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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. $10$
edited | 6k views
0

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

edited
+3
But then it should be called 3-64 decoder where is the six ??
+12
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
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+24

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

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

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

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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

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
Is this formula really true. Where does it come from, give some reference man.
ans 10
+1
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

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