Semiconductors

JEE Main level, single correct, medium

28 of 209

Question 28 of 209

The maximum rated power of the LED is 22 mW and it is used in the circuit with input voltage of 55 V as shown in the figure below. The current through resistance RSR_S is 0.50.5 mA. The minimum value of the resistance of RSR_S, to ensure that the LED is not damaged is _______ kΩ\Omega.

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Answer: (B) 22

Since the reverse-biased diode branch acts as an open circuit, the entire current through the series resistor RSR_S flows through the LED:
ILED=IRS=0.5 mAI_{\text{LED}} = I_{R_S} = 0.5\text{ mA}

At the boundary condition where the LED dissipates its maximum rated power of 2 mW2\text{ mW}:
PLED=VLED⋅ILEDP_{\text{LED}} = V_{\text{LED}} \cdot I_{\text{LED}}

VLED=PLEDILED=2 mW0.5 mA=4 VV_{\text{LED}} = \dfrac{P_{\text{LED}}}{I_{\text{LED}}} = \dfrac{2\text{ mW}}{0.5\text{ mA}} = 4\text{ V}

Applying Kirchhoff's voltage law to the loop:
Vinput−VRS−VLED=0V_{\text{input}} - V_{R_S} - V_{\text{LED}} = 0

5−(IRS⋅RS)−4=05 - (I_{R_S} \cdot R_S) - 4 = 0

IRS⋅RS=1 VI_{R_S} \cdot R_S = 1\text{ V}

RS=1 V0.5×10−3 A=2000 Ω=2 kΩR_S = \dfrac{1\text{ V}}{0.5 \times 10^{-3}\text{ A}} = 2000\ \Omega = 2\text{ k}\Omega

Hence, the correct option is (2) 2(2)\ 2.

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