So.. I know how to build a BCD to 7-segment using 3×8 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 3×8 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 3×8 decoder, a 7 segment and minimal gates.
I'm really confused here. Please help!
Best Answer
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.