# Electronic Devices

 Question 1
A p-type semiconductor with zero electric field is under illumination (low level injection) in steady state condition. Excess minority carrier density is zero at $x=\pm 2l_n$, where $l_n=10^{-4} cm$ is the diffusion length of electrons. Assume electronic charge, $q=-1.6 \times 10^{-19}C$. The profiles of photo-generation rate of carriers and the recombination rate of excess minority carriers (R) are shown. Under these conditions, the magnitude of the current density due to the photo-generated electrons at $x=+2l_n$ is _________ $mA/cm^2$ (rounded off to two decimal places).

 A 0.44 B 0.59 C 0.77 D 0.83
GATE EC 2022      PN-Junction Diodes and Special Diodes
Question 1 Explanation:
Given,
\begin{aligned} \delta _n(x)&=R\tau _n=10^{20}e^{-|x|/l_n}\cdot \tau _n\\ \delta (l_n)&=10^{20}e^{-1}\tau _n \;\;\;...(i)\\ for \;\; l_n\leq x\leq 2l_n \end{aligned}
Continuity equation is given by
$D_n\frac{\partial^2 \delta _n}{\partial x^2}+G-R \;\;\;...(ii)$
G and R both are zero - for $[l_n\leq x\leq 2l_n]$
Hence Equation (i) reduced to $D_n \frac{\partial^n \delta _n }{\partial x^2}=0$
$\Rightarrow \delta _n(x)=Ax+B$
For calculating A and B we use Boundary condition
$\delta _n(2l_n)=0\; \Rightarrow A=\frac{-B}{2l_n}$
$\therefore \delta _n(x)=\frac{-B}{2l_n}x+B=B\left [ 1-\frac{x}{2l_n} \right ] \;\;\;...(iii)$
At $x=l_n$
\begin{aligned} 10^{20}e^{-1}\tau _n&=B\left [ 1-\frac{l_n}{2l_n} \right ] \\ B&= 2 \times 10^{20}e^{-1}\tau _n\\ \therefore \; \delta _n(x) &=2 \times 10^{20} e^{-1}\tau _n\left [ 1-\frac{x}{2l_n} \right ] \end{aligned}
for $l_n \leq x \leq 2l_n$
Electron diffusion current density is given by $|J_n|_{diff}=qD_n\frac{d\eta }{dx}=qD_n \times 2 \times 10^{20} \times e^{-1} \times \tau _n\left [ 0-\frac{1}{2l_n} \right ]$
$=\frac{1.6 \times 10^{-19} \times l_n ^2 \times 2 \times 10^{20} \times e^{-1}}{2l_n}$
$=1.6 \times 10^{-19} \times l_n \times 10^{20} \times e^{-1}$
$=1.6 \times 10 \times 1 \times 10^{-4} \times e^{-1}$
$=0.59mA/cm^2$
 Question 2
Select the CORRECT statement(s) regarding semiconductor devices.
 A Electrons and holes are of equal density in an intrinsic semiconductor at equilibrium. B Collector region is generally more heavily doped than Base region in a BJT. C Total current is spatially constant in a two terminal electronic device in dark under steady state condition. D Mobility of electrons always increases with temperature in Silicon beyond 300 K.
GATE EC 2022      Basic Semiconductor Physics
Question 2 Explanation:
At equilibrium $n = p = n_i$ for intrinsic semiconductor
Collector region is generally lightly doped then base region in BJT. Hence option B is wrong.
By increasing temperature above 300K, mobality of electrons decreases hence option (D) is also wrong
 Question 3
An ideal MOS capacitor (p-type semiconductor) is shown in the figure. The MOS capacitor is under strong inversion with $V_G=2V$. The corresponding inversion charge density $Q_{IN}$ is $2.2\mu C/cm^2$. Assume oxide capacitance per unit area as $C_{OX}=1.7\mu F/cm^2$. For $V_G=4V$, the value of $Q_{IN}$ is ______$\mu C/cm^2$ (rounded off to one decimal place).

