Hello, an elementary question here.
I have attached a schematic of a typical BJT amp circuit.

The following is my understanding:
The electret mic is biased by R1 to make all parts of the signal have the same positive polarity. The signal then passes through the AC coupling capacitor C4 which removes the DC bias.
My question is: now that the DC bias has been removed, has that left us with an AC signal that is centred on 0V ? If so how could the negative part of it be amplified ?

Thanks for clarification.

 

No, the voltage following the capacitor is not centred on 0V. That would only be the case if the capacitor were simply returned to a resistor connected to 0V.

If you look at the schematic, you will see that the capacitor actually returns to a transistor base, which is biased by a resistor connected to the 5V supply. The signal swing will therefore not centre about zero, but about the base bias voltage.

Edit: The microphone's DC bias has been blocked, but the transistor is biased separately. We might argue about the efficacy of simple biasing from the supply, but that is not the subject here. The point is that the capacitor blocks DC so that circuits can be biased independently, but passes AC signals through.

 

The electret microphone has an internal N-channel JFET which provides a reasonable output impedance and some gain. It is a common-source amplifier, and resistor R1 acts as the drain load resistor, providing bias current and voltage to the JFET.

 

The following is my understanding:
The electret mic is biased by R1 to make all parts of the signal have the same positive polarity.

If all parts of the signal have the same polarity then something is rectifying the audio which would cause severe distortion. But actually the audio is AC which swings up and down around the DC bias voltage.