NEET Practice Q & A

Hint

(d) Concept:
Relation between Current Density and Electric Field:
Current Density:
The amount of electric current travelling per unit cross-section area is called as current density and expressed in amperes per square meter.
Formula for Current Density is given as,
\(
\mathrm{J}=\mathrm{I} / \mathrm{A}
\)
Where,
I = current flowing through the conductor in Amperes
\(A=\) cross sectional area in \(\mathrm{m}^2\).
Consider a conductor of length I and area of cross-section A. Let its two ends be raised to potentials \(\mathrm{V}_1[latex] and [latex]\mathrm{V}_2\left(\mathrm{~V}_1>\mathrm{V}_2\right)\).
The electric current density is given by the formula:
\(
j=\sigma E=n e v_d
\)
Calculation:
Given:
Density of electrons, \(\mathrm{n}_{\mathrm{e}}=10^{19} \mathrm{~m}^{-3}\)
Charge of electron, \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\)
Mobility of electron, \(\mu_e=1.6 \mathrm{~m}^2 /(\mathrm{V} . \mathrm{s})\)
The electric current density is given by the formula:
\(
\mathrm{j}=\sigma \mathrm{E}=\mathrm{nev}_{\mathrm{d}}
\)
For intrinsic semiconductors (no impurities), the number of electrons will be equal to the number of holes.
\(
\sigma=\mathrm{ne} \frac{\mathrm{v}_{\mathrm{d}}}{\mathrm{E}}=\mathrm{ne} \mu
\)
For N -type semiconductor, electrons are majority carriers.
Conductivity, \(\sigma \approx[latex] ne [latex]\mu\)
Resistivity, \(\rho=\frac{1}{\sigma}=\frac{1}{\mathrm{n}_{\mathrm{e}} \mathrm{e} \mu_{\mathrm{e}}} \dots(1)\)
By substituting the given values in the equation (1),
\(
\begin{aligned}
& \rho=\frac{1}{10^{19} \times 1.6 \times 10^{-19} \times 1.6} \\
& \rho=0.4 \Omega \mathrm{~m}
\end{aligned}
\)

Hint

(c) The temperature coefficient of resistance is equal to the change in resistance of the wire of resistance one ohm at \(0^{\circ} \mathrm{C}\) when the temperature changes by \(1^0 \mathrm{C}\).
That is denoted by \(\alpha\).
\(
\alpha=\frac{R_t-R_o}{R_o t}
\)
Where,
\(\alpha=\) temperature coefficient
Let’s consider a case,
If \(R_o=1\) and \(\mathrm{t}=1^0 \mathrm{C}\) then, \(\alpha=R_t-R_o\)
In general, for two different temperature and resistance, we can write the equation as,
\(
\alpha=\frac{R_2-R_1}{R_1 t-R_2 t}
\)
From this equation, we can say that as the temperature of a semiconductor increases, resistance decreases.
Then the value \(R_2-R_1\) becomes a negative value. Therefore the entire value becomes negative in the case of the semiconductor.
Thus, the temperature coefficient of resistance \((\alpha)\) of a semiconductor is always negative.

Hint

(b) Energy bands in solids are a consequence of (b) Pauli exclusion principle.

Explanation: The Pauli exclusion principle states that no two electrons in an atom can have the same set of quantum numbers, which leads to the formation of distinct energy levels when atoms come together to form a solid.

Why other options are incorrect:
(a) Ohm’s law:

Ohm’s law relates current, voltage, and resistance in a conductor. It doesn’t explain the formation of energy bands.
(c) Bohr’s theory:

Bohr’s theory describes the quantized energy levels of electrons in individual atoms, but it doesn’t explain how these levels merge into bands when atoms form a solid.
(d) Heisenberg’s uncertainty principle:

Heisenberg’s uncertainty principle deals with the limitations in simultaneously knowing the position and momentum of a particle. While it’s a fundamental principle in quantum mechanics, it doesn’t directly explain the formation of energy bands.

Hint

(b) Compare the change in current through the diode and the resistor when the voltage is increased.
Step 1: Initially, the readings of A1 and A2 are the same, meaning the current through the diode ( \(I_D\) ) and the current through the resistor ( \(I_R\) ) are equal:
\(I_D=I_R\)

Step 2: The current through the diode increases exponentially with voltage:
\(I_D=I_0\left(e^{\frac{c V}{n k T}}-1\right)\)
Where \(I_0\) is the reverse saturation current, \(\boldsymbol{e}\) is the electron charge, \(V\) is the voltage, \(n\) is the ideality factor, \(k\) is Boltzmann’s constant, and \(T\) is the temperature.
When the voltage increases, the exponential term dominates, and the current through the diode increases significantly.

Step3: The current through the resistor increases linearly with voltage according to Ohm’s law:
\(I_R=\frac{V}{R}\)
When the voltage increases, the current through the resistor also increases, but linearly.

Step4: Therefore, the reading of A1 (diode current) will be greater than the reading of A2 (resistor current).

Hint

(d) When an electron diffuses from \(n \rightarrow p\), it leaves behind an ionised donor on \(n\)-side. This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. As the electrons continue to diffuse from \(n \rightarrow p\), a layer of positive charge (or positive space-charge region) on n -side of the junction is developed.

