GATE 2024 EC Q.61: solution and answer

GATE 2024 · ECQ.61 · 2 marks · NAT

Consider a MOS capacitor made with p-type silicon. It has an oxide thickness of 100 nm, a fixed positive oxide charge of $1 0^{- 8} C/cm^{2}$ 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 $8.85 \times 1 0^{- 14} F/cm$ . If the flatband voltage is 0 V, the work function of the p-type silicon (in eV, rounded off to two decimal places) is ______.

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Answer and solution

ANSWER

The work function of the p-type silicon is φ_s ≈ 4.32 eV.

Official answer: 4.10 to 4.50Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

Convert oxide thickness to cm and compute oxide permittivity: $t_{o x} = 100 nm = 1 \times 1 0^{- 5} cm$ , and $ε_{o x} = ε_{r} ε_{0} = 4 \times 8.85 \times 1 0^{- 14} F/cm = 3.54 \times 1 0^{- 13} F/cm$ .
Oxide capacitance per unit area: $C_{o x} = \frac{ε _{o x}}{t _{o x}} = \frac{3.54 \times 1 0 ^{- 13}}{1 \times 1 0 ^{- 5}} = 3.54 \times 1 0^{- 8} F/cm^{2}$
Flat-band condition with a fixed positive oxide charge $Q_{o x}$ at the oxide-silicon interface: $V_{F B} = ϕ_{m s} - \frac{Q _{o x}}{C _{o x}} = 0, ϕ_{m s} = ϕ_{m} - ϕ_{s}$
Solve for the silicon work function: $ϕ_{s} = ϕ_{m} - \frac{Q _{o x}}{C _{o x}}$
Compute the charge term: $\frac{Q _{o x}}{C _{o x}} = \frac{1 \times 1 0 ^{- 8} C/cm ^{2}}{3.54 \times 1 0 ^{- 8} F/cm ^{2}} = 0.2825 V$
Substitute: $ϕ_{s} = 4.6 - 0.2825 = 4.3175 eV$
Rounded to two decimal places: $ϕ_{s} \approx 4.32 eV$

KEY IDEA

The flatband voltage relates the metal-semiconductor work function difference to the effect of oxide charge: $V_{F B} = ϕ_{m s} - \frac{Q _{o x}}{C _{o x}}$ , with $ϕ_{m s} = ϕ_{m} - ϕ_{s}$ . Setting $V_{F B} = 0$ gives $ϕ_{s} = ϕ_{m} - Q_{o x} / C_{o x}$ directly, once $C_{o x}$ is found from the oxide thickness and permittivity.

WATCH OUT FOR

Forgetting to convert the oxide thickness from nanometres to centimetres (100 nm = $1 \times 1 0^{- 5}$ cm) before computing $C_{o x}$ , or mixing up the sign in $V_{F B} = ϕ_{m s} - Q_{o x} / C_{o x}$ and adding the charge term instead of subtracting it, which shifts the answer outside the accepted key range.

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GATE 2024 EC Q.61: solution and answer

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