Avoiding accidental earthing, voltage rejection, and voltage dividers for a motor control output circuit
I don't understand why the motor controller output in pic 1 is after R4 - even if my plc was 24V, surely in this position it will be getting a 0V signal? 


R4 and the Zener diode provide a 24 V clamp to protect downstream devices. When S1/S2/S3 is less than 24 V the voltage will be available on OUT_24V_MAX with a source impedance of 4k7. If the voltage exceeds 24 V, and the extract says it can rise to 56 V, then the Zener clamps the voltage to protect the downstream devices.


  Anyway to connect my 10V PLC through a voltage divider which of my proposed circuits below that is correct, if either?


Your Circuit 2 outputs 40% of the S1/S2/S3 signal. At 56 V this will be 22 V going to your 0 - 10 V analog input. This is unlikely to end well.

Your Circuit 1 is a better solution but you have forgotten to factor in the R4 value into your divider. We'll need to recalculate and make some assumptions:


S1/S2/S3 nominal maximum = 24 V.
S1/S2/S3 absolute maximum = 56 V.
PLC analog input nominal maximum = 10 V.
PLC analog input absolute maximum = 15 V. (You'll have to check the specifications to choose the correct value for this.)








simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. The two cases to be considered. In (b) the Zener won't be in breakdown so it is effectively out of circuit.

From (a) we can write the equation
$$ 24 \frac {R_3}{R_2+R_3} = 15 \tag 1 $$

From (b) we can write:
$$ 24 \frac {R_3}{4k7+R_2+R_3} = 10 \tag 2 $$

The simultaneous equations are:
$$ 24 R_3 = 15 R_2 + 15 R_3 \tag 3 $$
$$ 24 R_3 = 10 \times 4k7 + 10 R_2 + 10 R_3 \tag 4 $$

which can be solved to show

$$ R_2 + R_3 = 9.4k \tag 5$$

Now we know that the whole R1 + R2 + R3 chain will be 4k7 + 9k4 = 14k1.
Since the ratio required to drop 24 V to 10 V is \$ \frac {10}{24} = 0.417 \$ you can calculate the value of R3 and, after that, R2.


  Also the manual states 'The accidental earthing of the outputs or connecting them to a capacitive load will result in them being destroyed.' Does this just mean connecting them directly to ground without a load (e.g. a resistor) in between? 


Yes. A short circuit will overload and destroy T8. A large capacitive load will initially present as a short-circuit while it is charging up. Excessive current may result and again T8 will be the victim. R4 will provide some protection against this or a short.


  i.e. both my circuits below would be OK?


Both will protect T8. Circuit 1 will protect the PLC.



Update: the LabJack PLC specifications state that the analog input can tolerate -20 to +20 without damage.