Digital Logic - Karnaugh Map

Question

The initial problem starts like this. There are 6 states. At each state when w=1 move to the next state, when w=0 then stay at the current state. At each state display a number using a standard 7 led display (BCD). Those numbers are 8 -> 1 -> 9 -> 4 -> 2 -> 2.

So here is my attempt at this problem. I start with a state table: From left to right y2,y1,y0

    w=0  w=1  a  b  c  d  e  f  g
000|000  001  1  1  1  1  1  1  1
001|001  010  0  1  1  0  0  0  0
010|010  011  1  1  1  1  0  1  1
011|011  100  0  1  1  0  0  1  1
100|100  101  1  1  0  1  1  0  1
101|101  000  1  1  0  1  1  0  1

Then Yo Y1 & Y2 equations are made using karnaugh maps

    y1.y0               _        _
w.y2 00 01 11 10   Y0 = w.y0 + w.y0
  00 0  1  1  0    
  01 0  1  d  d
  11 1  0  d  d
  10 1  0  0  1

    y1.y0               _        _  _            _
w.y2 00 01 11 10   Y1 = w.y1 + w.y2.y1.y0 + w.y1.y0
  00 0  0  1  1
  01 0  0  d  d
  11 0  0  d  d
  10 0  1  0  1

    y1.y0               _      _  _  _
w.y2 00 01 11 10   Y2 = w.y2 + y2.y1.y0 + w.y1.y0
  00 0  0  0  0
  01 1  1  d  d
  11 1  0  d  d 
  10 0  0  1  0

Then the outputs need addition maps created.

    Y1.Y0                     _  _ 
  Y2    00 01 11 10  a = Y2 + Y0.Y2
      0 1  0  0  1
      1 1  1  d  d


    Y1.Y0              
  Y2    00 01 11 10  b = 1
      0 1  1  1  1
      1 1  1  d  d


    Y1.Y0                _
  Y2    00 01 11 10  c = Y2
      0 1  1  1  1
      1 0  0  d  d


    Y1.Y0                     _  _  
  Y2    00 01 11 10  d = Y2 + Y0.Y2 
      0 1  0  0  1
      1 1  1  d  d


    Y1.Y0                     _  _  _
  Y2    00 01 11 10  e = Y2 + Y0.Y1.Y2
      0 1  0  0  0
      1 1  1  d  d


    Y1.Y0                _  _
  Y2    00 01 11 10  f = Y2.Y0 + Y1
      0 1  0  1  1
      1 0  0  d  d


    Y1.Y0                          _  _
  Y2    00 01 11 10  g = Y1 + Y2 + Y1.Y0
      0 1  0  1  1
      1 1  1  d  d

Currently I am using a 3 bit D flip flop counter to create the 6 inputs.

The display shows.

 _        _        _
|_|   |  |_|  |_|   |
|_| |     _|    | |_   _

Is there a mistake with the logic or is it possible that the counter could be creating this problem?

Answer 1

As I understand, your state machine has 6 states? And in last two states the same digit need to be displayed?

I think it is possible to made it with T flip-flops.

1) Connect each output of T flip-flop to input of next, three flip-flops needed to hold 6 states.

2) You need the "reset circuit" that resets all triggers when combination on outputs is equal to 110 (6). So, the output on w=1 of T flip-flops will be next: 000 w 001 w 010 w 011 w 100 w 101 w 110->000* w 001 etc (*reset moves flip-flops to init state). This is the first function: RST.

3) You need to create an encoder do convert codes from 0 to 5 to 7 signals to LED display.

So, the biuilt ruth table will looks like next:

#TABLE: t3,t2,t1 => a,b,c,d,e,f,g,RST
 000 => 11111110
 001 => 01100000
 010 => 11110110
 011 => 01100110
 100 => 11011010
 101 => 11011010
 110 => 00000001
 111 => 00000000

Create 8 K-Maps and minimize them, or use any other minimization method. I got this results:

a = t3 !t2 | !t3 !t1;
b = !t3 | t3 !t2;
c = !t3;
d = t3 !t2 | !t3 !t1;
e = !t2 !t1 | t3 !t2;
f = !t3 t2 | !t3 !t1;
g = !t3 t2 | t3 !t2 | !t3 !t1;
RST = t3 t2 !t1;

Answer 2

Typing out the entire question again assisted myself in figuring out the part that was done wrong.

The problem was in the Y2 Karnaugh Map.

By looking at the outputs I was able to see which pins were not working and trace them back to the source of the error.