A Zener of breakdown voltage $\mathrm{V}_{\mathrm{Z}}=8 \mathrm{~V}$ and maximum zener current, $\mathrm{I}_{\mathrm{ZM}}=10 \mathrm{~mA}$ is subjected to an input voltage $\mathrm{V}_{\mathrm{i}}=10 \mathrm{~V}$ with series resistance $\mathrm{R}=100 \Omega$. In the given circuit $\mathrm{R}_{\mathrm{L}}$ represents the variable load resistance. The ratio of maximum and minimum value of $R_{L}$ is

Semiconductors — JEE Main · 2022 · 28 Jun 2022 · Shift 2 Question with Solution

A Zener of breakdown voltage

V_{Z} = 8 V

and maximum zener current,

I_{ZM} = 10 mA

is subjected to an input voltage

V_{i} = 10 V

with series resistance

R = 100Ω

. In the given circuit

R_{L}

represents the variable load resistance. The ratio of maximum and minimum value of

R_{L}

Answer: 2

The current through the series resistance is:

I = \frac{V _{i} - V _{Z}}{R} = \frac{10 - 8}{100} = \frac{2}{100} = 20 mA

This current splits between the Zener and the load:

I = I_{Z} + I_{L}

For maximum

R_{L}

: Zener current is maximum (

I_{Z} = 10

mA), so load current is minimum:

I_{L} = 20 - 10 = 10 mA

R_{L_{m a x}} = \frac{V _{Z}}{I _{L}} = \frac{8}{10 \times 1 0 ^{- 3}} = 800Ω

For minimum

R_{L}

: Zener current is minimum (ideally

I_{Z} = 0

), so load current is maximum:

I_{L} = 20 mA

R_{L_{m i n}} = \frac{V _{Z}}{I _{L}} = \frac{8}{20 \times 1 0 ^{- 3}} = 400Ω

Ratio:

\frac{R _{L_{m a x}}}{R _{L_{m i n}}} = \frac{800}{400} = 2

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer