PN-Junction Diodes and Special Diodes

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
A
0.84 \times 10^{19}
B
5.57 \times 10^{19}
C
1.04 \times 10^{19}
D
83.6 \times 10^{19}
GATE EC 2020   Electronic Devices
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}.
A
-5.7
B
-3.8
C
-1.2
D
-0.4
GATE EC 2020   Electronic Devices
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?
A
With all other parameters unchanged, U_{max} decreases if the intrinsic carrier density is reduced.
B
U_{max} occurs at the edges of the depletion region in the device.
C
U_{max} depends exponentially on the applied bias.
D
With all other parameters unchanged,U_{max} increases if the thermal velocity of the carriers increases.
GATE EC 2020   Electronic Devices
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]
A
35.83
B
22.48
C
44.86
D
56.32
GATE EC 2019   Electronic Devices
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.
A
0.124
B
0.231
C
0.426
D
0.369
GATE EC 2019   Electronic Devices
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}
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
A
R=\frac{\eta \times \lambda }{1.24}
B
R=\frac{ \lambda }{\eta \times 1.24}
C
R=\frac{ \lambda \times 1.24}{ \eta}
D
R=\frac{1.24}{ \lambda \times \eta}
GATE EC 2019   Electronic Devices
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}
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
A
B
B
C
C
D
D
GATE EC 2019   Electronic Devices
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 _________________.
A
0.5
B
0.12
C
0.2
D
0.63
GATE EC 2018   Electronic Devices
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}
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 _______.
A
0.6
B
0.7
C
0.5
D
0.9
GATE EC 2018   Electronic Devices
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}
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?
A
V_{R} \gt V_{G} \gt V_{B}
B
V_{R} \lt V_{G} \lt V_{B}
C
V_{R} = V_{G} =V_{B}
D
V_{R} \gt V_{G} \lt V_{B}
GATE EC 2018   Electronic Devices
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}
There are 10 questions to complete.