14-segment display for Arduino

[jogin]

I want to connect a 14-segment LED display to Arduino. Don't know how to do it?

[Nuke]

the way I see it is that you connect 7 segments to one port and 7 segments to the other port. Or you use a double expander from Michrochip. You connect it via I2C. I can still think of multiplexing two groups of outputs, each with 7 segments. This will require an external multiplexer with 1x 7 inputs and 2x 7 outputs. You can connect 7 inputs to the 8-bit port, for example D0 to D6 and to D7 the switching of the multiplexer.
Then there is Charlieplexing developed by Maxim. There are extra circuits for this that can display ASCII characters. It also depends on how many places the display should have.

[solaris]

here is an example of a 6x 14 segment display with arduino. They also have a digital clock program there.
http://timewitharduino.blogspot.com/201 ... x6955.html

[jogin]

I need to connect a total of 9 digits, each with 14 segments. That would be 9 double expanders. I also looked something up about Charlieplexing, but didn't get it at all. I don't know how to connect it at all. So what if I connect 7 segments to port D and 7 segments to port B and put a 4-bit demultiplexer 0000 to 1000 on port C and scan the individual digits?

[solaris]

I don't know what the display is for, but the clock has the advantage of generating its own data. When the MCU will be getting data from somewhere else, such as over a serial line, it will have to stop the multiplex display on every interrupt. The display will then be dead. You will have to include an extra procedure in the display service program, which will turn off the entire display in the event of any interruption. For that reason, it is better to use I2C two-port expanders for each digit.

[Nuke]

with 9 digits, I see the simplest solution to use the 8x double expander from Michrochip, MCP23016. You only have to use 8 of them, because you can't connect more of them to one I2C line. This means that the 9th digit will always use 2 bits from each expander with address 0 to 6. You will have two free bits left on the last expander (with address 7).