question_answer

In the given circuit, the Zener diode has breakdown voltage \[{{V}_{z}}=3\] volt and the maximum power limit \[{{P}_{\max }}=18\,mW.\] Choose the correct option.

A) If \[{{V}_{B}}=12\] volt, the power dissipated in Zener diode will exceed the maximum power limit, specified for it.

B) If \[{{V}_{B}}=12\] volt, the power dissipated in Zener diode will not exceed the maximum power limit, specified for it.

C) If \[{{V}_{B}}=15\]volt, the power dissipated in Zener diode will exceed the maximum power limit, specified for it.

D) If \[{{V}_{B}}=20\]volt, the power dissipated in Zener diode will not exceed the maximum power limit, specified for it.

Correct Answer: B

Solution :

[b] \[1000I+1000(I-{{I}_{Z}})={{V}_{B}}\] \[\Rightarrow \,\,\,\,\,2I-{{I}_{Z}}={{V}_{B}}\times {{10}^{-3}}\]                 ...(i)

\[-500{{I}_{z}}-{{V}_{z}}+1000(I-{{I}_{Z}})=0\] \[-1500{{I}_{z}}+1000\,I={{V}_{z}}\] \[=1.5{{I}_{z}}+I={{V}_{z}}\times {{10}^{-3}}\] \[=1.5{{I}_{z}}+I=3\times {{10}^{-3}}\] \[2I-3{{I}_{z}}=6\times {{01}^{-3}}\]                          ? (ii) With equations (i) and (ii), we have \[2{{I}_{z}}=\left( {{V}_{B}}-6 \right)\times {{10}^{-3}}\] \[\Rightarrow \,\,{{I}_{Z}}=\frac{{{V}_{B}}-6}{2}mA\] \[{{P}_{Z}}={{V}_{Z}}{{I}_{Z}}=\frac{3{{V}_{B}}-18}{2}\le 18\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,{{V}_{B}}\le 18\] Volt