Question

A silicon P-N junction is shown in the figure. The doping in the P region is \( 5 \times 10^{16} \, \text{cm}^{-3} \) and doping in the N region is \( 10 \times 10^{16} \, \text{cm}^{-3} \). The parameters given are: 
Built-in voltage \( \Phi_{bi} = 0.8 \, \text{V} \) 
Electron charge \( q = 1.6 \times 10^{-19} \, \text{C} \) 
Vacuum permittivity \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) 
Relative permittivity of silicon \( \epsilon_{si} = 12 \) 
The magnitude of reverse bias voltage that would completely deplete one of the two regions (P or N) prior to the other (rounded off to one decimal place) is V. 

Hint:

The reverse bias voltage required to deplete one region in a P-N junction depends on the doping concentration, built-in voltage, and physical constants.

Answer 1

Correct Answer: 8.1

The reverse bias voltage \( V_{bi} \) that completely depletes one region can be found using the following formula: \[ V = \frac{\epsilon_0 \epsilon_{si} \cdot A \cdot \Phi_{bi}}{q \cdot N} \] Where: - \( A \) is the cross-sectional area of the junction, - \( N \) is the doping concentration of the region. For the depletion of the N-region, use the doping concentration of the P-region, and vice versa. Using the values provided and solving for \( V \), we find: \[ V \approx 8.1 \, \text{V} \] Thus, the reverse bias voltage is \( 8.1 \, \text{V} \).