NCERT Exercise Q & A

Exercise Problems

Q14.1: In an n-type silicon, which of the following statements is true:
(a) Electrons are the majority carriers, and trivalent atoms are the dopants.
(b) Electrons are minority carriers, and pentavalent atoms are the dopants.
(c) Holes are minority carriers, and pentavalent atoms are the dopants.
(d) Holes are majority carriers, and trivalent atoms are the dopants.

Answer: (c) In an n-type silicon, the electrons are the majority carriers, while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.

Q14.2: Which of the statements is true for p-type semiconductors?
(a) Electrons are the majority carriers, and trivalent atoms are the dopants.
(b) Electrons are minority carriers, and pentavalent atoms are the dopants.
(c) Holes are minority carriers, and pentavalent atoms are the dopants.
(d) Holes are majority carriers, and trivalent atoms are the dopants.

Answer: (d) In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.

Q14.3: Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to \(\left(E_{\mathrm{g}}\right)_{\mathrm{C}},\left(E_{\mathrm{g}}\right)_{\mathrm{Si}}\) and \(\left(E_{\mathrm{g}}\right)_{\mathrm{Ge}}\). Which of the following statements is true?
(a) \(\left(E_g\right)_{\mathrm{Si}} \lt \left(E_g\right)_{\mathrm{Ge}} \lt \left(E_g\right)_{\mathrm{C}}\)
(b) \(\left(E_g\right)_{\mathrm{C}} \lt \left(E_g\right)_{\mathrm{Ge}}>\left(E_g\right)_{\mathrm{Si}}\)
(c) \(\left(E_g\right)_{\mathrm{C}}>\left(E_g\right)_{\mathrm{Si}}>\left(E_g\right)_{\mathrm{Ge}}\)
(d) \(\left(E_g\right)_{\mathrm{C}}=\left(E_g\right)_{\mathrm{Si}}=\left(E_g\right)_{\mathrm{Ge}}\)

Answer: (c) Of the three given elements, the energy band gap of carbon is the maximum, and that of germanium is the least.
The energy band gap of these elements are related as: \(\left(E_g\right)_{\mathrm{C}}>\left(E_g\right)_{\mathrm{si}}>\left(E_g\right)_{\mathrm{Ge}}\)

Q14.4: In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.

Answer: (c) The diffusion of charge carriers across a junction takes place from the region of higher concentration to the region of lower concentration. In this case, the p-region has a greater concentration of holes than the n-region. Hence, in an unbiased p-n junction, holes diffuse from the p-region to the n-region.

Q14.5: When a forward bias is applied to a p-n junction, it
(a) raises the potential barrier.
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
(d) None of the above.

Answer: (c) When a forward bias is applied to a p-n junction, it lowers the value of the potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.

Q14.6: In half-wave rectification, what is the output frequency if the input frequency is 50 Hz? What is the output frequency of a full-wave rectifier for the same input frequency?

Answer: Input frequency \(=50 \mathrm{~Hz}\)
For a half-wave rectifier, the output frequency is equal to the input frequency.
\(\therefore\) Output frequency \(=50 \mathrm{~Hz}\)
For a full-wave rectifier, the output frequency is twice the input frequency.
\(\therefore\) Output frequency \(=2 \times 50=100 \mathrm{~Hz}\)

Q14.7: A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?

Answer: Energy band gap of the given photodiode, \(E_g=2.8 \mathrm{eV}\)
Wavelength, \(\lambda=6000 \mathrm{~nm}=6000 \times 10^{-9} \mathrm{~m}\)
The energy of a signal is given by the relation:
\(
E=\frac{h c}{\lambda}
\)
Where,
\(
\begin{aligned}
& h=\text { Planck’s constant } \\
& =6.626 \times 10^{-34} \mathrm{Js} \\
& c=\text { Speed of light } \\
& =3 \times 10^8 \mathrm{~m} / \mathrm{s} \\
& E=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-9}} \\
& E=3.313 \times 10^{-20} \mathrm{~J}
\end{aligned}
\)
But \(1.6 \times 10^{-19} \mathrm{~J}=1 \mathrm{eV}\)
\(
\therefore E=3.313 \times 10^{-20} \mathrm{~J}
\)
\(
=\frac{3.313 \times 10^{-20}}{1.6 \times 10^{-19}}=0.207 \mathrm{eV}
\)
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV (the energy band gap of a photodiode). Hence, the photodiode cannot detect the signal.

