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I have an idea of how to get started, but not sure where to go from there.

Since I am using two 3-8 decoders to develop a 4-to-16 decoder, I want to use 4 inputs out of the two 3-8 decoders. So I'll use all three of the first and the first of the second, and connect the last two inputs to ground, since they won't be used.

Any pointers on where to go from here are appreciated.

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  • \$\begingroup\$ Did you search this site? I think I have seen this question a few times before. And your approach (as far as I can tell from your wording) will not work. \$\endgroup\$ – Wouter van Ooijen Feb 28 '15 at 9:28
  • \$\begingroup\$ @WoutervanOoijen yes, it seems we get these questions almost as often as "How do I connect an LED to my Arduino". \$\endgroup\$ – tcrosley Feb 28 '15 at 10:07
  • \$\begingroup\$ Your question is a bit strange, if you plan on using just the two decoder chips what are the 16 two-input AND gates for? The circuit could be made using just two 3-8 decoders, or possibly one 3-8 decoder and the 16 AND gates. But it depends on the actual decoder chip being used. \$\endgroup\$ – Nedd Feb 28 '15 at 10:21
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you have to design a 4x16 decoder using two 3x8 decoders. here is the schematic that may help you.

schematic

simulate this circuit – Schematic created using CircuitLab

the two squares are two 3x8 decoders with enable lines. the three selection lines of each decoders are connected together as common line(X,Y,Z) , the enable lines are ACTIVE LOW, they are also connected together with a common line W , but the second one having a NOT gate connected within. So, there are now 4 selection inputs i.e W,X,Y,Z. For the values 0000 to 0111 ,the first decoder will turn on giving the decoded outputs 0 to 7 , and for 1000 to 1111 , the second decoder will turn on , giving decoded output 8 to 15. How? Because for the first 8 combinations, the W bit is 0 , so it is a 1 for the first decoder, and enable line is on(ACTIVE LOW) , but it goes through a NOT GATE and then to the ACTIVE LOW enable port of the second decoder, so it remains 0 , so the second decoder doesn't activate. then for the next 8 combinations, the W line is 1 , which becomes 0 inside the 1st decoder(ACTIVE LOW) and the 1st decoder turns off, but it goes through a NOT gate and then ACTIVE LOW enable port of 2nd decoder, becomes a 1 and activates the next outputs 8 to 15. so you get a total 16 outputs (0-15)

If you use two 3x8 decoders you need not use 16 AND gates, and If you want to use AND gates only, then think about the internal circuitry of a decoder, think of this diagram , and you can easily construct using AND gates.

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If using just two 3-8 decoder chips: You would need to connect the first 3 data lines in parellel to the two decoder chips, then use the remaining high bit as an enable to the higher decoder chip and a disable the lower decoder chip. For example the 74HC238 decoder chip has three enable pins (two low and one high), your higher bit input would connect to ~G2A of the first decoder chip and to G1 of the higher decoder chip. (The other remaining enable pins also need to be connected to the appropriate high or low to allow for normal operation). A 74HC138 decoder chip could also be used, the decoded output is just inverted.

http://www.st.com/st-web-ui/static/active/en/resource/technical/document/datasheet/CD00000285.pdf

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