If we assume that allowable inputs a, b and c are applied to the following logic gates, what is the output in terms of a,b and c?

enter image description here

I ran into the above problem, but I'm not sure if it makes sense. Because some NOT logic gates have two inputs which is not what I expect. I wonder if the problem is wrong or I'm missing something.

I need to calculate the output of the logic gates in terms of a, b and c with logic operators like: \$\bullet \$ for AND, \$+\$ for OR, \$\overline{a}\$ for NOT \$a\$

I appreciate if anybody can help.

  • 1
    \$\begingroup\$ You're missing the session of your course that explained what those gates are. \$\endgroup\$ Mar 20, 2017 at 17:35
  • \$\begingroup\$ The additional input controls whether the output is connected to H or L at all. That undefined state is called Hi-Z and it's required for bus systems where multiple outputs are connected to one bus line. \$\endgroup\$
    – Janka
    Mar 20, 2017 at 18:32
  • 2
    \$\begingroup\$ Your mysterious gate is a tri-state inverting buffer. Here is an example of such chip: assets.nexperia.com/documents/data-sheet/74HC_HCT125.pdf (although this one has the ENABLE input inverted relative to yours, and does not invert the output) \$\endgroup\$
    – bobflux
    Mar 20, 2017 at 18:43
  • \$\begingroup\$ @peufeu I'm going to study the document. \$\endgroup\$
    – Megidd
    Mar 20, 2017 at 18:44
  • \$\begingroup\$ I can't find the part# which corresponds to your "inverter", here is a closer one: inverting output (like yours), and active-low OUTPUT_ENABLE (OE) signal (unlike yours): ti.com/lit/ds/symlink/sn74lvc1g240.pdf \$\endgroup\$
    – bobflux
    Mar 20, 2017 at 18:51

1 Answer 1


As a theoretical logic diagram this is pure nonsense. Inverter has only one input and connecting 2 outputs together means nothing.

As practical circuit diagram this can be valid. the special 2 input inverters can really be disconnect-able by the disabling signals. In practice we say "they have 3-state outputs". The 3rd state is called "high-Z" and that means "disconnected by internal electronic switch."

The rightmost a and c should not be the same as the leftmost a and c.

But if you really have this as written and 0 means "disabled" for the special inverters, then you have the following truth table:

a b c .....out

0 0 0 .....undefined (=no proper input to the rightmost inverter)

0 0 1......0

0 1 0 .....undefined

0 1 1 .....0

1 0 0......0

1 0 1 .....smoke (=a short circuit)

1 1 0.....0

1 1 1 ....1

  • \$\begingroup\$ Thanks, if we assume we have special 2-input inverters, I wonder if it is possible to calculate the output in terms of a, b and c and the logic operations between them. \$\endgroup\$
    – Megidd
    Mar 20, 2017 at 18:26
  • \$\begingroup\$ @user4838962 no boolean equation is equivalent with this truth table, because boolean equations have only variables that can be 0 or 1. In addition the operators Joint and DisableInv are missing. But nothing prevents you to expand the Boolean algebra to have additional states Hi-Z, Undefined and Smoke and a new operators Joint and DisableInv \$\endgroup\$
    – user136077
    Mar 20, 2017 at 18:50
  • \$\begingroup\$ Thanks, I wonder if joint operator is the same as OR operator. I feel like because a joint is where two wires merge, if any of the wires is on, the joint output is on. Therefore joint is acting like OR gate. I'm not sure. \$\endgroup\$
    – Megidd
    Mar 20, 2017 at 19:04
  • \$\begingroup\$ @user4838962 Joint = OR if the outputs cannot actively pull down, only connect to + voltage. This is very rare, It has name "uncommitted emitter output type" More common is so called "open collector type output" that only pulls down to Ground. For them Joint = AND. For TTL outputs the Joint is forbidden because 0 Joint 1 = 1 Joint 0 = Smoke. \$\endgroup\$
    – user136077
    Mar 20, 2017 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.