Past NEET Entrance Papers

Hint

(d) A : Solar cell characteristics

B : In a reverse-bias \(p-n\) junction diode, the current measured in \((\mu \mathrm{A})\), is due to minority charge carrier.

Hint

(b) \(Y=\bar{A} \cdot \bar{B}+A \bar{B}=\bar{B} (\bar{A} + A)=\bar{B}\)

Hint

(c) AND gate

\(
\begin{aligned}
Y & =\overline{\bar{A}+\bar{B}} \\
& =\bar{\bar{A}} \cdot \bar{\bar{B}}=A \cdot B
\end{aligned}
\)

Hint

(d) To Balance Bridge
\(\frac{P}{Q}=\frac{R}{S}\)
Here
\(
\begin{gathered}
P=10 \Omega \\
Q=10 \Omega \\
R=15 \Omega \\
S=5+R_{\text {diode }}
\end{gathered}
\)
The diode can conduct and have resistance \(R_D=10 \Omega\) because diode have dynamic resistance. In that case bridge will be balanced.

Hint

(c) A solar cell, also known as a photovoltaic cell, is a type of diode that converts light into electricity using the photovoltaic effect. It’s different from a forward-biased Zener diode (which is a voltage regulator) or a light-emitting diode (which converts electricity into light), and it doesn’t require external biasing like some photodiodes.

Hint

(b) For \(\mathrm{A}=1, \mathrm{~B}=1\) : Both AND gates output 1 , so \(\mathrm{Y}=1\).
For \(\mathrm{A}=0, \mathrm{~B}=1\) : The first AND gate outputs 0 , but the second AND gate outputs 1 , so \(\mathrm{Y}=1\).
For \(\mathrm{A}=1, \mathrm{~B}=0\) : The first AND gate outputs 1 , but the second AND gate outputs 0 , so \(\mathrm{Y}=1\).
For \(\mathrm{A}=0, \mathrm{~B}=0\) : Both AND gates output 0 , so \(\mathrm{Y}=0\).

Hint

(d) A Zener diode works as a voltage regulator in reverse bias and behaves as a simple pn junction diode in forward bias. In other words, a Zener diode is primarily designed to be used in reverse bias, where it regulates voltage by maintaining a constant voltage across it once the reverse voltage reaches its breakdown voltage. However, in forward bias, it behaves like a regular diode, allowing current to flow in one direction.

Hint

(d) The I-V characteristics of solar cell is

Hint

(b) When the output of an OR gate is applied as input to a NOT gate, the combination acts as a NOR gate.

Hint

(c) \(Y=\overline{\bar{A} \bar{B}}=A+B\) (OR Gate)

Hint

(d) The component that removes the AC ripple from the rectified output in a full wave rectifier circuit is the capacitor.

Explanation: A capacitor acts as a filter, storing energy during the peaks of the rectified voltage and releasing it during the troughs, thus smoothing out the voltage fluctuations and reducing the ripple.

Hint

(d) Statement I: Photocell/solar cell convert light energy into electric energy/current.
Statement II : We use Zener diode in reverse-bias condition, when the reverse-bias voltage is more than breakdown voltage, it act as stablizer.

Hint

(c)

\(
\begin{aligned}
& Y=\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{~B}}}=\overline{\overline{\mathrm{A}}}+\overline{\overline{\mathrm{B}}} \\
& =(\mathrm{A}+\mathrm{B}) \text { OR Gate }
\end{aligned}
\)

Hint

(d) \(Y= \overline{\bar{A}+\bar{B}}=A \cdot B\) which is an AND gate.

Hint

(d) A p-type extrinsic semiconductor is obtained when Germanium is doped with boron (a trivalent impurity).

Hint

(b) Electrical conductivity is a measure of a material’s ability to conduct an electric current. The higher the conductivity, the lower the resistivity. So, if we are looking for the material with the smallest resistivity, we are looking for the material with the highest electrical conductivity.

Among the options given, silver (Option b) is known to have the highest electrical conductivity and thus, the smallest resistivity. Germanium (Option a) and silicon (Option d) are semiconductors, meaning they have moderate conductivity that can be manipulated, while glass (Option c) is generally a poor conductor, or an insulator. Therefore, silver is the correct answer.

