Electronics and Communication Engineering - Digital Electronics - Discussion

Discussion Forum : Digital Electronics - Section 1 (Q.No. 34)
34.
The counter in the given figure is
Mod 3
Mod 6
Mod 8
Mod 7
Answer: Option
Explanation:

When counter is 110 the counter resets. Hence mod 6.

Discussion:
13 comments Page 1 of 2.

Vishnu said:   4 years ago
Here is the trick;

a(lsb) b c(msb)
a and b set to clear
So 110==4+2+0=mod-6.

Karthikraj said:   4 years ago
We get a high o/p at MSB ff is high. To get this B, C should be 1 and A is 0 to clock the middle ff. To get high at o/p is itself a reset state. Hence 110 is a toggle for reset. B=C=0 or B=0 and C=1 or vice versa are not possible. Hence the count is the answer i.e. mod-6.

Pryanka yadav said:   6 years ago
Given Mod-6 counter is correct.
(2)

Ashish said:   7 years ago
It is asynchronous counter with A as LSB (as clk is applied to A only).
Hence only up counting or down counting takes place.
Mod 6 is correct.

Bhanu said:   7 years ago
Mod 3 is correct answer. Because MSB is always at right side in counters topic itself.

Vijayalakshmi said:   7 years ago
JK flip flop state is 2N:1.
N=3 flip flop,so mod=2N=2*3=6.
Ans:B.

Darshith said:   8 years ago
I think its Better to Mention Which one is MSB and LSB in the question.

Jay Bhatt said:   8 years ago
Answer should be MOD 7. Because if it starts from 000 (A,B and C respectively) it will go as below,
ABC
000
100
010
110
001
101
011
back to
000

Hence it is Mod 7.
(1)

Lal said:   9 years ago
I think the states will be like this,

000
101
011
100
101
110 (but the flipflops would reset at this state) so it is a mod 6 counter.

Rakesh said:   9 years ago
Of course @Vijaya Venkat,

The answer follows like this I think.

000, 001, 010, 011 then counter resets. So the answer must be mod 3.


Post your comments here:

Your comments will be displayed after verification.