Question 1 |
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 1 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 2 |
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 2 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 3 |
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 4 |
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 4 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 5 |
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 5 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 6 |
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 6 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 7 |
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 8 |
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 8 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}
Question 9 |
A solar cell of area 1.0 cm^{2} , operating at 1.0 sun intensity, has a short circuit current of
20 mA, and an open circuit voltage of 0.65 V. Assuming room temperature operation and
thermal equivalent voltage of 26mV, the open circuit voltage (in volts, correct to two
decimal places) at 0.2 sun intensity is _______.
0.6 | |
0.7 | |
0.5 | |
0.9 |
Question 9 Explanation:
For solar cell,
\begin{aligned} V_{\mathrm{OC}} &=V_{T} \ln \left(\frac{I_{S C}}{I_{O}}\right) \\ V_{\mathrm{OC} 2}-V_{\mathrm{OC} 1} &=V_{T} \ln \left(\frac{I_{S C 2}}{I_{S C}}\right)=V_{T} \ln \left(\frac{0.20}{1.0}\right) \\ V_{\mathrm{OC} 2} &=V_{\mathrm{OC} 1}-0.026 \mathrm{ln}(5) \\ &=0.65-0.041845=0.608 \mathrm{V} \end{aligned}
\begin{aligned} V_{\mathrm{OC}} &=V_{T} \ln \left(\frac{I_{S C}}{I_{O}}\right) \\ V_{\mathrm{OC} 2}-V_{\mathrm{OC} 1} &=V_{T} \ln \left(\frac{I_{S C 2}}{I_{S C}}\right)=V_{T} \ln \left(\frac{0.20}{1.0}\right) \\ V_{\mathrm{OC} 2} &=V_{\mathrm{OC} 1}-0.026 \mathrm{ln}(5) \\ &=0.65-0.041845=0.608 \mathrm{V} \end{aligned}
Question 10 |
Red (R), Green (G) and Blue (B) Light Emitting Diodes (LEDs) were fabricated using p-n
junctions of three different inorganic semiconductors having different band-gaps. The builtin
voltages of red, green and blue diodes are V_{R}, V_{G} \; and \; V_{B} , respectively. Assume donor and acceptor doping to be the same (N_{A} \; and \; N_{D} , respectively) in the p and n sides of all the three diodes. Which one of the following relationships about the built-in voltages is TRUE?
V_{R} \gt V_{G} \gt V_{B} | |
V_{R} \lt V_{G} \lt V_{B} | |
V_{R} = V_{G} =V_{B} | |
V_{R} \gt V_{G} \lt V_{B} |
Question 10 Explanation:
\lambda_{R} \gt \lambda_{G} \gt \lambda_{B}
Energy gap, E_{g} \propto \frac{1}{\lambda}
So, E_{g R} \lt E_{g G} \lt E_{g B}
Materials with high energy gap will have high built-in voltages, when doping concentrations are same.
So, V_{R} \lt V_{G} \lt V_{B}
Energy gap, E_{g} \propto \frac{1}{\lambda}
So, E_{g R} \lt E_{g G} \lt E_{g B}
Materials with high energy gap will have high built-in voltages, when doping concentrations are same.
So, V_{R} \lt V_{G} \lt V_{B}
There are 10 questions to complete.
