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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. $10$
in Digital Logic by Veteran (52.3k points)
edited by | 7.5k views
0
@Arjun Sir, please help to solve these type of questions.

5 Answers

+35 votes
Best answer

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

by Loyal (8.1k points)
edited by
+4
But then it should be called 3-64 decoder where is the six ??
+18
When the first decoder will select any one of the 8 decoders attached with their enable lines  then rest of the decoder will not be counted and first decoder will have 3 inputs leaving enable line and the second decoder will have remaining 3 line hence, 6 inputs
0
Please explain with a diagram
+44

If still not clear, here's the diagram just for you :)

0
Correct as the output depends on 6 inputs
0
I0, I1, I2 are most significant bits( input sequence- I0 I1 I2 I3 I4 I5 )
0

conceptual answer @partha sarkar

+18 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

by Boss (45.5k points)
+2
Can someone explain what I have written wrong in this Answer .I have got a downvote on it .Whoever gave it (downvote) please explain the correct method to solve it along with answer...
+2
I think Ur not wrong
+14 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

by Boss (33k points)
+2
Is this formula really true. Where does it come from, give some reference man.
0 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]

 

by (75 points)
–9 votes
ans 10
by Loyal (5.2k points)
+3
a simple way of doing these type of question

just pick up maximum values from from both given and want to design.

in above question:

given  3-to-8

design 6-to-64

therefore 8 and 64

now

divide (64/8)=8

again till we reaached 1 or less then 1

8/8=1

then add both of them 8+1=9 your answer..
Answer:

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