Class A Amplifier

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Derivation 1

$$P_{avg,load}=\cfrac{\int_0^TP\cdot{}dt}{T}$$ $$=\cfrac{\int_0^T\cfrac{V_{out}^2}{R_L}\cdot{}dt}{T}$$ $$=\cfrac{\int_0^T\cfrac{V_p^2\sin{}^2(\omega{}_ot)}{R_L}dt}{T}$$ $$=\cfrac{\cfrac{V_p^2}{2R_L}\int_0^T2\sin{}^2(\omega{}_ot)dt}{T}$$ $$=\cfrac{\cfrac{V_p^2}{2R_L}\int_0^T(1-\cos{}(2\omega{}_ot))dt}{T}$$ $$=\cfrac{\cfrac{V_p^2}{2R_L}T-\left[\cfrac{\sin{}(2\omega{}_ot)}{2\omega{}_o}\right]_0^T}{T}$$ $$=\cfrac{V_p^2}{2R_L}$$ $$V_p=\cfrac{V_{CC}}{2}$$ $$\implies{}P_{avg,load}=\cfrac{V_{CC}^2}{8R_L}$$

Derivation 2

$$P_{avg,sup}=\cfrac{\int_0^TP\cdot{}dt}{T}$$ $$=\cfrac{\int_0^TV_{CC}I_{R}dt}{T}$$ $$=\cfrac{\int_0^TV_{CC}\cfrac{V_{CC}-V_{out}}{R_L}dt}{T}$$ $$V_{out}=\cfrac{V_{CC}}{2}+V_p\sin{}(\omega{}_ot)$$ $$V_p=\cfrac{V_{CC}}{2}$$ $$P_{avg,sup}=\cfrac{V_{CC}^2}{2R}$$

Derivation 3

$$P_{avg,sup}=\cfrac{\int_0^TP\cdot{}dt}{T}=\cfrac{\int_0^Ti_R^2R_Ldt}{T}$$ $$=\cfrac{\int_0^TI_p^2R_Ldt}{T}$$ $$=\cfrac{1}{2}I_p^2R_L$$ Let's understand the DC condition of the circuit. The Inductor is behaving as a short, so the left terminal of the capacitor is at V_CC. The resistor has pulled down the capacitor's right terminal to ground. So, the capacitor is now charged to V_CC. When the transistor has pulled down collector node down to zero, left terminal of the resistor is at -V_CC. So, the current through the resistor is -V_CC/R_L. When the transistor's input voltage is low, transistor is near to cut-off region. Entire inductor current flows through the capacitor and resistor. The magnitude of the inductor current (I_L) should be > V_CC/R_L to support symmetric swing across the load. The collector voltage reaches 2V_CC. Therefore, the power delivered to the load is: $$P_{avg,load}=\cfrac{1}{2}I_p^2R_L=\cfrac{V_{CC}^2}{2R_L}$$

Derivation 4

$$P_{avg,sup}=\cfrac{\int_0^TP\cdot{}dt}{T}$$ The current drawn from the supply is always I_L. So, the power drawn is V_CCI_L = V_CC²/R_L