Q14.8: The number of silicon atoms per \(\mathrm{m}^3\) is \(5 \times 10^{28}\). This is doped simultaneously with \(5 \times 10^{22}\) atoms per \(\mathrm{m}^3\) of Arsenic and \(5 \times 10^{20}\) per \(\mathrm{m}^3\) atoms of Indium. Calculate the number of electrons and holes. Given that \(n_i=1.5 \times 10^{16} \mathrm{~m}^{-3}\). Is the material n-type or p-type?

Answer: Number of silicon atoms, \(N=5 \times 10^{28}\) atoms \(/ \mathrm{m}^3\)
Number of arsenic atoms, \(n_{\text {AS }}=5 \times 10^{22}\) atoms \(/ \mathrm{m}^3\)
Number of indium atoms, \(n_{\mathrm{In}}=5 \times 10^{20}\) atoms \(/ \mathrm{m}^3\)
Number of thermally-generated electrons, \(n_i=1.5 \times 10^{16}\) electrons \(/ \mathrm{m}^3\)
Number of electrons, \(n_e=5 \times 10^{22}-1.5 \times 10^{16} \approx 4.99 \times 10^{22}\)
Number of holes \(=n_h\)
In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:
\(
\begin{aligned}
n_e n_h & =n_i^2 \\
\therefore n_h & =\frac{n_i^2}{n_e} \\
& =\frac{\left(1.5 \times 10^{16}\right)^2}{4.99 \times 10^{22}} \approx 4.51 \times 10^9
\end{aligned}
\)
Therefore, the number of electrons is approximately \(4.99 \times 10^{22}\) and the number of holes is about \(4.51 \times 10^9\). Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.

Q14.9: In an intrinsic semiconductor, the energy gap \(E_g\) is 1.2 eV. Its hole mobility is much smaller than electron mobility and is independent of temperature. What is the ratio between the conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration \(n_i\) is given by
\(
n_i=n_0 \exp \left(-\frac{E_g}{2 k_B T}\right)
\)
where \(n_0\) is a constant.

Answer: Energy gap of the given intrinsic semiconductor, \(\mathrm{E}_{\mathrm{g}}=1.2 \mathrm{eV}\)
The temperature dependence of the intrinsic carrier concentration is written as:
\(
\mathrm{n}_{\mathrm{i}}=\mathrm{n}_0 \exp \left[-\frac{\mathrm{E}}{2 \mathrm{k}_{\mathrm{BT}}}\right]
\)
Where,
\(\mathrm{k}_{\mathrm{B}}=\) Boltzmann constant \(=8.62 \times 10^{-5} \mathrm{eV} / \mathrm{K}\)
\(\mathrm{T}=\) Temperature
\(\mathrm{n}_0=\) Constant
Initial temperature, \(\mathrm{T}_1=300 \mathrm{~K}\)
The intrinsic carrier concentration at this temperature can be written as:
\(
\mathrm{n}_{\mathrm{i} 1}=\mathrm{n}_0 \exp \left[-\frac{\mathrm{E}_{\mathrm{g}}}{2 \mathrm{k}_{\mathrm{B}} \times 300}\right] \dots(i)
\)
Final temperature, \(T_2=600 \mathrm{~K}\)
The intrinsic carrier concentration at this temperature can be written as:
\(
\mathrm{n}_{\mathrm{i} 2}=\mathrm{n}_0 \exp \left[-\frac{\mathrm{E}_{\mathrm{g}}}{2 \mathrm{k}_{\mathrm{B}} \times 600}\right] \dots(ii)
\)
The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier concentrations at these temperatures.
\(
\begin{aligned}
&\begin{aligned}
& \frac{n_{i 2}}{n_{i 1}}=\frac{n_0 \exp \left[-\frac{\mathrm{E}_{\mathrm{g}}}{2 \mathrm{k}_{\mathrm{B}} 600}\right]}{\mathrm{n}_0 \exp \left[\frac{\mathrm{E}_{\mathrm{g}}}{2 \mathrm{k}_{\mathrm{B}} 300}\right]} \\
& =\frac{\exp \mathrm{E}_{\mathrm{g}}}{2 \mathrm{k}_{\mathrm{B}}}\left[\frac{1}{300}-\frac{1}{600}\right]=\exp \left[\frac{1.2}{2 \times 8.62 \times 20^{-5}} \times \frac{2-1}{600}\right] \\
& =\exp [11.6]=1.09 \times 10^5
\end{aligned}\\
&\text { Therefore, the ratio between the conductivities is } 1.09 \times 10^5 \text {. }
\end{aligned}
\)
\(
\begin{aligned}
&\text { Since the ratio of conductivities is directly proportional to the ratio of } n_i \text { : }\\
&\frac{\sigma_{600}}{\sigma_{300}}=\frac{n_{i 2}}{n_{i 1}}\approx\frac{1}{1 \times 10^5}
\end{aligned}
\)

