Question 4.SP.25: The amplifier of Example 4.7 has plate current iP = IP + ip ......
The amplifier of Example 4.7 has plate current
i_P = I_P + i_p = 8 + \cos ωt \text{mA}
Determine (a) the power delivered by the plate supply voltage V_{PP}, (b) the average power delivered to the load R_L, and (c) the average power dissipated by the plate of the triode. (d) If the tube has a plate rating of 2 W, is it being properly applied?
(a) The power supplied by the source V_{PP} is found by integration over a period of the ac waveform:
P_{P P} = \frac{1}{T} \int_0^T V_{P P} i_P d t = V_{P P} I_P = (300)\left(8 \times 10^{-3}\right) = 2.4 W
(b) P_L = \frac{1}{T} \int_0^T i_P^2 R_L d t = R_L \left(I_P^2 + I_p^2\right) = 10 \times 10^3 \left[\left(8 \times 10^{-3}\right)^2 + \left(\frac{1 \times 10^{-3}}{\sqrt{2}}\right)^2\right] = 0.645 W
(c) The average power dissipated by the plate is
P_P = P_{P P} – P_L = 2.4 – 0.645 = 1.755 W
(d) The tube is not properly applied. If the signal is removed (so that i_P = 0), then the plate dissipation increases to P_P = P_{P P} = 2.4 W, which exceeds the power rating.