Electronic – damage a hall effect sensor with too much current

The measured current has to flow through the package to produce any result.

Therefore, you have to cut the wire, and solder one end to one of the terminals, and the other end to the other terminal.

The nominal output is 1/2 Vcc, assuming you are using the birirectional variant. The datasheet states:

_{Datasheet, page 15}

Quiescent output voltage (V_IOUT(Q)). The output of the device when the primary current is zero. For bidirectional devices, it nominally remains at VCC ⁄ 2. Thus, VCC = 5 V translates into V_IOUT(QBI) = 2.5 V. For unidirectional devices, it nomi- nally remains at 0.1 × VCC. Thus, VCC = 5 V translates into V_IOUT(QUNI) = 0.5 V. Variation in V_IOUT(Q) can be attributed to the resolution of the Allegro linear IC quiescent voltage trim, magnetic hysteresis, and thermal drift.

^{_{QUNI means Quiescent (e.g. no current flow), unidirectional variant. QBI means Quiescent, Bi-Directional variant. VIOUT is the name of the output pin. See the pinout on page 1.}}

Since you say you are getting ~1.6V, I would guess you're using the bidirectional variant, and powering it from 3.3V.

I don't know how you're having trouble figuring out the current, it's very simple:

_{Datasheet, page 15}

Sensitivity (Sens). The change in device output in response to a 1 A change through the primary conductor. The sensitivity is the product of the magnetic circuit sensitivity (G / A) and the linear IC amplifier gain (mV/G). The linear IC amplifier gain is pro- grammed at the factory to optimize the sensitivity (mV/A) for the half-scale current of the device.

There are multiple versions, with different sensitivity.

From page 2 of the datasheet:

The sensitivity is given in mV/A. Therefore, with a device with a 40mV/A sensitivity, 1A of current through the device will result in the output voltage increasing by 40mV.

The overall equation is:

$$V_{IOUT} = ( Vcc * OffsetScaling ) + ( Scaling Factor In Volts * Amps ) $$

\$OffsetScaling\$:

• Bidirectional version = 0.5
• Unidirectional version = 0.1

\$Scaling Factor In Volts\$:

• This is the scaling factor given in the above table, converted to volts.

Therefore, for the bi-directional variant, with 40mv/A sensitivity, the output voltage will be: $$V_{IOUT} = Vcc*0.5+ 0.040 * A$$ The uni-directional variant with 40mv/A sensitivity would be: $$V_{IOUT} = Vcc* 0.1 + 0.040 * A$$

^{_{0.040 is 40mV in Volts. Change to match your device sensitivity.}}

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?