 A 4.8 B 5.6 C 8.2 D 9.7
GATE EC 2022      BJT and FET Basics
Question 3 Explanation:
\begin{aligned} Q_{IN}&=-CO_X(V_G-V_T)\\ Q_{IN_1}&=-CO_X(V_{G1}-V_T)\;\;\;...(i)\\ Q_{IN_2}&=-CO_X(V_{G2}-V_T)\;\;\;...(ii)\\ &(ii)-(i)\\ Q_{IN_2}-Q_{IN_1}&=-CO_X(V_{G2}-V_{G1})\\ Q_{IN_2}-(-2.2\mu c/cm^2)&=-1.7\mu c/cm^2(4-2)\\ Q_{IN_2}&=2.2\mu c/cm^2-3.4\mu c/cm^2\\ &=-5.6\mu c/cm^2 \end{aligned}
 Question 4
In a non-degenerate bulk semiconductor with electron density $n=10^{16}cm^{-3}$, the value of $E_C-E_{Fn}=200meV$, where $E_C$ and $E_{Fn}$ denote the bottom of the conduction band energy and electron Fermi level energy, respectively. Assume thermal voltage as 26 meV and the intrinsic carrier concentration is $10^{10}cm^{-3}$. For $n=0.5 \times 10^{16}cm^{-3}$, the closest approximation of the value of ($E_C-E_{Fn}$), among the given options, is ______.
 A 226 meV B 174 meV C 218 meV D 182 meV
GATE EC 2022      Basic Semiconductor Physics
Question 4 Explanation:
Here we have to find the value of $E_c-E_{fn}$
As we know,
$E_C-E_F=kT \ln\left ( \frac{N_c}{n} \right ) \;\;\;...(i)$
$E_C-E_{F1}=kT \ln\left ( \frac{N_c}{n_1} \right ) \;\;\;...(ii)$
$E_C-E_{F2}=kT \ln\left ( \frac{N_c}{n_2} \right ) \;\;\;...(iii)$
Equation (ii) - Equation (iii)
$(E_C-E_{F1})-(E_C-E_{F2})=kT \ln \left ( \frac{\frac{N_c}{n_1}}{\frac{N_c}{n_2}} \right )=kT \ln \frac{n_2}{n_1}$
$\Rightarrow 200meV-(E_C-E_{F2})=26meV \times \ln \left ( \frac{0.5 \times 10^{16}}{1 \times 10^{16}} \right )$
$200meV-(E_C-E_{F2})=+26meV \ln (0.5)=-18$
$(E_C-E_{F2})=200+8=218meV$
 Question 5
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 _____
 A $0.37 \times 10^{14} cm^{-3}$ B $0.63 \times 10^{13} cm^{-3}$ C $3.7 \times 10^{14} cm^{-3}$ D $0^{3} cm^{-3}$
GATE EC 2022      Basic Semiconductor Physics
Question 5 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 $x=2\mu m$
$\delta _n(2\mu m)=10^{14}e^{-2/2}=10^{14}e^{-1}=0.37 \times 10^{14}$
 Question 6
For the transistor $M_{1}$ in the circuit shown in the figure, $\mu_{n} C_{\text{ox}} = 100\:\mu A/V^{2}$ and $(W/L)=10$, where $\mu_{n}$ is the mobility of electron, $C_{\text{ox}}$ is the oxide capacitance per unit area , W is the width and L is the length.

