this post was submitted on 19 Oct 2023
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Does multiplexing a usb power cable from 2 cables into 1 impact energy efficiency? (sh.itjust.works)
submitted 1 year ago by SoySaucePrinterInk@sh.itjust.works to c/askelectronics@discuss.tchncs.de
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I have two led strips used for 2x27" monitor bias lighting. Each have their own USB cables for power. These two USB cables are plugged into a 2-port wall charger for a phone. I would like to use a 2x female to 1x male adapter to join the two USB cables into one, then plug it into a much smaller 1-port USB wall charger. The reason is due to my space constraints.

Will doing this impact its energy efficiency, ie using an adapter like this?

https://www.aliexpress.com/item/1005004340044483.html

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[–] ChaoticNeutralCzech@feddit.de 5 points 1 year ago* (last edited 1 year ago)

The strips use 3V white LEDs, power to which is delivered via resistors or linear current regulators. Unless you see any inductors, there is no buck converter from 5 V to 3 V.

Why does this matter? Well, with resistors and linear regulators, 𝐼~in~ (current in) pretty much equals 𝐼~out~ (current out). So the efficiency πœ‚ = 𝑃~in~ / 𝑃~out~ = (𝐼 Γ— 3 V) / (𝐼 Γ— 5 V) = 60 %. Extra cable resistance will reduce the current, brightness and power, but still exactly 40% of power leaving the USB charger will be wasted before it gets to the LEDs.

However, I would advise against multiple, cheap USB connectors in the circuit: when moving the setup, their resistance changes somewhat and you would get blinking. The worst thing that could happen is a switch, such low voltage cannot spark over an oxide layer and eventually even small movement will blink the lights. I would get a good thick USB cable and solder it directly to one of the strips instead of whatever it came with, connecting the other with some thick wires.

So it does not matter, if you want better efficiency, use 12V (75%) or 24V strips (87%), or get just an LED array without resistors that needs a constant current driver (theoretically 100% but CC PSUs are slightly less efficient). Or make a constant current driver by fine-tuning the voltage of a PSU (by adjusting the feedback resistive divider) to 0.5-1 V above the LEDs’ voltage drop, then using an appropriate resistor to limit the current.