GATE ECE · Electronic Devices

Pn Junction And Special Diodes – GATE ECE Electronic DevicesPractice Questions & PYQs

Generate GATE-level questions on PN Junction and Special Diodes. Focus on: 1. P-N junction characteristics: Diode current equation and Junction capacitance. 2. Special Diodes: Zener diode, Tunnel diode, LED, Photo diode, and Solar cell. 3. Breakdown mechanisms: Zener and Avalanche breakdown.

56 questions · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2025NAT

An ideal p-n junction germanium diode has a reverse saturation current of $10 \mu \mathrm{~A}$ at 300 K . The voltage (in Volts, rounded off to two decimal places) to be applied across the junction to get a forward bias current of 100 mA at 300 K is __________ . (Consider the Boltzmann constant $\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$ and the charge of an electron $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ .)

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📖 Explanation

$I = I_s \left( e^{\frac{eV}{k_BT}} - 1 \right)$ Where: $I = 100\ \text{mA} = 0.1\ \text{A}$ $I_s = 10\ \mu\text{A} = 10 \times 10^{-6}\ \text{A}$ $T = 300\ \text{K}$ $e = 1.6 \times 10^{-19}\ \text{C}$ $k_B = 1.38 \times 10^{-23}\ \text{J/K}$ Approximate since $I \gg I_s$ , so: $I \approx I_s e^{\frac{eV}{k_BT}}$ Taking natural logarithm on both sides: $\frac{eV}{k_BT} = \ln\left(\frac{I}{I_s}\right)$ $V = \frac{k_BT}{e} \ln\left(\frac{I}{I_s}\right)$ Now plug in the values: $\frac{k_BT}{e} = \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} = 0.025875\ \text{V}$ $\ln\left(\frac{0.1}{10 \times 10^{-6}}\right) = \ln(10000) = \ln(10^4) = 4 \ln(10) \approx 4 \times 2.3026 = 9.2104$ So, $V = 0.025875 \times 9.2104 \approx 0.2382\ \text{V}$

Q2GATE 2024NAT

The photocurrent of a PN junction diode solar cell is $1 \mathrm{~mA}$ . The voltage corresponding to its maximum power point is $0.3 \mathrm{~V}$ . If the thermal voltage is $30 \mathrm{mV}$ , the reverse saturation current of the diode (in $\mathrm{nA}$ , rounded off to two decimal places) is ________

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📖 Explanation

Note: In this question voltage corresponds to max. power, $V_{m}$ is given i.e., $\quad V_{m}=0.3$ Volt ( $V_{O C}$ is not given be careful)

\begin{aligned} I_{0} & =\frac{I_{L}}{\left(1+\frac{V_{m}}{V_{T}}\right) e^{V_{m} / V_{T}}-1}=\frac{1 \mathrm{~mA}}{\left(1+\frac{0.3}{0.03}\right) e^{0.3 / 0.03}-1} \\ & =4.127 \mathrm{nA} \simeq 4.13 \mathrm{nA} \end{aligned}

Q3GATE 2022NAT

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).

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📖 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 $|J_n|_{diff}=qD_n\frac{d\eta }{dx}=qD_n \times 2 \times 10^{20} \times e^{-1} \times \tau _n\left[ 0-\frac{1}{2l_n} \right]$ $=\frac{1.6 \times 10^{-19} \times l_n ^2 \times 2 \times 10^{20} \times e^{-1}}{2l_n}$ $=1.6 \times 10^{-19} \times l_n \times 10^{20} \times e^{-1}$ $=1.6 \times 10 \times 1 \times 10^{-4} \times e^{-1}$ $=0.59mA/cm^2$

Q4GATE 2021NAT

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.

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📖 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}

Explanation Figure 2

Q5GATE 2020MCQ

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?

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Q6GATE 2020MCQ

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

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📖 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}

Q7GATE 2020MCQ

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 ____ $\times 10^{20} F^{-2} V^{-1}$ .

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📖 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.

Explanation Figure 2

Q8GATE 2019NAT

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. $[ $k=1.38 × 10^{-23}JK^{-1}$ , $h=6.625 × 10^{-34}J-s$ , $q=1.602 × 10^{-19}C$ ]$

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📖 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}

Q9GATE 2019MCQ

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?

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Q10GATE 2019NAT

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 ______ $\mu m$ .

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📖 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}

Q11GATE 2019MCQ

The quantum efficiency ( $\eta$ ) and responsivity (R) at a wavelength $\lambda (\in \; \mu m)$ in a p-i-n photodetector are related by

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📖 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}

Q12GATE 2018NAT

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 _________________.

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📖 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}

Q13GATE 2018MCQ

In a p-n junction diode at equilibrium, which one of the following statements is NOT TRUE?

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📖 Explanation

$\longrightarrow$ Hole diffusion $\longleftarrow$ Electron diffusion $\longleftarrow$ Hole drift $\longrightarrow$ Electron drift $\longrightarrow$ Hole diffusion current $\longrightarrow$ Electron diffusion current $\longleftarrow$ Hole drift current $\longleftarrow$ Electron drift current

Explanation Figure 1

Q14GATE 2018NAT

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 _______.

