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34 votes
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How many $3$-to-$8$ line decoders with an enable input are needed to construct a $6$-to-$64$ line decoder without using any other logic gates?

  1. $7$
  2. $8$
  3. $9$
  4. $10$
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5 Answers

Best answer
59 votes
59 votes

Answer is C:

To get $6:64$ we need $64$ $o/p$

We have $3:8$ decode with $8$ $o/p$. So, we need $64/8=8$ decoders.

Now, to select any of this $8$ decoder we need one more decoder.

Total$=$ $8+1= 9$ decoders

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26 votes
26 votes

Answer : C

In first Level we need 64 / 8 = 8 Decoder

In second Level to cover 8 select lines Which are coming out from 8 decoder we need  8 / 8 = 1 Decoder

Total =8+1= 9 decoders

19 votes
19 votes

Quickest way to solve this type of problems will b:

Number of m bit MUX/DeMUX/Decoder/Encoder to construct N bit  MUX/DeMUX/Decoder/Encoder is:

ceil (N-1)/(M-1)

In given problem,  (N-1)/(M-1)= 64-1/8-1 =63/7 =9

1 votes
1 votes

Answer - (C) 9

In general to construct a log2m x m decoder using log2n x n decoders,

  • Number of levels needed, k = ceil(lognm)

  • Total number of devices needed = ((m)/(n-1))*(1-1/Nk)

Here m=64 and n=8. So k =2 and no. of devices = 9.

[log2n means logn base 2, similarly lognm means logm base n]

 

Answer:

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