AndyUK's Build - 1.2x1.0m Gantry

[AndyUK]

Hi All,

So, I thought I had the machine all ready to make its first self-powered movements this evening, however I've stumbled!

I'm pretty sure the problem is that the drives are not enabling, because they worked before I reprogrammed the enable to be active high rather than active low, so one of the things I'll do to test is reprogram a drive enable to be active low again and retest.

I thought I'd throw this out to people to see if you can see where I've gone wrong. The drives are wired as per the EM806 manual (below) and my circuit diagram two posts ago.



So, each drive has a line from +24V on the UB1 to ENABLE + on the drive via a 2k ohm resistor. The ENABLE - lines are then commoned, and fed into the OSSD output on the UB1. This is the output of the safety circuit, which I can override with a button (so I know the cause isn't limits etc).

The UB1 manual says this about the outputs:



And lists the OSSD as just another output that is controlled by the safety circuit:



My thoughts were, that when the output is off (i.e. safety circuit broken), no current can flow through the enable optocoupler, so its a logic low. When the output is on current can flow, logic high, drive enabled.

The drives target 10mA on those optocouplers, so 40mA approximately should be within the 70mA sink of the output.

I know the OSSD output is doing its thing because I can see the LED come on when I press the override.

Have I missed something?

[Neale]

My thoughts were, that when the output is off (i.e. safety circuit broken), no current can flow through the enable optocoupler, so its a logic low. When the output is on current can flow, logic high, drive enabled.

I haven't ploughed my way through the UB1 manual, but in general terms it looks like the internal "switch" in the UB1 shorts the pin to earth (in which case current can flow) or not, in which case no current flows. What might be causing confusion is that when the switch is on, current flows, and the output voltage on the pin is more-or-less at earth potential. When the switch is off, the output pin is high voltage (as the resistor and enable output path on the driver form a pull-up circuit). So is your logic the wrong way round? :Low and high usually refer to the voltage on the pin, not current flow.