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Author Topic: Reading chess board with reed switches  (Read 3408 times)
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Belo Horizonte, MG, BR
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Hm... then sounds like the same amount of soldering trouble, but cheaper? I'm failing to see other benefits. I find soldering one reed switch to each diode and dealing with a single IC easier to do.

Thanks guys!
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That would be fine. A 595 for the rows, and 8 arduino inputs for the columns (with the internal pullup resistors turned on).

As for price:
For the Matrix method
Diodes are about 5p each, and the 74hc595 is around 41p,

64*5p + 41p = £3.60

For the parallel to serial method
Resistors are about 0.25p each, 74hc165s are around 31p each.

8*31p + 64*0.25p = £2.64


As for soldering, that is really up to you.
Disadvantages of 2nd:
- Have to solder 8 IC sockets rather than 1.
- Have to solder 64 resistors.
- You have 65 wires coming off the chess board rather than 17. Though if you put the electronics under the chess board this isn't an issue
Advantages:
- Dont have to solder 64 diodes
- Dont have to solder the reed switches into a matrix of rows and columns. Instead all reed switches go to an input, and to GND.

Its neck and neck really. The only real difference is that the matrix uses 9 pins, while the parallel to serial method uses 3.
Either is fine
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~Tom~

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Have you seen this project for chess on an arduino
http://www.andreadrian.de/schach/#Selbstbau_Schachcomputer_SHAH
It is in German however.
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I am trying to build a reed switch based chess board too. The scanning matrix aproach seems the most elegent solution.
 However there is another way. (I think, I am new to Arduinio and may have missed something) Have you seen the Centipede Shield? See http://macetech.com/store/index.php?main_page=product_info&products_id=23. This is a shield with 64 pins that links to the Arduino boad using only a few pins. It also uses ribbon style connectors so that would simplify wiring. I assume you would wire one end of all the reed switches to ground and the other 64 to ribbon ends.
I assume that this would simplify processing as you would just need to scan for a change of state. (Would you still need a resistor or diode on the Reeds?)

If you looked for two move combinations, ie the "From" move followed by the "To" move, then the first move would always be a change of state from closed (Occupied square) to open (Piece gone). The second move would be to an empty square (Open to closed) unless its a capute then its to a closed and you can assume you will get a combination as the old piece is first lifted (closed to open) and the replaced (Open to closed). But all are on the same square and the final state is closed (occupied).

Any practical observations most welcome.

Max
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It says about that shield.
Quote
Each I/O port corresponds to one MCP23017 IC.
So those chips have a built in pull up resistor so there is no need to add one to your reed switch.
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