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).

GATE EC 2022      PN-Junction Diodes and Special Diodes
Question 1 Explanation: 
\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}
Question 2
Select the CORRECT statement(s) regarding semiconductor devices.
Electrons and holes are of equal density in an intrinsic semiconductor at equilibrium.
Collector region is generally more heavily doped than Base region in a BJT.
Total current is spatially constant in a two terminal electronic device in dark under steady state condition.
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).

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 ______.
226 meV
174 meV
218 meV
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
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 mand 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 _____
0.37 \times 10^{14} cm^{-3}
0.63 \times 10^{13} cm^{-3}
3.7 \times 10^{14} cm^{-3}
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
For x \gt 0, G_n is also zero
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.
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.
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,
\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
7.37\times 10^{15}\:cm^{-3}
4.37\times 10^{15}\:cm^{-3}
2.37\times 10^{15}\:cm^{-3}
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
-g_{m2}\left ( r_{01}\left | \right |r_{02}\right )
-g_{m2}\left ( \frac{1}{g_{m1}}\left | \right |r_{02} \right )
-g_{m1}\left ( \frac{1}{g_{m2}}\left | \right |r_{01}\left | \right |r_{02} \right )
-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
\frac{\Delta }{qL}
\frac{2\Delta }{qL}
\frac{\Delta }{2qL}
\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.