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Can anyone please help with the following. I am in the process of connecting a Schneider Electric Servo Drive for a Cnc project.

I want to wire a fault output from one of the 5 drive outputs. (Do1-Do5).

The manual says" Use an appropriate external protective circuit or device to reduce the risk of inductive direct current load damage. "

The output will simply be disconnecting the enable to the drive if a fault is detected within the drive. I am not switching any other load.

The question is do I need to protect this circuit as described in the manual or can I wire directly?

I have attached a screenshot of the circuit from the manual.

Thanksenter image description here

Michael
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1 Answers1

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If I understand correctly, you will be using one of the 5 driver output to drive a relay which will connect/disconnect the enable of the driver.

The function of the relay is irrelevant, a relay is an inductive load, hence you will need to "Use an appropriate external protective circuit or device to reduce the risk of inductive direct current load damage ". Even if you just drive the relay to make it click, without serving any other purpose, you will need to protect the servo driver output.

An inductor (the relay coil) can't change it's current instantly, the purpose of the protection circuit is to provide a path for the current when the coil is switched off. Otherwise, a voltage spike will occur possibly damaging the output of the servo driver.

Gustav
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  • Can you provide a simple schematic to show how to imliment this? – Michael Nov 11 '18 at 21:33
  • Well, if you are using a digital output of the servo driver I suppose it will be a 24VDC output, then you will need to use a 24Vdc rated relay, so the easiest protection is a flyback diode. [Here](https://electronics.stackexchange.com/questions/100212/why-dont-relays-incorporate-flyback-diodes) you can find different solutions. – Gustav Nov 11 '18 at 21:58
  • Thank you Gustav for your time and explanation. Makes more sense now. – Michael Nov 15 '18 at 03:51