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For IC 7485 truth table(given below), why is the the output A>B = 1 and A<B = 1 when you have got the inputs A(3,2,1,0)=B(3,2,1,0) with 0 for all cascading inputs(the last row). What exactly is the use of cascading input? (I only know they can be used to merge two 7485's into 1 as a 8 bit comparator) but how does it affects the functionality? Sorry if I am wrong with my assumptions!

IC 7485 Truth Table

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The cascading inputs come from the outputs of another 7485. If you wish to create a 16 bit magnitude comparator, the other 7485 connects to the least significant byte (LSB).

As for the last line of the truth table, the outputs indicate that the 'A' inputs are simultaneously greater and less than the 'B' inputs, which is contradictory. But notice that the cascading inputs indicate that the 'A' LSB is neither greater than nor less than nor equal to the 'B' LSB which is also contradictory, ie an illegal state.

Similarly, the penultimate line suggests that the 'A' LSB inputs are simultaneously greater and less than the 'B' LSB inputs which is also contradictory, so the last two lines indicate an error condition.

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  • \$\begingroup\$ Nice answer, but why do we only consider Cascaded Inputs for A=B? And "why not" for A>B / A<B ? \$\endgroup\$
    – Zaid Khan
    Sep 21 '13 at 13:50
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The last two rows of the truth table are degenerate in that there should be no way to generate them with proper input. They are there for completeness, in order to show that all possible inputs have been accounted for.

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    \$\begingroup\$ "degenerate" ?? Thanks for the answer but can you please elaborate! \$\endgroup\$
    – Zaid Khan
    Sep 20 '13 at 11:53
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    \$\begingroup\$ Not really sure what else you want me to tell you that "there should be no way to generate them with proper input" doesn't cover... \$\endgroup\$ Sep 20 '13 at 12:06
  • \$\begingroup\$ It's interesting that the last two lines flip the polarities of the inputs. I wonder if, when all inputs were equal it was easier to make A>B be A=B nor A<B than to make it be A>B and not A=B or simply A>B? \$\endgroup\$
    – supercat
    Feb 11 '14 at 16:45

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