Hint

(d) In half-wave rectification, if the input frequency is 60 Hz , the output frequency will also be 60 Hz . A half-wave rectifier only allows one half-cycle of the AC input to pass, so the output frequency remains the same as the input frequency.

Hint

(d) The resistance increases for conductors but decreases for semiconductors as temperature increases. This is because the increase in temperature causes more collisions between electrons and atoms in conductors, increasing resistance. In semiconductors, more electrons gain energy and become available to carry current, reducing resistance.

Elaboration:
Conductors:
In conductors, increased temperature leads to more atomic vibrations, which hinders the flow of electrons, increasing resistance.

Semiconductors:
In semiconductors, higher temperatures provide more electrons with enough energy to move to the conduction band, increasing conductivity and reducing resistance. 

Hint

(a) Potential drops across the p-n junctions will be same if either both junctions are forward biased or both junction are reverse biased.

In figure (a) and (c), both junctions are forward biased therefore both have same potential.

In figure (b) first junction is forward biased and the second junction is reverse biased, so both junctions have different potential difference.

Hint

(d)

\(
\begin{aligned}
& C=(\overline{A \cdot B}) \cdot(\overline{\bar{A} \cdot B}) \\
& \Rightarrow C=\overline{A \cdot B+\bar{A} \cdot B} \\
& \Rightarrow C=\overline{(A+\bar{A}) B} \\
& \Rightarrow C=\bar{B}
\end{aligned}
\)

Hint

(a) \(\begin{aligned} y & =\overline{\overline{A+B}} \\ y & =A+B\end{aligned}\) which is an OR gate.

Hint

(a) When Switch S is open, basically the circuit behaves as a full-wave rectifier.

When S is closed, D5 turns on during the negative half of the input signal and the output voltage across load \(R_L\) = 0 V.  During the positive half cycle, D5 is OFF and positive half cycle will appear across the load \(R_L\). Hence option a is correct.

Hint

(c) Zener diode is special-purpose p-n junction diode, which is generally operated in reverse bias for its operation of voltage regulation. It is a heavily doped p-n junction. Due to the large doping concentration depletion width is narrower.

Hint

(a) The given circuit clearly indicates that the bulb will glow when either of the switches are closed.
The corresponding truth table will be
\(
\begin{array}{|c|c|c|}
\hline \text { Input A } & \text { Input B } & \text { Output } \mathrm{Y} \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 1 \\
\hline
\end{array}
\)
This suggests that it will correspond to OR gate.

Hint

(a)

\(
\begin{aligned}
&\text { The current through a semiconductor is }\\
&\begin{aligned}
& \mathrm{I}=\mathrm{neAv}_{\mathrm{d}} \\
& \mathrm{I}=\mathrm{neA} \mu \mathrm{E} \quad\left(\text { as } \mathrm{v}_{\mathrm{d}}=\mu \mathrm{E}\right) \\
& \frac{I_n}{I_p}=\frac{n_e e A \mu_e E}{n_h e A \mu_h E} \\
& \frac{I_n}{I_p}=\frac{\mu_e}{\mu_h} \\
& \because \mu_{\mathrm{e}}>\mu_{\mathrm{h}} \\
& \Rightarrow I_n>I_p
\end{aligned}
\end{aligned}
\)

Hint

(a) Statement A: A Zener diode is always connected in reverse bias to use it as a voltage regulator.
Explanation: A Zener diode is specifically designed to operate in reverse bias. When connected in reverse bias, it maintains a constant output voltage (known as the Zener voltage) across its terminals, even when the input voltage varies. This property makes it ideal for use in voltage regulation applications. Therefore, this statement is True.

Statement B: The potential barrier of a p-n junction lies between 0.1 to 0.3 V, approximately.
Explanation: The potential barrier of a p-n junction is typically in the range of 0.6 to 0.7 volts for silicon diodes, and can be higher for other semiconductor materials. The range of 0.1 to 0.3 volts is not accurate for \(p-n\) junctions. Therefore, this statement is False.

