BJT and FET Basics

Question 1
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   Electronic Devices
Question 1 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 2
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   Electronic Devices
Question 2 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 3
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   Electronic Devices
Question 3 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 4
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   Electronic Devices
Question 4 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 5
The base of an npn BJT T1 has a linear doping profile N_B(x) as shown below. The base of another npn BJT T2 has a uniform doping N_B of 10^{17} cm^{-3}. All other parameters are identical for both the devices. Assuming that the hole density profile is the same as that of doping, the common-emitter current gain of T2 is
approximately 2.0 times that of T1
approximately 0.3 times that of T1
approximately 2.5 times that of T1
approximately 0.7 times that of T1
GATE EC 2020   Electronic Devices
Question 5 Explanation: 

As per GATE official answer key MTA (Marks to ALL)
\frac{\beta _{1}}{\beta _{2}}=\frac{\int_{0}^{W}N_{A_{2}}(x)dx}{\int_{0}^{W}N_{A_{1}}(x)dx}=\frac{W\times 10^{17}}{\frac{1}{2}\times W\times (10^{17}-10^{14})}=\frac{2\times 10^{17}}{10^{17}+10^{14}}\simeq 2

There are 5 questions to complete.