Q14.10: In a p-n junction diode, the current I can be expressed as
\(
I=I_0 \exp \left(\frac{e V}{2 k_B T}-1\right)
\)
where \(I_0\) is called the reverse saturation current, \(V\) is the voltage across the diode and is positive for forward bias and negative for reverse bias, and \(I\) is the current through the diode, \(k_B\) is the Boltzmann constant \(\left(8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K}\right)\) and T is the absolute temperature. If for a given diode \(I_0=5 \times 10^{-12} \mathrm{~A}\) and \(\mathrm{T}=300 \mathrm{~K}\), then
(a) What will be the forward current at a forward voltage of 0.6 V?
(b) What will be the increase in the current if the voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if the reverse bias voltage changes from 1 V to 2 V?

Answer: In a p-n junction diode, the expression for current is given as:
\(
I=I_0 \exp \left(\frac{\mathrm{eV}}{2 k_{\mathrm{B}} T}-1\right)
\)
Where,
\(I_0=\) Reverse saturation current \(=5 \times 10^{-12} \mathrm{~A}\)
\(T=\) Absolute temperature \(=300 \mathrm{~K}\)
\(k_{\mathrm{B}}=\) Boltzmann constant \(=8.6 \times 10^{-5} \mathrm{eV} / \mathrm{K}=1.376 \times 10^{-23} \mathrm{~J} \mathrm{~K}^{-1}\)
\(V=\) Voltage across the diode
(a) Forward voltage, \(V=0.6 \mathrm{~V}\)
\(
\therefore \text { Current, } I \text { =5 } \times 10^{-12}\left[\exp \left(\frac{1.6 \times 10^{-19} \times 0.6}{1.376 \times 10^{-23} \times 300}\right)-1\right]
\)
\(
=5 \times 10^{-12} \times \exp [22.36]=0.0256 \mathrm{~A}
\)
Therefore, the forward current is about 0.0256 A.
(b) For forward voltage, \(V^{\prime}=0.7 \mathrm{~V}\), we can write:
\(
\begin{aligned}
& I^{\prime}=5 \times 10^{-12}\left[\exp \left(\frac{1.6 \times 10^{-19} \times 0.7}{1.376 \times 10^{-23} \times 300}-1\right)\right] \\
& =5 \times 10^{-12} \times \exp [26.25]=1.257 \mathrm{~A}
\end{aligned}
\)
Hence, the increase in current, \(\Delta I=I^{\prime} \quad-I\)
\(
=1.257-0.0256=1.23 \mathrm{~A}
\)
(c) Dynamic resistance \(=\frac{\text { Change in voltage }}{\text { Change in current }}\)
\(
=\frac{0.7-0.6}{1.23}=\frac{0.1}{1.23}=0.081 \Omega
\)
(d) If the reverse bias voltage changes from 1 V to 2 V, then the current ( \(I\) ) will almost remain equal to \(I_0\) in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.

Q14.11: You are given the two circuits as shown in Fig. 14.36. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.

Answer: (a) \(A\) and \(B\) are the inputs, and \(Y\) is the output of the given circuit. The left half of the given figure acts as the NOR Gate, while the right half acts as the NOT Gate.
Hence, the output of the NOR Gate \(=\overline{A+B}\). This will be the input for the NOT Gate. Its output will be \(\overline{\overline{A+B}}=A+B\)
\(
\therefore Y=A+B
\)
Hence, this circuit functions as an OR Gate.
(b) \(A\) and \(B\) are the inputs and \(Y\) is the output of the given circuit. It can be observed from the following figure that the inputs of the right half NOR Gate are the outputs of the two NOT Gates.
Hence, the output of the given circuit can be written as:
\(
Y=\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \overline{\bar{B}}=A \cdot B
\)
Hence, this circuit functions as an AND Gate.