Hint

(c)

\(
\begin{aligned}
& Y=A B+\overline{B C} \\
& =A B+\bar{B}+\bar{C}=B+\bar{B}+A+\bar{B}+\bar{C}=1+A+\bar{B}+\bar{C}=1
\end{aligned}
\)

Hint

(c) \(Y=\overline{\bar{A}+\bar{B}}=\overline{\overline{A . B}}=\text { A.B }\)

Hint

(a) Reverse bias only.
Explanation:
In a p-n junction diode, reverse bias causes the majority charge carriers (holes in the p-type region and electrons in the n-type region) to be pulled away from the junction, resulting in an increase in the width of the depletion region.
Forward bias, on the other hand, decreases the width of the depletion region by allowing charge carriers to flow across the junction.

Hint

(c) The answer is insulators and semiconductors. Metals are excellent conductors, while insulators do not allow electrons to flow easily, and semiconductors have conductivity between the two. For metals temperature coefficient of resistance is positive while for insulators and semiconductors, temperature coefficient of resistance is negative

Hint

(d) For forward biasing potential of \(p\) type is greater than potential of \(n\) type junction.

Hint

(c) A Universal Gate is a gate by which every other gate can be realized.
AND, OR, NOT, etc. are basic gates.
NAND, NOR, etc. are the universal gate.
That means we can implement any logic function using NAND and NOR gates without the need of AND, OR, or NOT gates.

Hint

(a) To create N-type semiconductor, intrinsic semiconductor is doped by pentavalent impurity (Phosphorus)

Hint

(c) Holes are the majority carriers and trivalent atoms are the dopants.

Hint

(d)

\(
\begin{array}{|l|l|l|}
\hline A & B & Y \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}
\)

This is the truth table of an NAND gate.

Hint

(b) Planck’s constant: \(h=6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\)
Speed of light: \(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Conversion factor: \(1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\)
Relationship between energy and wavelength: \(E=h \frac{c}{\lambda}\)
\(
\begin{aligned}
& E_g=1.9 \mathrm{eV} \times 1.602 \times 10^{-19} \mathrm{~J} / \mathrm{eV} \\
& E_g=3.0438 \times 10^{-19} \mathrm{~J}
\end{aligned}
\)
\(
\begin{aligned}
& \lambda=\frac{h c}{E_g} \\
& \lambda=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{3.0438 \times 10^{-19} \mathrm{~J}} \\
& \lambda=6.539 \times 10^{-7} \mathrm{~m} \\
& \lambda \approx 654 \times 10^{-9} \mathrm{~m} \\
& \lambda=654 \mathrm{~nm}
\end{aligned}
\)
The wavelength of the emitted light is 654 nm.

Hint

(A)

\(
\begin{array}{|c|c|c|}
\hline \text { A } & {B} & f{Y} \\
\hline 0 & 0 & 1 \\
\hline 1 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}
\)

This is truth table of a NOR gate.

Hint

(b) \(Y=(A \cdot \bar{B}+\bar{A} \cdot B)\)

Hint

(d) The correct answer is affects the overall V-I characteristics of a p-n junction.

Explanation:
When a p-n junction diode is heated, the number of free charge carriers (electron-hole pairs) increases. This increase in charge carriers affects both the forward and reverse bias characteristics of the diode. The change in temperature influences the resistance of the diode, leading to a shift in the V-I (voltage-current) curve.

Hint

(a) In forward bias, the p-type semiconductor is at a higher potential w.r.t. n -n-type semiconductor.

Hint

(b) \(Y=\overline{A+B}\)

Hint

(d) In forward bias, the p-type semiconductor is at a higher potential w.r.t. n -n-type semiconductor.

Hint

(a) Current will not flow through \(D_1\) as it is reverse biased. Current will flow through cell, \(R_1, D_2\) and \(R_3\).
\(
\therefore i=\frac{10}{2+2}=2.5 A
\)

Hint

(c) \(Y=(A+B) C=1\)

Hint

(c) 

\(
Y=\overline{(A B) C}
\)
\(
\begin{aligned}
& \text { for } A=B=C=0, y=1 \\
& \text { for } A=B=C=1, y=0
\end{aligned}
\)

Hint

(a) For an ideal diode, potential drop across diode is zero. Since the current flows from higher potential to lower potential, the direction of a current will be from \(A\) to \(B\).
Current in the diode, \(I=\frac{\Delta V}{R}\)
\(
\Rightarrow I=\frac{V_A-V_B}{R}=\frac{4-(-6)}{1000}=\frac{10}{1000}=10^{-2} \mathrm{~A}
\)

Hint

(d) The potential difference across the resistance \(R\) is
\(
V=3.5 \mathrm{~V}-0.5 \mathrm{~V}=3 \mathrm{~V}
\)
By Ohm’s law,
The current in the circuit is
\(
I=\frac{V}{R}=\frac{3 \mathrm{~V}}{100 \Omega}=3 \times 10^{-2} \mathrm{~A}=30 \times 10^{-3} \mathrm{~A}=30 \mathrm{~mA}
\)

Hint

(d) Here P-N junction diode rectifies half of the ac wave, i.e., acts as half half-wave rectifier. During the positive half cycle
Diode \(\rightarrow\) forward biased output across will be

During – ve half cycle Diode \(\rightarrow\) reverse biased output will be 0V as (current across \(R_L\)=0 and hence no output voltage).

Hint

(c)

\(
\begin{aligned}
&Y=\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}=A \cdot B\\
&\text { Thus, the combination represents AND gate. }
\end{aligned}
\)

Hint

(a) The \(V-I\) characteristic for a solar cell is as shown the figure.

Hint

(d) The barrier potential depends on type of semiconductor (For \(\mathrm{Si}, V_b=0.7 \mathrm{~V}\) and for Ge, \(V_b=0.3 \mathrm{~V}\) ), amount of doping and also on the temperature.

Hint

(b) In an n-type semiconductor, electrons are the majority charge carriers and holes are the minority charge carriers, and pentavalent atoms are dopants.

Hint

(b) The output of the given logic circuit is
\(
X=\overline{\overline{A . B}}=A . B
\)

Hint

(a) The output of the given logic gate is
\(
C=\overline{\overline{A B}}=A \cdot B
\)
Hence, the resulting gate is a AND gate.

Hint

(b) \(\text { The higher hole concentration is in } p \text {-region than that in } n \text {-region. }\)

Hint

(d) In the given circuit the upper diode \(D_1\) is forward biased and the lower diode \(D_2\) is reverse biased. So, the current supplied by the battery is
\(
I=\frac{5 \mathrm{~V}}{10 \Omega}=\frac{1}{2} \mathrm{~A}=0.5 \mathrm{~A}
\)

Hint

(c)

\(
\begin{aligned}
&\text { The electronic configuration of Carbon (C) and Silicon (Si) is given by }\\
&\begin{aligned}
& { }_6 \mathrm{C}^{12}=2,4 (1 s^2 2 s^2 2 p^2) \\
& { }_{14} \mathrm{Si}^{28}=2,8,4 (1 s^2 2 s^2 2 p^6 3 s^2 3 p^2)
\end{aligned}
\end{aligned}
\)
Hence, the four bonding electrons of C and Si respectively lie in second and third orbit.

Hint

(a) The truth table of the given waveform is as shown in the table.
\(
\begin{array}{|c|c|c|c|}
\hline \text { Time interval } & \text { Input } A & \text { Input } B & \text { Output } C \\
\hline 0 \text { to } t_1 & 0 & 0 & 0 \\
\hline t_1 \text { to } t_2 & 1 & 0 & 1 \\
\hline t_2 \text { to } t_3 & 1 & 1 & 1 \\
\hline t_3 \text { to } t_4 & 0 & 1 & 1 \\
\hline t_4 \text { to } t_5 & 0 & 0 & 0 \\
\hline t_5 \text { to } t_6 & 1 & 0 & 1 \\
\hline >t_6 & 0 & 1 & 1 \\
\hline
\end{array}
\)
The logic circuit is OR gate.

