ABB pulse output to discrete input module

[KirkC]

I want to read the pulse output from an ABB mag flow meter with a Modicon BMXDDI1602 discrete input module (positive 24VDC sinking). I will only get a pulse about once per minute so I don't need an HSC module.

The ABB book (see attached) shows a diagram for the pulse output, passive, optocoupler. I think I need this one, not the active one. The diagram shows a resistor, R(b). I do not understand what this resistor is for or what value it should have. The book gives a calculation for the value. I'm guessing the U(ce) is the 24 volts, but what is I(ce)? Doesn't the discrete input module itself act as a load?

Thanks for helping!

 

[drbitboy]

I think it's standard transistor jargon cf. http://learn.parallax.com/tutorials...n-shield-arduino/chapter-6-light-sensitive-15


I(ce) => current (I) from collector (C) to emitter (E).


U is voltage. So Ohms law [Voltage = Current times Resistance] rearranges to R = U/I

The resistor keeps too much current from flowing, because when the light hits the phototransistor, its resistance goes from high to very low.





Am I close?

 

[parky]

If you notice in the manual there is a * next to RB so probably on the bottom of the page there is a note indicated by * that should give you some info but typically a DC 24v input will probably only need max of 4ma so you could work out the resistance from that. Although the input circuit will have quite a high resistance so probably could be either ignored or for extra safety a small value of say 500 ohms.

 

[Lare]

RB resitor on drawing represinting also your device. It can be PLC, relay, counter unit.

If it is between current/resistance values allready, then you don't need extra resistor or relays between.