NCERT Exemplar Q & A

Hint

(d) We know that \(\sigma=\frac{n e^2 \tau}{m}\),
So, \(\sigma \propto n \tau\)
where, \(n=\) number density and \(\tau=\) relaxation time
In semiconductors, conductivity increases with an increase in temperature, because the number density of current carries increases, relaxation time decreases but effect of decrease in relaxation is much less than increase in number density.

Hint

(b) The height of the potential barrier decreases when the p-n junction is forward biased; it opposes the potential barrier junction. When a p-n junction is reverse biased, it supports the potential barrier junction, resulting increase in potential barrier across the junction.

Hint

(b) A symbol of the diode is represented like this: In this problem first we have to check the polarity of the diodes. -10 V is the lower voltage in the circuit. Now \(p\)-side of \(p\)-n junction \(D_1\) is connected to lower voltage and \(n\)-side of \(D_1\) to higher voltage. Thus \(D_l\) is reverse biased.
Now, let us analyse 2nd diode of the given circuit. The \(p\)-side of \(p\) – \(n\) junction \(D_2\) is at higher potential and \(n\)-side of \(D_2\) is at lower potential. Therefore \(D_2\) is forward biased.
Hence, current flows through the junction from \(B\) to \(A\).

Hint

(d) As p-n junction diode will conduct during positive half cycle only, during negative half cycle diode is reverse biased. During this diode will not give any output. So, potential difference across capacitor \(C=\) peak voltage of the given AC voltage
\(
=V_0=V_{\mathrm{rms}} \sqrt{2}=220 \sqrt{2} \mathrm{~V}
\)

Hint

(b) Concept of holes in the semiconductor:
1. When an electron is removed from a covalent bond, it leaves a vacancy behind. An electron from a neighbouring atom can move into this vacancy, leaving the neighbour with a vacancy. In this way the vacancy formed is called a hole (or cotter), and can travel through the material and serve as an additional current carriers.
2. A hole is considered as a seat of positive charge, having magnitude of charge equal to that of an electron.
3. Holes acts as a virtual charge, although there is no physical charge on it.
4. Effective mass of hole is more than an electron.
5. Mobility of hole is less than an electron.

Hint

(c) When the diode is forward biased during positive half cycle of input AC voltage, the resistance of \(p-n\) junction is low. The current in the circuit is maximum. In this situation, a maximum potential difference will appear across resistance connected in a series of circuit. This result into zero output voltage across p-n junction.
And when the diode is reverse biased during negative half cycle of \(A C\) voltage, the \(p-n\) junction is reverse biased. The resistance of \(p-n\) junction becomes high which will be more than resistance in series. That is why, there will be voltage across p-n junction with negative cycle in output, hence option (c) is correct

Hint

(b) Let us consider the fig. (b) given above in the problem, suppose the potential difference between \(A\) and \(B\) is \(V_{A B}\).
\(
\begin{aligned}
& \text { Then, } \quad V_{A B}-0.3=\left[\left(r_1+r_2\right) 10^3\right] \times\left(0.2 \times 10^{-3}\right) \\
& {\left[\because V_{A B}=i r\right]} \\
& =\left[(5+5) 10^3\right] \times\left(0.2 \times 10^{-3}\right) \\
& =10 \times 10^3 \times 0.2 \times 10^{-3}=2 \\
& \Rightarrow \quad V_{A B}=2+0.3=2.3 \mathrm{~V}
\end{aligned}
\)

Hint

(c) In this problem the input \(C\) of OR gate and which is an output of AND gate. So, ” \(C\) equals \(A\) AND \(B\) ” or \(C=A \cdot B\) and ” \(D\) equals Not \(A\) AND \(B\) ” or \(D=\bar{A} \cdot B\)
and ” \(E\) equals \(C\) AND \(D\) ” or \(E=C+D=(A \cdot B)+(A \cdot B)\)

Hint

(a, c) In valence band electrons are not capable of gaining energy from external electric field. While in conduction band the electrons can gain energy from external electric field. When electric field is applied across a semiconductor, the electrons in the conduction band (which is partially filled with electrons) get accelerated and acquire energy. They move from lower energy level to higher energy level. While the holes in valence band move from higher energy level to lower energy level, where they will be having more energy.

Hint

(a, b, d) On account of difference in concentration of charge carrier in the two sections of P-N junction, the electrons from N-rcgion diffuse through the junction into P-region and the hole from P-region diffuse into N -region.
Due to diffusion, neutrality of both N -and P-type semiconductor is disturbed, a layer of negative charged ions appear near the junction in the P-crystal and a layer of positive ions appears near the junction in N crystal. This layer is called depletion layer.
The thickness of depletion layer is 1 micron \(=10^{-6} \mathrm{~m}\).
Width of depletion layer \(\propto\) 1/Dopping
Depletion is directly proportional to temperature.
Important point: The P-N junction diode is equivalent to capacitor in which the depletion layer acts as a dielectric.

Hint

(b, d) Symbolically zener diode represents like this:
In the forward bias, the zener diode acts as an ordinary diode. It can be used as a voltage regulator.

A zener diode when reverse biases offers constant voltage drop across in terminals as unregulated voltage is applied across circuit to regulate. Then during regulation action of a Zener diode, the current through the series resistance \(R_s\) changes and resistance offered by the Zener changes. The current through the Zener changes but the voltage across the Zener remains constant.

Hint

(a, c, d) Ripple factor may be defined as the ratio of r.m.s. value of the ripple voltage to the absolute value of the DC component of the output voltage, usually expressed as a percentage. However ripple voltage is also commonly expressed as the peak-to-peak value. Ripple factor ( \(r\) ) of a full wave rectifier using capacitor filter is given by
\(
r=\frac{0.236 R}{\omega L}
\)
Where, \(L\) is inductance of the coil and \(\omega\) is the angular frequency.
or Ripple factor can also be given by
\(
\begin{gathered}
r=\frac{1}{4 \sqrt{3} v R_L C_V} \\
\text { i.e., } r \propto \frac{1}{R_L} \Rightarrow r \propto \frac{1}{C}, r \propto \frac{1}{V}
\end{gathered}
\)
Ripple factor is inversely proportional to \(R_L, C\) and \(v\).
Thus to reduce \(r\), \(R_L\) should be increased, input frequency \(v\) should be increased and capacitance \(C\) should be increased.

Hint

(a, d) Reverse biasing: Positive terminal of the battery is connected to the N -crystal and negative terminal of the battery is connected to P-crystal.
(i) In reverse biasing width of depletion layer increases
(ii) In reverse biasing resistance offered \(R_{\text {Reverse }}=10^5 \Omega\)
(iii) Reverse bias supports the potential barrier and no current flows across the junction due to the diffusion of the majority carriers.
(A very small reverse current may exist in the circuit due to the drifting of minority carriers across the junction)
(iv) Break down voltage: Reverse voltage at which break down of semiconductor occurs. For Ge it is 25 V and for Si it is 35 V .
So, we conclude that in reverse biasing, ionization takes place because the minority charge carriers will be accelerated due to reverse biasing and striking with atoms which in turn cause secondary electrons and thus more number of charge carriers.
When doping concentration is large, there will be a large number of ions in the depletion region, which will give rise to a strong electric field.

