Semiconductors

JEE Main level, numerical answer, medium

30 of 209

Question 30 of 209

A diode has Zener voltage of 1010 V and maximum power dissipation of 0.50.5 W, then the minimum resistance to be used in series with this diode for safety when it is connected to a 2525 V power supply is ______ Ω\Omega.

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Answer: 300

Given Zener voltage VZ=10V_Z = 10 V and maximum power dissipation Pmax=0.5P_{max} = 0.5 W.

The maximum safe current through the Zener diode is given by Imax=PmaxVZ=0.510=0.05I_{max} = \dfrac{P_{max}}{V_Z} = \dfrac{0.5}{10} = 0.05 A.

The voltage drop across the series resistor is VR=VS−VZ=25−10=15V_R = V_S - V_Z = 25 - 10 = 15 V.

For safety, the current through the circuit should not exceed ImaxI_{max}. Thus, the minimum resistance RR is R=VRImax=150.05=300R = \dfrac{V_R}{I_{max}} = \dfrac{15}{0.05} = 300 Ω\Omega.

Answer: 300300

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