A sequential circuit has two D flip-flops A and B, two inputs X and Y, and one output Z. The flip-flop input equations and the circuit output are as follows.

D

_{A}= X' Y + X AD

_{B}= X' B + X AZ = B

Note that this question was originally entitled "Count-down Counter" in error. This question is simply about flip-flop circuits and state tables.

a. Draw the logic diagram of the circuit.

b. Tabulate the state table.

Design a 2-bit count-down counter. This is a sequential circuit with two flip-flops and one input X. When X=0, the state of the flip-flops does not change. However, when X=1, the state sequence is 11, 10, 01, 00, then back to 11 and the sequence repeats. (Under normal use, X=1 is applied for a short time during which the circuit counts down by one to obtain the final stored result. Additional pulses of X=1 continue the counting down operation.)

The state table is shown below:

This leads to the Boolean logic expressions:

A

^{+}= X’A + X(AB + A’B’) B^{+}= X’B + XB’These expressions can be drawn as block circuit diagrams using AND-OR gates, or including XOR and XNOR gates. The A and B inputs on the right hand side are the original stored values in the flip-flops and the resulting expression values are

fed backto the inputs of the flip-flops to produce the final outputs A^{+}and B^{+}. For this type of circuit the D type flip-flop is quite suitable.The block diagram is left as a straightforward exercise for students.

Design a sequential circuit with two J-K flip-flops A and B and two inputs E and X. If
E=0, the circuit remains in the same state regardless of the value of X. When E=1 and X=1, the
circuit goes "up" through the state transitions from 00 to 01 to 10 to 11 and back to 00, then
repeating. When E=1 and X=0, the circuit goes "down" through the state transitions from 00 to 11 to
10 to 01 back to 00, then repeating. (Note that if the E=1 and X inputs are applied only
long enough to produce a single state transition, then a new set of inputs E=1 and the
*complement value* of the previous X input is applied, the counter can be used to count
up/down first, then count down/up).

The state table is shown below:

This leads to the Boolean logic expressions:

A

^{+}= EX’(AB+A’B’) + EX(AB’+A’B) + E’A B^{+}= EB’ + E’BThese expressions can be drawn as block circuit diagrams using AND-OR gates, or including XOR and XNOR gates. The A and B inputs on the right hand side are the original stored values in the flip-flops and the resulting expression values are

fed backto the inputs of the flip-flops to produce the final outputs A^{+}and B^{+}.It is required that you use J-K flip-flops for this question. By connecting the expression outputs to the J-input, and the

complementof the expression to the K-input, you basically transform the JK to a D type flip-flop.The block diagram is left as a straightforward exercise for students.

© All information on this website is Copyright © 2009 by Robert D. Kent. All rights reserved.