Similarly, when a hole diffuses from \(\mathrm{p} \rightarrow \mathrm{n}\) due to the concentration gradient, it leaves behind an ionised acceptor (negative charge) which is immobile. As the holes continue to diffuse, a layer of negative charge (or negative space-charge region) on the p-side of the junction is developed. This space-charge region on either side of the junction together is known as depletion region as the electrons and holes taking part in the initial movement across the junction depleted the region of its on diffusion free charges (Fig. D).

Hint

(c) The given frequency of the input \(=50 \mathrm{~Hz}\)
The fundamental frequency of the output ripples \(=2\) times the input frequency \(=2 \times 50=100 \mathrm{~Hz}\).

Hint

(a) To obtain a P-type silicon semiconductor, you need to dope pure silicon with a trivalent impurity like boron, aluminum, or gallium.

Explanation: When a trivalent impurity is added to silicon, it creates “holes” in the crystal lattice, allowing for positive charge carriers to move, which is characteristic of a P-type semiconductor.

Hint

(c)

\(
\begin{aligned}
&\text { By Kirchhoff’s node law, }\\
&I_T=I_Z+I_L
\end{aligned}
\)
\(
I_T=\frac{(100-40)}{3K}=20 \mathrm{~mA}
\)
\(
I_L=\frac{40}{6K}=\frac{20}{3} \mathrm{~mA}
\)
\(
I_Z=20-\frac{20}{3}= \frac{40}{3} \mathrm{~mA}
\)

Hint

(b) Diode D1 and D4 will be forward biased (D2 and D3 will be reverse biased) and replaced with 50 Ohm resistor each.

\(
\begin{aligned}
I & =\frac{V}{R_t} \\
I & =\frac{12 \mathrm{~V}}{300 \Omega} \\
I & =0.04 \mathrm{~A}
\end{aligned}
\)

Hint

(d) \(X=1, Y=0\)

Hint

(d)

\(
Y=A\bar{B}
\)

Hint

(d)

Calculate the energy of the photon with the given wavelength and equate it to the band gap energy. Then, solve for \(n\).

Calculate the energy of the photon:

The energy of the photon is given by:
\(E=\frac{h c}{\lambda}\)

Substituting the given values:
\(E=\frac{\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{2480 \times 10^{-9} \mathrm{~m}}\)
\(E=\frac{19.878 \times 10^{-26}}{2480 \times 10^{-9}} \mathrm{~J}\)
\(E=8.015 \times 10^{-20} \mathrm{~J}\)

Convert the energy to electron volts (eV)
Using the conversion factor \(1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\) :
\(E=\frac{8.015 \times 10^{-20} \mathrm{~J}}{1.602 \times 10^{-19} \mathrm{~J} / \mathrm{eV}}\)
\(E=0.500 \mathrm{eV}\)

Equate the energy to the band gap and solve for \(\boldsymbol{n}\)
The band gap is given as \(n \times 10^{-2} \mathrm{eV}\) :
\(0.500 \mathrm{eV}=n \times 10^{-2} \mathrm{eV}\)
\(n=\frac{0.500}{10^{-2}}\)
\(n=50\)

Hint

(b) Step 1: Express the ratio of currents in terms of concentrations and drift velocities.
The electron current is \(I_e=n_e e A v_e\).
The hole current is \(I_h=n_h e A v_h\).
The ratio of currents is \(\frac{I_e}{I_h}=\frac{n_e e A v_e}{n_h e A v_h}\).
Simplify the ratio: \(\frac{I_e}{I_h}=\frac{n_e v_e}{n_h v_h}\).
Step 2: Solve for the ratio of drift velocities.
Rearrange the equation from Step 1: \(\frac{v_e}{v_h}=\frac{I_e}{I_h} \cdot \frac{n_h}{n_e}\).
Substitute the given ratios: \(\frac{v_e}{v_h}=\frac{7}{4} \cdot \frac{5}{7}\).
Simplify: \(\frac{v_e}{v_h}=\frac{5}{4}\).

Step 3: Find \(n\).
We are given that \(\frac{v_e}{v_h}=\frac{n}{4}\).
From Step 2, we found that \(\frac{v_e}{v_h}=\frac{5}{4}\).
Equate the two expressions: \(\frac{n}{4}=\frac{5}{4}\).
Solve for \(n: n=5\).
Solution
The value of \(n\) is 5 .

Hint

(d) Given, voltage across Zener diode, \(V_Z=6 \mathrm{~V}\) Since, Zener diode provide stabilised supply of 6 V.

In this case, the value of \(R\) should be chosen such that negligible current flows through the Zener diode.
\(
\begin{array}{lrl}
\therefore & & I=I_Z+I_L=0+I_L \\
& & I=\frac{30}{R+10^3} \\
\text { But } & & I=6 \mathrm{~mA}=6 \times 10^{-3} \mathrm{~A} \\
\therefore & & 6 \times 10^{-3}
\end{array}=\frac{30}{R+10^3} \Rightarrow R=4 \times 10^3 \Omega=4 \mathrm{k} \Omega \text { llor }
\)

Hint

(b) A p-n junction photodiode can be operated under photovoltaic conditions similar to that of a solar cell. The current-voltage characteristics of a photodiode and solar cell under illumination are also similar.
When light energy falls on the solar cell, then it converts solar energy into electrical energy.
The area of solar cell should be larger, so that it could absorbed more amount of sunlight and hence more amount of electrical energy is obtained.
Hence, Assertion and Reason both are correct but Reason is not the correct explanation of Assertion.

