Question 3: Semiconductors and Logic Gates

Question

The maximum rated power of the LED is 2 mW and it is used in the circuit with input voltage of 5 V as shown in the figure below. The current through resistance ${R}_S$ is 0.5 mA .
The minimum value of the resistance of ${R}_S$, to ensure that the LED is not damaged is $\_\_\_\_$ $\mathrm{k} \Omega$.
![](https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1mo5o9w11/70b9de56-1036-4030-b152-bd2e1ded389a/72100950-3be1-11f1-aafb-6f2b8ad54979/file-6y3zli1mo5o9w12.png?format=png)

Accepted Answer

The input voltage is $\mathrm{V}_{\mathrm{s}}=5 \mathrm{~V}$. $\$
The total current in shunt resistance is $\mathrm{I}_{\mathrm{s}}=0.5 \mathrm{~mA}=0.5 	imes 10^{-3} \mathrm{~A}$. $\$
The power rating of LED is $\mathrm{P}_{\mathrm{LED}}=2 \mathrm{~mW}=2 	imes 10^{-3} \mathrm{~W}$. $\$
The resistance added parallel to LED is R $=1 \mathrm{k} \Omega=1000 \Omega$. $\$ 	ext{ } $\$
As the diode in branch containing $R$ is in reverse biased, so current through it is zero. $\$
Hence, the current through LED is same as that through $\mathrm{R}_{\mathrm{S}}, $\$
\mathrm{I}_{	ext {LED }}=0.5 \mathrm{~mA}=0.5 	imes 10^{-3} \mathrm{~A}$ $\$ 	ext{ } $\$
The power rating of LED is $2 	imes 10^{-3} \mathrm{~W}$, $\$
Since $\mathrm{V}_{\mathrm{L}}=\mathrm{I}_{\mathrm{R}} 	imes \mathrm{R}$, we have $\mathrm{I}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{L}}}{1000}$. $\$	ext{}$\$
To find the maximum safe operating voltage for the LED, we use the power formula:
$ \mathrm{P}=\mathrm{V} 	imes \mathrm{I} $ 
The power consumed by the LED is $\mathrm{P}_{\mathrm{LED}}=\mathrm{V}_{\mathrm{L}} 	imes \mathrm{I}_{\mathrm{LED}}$.
$ 2 	imes 10^{-3}=V_L 	imes 0.5 	imes 10^{-3} \Rightarrow V_L=4 V $ $\$
From the circuit, the voltage drop across $R_S$ is $V_{R_S}=V_S-V_L$.
$ V_{R_S}=5 \mathrm{~V}-4 \mathrm{~V}=1 \mathrm{~V} $ $\$
By Ohm's Law $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{V}_{\mathrm{R}_{\mathrm{S}}}}{\mathrm{I}_{\mathrm{s}}}=\frac{1}{0.5 	imes 10^{-3}}=2 	imes 10^3 \Omega=2 \mathrm{k} \Omega$ $\$ 	ext{ } $\$
Therefore, the minimum value of the resistance $\mathrm{R}_{\mathrm{S}}$ to ensure that LED is not damaged is $2 \mathrm{k} \Omega$. $\$
Thus, the correct option is (B).

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6

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2

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4

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5

Semiconductors and Logic Gates

The maximum rated power of the LED is 2 mW and it is used in the circuit with input voltage of 5 V as shown in the figure below. The current through resistance {R}_S is 0.5 mA . The minimum value of the resistance of {R}_S, to ensure that the LED is not damaged is \_\_\_\_ \mathrm{k} \Omega.

Choose the correct answer: