Q41GATE 2007MCQ

A $p^{+}n$ junction has a built-in potential of 0.8 V. The depletion layer width a reverse bias of 1.2 V is 2 $\mu m$ . For a reverse bias of 7.2 V, the depletion layer width will be

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📖 Explanation

Junction potential = Built-in potential + reverse bias voltage $V_{j}=V_{0}+V_{R}$ Now for abrupt pn-junction depletion width

\begin{aligned} W &\propto V_{j}^{1 / 2} \\ W&=K V_{j}^{1 / 2} \\ 2 \mu m&=K(0.8+1.2)^{1 / 2} &\ldots(i)\\ x&=K(0.8+7.2)^{1 / 2} &\ldots(ii)\\ \text{From (i) \& (ii)} x&=4 \mu \mathrm{m} \end{aligned}

Q42GATE 2007MCQ

Group I lists four different semiconductor devices. match each device in Group I with its charactecteristic property in Group II

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Q43GATE 2007MCQ

Group I lists four types of p-n junction diodes. Match each device in Group I with one of the option in Group II to indicate the bias condition of the device in its normal mode of operation.

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Q44GATE 2007MCQ

In a $p^{+}$ n junction diode under reverse biased the magnitude of electric field is maximum at

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Q45GATE 2006MCQ

In the circuit shown below, the switch was connected to position 1 at $t \lt 0$ and at t = 0, it is changed to position 2. Assume that the diode has zero voltage drop and a storage time $t_{s}$ . For $0 \lt t \leq t_{s}, v_{R}$ is given by (all in Volts)

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📖 Explanation

When diode is instantaneously switched from a conduction state it needs time to return to nonconduction state, so behaves short circuited for a small time period in reverse direction. Storage time: The time period for which diode remains in conduction state even in the reverse direction so, $v_{R} = -5V$ For $t \lt 0,$ For $0 \lt t \lt t_{s},$

Q46GATE 2006MCQ

Find the correct match between Group 1 and Group 2

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Q47GATE 2006MCQ

The values of voltage ( $V_{D}$ ) across a tunnel-diode corresponding to peak and valley currents are $V_{p}$ and $V_{v}$ respectively. The range of tunnel-diode voltage $V_{D}$ for which the slope of its $I-V_{D}$ characteristics is negative would be

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Q48GATE 2005MCQ

A Silicon PN junction diode under reverse bias has depletion region of width $10\mu$ m. The relative permittivity of Silicon, $\varepsilon _{r}$ = 11.7 and the permittivity of free space $\varepsilon _{0}=8.85 \times 10^{-12} F/m$ . The depletion capacitance of the diode per square meter is

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📖 Explanation

\begin{aligned} C&=\frac{\epsilon_{0} \epsilon_{r} A}{d} \\ \frac{C}{A}&=\frac{8.85 \times 10^{-12} \times 11.7}{10 \times 10^{-6}}=10.35 \mu F \end{aligned}

Q49GATE 2005MCQ

A Silicon PN junction at a temperature of $20^{\circ}$ C has a reverse saturation current of 10 pico-Ameres (pA). The reserve saturation current at $40^{\circ}$ C for the same bias is approximately

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📖 Explanation

we know that

\begin{aligned} I_{Q_{T_{2}}} &=I_{\left(Q_{T_{1}}\right)} \times 2^{\frac{\left(T_{2}-T_{1}\right)}{10}} \\ I_{q_{40^{\circ} C}} &=I_{q_{20^{\circ} C}} \times 2^{2} \\ I_{q_{40^{\circ} C}} &=40 \mathrm{pA} \end{aligned}

Q50GATE 2004MCQ

In an abrupt p-n junction, the doping concentrations on the p-side and n-side are $N_{A}=9\times 10^{16}/cm^{3}$ and $N_{D}=1\times 10^{16}/cm^{3}$ respectively. The p-n junction is reverse biased and the total depletion width is 3 $\mu$ m. The depletion width on the p-side is

