Digital Input Modules [Sinking]

hi All,

I have a Digital Sinking Input Module from Allen Bradley. The Typical Input impedance at 24V is given as 5.1kohms.

I wanted to interface a 5V signal to this module so I am using an Optocoupler to do it. My problem is with the value of resistor to be chosen from Emitter to GND of this optocoupler.

5v ---2k --LED(optocoupler)---GND

24V --- Collector ----Emitter---- PLC_Input[Sinking]
Emitter ---Resistor --- GND

I am confused if I should use 5K [to match the input impedance of the PLC input module] or to use 22K [so that parallel resistor value comes down to abut 5K, which is the input impedance of the module]

Any suggestions will be welcome

Thanks,

sunnyvaleplc

More information, like a manufacturer and part number, for the optocoupler would help.

hi Bernie,

The Optocoupler part Number is PC8171 from Sharp.

Thanks

5v ---2k --LED(optocoupler)---GND

24V --- Collector ----Emitter---- PLC_Input[Sinking]
Emitter ---Resistor --- GND

Click to expand...

Assuming 'GND' in the third line is the common of the 24 VDC power supply then it should be connected to the common terminal on the input module.

I'm not sure what you are doing with the 'Emitter --- Resistor". They shouldn't be needed at all.

So the hookup relative to the output side of the optocoupler and the 24VDC power supply would be:

24VDC -> Collector
Emitter -> PLC Input
24VDC Gnd -> PLC Input card common

hi Bernie,

Thanks for the reply. Yes, the optocoupler without any resistor will work, since now all the current is actually passing through the Input impedance of the Digital Input (5.1K ohms), so there is no loading.

However in my case, I am actually interfacing to a board that already has this optocoupler with a 5K ohm resistor from Emitter to GND and I get the Emitter point of the opto coupler to put into the digital input.

I wanted to be sure that this current design will not work before I raise red flags. 5K in parallel to 5.1K is 2.5K so there will be loading on the input.

Either the current design needs to remove the resistor all together like you suggested, or raise it to a much higher value to avoid the loading.

Am i right?

Thanks

The optocoupler rating shows 50 ma. Your 2.5K would only be about 10 ma. I don't thing you need to cut the resistor.