The channel length modulation coefficient is ignored. If the gate-to-source voltage $V_{\text{GS}}\text{ is } 1\:V$ to keep the transistor at the edge of saturation, then the threshold voltage of the transistor (rounded off to one decimal place) is _______ V.
 A 0.3 B 0.5 C 1.2 D 1.5
GATE EC 2021      BJT and FET Basics
Question 6 Explanation:
\begin{aligned} I_{D S}&=\frac{\mu_{n} C_{O x}}{2} \times \frac{W}{L}\left(V_{G S}-V_{T}\right)^{2} \\ \text{Given}\qquad V_{G S}&=1 \mathrm{~V} \\ I_{D S}&=\frac{1}{2}\left(1-V_{T}\right)^{2} \end{aligned}

\begin{aligned} V_{D S}=3-20 \times I_{D S} \\ V_{D S}=3-\frac{20}{2}\left(1-V_{T}\right)^{2} \\ V_{D S}=3-10\left(1-V_{T}\right)^{2} \end{aligned}
MOSFET operates in saturation if
\begin{aligned} V_{D S} &\geq V_{G S}-V_{T}\\ \text{So, we take},\qquad \quad V_{D S}&=V_{G S}-V_{T} \\ V_{G S}-V_{T}&=3-10\left(1-V_{T}\right)^{2} \\ 1-V_{T}&=3-10\left(1-V_{T}\right)^{2}\\ \text{Let,}\qquad 1-V_{T} &=x \\ 3-10 x^{2} &=x \\ 10 x^{2}+x-3 x &=0 \\ \text{We get,}\qquad x &=-\frac{1 \pm \sqrt{1+120}}{20}\\ x &=-\frac{1 \pm 11}{20}\\ \therefore \qquad \quad x&=0.5 \text { and }-0.6\\ x&=0.5\\ \Rightarrow \qquad \quad 1-V_{T} & =0.5 \\ \Rightarrow \qquad \quad V_{T} & =0.5 \mathrm{~V} \\ x & =-0.6 \\ \Rightarrow\qquad \quad 1-V_{T} & =-0.6 \\ \Rightarrow\qquad \quad V_{T} & =1.6 \mathrm{~V} \\ \text { But }\qquad \quad V_{G S} & \gt V_{T} \\ \text { or }\qquad \quad V_{T} & \lt V_{G S} \\ \text { i.e., }\qquad \quad V_{T} & \lt 1 \\ \therefore\qquad \quad V_{T} & =0.5 \mathrm{~V} \end{aligned}
 Question 7
A silicon P-N junction is shown in the figure. The doping in the P region is $5\times10^{16}\:cm^{-3}$ and doping in the N region is $10\times10^{16}\:cm^{-3}$. The parameters given are
Built-in voltage $\left ( \Phi _{\text{bi}} \right ) = 0.8\:V$
Electron charge $(q) = 1.6\times10^{-19} C\:$
Vacuum permittivity $\left ( \varepsilon _{0} \right ) = 8.85\times10^{-12}\:F/m$
Relative permittivity of silicon $\left ( \varepsilon _{\text{Si}} \right ) = 12$

he magnitude of reverse bias voltage that would completely deplete one of the two regions (P or N) prior to the other (rounded off to one decimal place) is _________V.
 A 2.4 B 8.2 C 6.4 D 9.7
GATE EC 2021      PN-Junction Diodes and Special Diodes
Question 7 Explanation:

