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[on:]

I would like to use a 4 digit 7 segment display for a project at university and we have been given an arduino nano.

Using the display in the link below would it be possible to connect this to the arduino nano and how many of the pins would it take up?

https://www.sparkfun.com/products/11408

[Reply #1 on:]

how many of the pins would it take up?

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.

[Reply #2 on:]

Thanks. I take it that they run off digital pins?

I also need to connect an accelerometer to the arduino nano, would I have enough pins left?

[Reply #3 on:]

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.

[Reply #4 on:]

The accelerometer I'd like to use is http://proto-pic.co.uk/triple-axis-accelerometer-breakout-mma7361/. I think it'll be I2C...

Does it matter which shift register I use?

[Reply #5 on:]

The 74HC595 is the standard choice for that sort of application.

[Reply #6 on:]

The 74HC595 is the standard choice for that sort of application.

Thanks. When connecting both the display and shift register to the breadboard / arduino will I need any resistors or other components?

[Reply #7 on:]

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.

[Reply #8 on:]

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?

http://uk.rs-online.com/web/p/darlington-transistors/4368451/

[Reply #9 on:]

No, just 7 or 8 resistors - they only go in the segment lines, not the common cathode lines.
Yes, that's the correct part.

[Reply #10 on:]

Thanks for the help

Would it be possible to use this buffer instead?

http://proto-pic.co.uk/darlington-driver-8-channel-uln2803-dip/

[Reply #11 on:]

I would like to use a 4 digit 7 segment display for a project at university and we have been given an arduino nano.

... will I need any resistors or other components?

... Is this the buffer I am looking for?

... Would it be possible to use this buffer instead?

What is your major?  If it's technology or engineering we don't want to do too much of your work for you.

Don

[Reply #12 on:]

I'm a mechanical student and have no idea on electronics.

We're going to be getting some help from an electrician at university, but I'd thought some research wouldn't go a miss first.

[Reply #13 on:]

Would it be possible to use this buffer instead?

Yes.