[SOLVED] why using CMOS 4069 inverter as ampilifer?

The inverter is made with one N-channel Mosfet connected to a P-channel Mosfet as a push-pull amplifier. Both Mosfets have similar spec's so if one has a gate voltage of 2V that causes it to conduct 2mA then the other one also conducts 2mA when its gate voltage is 2V. Its output is about at half the supply voltage when its input is at half the supply voltage.

Now connect a resistor from its output to its input creating DC negative feedback. If something causes the output voltage to drop then the resistor causes the input voltage also to drop and the gain of the Mosfets causes the output to go up to half the supply voltage where it should be. The same if the output tries to be higher than it should be, the negative feedback and the voltage gain causes the output to be at half the supply voltage.

Since the input is driven from a low impedance and is capacitor-coupled then there is only a small amount of negative feedback for AC signals then the AC voltage gain is high.

Mryemeni said:

What I don't understand is how to choose the value of the resistance in the feedback?,
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Regarding the value of the feedback resistor there is one obvious argument that was NOT mentioned up to now:
Because the CMOS inverter acts as (non-ideal) CURRENT source each resistor at its output is a LOAD that determines the gain. And this applies also to the feedback resistor because the signal current goes through the signal source at the input. Thus: Large feedback resistor results in larger gain.

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Audioguru said:

.... then there is only a small amount of negative feedback for AC signals then the AC voltage gain is high.

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I rather would say: .........then the AC voltage gain can be made relatively high in case the feedback resistor is large.