12 Vdc Intel fan tachometer output to 24 Vdc PLC input
PLC 24 V digital inputs come in two varieties: 


Current sink requires a 24 V feed to power them on. (I'm assuming that this is what you're using.)
Current source requires a connection to ground to sink the current sourced by the input.








simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Open collector output to current sinking PLC input.

Assuming that you are using a current sinking input, the best way to do this would be to use a converter unit or industrial opto-isolator but using a pull-up resistor is not uncommon. The values in the calculation below came from the PLC spec in another question I answered. Yours should be similar but it would be worth checking to ensure reliable information.

R4 value 

The PLC is guaranteed to turn on at 3 mA / 11 V on the input. Let's give it 12 V to make the maths easier. R4 then has to drop 12 V at 3 mA. \$R = \frac {V}{I} = \frac {12}{3m} = 4~k\Omega \$. The next lower value is 3k9. 

Q2 current

Now check the max current through the fan transistor shown as 30 mA max. When Q2 is on R4 will be have 24 V across it. \$ I_{max} = \frac {V}{R} = \frac {24}{4k} = 6~mA \$. We're fine. In fact we could decrease R4 to 3k3 for extra PLC margin if we wanted.

R4 power dissipation

Finally check the power rating of R4. Worst case will be if the fan stops with Q2 on. \$ P = VI = 24 \cdot 6m = 150~mW \$. A 1/4 W resistor would suffice.

Re-run the calculations for your components and you should be able to make it work.

If, by any chance, you have a current sourcing input then just make a direct connection to your open-collector transistor.