The maximum rated power of the LED is $2 \text{ mW}$ and it is used in the circuit with input voltage of $5 \text{ V}$ as shown in the figure below. The current through resistance $R$ is $0.5 \text{ mA}$. The minimum value of the resistance $R_s$, to ensure that the LED is not damaged is ______ $\text{k}\Omega$.
In the provided circuit, the input voltage is $5 \text{ V}$, a series resistor $R_s$ is connected, and the circuit then branches into two parallel paths: one with a resistor $R = 1 \text{ k}\Omega$ in series with a Zener diode, and another with the LED.
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The total current is the sum of the LED current and the current through $R$. Use $P = VI$ for the LED and assume a standard operating voltage of $2\text{V}$ to find the current.
In this circuit, the LED is operating at its maximum power limit of $P_{max} = 2 \text{ mW}$ to determine the minimum value of the series limiting resistor $R_s$.
Step 1: Determine the node voltage $V$. Standard GaAs LEDs (common in these problems) operate at a forward voltage of $V_f \approx 2 \text{ V}$. In the parallel branch, we have a resistor $R = 1 \text{ k}\Omega$ with current $I_R = 0.5 \text{ mA}$. The voltage across $R$ is $V_R = I_R \times R = 0.5 \text{ mA} \times 1 \text{ k}\Omega = 0.5 \text{ V}$. Since the LED and the branch $(R + \text{Zener})$ are in parallel, the node voltage is $V = V_{LED} = 2 \text{ V}$. (This also implies the Zener voltage is $V_Z = 2 - 0.5 = 1.5 \text{ V}$, which is a realistic value).
Step 2: Calculate the maximum current through the LED: $$I_{LED} = \frac{P_{max}}{V_{LED}} = \frac{2 \text{ mW}}{2 \text{ V}} = 1 \text{ mA}$$
Step 3: Calculate total current from the source: $$I_s = I_{LED} + I_R = 1 \text{ mA} + 0.5 \text{ mA} = 1.5 \text{ mA}$$
Step 4: Use the voltage divider / KVL principle for the input section: The voltage drop across $R_s$ is $V_{Rs} = V_{in} - V_{node} = 5 \text{ V} - 2 \text{ V} = 3 \text{ V}$. According to Ohm's Law: $$R_s = \frac{V_{Rs}}{I_s} = \frac{3 \text{ V}}{1.5 \text{ mA}} = 2 \text{ k}\Omega$$ Therefore, the minimum value of $R_s$ required to keep the LED power at or below $2 \text{ mW}$ is $2 \text{ k}\Omega$.
