Hall Effect Sensor to a +24VDC PLC Input
The A1217 doesn't seem to have a datasheet available anywhere. From your comments it seems to be an automotive cog tooth sensor.

The application note you posted shows the sensor fed from a current limiting resistor Rp = 182 Ω. The Up = 13 V is a 12 V battery average voltage.


With a 2-wire sensor it needs some current to power it even when "off". That means the only way to signal "on" is to pass more current. We can see this from the test circuit information:
When the sensor turns off little current will flow so the voltage at (4) will increase (because there will be less voltage dropped across Rp). Taking the average value of UH as 12 V means that 1 V is dropped across Rp and we can work out the current as \$ I_H = \frac {V_H}{R} = \frac {1}{168} = 6~\mathrm mA \$.
When the sensor turns on the voltage at (4) will drop. Taking the average value of UL as 8.2 V we get 4.8 V dropped across Rp and we can work out the current as  \$ I_L = \frac {V_H}{R} = \frac {4.8}{168} = 29~\mathrm mA \$.


Meanwhile, if your PLC input is like this one then you will need &lt; 1 mA (5 V) to guarantee a '0' and > 2.5 mA (15 V) to guarantee a '1'. Given those specs it appears that the PLC has about 5k - 6k input resistance.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Connecting the sensor as a high-side switch directly to the PLC with a shunt resistor just might work.

The idea in Figure 1 is that when the sensor is off it will pass 6 mA. With the PLC input in parallel with 820 Ω you'll get about 700 Ω total load and the voltage at the input will be \$ V_{IN} = IR = 6m \times 700 = 4.2~\mathrm V \$. When the sensor turns on it will pass a higher current. 

Assuming 29 mA flows then \$ V_{IN} = IR = 29m \times 700 = 20.3~\mathrm V \$. Now this is unlikely as the sensor voltage would be reduced below the 8.2 V we saw in our earlier calculations. It might, however, pass enough current to reliably exceed the 15 V turn-on point.

Please try it out and report back on the voltages obtained.