I'm looking at the 4-Way 2.1mm DC Barrel Jack Splitter and I'm wondering how much ampere it can safely transfer. I want to use the splitter for powering NeoPixel LEDs.
The Adafruit website says "Jack-side cable is 22AWG. Plug side cables are 24AWG."
I found an AWG to maximum amps table. I turns out 22AWG corresponds to 0.92A and 24AWG corresponds to 0.577A. So this means it can transfer up to 0.92A of power in total, while each of the four connectors can transfer up to 0.577A.
I just want to know: Is this the right reasoning or can it transfer more? 0.92A seems a bit low.
You're thinking in the right direction. For that kind of break-out cable, it's a "whichever comes first" limit.
If the limit for a single piece of 22-gauge is 600mA, five lines running in parallel could carry a total of 3A.
If the limit for a single piece of 24-gauge is 900mA, that controls the total amount of current that can flow through the five parallel 22-gauge lines it feeds.
You're using the wrong column from the table though.. the 600mA and 900mA limits are for power transmission, which means cables longer than about 5m. Short cables fall under the heading of chassis wiring, and the limits there are much higher.. 3.5A for 22-gauge and 7A for 24-gauge.
The actual limiting factor is the resistance of the wire. The resistance of 24-gauge wire about 85 ohms per 1000m. A 5m @ 24-gauge cable would have an end-to-end resistance of about 0.4 ohms, and you'd lose about 1.5V sending 3.5A through it. The total power loss (which turns into heat) would be about 5W.
A 10cm piece of 24-gauge has an end-to-end resistance of about 8.5 milliohms. The voltage drop at 3.5A is about 30mV, and the power loss is about 100mW.. small enough that you don't have to worry about it.
The single run of 24-gauge is still the one that sets the maximum current limit for the cable, and at 7A that would come to about 1.2A per 24-gauge tail.
