Modern Physics 6 Question 4
4. 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?
(Main 2019, 10 April II)
(a) $1.5 mA, 8.5 mA$
(b) $1 mA, 8.5 mA$
(c) $0.5 mA, 8.5 mA$
(d) $0.5 mA, 6 mA$
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Solution:
- In given voltage regulator circuit,
Zener breakdown voltage, $V _Z=6 V$
So, across $R _L$, potential drop is always $6 V$.
So, current through load resistance is
$$ i _L=\frac{V _Z}{R _L}=\frac{6}{4 \times 10^{3}}=1.5 \times 10^{-3} A $$
Now, when $V _B=8 V$
Potential drop across $R _1=8-6=2 V$
So, current through $R _1$ is $i _1=\frac{V}{R _1}=\frac{2}{1 \times 10^{3}}=2 \times 10^{-3} A$
So, current through Zener diode is
$$ \begin{aligned} i _Z & =i _1-i _L=2 \times 10^{-3}-1.5 \times 10^{-3} \\ & =0.5 \times 10^{-3} A=0.5 mA \end{aligned} $$
Similarly, when $V _B=16 V$
$$ \begin{array}{rlrl} & & V _{R _1} & =16-6=10 V \\ \therefore & i _1 & =\frac{10}{1 \times 10^{3}}=10 \times 10^{-3} A \end{array} $$
Hence,
$$ \begin{aligned} i _Z & =i _1-i _L=10 \times 10^{-3}-1.5 \times 10^{-3} \\ & =8.5 \times 10^{-3} A=8.5 mA \end{aligned} $$