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📖 Explanation

\begin{aligned} \frac{\omega_{n}}{\omega_{p}}&=\frac{N_{A}}{N_{D}} \qquad \omega \simeq \omega_{n} \quad\left(\because \omega_{n}\gt \gt \omega_{p}\right) \\ \omega_{p}&=\frac{\omega_{n} \times N_{D}}{N_{A}}=\frac{3 \mu m \times 10^{16}}{9 \times 10^{16}}=0.3 \mu \mathrm{m} \end{aligned}

Q51GATE 2004MCQ

Consider an abrupt p-n junction. Let $V_{bi}$ be the built-in potential of this junction and $V_R$ be the applied reverse bias. If the junction capacitance ( $C_{1}$ ) is 1 pF for $V_{bi}+V_{R}$ =1V, then for $V_{bi}+V_{R}$ =4V, $C_{j}$ will be

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📖 Explanation

\begin{aligned} C_{j} & \propto V_{i}^{-1 / 2} \\ \frac{C_{i 2}}{C_{i_{1}}} &=\sqrt{\frac{V_{1}}{V_{2}}}=\sqrt{\frac{1}{4}}=\frac{1}{2} \\ C_{i_{2}} &=\frac{C_{h}}{2}=\frac{1 \mathrm{pF}}{2}=0.5 \mathrm{pF} \end{aligned}

Q52GATE 2003MCQ

Choose proper substitutes for X and Y to make the following statement correct Tunnel diode and Avalanche photo diode are operated in X bias ad Y bias respectively

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Q53GATE 2003MCQ

Match items in Group 1 with items in Group 2, most suitably.

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Q54GATE 2003MCQ

A particular green LED emits light of wavelength 5490 $^{\circ}A$ . The energy bandgap of the semiconductor material used there is (Plank's constant = $6.626\times 10^{-34}J-s$ )

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📖 Explanation

\begin{aligned} E_{g} &=\frac{1.24}{\lambda(\mu \mathrm{m})} \mathrm{eV} \\ &=\frac{1.24}{5490 \times 10^{-4} \mu \mathrm{m}}=2.26 \mathrm{eV} \end{aligned}

Q55GATE 2003MCQ

At 300 K, for a diode current of 1 mA, a certain germanium diode requires a forward bias of 0.1435 V, whereas a certain silicon diode requires a forward bias of 0.718 V. Under the conditions state above, the closest approximation of the ratio of reverse saturation current in germanium diode to that in silicon diode is

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📖 Explanation

$\eta=1$ for Germanium $\eta=2$ for Silicon at low value of current

\begin{aligned} I&=I_{O S i}\left(e^{\frac{V_{D 1}}{\eta V_{T}}}-1\right) &\ldots(i)\\ I&=I_{O G e}\left(e^{\frac{V_{D 2}}{\eta V_{T}}}-1\right)&\ldots(ii) \end{aligned}

L.H.S. of (i) = (ii) $\Rightarrow I_{O_{S i}}\left(e^{\frac{V_{D 1}}{\eta V_{T}}}-1\right)=I_{O_{G e}}\left(e^{\frac{V_{D 2}}{\eta V_{T}}}-1\right)$ $\frac{I_{O_{Ge}}}{I_{O_{Si}}}=\frac{e^{\frac{V_{D_{1}}}{\eta V_{T}}}-1}{e^{\frac{V_{D_{2}}}{\eta V_{T}}}-1}=\frac{e^{\frac{0.718}{2 \times 26 \times 10^{-3}}}-1}{e^{\frac{0.1435}{26 \times 10^{-3}}}-1}=4 \times 10^{3}$

Q56GATE 2002MCQ

In Figure shown below, a silicon is carrying a constant current of 1 mA. When the temperature of the diode is $20^{\circ}$ C, $V_{D}$ is found to be 700 mV. If the temperature rises to $40^{\circ}$ C, $V_{D}$ becomes approximately equal to

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📖 Explanation

For either Si or Ge $\frac{d V}{d T} \simeq-2 \mathrm{mV} /^{\circ} \mathrm{C}$ in order to maintain a constant value of I.

\begin{aligned} T_{2}-T_{1}&=40-20=20^{\circ} \mathrm{C} \\ -2 \times 20 \mathrm{mV}&=40 \mathrm{mV} \\ \text { Therefore, } \quad V_{D}&=700-40=660 \mathrm{mV} \end{aligned}