When a silicon `PN` junction is in forwards biased condition with series resistance, it has knee voltage of `0.6 V`. Current flow in it is `5 mA`, when `PN` junction is connected with `2.6 V` battery, the value of series resistance is

Options

A. `100 Omega`
B. `200 Omega`
C. `400 Omega`
D. `500 Omega`

Answer

To solve the problem of finding the value of the series resistance in a silicon PN junction diode circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Knee voltage (V_k) = 0.6 V
- Current (I) = 5 mA = 5 × 10^(-3) A
- Total voltage from the battery (V_battery) = 2.6 V

2. **Identify the Voltage Across the Series Resistance**:
- The voltage across the series resistance (V_R) can be calculated by subtracting the knee voltage from the total voltage supplied by the battery:
\[ V_R = V_battery - V_k = 2.6 V - 0.6 V = 2.0 V \]

3. **Use Ohm's Law to Find the Resistance**:
- According to Ohm's law, the relationship between voltage (V), current (I), and resistance (R) is given by:
\[ V = I \cdot R \]
- Rearranging this formula to find resistance gives us:
\[ R = \frac{V_R}{I} \]

4. **Substitute the Values**:
- Now, substituting the values we have:
\[ R = \frac{2.0 V}{5 \times 10^{-3} A} \]
- Calculating this gives:
\[ R = \frac{2.0}{0.005} = 400 \, \Omega \]

5. **Conclusion**:
- The value of the series resistance is **400 Ω**.