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📖 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}

Q15GATE 2018NAT

A p-n step junction diode with a contact potential of 0.65V has a depletion width of 1 $\mu$ m at equilibrium. The forward voltage (in volts, correct to two decimal places) at which this width reduces to 0.6 $\mu$ m is _______.

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📖 Explanation

\begin{aligned} W_{\mathrm{dep}} &=\sqrt{\frac{2 \varepsilon}{q}\left(\frac{1}{N_{A}}+\frac{1}{N_{D}}\right)\left(V_{b i}-V_{A K}\right)} \\ \frac{\sqrt{\left(0.65-V_{A K}\right)}}{\sqrt{0.65}} &=\frac{0.6 \mu \mathrm{m}}{1 \mu \mathrm{m}}=0.6 \\ 1-\frac{V_{A K}}{0.65} &=0.36 \\ V_{A K} &=0.65(1-0.36) \\ &=0.65 \times 0.64=0.416 \mathrm{V} \simeq 0.42 \mathrm{V} \end{aligned}

Q16GATE 2018MCQ

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?

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📖 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, $V_{R} \lt V_{G} \lt V_{B}$

Q17GATE 2017NAT

As shown, two Silicon (Si) abrupt p-n junction diodes are fabricated with uniform donor doping concentration of $N_{D1}=10^{14}cm^{-3}$ and $N_{D2}=10^{16}cm^{-3}$ in the n-regions of the diodes, and uniform acceptor doping concentration of $N_{A1}=10^{14}cm^{-3}$ and $N_{A2}=10^{16}cm^{-3}$ in the pregions of the diodes, respectively. Assuming that the reverse bias voltage is $\gt \gt$ built-in potentials of the diodes, the ratio $\frac{C_{2}}{C_{1}}$ of their reverse bias capacitances for the same applied reverse bias, is __________.

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📖 Explanation

\begin{aligned} W&=\sqrt{\frac{2 \varepsilon V_{R}}{e}\left(\frac{1}{N_{A}}+\frac{1}{N_{D}}\right)} \\ \because \; &V_{R}\gt \gt V_{bi}\\ \text{Under reverse } &\text{biased condition,}\\ C&=\frac{\varepsilon A}{W} \quad \Rightarrow C \propto \frac{1}{W}\\ \text{So}\quad\frac{C_{2}}{C_{1}} &=\frac{W_{1}}{W_{2}} \\ &=\sqrt{\frac{\frac{2 \varepsilon V_{R}}{e}\left(\frac{2}{10^{14}}\right)}{\frac{2 \varepsilon V_{R}}{e}\left(\frac{2}{10^{16}}\right)}}=10 \end{aligned}

Q18GATE 2017MCQ

An $n^{+} -n$ Silicon device is fabricated with uniform and non-degenerate donor doping concentrations of $N_{D1}=1 \times 10^{18}cm^{-3}$ and $N_{D2}=1 \times 10^{15}cm^{-3}$ corresponding to the $n^{+}$ and n regions respectively. At the operational temperature T, assume complete impurity ionization, kT/q = 25 mV, and intrinsic carrier concentration to be $n_{i}=1 \times 10^{10}cm^{-3}$ . What is the magnitude of the built-in potential of this device?

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📖 Explanation

\begin{aligned} V_{0} &=V_{T} \ln \frac{n_{1}}{n_{2}}=V_{T} \ln \frac{N_{D 1}}{N_{D 2}} \\ &=0.25 \ln \frac{10^{18}}{10^{15}}=0.025 \ln (1000) \\ &=0.173 \mathrm{V} \end{aligned}

Q19GATE 2017NAT

For a particular intensity of incident light on a silicon pn junction solar cell, the photocurrent density ( $J_{L}$ ) is 2.5 mA/ $cm^{2}$ and the open-circuit voltage ( $V_{oc}$ ) is 0.451 V. consider thermal voltage ( $V_{T}$ ) to be 25mV. If the intensity of the incident light is increased by 20 times,assuming that the temperature remains unchanged. $V_{oc}$ (in volts) will be ______.

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📖 Explanation

\begin{aligned} V_{\mathrm{OC}} &=V_{T} \ln \frac{J_{L}}{J_{S}} \quad J_{L} \propto \text { Light intensity } \\ V_{\mathrm{OC} 2}-V_{\mathrm{OC} 1} &=V_{T} \ln \frac{J_{L 2}}{J_{L 1}} \\ V_{\mathrm{OC} 2}-V_{\mathrm{OC} 1} &=25 \ln 20 \cong 75 \mathrm{mV}=0.075 \mathrm{V} \\ V_{\mathrm{OC} 2} &=0.451 \mathrm{V}+0.075 \mathrm{V}=0.526 \mathrm{V} \end{aligned}

Q20GATE 2017MCQ

An abrupt pn junction (located at x = 0) is uniformly doped on both p and n sides. The width of the depletion region is W and the electric field variation in the x-direction is E(x). Which of the following figures represents the electric field profile near the pn junction?

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