Hint

Let the electric field be \(E\). The force on an electron is \(e E[latex]. As the electron moves through a distance [latex]d\), the work done on it is \(e E d\). This is equal to the energy transferred to the electron. As the electron travels an average distance of \(4 \times 10^{-8} \mathrm{~m}\) before a collision, the energy transferred is \(e E\left(4 \times 10^{-8} \mathrm{~m}\right)\). To get 1 eV energy from the electric field,
\(
e E\left(4 \times 10^{-8} \mathrm{~m}\right)=1 \mathrm{eV}
\)
\(
E=2.5 \times 10^7 \mathrm{Vm}^{-1}
\)

Hint

(b) The conductivity of the semiconductor is
\(
\sigma=e\left(n_e \mu_e+n_h \mu_h\right)
\)
\(
\begin{aligned}
&\begin{aligned}
&=\left(1.6 \times 10^{-19} \mathrm{C}\right)\left[\begin{array}{rl}
( & 8
\end{array} \times 10^{19} \mathrm{~m}^{-3}\right) \times\left(2.3 \mathrm{~m}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}\right) \\
&\left.+\left(5 \times 10^{18} \mathrm{~m}^{-3}\right) \times\left(10^{-2} \mathrm{~m}^{-2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}\right)\right] \\
& \approx 2.94 \mathrm{Cm}^{-1} \mathrm{~V}^{-1} \mathrm{~s}^{-1}
\end{aligned}\\
&\text { The resistivity is } \rho=\frac{1}{\sigma}=\frac{1}{2 \cdot 94} \mathrm{~m} \mathrm{Vs} \mathrm{C}^{-1} \approx 0 \cdot 34 \Omega \mathrm{~m} \text {. }
\end{aligned}
\)

Hint

(c)

(i) The current at 1 volt is 10 mA and at 1.2 volt it is 15 mA. The dynamic resistance in this region is
\(
R=\frac{\Delta V}{\Delta i}=\frac{0 \cdot 2 \mathrm{volt}}{5 \mathrm{~mA}}=40 \Omega
\)
(ii) The current at 2 volt is 400 mA and at 2.1 volt it is 800 mA. The dynamic resistance in this region is
\(
R=\frac{\Delta V}{\Delta i}=\frac{0.1 \mathrm{volt}}{400 \mathrm{~mA}}=0.25 \Omega
\)

Hint

(d) (i) The impurity provides impurity levels close to the conduction band and a number of electrons from the impurity level will populate the conduction band. Thus, the majority carriers are electrons and the material is \(n\)-type.
(ii) According to the question, \(k T=30 \mathrm{meV}\)
\(
\begin{aligned}
T & =\frac{30 \mathrm{meV}}{k} \\
& =\frac{0.03 \mathrm{eV}}{8.62 \times 10^{-5} \mathrm{eV} \mathrm{~K}^{-1}}=348 \mathrm{~K}
\end{aligned}
\)

Hint

(b)

(i) The energy of the photon is \(E=\frac{h c}{\lambda}\)
\(
=\frac{1242 \mathrm{eV} \mathrm{~nm}}{589 \mathrm{~nm}}=2 \cdot 1 \mathrm{eV} .
\)
Thus the band gap is \(2 \cdot 1 \mathrm{eV}\). This is also the minimum energy \(E\) required to push an electron from the valence band into the conduction band. Hence, the minimum energy required to create a hole-electron pair is \(2 \cdot 1 \mathrm{eV}\).
(ii) At
\(
\begin{aligned}
T & =300 \mathrm{~K}, \\
k T & =\left(8.62 \times 10^{-5} \mathrm{eV} \mathrm{~K}^{-1}\right)(300 \mathrm{~K}) \\
& =25.86 \times 10^{-3} \mathrm{eV}
\end{aligned}
\)
Thus, \(\quad \frac{E}{k T}=\frac{2.1 \mathrm{eV}}{25.86 \times 10^{-3} \mathrm{eV}}=81\).
So it is difficult for the thermal energy to create the hole-electron pair but a photon of light can do it easily.

Hint

(d)

To create a hole, an electron from the valence band should be given sufficient energy to go into one of the acceptor levels. Since the acceptor levels are 57 meV above the valence band, at least 57 meV is needed to create a hole.
If \(\lambda\) be the wavelength of light, its photon will have an energy \(h c / \lambda\). To create a hole,
\(
\frac{h c}{\lambda} \geq 57 \mathrm{meV}
\)
\(
\begin{aligned}
\lambda & \leq \frac{h c}{57 \mathrm{meV}} \\
& =\frac{1242 \mathrm{eV} \mathrm{~nm}}{57 \times 10^{-3} \mathrm{eV}}=2.18 \times 10^{-5} \mathrm{~m}
\end{aligned}
\)

Hint

(c) The number of charge carriers in an intrinsic semiconductor is double the number of hole-electron pairs. If \(N_1\) be the number of charge carriers at temperature \(T_1\) and \(N_2\) at \(T_2\), we have
\(
N_1=N_0 e^{-\Delta E / 2 k T_1}
\)
and \(\quad N_2=N_0 e^{-\Delta E / 2 k T_2}\).
The percentage increase as the temperature is raised from \(T_1\) to \(T_2\) is
\(
\begin{aligned}
f & =\frac{N_2-N_1}{N_1} \times 100=\left(\frac{N_2}{N_1}-1\right) \times 100 \\
& =100\left[e^{\frac{\Delta E}{2 k}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)-1}\right]
\end{aligned}
\)
\(
\frac{\Delta E}{2 k}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)
\)
\(
\begin{aligned}
& =\frac{0.68 \mathrm{eV}}{2 \times 8.62 \times 10^{-5} \mathrm{eV} \mathrm{~K}^{-1}}\left(\frac{1}{300 \mathrm{~K}}-\frac{1}{320 \mathrm{~K}}\right) \\
& =0.82
\end{aligned}
\)
Thus, \(\quad f=100 \times\left[e^{0.82}-1\right] \approx 127\).
Thus, the number of charge carriers increases by about \(127 \%\).

