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So.. I know how to build a BCD to 7-segment using 3x8 decoder but the question provides 4-digit binaries and asks to only input 3 of those.. my question is that won't two different digits be regarded as one.. If that's confusing, here's the question:

Consider a BCD digit \$X_3X_2X_1X_0\$ with \$X_3\$ being the MSB. Design a BCD to 7-segment converter using two 3x8 decoders and minimum number of gates, such that \$X_3\$, \$X_2\$ and \$X_0\$ are applied at the inputs of the decoders. The converter should be compatible with a common cathode display.

So.. in this case 0 is 0000 and 2 is 0010 but if we don't input \$X_1\$, we're only giving out 0000 in both cases.. so how would the decoder know which is 0 and which is 2? Do I use \$X_1\$ as enabler? And we're not supposed to use anything other than 3x8 decoder, a 7 segment and minimal gates. I'm really confused here. Please help!

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By my logic, \$X_{1}\$ must be used as an enable on the decoders, directly to one and through a not gate to the other. This is because without using it as an enable, both decoders will have the same inputs and thus the same outputs, which is redundant.

I think the trick to this question is the realization that since \$X_{3}\$ is NOT the enable (A more common case), then every two decimal digits will produce a high output on a different decoder. Below are two images showing the decimal BCD codes that will produce a high value on each output.

BCD codes for "normal" orderBCD codes for your question

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    \$\begingroup\$ Please don't turn this into a homework solution service. We try to encourage people to try to solve their own problems and only give hints along the way. \$\endgroup\$ – Elliot Alderson Feb 10 at 16:58
  • \$\begingroup\$ @ElliotAlderson Could you clarify exactly what a "hint" constitutes? I thought I did a fairly good job of clarifying the problem given by answering the OP's specific confusion ("how would the decoder know which is 0 and which is 2? Do I use X1 as enabler?"), while avoiding providing a full worked solution. There is a lot left to the full problem solution. \$\endgroup\$ – Brandon Hill Feb 10 at 18:24
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    \$\begingroup\$ I would have stopped at the first paragraph. \$\endgroup\$ – Elliot Alderson Feb 10 at 19:38
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Correct, You'll need to use X1 in combination with the 3x8 decoder output to generate some of the 7 segment display outputs.

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  • \$\begingroup\$ Remember any decoder output could be combined with X1 so that an output from the 3-8 decoder that indicates the input is either a 2 or a 0 (let's call it the Op0 line) When you make this combination you then have Op0&X1 indicates we should display a 2 while Op0&~X1 indicates we should display a 0. This removes the need for an enable or multiple decoders. \$\endgroup\$ – Andrew Macrae Feb 10 at 8:13

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