Controller compatibility.

[alexfisher]

I want to replace my motion controller with the UC300ETH-5LPT. I need to confirm that my stepper motor drivers (http://www.arceurotrade.co.uk/projects/ ... Manual.pdf) will work with the UC300ETH-5LPT, and their correct connection.

My drivers accept 5v logic from a current sinking (NPN) interface, 15mA typical. Inputs are isolated from the case (ground) and the logic input terminals and can be connected as Common Positive as shown in the wiring diagram (or Common Negative).

Q1: Is the UC300ETH-5LPT compatible with my drivers?
Q2: Is my interpreted connection correct?
Q3: Is the series resistor R value > 50ohm?

[ger21]

I believe that you only need that series resistor if you're using 12V or 24V step signals, per the Leadshine datasheet.
http://www.leadshine.com/UploadFile/Down/M542d.pdf

[cncdrive]

Yes, they are compatible.
The UC300ETH-5LPT outputs can drive the optocouplers out without problem.
Both connection methods on the 2 drawings will work.
If you decide to connect as per the first drawing then you can get the +5V from the powersupply you powering the UC300ETH with which voltage you can also access from the 26.pin of the IDC26 connectors.

And if you connect as per your second drawing then the GND is pins 18. to 25. in each IDC26 connectors.

[alexfisher]

Thanks ger21 for the Leadshine link, I only knew of my driver as a 'generic type' (having had it's markings erased by Arceurotrade).

My driver manual shows an internal series resistance for the opto-isolators of 200 Ohm.
5V / 200ohm = 25mA, exceeding the ABSOLUTE MAXIMUM 20mA output pin current.

CNCDRIVE, thanks for confirming the circuit diagram options. What about the UC300ETH-5LPT output pin current limiting series resistor R?
My controller has just arrived - don't want to fry it! Alex.

[A_Camera]

My driver manual shows an internal series resistance for the opto-isolators of 200 Ohm.
5V / 200ohm = 25mA, exceeding the ABSOLUTE MAXIMUM 20mA output pin current.

That's not how you calculate the current...

(5-LED voltage drop)/200 = current

If you don't know your data then you can take 1.4V as a good guess of LED voltage drop, so the current is going to be (5 - 1.4) / 200 = 18mA.

[cncdrive]

Yes, A_Camera is right, you should deduct the voltage drop on the LED which is about 1.4-1.8 Volts depending on the type of LED, the optocoupler's datasheet can tell you the exact value.
And so 200R series resistance with the opto LED will not exceed the maximum sink/source rating of the UC300ETH-5LPT outputs, you can connect it.