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?
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, 𝐼in (current in) pretty much equals 𝐼out (current out). So the efficiency 𝜂 = 𝑃in / 𝑃out = (𝐼 × 3 V) / (𝐼 × 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.
I’m slightly confused by your question. Are you trying to power two thingies off a twofer adapter, or are you trying to multiplex something?
It sounds like you’re just trying to do power so:
See how much each strip draws (either via data sheet of amps per meter times 0.69, or measure it) and if that amount exceeds the capacity of the adapters you want to use, it’ll either not work or shorten the life of the gizmos (or if it’s really cheap crap or you’ve gone way the heck over the capacity of any component it might catch fire a bit).
Note the current on the datasheet wont be linear per length, but for 27” chonks if should be close enough.
Edit: I see now your LEDs are 3V. Personally I’d recommend replacing them with 12 or 24V, or at least 5V to run properly on the 5V thing you’re plugging them into. Why did you go with 3V product?
I am assuming the LEDs are white (blue with yellow phosphor), which always have a roughly 3V voltage drop, this is just a physical fact. Some chips like those in LED bulbs have several LEDs in series for a voltage drop of 6/9/12/15/18 V but this is not the case here.
The vast majority of 5V strips have no step-down switching power supply (aka constant current buck converter) to reduce 5 V from the power rails to 3 V, instead just driving the LEDs with a resistor in series – it drops 2 V and if it’s a 100Ω resistor (usually labeled “101”), it lets 20 mA to the LED as per Ohm’s law. In practice, multiple LEDs are often in parallel to one resistor to save cost, in which case the current divides among them.
The strip may be RGB, in which case the LED voltages are 🔴1.8 V 🟢2.4-3.0 V 🔵3 V and pretty much the same applies. There may be an external controller but it usually just uses PWM to pulse the 5V rail of each color rather than adjusting the current. Individually cntrollable LED strips have a chip driving each LED, and there is just one power rail and serial data line between them.
By the way, the resistance of the strip’s power rails may be significant. If this is the case, do this: https://www.youtube.com/watch?v=wCsDZK0tJvU&t=393
Wow awesome, thank you for the info and resources! I have to follow the links later but I really appreciate the reply.
I don’t think OP’s strips have pixel control since in my experience you wouldn’t have a separate USB power line that bypasses the controller like OP is describing, but now am curious and hope they follow up.
There are two questions to answer:
- which current and voltage do the LED strips need (5V, 2A would be something you could support with USB 2).
- what output does the charger provide as a maximum? Is it enough to power both strips (5V and two times the A).
If these match it should work. Another topic could be the cable itself. Theoretically it could start to burn, if you try to channel to much current through it, but with simple USB I doubt it. If it is getting hot after some time, scrap your setup. This would be a fire hazard.
Fire hazard? Not really. The charger will not provide more than 5 V, 3.1 A or whatever its rating is. Even if the strip fails short circuit (unlikely, most LEDs blow open and there is at least a resistor in series anyway), nothing will receive more than 15 W of power. A 1m cable can safely dissipate that.
Worst case scenario, an LED fails short and its series resistor will receive over 6x its intended power (with 5 V across it instead of 2 V, and its current will increase correspondingly). It is soldered on power-dissipating flexible PCB and will probably not blow open. It will get the area somewhat hot and potentially melt the plastic in the back of the monitor while the rest of the strip keeps glowing but more dimly. Hard to tell if it could get over flammable temperatures with moderate heatsinking and only a few watts of power.
To keep safe, I would deliberately add resistance before the LED strip, using something like a 1Ω 5W resistor (or a shitty long cable). That way, the voltage drops significantly in case of a short. Also, the LEDs will run at lower-than-intended current, which prolongs their life and decreases risk of failure.
Edit: Some microwaves have a 0.8Ω 25W resistor as part of inrush, at least in 230V regions. Feel free to use that, it will happily handle a semi-short circuit. Or you know, an automotive fuse.
The safest option is replacing the whole setup with an LED strip that has no resistors (bare LEDs) and a constant current driver.
Well, it is nice OP got an answer from someone, who is clearly more knowledgeable than I am. Thanks!
The LED strips use SMD 3528 LEDs which need 5V and the wattage is listed at 11.52W/min. The amperage isn’t listed but for those LEDs, I’m seeing 5Ah online. The charger provides 40W
Watts in a resistive example like yours is Volts x Amps. I would have been able to much better answer this question a year ago so forgive me if I’m misremembering the specs but I’ll answer since nobody else has. Two things that suggest to me this might be a bad idea:
- Charger is 40W, that’s probably usb PD (I don’t know anything about QC so maybe I’m wrong). PD supplies more than 15W (5V x 3A) by stepping up the voltage, not the amperage. While stepping up either would likely be bad or very bad for some part of your circuit, don’t worry about that though; without the powered device telling it to, PD won’t activate. It should max out at 15W… I think. It depends on the resistance on the CC lines and using a splitter could screw up the resistance that tells the power supply which USB version to support so it can go up to 3A (15W). Sorry, it’s been a while since I’ve worked with USB power. 2 strips of 11W will need more power than that. Basically my concern is you won’t get adequate power out of the charger for one reason or another.
- Where are you getting the 11.52W/min number? Watts don’t have a time unit and that much precision sketches me out. Almost as if someone who isn’t adequately educated measured the power straight off a multimeter once and just wrote that on a product page. Is the LED strip from a reputable manufacturer?
He meant watts per meter, not minute – the strips can be cut and rejoined. As for the 5 Ah, no clue.
The circumference of a monitor is more than 1 m, so a charger of 3 A at least will be necessary for each. This is why I prefer higher-voltage strips where less current is required and higher resistance is tolerable. Anyway, the power is quite high and this could cause overheating problems.
You’re confused. W/m means watts per meter, and the “5Ah” is probably actually 5 A, or the current you can push through the strip (limiting the length to below 2 m).
