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 |
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 2 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 3 |
A pn junction solar cell of area 1.0 cm^{2}, illuminated uniformly with 100 mW cm^{-2},
has the following parameters: Efficiency = 15%, open circuit voltage = 0.7 V, fill
factor = 0.8, and thickness = 200 \mu m, The charge of an electron is 1.6 \times 10^{-19}C. The
average optical generation rate (in cm^{-3} s^{-1}) is
0.84 \times 10^{19} | |
5.57 \times 10^{19} | |
1.04 \times 10^{19} | |
83.6 \times 10^{19} |
Question 3 Explanation:
\begin{aligned}\eta &=\frac{\left ( FF \right )V_{OC}I_{SC}}{P_{in}} \\ 0.15&=\frac{0.8\times 0.7\times I_{SC}}{100mW} \\ I_{SC}&=\frac{15}{0.56}mA \\ G_{L}&=\frac{I_{SC}}{q \times \text{ Area } \times \text{ thickness } } \\ &=\frac{15 \times 10^{-3}}{0.56 \times 1.6 \times 10^{-19} \times 1\times 200\times 10^{-4}} \\ &=\frac{15}{0.56\times 32}\times 10^{19}\\ &=0.837\times 10^{19} \end{aligned}
Question 4 |
A one-sided abrupt pn junction diode has a depletion capacitance CD of 50 pF at a reverse
bias of 0.2 V. The plot of 1/C_D^2 versus the applied voltage V for this diode is a straight
line as shown in the figure below. The slope of the plot is ____\times 10^{20} F^{-2} V^{-1}.
-5.7 | |
-3.8 | |
-1.2 | |
-0.4 |
Question 4 Explanation:
As per GATE official answer key MTA (Marks to ALL)
Depletion or transition capacitance is,
C_{D}=\frac{A_{\epsilon }}{W}
For one-sided PN junction \left (EX:P^{+}N Junction \right )
W=\sqrt{\frac{2\epsilon V_{B}}{eN_{D}}}=\sqrt{\frac{2\epsilon \left ( V_{bi}-V \right )}{eN_{D}}}
where V is anode to cathode applied potential.
\Rightarrow \, \, \, \, C_{D}=\frac{A\epsilon }{\sqrt{\frac{2\, \epsilon \left ( V_{bi}-V \right )}{eN_{D}}}}
\Rightarrow \, \, \, \, \frac{1}{C_{D}^{2}}=\frac{2}{A^{2}\epsilon eN_{D}}\left ( V_{bi}-V \right )
\frac{1}{C_{D}^{2}}\, becomes \, zero \, at V=V_{bi}
From above graph,y=\frac{1}{C_{D}^{2}}=0\, at\, x_{1}=V_{bi}
And \, \, \, y_{2}=\frac{1}{C_{D}^{2}}=4\times 10^{20}\, at\, x_{2}=-0.2V
Slope=\frac{y_{2}-y_{1}}{x_{2}-X_{1}}=\frac{4\times 10^{20}-0}{-0.2-V_{bi}}
V_{bi} is not provided,slope cannot be found.
Question 5 |
Consider the recombination process via bulk traps in a forward biased pn homojunction
diode. The maximum recombination rate is U_{max}. If the electron and the hole capture
cross-section are equal, which one of the following is False?
With all other parameters unchanged, U_{max} decreases if the intrinsic carrier density
is reduced. | |
U_{max} occurs at the edges of the depletion region in the device. | |
U_{max} depends exponentially on the applied bias. | |
With all other parameters unchanged,U_{max} increases if the thermal velocity of the
carriers increases. |
Question 6 |
In an ideal pn junction with an ideality factor of 1 at T=300 K, the magnitude of the reverse-bias voltage required to reach 75% of its reverse saturation current, rounded off to 2 decimal places, is ____mV.
[k=1.38 \times 10^{-23}JK^{-1}, h=6.625 \times 10^{-34}J-s, q=1.602 \times 10^{-19}C]
[k=1.38 \times 10^{-23}JK^{-1}, h=6.625 \times 10^{-34}J-s, q=1.602 \times 10^{-19}C]
35.83 | |
22.48 | |
44.86 | |
56.32 |
Question 6 Explanation:
\begin{aligned} V_{T} &=\frac{k T}{q}=\frac{1.38 \times 10^{-23} \times 300}{1.602 \times 10^{-19}} \mathrm{V} \\ &=25.843 \mathrm{mV} \\ I &=I_{0}\left(e^{V / V_{T}}-1\right)=-\frac{3}{4} I_{0} \\ \Rightarrow\quad V &=V_{T} \ln 0.25=-35.83 \mathrm{mV} \\ V_{R} &=|V|=35.83 \mathrm{mV} \end{aligned}
Question 7 |
A Germanium sample of dimensions 1cmx1cm is illuminated with a 20 mW, 600 nm laser light source as shown in the figure. The illuminated sample surface has a 100 nm of loss-less Silicon dioxide layer that reflects one-fourth of the incident light. From the remaining light, one-third of the power is reflected from the Silicon dioxide-Germanium interface, one-third is absorbed in the Germanium layer, and one-third is transmitted through the other side of the sample. If the absorption coefficient of Germanium at 600 nm is 3 \times 10^4 cm^{-1} and the bandgap is 0.66 eV, the thickness of the Germanium layer, rounded off to 3 decimal places, is ______ \mu m.
