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 _______ k$$\Omega$$.

The Middle Diode: It points upwards against the $$5\text{ V}$$ downward current flow. It is reverse-biased (acts as an open circuit, carrying $$0\text{ mA}$$).

The LED: It points downwards with the current flow. It is forward-biased and carries the entire current from $$R_S$$.

$$\text{Therefore, } I_{\text{LED}} = I_S = 0.5\text{ mA}$$

$$V_{\text{LED}} = \frac{P_{\text{LED}}}{I_{\text{LED}}} = \frac{2\text{ mW}}{0.5\text{ mA}} = 4\text{ V}$$

$$V_{\text{in}} = V_{R_S} + V_{\text{LED}}$$

$$5\text{ V} = (I_S \cdot R_S) + 4\text{ V}$$

$$1\text{ V} = 0.5\text{ mA} \times R_S$$

$$R_S = \frac{1\text{ V}}{0.5 \times 10^{-3}\text{ A}} = 2000\ \Omega = 2\ \text{k}\Omega$$