How ground loops smoke your PC and what to look for.

[PMinMO]

One of the issues with non isolation is where is 0 volts. Sometimes I like to use extremes to illustrate a point. Fundamentally, wire has resistance and ohms law applies.
DON"T DO WHAT I DEPICT HERE STRICTLY HYPOTHETICAL TO EXPLAIN
Consider you have one wire from the motor power supply negative to drive X, from drive X to drive Y, from drive Y to drive Z then to the PC. Each drive handle 6A worth of motor. Each drive wire segment is .2 ohms. Ohms law is E=I*R so what you have in drive currents is 6A from the Z segment to the Y segment, 12A in the Y segment to X segment and 18A in the X segment to the Power supply. That means that the 0V at the power supply negative terminal, is actually 3.6V at the first drives negative terminal, 6V at the second drives negative terminal, and 7.2V at the Z drives negative terminal, which is wired to the PC PPort common. In all probablility the PC common is tied to safety gnd, and in good practive the power supply negative terminal is wired to safety gnd (I've never seen one that hasn't and I suspect you can't and be UL approved). So now now have a loop via you ac wiring safety gnd. The wire segment from the Z drive to the pc is probably going to pull a hefty current as there maybe parallel wiring in the cable and connections to/from the db25. You think this is just signal wiring, NBD and the next thing you know you power up, smell smoke, see you parallel cable wire smoking or fused open cause it was too small to handle the current.
If the drives are optically isolated and treated properly that last wire segment (Dark green)doesn't exist.

[creep]

Hi there I hopping you might be able to help me.

I'm trying to learn how to run my CNC machine using Mach2 or 3 and the original steppers and drivers, I been given a circuit diagram for a opto isolator board that works supposedly see here, http://www.cnczone.com/forums/attachment.php?attachmentid=12169

After reading your post I'm concered about the lack of isolation in the circuit around the limit switches, if you wouldn't mind could you have a look and advise either good or bad.

Also the circuit uses 7407, MCT271 and TIL128, I can't find any MCT271 or TIL128 for sale in the UK so I looking for replacements, I found a equivalent list listing TLP624 for the MCT271 and PC829 for the TIL128, would these be suitable, I notice you recomend the TLP624 on your website. I presume if I use a different chip to the ones specifide the restor value will have to be changed to make it work properly, how do you calculate the required resistance.

Sorry for all the questions, just want to learn.

Chris

[PMinMO]

Your drawing is kind of tough for me to follow, but it looks like the switches are wired via the same 5V as the buffers between the optos and Db25. Thats fine. As long as your 5V rtn is isolated not a problem. What are you using for your 5V?

Phil

[creep]

Hi

This is near enough the same circuit but is alot clearer to see http://www.cnczone.com/forums/attachmen ... entid=8239

Not sure what I'll use for the 5v supply, there is a 5v supply on the machine but I guess this is wrong way of doing it as its not isolated if I read what you say correctly, I presume it is possible to hack a 5v supply form a old PC or even use the PC's 5v supply some how

Thanks for looking

Chris

[PMinMO]

Yes that layout is ok. There really isn't a need for opto isolators on the switch inputs as they should utilize the same isolated 5V and the pc side of the interface. If the PC has usb, you can get your 5V from it. Your can also get it from an unsed PC diskdrive connector. The easiest is a simple 5VDC wallwart. like the 1200ma from here:
http://www.allelectronics.com/matrix/DC_Wall_Transformers.html
or
http://www.mpja.com/productview.asp?product=15816+PD

[Creep]

Hi

Been looking at the PC I will probably be using and it has spare 5v leads in the case (for supplying a extra CD drive), although the 5v supply you mention may be easier.

The current limit switches run on a 24v circuit in with the E Stop http://www.trialstalk.co.uk/images/stories/Ebay/picture%201.pdf
I'll disconnect them and try them with 5v to make sure they still work.

Any input on the chip replacments.

Thanks again

Chris

[PMinMO]

If your referring to the opto isolators, they are pin compatable, but not current transfer ratio which affects the resistor sizes. CTR is in a %, so if you have 16ma in the diode emitter then a CTR of 100% would mean it would trie to sink or source 16ma, a 50% ctr would mean 8ma. But you can never sink or source more current that the resitor in the output leg devided by the voltage available.

[PMinMO]

Above is a simple representation of how to start wire the motor to drive negative and the electrical isolation of a drive that has optoisolated inputs.