Digital Electronics - Shift Registers - Discussion
Discussion Forum : Shift Registers - General Questions (Q.No. 11)
11.
A modulus-12 ring counter requires a minimum of ________.
Discussion:
11 comments Page 1 of 2.
Insha said:
5 years ago
Can anyone explain it briefly with the concept?
Ritvik Khandelwal said:
6 years ago
Because every register contain a flip flop.
Simon said:
8 years ago
The right answer is 12. Because in ring counter n ff's generate n states. In johnson n ff's generate 2n states. What you guys answering is for mod N counter it needs n ff's (N<=2^n).
Sahithi said:
1 decade ago
May I known why is it so @Velkuargn. Please explain.
Velkuargn said:
1 decade ago
For mod-12 counter only 4 FF.
For mod- ring counter 12 FF.
For mod- ring counter 12 FF.
Aruna said:
1 decade ago
12 flip flops because ring counter uses states n so here modulus is 12.
Pooja singh said:
1 decade ago
It should be 4ff as per the condition N<2^n. But here in options 6 is the nearst answer.
Pydiraju said:
1 decade ago
In the ring counter n flip-flops are used for the mod-n.
Mahesh dadhich said:
1 decade ago
It should be 6 ff because there is condition that N<2^n where n is smallest number for flip flop.
Venkat said:
1 decade ago
Please explain me the answer.
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