You are not using enough voltage.
The page you referenced notes that there multiple LEDs in series.
They say:
- Large (58mm) 7-segment LED display
Digits color can be either red or green.
Type: 12101BEG
If max: 30mA
If peak 1/10 dutycycle, 0.1ms: 100mA
Vf red: 7.2-8.8V
Vf green: 8.8-10.0V
Each segment consists of four LEDs connected in series.
In most cases simply trying every combination in turn will work.
There is a small chance that you will destroy segment LEDs with reverse voltage.
Supply two leads from psu via a resistor so you can easily swap both leads.
Try each combination in turn.
A1 A2 A3 A4 A5 B1 B2 .......
The red wire is hooked up to the 5v output on the arduino and the blue to the ground. As can be seen the decimal point has been lit up in green, and if I move the red wire four pins to the left the decimal point turns red instead.
As they say CA = common Anode = common +ve you may have
2 x Anode's
+ 8 x Cathodes.
So Bottom right may be common green anode and
bottom left = common red anode.
SO increase Vsupply to say 8V, use a 10k resistor for initial safety, put V+ on bottom right and try all pins with v-. If that works then put V+ via 10l on bottom left and repeat.
The supplier should be able to provide a datasheet.
Okay, I can see the changes.
As shown now, the LEDs on the receiving IC will block the data high as they are reversed. The LEDs on the transmitting side aren't a good idea either. If you want LEDs you need to put them from the data line to ground through a resistor, not in series with the connection to cable socket.
Just seen your picture - it looks like you have removed the LEDs, good (I was just about to suggest this.. :-) )
So now it looks as if you have direct connections from IC1 to IC2. If this is the case then if the code (looks reasonable at a glance) and IC wiring are correct then it should work.
If you can confirm with a multimeter that the input pins are seeing a high (or low) voltage and the read value is different, then this would confirm the issue is one or the other of the above. Maybe just apply a known voltage directly and see if you can read that okay)
However, if you are reading different values when the pullups are on/off then that would seem to indicate the read is correct. Try reading a direct voltage and post results, I'm just checking the datasheet for the ICs, will add more shortly.
EDIT - about the pullups:
You can use the internal pullups if you don't mind the line "relaxing" to high when not driven (i.e. default state 1) These are often used for interfacing with open drain buses, or for button to ground, etc to save an external pullup.
If you want to have the lines default state low though, (as is the case for you) you need a pulldown to stop the high impedance floating. Since the IC in question doesn't have internal pulldowns, you need to add them externally.
EDIT - Doh! I've just seen the problem...
In your code you set 1 pin at a time to output and all the rest to inputs. This means that if you have internal pullups on, the undriven pins will default to high! When a pin is set to input, it is high impedance, so effectively it's like disconnecting that end of the line, and the weak pullups will pull the receiving end high.
You need to keep all pins as outputs, and just set one high at a time, this will keep all the pins driven - try this with the pullups on, it should work.
If you know the lines will be driven correctly all the time, you don't need the pullups, but it doesn't hurt to keep them on.
Here is the relevant code (in the intialise registers function):
i2c_start_wait(SLAVE_ADDRESS(0x4E)+I2C_WRITE); // Address Slave 1
i2c_write(0x00); // Set memory pointer to the IODIRA register (IODIRA address is 0x00 - see Page 9)
i2c_write(~(Value)); // Set only one pin at a time as an output and everything else as inputs
i2c_stop();
Change it to:
i2c_start_wait(SLAVE_ADDRESS(0x4E)+I2C_WRITE); // Address Slave 1
i2c_write(0x00); // Set memory pointer to the IODIRA register (IODIRA address is 0x00 - see Page 9)
i2c_write(0x00); // Set all pins as outputs
i2c_stop();
Best Answer
Connecting separate pads of the crystal is a bad idea, but you don't need to.
It's a bad idea because (on some packages) the metal case around the crystal is connected to the spare pins, and so connecting them would short the crystal through the case.
You don't need to because the pinout is standardized for this package—the crystal is on pins 1 and 3. The datasheet you linked for the ECX-32 is confusing, because it shows the schematic from underneath, but it still follows the standard.