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How does a demultiplexer ignore/discard/block the non-required outputs? A demultiplexer channels the input to one of the outputs, but there are several outputs. When one output is selected (depending on the state of the select line(s)) then rest of the outputs are 0/low (am I right?). Now, if the input is 0/low then the selected and non-selected outputs all becomes 0/low. How does the demultiplexer perceive the final result in such a case??

Please explain in details (with reference to 1:2 demux if possible). Thank You!

samm
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  • I don't understand your question. The demultiplexer implements a specific function from its inputs to its outputs. Are you asking what outputs are generated for some given input? Are you asking what, electronically, happens to produce that output? Something else? – David Richerby May 30 '17 at 13:34

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It's just AND gates. So, assuming $in$ is the input, $sel$ is your selector bit, and $out_1$ and $out_2$ are your two possible outputs, you would get:

$out_1 = in \land \lnot sel$

$out_2 = in \land sel$

The idea is that, if $in$ is $0$, out must always be $0$. If $in$ is $1$, then we have to output it to the proper gate. If you had a larger demux, you would simply have more selector bits. So, for a demux with 16 outputs, we would need 4 selector bits: $sel1$, $sel2$, $sel4$, and $sel8$. Output 10 would thus be a 5-input AND gate:

$out_{10} = in \land (sel_8 \land \lnot sel_4 \land sel_2 \land \lnot sel_1)$

The other outputs would be blocked because they would have different combinations of $sel1$, $sel2$, $sel4$, and $sel8$ ANDed into their outputs.

Ben I.
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  • But how the other outputs would be blocked? My question was exactly that! When the input is '0', then the selected output (based on the state of the select lines), as well as the all the other non-selected outputs would be '0'. How would one understand which output is actually being selected, in such a case. – samm May 31 '17 at 09:21
  • Take $out_{10}$ as the example. It can only go through if we feed exactly binary ten ($1010$) into the selector bits (because we require $sel_8$, and we require NOT of $sel_4$, etc.) The other outputs are all $0$ at that moment because we haven't fed their exact pattern into the selector bits. – Ben I. May 31 '17 at 10:27