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.