Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 1 (Q.No. 34)
34.
The counter in the given figure is
Answer: Option
Explanation:
When counter is 110 the counter resets. Hence mod 6.
Discussion:
13 comments Page 1 of 2.
Sharon said:
1 decade ago
Expand above expression?
Vijaya Venkat said:
1 decade ago
Actually A is LSB and C is MSB. Here clear is active low so that we have to give clr=0. For that B and C should be 1 and 1. Because the output of NAND must 0.
Deblina mukherjee said:
9 years ago
Not getting please elaborate, how 6?
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.
The answer follows like this I think.
000, 001, 010, 011 then counter resets. So the answer must be mod 3.
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.
000
101
011
100
101
110 (but the flipflops would reset at this state) so it is a mod 6 counter.
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.
ABC
000
100
010
110
001
101
011
back to
000
Hence it is Mod 7.
(1)
Darshith said:
8 years ago
I think its Better to Mention Which one is MSB and LSB in the question.
Vijayalakshmi said:
7 years ago
JK flip flop state is 2N:1.
N=3 flip flop,so mod=2N=2*3=6.
Ans:B.
N=3 flip flop,so mod=2N=2*3=6.
Ans:B.
Bhanu said:
7 years ago
Mod 3 is correct answer. Because MSB is always at right side in counters topic itself.
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.
Hence only up counting or down counting takes place.
Mod 6 is correct.
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