limiter built-in. We have fuses in equipment for much the same reason.