Given: $\quad N_{A}=5 \times 10^{16} \mathrm{~cm}^{-3} \quad ; \quad N_{D}=10 \times 10^{16} \mathrm{~cm}^{-3}$
Built-in potential, $\quad \phi_{\mathrm{bi}}=0.8 \mathrm{~V}$
Electron charge, $q=1.6 \times 10^{-19} \mathrm{C}$
Vacuum permittivity, $\epsilon_{o}=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}=8.85 \times 10^{-14} \mathrm{~F} / \mathrm{cm}$
Relative permittivity silicon,
$\epsilon_{\mathrm{si}}=12$
$\Rightarrow$ Doping on both sides is comparable, so smaller region would deplete first.
So, depletion region width on N -side $=x_{n}=0.2 \mu \mathrm{m}$
\begin{aligned} \Rightarrow \qquad x_{n}&=0.2 \times 10^{-4} \mathrm{~cm}\\ x_{n}&=\sqrt{\frac{2 \epsilon_{S i}}{q}\left(\frac{N_{A}}{N_{D}}\right)\left(\frac{1}{N_{A}+N_{D}}\right)\left(\phi_{b i}+V_{R}\right)} \end{aligned}
where, $V_{R} \rightarrow$ Magnitude of reverse bias potential
\begin{aligned} \Rightarrow \qquad\qquad 0.2 \times 10^{-4}&=\sqrt{\frac{2 \times 12 \times 8.85 \times 10^{-14}}{1.6 \times 10^{-19}} \cdot \frac{5 \times 10^{16}}{10 \times 10^{16}} \cdot \frac{1}{\left(15 \times 10^{16}\right)}\left(\phi_{b i}+V_{R}\right)} \\ \Rightarrow \qquad\qquad \phi_{b j}+V_{R}&=9.039\\ \Rightarrow \qquad\qquad V_{R}&=9.039-0.8\\ \Rightarrow \qquad\qquad V_{R}&=8.239 \mathrm{~V} \end{aligned}
 Question 8
For an n-channel silicon $\text{MOSFET}$ with $10\:nm$ gate oxide thickness, the substrate sensitivity $\left ( \partial V_{T}/\partial \left | V_{BS} \right | \right )$ is found to be $50\:mV/V$ at a substrate voltage $\left | V_{BS} \right |=2V$, where $V_{T}$ is the threshold voltage of the $\text{MOSFET}$. Assume that, $\left | V_{BS} \right | \gg 2\Phi _{B}$, where $q\Phi _{B}$ is the separation between the Fermi energy level $E_{F}$ and the intrinsic level $E_{i}$ in the bulk. Parameters given are
Electron charge $(q) = 1.6 \times 101^{-19}\:C$
Vacuum permittivity $(\varepsilon _{0}) = 8.85 \times 10^{-12}\: F/m$
Relative permittivity of silicon $(\varepsilon _{Si}) = 12$
Relative permittivity of oxide $(\varepsilon _{ox}) = 4$
The doping concentration of the substrate is
 A $7.37\times 10^{15}\:cm^{-3}$ B $4.37\times 10^{15}\:cm^{-3}$ C $2.37\times 10^{15}\:cm^{-3}$ D $9.37\times 10^{15}\:cm^{-3}$
GATE EC 2021      BJT and FET Basics
Question 8 Explanation:
Given, N -channel MOSFET
\begin{aligned} t_{o x}&=10 \mathrm{~nm}=10 \times 10^{-7} \mathrm{~cm} & \\ \frac{\partial V_{T}}{\partial\left|V_{B S}\right|}&=50 \mathrm{mV} / \mathrm{V}, & \left|V_{B S}\right|=2 \mathrm{~V} \\ \qquad q&=1.6 \times 10^{-19} \mathrm{C} & \left|V_{B S}\right|>>2 \phi_{B} \end{aligned}

