Question

A p--n junction has acceptor impurity concentration of \(10^{17}\,cm^{-3}\) in the p-side and donor impurity concentration of \(10^{16}\,cm^{-3}\) in the n-side. What is the contact potential at the junction (\(T=\) thermal energy, intrinsic semiconductor concentration \(n_i = 1.4\times 10^{10}\,cm^{-3}\)) ?

• A. \((kT/e)\ln(4\times 10^{12})\)
• B. \((kT/e)\ln(2.5\times 10^{23})\)
• C. \((kT/e)\ln(10^{23})\)
• D. \((kT/e)\ln(10^{9})\)

Hint:

Built-in potential: \(V_0=\frac{kT}{e}\ln\left(\frac{N_A N_D}{n_i^2}\right)\). Higher doping increases junction potential.

Answer 1

The Correct Option is A

Step 1: Use formula for built-in (contact) potential.
\[ V_0 = \frac{kT}{e}\ln\left(\frac{N_A N_D}{n_i^2}\right) \]
Step 2: Substitute given values.
\[ N_A = 10^{17},\quad N_D = 10^{16},\quad n_i = 1.4\times 10^{10} \]
Step 3: Compute the ratio inside log.
\[ \frac{N_A N_D}{n_i^2} = \frac{10^{17}\times 10^{16}}{(1.4\times 10^{10})^2} \]
\[ = \frac{10^{33}}{1.96\times 10^{20}} = 5.1 \times 10^{12} \approx 4\times 10^{12} \]
Step 4: Final expression.
\[ V_0 = \frac{kT}{e}\ln(4\times 10^{12}) \]
Final Answer:
\[ \boxed{\left(\frac{kT}{e}\right)\ln(4\times 10^{12})} \]