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).
4.8 | |
5.6 | |
8.2 | |
9.7 |
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.
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.
0.3 | |
0.5 | |
1.2 | |
1.5 |
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}
\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
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} |
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}
\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
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 ) |
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}
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 |
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.