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When I was practicing logic circuit design using PMOS and NMOS , It's said that we have to use all PMOS for Pull Up Network (PUN ) and NMOS for Pull Down Network (PDN). But we know that for input voltage 0 , PMOS will allow current flow and for input voltage \$ V_{DD} \$ or logic value 1 , NMOS will allow current flow.If this is the case , why can't I solve this function $$ f=\bar{x_1}.x_2 $$ like this:

schematic

simulate this circuit – Schematic created using CircuitLab

What kind of problem this circuit can create?? Is it any problem for bias voltage ?

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Imagine a load resistor (say 10K) from output to Vdd, and make X_1 and X_2 low. Ideally, output should be 0.0V. You can figure out what it is (not 0V).

Now imagine a 10K load resistor from output to GND and make X_1 low and X_2 high. Ideally, output should be Vdd. You can figure out what it is (not Vdd).

schematic

simulate this circuit – Schematic created using CircuitLab

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"But we know that for input voltage 0 , PMOS will allow current flow and for input voltage VDD or logic value 1 , NMOS will allow current flow."

That is not true in general, it holds only when the NMOS source is at 0 and the PMOS source is at 1, which is not (always) the case in your circuit.

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if you interchange the positions of PMOS and NMOS transistor in a CMOS inverter, it will become a buffer,and the buffer is not so good, because maximal output high level will be VCC - Vtt(Vtt is NMOSFET's threshold voltage), minimal output low level will be VSS + Vtp(Vtp is PMOSFET's threshold voltage).this is the reason that in digital electronics a buffer is made of two inverters, because the output voltage will close VCC or GND.

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