Disabling a Third Party Drive/Motor thru a Relay

[corvette]

I can’t seem to get my INHOUT signal to disable my drive/motor.  I have the PCI-7342 Stepper/Servo Motion Controller with the UMI-7764 connected to a third party drive and servo motor.  My third party drive requires a constant 24 volt signal to enable my drive/motor.   I have the INHOUT connected to a relay in order to control this 24 volt signal.  When there is 5 volts at my relay this will enable my drive, when the 5 volts goes low then my drive will be disabled.  I have the drive wired up per the attached schematic.  Any suggestions would be appreciated.

Thanks,

Nick Argyros

nicka@idinet.com

[Giridhar_Rajan]

Hi,

May be the INH is not able to source enough current to drive the relay. Check the datasheets of 7342 or contact NI to know the maximum sourcing current of INH. The maximum sourcing current should be more than the current requirement of the 24V relay. If not, then use a power transistor (preferably SL100) to give it the required current drive. Also be sure to provide some kind of isolation for protection of your card.

Now lets think of another approach to your problem. You can use the 24V digital I/Os to programmatically control the output through a pin whenever your want to enable/disable your drive.

Regards,

Giridhar Rajan

Automation Engineer

Cruiser Controls

Mumbai, India

[Jochen]

As described in the user manual of the 734x boards the Inhibit output is an open collector output, pulled up to 5V with a 3.3 kOhm resistor. That means that without further circuits the output can only drive about 1.5 mA. You can add a resistor in parallel to the 3.3 kOhm (from 5V to the INHOUT terminal).

The maximum amount of current that the  INHOUT output can sink  is 64 mA. Thus make sure that the parallel connection of your resistor with 3.3 kOhm in combination with the serial resistance of the relay result in a current not bigger than that amount.

Best regards,

Jochen

[Buechsenschuetz]

If you use the source current of the INH output it probably will not be able to drive a 5V relay. Jochen's calculations are valid for the limiting current, i.e. the current where the output voltage will be zero. With a resistive output the output voltage will depend on the current needed by the relay coil and will never reach 5V.

It is better practice to use the 64mA sink capability of the INH output. But of course in this case the relay will be activated when the output is low. In your case you may wire the relay between +5V (do NOT use the +5V pin of the 734x connector!!! - you may use a +5V terminal of the UMI box provided your UMI power supply can supply the UMI box plus the relay current) and INH, and use the N.O. contact of the relay.

If this is not desired I would suggest to add an extra output transistor (any small signal transistor can sink 100mA or more) connected as an inverter or a driver IC such as an ULN2803. A single transistor has a saturation voltage of less than 0.5V, and most, if not all, relays rated at 5VDC will switch reliably at 4.5V. Check the relay specs for the 'must on' voltage.

Never, I said NEVER, connect a relay to a semiconductor device without a snubber diode. The diode should be connected across the coil terminals (and close to them), with the cathode to the (+) terminal. It will short the inducted high voltage when the relay is de-energized. Some relays have integrated snubber diodes.

Also, I see from your diagram that the emergency switch can be operated at a certain relay position only. At least in Germany this would be illegal since the emergency switch has to be independent from any other control circuitry.