Multiple choices can be correct. Mark all of them.
For the initial state of 000, the function performed by the arrangement of the J-K flip-flops in figure is:
Circuit behaves as shift register and mod6 counter
Clock cycle output
EDIT- This is Johnson counter which is application of Shift Register. And Johnson counter is mod 2N counter.
can you please explain how you got the states(100,110,111,011,001,000)
initial state is 000 so for the clock 1
i am getting Flip flop 1: J =1(Q3`) k=0(Q3) Q1=1
Flip flop 2: J=1(Q1) k=0(Q1`) Q2=1
Flip flop 3: J=1(Q2) k=0(Q2`) Q3=1
Flip flop 1: J =0(Q3`) k=1(Q3) Q1=0
Flip flop 2: J=0(Q) k=1(Q`) Q2=0
Flip flop 3: J=0(Q) k=1(Q`) Q3=0
same repeating ..
A. Shift Register
C. Mod-6 Counter
Given circuit diagram is for a Mod-6 Johnson Counter.
this(link) is how a Mod 6 counter looks like. Anything that has 6 states does not imply that it's a Mod 6 counter.
Yes, it is a Johnson counter, but not Mod 6 counter. Also, it shifts the binary number $000111$ so, we are calling it a Shift register.
@amar there are lot of e - resources over the web, that support that we can say it is mod6 counter
" The sequence for a Mod N counter may follow the binary count or may be any other arbitrary sequence. "
Between, this is from M. Mano, Section 6.5, 5th edition, pg 278
According to me, I think that A, C and D are the answer. Pleas, find the attached screen shot.
this is not D-FF. in JK FF, Q(next) = JQ' + K'Q then how can you consider that Q0N=Q1, Q1N=Q2....??
Raju Kalagoni When i/p's of JK-ff are complement to each other then it is D-ff.
Thankyou @ manojsahu @ shashank023
option : C
please mention if any mistake is found..
It is "Johnson counter" which is MOD2n counter where "n" number of FF so for this example its "6" .....if you want to visualize how 6 coming then draw a the table and see the flow of state will be like this here its "shifting" from one state to another
Answer:- Mod6, shift register