\begin{aligned} \epsilon_{o} &=8.85 \times 10^{-14} \mathrm{~F} / \mathrm{cm} \\ \epsilon_{r_{\mathrm{Si}}} &=12 \\ \epsilon_{r_{O x}} &=4 \end{aligned}
Threshold voltage, including body effect,
$V_{T}=\phi_{m s}+\frac{\sqrt{2 \epsilon_{s i} q N_{A}\left(2 \phi_{B}-V_{B S}\right)}}{C_{o x}}+2 \phi_{B}$
\begin{aligned} \text{In question, we need, } \left|V_{B S}\right|&=\left|V_{S B}\right| \\ \therefore \qquad \qquad V_{T}&=\phi_{m s}+\frac{\sqrt{2 \epsilon_{s i} q N_{A}\left(2 \phi_{B}+V_{S B}\right)}}{C_{o x}}+2 \phi_{B}\\ \Rightarrow \qquad \qquad V_{T}&=\phi_{m s}+\frac{\sqrt{2 \epsilon_{s i} q N_{A}\left(2 \phi_{B}+\left|V_{S B}\right|\right)}}{C_{O x}}+2 \phi_{B}\\ \therefore \qquad\qquad \frac{\partial V_{T}}{\partial\left|V_{B S}\right|}&=0+\frac{\sqrt{2 \epsilon_{s i} q N_{A}}}{C_{O x}} \cdot \frac{1}{2 \sqrt{2 \phi_{B}+\left|V_{S B}\right|}}+0\\ \Rightarrow \qquad \qquad 50 \times 10^{-3}&=\frac{\sqrt{2 \times 8.85 \times 10^{-14} \times 12 \times 1.6 \times 10^{-19} \mathrm{~N}_{A}}}{\epsilon_{\alpha x} / t_{0 x}} \cdot \frac{1}{2 \sqrt{\left|V_{S B}\right|}}\\ \left(\frac{50 \times 10^{-3} \times 4 \times 8.85 \times 10^{-14}}{10 \times 10^{-7}}\right)^{2}&=\frac{2 \times 8.85 \times 10^{-14} \times 12 \times 1.6 \times 10^{-19}}{4 \times 2}\\ &\left[\because\left|V_{S B}\right| \gt \gt 2 \phi_{B}\right]\\ \Rightarrow \qquad \qquad N_{A}&=7.375 \times 10^{15} \mathrm{~cm}^{-3} \end{aligned}
 Question 9
In the circuit shown in the figure, the transistors $M_{1}$ and $M_{2}$ are operating in saturation. The channel length modulation coefficients of both the transistors are non-zero. The transconductance of the $\text{MOSFETs} M_{1}$ and $M_{2}$ are $g_{m1}$ and $g_{m2}$ , respectively, and the internal resistance of the $\text{MOSFETs} M_{1}$ and $M_{2}$ are $r_{01}$ and $r_{02}$ , respectively.

Ignoring the body effect, the ac small signal voltage gain $\left ( \partial V_{out}/\partial V_{in} \right )$ of the circuit is
 A $-g_{m2}\left ( r_{01}\left | \right |r_{02}\right )$ B $-g_{m2}\left ( \frac{1}{g_{m1}}\left | \right |r_{02} \right )$ C $-g_{m1}\left ( \frac{1}{g_{m2}}\left | \right |r_{01}\left | \right |r_{02} \right )$ D $-g_{m2}\left ( \frac{1}{g_{m1}}\left | \right |r_{01}\left | \right |r_{02} \right )$
GATE EC 2021      BJT and FET Basics
Question 9 Explanation:
MOSFET $M_2$ acts as common source amplifier.

Drain to gate connected MOSFET $M_1$ acts as load.

For given circuit, AC equivalent is as shown.

Replace $M_2$ with small signal model

\begin{aligned} \frac{V_{\text {out }}}{V_{\text {in }}} &=\frac{-g_{m 2} V_{g s}\left(r_{\infty} \| R_{\text {eq }}\right)}{V_{g s}} \\ A_{V} &=-g_{m 2}\left( \frac{1}{g_{m 1}}|| r_{o1} || r_{o 2} \right) \end{aligned}
 Question 10
The energy band diagram of a p-type semiconductor bar of length L under equilibrium condition (i.e.. the Fermi energy level $E_{F}$ is constant) is shown in the figure. The valance band $E_{V}$ is sloped since doping is non-uniform along the bar. The difference between the energy levels of the valence band at the two edges of the bar is $\Delta$.

If the charge of an electron is q, then the magnitude of the electric field developed inside this semiconductor bar is
 A $\frac{\Delta }{qL}$ B $\frac{2\Delta }{qL}$ C $\frac{\Delta }{2qL}$ D $\frac{3\Delta }{2qL}$
GATE EC 2021      Basic Semiconductor Physics
Question 10 Explanation:
The built-in electric field is due to non-uniform doping (the semiconductor is under equilibrium)

\begin{aligned} E &=\frac{1}{q}\frac{ d E_{v}}{d x} \\ &=\frac{1}{q} \frac{\Delta}{L} \\ &=\frac{\Delta}{q L} \end{aligned}

There are 10 questions to complete.