Hint

(a) The Boolean expression of the given circuit is
\(
Y=(A+B) \cdot C
\)
From the above truth table it is clear that \(Y=1\), when \(A=1, B=0\) and \(C=1\)

Hint

(c) (ii)-AND, (iii)- NOT and (iv)-NAND

Hint

(b) When a small amount of antimony (pentavalent) is added to germanium (tetravalent) crystal: then crystal becomes \(n\)-type semiconductor. In \(n\)-type semiconductor electrons are the majority charge carriers and the holes are the minority charge carriers.

Hint

(b) In forward biasing, the positive terminal of the battery is connected to \(p\)-side and the negative terminal to \(n\)-side of \(p\) – \(n\) junction. The forward bias voltage opposes the potential barrier. Due to it, the depletion region becomes thin.

Hint

(a)

The voltage drop across \(1 \mathrm{k} \Omega=V_Z=15 \mathrm{~V}\)
The current through \(1 \mathrm{k} \Omega\) is
\(
I^{\prime}=\frac{15 \mathrm{~V}}{1 \times 10^3 \Omega}=15 \times 10^{-3} \mathrm{~A}=15 \mathrm{~mA}
\)
The voltage drop across \(250 \Omega=20 \mathrm{~V}-15 \mathrm{~V}=5 \mathrm{~V}\)
The current through \(250 \Omega\) is
\(
I=\frac{5 \mathrm{~V}}{250 \Omega}=0.02 \mathrm{~A}=20 \mathrm{~mA}
\)
The current through the zener diode is
\(
I_{\mathrm{z}}=I-I^{\prime}=(20-15) \mathrm{mA}=5 \mathrm{~mA}
\)

Hint

(c) p-n junction is said to be forward biased when \(p\) side is at a higher potential than \(n\) side. It is for circuit (a) and (c).

Hint

(a) \(p\)-type semiconductor is obtained when Si or Ge is doped with a trivalent impurity like aluminium ( Al ), boron ( B ), indium ( In ) etc, Here, \(n_i=1.5 \times 10^{16} \mathrm{~m}^{-3}, \quad n_h=4.5 \times 10^{22} \mathrm{~m}^{-3}\)
As we know \(n_e n_h=n_i^2\)
\(
n_e=\frac{n_i^2}{n_h}=\frac{\left(1.5 \times 10^{16} \mathrm{~m}^{-3}\right)^2}{4.5 \times 10^{22} \mathrm{~m}^{-3}}=5 \times 10^9 \mathrm{~m}^{-3}
\)

Hint

(b) In an n-type semiconductor, electrons are the majority carriers and holes are the minority carriers.

Hint

(c) It is clear from given logic circuit, that out put \(Y\) is low when both the inputs are high, otherwise it is high. Thus logic circuit is NAND gate.
\(
\begin{array}{|c|c|c|}
\hline A & B & Y \\
\hline 1 & 1 & 0 \\
\hline 0 & 0 & 1 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline
\end{array}
\)

Hint

(d) Band gap \(=2.5 \mathrm{eV}\)
The wavelength corresponding to 2.5 eV
\(\)
=\frac{12400 \mathrm{eV} Å}{2.5 \mathrm{eV}}=4960 Å.
\(\)
4000 Å can excite this.

The p-n photodiode can detect wavelengths shorter than 4960 Å . Therefore, it can detect signals in the visible spectrum and some ultraviolet light.

Hint

(c) (iv)-OR gate, (ii)-NOT gate, (i)-NAND gate

Hint

(b) Band gap \(=2 \mathrm{eV}\).
Wavelength of radiation corresponding to this energy,
\(
\lambda=\frac{h c}{E}=\frac{12400 \mathrm{eVA}}{2 \mathrm{eV}}=6200 Å
\)
The frequency of this radiation
\(
\begin{aligned}
= & \frac{c}{\lambda}=\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{6200 \times 10^{-10} \mathrm{~m} / \mathrm{s}} \\
\Rightarrow f & =5 \times 10^{14} \mathrm{~Hz} .
\end{aligned}
\)

Hint

(c) \(Y=\overline{\overline{\overline{A+B}}}=\overline{A+B}.\)

Hint

(d) \(Y=\overline{\overline{A+B}}=A+B .\)

Hint

(a) In this Band diagram, the number of holes in the valence band are more as compared to number of electrons in the conduction band. So, it is a p-type semiconductor.

