A 12 V machine and a 3.3 V pin
The coin acceptor is a 12 V machine, and the wire it signals on swings the whole 12 V. A microcontroller input pin is built for 3.3 V or 5 V. The little board in the box exists to stand between those two numbers.
The acceptor is a machine, not a sensor
Inside the black plastic there is a coil, a motor-driven flap and a small computer with a two-digit display. It weighs a coin, measures its diameter and its metal, and if it matches something it has been taught, it lets it through and pulses an output wire. All of that runs on 12 V.
It came out of arcade cabinets and vending machines, where 12 V is what is already in the enclosure. It has no idea that microcontrollers exist.
The pulse is the right shape and the wrong size
The COIN wire carries exactly the information you want: a short burst of pulses, and the number of them says which coin went in. The trouble is the voltage it says it at.
An input pin on an ESP32 is rated to about 3.6 V. Put 12 V on it and current flows into the chip's supply through a protection diode that exists to shrug off static, not to carry a signal. Sometimes the pin dies. More often it half dies, and the board keeps working well enough that you go looking for the bug in your sketch.
What the module does, in one sentence
It takes 5 V in over USB-C, makes the 12 V the acceptor needs, and hands your microcontroller the same pulses at 3.3 V or 5 V.
That is three jobs, and skipping the module means solving all three yourself: a separate 12 V supply, a level shifter on the signal, and a shared ground between them. The board in the box is about the size of a postage stamp and does all of it from one cable.
Which of the two voltages
The module leaves the factory set to 3.3 V, which is what an ESP32, an ESP32-S3, an ESP32-C3 or a Pico wants. An Arduino UNO wants 5 V, and there is a selector on the back of the module for it. That is the one setting to check before wiring anything, and wire it in this order covers where it is and what happens if you get it wrong.
When it does not work
No. The acceptor's COIN wire idles at 12 V and a 3.3 V input pin is rated for about 3.6 V at the absolute most. Current goes into the chip through a protection diode that was never meant to carry it, and the damage is often partial rather than total, so the board keeps half working and you spend an evening blaming your code.
A divider would get the voltage down, but it would not give you the acceptor's 12 V supply, and that is the other half of what the module does. One USB-C cable in, 12 V out to the acceptor, logic-level pulses out to your board. A divider solves a third of the problem.
White is COIN, the pulse train that says which coin. Grey is COUNTER, one pulse per accepted coin. Red is the 12 V supply and black is ground. All four go to the module's COIN ACCEPTOR header, which is printed in the same order as the acceptor's own socket.
No, and that is the point of it. Same count, same width, same timing — only the voltage changes. Whatever you program into the acceptor is what your sketch counts.
The two boards, the harness, the screws, and the two headers you will be reading all week.
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