Adding a 2k ohm resistor to drop 24v to 5v question

[Black Forest]

On the drives for my CNC mill they require a 2k ohm resistor to drop the voltage from 24v to 5v for the enable and alarm signals. "They" all say to add a resistor to the line. Do I just solder the resistor inline to the wires? Or is there a better way to add the resistor?

[polaraligned]

I would assume it goes inline, but hard to say without seeing the circuit. Could go inline or it could go somewhere else forming a voltage divider. Those are crappy directions.

[eKretz]

Yeah that's weird. Not clear at all. Generally a resistor is not used inline to drop voltage, but to limit current. Either that's the goal or the resistor is used in a different way, like the voltage divider mentioned above or something similar. I'd call the machine maker for tech service to confirm.

[mikey553]

I happen to have a couple of stepper motor drivers and they also can use control voltage from 5 to 24 V. The same recommendation about adding a current limiting resistor for 12 V and 24 V control voltage is also given. 5 V control voltage does not require any resistors.

You add the resistors in series. Use 1/8 W or bigger resistors. Soldering one end of the resistor to the wire and clamping the other end to the control terminal will provide the best electrical connection. Do not forget to isolate the bare wires.

[Black Forest]

I happen to have a couple of stepper motor drivers and they also can use control voltage from 5 to 24 V. The same recommendation about adding a current limiting resistor for 12 V and 24 V control voltage is also given. 5 V control voltage does not require any resistors.

You add the resistors in series. Use 1/8 W or bigger resistors. Soldering one end of the resistor to the wire and clamping the other end to the control terminal will provide the best electrical connection. Do not forget to isolate the bare wires.

Do I put a resistor on each of the two wires?

[mikey553]

Don't you have a connection diagram? Even the most inexpensive stepper drivers have them. In my case I can connect the driver to a controller as common-anode or common-cathode. With common cathode you connect all negative (-) driver terminals to a controller ground without any resistors and add a resistor to each of the positive (+) driver terminals.

Make sure you consult your documentation or the manufacturer if you are not sure about the connections. Otherwise a magic smoke may escape...

[ed_h]

Well, it is a voltage divider, but the other resistor is the load itself. Dropping 24V to 5V with a 2K series resistor, the load would be 530 ohms or so.

The dropping resistor would need to dissipate a little under 200 mW, so should be at least a 1/4 watt unit. I'd use a 1/2 W.

Yeah, they are pretty crappy instructions.

Ed

[MaxHeadRoom]

The way it appears to me is the internal electronic regulator, Zener, components etc, and is set for 5vdc operation, if feeding the unit with 12vdc use a 1k series resistor, if 24vdc is used, use a 2k series resistor.
In order to drop the voltage, and limit the current.

[Black Forest]

THis wiring diagram is basically what I am following. It is Centroid's Acorn board. As of now I only have 5v connected to the pul+ and dir+ and then the signal wires back to a DB 25 breakout board mounted on the Acorn.

[MaxHeadRoom]

It looks to me as the normal 5v input in all probability goes into an Opto isolator, and the internal series resistor would be sized accordingly for 5vdc, when using the other voltages, the resistor has to be added to conform to the opto current.
If you trace the inputs and do a little reverse-engineering it may disclose whether it is the case!

[Black Forest]

It looks to me as the normal 5v input in all probability goes into an Opto isolator, and the internal series resistor would be sized accordingly for 5vdc, when using the other voltages, the resistor has to be added to conform to the opto current.
If you trace the inputs and do a little reverse-engineering it may disclose whether it is the case!

Max, are you referring to the drive or to the Acorn board? I have no idea what to trace or how to do it. Sorry.

[MaxHeadRoom]

In your OP you mention the input to the drives?
I meant if you could see the circuit components, IC etc, connected to the relative terminal input, you may identify the nature of the input, e.g. opto etc.
As it appears to me that the resistor is simply connected in series with the relevant input when the higher voltages are used.

[Black Forest]

In your OP you mention the input to the drives?
I meant if you could see the circuit components, IC etc, connected to the relative terminal input, you may identify the nature of the input, e.g. opto etc.
As it appears to me that the resistor is simply connected in series with the relevant input when the higher voltages are used.

OK, got it! I am able to jump start my tractor with jumper cables! That is about it for my electricity knowledge.

[mikey553]

Your wiring diagram shows a common anode connection of the stepper drives. All positive PUL, DIR and ENA terminals are connected together and go to V1 on the Meanwell RD-35B, which is +5VDC. Negative PUL, DIR and DENA terminals are connected to the Acorn board, which controls all stepper drives.

So it looks like you have 5V control voltage for the drives and do not need any resistors in series. Where did you get the idea that you need them?
What is the Meanwell device? Is it just a power supply to the Acorn and stepper drives?

[mikey553]

Here is a link to the Leadshine page for the drive you have. From here you can access the drive manual, which shows the connection diagrams.
https://leadshineindia.com/products/...-drive-cs-d508