Hint

\(
\begin{aligned}
&\text { (a) According to give logic diagram, }\\
&\begin{aligned}
& & Y^{\prime} & =\left(A^{\prime} B^{\prime}\right)^{\prime}=\left(A^{\prime}\right)^{\prime}+\left(B^{\prime}\right)^{\prime}=A+B \\
\therefore & & Y & =Y^{\prime}=(A+B)^{\prime}=\text { output of NOR gate }
\end{aligned}
\end{aligned}
\)

Hint

(c) Current passing in the circuit,
\(
I=\frac{P}{V}=\frac{100 \times 10^{-3}}{0.5}=0.2 \mathrm{~A}
\)
Value of connected series resistance,
\(
\begin{aligned}
R & =\frac{1.5-0.5}{0.2} \Rightarrow R=\frac{1}{0.2} \\
\therefore \quad R & =5 \Omega
\end{aligned}
\)

Hint

(c) Given, one indium atom to be doped in \(5 \times 10^7\) silicon atoms.
Number density of silicon
\(
\begin{aligned}
& =5 \times 10^{28} \text { atom m } \mathrm{m}^{-3} \\
& \approx 5 \times 10^{22} \text { atom } \mathrm{cm}^{-3}
\end{aligned}
\)
Number of acceptor atom \(\mathrm{cm}^{-3}\)
\(
=\frac{5 \times 10^{22}}{5 \times 10^7}=1 \times 10^{15} \mathrm{~cm}^{-3}
\)
Hence, number of acceptor atom \(/ \mathrm{cm}^3\) is \(1 \times 10^{15} \mathrm{~cm}^{-3}\).

Hint

(a) So, output \(Y=\overline{\bar{A} \cdot \bar{B} \cdot \bar{C}}=\stackrel{=}{A}+\stackrel{=}{B}+\stackrel{=}{C}=A+B+C\)
If \(A=0, B=1, C=1\)
\(
\therefore \quad Y=0+1+1=0+1=1
\)

Hint

(d) If \(Y\) represents the given Boolean expression \(P+\bar{P} Q\), then
\(
\begin{array}{rlr}
Y & =P+\bar{P} Q=P(1)+\bar{P} Q & \\
& =P(1+Q)+\bar{P} Q & (\because 1=1+X) \\
& =P+P Q+\bar{P} Q & \\
& =P+Q(P+\bar{P}) & \\
& =P+Q & (\because \bar{X}+X=1) \\
& =\text { output of OR gate } &
\end{array}
\)
Hence, given expression represents OR gate.

Hint

(a) The overall conductivity of a semiconductor,
\(
\begin{gathered}
\sigma=n_e e \mu_e+n_p e \mu_p \\
\sigma=e\left[\mu_e n_e+n_p \mu_p\right]
\end{gathered}
\)
Also, \(n_e n_p=n_i^2\) [for an intrinsic semiconductor]
\(
\Rightarrow \quad n_e=\frac{n_i^2}{n_p}
\)
Now,
\(
\sigma=e\left[\frac{n_i^2}{n_p} \mu_n+\mu_p n_p\right]
\)
On differentiating, \(\frac{d \sigma}{d n_p}=e\left[-\frac{n_i^2}{n_p^2} \mu_n+\mu_p\right]=0\)
\(
\Rightarrow \quad \mu_p=\frac{n_i^2}{n_p^2} \mu_n \Rightarrow n_p=n_i \sqrt{\frac{\mu_n}{\mu_p}}
\)

Hint

(d) The common application of Zener diode is stabilisation or voltage regulation.

Hint

(b) Gate I \(\rightarrow\) NAND gate
Gate II \(\rightarrow\) AND gate
Above arrangement represents NAND gate.

Hint

(a) The output of NOR gate
\(
\overline{A+B}=Y
\)
When, \(A=B\), then \(\overline{A+A}=\bar{A}=\) NOT gate

Hint

(a) Given, energy gap, \(E_g=1.9 \mathrm{eV}=1.9 \times 1.6 \times 10^{-19} \mathrm{~V}\)
We know that, \(E_g=\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{E_g}\)
Wavelength of the emitted light,
\(
\begin{aligned}
& \lambda=\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.9 \times 1.6 \times 10^{-19}} \\
& \lambda=6.5 \times 10^{-7} \mathrm{~m}
\end{aligned}
\)

Hint

(d) Output of NOR gate, \(Y^{\prime}=\overline{A+B}\)
Output of NAND gate,
\(
\begin{aligned}
Y^{\prime \prime} & =\overline{(A+B) \cdot(A+B)}=\overline{\overline{(A+B)}} \cdot \overline{\overline{(A+B)}} \\
& =(A+B)(A+B)=A+B
\end{aligned}
\)
Output of NOT gate,
\(
Y^{\prime \prime \prime}=\overline{A+B}
\)
\(\therefore\) The boolean expression is for NOR gate.

Hint

(c) The output of digital circuit can be given as \(Y=\overline{A B}+\bar{C}\).
It will be same as \(Y=\bar{A}+\bar{B}+\bar{C}\).

Hint

(c) The output of \(G_1=A+B\)
Output of \(G_2=\overline{A \cdot B}\)
and output of \(G_3=(A+B)(\overline{A \cdot B})=A \bar{B}+B \bar{A}\)
This output is same as the output of an XOR gate.

Hint

(b) As the lower diode attached to \(100 \Omega\) resistance is in reverse biased, so it is non-conducting (we can remove that).
\(
\therefore \text { Current through the circuit, } I=\frac{V}{R}=\frac{10}{200}=0.05 \mathrm{~A}
\)

Hint

(d) The mobility of electron is greater than holes because electron require less energy to move.

