GATE CSE 2007 | Question: 8, ISRO2011-31

Question

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?

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Comments

@Prashant_Dubey, Here if we realize 6 X 64 using 4 X 16 then to achieve it by using just 4 such decoders then we need other logic gates for the purpose of Enabling the necessary Decoder.
If other Logic Gates are prohibited to be used then we need 5 such decoders.

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Similar concept question Asked in ISRO 2025  
How many 3 to 8 line decoders are required for a 1 of 32 decoder?

1 , 3 , 4 , 8

Set A  Q 50.

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aap toh smart ho, konsa college mila bhaiya aapko ?

Answer 1

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

Comments

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 )

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conceptual answer @partha sarkar

Answer 2

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

Comments

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...

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I think Ur not wrong

Answer 3

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

Comments

@Deepak Poonia sir is this formula correct?

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this formula is not correct, because a 4x16 decoder can be constructed from only 2 3x8 decoders with enable input

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@Digvi_sp , in your implementation , you are using one of the input lines as enable input.

Answer 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 Answer: C

a total of 9 decoders are required.

(MSB) 0 0 1 1 0 0 (LSB) 

           5 4 3 2 1 0

here all the output is in normal order i.e 0, 1, 2 ...  to 63, for example input 001100, you'll see the 12th line selected.

Comments

where is this 0,1,2 coming from? Am I the only who doesn't understand this? Shouldn't all the input go to two 3x8 decoders first, so it can be 6 inputs?

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how do you plan to give 6 inputs to 3x8 decoder ? 3x8 will only have 3 inputs and 1 enable. if you think that we should somehow incorporate n no. of decoder such that all the 6 inputs go to these n decoders then thats not possible hence the above approach

 

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To build a 6-to-64 decoder, I can partition the 6 input bits into two groups:

• 3 most significant bits (MSBs). 3,4,5
• 3 least significant bits (LSBs)   0,1,2

  First level - Address decoding: The 3 MSBs can be decoded using one 3-to-8 decoder. This produces 8 outputs, each representing one of the 8 "blocks" of 8 outputs in the final 64 outputs.

  Second level - Final output generation: For each of the 8 blocks, I need another 3-to-8 decoder that:

• Takes the 3 LSBs as inputs
• Uses one of the 8 outputs from the first decoder as its enable signal
• Produces 8 outputs

• So I need 8 decoders for the second level (one for each block).

  Total count:

• 1 decoder for the first level (to decode the 3 MSBs)
• 8 decoders for the second level (to decode the 3 LSBs for each block)
• Total: 1 + 8 = 9 decoders

Answer 5

Let me provide a general formula to solve such questions. Suppose we want to construct an $n×2^{n}$ decoder using $m×2^{m}$ decoders, assume that $n>m$. Then the general formula is $1+2^{n-m}$. Some output lines which decode the last few bits might remain unused, like when we construct $3×8$ decoder using $2×4$ decoders but the functionality works fine.

Answer 6

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]