# Electronic Devices

 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      PN-Junction Diodes and Special Diodes
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
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 A approximately 2.0 times that of T1 B approximately 0.3 times that of T1 C approximately 2.5 times that of T1 D approximately 0.7 times that of T1
GATE EC 2020      BJT and FET Basics
Question 2 Explanation:

As per GATE official answer key MTA (Marks to ALL)
$\frac{\beta _{1}}{\beta _{2}}=\frac{\int_{0}^{W}N_{A_{2}}(x)dx}{\int_{0}^{W}N_{A_{1}}(x)dx}=\frac{W\times 10^{17}}{\frac{1}{2}\times W\times (10^{17}-10^{14})}=\frac{2\times 10^{17}}{10^{17}+10^{14}}\simeq 2$
 Question 3
The band diagram of a p-type semiconductor with a band-gap of 1 eV is shown. Using this semiconductor, a MOS capacitor having $V_{TH}$ of $-0.16 V, C'_{ox}$ of 100 $nF/cm^2$ and a metal work function of 3.87 eV is fabricated. There is no charge within the oxide. If the voltage across the capacitor is $V_{TH}$. the magnitude of depletion charge per unit area (in $C/cm^2$) is A $1.7 \times 10^{-8}$ B $0.52 \times 10^{-8}$ C $1.41 \times 10^{-8}$ D $0.93 \times 10^{-8}$
GATE EC 2020      BJT and FET Basics
Question 3 Explanation:
MOS capacitance

\begin{aligned}\phi_{m}&=3.87,\phi _{B}=4.8,\phi _{ms}=-0.93\\V_{T}&=\phi _{ms}-\frac{{Q}'_{ox} }{{C}_{ox}}-\frac{{Q}'_{d}}{{C}_{ox}}+2\phi_{F_{p}}\\ \phi _{F_{p}}&=E_{i}-E_{F}=0.5-0.2=0.3\\ -0.16&=-0.93-0-\frac{{Q}'_{d}}{C_{ox}}+2\times 0.3 \\ \frac{{Q}'_{d}}{C_{ox}}&=0.6+0.16-0.93=-0.17\\ Q_{b}&=-0.17\times C_{ox}\\ &=-0.17\times 100\times 10^{-9} \\ & =-1.7\times 10^{-8}C/cm^{2}\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 ____$\times 10^{20} F^{-2} V^{-1}$. A -5.7 B -3.8 C -1.2 D -0.4
GATE EC 2020      PN-Junction Diodes and Special Diodes
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?
 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      PN-Junction Diodes and Special Diodes
 Question 6
A single crystal intrinsic semiconductor is at a temperature of 300 K with effective density of states for holes twice that of electrons. The thermal voltage is 26 mV. The intrinsic Fermi level is shifted from mid-bandgap energy level by
 A 18.02 meV B 9.01 meV C 13.45 meV D 26.90 meV
GATE EC 2020      Basic Semiconductor Physics
Question 6 Explanation:
$\frac{E_{c}+E_{v}}{2}-E_{F_{i}}=\frac{KT}{2}\ln \left ( \frac{N_{C}}{N_{V}}\right )\, \, \, \, \, \, \left ( \because N_{C}=\frac{N_{V}}{2} \right )$
$=\frac{0.026}{2}\ln 0.5=-9.01\, meV$
 Question 7
Consider a long-channel MOSFET with a channel length 1 $\mu m$ and width 10 $\mu m$. The device parameters are acceptor concentration $N_A=5 \times 10^{16}cm^{-3}$, electron mobility $\mu_n=800cm^2/V-s$, oxide capacitance/area $C_{ox}=3.45 \times 10^{-7}F/cm^{2}$, threshold voltage $V_T=0.7V$. The drain saturation current ($I_{Dsat}$) for a gate voltage of 5 V is _____mA(rounded off to two decimal places).
$[\varepsilon _0=8.854 \times 10^{-14}F/cm,\varepsilon _{Si}=11.9]$
 A 12.45 B 18.36 C 25.52 D 36.48
GATE EC 2019      BJT and FET Basics
Question 7 Explanation:
\begin{aligned} I_{D(\text { sat })} &=\frac{1}{2} \mu_{n} C_{\text {ox }} \frac{W}{L}\left(V_{G S}-V_{T}\right)^{2} \\ &=\frac{1}{2} \times 800 \times 3.45 \times 10^{-7} \times \frac{10}{1}(5-0.7)^{2} \mathrm{A} \\ &=25.5162 \mathrm{mA} \simeq 25.52 \mathrm{mA} \end{aligned}
 Question 8
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      PN-Junction Diodes and Special Diodes
Question 8 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 9
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      PN-Junction Diodes and Special Diodes
Question 9 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 10
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      PN-Junction Diodes and Special Diodes
Question 10 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}

There are 10 questions to complete.