# PN-Junction Diodes and Special Diodes

 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).

 A 0.44 B 0.59 C 0.77 D 0.83
GATE EC 2022   Electronic Devices
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 $|J_n|_{diff}=qD_n\frac{d\eta }{dx}=qD_n \times 2 \times 10^{20} \times e^{-1} \times \tau _n\left [ 0-\frac{1}{2l_n} \right ]$
$=\frac{1.6 \times 10^{-19} \times l_n ^2 \times 2 \times 10^{20} \times e^{-1}}{2l_n}$
$=1.6 \times 10^{-19} \times l_n \times 10^{20} \times e^{-1}$
$=1.6 \times 10 \times 1 \times 10^{-4} \times e^{-1}$
$=0.59mA/cm^2$
 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.
 A 2.4 B 8.2 C 6.4 D 9.7
GATE EC 2021   Electronic Devices
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
 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 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 ____$\times 10^{20} F^{-2} V^{-1}$.
 A -5.7 B -3.8 C -1.2 D -0.4
GATE EC 2020   Electronic Devices
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   Electronic Devices
 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.
[$k=1.38 \times 10^{-23}JK^{-1}$, $h=6.625 \times 10^{-34}J-s$, $q=1.602 \times 10^{-19}C$]
 A 35.83 B 22.48 C 44.86 D 56.32
GATE EC 2019   Electronic Devices
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 ______ $\mu m$.
 A 0.124 B 0.231 C 0.426 D 0.369
GATE EC 2019   Electronic Devices
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}
 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
 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 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}
 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 A B B C C D D
GATE EC 2019   Electronic Devices
 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 _________________.
 A 0.5 B 0.12 C 0.2 D 0.63
GATE EC 2018   Electronic Devices
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}
There are 10 questions to complete.