Hint

(b) The number of charge carriers before doping is equal to the number of holes plus the number of conduction electrons. Thus, the number of charge carriers per cubic metre before doping
\(
=2 \times 7 \times 10^{15}=14 \times 10^{15}
\)
Since antimony is doped in a proportion of 1 in \(10^7[latex], the number of antimony atoms per cubic metre is [latex]10^{-7} \times 5 \times 10^{28}=5 \times 10^{21}\). As half of these atoms contribute electrons to the conduction band, the number of extra conduction electrons produced is \(2.5 \times 10^{21}\) per cubic metre. Thus, the number of charge carriers per cubic metre after the doping is
\(
\begin{aligned}
& 2.5 \times 10^{21}+14 \times 10^{15} \\
\approx & 2.5 \times 10^{21} .
\end{aligned}
\)
The factor by which the number of charge carriers is increased
\(
=\frac{2.5 \times 10^{21}}{14 \times 10^{15}}=1.8 \times 10^5 .
\)
In fact, as the \(n\)-type impurity is doped, the number of holes will decrease. This is because the product of the concentrations of holes and conduction electrons remains almost the same. However, this does not affect our result as the number of holes is anyway too small as compared to the number of conduction electrons.

Hint

(a)

\(
\begin{aligned}
&\text { (i) The electric field is } E=V / d\\
&=\frac{0.50 \mathrm{~V}}{5.0 \times 10^{-7} \mathrm{~m}}=1.0 \times 10^6 \mathrm{Vm}^{-1}
\end{aligned}
\)
(ii)

Suppose the electron has a speed \(v_1\) when it enters the depletion layer and \(v_2\) when it comes out of it (figure above). As the potential energy increases by \(e \times 0.50 \mathrm{~V}\), from the principle of conservation of energy,
\(
\frac{1}{2} m v_1^2=e \times 0.50 \mathrm{~V}+\frac{1}{2} m v_2^2
\)
\(
\begin{aligned}
\frac{1}{2} \times(9.1 & \left.\times 10^{-31} \mathrm{~kg}\right) \times\left(5.0 \times 10^5 \mathrm{~m} \mathrm{~s}^{-2}\right)^2 \\
& =1.6 \times 10^{-19} \times 0.5 \mathrm{~J}+\frac{1}{2}\left(9.1 \times 10^{-31} \mathrm{~kg}\right) v_2^2
\end{aligned}
\)
\(
\begin{aligned}
1.13 \times 10^{-19} \mathrm{~J}=0.8 & \times 10^{-19} \mathrm{~J} \\
& +\left(4.55 \times 10^{-31} \mathrm{~kg}\right) v_2^2
\end{aligned}
\)
Solving this, \(v_2=2.7 \times 10^5 \mathrm{~m} \mathrm{~s}^{-1}\).

Hint

(d) The reverse-biased current is caused mainly due to the drift current. The drift current in a \(p-n\) junction is caused by the formation of new hole-electron pairs and their subsequent motions in the depletion layer. When the junction is exposed to light, it may absorb energy from the light photons. If this energy supplied by a photon is greater than (or equal to) the band gap, a hole-electron pair may be formed. Thus, the reverse-biased current will increase if the light photons have energy greater than (or equal to) the band gap. Hence the band gap is equal to the energy of a photon of 600 nm light which is
\(
\frac{h c}{\lambda}=\frac{1242 \mathrm{eV} \mathrm{~nm}}{600 \mathrm{~nm}}=2.07 \mathrm{eV} .
\)

Hint

(c) Electric conduction in a semiconductor takes place due to both electrons and holes.

Explanation: In semiconductors, current is carried by both free electrons (negatively charged) and “holes” (which are effectively positive charge carriers representing the absence of an electron).

Key points:
Electrons: Electrons can become free to move when they gain enough energy to jump from the valence band to the conduction band.
Holes: When an electron leaves its position in the valence band, it creates a “hole” (a positively charged vacancy) that can also move as other electrons fill that space, creating a new hole.
Therefore, both electrons and holes contribute to the electric current flow in a semiconductor.

Hint

(b) As we increase the temperature, additional electron-hole pairs are created in the semiconductor.
As a result, the number of charge carriers ( \(n\) ) increases.
Now, drift velocity ( \(\mathrm{v}_{\mathrm{d}}\) ) is given by
\(
\mathrm{v}_{\mathrm{d}}=\mathrm{eE} \tau / \mathrm{m}
\)
As temperature \(\uparrow\) relaxation time \((\tau) \downarrow \therefore \mathrm{vd} \uparrow\)

Explanation:

Increased number of charge carriers ( \({n}\) ): When the temperature of a semiconductor increases, more electrons gain enough energy to break free from their bonds, creating more electron-hole pairs. This leads to an increase in the number of charge carriers ( \(n\) ).

Decreased average drift speed (\(v\)): However, with more charge carriers present, they also collide more frequently with each other and the atoms in the semiconductor lattice, which slows down their average drift speed (v).

Hint

(b) For an intrinsic semiconductor, the number of holes \(\left(n_p\right)\) is equal to the number of conduction electrons ( \(n_e\) ). Therefore, the correct answer is (b) \(n_p=n_e\).

Explanation: In an intrinsic semiconductor, electrons are excited from the valence band to the conduction band, leaving behind holes in the valence band. Since each electron that moves to the conduction band creates a corresponding hole, the number of electrons in the conduction band will always be equal to the number of holes.

Hint

(d) In an extrinsic semiconductor, the number of holes \(\left(n_p\right)\) and conduction electrons ( \(n_e\) ) are not equal, meaning \(n_p \neq n_e\). Extrinsic semiconductors are created by doping a semiconductor with impurities, which alters the number of charge carriers. In N-type semiconductors, there are more electrons than holes, and in P-type semiconductors, there are more holes than electrons.

Hint

(c) A p-type semiconductor is uncharged.
Explanation: While a p-type semiconductor has an abundance of “holes” (which act as positive charge carriers due to the doping process), it remains overall electrically neutral. This is because the number of protons in the semiconductor’s atoms still equals the number of electrons, maintaining overall charge neutrality. The presence of holes just affects its electrical conductivity properties, not its overall charge.

Hint

(a) When an impurity is doped into an intrinsic semiconductor, the conductivity of the semiconductor increases.

Explanation: Doping adds extra charge carriers to the semiconductor, which significantly enhances its ability to conduct electricity.
Key points about doping:
N -type doping: Adds electrons to the semiconductor by introducing a donor impurity like arsenic.
P-type doping: Adds holes to the semiconductor by introducing an acceptor impurity like boron.

Hint

(a) If the two ends of a p-n junction are joined by a wire, there will not be a steady current in the circuit.

Explanation: A p-n junction has a built-in potential barrier that prevents the flow of current when no external voltage is applied. When the two ends are connected by a wire, there is no external voltage to overcome this barrier, so the diffusion current of majority carriers from the \(p[latex]-side to the [latex]n\)-side is balanced by the drift current of minority carriers from the \(n\)-side to the \(p\) side, resulting in no net current.

Hint

(a) from the \(n\)-side to the \(p\)-side.
Explanation:
Drift current: This current is caused by the movement of charge carriers (electrons and holes) due to an electric field.
In a p-n junction:
The electric field is created by the depletion region, which forms when the p-type and n-type semiconductors are joined.
Direction: The drift current always moves from the \(n\)-side to the \(p\)-side due to the electric field pushing electrons from the \(n\)-side and holes from the \(p\)-side.

Hint

(b) Due to concentration differences, holes try to diffuse from the \({p}\)-side to the n-side.
But the electric field at the junction exerts a force on the holes towards the left as they come to the depletion larger.
Only those holes which start moving towards the right with high kinetic energy are able to cross the junction. Similarly, diffusion of electrons takes place from right to left.
This diffusion results in an electric current from the \({p}\)-side to the \({n}\)-side known as diffusion current.

Diffusion Current: This current arises due to the concentration gradient of charge carriers (electrons and holes) across the p-n junction.

Hint

(a) if the junction is forward-biased. Here’s why:
Explanation:
Forward Bias: When a p-n junction is forward-biased, the positive terminal of the battery is connected to the \(p\)-side, and the negative terminal is connected to the \(n\)-side. This reduces the potential barrier (depletion region), allowing more majority charge carriers (electrons in the \(n\)-side and holes in the \(p\)-side) to move across the junction. This increased movement of majority carriers results in a larger diffusion current.

Diffusion Current: Diffusion current arises from the movement of charge carriers from regions of higher concentration to regions of lower concentration. In a forward-biased p-n junction, the majority carriers are pushed towards the junction, leading to a high diffusion current.

Drift Current: Drift current is caused by the movement of charge carriers due to an electric field. In forward bias, the electric field created by the applied voltage opposes the diffusion of majority carriers but is much weaker than the diffusion force in forward bias, and the drift current is smaller than the diffusion current.

Hint

(b) The potential difference across the two p-n junctions are equal in circuit 2 and circuit 3.

Explanation:
Circuit 2: In this circuit, both p-n junctions are forward-biased. When a diode is forward-biased, it offers very little resistance, effectively acting like a closed switch. Therefore, the potential difference across each junction will be nearly equal and determined by the voltage of the battery and the forward resistance of the diodes.

Circuit 3: In this circuit, both p-n junctions are reverse-biased. When a diode is reverse-biased, it blocks current flow, effectively acting like an open switch. However, there will be a small leakage current that causes a voltage drop across each junction. Since both junctions are reverse-biased in the same way, the voltage drops will be equal.

Circuit 1: In this circuit, one junction is forward-biased and the other is reverse-biased. This results in unequal voltage drops across the two junctions, as the forward-biased diode offers very low resistance and the reverse-biased diode blocks current flow.

Hint

(b) The charge on a discharging capacitor is given by \(Q(t)=Q_0 e^{-t / \tau}\).
The time constant for an RC circuit is \(\tau=R C\).
A forward-biased diode acts as a short circuit.
A reverse-biased diode acts as an open circuit.
Analyze circuit A.
Identify the diode’s bias: The diode in circuit A is forward-biased.
Determine the diode’s behavior: It acts as a short circuit.
Calculate the initial charge: \(Q_A(0)=C V\).
Determine the charge at \(t=C R: Q_A(C R)=Q_A(0) e^{-C R / C R}=C V e^{-1}=\frac{C V}{e}\).
Analyze circuit B.
Identify the diode’s bias: The diode in circuit B is reverse-biased.
Determine the diode’s behavior: It acts as an open circuit.
Determine the charge at \(t=C R\) : Since no current flows, the charge remains constant at its initial value, \(Q_B(C R)=C V\).
The charges on capacitors A and B at time \(t=C R\) are \(\frac{V C}{e}\) and \(V C\) respectively.

Hint

(c) A hole diffuses from the p side to the n side in a \(\mathrm{p}-\mathrm{n}\) junction; that is, an electron moves from the n side to the p side. This implies that a bond is broken on the n side. As the electron travels towards the p side, which is rich in holes, it combines with a hole. A hole is created because of the deficiency of one electron. So, when an electron combines with a hole, it completes that bond.

Hint

(a, c, d) In semiconductors, the valence band is full at 0 K, but the conduction band is empty. So, no free electron is available for conduction at 0 K.
As the temperature increases, covalent bonds that provide free charge carriers for conduction in a semiconductor break.
Conductors have many free electrons in the conduction band, even at low temperatures, because their valence and conduction bands are very close or overlap, while semiconductors have a larger energy gap.

Hint

(b) holes and conduction electrons systematically go from the p -side to n -side and from the n -side to p -side, respectively
(c) there is no net charge transfer between the two sides
(d) there is a constant electric field near the junction

Because of the difference in the concentration of charge carriers in the \(\mathrm{p}-\mathrm{n}\) junction, holes from the p side move to the n side and electrons from the n side move to the pside. This motion of charge carriers gives rise to diffusion current. Because of this, a negative space charge region is formed in the p region and a positive space region is formed in the n region. This sets up an electric field across the junction. Thus, there is a constant electric field near the junction. This electric field further opposes the diffusion of majority charge carriers across the junction. As a result, an electron from the p region starts moving to the n region and a hole from the n region starts moving to the p region. This sets up drift current. Thus, there is a systematic flow of charge carriers across the junction. Also, there is no net charge transfer between the two sides.

Hint

(a) new holes and conduction electrons are produced continuously throughout the material
(d) holes and conduction electrons recombine continuously throughout the material except in the depletion region

In a \(\mathrm{p}-\mathrm{n}\) junction diode, diffusion current flows because of the diffusion of holes from the p side to the \({\mathrm{n}}\) side and of electrons from the n side to the p side. The current flowing in the diode due to the diffusion of charge carriers across the junction is called the diffusion current. The current flowing in the diode due to the movement of minority carriers across the junction due to their thermal energy is called the drift current. In an unbiased diode, the net current flowing across the junction is zero due to the cancellation of the drift current by the diffusion current. For the flow of diffusion and drift currents, holes and electrons are produced continuously throughout the material. When a hole crosses the junction, it combines with an electron on the n side. As the depletion region is devoid of free charge carriers, this recombination never takes place inside the depletion region.

Hint

(b, d) A p-type semiconductor is formed by doping an intrinsic semiconductor with a trivalent atom (atom having valency 3 ).
Boron (B): Group 13, suitable for creating p-type semiconductors.
Aluminium (AI): Also in group 13, effective for p-type doping.

Hint

(a) increasing the temperature
(b) doping acceptor impurities
(c) doping donor impurities
(d) irradiating ultraviolet light on it

We know that the conductivity of any semiconductor can be increased by increasing the number of charge carriers. All the given methods are effective in increasing the number of free charge carriers. Hence, all options are correct.

Hint

(d) As a p-n junction allows the flow of current in forward bias and stops the current in reverse bias (almost negligible reverse leakage current flows in the reverse-biassed \(\mathrm{p}-\mathrm{n}\) junction), the device should be a \(\mathrm{p}-\mathrm{n}\) junction. Other options are examples of semiconductors that allow moderate current to flow and that do not have any effect of changing the polarity of the battery.

