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Ardbox Analog HF and RS-485 and disabled pins - confusion.

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Richard Fabo

 

(I asked this also as a ticket, but as forum's question is this more suitable).

I want to use RS-485 HS on Ardbox Analog HF.

The documentation says the following:

Manual: Rev. 0: 30-08-19 (also part on https://www.industrialshields.com/technical-features-industrial-plc-arduino-ardbox-20-ios-analog-hf-modbus):

Page 23 + 24:
Top zone: choice between RS-485 and I0.2 + I0.3
(one option excludes the other).

Page 22:
Left zone: choice between RS-485 and A0.1 (Q0.1) + A0.2 (Q0.2)
(one option excludes the other).

but page 27 says: Avalaible all 10 digital outputs.
(which is probably mistake, as A0.1 (Q0.1) + A0.2 (Q0.2) are disabled)

and:

On the pinout here: https://www.industrialshields.com/web/image/biztech.product.images/58/image?unique=9312080
is marked (left bottom corner), that
Arduino Pin 0 goes to A - which is actually Q0.9
Arduino Pin 11 goes to B - which is actually Q0.1
Arduino Pin 10 goes to Y - which is actually Q0.2
Arduino Pin 1 goes to Z - which is actually Q0.8
(so not the I0.2 and I0.3)

On the other hand, in manual there is a remark that hardware jumper switches between RS-485 and RS-232 and Arduino pins 0, 1 and 4, 8 which belongs to Q0.8, Q0.9 and I0.2 and I0.3
(so not the Q0.1 abd Q0.2).


And one more inconsistency: It is mentioned in the manual that although the Top zone talks about Q0.8 and Q0.8, I0.2 and I0.3 must be taken into account. This is clear.
However, as far as the Left zone is concerned, there is (physically) written: D19 / I0.8 and D18 / I0.9.

So the question: how is it really? Which DI or DO are inactive when using RS-485 HS?
I understand that development and changes during the time caused this uncertainty.

And one more interesting question:
In the blog: https://www.industrialshields.com/blog/industrial-shields-blog-1/post/learning-the-basics-about-rs485-of-an-industrial-plc-195
it is mentioned that Half Duplex is connected to A + B, Full duplex of course adds Y + Z.

Our device only uses Half Duplex. So if we knew that which pins were connected to connections A and B, then we could (apparently) use the free pins of Y and Z (if the dip switch was in the correct position) as DI and DO ... It's like this possible?
The premise is to know where A, B, Y, Z are connected, because it seems that the description from the pinout (see above) is probably not correct ...

Thanks for the answers. I will process the correct information on our website in the local language ...

Richard Fabo

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Andrei Postolache
Best Answer

Hello Richard,


This mistake is due to the previous model of Ardbox Analog, now you can have a look to the last edition of the User Guide.

Sorry for this misunderstood.


Thank you.

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