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This circuit uses the capacitance transformer method to find the load at the collector. The input impedance of transistor is magnified by the two capacitors. This appears at the load.
Normally, if the base bias resistors are decoupled, the input resistance is just re, 0.026v/Ie ohms. This would be 2.6 ohms in this circuit, however, in this circuit, they are not decoupled.
The input resistance is roughly 1/(1/Rx + gmxRpi/(Rpi+RBB)). where RBB is the parallel resistance of the 2K2 and the 3K3 resistors. Rx is the parallel resistance of (Rpi+RBB) and RE (470 ohms). In this circuit, Rx is about 362 ohms. HFE = 100. Therefore, Rpi = 260 ohms. RBB in the simulation is about 1320 ohms.
I have called the resistance seen at the collector, R'. In this case, it is about 212 ohms. (This is calculated by taking the input resistance of the transistor as 15 ohms. In fact, it is lower due to internal capacitance). This value is then put through the capacitive transformer process. 212 ohms appears as the load resistance.
The open loop gain can be approximated by gmxR'xRpi/(Rpi+RBB). The feedback fraction B is approximately C1/(C1+C2). Therefore, BAo is approximately gmxR'xRpixC1/(C1+C2)x(Rpi+RBB). In this case, it comes to a value of 2.98. This is high, but it does work on the simulator.
The quiescent collector current is about 10 mA, giving a gm of 0.01/0.026 = 0.385.
(The 120pf represents the tuning capacitance).
I find the return ratio method far better for designing oscillators. Sometimes, it is useful not to decouple the base resistors. A signal can then be applied to the base to make a mixer oscillator circuit.
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