Explanation: Electrons are lighter and experience less resistance when moving through the crystal lattice compared to holes, which are essentially the absence of an electron and are more influenced by the positively charged nucleus. 

Hint

(d) As the diode is forward biased, so it takes only positive value. Hence, option (d) is correct.

Hint

(b) In the given circuit, \(D_1\) is forward biased and \(D_2\) is reverse biased, so \(D_1\) works like a closed switch and \(D_2\) works like an open circuit, so circuit becomes
\(
\begin{aligned}
&\text { Now, current in the circuit, }\\
&\begin{array}{ll}
& I=\frac{E}{R_S}=\frac{12}{4+3} \\
\Rightarrow & I=\frac{12}{7} \\
\Rightarrow & I=1.71 \mathrm{~A}
\end{array}
\end{aligned}
\)

Hint

(c) The conductivity of the intrinsic semicondutor is independent of fermi energy. It depends upon band gap, temperature as well as effective mass of charge carriers.

Fermi Energy:
In an intrinsic semiconductor, the Fermi level (Fermi energy) lies roughly in the middle of the band gap. The Fermi level itself does not directly determine conductivity; rather, it indicates the energy level at which the probability of finding an electron is \(50 \%\). The conductivity is determined by the number of electrons that have enough energy to jump into the conduction band and the number of holes left behind.

Hint

(d) \(Y_1=\) Output due to NAND gate \(1=\overline{A \cdot A}=\bar{A}\)
\(Y_2=\) Output due to NAND gate \(2=\overline{B \cdot B}=\bar{B}\)
Output due to NAND gate 3
\(
Y=\overline{Y_1 \cdot Y_2}=\overline{Y_1}+\overline{Y_2}=\overline{\bar{A}}+\overline{\bar{B}}=A+B
\)
This is OR gate. Hence, the combination is equivalent to an OR gate.

Hint

(c) In an insulator, the conduction band (CB) is empty and the valence band (VB) is filled with electrons. This means there are no free electrons in the conduction band to carry an electrical current, while the valence band, the highest energy band containing electrons, is completely occupied.

Hint

(c) The resistivity of semiconductors is greater than metals.

Explanation: While semiconductors have lower resistivity than insulators, their resistivity is still higher than that of metals.

Hint

(d) The \(n\)-type semiconductor region has (negative) electrons as majority charge carriers and an equal number of fixed positively charged donor ions. Again, the material as a whole is neutral. That is a reason, that atom is electrically neutral.

Hint

(c) The wrong statement with reference to a solar cell is that, it uses materials with band gap of 5 eV. Because, for solar cell, band gap is less than 3 eV.

Hint

(b) Initial current, \(I_{\mathrm{in}}=0\)
Final current, \(I_f=\frac{(6-3)}{150}=0.02 \mathrm{~A}\)
So, change in current, \(\Delta I=I_f-I_{\mathrm{in}}=0.02-0\) \(=0.02 \mathrm{~A}=20 \mathrm{~mA}\)

Hint

(b) If the two inputs of a NAND gate are joined together, it works as a NOT gate. Now, if the inputs \(A\) and \(B\) are inverted by using two NOT gates (obtained from two NAND gates) and the resulting output \(\bar{A}\) and \(\bar{B}\) are fed to a third NAND gates.
So, three NAND gates are required to make an OR gate.

Hint

\(
\text { (b) By absorption law, output } Y=\bar{A}+\bar{A} \cdot B
\)
\(
Y=\bar{A} \cdot B+\bar{A}=\bar{A} \cdot(B+1)=\bar{A} \cdot 1=\bar{A}
\)

Hint

(d) According to the logical relationship, Output, \(Y=\overline{\bar{A}+\bar{B}}=\bar{A} \cdot \bar{B}=A \cdot B\)
So, it is AND gate.

Hint

(c) In this case, diode P side is 0 V and the N side is connected to 5 V. So the diode is reverse biased.

Hint

(d) \(Y_1=A B \quad Y_2=\bar{Y}_1 \cdot Y_1=\bar{Y}_1\)

Hint

(c) In a p-type semiconductor, the acceptor energy level lies near the valence band.

Explanation:
In p-type semiconductors, acceptor impurities (like Boron in Silicon) introduce energy levels near the valence band. These acceptor levels are slightly above the valence band and facilitate the movement of electrons from the valence band to these acceptor levels, creating holes (positive charge carriers) in the valence band.

Hint

(d) When a p-n junction is reverse biased, the current through the junction is mainly due to only drift of charges.

Explanation:
Reverse biasing:
In reverse bias, the applied voltage creates a strong electric field across the p-n junction, which pushes the minority carriers (electrons from the p-side and holes from the \(n\)-side) to drift across the junction.

Hint

(c) \(\bar{A}+A\) should always be equal to 1. As, addition of inputs like 1 to 0 or 0 to 1 always gives 1.

Hint

(c) The depletion layer in the \(p-n\) junction region is caused due to the diffusion of carriers on either side of the junction.

Explanation:
Diffusion of carriers:
In a p-n junction, the majority carriers (holes in the p-type region and electrons in the n-type region) naturally diffuse across the junction due to their concentration gradients. This diffusion causes electrons to move from the n-side to the p-side and holes to move from the p-side to the n-side.

