GATE 2020 EC Q.32: solution and answer

GATE 2020 · ECQ.32 · 2 marks · MCQ

The band diagram of a p-type semiconductor with a band-gap of 1 eV is shown. Using this semiconductor, a MOS capacitor having $V_{T H}$ of $- 0.16 V$ , $C_{o x}^{'}$ of $100 nF/cm^{2}$ and a metal work function of $3.87 eV$ is fabricated. There is no charge within the oxide. If the voltage across the capacitor is $V_{T H}$ , the magnitude of depletion charge per unit area (in $C/cm^{2}$ ) is

GATE 2020 EC question 32: Energy band diagram of a p-type semiconductor, vertical energy axis. — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
1.70 x 10^-8
(B)
0.52 x 10^-8
(C)
1.41 x 10^-8
(D)
0.93 x 10^-8

Answer and solution

ANSWER

The magnitude of depletion charge per unit area is $1.70 \times 1 0^{- 8} C/cm^{2}$ (Option A).

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

From the band diagram, the electron affinity is $χ = 4 eV$ (vacuum level to $E_{C}$ ).
Using $E_{C} - E_{i} = 0.5 eV$ , $E_{F s} - E_{V} = 0.2 eV$ , and band-gap $= 1 eV$ : $E_{i} - E_{V} = 0.5 eV$ , so $E_{i} - E_{F s} = 0.5 - 0.2 = 0.3 eV$ .
Semiconductor work function: $Φ_{s} = χ + (E_{C} - E_{i}) + (E_{i} - E_{F s}) = 4 + 0.5 + 0.3 = 4.8 eV$ .
Flat-band voltage: $V_{F B} = Φ_{m} - Φ_{s} = 3.87 - 4.8 = - 0.93 V$ .
Bulk potential magnitude: $ϕ_{F} = \frac{E _{i} - E _{F s}}{q} = 0.3 V$ , so $2 ϕ_{F} = 0.6 V$ .
Threshold-voltage relation for a p-substrate MOS capacitor with no oxide charge: $V_{T H} = V_{F B} + 2 ϕ_{F} + \frac{Q _{d e p}}{C _{o x}^{'}}$
Substituting: $- 0.16 = - 0.93 + 0.6 + \frac{Q _{d e p}}{C _{o x}^{'}} \Rightarrow \frac{Q _{d e p}}{C _{o x}^{'}} = 0.17 V$ .
$Q_{d e p} = 0.17 \times 100 nF/cm^{2} = 17 nC/cm^{2} = 1.70 \times 1 0^{- 8} C/cm^{2}$ , matching option A.

KEY IDEA

MOS threshold voltage relation for a p-type substrate: $V_{T H} = V_{F B} + 2 ϕ_{F} + \frac{Q _{d e p}}{C _{o x}^{'}}$ , where $V_{F B} = Φ_{m} - Φ_{s}$ is obtained directly from the semiconductor band diagram (electron affinity plus the gap from $E_{C}$ down to $E_{F s}$ ).

WATCH OUT FOR

Misreading the band diagram — adding instead of subtracting the 0.2 eV arrow (which points from $E_{V}$ up to $E_{F s}$ ), giving $E_{i} - E_{F s} = 0.8$ eV instead of $0.3$ eV, or forgetting that the semiconductor work function must include the full gap from $E_{C}$ to $E_{F s}$ (not just to $E_{i}$ ) when computing $Φ_{s}$ .

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 7 MOS devices questions, with notes

GATE 2020 EC Q.32: solution and answer

Answer and solution

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