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 $\_\_\_\_$ $\mathrm{k} \Omega$.

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 $=1 \mathrm{k} \Omega=1000 \Omega$.
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 $\mathrm{R}_{\mathrm{S}}, \mathrm{I}_{	ext {LED }}=0.5 \mathrm{~mA}=0.5 	imes 10^{-3} \mathrm{~A}$
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 $\mathrm{I}_{\mathrm{R}}=\frac{\mathrm{V}_{\mathrm{L}}}{1000}$.
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 $\mathrm{R}_{\mathrm{S}}=\frac{\mathrm{V}_{\mathrm{R}_{\mathrm{S}}}}{\mathrm{I}_{\mathrm{s}}}=\frac{1}{0.5 	imes 10^{-3}}=2 	imes 10^3 \Omega=2 \mathrm{k} \Omega$
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).

Answer

6

Answer

2

Answer

4

Answer

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: