Physics - Semiconductors Question with Solution | TestHub

In Zener diode,

$\Rightarrow \frac{{\mathrm{V}}_{\max }-{\mathrm{V}}_{\mathrm{Z}}}{{\mathrm{R}}_1}\ge {\mathrm{I}}_{\mathrm{knee}}+{\mathrm{I}}_{\mathrm{R}}$

${\mathrm{R}}_1$ is the series resistance of $50 \mathrm{\Omega }$ and ${\mathrm{I}}_{\mathrm{R}}$ is the current through the resistance $\mathrm{R}$

$\Rightarrow \frac{10-6}{50}\ge 5\mathrm{m}+{\mathrm{I}}_{\mathrm{R}}$

$\Rightarrow \mathrm{ }{\mathrm{I}}_{\mathrm{R}}\le 80 \mathrm{mA}-5 \mathrm{mA}$

$\because {\left({\mathrm{I}}_{\mathrm{R}}\right)}_{\max }=75 \mathrm{mA}$     (the maximum current through $\mathrm{R}$)

So the minimum value of the resistance $\mathrm{R}$ is

${\mathrm{R}}_{\min }=\frac{\mathrm{V}}{{\left({\mathrm{I}}_{\mathrm{R}}\right)}_{\max }}=\frac{6 \mathrm{V}}{75\mathrm{ }\mathrm{mA}}=\frac{6000}{75}=80\mathrm{ }\mathrm{\Omega }$