Question 4.17: Finding the power dissipation of a diode A diode is operated...

From Eq. (4.50), the junction power dissipation at the Q-point is

P_{ D }=I_{ D } V_{ D }                                  (4.50)

P_{ D }=I_{ D } V_{ D }=1 A \times 0.7 V =0.7 W

From Eq. (4.51), the thermal resistance from junction to ambient is

\Delta T=T_{ j }-T_{ a }=\theta_{ ja } P_{ D }                             (4.51)

\theta_{ ja }=\frac{\Delta T}{P_{ D }}=\frac{T_{ jm }-T_{ a }}{P_{ D }}=\frac{200-50}{1}=150^{\circ} C / W

(a) From Eq. (4.51), the junction temperature at the Q-point is

T_{ j }=T_{ a }+\theta_{ ja } P_{ D }=25+150 \times 0.7=130^{\circ} C

(b) From Eq. (4.53),

P_{ D }= \begin{cases}P_{ Dm }\left(\text { at } T_{ j }=25^{\circ} C \right) & \text { for } T_{ a } \leq 25^{\circ} C \\ -\frac{T_{ a }}{\theta_{ ja }}+\frac{T_{ jm }}{\theta_{ ja }} & \text { for } T_{ a }>25^{\circ} C \end{cases}                                        (4.35)

P_{ Dm }\left(T_{ a }=25^{\circ} C \right)=\frac{T_{ jm }-T_{ a }}{\theta_{ ja }}=\frac{200^{\circ} C -25^{\circ} C }{150^{\circ} C / W }=1.17 W

(c) For T_{ a }=75^{\circ} C ,

P_{ Dm }\left(T_{ a }=75^{\circ} C \right)=\frac{200^{\circ} C -75^{\circ} C }{150^{\circ} C / W }=833 mW