Electronics and Communication Engineering - Digital Electronics - Discussion


The counter in the given figure is

[A]. Mod 3
[B]. Mod 6
[C]. Mod 8
[D]. Mod 7

Answer: Option B


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

Sharon said: (Mar 16, 2015)  
Expand above expression?

Vijaya Venkat said: (May 31, 2015)  
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: (Aug 8, 2016)  
Not getting please elaborate, how 6?

Rakesh said: (Aug 21, 2016)  
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.

Lal said: (Sep 16, 2016)  
I think the states will be like this,

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

Jay Bhatt said: (Aug 18, 2017)  
Answer should be MOD 7. Because if it starts from 000 (A,B and C respectively) it will go as below,
back to

Hence it is Mod 7.

Darshith said: (Dec 4, 2017)  
I think its Better to Mention Which one is MSB and LSB in the question.

Vijayalakshmi said: (Oct 20, 2018)  
JK flip flop state is 2N:1.
N=3 flip flop,so mod=2N=2*3=6.

Bhanu said: (Jan 11, 2019)  
Mod 3 is correct answer. Because MSB is always at right side in counters topic itself.

Ashish said: (Jan 22, 2019)  
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.

Pryanka Yadav said: (Jan 10, 2020)  
Given Mod-6 counter is correct.

Karthikraj said: (Apr 23, 2021)  
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.

Vishnu said: (May 8, 2021)  
Here is the trick;

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

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