Hint

(d) For forward bias p-side should be positive with respect to n-side. Option (d) satisfies this condition.

Hint

(a) The truth table corresponding to waveform is given by
\(
\begin{array}{|l|l|l|}
\hline A & B & C \\
\hline 1 & 1 & 1 \\
\hline 0 & 1 & 0 \\
\hline 1 & 0 & 0 \\
\hline 0 & 0 & 0 \\
\hline
\end{array}
\)
\(\therefore\) The given logic circuit gate is AND gate.

Hint

(b) When \(p\)-side of junction diode is connected to positive of battery and \(n\)-side to the negative, then junction diode is forward biased. In this condition, more number of electrons enter in \(n\)-side from battery, thereby increasing the number of donor on \(n\)-side.

Hint

(b) Band gap of carbon is 5.5 eV while that of silicon is 1.1 eV.
\(
\left(E_g\right)_{\mathrm{C}}>\left(E_g\right)_{\mathrm{S}_1}
\)

Hint

(a) In semiconductor, resistivity decreases with increase in temperature. With the increase in temperature, more and more electrons jump to the conduction band. So, conductivity increases and resistivity decreases.

Hint

(b) Zener diode is used for stabilisation while \(p-n\) junction diode is used for rectification.

Hint

(b) In a photocell, photoelectromotive force, is the force that stimulates the emission of an electric current when photovoltaic action creates a potential difference between two points, and the electric current depends on the intensity of incident light.

Hint

(a)

\(
\begin{array}{|l|l|l|}
\hline A & B & Y \\
\hline 0 & 0 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 1 \\
\hline
\end{array}
\)
From truth table we can observe that if either of input is one then output is one. Also if both the inputs are one then also output is one.

Hint

(a) Hence, we can say that the dc component of the output voltage is given by:
\(
V_{dc}=\frac{\int_0^\pi 10 \sin (\omega t)}{2 \pi}
\)
We have to put the values to get the following expression:
\(
\begin{aligned}
& =\frac{10}{2 \pi}(-\cos \theta)_0^\pi=\frac{2 \times 10}{2 \pi} \\
& =10 / \pi V
\end{aligned}
\)
So, we can say that the peak voltage in the output of a half-wave diode rectifier fed with a sinusoidal signal without filter is 10 V will have the dc component of the output voltage as \(10 / \pi ~V\).

Note: \(V_{d c}=\frac{V_{\text {peak }}}{\pi}\)

Hint

(b) A diode is said to be reverse biased if \(p\)-type semiconductor of \(p-n\) junction is at low potential with respect to \(n\)-type semiconductor of \(p-n\) junction. It is so for circuit (b).

Hint

(d) In semiconductors at room temperature the electrons get enough energy so that they are able to over come the forbidden gap. Thus at room temperature the valence band is partially empty and conduction Band is partially filled. Conduction band in semiconductor is completely empty only at 0 K.

Hint

(a) \(Y=\overline{\overline{\mathrm{A} \cdot \mathrm{~B}}}={\mathrm{A} \cdot \mathrm{~B}}\)

Hint

(a) Barrier potential of a p-n junction diode does not depend on diode design.

Hint

(d) In a full wave rectifier, the fundamental frequency in ripple is twice of input frequency (100 Hz).

Hint

(b) In reverse biasing, the conduction across the \(p-n\) junction takes place due to minority carriers, therefore, the size of the depletion region (potential barrier) rises.

Hint

(a) In forward biasing, the resistance of \(p-n\) junction diode is very low to the flow of current. So practically all the voltage will be dropped across the resistance \(R\), i.e. voltage across \(R\) will be \(V\). In reverse biasing, the resistance of \(p-n\) junction diode is very high. So the voltage drop across \(R\) is zero.

Hint

(a) A NAND gate’s output is a logical “0” (low) only when all its inputs are logic “1” (high). Otherwise, the output is logic “1”. It’s essentially the complement of an AND gate.

