2-bit counter:

• Let's try to design a bit different 2-bit counter:

-   Again, counter outputs will be 00, 01, 10 and 11.

-   Now, there is single input, X. When X=0, counter value should increment on each clock cycle. But when X=1, value should decrement on the successive cycles.

• We will need 2 flip-flops again. Here are the 4 possible states:

Present State           Inputs        Next State

Q1            Q0               X             Q1              Q0

0                0                0               0            1

0                0                1               1             1

0                1                0               1            0

0                1                1               0            0

1                0                0               1             1

1                0                1               0            1

1                1                0               0            0

1                1                1               1            0

• If we are using D flip-flops, then D inputs will be the same as in the desirednext states.
• The equations for D flip-flop inputs are shown at the right.
• Does D0 = Q0’ make sense

D₁ = Q₁⊕Q₀⊕X

D₀ = Q₀'

• If we are using JK flip-flops instead, then we have to calculate the JK inputs for each flip-flop.
• Look at present and desired next state, and use excitation table on the right.

Present State    Inputs    Next State            Flip flop inputs

Q1        Q0         X         Q1         Q0       J1       K1      J0        K0

0           0           0          0         1         0         x         1         x

0           0            1           1          1       1         x         1         x

0           1            0           1         0        1         x         x         1

0           1            1           0         0        0         x         x         1

1            0           0           1          1       x         0         1         x

1            0            1           0         1       x         1         1         x

1            1            0           0         0       x         1         x         1

1            1            1           1         0       x         0         x         1

We can then find the equations for all the 4 flip-flop inputs, in terms of present state and inputs. Now, it turns out J₁ = K₁ and J₀ = K₀.

J₁ = K₁ = Q₀’ X + Q₀ X’

J₀ = K₀ = 1

Johnson Counter

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