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Design a 3-bit counter using D-flip flops such that not more than one flip-flop changes state between any two consecutive states.
asked in Digital Logic by Veteran (413k points)
edited by | 824 views
0
0
Can we use Johnson Counter here??
0
Yes.

4 Answers

+19 votes
Best answer

State diagram will be as  (remember concept of gray code) 

State table and $3-bit$ synchronous counter with D FFs, will be as 

$\begin{array}{c|c|c}
\text{Present State}&\text{Next State}&\text{FF Inputs}\\\hline
ABC&\bar A \bar B \bar C & D_AD_BD_C\\\hline
001&011&011\\000&001&001\\010&110&110\\011&110&110\\100&000&000\\101&100&100\\110&111&111\\
\end{array}$

$D_{a}= AC+B\bar{C}$
$D_{b}= \bar{A}C+B\bar{C}$
$D_{c}= \bar{A}\bar{B}+AB=A\odot B$

answered by Veteran (56k points)
edited by
0
a johnson counter would also count the same
0
@Tushar, But Johnson's counter can count only 6 states with 3 flipflops(3 bits output), right? how can it count the same?
0

oh yeah i had got that now  @chauhansunil20th thanks for pointing out 

+1
don't hide your comments, it breaks the sequence of thread, other might have the same doubt.
+4 votes

We count in Gray Code :

answered by Boss (30.5k points)
+3 votes
  Present  state       Next state  
Q2 Q1 Q0 D2 D1 D0 Q2 Q1 Q0
0 0 0 0 0 1 0 0 1
0 0 1 0 1 1 0 1 1
0 1 1 0 1 0 0 1 0
0 1 0 1 1 0 1 1 0
1 1 0 1 1 1 1 1 1
1 1 1 1 0 1 1 0 1
1 0 1 1 0 0 1 0 0
1 0 0 0 0 0 0 0 0

D2=Q1Q0'+Q2Q0

D1=Q2'Q1+Q1Q0'

D0=Q2'Q1'+Q2Q1

answered by Boss (30.9k points)
0
I am getting

D2= Q1Q0' + Q2Q0

D1= Q2'Q0 + Q1Q0'

D0= Q2'Q1' + Q2Q1. Could you please check?
0
how to get Do,D1,D2 coloum pllzzz explain
0 votes

use gray code for 1 bit change

answered by Active (2.4k points)

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