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A binary ripple counter is created by cascading N-number of flip-flops. The output frequency of every flip flop is named after its position i.e. then the MOD of the counter is ________.

(Assume N > 6)

A.) 8

B.) 11

C.) 14

D.) None of these

in Digital Logic by Junior (829 points)
edited by | 249 views
+4
$11?$

Edit :
Reasoning :
Every flop will reduce frequency by $2$ and cascaded flipflops are likely to depend on previous one .
$\therefore$ let initial  frq $=F$

Then : $\Large\frac{\frac{F}{2^N}}{\frac{F}{2^{\frac{N+1}{2}}}}$ $= \Large \frac{\frac{F}{2^6}}{\frac{F}{2^1}}$

$2N-N-1=10 \\N=11.$

$N=11$ which is total no of flipflop present in system,now the mod of this counter will be $2048$ with states transition as $0\rightarrow1....2047\rightarrow 0$
0
Answer is D

I think you missed this

$2^{11}=2048$

please share your approach
0
@ashish Please check and comment!
+1
2048 are no of states which can generate by the counter.

Mod 11 is correct
+1
@saxena0612 you got N=11

that is no. of flip flops

each flip flop is a MOD 2 counter

so N flip flops will act as

2*2*2*........(N times) = 2^N

please tell me where i am wrong
+2

@ashish pal

your reasoning and your answer are exacly correct.

@Avdesh Singh Rana

2048 are no of states which can generate by the counter.

Mod 11 is correct

the no. of distinct state generated by the counter is known as its mod, therefore it is $mod2^{11}$ counter

0
Yes ! I was holding the total no of flipflops but here mod is asked .States will be $2^{11}$ and mod will be 2048 Thanks nitish.

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