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I'm trying to connect up an integrated control DC motor with a 24V output (Motor manual - relevant page p32) with my 10V PLC (PLC Manual), and looking for a little help please.

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

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

Many thanks for your help!

Oli

circuit given in manual THE CIRCUIT GIVEN IN THE MANUAL

Proposed circuit 1 MY PROPOSED CIRCUIT 1

Proposed circuit 2 MY PROPOSED CIRCUIT 2

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  • \$\begingroup\$ (1) What is a 10V PLC? Do you mean a PLC 0 - 10 V analog input? (2) Can you include a link to the manuals for the motor and PLC. (3) Does "pic 1" mean "Picture 1" or is it a device in your circuit. (4) What's the difference between your proposed circuits 1 and 2? (5) You can't get negative voltages out of that circuit. Are you expecting to? \$\endgroup\$
    – Transistor
    Commented May 19, 2018 at 11:52
  • \$\begingroup\$ I've spotted the difference. Answer in progress. \$\endgroup\$
    – Transistor
    Commented May 19, 2018 at 11:59
  • \$\begingroup\$ @Transistor Hi, manuals added above, cheers \$\endgroup\$ Commented May 19, 2018 at 13:05

1 Answer 1

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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.)

schematic

simulate this circuit – 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.

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  • \$\begingroup\$ Hi Transistor, one more thing - Should the outputs above be grounded to the PLC ground, or the SMPS which is supplying the 24V, or does it not matter? Many thanks \$\endgroup\$ Commented May 22, 2018 at 20:29
  • \$\begingroup\$ First make sure that the ground of PLC is connected to the SMPS ground or this won't work. With this circuit it won't matter which end R3 is grounded at. \$\endgroup\$
    – Transistor
    Commented May 22, 2018 at 20:42
  • \$\begingroup\$ The power for the PLC is via USB from a PC. Are you saying ALSO connect the PLC ground to the SMPS ground? \$\endgroup\$ Commented May 22, 2018 at 21:22
  • \$\begingroup\$ Yes. The voltage on the motor controller output is referenced to its ground. The voltage read by the PLC is referenced to its ground. Both grounds need to be connected otherwise there is no return path from the PLC input to the motor controller. You can't have just one wire from one circuit to another. You must have a return path. \$\endgroup\$
    – Transistor
    Commented May 22, 2018 at 21:30

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