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Bundypro
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« on: October 26, 2010, 07:35:27 AM »

Can someone explain to me why you would pull a pin high to +5v? 

The reason I ask:

I'm trying to setup an auto zero for my Z-axis.  I set the input to active low but when I run any script Mach never gets a signal to ground to trip the script.  I'm going directly to the port, no BOB.

What's going on here?

I've seen a few setup this way but I don't know why.
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Tweakie.CNC
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« Reply #1 on: October 26, 2010, 11:57:25 AM »

I think Brian has his problem sorted now but for those that follow pulling pins high to +5Volts, with a high value resistor 10K or so helps the following logic (and Mach) see the transition, from high to low, when the pin is taken low by a probe or other device – this arrangement is configured within Mach as ‘active low’. Pins could quite equally be pulled low (10K or so to GND) and this arrangement would then be configured within Mach as ‘active high’.
Once upon a time the output from all logic gates was either high or low and a transition was always obvious but then along came tri-state logic which could be either high, low or high impedance (open circuit) which enabled the inputs and outputs of many devices to be physically connected to the same bus and with the use of enable signals communicate to each other without loading those that were just waiting their turn.
Generally the manufacturers of the controllers, breakout boards etc. fit the necessary resistors to the inputs as standard and some even provide the option for pulling high or low as required. As always, their recommendations / instructions should be followed in preference to anything I have mentioned here.

Tweakie.

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