Can someone explain to me how input sourcing works inside the Leadshine MX3660?
More specifically, why does this schematic in the manual have 2 ground connections? One is from the auxiliary 12 volt power for proximity switch and the other is to the ground of an unused input channel.
Is the additional power supply required because the unit cannot supply enough amperage to run specific switches? Why not ground everything to the same input?
It looks like the fig 18 in the MX3660 manual is wrong !
see page 19 in PDF manual V1.2
part of diagram from MX3660 manual
in the MX4660 manual , this shows the internal opto-isolator circuit
this is what I think the MX3660 input wiring should be
this is what I would expect the missing MX3660 internal opto-isolator input circuit to be
if you assume the internal +12V supply is connected , via a 1K resistor , to the anode of opto-isolator's LED
the input pin will source a current of about 10mA when the input pin is connected to the ground pin
(about 2V will be dropped across the LED , I = 10V / 1000 ohms = 10 mA )
the open circuit voltage between the input pin and the ground pin will be approximately 12V depending on the current taken by your volt meter
John
Last edited by john-100; 12-06-2017 at 08:26 AM.
Reason: add ref to 1K opto-isolator's LED current limiting resistor
Fantastic! This makes way more sense now. Thank you.
For my knowledge can someone explain the function of the capacitor, resistor and diode in parallel on the internal side? Is it to expedite the turn on speed on the octocoupler?
LED's are not very good rectifiers as the reverse breakdown voltage is about 5 or 6 V and can easily be damaged
if the input pin is connected to +12V from your proximity switch for instance, while the power to the MX3600 is off
the diode will conduct instead of the LED breaking down
the capacitor across the LED is an alternative position for a noise filter capacitor
instead of using a capacitor between the input pin and ground
the resistor in parallel with the LED & capacitor provides a discharge path
the resistor must be chosen so it does not divert too much current from the 1K resistor supplying the LED current
but not take too long to discharge the capacitor when the limit switch opens
I think your thinking of the speed up capacitor in this circuit
For my application I have a Z touch plate (that allows current to flow when tool touches plate) and touch probe (that activates inverse | current cut when activated).
With my setup both need to be connected to the same input pin for my controller software to recognize activation.
I'm trying to avoid constantly flipping the input from active high to low when changing tools.
Is it possible to make a circuit that would allow current to flow when a switch is open and not flow when the switch is closed?
This seems possible with an external power source, but is it possible using the source feed of one input of the MX3660?
the N/C probe grounds the input of the inverter its connected to and switches off the open collector output
the 1K pull up resistor connected to the open circuit touch probe
switches on the first of a pair of inverters
the second inverter is wire ORed with the inverter connected to the N/C probe
the OR junction is pulled high by a 1K resistor and is buffered by another pair of inverters
the open collector out put that connects to the MX3660 input is switched off
when either the N/C probe opens or the touch probe is grounded
the MX3660 input is grounded so it will be set up as active LOW
and the LED indicator will be switched on
the circuit uses a 74HC06 open collector inverter
John
PS
a 12V version could be made using a CMOS open drain inverter like MC14069
the resistors will need to be 2.5 to 3 times higher
Last edited by john-100; 12-06-2017 at 04:44 PM.
Reason: add descripton of how the circuit works and ref to 12V version