Hint

(a) On account of difference in concentration of charge carriers in the two sections of \(p-n\) junction, the electrons from \(n\) region diffuse through the junction into \(p\)-region and the holes from \(p\)-region diffuse into \(n\) region.
Since the hole is a vacancy of an electron, when an electron from \(n\) region diffuses into the \(p\)-region, the electron falls into the vacancy, i.e. it completes the covalent bond. Due to migration of charge carriers across the junction, the \(n\)-region of the junction will have its electrons neutralized by holes from the p-region, leaving only ionised donor atoms (positive charges) which are bound and cannot move. Similarly, the \(p\) region of the junction will have ionised acceptor atoms (negative charges) which are immobile.
The accumulation of electric charges of opposite polarities in the two regions of the junction gives rise to an electric field between these regions as if a fictitious battery is connected across the junction with its positive terminal connected to \(n\) region and negative terminal connected to \(p\) region. Therefores in a \(p-n\) junction high potential is at N side and low potential is at \(P\) side.

Hint

(b) \(D_1 \rightarrow\) reverse biased \& \(D_2 \rightarrow\) forward biased. Equivalent circuit is

\(
I=\frac{5 \mathrm{~V}}{(30+20) \Omega}=\frac{5}{50} \mathrm{~A} .
\)

Hint

(b) AND gate: Output is 1 when both inputs are 1.

Hint

(a) A diode is said to be forward biased if \(p\) type semiconductor of \(p-n\) junction is at positive potential with respect to \(n\)-type semiconductor of \(p-n\) junction.

Hint

(c) In an unbiased p-n junction, the depletion layer primarily consists of static ions (ionized dopant atoms). These ions, which are immobile, are left behind after charge carriers (electrons and holes) diffuse across the junction.

Hint

(c) A Zener diode can be used as a voltage regulator or voltage stabilizer to provide a constant voltage from a source whose voltage ( AC) may fluctuate over a wide range. 

Hint

(b) This is an EX-OR gate. The output of EX-OR is 1 when the inputs are different(A=1, B=0 or B=1 or A=0), otherwise the output is 0.

Hint

(d) A p-n junction diode is primarily used as a rectifier.
Explanation:
A diode acts as a rectifier because it allows current to flow primarily in one direction (forward bias) and blocks it in the other (reverse bias). This unidirectional current flow is crucial for converting alternating current (AC) to direct current (DC).

Hint

(c) In a p-type semiconductor, the majority charge carriers are holes.
In an n-type semiconductor, the majority charge carriers are electrons.

Hint

(b) In forward biasing, the conduction across \(p-n\) junction takes place due to migrations of majority carriers (i.e. electrons from \(n\)-side to \(p\)-side and holes from \(p\)-side to \(n\)-side), hence the size of depletion region decreases.

Hint

(d) In a semiconductor, when an electron leaves its place, a positive charge is left behind and it is known as hole. Hence, holes are created becauase of missing electrons.

Hint

(b) To create an n-type semiconductor, we have to dope with a pentavalent impurity (P (phosphorus) is pentavalent).

Hint

(d) concentration of positive and negative charges near the junction.
The potential barrier in a p-n diode is created due to the diffusion of majority charge carriers (electrons from n-type to p-type and holes from p-type to n type) across the junction. This diffusion leaves behind immobile ions in the respective regions, creating a depletion region. These immobile ions, being positively or negatively charged, near the junction cause a concentration of positive and negative charges, which in turn establishes an electric field that opposes further diffusion and creates the potential barrier.

Hint

(c) On reversing the polarity of the battery, the \(p-n\) junction is reverse-biased. As a result of which its resistance becomes high, and the current through the junction drops to almost zero.

Hint

(b) NAND

Hint

(c) The given truth table is for an NAND gate.

Hint

(a, d) The output of NAND and EX-NOR gate is 1.

