Arduino 24v I/o shield question

1 Mar 2018

I’m trying to figure a few things out and I’m running into a roadblock that’s frustrating me pretty decent.

i have a ruggeduino and a 24v industrial I/o shield both from rugged circuits. I’ve read all the info available about this shield and I still just can’t figure out how to address the pins.

For example it has banks of digital inputs(B0-B9) and banks of digital outputs(P1-P8), how would I go about something like: int B1Pin = ?? .... pinMode(B1Pin, INPUT).... digitalRead(B1Pin) & digitalWrite(P3Pin)? Or something like that

Top Comments

mcb1 over 7 years ago in reply to masterbri

please bear with me, I’m still learning!
We're all still learning, but just at different stages of the process.

You framed your question very well which meant the readers were able to provide answers, or suggestions easily.

Mark
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masterbri over 7 years ago

I was reading up on this subject some earlier today and came across this piece: http://tronixstuff.com/2011/08/26/tutorial-maximising-your-arduinos-io-ports/

i think this explains exactly what I was after. Using the binary, hexadecimal or decimal equivalent of which I/o I’m trying to use at any given time.

thank you all for your help!!
please bear with me, I’m still learning!
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mcb1 over 7 years ago in reply to masterbri

There is an example sketch on the 24v I/O board page, near the bottom.
It only talks about a single input but should be enough.

You won't need the PinMode commands as it's using I2C via A4, A5 of the Arduino.

The Ruggeduino has 24v inputs due to the resistor and zener.

Mark
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genebren over 7 years ago in reply to masterbri

Brian,

It appears that the 'address' is set by the following jumper:

Digital Inputs and Outputs are driven by an I2C I/O Expander (MCP23017) on Arduino Pins A4 and A5. The I2C Address of the expander may be configured by shorting the jumper J1 (I/O ADDR). The default address is 0b0000001, with the jumper shorted the address becomes 0b0000000.

From there, the I2C command, encodes the address in the first byte of the message. It appears (not stated in the write-up, thanks dougw), that the second byte would command the outputs. It looks as if the 8 bits of this message control the 8 outputs, so that all outputs could be commanded in a single command. The ordering of the bits was not specified, so I would assume that the higher order bits control the higher numbered outputs.

Good Luck!
Gene
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masterbri over 7 years ago in reply to dougw

I get that, and thank you. But I’m just still not understanding how to address a specific I/o? are the individual I/o addressed as: 0B000001, 0B000002...? If so how do i tell it to use say a4 & 0B000001?
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