Good day, how can we transfer the AC voltage from the output of the capacitor towards the load (8 ohms)?

Notice that if we simply put the load in the output capacitor, the voltage there will be miniscule since AC passes through a capacitor, and 8 ohms and 6.2k will appear to be in parallel and the equivalent resistance is small and so the voltage across the load will be small also.

How can we transfer the full voltage swing across the load?

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How exactly did you arrive at 6.2k as the appropriate value for Re? – Brian Drummond Dec 4 '12 at 15:25
The voltage across the load will be whatever the transistor and the power supply can deliver, which is a question of rail voltage and PSU current capacity. Your reasoning on the point is fallacious. – EJP Dec 5 '12 at 2:09

How can we transfer the full voltage swing across the load?

The correct question is: How can we make the output impedance much less than 8 ohms?

A quick "back of the envelope" calculation for the small signal impedance looking into the emitter of Q3 is:

$r_e > \dfrac{V_T}{I_E}$

You want the output impedance to less than 1/10 the load in order to lose less than 10% in the output impedance.

Assuming $V_T = 25mV$, you want:

$I_E > 10 \ \dfrac{25mV}{8 \Omega} = 31.25mA$

For $V_E = 12.5V$, you'll need to reduce the value of R9 to less than or equal to 400 ohms.

Now, that's from a small signal perspective. From a large signal perspective, consider that this circuit will be able to pull the output voltage towards 25V easily through Q3 but, pulling down the output voltage to 0 volts will be through R9 rather than through Q3. Your output will not swing as far "down" as it will "up".

In your previous output circuits, you were using a push-pull configuration. Why did you switch to single-ended output?

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Im experimenting what are the advantages of each circuits – WantIt Dec 4 '12 at 22:17