Question 8.6: A class A amplifier stage has a thermal resistance Rθ = 2 °C......
A class A amplifier stage has a thermal resistance R_\theta=2\; ^{\circ} C/\text{mW}. Taking into account that I_{D S S}=10\text{ mA} , V_{D S, b r}=14\text{ V} and that the bias point is the optimum class A bias, estimate the temperature variation between two conditions: (a) zero input signal; (b) maximum input signal. Assume 20 dB operational gain.
The maximum class A power is:
P_{R F, M}=\frac{I_{D S S} V_{D S, b r}}{8}=\frac{10 \cdot 14}{8}=17.5\text{ mW};
we have P_{i n}=0 in case (a) and P_{i n}=17.5 / 100=0.175\text{ mW} in case (b), since G_{ \text{op} }=100. The DC power does not depend on the input signal level and is:
P_{D C}=\frac{I_{D S S} V_{D S, b r}}{4}=\frac{10 \cdot 14}{4}=35\text{ mW},
while the device dissipated power is:
P_{\text {diss }}=P_{D C}-P_{R F}=35-100 \cdot P_{i n, M} \cdot \frac{P_{i n}}{P_{i n, M}}=\left(35-17.5 \frac{P_{i n}}{P_{i n, M}}\right)\text{ mW} .
Correspondingly, the device temperature increase is:
\Delta T=R_\theta P_{\text{diss} }=2 \cdot\left(35-17.5 \frac{P_{i n}}{P_{i n, M}}\right)\;{}^{\circ} C =\left(70-35 \frac{P_{i n}}{P_{i n, M}}\right)\; {}^{\circ} C .
Thus the device cools down with increasing input signal power, going from 70\; {}^{\circ} C (a) to 35\; {}^{\circ} C (b).