Electronic – Reading a Open Loop Current sensor

Generally speaking, an open-loop Hall-effect sensor uses the Hall voltage directly to produce its output signal. This has the advantage of being simpler to implement, but can suffer from a certain amount of nonlinearity.

A closed-loop sensor has a coil that is actively driven to produce a magnetic field that opposes the field produced by the current being sensed. The Hall sensor is used as a null-detecting device, and the output signal is proportional to the current being driven into the coil, which is proportional to the current being measured. This method is more complex than the open-loop method, but it eliminates any nonlinearities associated with the Hall sensor itself, since it is being operated at just a single point in its range.

The HAL508 and Optek OH180U are both Hall Effect SWITCHES with hysteresis. The HAL has an open drain output ant the Optek has an open collector output, both on pin 3. When you look at the magnetic specs in the data sheets, after converting Gauss to milliTeslas, you'll see that the two sensors are very compatible.

Since switches are on/off, I assume that the function is to say whether the Hi Hat is open or closed, not to read out the linear position of the hi hat.

The circuit is rather confusing. Why you would tie an open drain high using the output of an op amp is simply beyond me. It's hard to believe that anyone would want to handle a switch in this fashion, even if you need to do inversion and/or level shifting. I suppose its possible that this funky circuit forms some sort of oscillator that when low pass filtered gives you a distance measurement, perhaps explaining the capacitor, but we'd really need to know what is being used as the output of this circuit to know.

Is it possible that the ORIGINAL original circuit used some sort of ratiometric Hall sensor, and not a switch?