I have gone through the desktop/benchtop configuration and testing and have run into a a question about the polarity viewed from the manual versus the schematics.
I have followed all the procedures in order as outlined in the ALLIN1DC mill installation manual V3.
I noticed in section 5.5.2 the manual drawing of the ALLIN1DC inputs showing black connecting to com INP 9-12 and yellow INP 11 Estop a +12 V DC.
My question is on the schematics this is just the opposite polarity for common and input point.
Which is correct for polarity…
Polarity
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AMDlloydsp
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Re: Polarity
The drawing at 5.5.2 appears to be incorrect -- but it's possible to be correct; below.
The schematics and pictorials show each of the configurable inputs to be set up with opto-isolators. Each is DRAWN as having an LED and a snubber diode in parallel, and in opposite polarity. IF that is the correct type of opto-isolator, as drawn, The 9-12(COM) lead must be positive with respect to the driven input when it is active.
In that case, the cathodes of the LEDs go separately to the configurable input leads, and the anodes go to the SIP resistor network used to establish the LED current for different voltages. An LED's cathode must be driven more negative than its anode in order to light up and activate the photodarlington in the opto-isolator.
HOWever... there are opto-isolators with two LEDs and series blocking diodes connected in "head to tail" configuration, which permit either polarity (or AC) to drive them. (there are also LEDs with reverse-voltage specs high enough that even the blocking diodes are not necessary)
So... I THINK the drawing is incorrect, and it's certainly not how the other schematics show it connected... but it's possible it's correct, and they just never updated the master schematics.
Best I can tell you is, that's not the way its shown on the master schematic, and not the way mine's wired!
Hope that helps.
Lloyd
The schematics and pictorials show each of the configurable inputs to be set up with opto-isolators. Each is DRAWN as having an LED and a snubber diode in parallel, and in opposite polarity. IF that is the correct type of opto-isolator, as drawn, The 9-12(COM) lead must be positive with respect to the driven input when it is active.
In that case, the cathodes of the LEDs go separately to the configurable input leads, and the anodes go to the SIP resistor network used to establish the LED current for different voltages. An LED's cathode must be driven more negative than its anode in order to light up and activate the photodarlington in the opto-isolator.
HOWever... there are opto-isolators with two LEDs and series blocking diodes connected in "head to tail" configuration, which permit either polarity (or AC) to drive them. (there are also LEDs with reverse-voltage specs high enough that even the blocking diodes are not necessary)
So... I THINK the drawing is incorrect, and it's certainly not how the other schematics show it connected... but it's possible it's correct, and they just never updated the master schematics.
Best I can tell you is, that's not the way its shown on the master schematic, and not the way mine's wired!
Hope that helps.
Lloyd
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Cliff
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Re: Polarity
Thank you so much for your explanation it is appreciated…
Re: Polarity
Inputs may be configured with either polarity. They work just as well either way.
All the inputs in a bank of four must use the same voltage and same polarity.
You can run different input banks at different voltages and with different polarities, to suit your application. Just install the correct SIP resistor for the voltage you are using, and connect the approriate side of your power supply to the common terminal for that bank.
All the inputs in a bank of four must use the same voltage and same polarity.
You can run different input banks at different voltages and with different polarities, to suit your application. Just install the correct SIP resistor for the voltage you are using, and connect the approriate side of your power supply to the common terminal for that bank.
-
AMDlloydsp
- Posts: 91
- Joined: Thu Oct 17, 2013 4:32 pm
- Allin1DC CNC Controller: Yes
- CNC Control System Serial Number: 0605130770
- DC3IOB: No
- CNC11: Yes
- CPU10 or CPU7: No
Re: Polarity
All the inputs in a bank of four must use the same voltage and same polarity.
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If there are, indeed 'bi-polar' LEDs in those optos, then... they need the same voltage, because they're all working through the same SIP resistor network. But they needn't be the same polarity, so long inputs of one polarity are driven from a supply isolated from the one driving the other polarity.
Draw it out as two batteries, rather than a single power supply, and it will become obvious.
I might add that it would be silly to do this just because you can, but it is possible to run individual ports within a group of four in opposite polarities simultaneously.
And THEN you run into that old electronics conundrum of one wire conducting in two directions, and thus carrying zero net current, even when current is obviously flowing! Or at least that's the way we presented it to trick and fool students, just for fun!
When you draw it out, you'll see the two power supplies are, themselves, in series with each other, and with the two sets of LEDs/resistors as a single load. (yeah, you have a 'center tap' configuration, but we're talking about perfectly balanced resistors and LEDs... right?<G>)
Lloyd
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If there are, indeed 'bi-polar' LEDs in those optos, then... they need the same voltage, because they're all working through the same SIP resistor network. But they needn't be the same polarity, so long inputs of one polarity are driven from a supply isolated from the one driving the other polarity.
Draw it out as two batteries, rather than a single power supply, and it will become obvious.
I might add that it would be silly to do this just because you can, but it is possible to run individual ports within a group of four in opposite polarities simultaneously.
And THEN you run into that old electronics conundrum of one wire conducting in two directions, and thus carrying zero net current, even when current is obviously flowing! Or at least that's the way we presented it to trick and fool students, just for fun!
When you draw it out, you'll see the two power supplies are, themselves, in series with each other, and with the two sets of LEDs/resistors as a single load. (yeah, you have a 'center tap' configuration, but we're talking about perfectly balanced resistors and LEDs... right?<G>)
Lloyd