Hint

(c) A digital signal has only two values of voltage variation with time. Its value is either zero or maximum (1).

Hint

(c) The diode \(D_1\) is in reversed biased and \(D_2\) is in forward biased. The resistance of \(D_1\) in in becomes infinite and of \(D_2\) is zero. Therefore, current in the circuit,
\(
I=\frac{\mathrm{emf}}{\text { total resistance }}=\frac{12}{4+2}=2 \mathrm{~A}
\)

Hint

\(
\begin{aligned}
& \text { (c) As, } V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}} \Rightarrow 220=\frac{V_0}{\sqrt{2}} \\
& \Rightarrow \quad V_0=311.1 \mathrm{~V}
\end{aligned}
\)

Hint

(c) The incorrect statement about semiconductors is At absolute zero temperature, it behaves like a conductor.

Explanation: At absolute zero temperature, a semiconductor behaves like an insulator because all the electrons are bound to their atoms and there are no free electrons available to conduct electricity.

Hint

(a) At room temperature, due to thermal vibrations, the few bonds of intrinsic semiconductor are broken and producing equal number of electrons and holes in the semiconductor.

Hint

(b) is not a rectifier circuit because it does not follow the typical configuration of diodes used in rectifier circuits.

Hint

(c) A semiconductor typically has four electrons in its valence shell. This characteristic allows semiconductors to form covalent bonds and exhibit properties between those of conductors and insulators.

Hint

(d) Copper is a metal.
Germanium is a semiconductor.
Both are cooled from room temperature to 80 K.
Metals have a positive temperature coefficient of resistance.
Semiconductors have a negative temperature coefficient of resistance.
Copper is a metal.
For metals, resistance decreases as temperature decreases.
Therefore, the resistance of copper decreases when cooled to 80 K .
Germanium is a semiconductor.
For semiconductors, resistance increases as temperature decreases. Therefore, the resistance of germanium increases when cooled to 80 K.
The resistance of copper decreases while that of germanium increases.

Hint

(c) The separation between the conduction band and valence band in a semiconductor is of the order of 1 eV .

Explanation:
Semiconductors have a band gap (energy difference between the valence and conduction bands) that is relatively small, allowing some electrons to jump from the valence band to the conduction band with moderate energy input. This allows for some electrical conductivity, which is why they are called semiconductors. A band gap of 1 eV is characteristic of many common semiconductors like silicon and germanium.

Hint

(d) zero.
Explanation:
In a good conductor, the valence band and conduction band essentially overlap, meaning there is no significant energy difference between them. This allows electrons to move freely between the bands, facilitating electrical conduction.

Hint

(a) At room temperature in semiconductors, the valence band is partially empty and the conduction band is partially filled. This is because some electrons in the valence band gain enough thermal energy to jump to the conduction band, leaving behind “holes” (vacancies) in the valence band. These holes and the electrons in the conduction band can then contribute to electrical conductivity.

Hint

(c) On increasing the temperature, the current in the circuit will increase because with rise in temperature, resistance of semiconductor decreases, hence overall resistance of the circuit decreases, which in turn increases the current in the circuit.

Hint

(c)\(\Delta E_{g(\text { Germanium })}=0.67 \mathrm{eV}\)

Hint

(b) The energy band gap of Silicon \((\mathrm{Si})\) at room temperature is a well-known physical constant. It is approximately 1.1 eV.

Hint

(c) the small width of the forbidden energy gap.
Explanation:
Semiconductor behavior: In a semiconductor like germanium (Ge), not all electrons are completely free to move, but they can be excited to a higher energy level with a relatively small amount of energy. This allows them to conduct electricity to a certain degree.

Hint

(b) Germanium has a smaller bandgap energy than silicon.
\(
E_{g, G e}<E_{g, S i}
\)
A smaller bandgap means less energy is needed to create electron-hole pairs.
Therefore, more electron-hole pairs are generated in materials with smaller bandgaps at a given temperature.
Since germanium has a smaller bandgap, more electron-hole pairs will be generated in germanium than in silicon at room temperature.
This implies less electron-hole pairs will be generated in silicon than in germanium.
Less number of electron-hole pairs will be generated in silicon than in germanium at room temperature.

Hint

(c) When the electrical conductivity of a semiconductor is solely due to the breaking of its covalent bonds, the semiconductor is said to be intrinsic.

Explanation: Intrinsic semiconductors are pure semiconductors (without any doping) where the electrical conductivity arises from thermally generated charge carriers (electrons and holes) formed when covalent bonds break.

Hint

(c) \(\text { The thickness of a } p-n \text { junction is of the order of } 10^{-6} \mathrm{~m} \text {. }\)

Hint

(a) At room temperature, a \(p\)-type semiconductor has large number of holes and few electrons, as holes are majority charge carriers.

Explanation:
p-type semiconductor:
In a p-type semiconductor, impurities are added that create “holes” – essentially vacancies where electrons should be. These holes act as positive charge carriers.

Few electrons:
While there are still a few free electrons present in the semiconductor, their number is significantly smaller compared to the number of holes.

Hint

(c) “The majority carriers in \(n\)-type semiconductors are holes”.

Explanation: In \(n\)-type semiconductors, the majority carriers are electrons, not holes.

Hint

(d) An \(n\)-type semiconductor is formed when a germanium crystal is doped with an impurity containing five valence electrons. 

