This amplifying stage is self-biased by connecting collector to base at DC (C3 is open). But if C3 didn't exist, it would give low gain at normal operation frequencies, 1+gm(R2+R3)/beta. Placing C3 causes a short-circuit path to ground at higher frequencies and increases gain to gm(R1||R4||R3). 

This amplifying stage is self-biased by connecting collector to base at DC (C3 is open). But if C3 didn't exist, it would give low gain at normal operation frequencies, 1+gm(R2+R3)/beta. Placing C3 causes a short-circuit path to ground at higher frequencies and increases gain to gm(R1||R4||R3). 

Thank you dear Iman

Simply speaking: at low frequencies, there is a negative feedback between the collector and base; at high frequencies, there is no negative feedback between them.

This negative DC feedback stabilizes quiescent point.

However, when we switch on the power voltage an operating point of the transistor accrues with a corresponding time constant.

IMO with the same success, we can connect C3 to V1 (VCC) instead to ground. Then, when we switch on the power supply, the output voltage will be close to zero?

Thank you all for the valuable response

Below is a possible solution

For what a problem?

Quick set of quiescent point after power connection.

Thank you

Clever trick - to use instead of a "pull down" or "pull up" capacitor the both...

In the midrange frequency the C3 acts as a short circuit so that, in the ac equivalent circuit of the amplifier, the resistor R2 will be connected between the input terminal and ground and the resistor R3 will be connected between output terminal and ground in parallel with R4

 Thank you  very much.

Answer in one sentence: It is the task of C3 to allow negative (stabilizing) feedback for DC only

Thank you very much professor.

Dear Rohith,

All Answers above are satisfactory. However, i have one question, what is the relation between 1/wC3 and R2 at the lowest operating frequency???

Best wishes

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