Q5GATE 2024NAT

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

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} C_{O X} & =\frac{\varepsilon_{O x}}{t_{O x}}=\frac{4 \times 8.85 \times 10^{-14}}{100 \times 10^{-9} \times 100} \\ & =4 \times 8.85 \times 10^{-14} \times 10^{5}=35.4 \times 10^{-9} \mathrm{~F} / \mathrm{cm}^{2} \\ V_{F B} & =\frac{Q_{O x}}{C_{O x}}+\phi_{m s} \\ 0 & =\frac{Q_{O x}}{C_{O x}}+\phi_{m s} \end{aligned}

$Q_{o x} \rightarrow$ Positive $\frac{Q_{O x}}{C_{o x}}=\frac{10^{-8}}{35.4 \times 10^{-9}}=\frac{10}{35.4} \text { volt }$ $\therefore \quad \frac{Q_{O x}}{C_{O x}} \rightarrow$ Negative

\begin{aligned} 0 & =\frac{-10}{35.4}+\phi_{\mathrm{ms}} \\ \phi_{\mathrm{ms}} & =\frac{10}{35.4} \\ \phi_{m}-\phi_{s} & =\frac{10}{35.4} \\ \phi_{s} & =\phi_{m}-\frac{10}{35.4}=4.6-\frac{10}{35.4}=4.317 \text { volt } \\ \therefore \quad 9 \phi_{s} & =4.317 \simeq 4.32 \mathrm{eV} \end{aligned}

Work function energy of semiconductor.

Q7GATE 2024NAT

A non-degenerate n-type semiconductor has 5% neutral dopant atoms. Its Fermi level is located at $0.25 \mathrm{eV}$ below the conduction band $\left(E_{C}\right)$ and the donor energy level $\left(E_{D}\right)$ has a degeneracy of 2. Assuming the thermal voltage to be $20 \mathrm{mV}$ . The difference between $E_{C}$ and $E_{D}$ (in $\mathrm{eV}$ , rounded off to two decimal places) is ______

🚀 Solve in Practice Mode

📖 Explanation

$5 \%$ neutral donor atoms/dopant atoms means $95 \%$ donor atoms are ionized. Concentration of electrons occupying the donor level is given as $n_{d}=\frac{N_{d}}{1+\frac{1}{g} \exp \left(\frac{E_{D}-E_{F}}{k T}\right)}$ and $n_{d}=N_{d}-N_{d}^{+}$ where $N_{d}^{+}$ is ionized donor atoms concentration. according to question

\begin{aligned} n_{d} & =5 \% \text { of } N_{d} \\ g & =2 \text { degeneracy factor } \end{aligned}

$\frac{5}{100} N_{d}=\frac{N_{d}}{1+\frac{1}{g} \exp \left(\frac{E_{D}-E_{F}}{k T}\right)}$ $\frac{1}{20}=\frac{1}{1+\frac{1}{g} \exp \left(\frac{E_{D}-E_{F}}{k T}\right)}$ $1+\frac{1}{g} \exp \left(\frac{E_{D}-E_{F}}{k T}\right)=20$ $\exp \left(\frac{E_{D}-E_{F}}{k T}\right)=38$

\begin{aligned} \frac{E_{D}-E_{F}}{k T} & =\ln (38) \quad\left\{\text { given } V_{T}=\frac{k T}{q}=20 \mathrm{mV}\right\} \\ E_{D}-E_{F} & =\ln (38) \times 0.020 \\ & =0.07275 \mathrm{eV} \end{aligned}

given: $\quad E_{C}-E_{F}=0.25 \mathrm{eV}$ $\therefore \quad E_{C}-E_{D}+E_{D}-E_{F}=0.25 \mathrm{eV}$

\begin{aligned} E_{C}-E_{D} & =0.25 \mathrm{eV}-\left(E_{D}-E_{F}\right) \\ & =0.25 \mathrm{eV}-0.07275 \mathrm{eV} \\ & =0.17725 \mathrm{eV} \simeq 0.18 \mathrm{eV} \end{aligned}

Q11GATE 2023NAT

In a semiconductor device, the Fermi-energy level is $0.35 \mathrm{eV}$ above the valence band energy. The effective density of states in the valence band at $T=300 \mathrm{~K}$ is $1 \times 10^{19} \mathrm{~cm}^{-3}$ . The thermal equilibrium hole concentration in silicon at $400 \mathrm{~K}$ is ___ $\_\times 10^{13} \mathrm{~cm}^{-3}$ . (rounded off to two decimal places). Given $\mathrm{kT}$ at $300 \mathrm{~K}$ is $0.026 \mathrm{eV}$ .

