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The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is $6\,V$ and the load resistance is $R_L= 4\,k\Omega .$ The series resistance of the circuit is $R_i = 1\,k\Omega .$ If the battery voltage $V_B$ varies from $8\,V$ to $16\,V,$ what are the minimum and maximum values of the current through Zener diode?
$0.5\,mA;\,\,0.6\,mA$
$1\,mA;\,\,8.5\,mA$
$1.5\,mA;\,\,8.5\,mA$
$0.5\,mA;\,\,8.5\,mA$
Solution
${{\text{V}}_{{\text{breaddwon }}}} = 6{\text{V}},$ ${{\text{R}}_{\text{L}}} = 4\,{\text{k}}\Omega ,$ ${{\text{R}}_{\text{i}}} = 1{\text{k}}\Omega $
${{\text{i}}_{\text{L}}} = \frac{6}{4} \times {10^{ – 3}} = $ $1.5 \times {10^{ – 3}} = 1.5{\text{mA}}$
$\mathrm{i}_{\mathrm{i}}=2 \times 10^{-3}$
${\text{i}} = {{\text{i}}_1} – {{\text{i}}_{\text{L}}} = 0.5{\text{mA}} – $ minimum current
$\mathrm{i}_{\mathrm{i}}=10 \times 10^{-3}=10 \mathrm{mA}$
$\operatorname{lmax}=8.5 \mathrm{mA}$
