Digital Logic Design (CS302)
You are required to design a 4-bit even up-counter using D flip flop by converting combinational
circuit to sequential circuit. The counter will only consider even inputs and the sequence of inputs
will be 0-2-4-6-8-10-0.
4-bit even up-counter using D flip flop
1. Draw the State diagram.
0101
0100
0010
0001
0000
1010
100
1
1000
0111
1101
1011
0011
0110
11111110
2. Generate State & Transition Table.
Q3 Q2 Q1 Q0 Q3 Q2 Q1 Q0 D3 D2 D1 D0
0 0 0 0 0 0 0 1 0 0 0 1
0 0 0 1 0 0 1 0 0 0 1 0
0 0 1 0 0 0 1 1 0 0 1 1
0 0 1 1 0 1 0 0 0 1 0 0
0 1 0 0 0 1 0 1 0 1 0 1
0 1 0 1 0 1 1 0 0 1 1 0
0 1 1 0 0 1 1 1 0 1 1 1
0 1 1 1 1 0 0 0 1 0 0 0
1 0 0 0 1 0 0 1 1 0 0 1
1 0 0 1 1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 1 1 0 1 1
1 0 1 1 1 1 0 0 1 1 0 0
1 1 0 0 1 1 0 1 1 1 0 1
1 1 0 1 1 1 1 0 1 1 1 0
1 1 1 0 1 1 1 1 1 1 1 1
1 1 1 1 0 0 0 0 0 0 0 0
3. Generate simplified Boolean Expression.
D3
Q1 Q0
Q3Q2
00 01 11 10
00 0 0 0 0
Present State Next State D Flip-Flop
01 0 0 1 0
11 1 1 0 1
10 1 1 1 1
3 3 2' 3 0' 3' 2 1 0D Q Q Q Q Q Q Q Q  
D2
Q1 Q0
Q3 Q2
00 01 11 10
00 0 0 1 0
01 1 1 0 1
11 1 1 0 1
10 0 0 1 0
2 2 0' 2' 1 0D Q Q Q Q Q 
D1
Q1 Q0
Q3 Q2
00 01 11 10
00 0 1 0 1
01 0 1 0 1
11 0 1 0 1
10 0 1 0 1
1 1' 0 1 0' 1 0D Q Q Q Q Q Q   
D0
Q1 Q0
Q3 Q2
00 01 11 10
00 1 0 0 1
01 1 0 0 1
11 1 0 0 1
10 1 0 0 1
0 0'D Q
Boolean Expression:
3 3 2 ' 3 0 ' 3' 2 1 0
2 2 0 ' 2 ' 1 0
1 1' 0 1 0 ' 1 0
0 0 '
D Q Q Q Q Q Q Q Q
D Q Q Q Q Q
D Q Q Q Q Q Q
D Q
  
 
   

4. Design the final Circuit diagram.

D-Flip-Flops(Digital Logic Design (CS302))

  • 1.
    Digital Logic Design(CS302) You are required to design a 4-bit even up-counter using D flip flop by converting combinational circuit to sequential circuit. The counter will only consider even inputs and the sequence of inputs will be 0-2-4-6-8-10-0. 4-bit even up-counter using D flip flop
  • 2.
    1. Draw theState diagram. 0101 0100 0010 0001 0000 1010 100 1 1000 0111 1101 1011 0011 0110 11111110
  • 3.
    2. Generate State& Transition Table. Q3 Q2 Q1 Q0 Q3 Q2 Q1 Q0 D3 D2 D1 D0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 3. Generate simplified Boolean Expression. D3 Q1 Q0 Q3Q2 00 01 11 10 00 0 0 0 0 Present State Next State D Flip-Flop
  • 4.
    01 0 01 0 11 1 1 0 1 10 1 1 1 1 3 3 2' 3 0' 3' 2 1 0D Q Q Q Q Q Q Q Q   D2 Q1 Q0 Q3 Q2 00 01 11 10 00 0 0 1 0 01 1 1 0 1 11 1 1 0 1 10 0 0 1 0 2 2 0' 2' 1 0D Q Q Q Q Q  D1 Q1 Q0 Q3 Q2 00 01 11 10 00 0 1 0 1 01 0 1 0 1 11 0 1 0 1 10 0 1 0 1 1 1' 0 1 0' 1 0D Q Q Q Q Q Q    D0 Q1 Q0 Q3 Q2 00 01 11 10 00 1 0 0 1
  • 5.
    01 1 00 1 11 1 0 0 1 10 1 0 0 1 0 0'D Q Boolean Expression: 3 3 2 ' 3 0 ' 3' 2 1 0 2 2 0 ' 2 ' 1 0 1 1' 0 1 0 ' 1 0 0 0 ' D Q Q Q Q Q Q Q Q D Q Q Q Q Q D Q Q Q Q Q Q D Q           4. Design the final Circuit diagram.