🚀 Solve in Practice Mode

📖 Explanation

Given, $E_{F}-E_{V}=0.35 \mathrm{eV} \quad \text { [Considering it is given at } 400 \mathrm{~K} \text { ] }$ Also, $V_{T_{1}}=K T_{1}=0.026 \mathrm{eV}$ at $T_{1}=300 \mathrm{~K}$ $\therefore$ $\frac{V_{T_{1}}}{V_{T_{2}}}=\frac{T_{1}}{T_{2}} \Rightarrow V_{T_{2}}=\frac{T_{2}}{T_{1}} \times V_{T_{1}}$ $\therefore$ $V_{T_{2}}=\frac{400}{300} \times 0.026$ $V_{T_{2}}=0.03466 \mathrm{eV}$ at $T_{2}=400 \mathrm{~K}$ Now, $N_{v}=1 \times 10^{19} / \mathrm{cm}^{3}$ at $T_{1}=300 \mathrm{~K}$ $N_{v} \propto T^{3 / 2}$ $\frac{N_{V_{2}}}{N_{V_{1}}}=\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2}$ $N_{V_{2}}=\left(\frac{T_{2}}{T_{1}}\right)^{3 / 2} N V_{1}$ $\left(\because T_{2}=400 \mathrm{~K}\right)$ $=\left(\frac{400}{300}\right)^{3 / 2} N_{V_{1}}$ $N_{V_{2}}=1.5396 \times 10^{19} / \mathrm{cm}^{3}$ Now, hole concentration at $400 \mathrm{~K}$ is given as

\begin{aligned} & p=N_{V} e^{-\left(E_{F}-E_{V}\right) / k T_{2}}=1.5396 \times 10^{19} \times e^{-0.35 \mathrm{eV} / 0.03466 \mathrm{eV}} \\ & p=63.36 \times 10^{13} \mathrm{~cm}^{-3} \end{aligned}

Q14GATE 2022MCQ

Consider a long rectangular bar of direct bandgap p-type semiconductor. The equilibrium hole density is $10^{17}cm^{-3}$ and the intrinsic carrier concentration is $10^{10}cm^{-3}$ . Electron and hole diffusion lengths are $2\mu m$ and $1\mu m$ , respectively. The left side of the bar ( $x=0$ ) is uniformly illuminated with a laser having photon energy greater than the bandgap of the semiconductor. Excess electron-hole pairs are generated ONLY at $x=0$ because of the laser. The steady state electron density at $x=0$ is $10^{14}cm^{-3}$ due to laser illumination. Under these conditions and ignoring electric field, the closest approximation (among the given options) of the steady state electron density at $x=2 \mu m$ , is _____

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📖 Explanation

From continuity equation of electrons $\frac{dn}{dt}=n\mu _n\frac{dE}{dx}+\mu _nE\frac{dn}{dx}+G_n-R_n+x_n\frac{d^2x}{dx^2} \;\;\;...(i)$ [Because $\vec{E}$ is not mentioned hence $\frac{dE}{dx}=0$ For $x \gt 0, G_n$ is also zero $n=\frac{n_i^2}{N_A}=\frac{10^{20}}{10^{17}}=10^3$ $n=n_0+\delta n=10^{3+}10^{14}=10^{14}$ at steady state, $\frac{db}{dt}=0$ Hence equation (i) becomes: $O=D_n\frac{d^2\delta n}{dx^2}-\frac{\delta n}{\tau _n}$ $\frac{d^2\delta n}{dx^2}=\frac{\delta n}{L_n^2} \;\;\;...(ii)$ From solving equation (ii) $\delta _n(x)=\delta _n(0)e^{-x/L_n}$ at $x=2\mu m$ $\delta _n(2\mu m)=10^{14}e^{-2/2}=10^{14}e^{-1}=0.37 \times 10^{14}$

Q16GATE 2021MCQ

A bar of silicon is doped with boron concentration of $10^{16} \text{cm}^{-3}$ and assumed to be fully ionized. It is exposed to light such that electron-hole pairs are generated throughout the volume of the bar at the rate of $10^{20} \text{cm}^{-3} s^{-1}$ . If the recombination lifetime is $100 \;\mu s$ , intrinsic carrier concentration of silicon is $10^{10} \text{cm}^{-3}$ and assuming $100\%$ ionization of boron, then the approximate product of steady-state electron and hole concentrations due to this light exposure is

🚀 Solve in Practice Mode

📖 Explanation

Boron $\rightarrow$ Acceptor type doping

\begin{aligned} N_{A} &=10^{16} \mathrm{~cm}^{-3} \\ g_{0 p} &=1020 \mathrm{~cm}^{-3} \mathrm{~s}^{-1} \\ \tau &=100 \mu \mathrm{s} \\ n_{i} &=10^{10} \mathrm{~cm}^{-3} \end{aligned}

Product of steady state electron-hole concentration =? At thermal equilibrium (before shining light)

\begin{array}{ll} \text { Hole concentration, } & p_{o} \simeq N_{A}=10^{16} \mathrm{~cm}^{-3} \\ \text { Electron concentration, } & n_{0}=\frac{n_{i}^{2}}{p_{0}}=\frac{10^{20}}{10^{16}}=10^{4} \mathrm{~cm}^{-3} \end{array}

After, illumination of light, Hole concentration, $p=p_{o}+\delta p$ Electron concentration, $\quad n=n_{o}+\delta n$ Due to shining light, excess carrier concentration,

\begin{aligned} \delta p &=\delta n=g_{o p} \cdot \tau=10^{20} \times 100 \times 10^{-6}=10^{16} \mathrm{~cm}^{-3} \\ \therefore \qquad p &=10^{16}+10^{16}=2 \times 10^{16} \mathrm{~cm}^{-3}\\ n&=10^{4}+10^{16} \simeq 10^{16} \mathrm{~cm}^{-3} \end{aligned}

So, product of steady state electron-hole concentration

\begin{aligned} &=n p=10^{16} \times 2 \times 10^{16} \\ &=2 \times 10^{32} \mathrm{~cm}^{-6} \end{aligned}

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