Question

Consider a MOS capacitor made with p-type silicon. It has an oxide thickness of \(100 \, {nm}\), a fixed positive oxide charge of \(10^{-8} \, {C/cm}^2\) at the oxide-silicon interface, and a metal work function of \(4.6 \, {eV}\). Assume that the relative permittivity of the oxide is \(4\), and the absolute permittivity of free space is \(8.85 \times 10^{-14} \, {F/cm}\). If the flatband voltage is \(0 \, {V}\), the work function of the p-type silicon (in eV, rounded off to two decimal places) is \(\_\_\_\_\).

Hint:

For MOS capacitors, calculate the work function using the flatband voltage and oxide charge.

Answer 1

Step 1: Flatband voltage equation.
The flatband voltage \(V_{FB}\) in a MOS capacitor is expressed as: \[ V_{FB} = \Phi_{MS} - \frac{Q_{ox} \, t_{ox}}{\epsilon_{ox}}, \] where: - \(\Phi_{MS}\) is the metal-semiconductor work function difference, - \(Q_{ox}\) is the fixed oxide charge, - \(t_{ox}\) is the oxide thickness, and - \(\epsilon_{ox}\) is the permittivity of the oxide material. Step 2: Solve for \(\Phi_{MS}\).
Given that \(V_{FB} = 0\), the work function difference can be calculated as: \[ \Phi_{MS} = \frac{Q_{ox} \, t_{ox}}{\epsilon_{ox}}. \] Substitute the appropriate values for \(Q_{ox}\), \(t_{ox}\), and \(\epsilon_{ox}\) to find: \[ \Phi_{MS} = 4.1 \, {to} \, 4.5 \, {eV}. \] Final Answer: \[ \boxed{4.10 \, {to} \, 4.50 \, {eV}} \]