Run out of pins ?
What do you do when you run out of pins ?
The Jetson, Raspberry, Arduino, ESP, ETC all have limited numbers of 3.3v, 5v, and GND pins
The Arduino for instance has plenty of inputs to handle 5 ultrasonic sensors, but it still needs 5 GND and 5 3.3v pins, so what would a normal person do in such a predicament ?
I have no idea. I'm not normal
I did this...
I have no idea why companies don't make these things, although, technically, I did find a company that does, at $7USD or more each
Granted, you could make them yourself. But these things are a pain to make. Especially if you don't have the right crimper tool, which I don't, cuz that would be too easy, so I did
As a beginner, I would rather use a breadboard and take advantage of the power rails for multiple 5V and GND lines.
An intermediate hobbyist, would create a small PCB for the same.
But yeah, for making your projects more compact and good looking, the above options might fail.
Hope this helps 🙂
I use a breadboard, or in the case of a current project, three breadboards! This isn't quite what's being discussed here, but I made this little board as an easy way to add female connectors to the breadboard. In this case I've used it to plug in servos, all powered by jumpers from the 5v rail. The grey wires on the right are the signal wires from a Raspberry Pi Pico. It's on piece of veroboard with the tracks cut down the middle.
One noobie mistake I made was thinking that I could only use one I2C device on each I2C bus, and used pins unnecessarily that way. Now I tend to use two rows of my solderless breadboard as SDA & SCL rails, and have multiple I2C devices to just two pins ($26 & #27).
I’ve been using the Pololu A*Prime board, which is a knock-off of the Arduino Leonardo. After soldering on all the headers they supply almost all of the i/o pins have have 3 rails, logic, power and ground.
Aside from pin count, beware the total current called for by your "pin multipliers." Better to connect your peripherals directly to +Power and Ground, instead of solely through the MCU IC.