A zener of breakdown voltage Vz = 8 V and maximum zener current, IZM = 10 mA is subjected to an input voltage Vi = 10 V with series resistance R = 100 Ω. In the given circuit RL represents the variable load resistance. The ratio of maximum and minimum value of RL is _____________. Numerical answer type. Work it out, then open the answer.

Answer: 2. Current through R: I = (10 - 8)/(100) = 20 mA, and I = IZ + IL. RL max when IZ = IZM = 10 mA: IL = 10 mA, RL,max = 810×10⁻³ = 800 Ω RL min when IZ ≈ 0: IL = 20 mA, RL,min = 820×10⁻³ = 400 Ω RL,maxRL,min = (800)/(400) = 2

A zener of breakdown voltage Vz = 8 V and maximum zener current, IZM… | JEE Main 2022 (28 Jun, Shift 2)

Q59JEE Main 2022 · 28 JUN · SHIFT 2Difficult

Special Purpose Diodes

A zener of breakdown voltage Vz = 8 V and maximum zener current, IZM = 10 mA is subjected to an input voltage Vi = 10 V with series resistance R = 100 $\Omega$ . In the given circuit RL represents the variable load resistance. The ratio of maximum and minimum value of RL is _____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

Current through $R$ : $I = \dfrac{10 - 8}{100} = 20$ mA, and $I = I_Z + I_L$ .
$R_L$ max when $I_Z = I_{ZM} = 10$ mA: $I_L = 10$ mA, $R_{L,max} = \dfrac{8}{10\times10^{-3}} = 800\,\Omega$
$R_L$ min when $I_Z \approx 0$ : $I_L = 20$ mA, $R_{L,min} = \dfrac{8}{20\times10^{-3}} = 400\,\Omega$
$\dfrac{R_{L,max}}{R_{L,min}} = \dfrac{800}{400} = 2$

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