8 (or 7 if you don't want the decimal point) for the segments plus 4 for the common connection so 12 in all.
You will also need to multiplex the display to make it look like there is something different on each display.
You have 20 IO pins available.
18 if you reserve D0, D1 for serial interfaciing.
Commit 11 or 12 to the display, leaves 6 or 7 free for the accelerometer.
Is that an analog outout device> I2C or SPI device?
If you need to free up pins, you can use a shift register to write the segments for the display, and gain 4 or 5 back that way (instead of 7/8 segments, have a datapin, clock pin, and latch pin for most shift registers). If you use a 2nd shift register daisy chained with the first, then you can gain the 4 common anode/cathode pins back also.
Yes, you will need 180 to 220 ohm current limit resistors, 1 for each segment.
You will need 100nF caps, one for each shift register's Vcc pin.
You will need a ULN2003 to buffer the common cathode signals - each cathode will have up to 160mA flowing thru it if all 8 segments are on.
The shift register, or arduino pin, will source current into the anode segments, the ULN2003 will sink current.
CrossRoads:
Yes, you will need 180 to 220 ohm current limit resistors, 1 for each segment.
You will need 100nF caps, one for each shift register's Vcc pin.
You will need a ULN2003 to buffer the common cathode signals - each cathode will have up to 160mA flowing thru it if all 8 segments are on.
The shift register, or arduino pin, will source current into the anode segments, the ULN2003 will sink current.
So thats about 13 resistors then? Is this the buffer I am looking for?