0.124 | |
0.231 | |
0.426 | |
0.369 |
Question 7 Explanation:
\begin{aligned} P_{\text {absorbed }} &=P_{\text {incident }}\left(1-e^{-\alpha T}\right) \\ \frac{1}{3} &=\frac{2}{3}\left(1-e^{-\alpha T}\right) \\ \frac{2}{3} e^{-\alpha T} &=\frac{1}{3} \end{aligned}
where \alpha=3 \times 10^{4} \mathrm{cm}^{-1}, absorption coefficient of Ge sample.
\begin{aligned} \therefore \quad T &=\frac{1}{\alpha} \ln (2)=\frac{1}{3 \times 10^{4}} \ln (2) \mathrm{cm} \\ &=0.231 \mu \mathrm{m} \end{aligned}
where \alpha=3 \times 10^{4} \mathrm{cm}^{-1}, absorption coefficient of Ge sample.
\begin{aligned} \therefore \quad T &=\frac{1}{\alpha} \ln (2)=\frac{1}{3 \times 10^{4}} \ln (2) \mathrm{cm} \\ &=0.231 \mu \mathrm{m} \end{aligned}
Question 8 |
The quantum efficiency (\eta) and responsivity (R) at a wavelength \lambda (in \; \mu m) in a p-i-n photodetector are related by
R=\frac{\eta \times \lambda }{1.24} | |
R=\frac{ \lambda }{\eta \times 1.24} | |
R=\frac{ \lambda \times 1.24}{ \eta} | |
R=\frac{1.24}{ \lambda \times \eta} |
Question 8 Explanation:
\begin{aligned} \eta&=\frac{I_{\text {out }}}{q} \times \frac{h f}{P_{\text {in }}} \\ R&=\frac{I_{\text {out }}}{P_{\text {in }}}\\ \text{So, }\quad R&=\eta \times \frac{q}{h f}=\eta \times \frac{q \lambda}{h c} \\ \end{aligned}
If \lambda is given in \mu \mathrm{m}, then
\begin{aligned} R &=\eta \lambda \times \frac{q \times 10^{-6}}{h c} \\ \frac{h c}{q \times 10^{-6}} & \simeq 1.24\\ \text{So, }\quad R&=\frac{\eta \lambda}{1.24} \end{aligned}
If \lambda is given in \mu \mathrm{m}, then
\begin{aligned} R &=\eta \lambda \times \frac{q \times 10^{-6}}{h c} \\ \frac{h c}{q \times 10^{-6}} & \simeq 1.24\\ \text{So, }\quad R&=\frac{\eta \lambda}{1.24} \end{aligned}
Question 9 |
Which one of the following options describes correctly the equilibrium band diagram at T=300 K of a Silicon pnp^+p^{++} configuration shown in the figure?
A | |
B | |
C | |
D |
Question 10 |
A junction is made between p^{-} Si with doping density N_{A1} = 10^{15} cm^{-3} and pSi with doping density N_{A2} = 10^{17} cm^{-3} .
Given: Boltzmann constant k = 1.38 \times 10^{-23} JK^{-1} , electronic charge q = 1.6 \times 10^{-19} C . Assume 100% acceptor ionization.
At room temperature (T = 300K), the magnitude of the built-in potential (in volts, correct to two decimal places) across this junction will be _________________.
Given: Boltzmann constant k = 1.38 \times 10^{-23} JK^{-1} , electronic charge q = 1.6 \times 10^{-19} C . Assume 100% acceptor ionization.
At room temperature (T = 300K), the magnitude of the built-in potential (in volts, correct to two decimal places) across this junction will be _________________.
0.5 | |
0.12 | |
0.2 | |
0.63 |
Question 10 Explanation:
Built-in potential,
\begin{aligned} V_{b i} &=\frac{k T}{q} \ln \left(\frac{N_{A 2}}{N_{A 1}}\right) \\ &=\frac{1.38 \times 3}{1.6 \times 100} \ln (100) \mathrm{V}=0.1192 \mathrm{V} \\ & \simeq 0.12 \mathrm{V} \end{aligned}
\begin{aligned} V_{b i} &=\frac{k T}{q} \ln \left(\frac{N_{A 2}}{N_{A 1}}\right) \\ &=\frac{1.38 \times 3}{1.6 \times 100} \ln (100) \mathrm{V}=0.1192 \mathrm{V} \\ & \simeq 0.12 \mathrm{V} \end{aligned}
There are 10 questions to complete.