Hint

(c) An \(n\)-type semiconductor is neutral.
Explanation:
” n ” stands for negative: In an \(n\)-type semiconductor, the majority charge carriers are electrons, which are negatively charged. However, this doesn’t mean the semiconductor is overall negatively charged.
Electrical neutrality: Despite the presence of more free electrons, there are also an equal number of positive charges (protons) within the semiconductor, resulting in a net neutral charge.

Hint

(d) The electrical conductivity of both intrinsic and \(p\)-type semiconductors increases with an increase in temperature.

Explanation:
Intrinsic semiconductors:
In these materials, the conductivity arises from thermally generated electron-hole pairs. When temperature increases, more electrons gain enough energy to jump from the valence band to the conduction band, creating more charge carriers (electrons and holes) and thus increasing conductivity.

P-type semiconductors:
Similar to intrinsic semiconductors, increasing temperature in p-type semiconductors also enhances conductivity. The added dopant atoms in p-type materials create an abundance of holes, but the conductivity is still significantly influenced by the number of free charge carriers. As temperature increases, more electrons are excited into the conduction band, and while some holes may be filled, the overall effect is an increase in conductivity.

Hint

(b) Pentavalent impurity atom is to be added to Ge crystal, so as to make a \(n\)-type semiconductor.

Hint

(b) 0.6 V .
Explanation:
A silicon diode’s barrier potential is typically around 0.6 to 0.7 volts. This is the voltage that needs to be applied across the diode to overcome the internal barrier and allow current to flow through it.

Hint

(d) A full-wave rectifier does not have lower efficiency, average DC, or average output voltage compared to a half-wave rectifier. In fact, it has higher efficiency, and higher average DC and output voltage. Therefore, the correct answer is none of these.

Hint

(d) A Zener diode is primarily used as a AC voltage stabilizer. While it can be used in conjunction with other components to create rectifiers, its main function is to maintain a constant voltage output even when the input voltage or load conditions change.

Hint

(a) The serious drawback of a semiconductor device is that they cannot be used with high voltage.

Explanation: Semiconductors have a limited tolerance for high voltages, which can damage them.

Hint

(d) Extrinsic semiconductors are also electrically neutral.

Explanation:

Intrinsic semiconductors:
These are pure semiconductors like silicon or germanium where the number of electrons is equal to the number of holes, resulting in no net charge, making them electrically neutral.

Extrinsic semiconductors:
These are created by adding impurities to an intrinsic semiconductor, which introduces additional charge carriers (electrons or holes). However, even with a large number of current carriers, the overall charge remains neutral because the added dopants are electrically neutral.

Hint

(b) majority carrier only. A potential barrier in a junction diode opposes the flow of majority carriers across the junction.

Explanation:
Majority carriers:
In a p-n junction diode, the majority carriers in the p-type material are holes, and the majority carriers in the n-type material are electrons.
Potential barrier:
When a \(p-n\) junction is formed, a depletion region (also called a space charge region) forms at the junction, creating a potential barrier.

Opposition to majority carriers:
This potential barrier opposes the flow of majority carriers (holes from the p-region and electrons from the \(n\)-region) from crossing the junction.

Minority carriers:
While the potential barrier opposes majority carriers, it actually aids the flow of minority carriers (electrons in the p-region and holes in the n-region).

Hint

(a) As, \(\quad i=n_e A v_d\)
So, \(\quad v_d=\frac{i}{n_e A}\)
So, when temperature is increased, \(n_e\) increases and \(v_d\) decreases.

Hint

(b) In forward biasing, the diffusion current increases and drift current remains constant, so net current is due to diffusion. In reverse biasing diffusion is not possible, so net current (very small) is due to the drift.

Hint

(b) At a specific reverse voltage (breakdown voltage) in \(p-n\) junction, a huge current flows in reverse direction known as avalanche current.

Hint

(c) In an unbiased p-n junction, holes diffuse from the p-region to the n-region primarily due to the concentration gradient, meaning the higher concentration of holes in the p -region compared to the n -region.

Explanation:
Concentration Gradient:
The fundamental driving force for diffusion is the tendency of particles (in this case, holes) to move from an area of high concentration to an area of low concentration. Since the p-region has a higher concentration of holes than the \(n\)-region, holes will naturally move across the junction from the \(p\)-side to the \(n\)-side.

Hint

(a) In a full wave rectifier, two diodes work alternately.

Explanation: A full wave rectifier uses two diodes to convert the AC input into DC, but they alternate which diode conducts during each half cycle, effectively utilizing both halves of the waveform.

Hint

(d) Resistance of semiconductor decreases with increase in temperature. Hence, its temperature coefficient is negative.

Explanation:

We know that the temperature coefficient of resistance is equal to the change in resistance of the wire of resistance one ohm at \(0^{\circ} \mathrm{C}\) when the temperature changes by \(1^{\circ} \mathrm{C}\).
It is denoted by \(\alpha\) and given as,
\(
\alpha=\frac{R_t-R_0}{R_0 t}
\)
Where, \(R_t\) is the resistance of the material at temperature \((\mathrm{t})\).
\(R_0\) is the resistance of the material at absolute zero temperature.
\(t\) is the temperature.
Let’s consider a case,
If \(R_0=1\) and \(t=1^{\circ} C\). Then, \(\alpha=R_t-1\).
In general, for two different temperature and resistance, we can write the equation as,
\(
\alpha=\frac{R_2-R_1}{R_1 t-R_2 t}
\)
From this equation, we can say that as the temperature of a semiconductor increases, resistance decreases. Then the value of \(R_2-R_1\) becomes a negative value. Therefore, the entire value becomes negative in the case of the semiconductor. Thus, the temperature coefficient of resistance ( \(\alpha\) ) of a semiconductor is always negative.