Hint

(b, c) The electron concentration increases due to the additional free electrons provided by the donor atoms.
The hole concentration decreases as a consequence of the increased electron concentration, maintaining the overall charge neutrality of the semiconductor.
Electron Contribution
Out of the five electrons from the donor atom, four are used to form covalent bonds with neighboring silicon atoms. The fifth electron is loosely bound and can easily be excited into the conduction band, thus increasing the electron concentration in the semiconductor. This can be expressed mathematically as:
\(
n=n_0+N_d
\)
where \(n\) is the electron concentration, \(n_0\) is the intrinsic electron concentration, and \(N_d\) is the concentration of donor atoms.
Impact on Hole Concentration
As the electron concentration increases due to the introduction of donor impurities, the total number of charge carriers (electrons and holes) remains constant in a neutral semiconductor. Therefore, when the number of electrons increases, the number of holes must decrease to maintain charge neutrality. This relationship can be represented by:
\(
p=\frac{n_i^2}{n}
\)
where \(p\) is the hole concentration and \(n_i\) is the intrinsic carrier concentration.
As \(n\) increases, \(p\) must decrease.

Hint

(c, d) NAND and NOR gates are universal gates and any other logic can be made by proper repetitive use of these gates.

Hint

(a) The energy band gap is given by \(E_g=h \nu=\frac{h c}{\lambda}\) where, \(h=\) Planck’s constant, \(c=\) speed of light and \(\lambda=\) wavelength at which solid absorbs energy.
On putting the values of \(h, c\) and \(\lambda\), we get
\(
\begin{aligned}
E_g & =\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^8\right)}{\left(10000 \times 10^{-10}\right)} \\
& =1.98 \times 10^{-19} \mathrm{~J}=\frac{1.98 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}=1.24 \mathrm{eV}
\end{aligned}
\)

Hint

(b) In an intrinsic semiconductor, \(n_e=n_h\) where, \(n_e=\) number of conduction electrons per unit volume and \(\quad n_h=\) number of holes per unit volume.
Given, \(\quad n_e=7 \times 10^{19}\) per \(\mathrm{m}^3\)
\(
\therefore \quad n_h=n_e=7 \times 10^{19} \text { per } \mathrm{m}^3
\)
So, total current carrier density
\(
n_e+n_h=7 \times 10^{19}+7 \times 10^{19}=14 \times 10^{19} \text { per } \mathrm{m}^3
\)
\(
\begin{aligned}
& \text { Now, total number of current carrier } \\
& \qquad \begin{array}{l}
=\text { Number density } \times \text { volume } \\
=\left(14 \times 10^{19}\right) \times\left(10^{-2} \times 10^{-2} \times 10^{-3}\right) \\
=1.4 \times 10^{13}
\end{array}
\end{aligned}
\)

Hint

(c) Given, \(n_i=1.5 \times 10^{16} \mathrm{~m}^{-3}, n_h=3 \times 10^{22} \mathrm{~m}^{-3}\)
For the semiconductor in thermal equilibrium, \(n_e n_h=n_i^2\) (Law of mass action)
\(
\Rightarrow \quad n_e=\frac{n_i^2}{n_h}=\frac{\left(1.5 \times 10^{16}\right)^2}{3 \times 10^{22}}=7.5 \times 10^9 \mathrm{~m}^{-3}
\)

Hint

(d) Here, number of conduction electrons, \(n_e=6.5 \times 10^{19} \mathrm{~m}^{-3}\) Volume of the sample, \(V=1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 2 \mathrm{~mm}=2 \times 10^{-7} \mathrm{~m}^3\) Number of holes in the sample \(\left(n_h\right)=\) Number of electrons in the sample \(\left(n_e\right)\)
\(
\begin{aligned}
& =n_e \times V \\
& =6.5 \times 10^{19} \times 2 \times 10^{-7} \\
& =1.3 \times 10^{13}
\end{aligned}
\)

Hint

(a) To create a hole, an electron of valence band has to be excited into one of the acceptor levels, which are 50 meV above the valence band. Hence, a minimum of 50 meV energy is needed to create a hole.
As,
\(
\begin{aligned}
E_{\min } & =\frac{h c}{\lambda_{\max }} \\
\lambda_{\max } & =\frac{h c}{E_{\min }}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{50 \times 1.6 \times 10^{-22}} \\
& =2.47 \times 10^{-5} \mathrm{~m}
\end{aligned}
\)

Hint

(b) Here, \(n_i=1.5 \times 10^{16} \mathrm{~m}^{-3}\)
Doping concentration of pentavalent As atoms
\(
=1 \mathrm{ppm}=1 \text { part per million }
\)
\(\therefore\) Number density of pentavalent as atoms,
\(
N_D=\frac{5 \times 10^{28}}{10^6}=5 \times 10^{22} \text { atom m }^{-3}
\)
Now, the thermally generated electrons ( \(n_i \propto 10^{16} \mathrm{~m}^{-3}\) ) are negligibly small as compared to those produced by doping, so
\(
n_e \simeq N_D=5 \times 10^{22} \mathrm{~m}^{-3}
\)
Also, \(\quad n_e n_h=n_i^2\) (Law of mass action)
\(
\therefore \quad n_h=\frac{n_i^2}{n_e}=\frac{1.5 \times 10^{16} \times 1.5 \times 10^{16}}{5 \times 10^{22}}=4.5 \times 10^9 \mathrm{~m}^{-3}
\)

Hint

(d) Conductivity of pure silicon,
\(
\sigma=n_i e \mu_e+n_i e \mu_h=n_i e\left(\mu_e+\mu_h\right)
\)
where, \(n_i\) is the total number of charge carrier, \(\mu_e\) and \(\mu_h\) are the mobility of electrons and holes, respectively.
Here,
\(
\begin{gathered}
n_i=1.072 \times 10^{10} \text { per } \mathrm{cm}^3 \\
\mu_h=1350 \mathrm{~cm}^2 / \mathrm{V}-\mathrm{s} \\
\mu_c=480 \mathrm{~cm}^2 / \mathrm{V}-\mathrm{s}
\end{gathered}
\)
\(
\begin{aligned}
\Rightarrow \quad \sigma & =\left(1.072 \times 10^{10}\right)\left(1.6 \times 10^{-19}\right)(1350+480) \\
& =3.14 \times 10^{-6} \mathrm{mho} / \mathrm{cm}
\end{aligned}
\)

Hint

(c) In \(n\)-type semiconductor, \(n_e \gg n_h\)
Let number density of donor atoms, \(N_0=n_e\)
Conductivity of semiconductor is given by \(\sigma=n_e e \mu_e\)
On putting the values, we get
\(
\begin{aligned}
& & 6 \times 10^2 & =n_e \times 1.6 \times 10^{-19} \times 3850 \times 10^{-4} \\
\Rightarrow & & n_e & =9.7 \times 10^{21} / \mathrm{m}^3
\end{aligned}
\)

