24v PLC to 5v Drive

sid76

Member
Join Date
Feb 2015
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Tamil Nadu
Posts
6
Hi

I am using a Delta sv PLC which gives a 24v pulse sink output which is connected in series with a 2.2Kohm resistor for both step and dir pulse input of the 5v stepper drive, my stepper motor run erratically, I have used shielded cable and the earth is good. I have been told that signal turn off time of 24 v plc is long for a 5v drive so i should be using a 470 ohm resistor between the step output and common of the plc or supply smps. is it correct
 
what brand/model the stepper drive? post the wiring diagram

if u still continues u will damage the controller
 
what brand/model the stepper drive? post the wiring diagram

if u still continues u will damage the controller

I was curious to know where the resistor value comes from.

The 2.2k resistor should keep the current down to a manageable level. But I was wondering how much current is required to drive the input. You might actually be starving it a bit. The documentation may tell us the requirements.

While a 470 ohm resistor may do the trick for you, the wattage will put you at about 1.25W. Use a 2W resistor.

Anotherher thing that has not been given is the frequency you are trying to accomplish.
 
The 24 Volt should be around 10 mA or higher so you will need a resistor of 2.2kiloohm for this. fine.
1 thing to get 5 volts is to put a resistor of 470 ohms in series with this 2.2 kOhm you will get around 5 volts on the knot.
you can use this to give it to the stepper driver.
However max speed is about 100 steps depending on the cycle time of the PLC.
Check how much current goes into the stepper driver (when optocoupler 2.2 k is good you wont need the 470 ohms.
Put a ammeter and 5 volts on an input and measure the current flowing.
 
Hi

Thanks for your reply, I have attached my circuit diagram for the stepper drive connection from my Delta PLC -DVP SV series( I am not very good at drawing ) and also attached a link for the stepper motor drive that I am using http://www.tinycontrols.com/tstep-484-Stepper-motor-driver.html.

If I connect a 470 ohms or 560 ohms 2 watts resistor between the Yo output of the PLC and the + output of the stepper smps will it bring down the 24V signal turn off time of the from PLC to the 5v of the stepper drive.

IMG-20150222-WA0006.jpg
 
Hi, if you look at the circuit diagram I have taken the 24 + ve output from smps to the step + and dir +ve through a 2.2Kohm of the drive and the pulse signal is given form the Y0 and Y1 of the PLC to the step(-) and Dir (-) of the drive, and you might notice the terminal 1,3 &5 are not connected to the ground.
 
According to the manual, we can glean two things from this one note:

1) the inputs are optimized for 5VDC
2) the inputs need 10mA to be properly driven. (Shooter was entirely correct on that, but it's good to have it verified in the documentation.)

Look at it this way. You are using 24V on the the entire input circuit. 5V is used for the actual input. That leaves 19V to drop across the external resistor.

19V / 2200ohms = 0.0086A, or 8.6mA. This is why I said you may be starving the drive.

19V / 10mA = 1900ohm. Using a 1.8Kohm resistor puts you at 10.5mA

The other thing I saw in the manual is to put a 100uF capacitor on the power terminals if the leads are over 12inches (approx 30cm). That may also be something to look into.

EDIT: I did see the note about using the 2.2k resistor. Just not quite sure that the math adds up.

10mA.jpg
 
Last edited:
I have connected my GND of the stepper drive to (-V) of the SMPS.

Kindly go through the attached document i got from the web and suggest me if it will work out for me without affecting the PLC. If yes the connection in the document is for source type PLC, so in my case being a sink type PLC should I connect the 470 or 560 ohms 2watt resistor between the Y0 output in my diagram and the 24v SMPS (+)
 

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