IT management

Question: Show that a Johnson counter with n flip-flops produces a sequence of 2n states. List the 10 states produced with five flip-flops and the boolean terms of each of the 10 AND gate outputs.

ANSWER

A Johnson counter is a sunchronous ring counter with theinverted output of the last flip flop connected back to the D inputof the first flip flop.

An n flip flop Jhonson counter sequence startingfrom all 0's,

First clock pulse  insertsa 1 to  the left most FF

Second clock pulse inserts 1 to the 1st FF and shifts 1 to thesecond FF

This sequence continues until all the n flip flops are 1 ,that is at the end of n clock pulses all the FFs are HIGH.This goes thrugh n states

The next clock pulse inserts a 0 to the left most and thesequence conitues until all the FFs are LOW. This goes through nstates.

Thus a total of n+n = 2n different states are present in an-bit Jhonson counter

With n=5, the truth table and the ouput of hte correspondingdecoding AND gate is as shown below

Count   A   B   C   D   E         ANDgate output

 1         0   0    0    0   0         A'E'

 2         1   0    0    0   0         AB'

 3         1   1    0    0   0         BC'

 4         1   1    1    0   0         CD'

 5         1   1    1    1   0         DE'

 6         1   1    1    1   1         AE

 7         0   1    1    1   1         A'B

 8         0   0    1    1   1         B'C

 9         0   0    0    1   1         C'D

 10      0    0    0   0    1         D'E

At the 11th clock pulse the output will be 00000 and thesequence continues.

Thus we see that there are 10 states for a 5 flip flop Jhonsoncounter.