Hint

(c) p-n junction. Reversing the polarity of the battery in a p-n junction causes it to be reverse-biased, significantly reducing the current flow.

Hint

(c) The band gap \(=1.5 \mathrm{e} \mathrm{V}\)
\(
\begin{aligned}
& =1.5 \times 1.6 \times 10^{-19} J \\
& =2.40 \times 1^{-19} J
\end{aligned}
\)
Wavelength of radiation emitted \((\lambda)=\frac{h c}{E}\)
\(
\begin{aligned}
& =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.4 \times 10^{-19}} \\
& =8.25 \times 10^{-7} \mathrm{~m} \\
& =8250 Å
\end{aligned}
\)
This wavelength belongs to infrared rays.

Hint

(b) When no external voltage is applied across a p-n junction, there would be an electric field pointing from the \(n\)-type to the \(p\)-type side across the junction.

Explanation:
Diffusion of charge carriers: In a p-n junction, electrons from the n-type region naturally diffuse into the p-type region, leaving behind positively charged ions in the \(n-\) type region. Similarly, holes from the p-type region diffuse into the n-type region, creating negatively charged ions in the p-type region.

Depletion region formation: This diffusion of charge carriers creates a region near the junction depleted of mobile charge carriers, called the depletion region.

Electric field development: Due to the separation of positive and negative charges, an electric field is established across the depletion region, pointing from the n-type side to the p-type side.

Hint

\(
\begin{aligned}
& \text { (d) }\left(E_g\right)_{\mathrm{Ge}}<\left(E_g\right)_{\mathrm{Si}}<\left(E_g\right)_{\mathrm{C}} \\
& {\left[\because\left(E_g\right)_{\mathrm{Ge}}=0.7 \mathrm{eV},\left(E_g\right)_{\mathrm{Si}}=1.1 \mathrm{eV},\left(E_g\right)_{\mathrm{C}}=5.5 \mathrm{eV}\right]}
\end{aligned}
\)

Hint

(a) When a diode is heavily doped, the Zener voltage will be low. This is correct.
Explanation: Heavy doping reduces the width of the depletion region, which leads to a lower Zener breakdown voltage.
(b) the avalanche voltage will be high: Avalanche breakdown is related to the Zener voltage. A lower Zener voltage due to heavy doping also means a lower avalanche voltage. This is incorrect.
(c) the depletion region will be thin: This is correct. Heavy doping results in a thin depletion region.
(d) the leakage current will be low: While the specific relationship between leakage current and doping level is complex, heavy doping doesn’t necessarily guarantee a low leakage current. Leakage current is a small reverse current that always flows through a diode even when it’s reverse biased. This is incorrect.

Hint

(b) From rectifier we get pulsating DC, which consists of AC ripples, to remove these unwanted ripples, we use filter circuits.

Hint

\(
\begin{aligned}
& \text { (d) For full wave rectifier, } \eta=\frac{81.2}{1+\frac{r_f}{R_L}} \\
& \Rightarrow \quad n_{\max }=81.2 \% \quad\left(\because r_f \ll R_L\right)
\end{aligned}
\)

Hint

(b) An XOR gate (Exclusive OR) is represented by symbol II, and a NOR gate (NOT OR ) is represented by symbol III.

Hint

(b) 

\(
\begin{aligned}
&\text { For a full-wave rectifier, the ripple frequency is twice the input frequency. }\\
&f_{\text {ripple }}=2 \times f_{\text {input }}
\end{aligned}
\)
\(
\begin{aligned}
&\text { The input frequency is } 50 \mathrm{~Hz} \text {. }\\
&f_{\text {ripple }}=2 \times 50 \mathrm{~Hz}= 100 \mathrm{~Hz}
\end{aligned}
\)

Hint

(c) Ripple frequency (half-wave) = Input AC frequency .
Ripple frequency (half-wave) = 50 Hz.
In a full-wave rectifier, the ripple frequency is twice the input \(A C\) frequency.
Ripple frequency (full-wave) \(=2 \times\) Input AC frequency.
Ripple frequency (full-wave) \(=2 \times 50 \mathrm{~Hz}=100 \mathrm{~Hz}\).

Hint

(c) As the temperature (T) increases, the conductivity ( \(\sigma\) ) increases.
Therefore, the resistivity ( \(\rho\) ) must decrease.
This can be expressed mathematically as:
\(\sigma \propto T\) (as T increases, \(\sigma\) increases)
Consequently, we can say:
\(\rho \propto \frac{1}{T} \quad\) (as T increases, \(\rho\) decreases)
Resistivity ( \(\rho\) ) on the \(y\)-axis and temperature (T) on the \(x\)-axis, the graph will be a rectangular hyperbola.

Hint

(b) Avalanche breakdown in a semiconductor diode occurs when reverse bias exceeds a certain value.

Explanation: Avalanche breakdown happens when a high reverse bias voltage is applied to the diode, causing a strong electric field that accelerates minority charge carriers to high speeds, resulting in the creation of more electron-hole pairs through impact ionization. This rapid increase in current is known as avalanche breakdown.

