# 7 segment counter, sequence:13567-13567 and so on

Discussion in 'Homework Help' started by B-Sparkes, Dec 30, 2010.

1. ### B-Sparkes Thread Starter New Member

Dec 28, 2010
3
0
I wonder if anyone can help me with this one?
I have to design a counter using flip flops to display on the seven segment display unit: 13567-13567 and so on.

I have to build it on multisim so what would i need to do this:

How many flip flops and what types?

what and if any gates would i use and what type?

and how would i build this?

If you can help me on this i would be very grateful?

Brian

2. ### beenthere Retired Moderator

Apr 20, 2004
15,815
293
Does that mean the count goes 1, 3, 5, 6, 7 and back to 1?

How many flip flips does it take to indicate the value of 7 in binary? Can you make a counter with flip flips that goes 0 - 7?

3. ### B-Sparkes Thread Starter New Member

Dec 28, 2010
3
0
Yes, im looking for a circuit that i can build on multisim that will produce one of these sequences:1 3 5 6 7 1 3 5 6 7 1 3 5 6 7 1 3 5 6 7 , 0 2 3 4 6 0 2 3 4 6 0 2 3 4 6, 2 4 6 0 7 2 4 6 0 7 2 4 6 7 0 etc exactly one of these would be perfect!

Thanks

Brian

4. ### Georacer Moderator

Nov 25, 2009
5,150
1,271
Since there is only one "jump" in the sequence, you can do the following:
Have a counter count from 2 to 7 and pass the result through a logic circuit that recognises the binary 2. Activate a quad 2-to-1 MUX with that result.
If the 2 is present, have the MUX conduct a constant binary 1. If not, have it give the result of the counter unaltered.

Is that clear?

5. ### djsfantasi AAC Fanatic!

Apr 11, 2010
3,519
1,282
There is a jump at 4 as well... But the same solution for jumping the 2 count can be applied to the 4 count...

6. ### Georacer Moderator

Nov 25, 2009
5,150
1,271
Well, yes and no:
You can wire the 4 so that it shows 3 or 5, but then that number will be displayed for 2 clock cycles.
Since there are two jumps I think it is convenient now to build an FSM, or use 3 8-to-1 MUXs along with a counter to build the sequence.