Hint

(b) In pure semiconductor, electron-hole pair concentration, \(n_h=n_e=7 \times 10^{15} \mathrm{~m}^{-3}\)
Total charge carrier,
\(
\begin{aligned}
n_{\text {total initial }} & =n_h+n_e=7 \times 10^{15}+7 \times 10^{15} \\
& =14 \times 10^{15}
\end{aligned}
\)
After doping, donor impurity,
\(
\begin{aligned}
& N_D=\frac{5 \times 10^{28}}{10^7}=5 \times 10^{21} \\
& n_e^{\prime}=\frac{N_D}{2}=2.5 \times 10^{21}
\end{aligned}
\)
\(
\begin{aligned}
& n_{\text {final }}=n_h+n_e^{\prime} \\
& n_{\text {final }} \approx n_e^{\prime} \approx 2.5 \times 10^{21} \quad\left(\because n_e^{\prime} \gg n_h\right)
\end{aligned}
\)
\(
\begin{aligned}
\text { Factor } & =\frac{n_{\text {final }}-n_{\text {initial }}}{n_{\text {initial }}} \\
& =\frac{2.5 \times 10^{21}-14 \times 10^{15}}{14 \times 10^{15}} \\
& \approx \frac{2.5 \times 10^{21}}{14 \times 10^{15}}=1.8 \times 10^5
\end{aligned}
\)

Hint

(c) The conductivity of a pure silicon is given by
\(
\sigma_i=n_i e\left(\mu_e+\mu_h\right)
\)
Also, the resistivity is reciprocal of conductivity.
\(
\begin{array}{ll}
\therefore & \rho_i=\frac{1}{\sigma_i}=\frac{1}{n_i e\left(\mu_e+\mu_h\right)} \\
\Rightarrow & n_i=\frac{1}{\rho_i e\left(\mu_e+\mu_h\right)}
\end{array}
\)
\(
\begin{aligned}
& =\frac{1}{3000 \times 1.6 \times 10^{-19}(0.15+0.030)} \\
& =1.157 \times 10^{16} / \mathrm{m}^3
\end{aligned}
\)
When \(10^{19}\) atoms of phosphorus (donor atoms) are added per \(\mathrm{m}^3\), we have
\(
\begin{array}{cc}
& n_e \gg n_i \\
\text { or } & n_e \gg n_h \\
\Rightarrow & n_e=10^{19}
\end{array}
\)
\(
\begin{aligned}
\rho & =\frac{1}{n_e e \mu_e} \\
& =\frac{1}{10^{19} \times 1.6 \times 10^{-19} \times 0.15} \\
& =4.17 \Omega-\mathrm{m}
\end{aligned}
\)

Hint

(a) Conductivity of semiconductor is given by
\(
\sigma=n_e e \mu_e+n_h e \mu_h
\)
Since semiconductor is intrinsic, so \(n_e=n_h=n_i\)
Conductivity,
\(
\begin{aligned}
\sigma & =n_i e\left(\mu_e+\mu_h\right) \\
& =1.6 \times 10^6 \times 1.6 \times 10^{-19}(0.4+0.2) \\
& =1.536 \times 10^{-13} \Omega^{-1} \mathrm{~m}^{-1}
\end{aligned}
\)
Current flowing, \(I=J A\)
where, \(J=\) current density \(=\sigma E=\sigma\left(\frac{V}{d}\right)\)
\(
\begin{aligned}
\therefore \quad I & =J A=\sigma\left(\frac{V}{d}\right) \cdot A \\
& =1.536 \times 10^{-13}\left(\frac{5}{1.5 \times 10^{-3}}\right)\left(3.5 \times 10^{-4}\right) \\
& =1.79 \times 10^{-13} \mathrm{~A}
\end{aligned}
\)
Heat produced, \(H=V I t=5 \times 1.79 \times 10^{-13} \times 120\)
\(
=10.74 \times 10^{-11} \mathrm{~J}
\)

Hint

(d) Conductivity of semiconductor is given by
\(
\begin{aligned}
\sigma & =n_e e \mu_e+n_h e \mu_h \\
n_e & =n_h=n_i \quad \text { (pure semiconductor) } \\
\sigma & =n_i e\left(\mu_e+\mu_h\right)=e\left(\mu_e+\mu_h\right) n_i \\
& =e\left(\mu_e+\mu_h\right) n_0 e^{-E_g / 2 k T} \\
\therefore \quad \frac{\sigma_{400}}{\sigma_{200}} & =\frac{e^{\frac{-E_g}{2 k \times 400}}}{\frac{-E_g}{e^{2 k \times 200}}}=e^{E_g / 800 k} \\
& =e^{1.12 \times 1.6 \times 10^{-19} / 800 \times 1.38 \times 10^{-23}}=e^{16.23}
\end{aligned}
\)

Hint

(a) The expected energy of electrons at absolute zero is called the Fermi energy. 

Explanation:
Fermi energy:
At absolute zero, all electrons occupy the lowest energy levels available. The highest energy level that an electron can occupy at absolute zero is called the Fermi energy. It’s the energy level at which electrons are most likely to be found at 0 K.

Hint

(a) The solid sample is a conductor.
Explanation: A conductor has a partially filled conduction band at 0 K because electrons can readily move within the band, allowing for electrical conduction even at absolute zero temperature.

Why other options are incorrect:
(b) Semiconductor: In a semiconductor, the conduction band is mostly empty at 0 K . At higher temperatures, some electrons gain enough energy to jump into the conduction band, enabling conductivity.
(c) Insulator: In an insulator, the valence band and conduction band are completely separated by a large energy gap. There are no electrons in the conduction band at any temperature, so it cannot conduct electricity.

Hint

(a) If the valence band and conduction band of a solid overlap at low temperature, the solid is likely to be a metal.

Explanation: In metals, the valence band and conduction band overlap, allowing electrons to easily move between them, which leads to high conductivity.

Hint

(d) In insulators, the forbidden energy gap is largest and it is of the order of 5 eV to 15 eV.

Hint

(b) In insulators, the forbidden energy band gap is very large, in case of semiconductor, it is moderate and in conductors, the energy gap is zero because in conductor, valence band and conduction band overlap to each other.

Hint

(b) Doping of intrinsic semiconductors is done to increase the concentration of majority charge carriers.

Explanation: Doping involves adding impurities to a pure semiconductor to alter its electrical properties by introducing either more electrons (n-type doping) or more holes (p-type doping), thereby increasing the concentration of the majority charge carrier.

Hint

(c) holes.
Explanation:
In a p-type semiconductor, the majority charge carriers are holes. A “hole” is considered a positive charge carrier because it represents the absence of an electron in the semiconductor’s crystal lattice. When an electron moves away from a particular location, it leaves behind a “hole” that can then be occupied by another electron, effectively allowing positive charge to move through the material.