Hint

(a) Given, \(E=6 \times 10^5 \mathrm{Vm}^{-1}\)
\(
\begin{aligned}
d & =500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m} \\
V & =?
\end{aligned}
\)
\(\therefore \quad\) Potential barrier, \(V=E d=\left(6 \times 10^5\right)\left(500 \times 10^{-9}\right)=0.30 \mathrm{~V}\)

Hint

\(
\begin{aligned}
&\text { (c) The electrical conductivity of germanium, }\\
&\begin{aligned}
\sigma & =\frac{1}{\rho}=e\left(\mu_e n_e+\mu_h n_n\right) \\
& =\left(1.6 \times 10^{-19}\right)[0.36+0.17]\left(2.5 \times 10^{19}\right) \\
& \quad\left(\because n_e=n_h=2.5 \times 10^{19} \mathrm{~m}^{-3}\right) \\
& =2.12 \mathrm{Sm}^{-1}
\end{aligned}
\end{aligned}
\)

Hint

(b) In \(1 \mathrm{~cm}^3\) of Si , there will be total \(\frac{6 \times 10^{28}}{10^6}=6 \times 10^{22}\) atoms Per \(6 \times 10^7\) silicon atoms, there is one indium atom. Therefore, total number of indium atoms will be \(\frac{6 \times 10^{22}}{6 \times 10^7}=10^{15}\)

Hint

(a) During the first half cycle when \(V_A>V_C[//latex], all the four diodes are forward biased. Hence, no current will flow through [latex]R_L\). During second half cycle when \(V_C>V_A\), all the four diodes are reverse biased. Again, no current will flow through \(R_L\).

Hint

(d) In forward biasing, both potential barrier \(V_B\) and the width of charge depleted region \(x\) decreases.

Hint

(a) Here, diode is in reverse biased condition. In reverse biasing, it acts as open circuit, hence no current flows.

Hint

(a) In the given circuit, diode is in reverse biasing, so it acts as open circuit. Hence, potential difference between \(A\) and \(B\) is 6 V.

Hint

(c) The average DC voltage \(\left(V_{D C}\right)\) for a half-wave rectifier is given by the formula:
\(
V_{D C}=\frac{V_0}{\pi}=\frac{100}{3.14} \simeq 31 \mathrm{~V}
\)

Hint

\(
\begin{aligned}
& \text { (c) Energy gap, } E_g=\frac{h c}{\lambda} \\
& \Rightarrow \quad \lambda=\frac{h c}{E_g}=\frac{12400 \mathrm{eV}-Å}{0.72 \mathrm{eV}}=17222 Å
\end{aligned}
\)

Hint

(b) \(D_1\) is reverse biased. Therefore, no current will flow through \(20 \Omega\) resistance. As, \(30 \Omega\) and \(20 \Omega\) resistance are in series, so total current will be \((5 / 50) \mathrm{A}\).

Hint

(b) Voltage across Zener diode is constant,

\(
\begin{gathered}
i_{1 \mathrm{k} \Omega}=\frac{15}{1 \times 10^3}=15 \mathrm{~mA} \\
i_{250 \Omega}=\frac{(20-15)}{250}=\frac{5}{250}=\frac{20}{1000}=20 \mathrm{~mA} \\
\therefore \quad i_{\text {Zener diode }}=(20-15)=5 \mathrm{~mA}
\end{gathered}
\)

Hint

\(
\text { (a) } Y=\overline{\bar{A}+\bar{B}}=A \cdot B \text {, i.e. it is a AND gate. }
\)

Hint

(b) In the positive half cycle of input AC signal diode, \(D_1\) is in forward biased and \(D_2\) is in reverse biased, so in the output voltage signal, \(A\) and \(C\) are due to diode \(D_1\). In negative half cycle of input AC signal, diode \(D_2\) is in forwards biased, hence it conducts, so output signals \(B\) and \(D\) are due to diode \(D_2\).

Hint

(d) The alarm can be triggered by the output of an OR gate. For both the temperatures above \(40^{\circ} \mathrm{C}\) and below \(20^{\circ} \mathrm{C}\).

Hint

(d) If the reverse bias voltage is greater than the \(V_Z\), then there is breakdown condition. In breakdown region, i.e. \(V_i>V_Z\) and for a long range of load \(\left(R_L\right)\), the voltage is constant.

Hint

\(
\text { (b) The potential across } R=17 \mathrm{~V}-9 \mathrm{~V}=8 \mathrm{~V}
\)

Hint

(b) In the given circuit, the Zener diode is used as a voltage regulating device.
Hence, the voltage across \(1 \mathrm{k} \Omega\) is 5 V.
Current flowing through \(1 \mathrm{k} \Omega\) resistor,
\(
I=\frac{5}{1 \times 10^3}=5 \times 10^{-3} \mathrm{~A}=5 \mathrm{~mA}
\)

Hint

(c) In the given circuit, diode \(D_1\) is reverse biased, so it will not conduct. Diodes \(D_2\) and \(D_3\) are forward biased, so they will conduct. The corresponding equivalent circuit is as shown in the figure

The equivalent resistance of the circuit is
\(
R_{e q}=\frac{(5+5) \times 20}{(5+5)+20}=\frac{10 \times 20}{10+20}=\frac{200}{30}=\frac{20}{3} \Omega
\)
Current through the battery, \(I=\frac{10 \mathrm{~V}}{\frac{20}{3} \Omega}=1.5 \mathrm{~A}\)

Hint

(d) Simple diode, zener diode, solar cell, light dependent resistance