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).
0.44 | |
0.59 | |
0.77 | |
0.83 |
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
\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.
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. |
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
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).
4.8 | |
5.6 | |
8.2 | |
9.7 |
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 |
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
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 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} |
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.
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 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}
\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.
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.
2.4 | |
8.2 | |
6.4 | |
9.7 |
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
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 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}
\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
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 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}
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
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} |
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}
\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.