Hint

(b) A \(p\)-type semiconductor is formed when a pure semiconductor like silicon or germanium is doped with a trivalent element (one with 3 valence electrons). Of the given options, (b) gallium to pure silicon is the correct answer because gallium is a trivalent element. Doping with a trivalent element creates “holes” in the semiconductor’s crystal lattice, which act as charge carriers, making it a p-type semiconductor, according to semiconductor physics experts.

Hint

(a) Aluminium is trivalent impurity which is doped with pure semiconductor to form \(p\)-type semiconductor.

Hint

(d) In a p-type semiconductor, the majority charge carriers are holes and the minority charge carriers are electrons.

Explanation:
p-type semiconductor:
In a p-type semiconductor, dopant atoms are added that create “holes” in the valence band, allowing for positive charge carriers to move through the material.

Majority and minority carriers:
The majority carriers are the most prevalent charge carriers in the semiconductor, which in this case are holes. The minority carriers are the less prevalent charge carriers, which in this case are electrons.

Hint

(b) The charge on hole is positive which is equivalent to charge on proton.

Hint

(b) For \(n[latex]-type semiconducting material, number of free electrons is very large than number of holes, i.e. [latex]n_e \gg n_h\).

Hint

(b) Both \(n\)-type and \(p\)-type germanium are neutral.

Explanation:
\(n\)-type germanium: While \(n\)-type germanium has extra electrons (negative charge carriers) due to doping, it remains electrically neutral because it also has an equal number of protons.
\(p\)-type germanium: Similarly, p-type germanium has “holes” (positive charge carriers) but is still overall neutral due to the presence of an equal number of electrons.

Hint

(a) When a semiconductor is heated, its resistance decreases.
Explanation: As temperature increases in a semiconductor, more electrons gain enough energy to jump from the valence band to the conduction band, which increases the number of free charge carriers and thus decreases the resistance.

Hint

(d) Temperature coefficient of semiconductor is negative.

Copper, aluminium, iron are conductors, while germanium is semiconductor.

Explanation: A negative temperature coefficient of resistance means that as the temperature of a material increases, its resistance decreases. Semiconductors like germanium typically have a negative temperature coefficient, while metals like copper, aluminum, and iron have a positive temperature coefficient, meaning their resistance increases with temperature.

Hint

(c) (i) Width of depletion layer, \(d=5 \times 10^{-7} \mathrm{~m}\)
\(
\text { Electric field, } E_B=\frac{V}{d}=\frac{0.5}{5 \times 10^{-7}}=10^6 \mathrm{Vm}^{-1}
\)
\(
\begin{aligned}
&\text { (ii) According to work-energy theorem, }\\
&\begin{aligned}
& \frac{1}{2} m_e v_i^2=e V+\frac{1}{2} m_e v_f^2 \\
& \text { Speed, } v_f=\sqrt{\frac{m_e v_i^2-2 e V}{m_e}} \\
& =\sqrt{\frac{9 \times 10^{-31} \times\left(5 \times 10^5\right)^2-2 \times 1.6 \times 10^{-19} \times(0.5)}{9 \times 10^{-31}}} \\
& =2.7 \times 10^5 \mathrm{~ms}^{-1}
\end{aligned}
\end{aligned}
\)

Hint

(c) Current through diode (or circuit), \(I=\frac{\text { Power }}{\text { Voltage }}\)
\(
\begin{aligned}
& \therefore \quad I=\frac{200 \times 10^{-3} \mathrm{~W}}{0.5 \mathrm{~V}}=0.4 \mathrm{~A} \\
& \text { and resistance, } R=\frac{\text { Net voltage }}{\text { Current }}=\frac{2-0.5}{0.4}=3.75 \Omega
\end{aligned}
\)

Hint

(d) The given circuit is in the form of a Wheatstone bridge,
Hence no current will from through diode, i.e. it will not offer any resistance. Hence, net resistance will be
\(
R_N=\frac{R_S \times R_S^{\prime}}{R_S+R_S^{\prime}}
\)
where \(R_S\) and \(R_S^{\prime}=(10+10)=20 \Omega\)
\(
\Rightarrow \quad R_N=10 \Omega
\)

Hint

(a) (i) Ge diode will start conducting before the silicon diode does so. The effective forward voltage across Ge diode is \((12-0.3) V=11.7 \mathrm{~V}\). This will appear as the output voltage across the load, i.e.
\(
V_o=11.7 \mathrm{~V}
\)
The current through \(R_L\),
\(
\begin{aligned}
i & =\frac{11.7}{5 \times 10^3} \mathrm{~A} \\
& =2.34 \mathrm{~mA}
\end{aligned}
\)
(ii) On reversing the connection of Ge diode, it will be reverse biased and conduct no current. Only Si diode will conduct. Therefore,
\(
\begin{aligned}
\qquad V_o & =(12-0.7) \mathrm{V}=11.3 \mathrm{~V} \\
\text { and current, } i & =\frac{11.3}{5 \times 10^3} \mathrm{~A}=2.26 \mathrm{~mA}
\end{aligned}
\)

Hint

(a) Rectification efficiency is the ratio of DC output power to the AC input power, i.e.
\(
\eta=\frac{\mathrm{DC} \text { output power }}{\mathrm{AC} \text { input power }} \times 100
\)
Here, DC output power \(=20 \mathrm{~W}\)
\(
\mathrm{AC} \text { input power }=60 \mathrm{~W}
\)
\(\therefore\) Rectification efficiency, \(\eta=\frac{20}{60} \times 100=33.3 \%\)
Also, power efficiency is given by
\(
\begin{aligned}
\mathrm{PE} & =\frac{\mathrm{DC} \text { output power }}{\mathrm{AC} \text { input power for half cycle }} \times 100 \% \\
& =\frac{20}{30} \times 100 \%=66.67 \%
\end{aligned}
\)

Hint

(b) Here, \(V=(50 \sin \omega t)\) volt
Comparing it with general equation \(V=V_0 \sin \omega t\), we get
\(
V_0=50 \text { volt, } R_L=650 \Omega, r_f=200 \Omega
\)
(i) Maximum output current, \(I_0=\frac{V_0}{r_f+R_L}\)
\(
=\frac{50}{(200+650)}=58 \mathrm{~mA}
\)
(ii) DC output current, \(I_{\mathrm{DC}}=\frac{I_0}{\pi}=\frac{58}{\pi}=18.5 \mathrm{~mA}\)
(iii) DC output power,
\(
=I_{\mathrm{DC}}^2 \cdot R_L=\left(18.5 \times 10^{-3}\right)^2 \cdot 650=0.22 \mathrm{~W}
\)
(iv) DC output voltage,
\(
=I_{\mathrm{DC}} \cdot R_L=18.5 \times 10^{-3} \times 650=12.02 \mathrm{~W}
\)

Hint

(c) In full wave rectification, output signal (ripple) frequency is double that of input frequency. So, output frequency is 200 Hz .