Hint

(c) Voltage drop across diode \(\left(V_D\right)=0.5 \mathrm{~V}\);
Maximum power rating of diode \((P)=100 \mathrm{~mW}\) \(=100 \times 10^{-3} \mathrm{~W}\) and source voltage \(\left(V_s\right)=1.5 \mathrm{~V}\).
The resistance of diode \(\left(R_D\right)\)
\(
=\frac{V_D^2}{P}=\frac{(0.5)^2}{100 \times 10^{-3}}=2.5 \Omega
\)
And current in diode \(\left(I_D\right)=\frac{V_D}{R_D}=\frac{0.5}{2.5}=0.2 \Omega\).
Therefore total resistance in circuit \((R)\)
\(
=\frac{V_x}{I_D}=\frac{1.5}{0.2}=7.5 \Omega .
\)
And the value of the series resistor \(=\) Total resistance of the circuit – Resistance of diode
\(
=7.5-2.5=5 \Omega \text {. }
\)

Hint

(b) \(For NOR gate, } Y=\overline{A+B}\)

Hint

(a) For a \(p\) type germanium semiconductor, it must be doped with a trivalent impurity atom. Since indium is a third group members therefore germanium must be doped in indium.

Hint

(b) When arsenic is added as an impurity to silicon, the resulting material is \(n\) type semiconductor.

Explanation:
Arsenic has five valence electrons, while silicon has four. When arsenic is added to silicon, it forms covalent bonds with the silicon atoms, leaving one extra electron per arsenic atom free to move around. This extra electron contributes to the conductivity of the material, making it an \(n\)-type semiconductor.

Hint

(a) In a conventional current flow in a Forward Biased PN Junction Diode, the current flows from the battery anode to the cathode.
In \(p\) region, the direction of the conventional current is the same as the flow of holes.
In \(n\) region direction of conventional current is opposite to the direction of the flow of electrons.
So, the figure correctly showing the conventional current inside the diode is option (a).

Hint

(a) Phosphorous (P) is pentavalent and silicon is tetravalent. Therefore, when silicon is doped with pentavalent impurity, it forms a n-type semiconductor.

Hint

(b) When \(p-n\) junction is reverse biased, the flow of current is due to drifting of minority charge carriers across the junction.

Hint

(c) The given circuit is a circuit of full wave rectifier.

Hint

(b) Diamond’s exceptional hardness is primarily due to its large cohesive energy. This energy arises from the strong covalent bonds between carbon atoms, forming a rigid three-dimensional network. The large cohesive energy signifies the significant amount of energy required to break these bonds and separate the atoms, contributing to diamond’s high hardness and melting point.

Hint

(d) When both copper and germanium are cooled from room temperature to 80 K , the resistance of copper will decrease while the resistance of germanium will increase.

Explanation:
Copper:
As a metal, copper has a positive temperature coefficient of resistance. This means its resistance generally increases with increasing temperature. When cooled, the thermal vibrations of the atoms decrease, leading to fewer collisions between electrons and atoms, which in turn reduces resistance.

Germanium:
Germanium is a semiconductor. Semiconductors have a negative temperature coefficient of resistance. This means their resistance decreases with increasing temperature. When cooled, the number of free charge carriers (electrons and holes) decreases, leading to a lower conductivity and thus an increase in resistance.

Hint

(c) van der Waals.
Explanation: Van der Waals forces are the weakest type of bonding because they are based on temporary, weak attractions between molecules due to fluctuations in electron distribution, making them significantly weaker than the other options.
lonic bonding: Involves the transfer of electrons between atoms, creating strong electrostatic attractions between positively and negatively charged ions.

Metallic bonding: Results from the delocalization of electrons in a metal, creating a “sea” of electrons that are attracted to the positively charged metal ions.

Covalent bonding: Involves the sharing of electrons between atoms, forming strong bonds between them.

Hint

(b) The depletion layer in the \(p-n\) junction region is caused by diffusion of charge carriers.

Hint

(b) This truth table is of identity, \(Y=A+B\), hence OR gate.

Hint

(d) Due to heating, when a free electron is produced than simultaneously a hole is also produced.

Hint

(a) For forward biasing of \(p-n\) junction, the positive terminal of external battery is to be connected to \(p\) semiconductor and negative terminal of battery to the \(n\) semiconductor.

Hint

(c) At absolute zero, silicon ( Si ) acts as an insulator.
Explanation: At absolute zero, all electrons are in their lowest energy states and do not have enough energy to move into the conduction band, making them unable to conduct electricity effectively, thus behaving like an insulator.