I have 3 logic input called: I1, I2, I3 and 1 ouput logic call OP. I need:

1/ When I3 set Low, OP is High, not depend on I1, I2 's state which mean even I1 or I2 state is Low or High, OP keep High.

2/ When I3 set High, I1 or I2 Low -> OP is High, I1 or I2 High -> OP is Low

Any IC with small footprint suggested or transistor's setup can do it?


P/S: This is my setup I1, I2 put in a OR Gate IC and ouput come with I3 into a AND Gate IC, but the final output OP is invert, I need 1 more Invert IC, so any IC can do it instead of 3 ICs?

Table 1. (For OP to edit. - Edited - Thanks Transistor)

I1  I2  I3   Out
 0   0   0 |  1
 0   0   1 |  1
 0   1   0 |  1
 0   1   1 |  0
 1   0   0 |  1
 1   0   1 |  0
 1   1   0 |  1
 1   1   1 |  0
  • \$\begingroup\$ Have you put any effort into this at all? How about a truth table? This site should not be used so that someone else can do your homework. Show some effort and others will help if you get stuck. \$\endgroup\$
    – Nedd
    Nov 10, 2019 at 7:59
  • \$\begingroup\$ @Nedd thanks, I edited my question \$\endgroup\$ Nov 10, 2019 at 8:14
  • 2
    \$\begingroup\$ I've added a truth table for you to edit. This will clarify your thinking and is much better than words for describing your required operation. Use the CircuitLab button on the editor toolbar to add a schematic of your circuit. Hit Save and Insert when done. \$\endgroup\$
    – Transistor
    Nov 10, 2019 at 14:47
  • \$\begingroup\$ To implement that you can use 8-input mux and tie off the inputs in the right way. That may be big, but it is only one IC :-) \$\endgroup\$
    – Oldfart
    Nov 10, 2019 at 16:10

1 Answer 1


There is no single gate that implements this.

You need an OR gate and a NAND gate. If you do not want to use two different chips, use two (SN)74LVC1G58s.


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