Hint

(a) Comparing \(V=220 \sin 314 t\) with general equation
\(
\begin{aligned}
& V=V_0 \sin \omega t, \text { we get } \\
& V_0=220 \mathrm{~V} \text { and } \omega=314 \mathrm{rads}^{-1}
\end{aligned}
\)
Also, it is given that \(R_L=2 \mathrm{k} \Omega=2000 \Omega, r_f=20 \Omega\)
(i) Peak value of current,
\(
I_0=\frac{V_0}{\left(r_f+R_L\right)}=\frac{220}{20+2000} \approx 109 \mathrm{~mA}
\)
(ii) DC value of current,
\(
I_{\mathrm{DC}}=\frac{2 I_0}{\pi}=\frac{2 \times 109}{3.14}=69.4 \mathrm{~mA}
\)
(iii) rms value of current, \(I_{\text {rms }}=\frac{I_0}{\sqrt{2}}=0.707 \times 109\)
\(
=77.06 \mathrm{~mA}
\)

Hint

(b) Current, \(I=\frac{V}{R}\) or \(R=\frac{V}{I}\)
Resistance of limiting resistor, \(R=\frac{(6-2) \mathrm{V}}{10 \times 10^{-6} \mathrm{~A}}=400 \mathrm{k} \Omega\)

Hint

(d) Use, energy, \(E=h \nu=\frac{h c}{\lambda}\)
The minimum wavelength of radiation,
\(
\begin{aligned}
\lambda=\frac{h c}{E} & =\frac{\left(6.4 \times 10^{-34}\right) \times\left(3 \times 10^8\right)}{1.5 \times 1.6 \times 10^{-19}} \\
& =8 \times 10^{-7} \mathrm{~m} \\
& =800 \mathrm{~nm}
\end{aligned}
\)

Hint

(c) Current through \(250 \Omega\) resistor,
\(
I_1=\frac{\text { Net voltage }}{\text { Resistance }}=\frac{(20-12)}{250}=32 \times 10^{-3} \mathrm{~A}=32 \mathrm{~mA}
\)
Current through \(1 \mathrm{k} \Omega\) resistor,
\(
I_2=\frac{12}{1 \times 10^3}=12 \times 10^{-3} \mathrm{~A}=12 \mathrm{~mA}
\)
So, current through the Zener diode,
\(
I_Z=I_1-I_2=32 \mathrm{~mA}-12 \mathrm{~mA}=20 \mathrm{~mA}
\)

Hint

(d) only fixed ions.

Explanation:
p-type semiconductor: Contains a majority of holes (positive charge carriers).
n-type semiconductor: Contains a majority of electrons (negative charge carriers).
Depletion region formation: When p-type and n-type semiconductors are joined, electrons from the n-type diffuse into the p-type, and holes from the p-type diffuse into the n-type. This diffusion leaves behind fixed positive ions in the p-type region and fixed negative ions in the n-type region, creating a depletion region with no free charge carriers.

Hint

(d) Concentration of positive and negative charges near the junction.

Explanation:
Formation of the depletion region:
When a p-type and n-type semiconductor materials are joined to form a p-n junction, the majority charge carriers (holes in p-type and electrons in ntype) diffuse across the junction due to their concentration gradient. This diffusion causes the region near the junction to become depleted of mobile charge carriers, leaving behind immobile positively charged donor ions in the \(n\)-region and negatively charged acceptor ions in the p-region.

Potential barrier creation:
This depletion region acts as a barrier to further diffusion of charge carriers. The fixed positive and negative ions in the depletion region create an electric field, which opposes the further movement of majority carriers across the junction. This electric field is what constitutes the potential barrier.

Hint

(d) The barrier potential of a p-n junction diode does not depend on the diode design. The barrier potential is primarily influenced by the type of semiconductor material, the doping density, and temperature.

Hint

(b) minority carriers.
Explanation:
In a p-n junction, the flow of current is primarily due to the movement of minority carriers across the junction. In a p-type semiconductor, the majority carriers are holes, and the minority carriers are electrons. In an n-type semiconductor, the majority carriers are electrons, and the minority carriers are holes. When a p-n junction is formed, the minority carriers from each side diffuse across the junction, creating a current. This diffusion of minority carriers is what controls the conduction of the p-n junction.

Hint

(b) In a forward-biased p-n junction diode, the p-end is connected to the positive terminal of the battery, and the n-end is connected to the negative terminal. This configuration allows current to flow through the diode. 

Hint

(b) decreases.
Explanation:
Forward bias: In forward bias, an external voltage is applied to a p-n junction diode in a way that opposes the built-in potential barrier.

Depletion region: The depletion region is the area near the junction where majority carriers are diffused away, leaving behind charged ions.

Effect of forward bias: When forward bias is applied, the external voltage pushes the majority carriers closer to the junction, narrowing the depletion region.

Hint

(c) When a \(p-n\) junction diode is forward biased, both the depletion layer and resistance decrease.

Explanation:
Depletion layer decrease: In forward bias, the applied voltage opposes the built-in potential barrier, causing the depletion region to narrow.

Resistance decrease: As the depletion layer shrinks, the resistance of the diode decreases, allowing more current to flow through the junction.

Hint

(a) If the two ends of a p-n junction diode are joined by a wire, then there will not be a steady current in the circuit because the diffusion and drift currents will cancel each other out.

Hint

(b) Reverse biasing a \(p-n\) junction diode increases the width of the potential barrier.

Explanation: When reverse biased, the applied voltage pushes the majority carriers further away from the junction, creating a wider depletion region which in turn increases the potential barrier that minority carriers must overcome to cross the junction.

Hint

(a) As, \(-6 \mathrm{~V}<-3 \mathrm{~V}\)
i.e. \(p\) is at low potential than \(n\).
\(\therefore\) Diode is reverse biased.

Hint

(c) When a large reverse voltage is applied across a \(p-n\) junction diode, a huge current flows in the reverse direction suddenly. This is called breakdown of \(p-n\) junction diode. At breakdown, the current through the diode increases dramatically, either due to avalanche multiplication of charge carriers or zener tunnelling.

Hint

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

Hint

\(
\text { (b) The } p-n \text { junction is used to convert } \mathrm{AC} \text { into } \mathrm{DC} \text { (rectifier). }
\)

Hint

(c) When a light falls on the junction, new hole-electron pairs are created. Numbers of produced electron-hole pairs depend upon numbers of photons. So photo-emf or current is proportional to intensity of light.

Hint

(d) Microampere.
Explanation:
Reverse saturation current: When a diode is reverse-biased, a very small current flows through it, called the reverse saturation current.

Microampere scale: This current is typically on the order of microamperes \((\mu \mathrm{A})\), meaning one millionth of an ampere.

Hint

(d) Any digital circuit can be realised by repetitive use of NOR gate or NAND gate. NOR gate and NAND gate are called universal gates because by using these gates we can realise other baisc gates like OR, AND and NOT gates.

Hint

(d) When the two inputs of NAND gates are shorted, \(Y=\overline{\mathrm{AA}}=\bar{A}\) which is equivalent to a NOT gate.