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+28 votes

Consider a 4-bit Johnson counter with an initial value of 0000. The counting sequence of this counter is 

  1. 0, 1, 3, 7, 15, 14, 12, 8, 0
  2. 0, 1, 3, 5, 7, 9, 11, 13, 15, 0
  3. 0, 2, 4, 6, 8, 10, 12, 14, 0
  4. 0, 8, 12, 14, 15, 7, 3, 1, 0
asked in Digital Logic by Boss (29.5k points)
edited by | 4.6k views
In case of synchronous counters choosing msb and lsb does not matter , it can be from any side because the operation is done in parallel per clock pulse but in case of asynchronous counter it matters because it is done serially . Since johnson counter is synchronous it wont matter here so answer can be both a and d.
Both options $A$ and $D$ are actually reverse orders of each other. So, if one can be done other can also be done by changing the order of MSB and LSB.
@mcjoshi for asynchronous counter what will be answer???

5 Answers

+12 votes
Best answer

$\text{Johnson Counter}$ is a switch‐tail ring counter in which a circular shift register with the complemented output of the last flip‐flop connected to the input of the first flip‐flop.

(D) is the correct answer!

answered by Boss (42.4k points)
selected by



these line of Mano specifies A) cannot be answer

so here we should forget about that taking which one is MSB & which one is LSB concept, right??
we should only focus on the features of Johnson counter which is specified in Morris Mano.
Is this the approach of this question???
+19 votes

option D

0000 - 0

1000 - 8

1100 - 12

and so on.

answered by Active (1.2k points)
why not (A)....?? if i consider q3 as MSB ... i got option (a) ... and here not mention what is MSB and LSB ... or  

i consider D0 = Q3' , D1=Q0, D2 =Q1, D3= Q2

  Q3N =D3, Q2N= D2 , Q1N= D1, Q0N= D0.
option A and D both are correct .
Isn't MSB defined in Johnson counter?
as per i know you can take any one as msb ( Q0 or Q3 ) , its depends on the user .
I have to see that, but GATE key had only D.

Arjun Sir I marked A in exam...lost markssadin wiki link it is given that clocl should be applied to Msb...but i dont think there is any such rule

yes sir in answer key ans is (D)... but we can do any method ....
@csegate2 nice , its a gif right ?? is it made by you ? or same old policy ctrl c + Ctrl v  :P
i dont know who you are , but I will find you, I will learn the secret of that pic :D ....

Listen everyone, I'll kill you I'm saying 

for mithilesh

The answer can be either A or D depending on which one you choose as MSB since nothing is given .
The persons answering it either A or D should be given mark.

What is the mistake in this? please explain.

I think correct answer is A

u r missing 14 only . rest is OK.
The flip flops should be arranged like D3->D2->D1->D0, in the same sequence as that of the table.

from morris mano 


If the answer would have been asked for ring counter then the answer is A. The johnson counter is switch‐tail ring counter so the complement of the output is given as input so answer is D. It would act like T-FF. Am i right?

+3 votes
Johnson counter is ring counter hence after period of clock cycles its output will repeat .

here 0000 is given as initial state and hence 1 is given as preset in first flipflop and then this one is propagated to consecutive flipflops producing









as output .
answered by (185 points)
+2 votes

Johnson counter /twisted ring counter :-

"N" bit then there will be "2N" states 

Option B easily eliminated as it has 9 state form rest we will have to check out.

Answer :- D 


answered by Loyal (8.3k points)
0 votes

Counting procedure of a Johnson's Counter

  1. Starting from all 0's , each shift operation inserts 1's from the left until the register is filled with all 1's.
  2. When the register has all 1s,each shift operation inserts 0's from the left until the register is filled with all 0's.
  3. Go to step 1.


0000 $\rightarrow$ ${\color{Red} 1}$000 $\rightarrow$ ${\color{Red} 1}{\color{Red} 1}$00 $\rightarrow$  ${\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$0 $\rightarrow$  ${\color{Red} 1}{\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$ $\rightarrow$  ${\color{Blue} 0}{\color{Red} 1}{\color{Red} 1}{\color{Red} 1}$ $\rightarrow$ ${\color{Blue} 0}{\color{Blue} 0}{\color{Red} 1}{\color{Red} 1}$ $\rightarrow$  ${\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}{\color{Red} 1}$ $\rightarrow$ ${\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}{\color{Blue} 0}$


0 $\rightarrow$ 8 $\rightarrow$ 12 $\rightarrow$ 14 $\rightarrow$ 15 $\rightarrow$ 7 $\rightarrow$ 3 $\rightarrow$ 1 $\rightarrow$ 0


$\therefore$ Option D is correct answer.

answered by Boss (16.1k points)

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