This is digital logic question. I think it's alright to post it here.
I'm trying to implement a 4 to 16 decoder using 2 to 4 decoder and 3 to 8 decoder.
What I did, I used 2x of 2-to-4 decoder and 1x 3-to-8 decoder. But I think there is a mistake in the 3-to-8 part. I hope you could point me out to it.
Here is what I did,
Note that I couldn't continue writing the full table. enter image description here
enter image description here
2 Answers 2
Q2 and Q3 will never be active at the same time, so it is useless to route them to the same decoder where one acts as enable.
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\$\begingroup\$ So the correct way is to connect Q2 to the 3to8 decoder and that's it? \$\endgroup\$Sobiaholic– Sobiaholic2012年11月11日 20:50:38 +00:00Commented Nov 11, 2012 at 20:50
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\$\begingroup\$ No, because the 3-to-8 must be active for A3=1, hence for its EN must be OR( Q2, Q3 ), and its third address input must be A2. Why do you want to use such an asymmetric set of chips? Five 2-to-4 decoders would be a more natural choice (or two 3-to-8 and one 1-to-two). \$\endgroup\$Wouter van Ooijen– Wouter van Ooijen2012年11月11日 21:19:32 +00:00Commented Nov 11, 2012 at 21:19
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\$\begingroup\$ @WoutervanOoijen this is a basic digital logic course. It's not my major, but we have to take it. About your suggestion, Yes! I was thinking inputting A2 is more correct than inputting Q2. But I'm still not sure why A2? My answer to my self is because A2 has 4 zeros and 4 ones which makes a new unique combinations. is my thought correct? \$\endgroup\$Sobiaholic– Sobiaholic2012年11月11日 22:55:16 +00:00Commented Nov 11, 2012 at 22:55
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\$\begingroup\$ 'more correct' is a weird phrase for digital logic! If you write the full logc table you can see that You can see that A3=1 is the condition for the 3-to-8 decoder, and that its output depends on A2. \$\endgroup\$Wouter van Ooijen– Wouter van Ooijen2012年11月12日 07:33:11 +00:00Commented Nov 12, 2012 at 7:33
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\$\begingroup\$ Thank you so much @WoutervanOoijen I get it now. I forgot that we're trying to run the 3to8 decoder. And to do that we used both Q3 and Q4 with a sum which makes it enabled. Thanks so much!!!! \$\endgroup\$Sobiaholic– Sobiaholic2012年11月13日 18:21:56 +00:00Commented Nov 13, 2012 at 18:21
You can do this with two 3-8 decoders and get by with having to use just two parts of the same type. Common 3-8 decoders come with enables for both high and low polarities to make expansion a piece of cake. Excuse my quick hack drawing.
enter image description here
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\$\begingroup\$ I appreciate it. However, this is a basic course i'm taking. So, I'm sure we haven't cover this in class. \$\endgroup\$Sobiaholic– Sobiaholic2012年11月13日 18:22:38 +00:00Commented Nov 13, 2012 at 18:22