This is the circuit
Can you please explain how did obtain the AC equivalent circuit
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Your question sounds like homework, so I will answer in rather general terms, to give you a guide.
To find the equivalent circuit, you must operate in the conventional manner, to determine the small signal model. Power supply nodes, correspond to GND for the signal, while for simplicity, you can assume at first that capacitors have negligible impedance (short-circuit).
Then, the input impedance \$z_{in}\$, according to the proposed circuit is the impedance that "sees" the signal source in this case is composed of resistors R1, R2, R3, R4 and \$h_{ic}\$ resistance equivalent of each transistor (note that the output is emmiter). You must read about BJT h-parameters model.
In the output circuit, the current source \$i_c\$ corresponds to the current flowing through the collectors. The \$r_e^\prime\$ impedance is the impedance on the emmiter circuit, reflected to the output circuit.
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This circuit won't work as class B or AB. Must be dual supply! When the PNP is conducting the other side of the load is at ground. It is trying to sink current to ground from ground!! You could build a rail splitter to provide a virtual ground at 1/2 V supply and reference the load from it.
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\$\begingroup\$ But this circuit will work with a single supply thanks to the "big" output capacitor in series with the load resistance. For a positive half cycle, the upper transistor (NPN) will deliver the current to the load by charging a capacitor. And for the negative half cycle, the lower transistor (PNP) will discharge the capacitor. And this is why we have a negative voltage at the load. i.stack.imgur.com/Sz6E3.jpg \$\endgroup\$ – G36 May 12 '19 at 16:15
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The upper transistor should be in cutoff, or close to it. Relying on the cap to supply load current is what I would call bad circuit design. Also, if you operate the circuit with dual supply you don't need the stupid cap. If the circuit is properly biased the output will be at ground potential with no input signal, so the speaker won't be subject to any DC bias.
