I have a quick question on 2-1 multiplexers

digital-logicmultiplexer

So say we have a 2-1 MUX with

  • \$D_0=GND\$,
  • \$D_1=Q_0\$ and
  • S=X.

Questions:

  • The output equation will be \$Q_0X\$, correct?
  • Now if we change \$D_0\$ to Vdd instead of GND, what happens?

If X=1 the output will be \$Q_0\$, but if X=0 the output will just be logical 1.

Would the equation be \$Q_0X+\bar{X}\$?

Best Answer

Yes, your logic is correct.

The output, Z, of the mux is Z = (D1 and S) or (D1 and S').

If D0 is held at GND then Z = (D1 and S). If D0 is held at +Vdd then Z = (D1 and S) or (S').