Digital Electronics - Combinational Logic Circuits - Discussion
Discussion Forum : Combinational Logic Circuits - General Questions (Q.No. 1)
1.
How many 3-line-to-8-line decoders are required for a 1-of-32 decoder?
Discussion:
14 comments Page 1 of 2.
Muhdin Abdo said:
1 decade ago
The 3-input data lines used as common lines for the 4-decoder of 8-out puts. Thus, one could have a total of 4*8=32 out puts. Hence one needs 4 a 3-line-to-8-line decoders.
Ravi kumar sinha said:
1 decade ago
In order to have 32 output lines you need to have 4 3*8 decorder and for the input you can connect enable and enable bar to four different different decorder so that we can different configuration for 1 input lines.
AYYAPPA said:
1 decade ago
Required outputs/given outputs == (32/8) = 4.
AJAY KUMAR said:
9 years ago
I think 1 to 32 means 5 by 32.
32/8 = 4 decoders sufficient but what about inputs?
Now inputs are 12 (4 decoders * inputs 3) , so again we need 2 decoders to complete the circuit.
32/8 = 4 decoders sufficient but what about inputs?
Now inputs are 12 (4 decoders * inputs 3) , so again we need 2 decoders to complete the circuit.
Ravi said:
9 years ago
Explain in the form of diagram eg. decoder.
Amareswari said:
9 years ago
We need one 2 to 4 decoder and four 3 to 8 decoders to implement 5 to 32 decoder.
MVP said:
8 years ago
Decoders having 2^m outputs and m inputs w.r.t this to obtain 32(2^5) output lines we need 5input lines.
How come you can make 1-32 decoder?
How come you can make 1-32 decoder?
Rajaram said:
8 years ago
Decoder can be used as demultiplexer also, so 1x32 demux is similar to 5x32 decoder. Here 3x8 decoders (4x8 =32 outputs) input is single line. So 2 input line is used in chip selection one of 4 chip.
(1)
Keerthi said:
8 years ago
Your explanation is simple and clear to understand @Ayyappa.
Rajashekar said:
8 years ago
(32÷8) + (4÷8) =4.5.
So, the answer is 4.
So, the answer is 4.
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