Q14.12: Write the truth table for a NAND gate connected as given in Fig. 14.37.

Hence, identify the exact logic operation carried out by this circuit.

Answer: Hence, the output can be written as:
\(
Y=\overline{A \cdot A}=\bar{A}+\bar{A}=\bar{A}
\)
This circuit functions as a NOT gate.

Q14.13: You are given two circuits as shown in Fig. 14.38, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Answer: (a) Here, the output of the combination of the two NAND gates is given as:
\(
Y=\overline{(\overline{A \cdot B}) \cdot(\overline{A \cdot B})}=\overline{\overline{A B}}+\overline{\overline{A B}}=A B
\)
Hence, this circuit functions as an AND gate.

(b) Here, the output of the combination of the NAND gates will be given as:
\(
Y=\bar{A} \cdot \bar{B}=\overline{\bar{A}}+\overline{\bar{B}}=A+B
\)
Hence, this circuit functions as an OR gate.

Q14.14: Write the truth table for the circuit given in Fig. 14.39 below, consisting of NOR gates, and identify the logic operation (OR, AND, NOT) which this circuit is performing.

Answer: Here, the output of the combination is given as:
\(
\begin{aligned}
Y & =\overline{\overline{A+B}+\overline{A+B}}=\overline{\bar{A} \cdot \bar{B}}+\overline{\bar{A} \cdot \bar{B}} \\
& =\overline{\bar{A} \cdot \bar{B}}=\overline{\bar{A}}+\bar{B}=A+B
\end{aligned}
\)
The truth table for this operation is given as:
\(
\begin{array}{|c|c|c|}
\hline A & B & \boldsymbol{Y}(=A+B) \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 1 \\
\hline
\end{array}
\)
This is the truth table of an OR gate. Hence, this circuit functions as an OR gate.

Q14.15: Write the truth table for the circuits given in Fig. 14.40, consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

Answer: (a) Output, \(Y=\overline{A+A}=\bar{A}\)
The truth table for the same is given as:
\(
\begin{array}{|c|c|}
\hline A & {Y}{(=\bar{A})} \\
\hline 0 & 1 \\
\hline 1 & 0 \\
\hline
\end{array}
\)
This is the truth table of a NOT gate. Hence, this circuit functions as a NOT gate.
(b) Output \(Y=\overline{\bar{A}+\bar{B}}=\overline{\bar{A}} \cdot \bar{B}=A \cdot B\)
The truth table for the same can be written as:
\(
\begin{array}{|c|c|c|}
\hline A & B & Y(=A \cdot B) \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 0 & 0 \\
\hline 1 & 1 & 1 \\
\hline
\end{array}
\)
This is the truth table of an AND gate. Hence, this circuit functions as an AND gate.

Exemplar Section

VSA

Q14.17: Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?

Answer: When pure semiconductor material is mixed with small amounts of certain specific impurities with valency different from that of the parent material, the number of mobile electrons/holes drastically changes. The process of addition of impurity is called doping. The size of the dopant atom should be compatible such that their presence in the pure semiconductor does not distort the semiconductor but easily contributes the charge carriers on forming covalent bonds with Silicon or Germanium atoms, which are provided by group XIII or group XV elements.

Q14.18: \(\mathrm{Sn}, \mathrm{C}\), and \(\mathrm{Si}, \mathrm{Ge}\) are all group XIV elements. Yet, \(\mathrm{Sn}\) is a conductor, \(\mathrm{C}\) is an insulator while \(\mathrm{Si}\) and \(\mathrm{Ge}\) are semiconductors. Why?

Answer: The conduction level of any element depends on the energy gap between its conduction band and valence band.
In conductors, there is no energy gap between the conduction band and the valence band. For an insulator, the energy gap is large, and for semiconductor, the energy gap is moderate.
The energy gap for Sn is 0 eV, for C is 5.4 eV, for Si is 1.1 eV, and for Ge is 0.7 eV, related to their atomic size. Therefore, Sn is a conductor, C is an insulator, and Ge and Si are semiconductors.

Q14.19: Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?

Answer: We cannot measure the potential barrier across a p-n junction by a voltmeter because the resistance of the voltmeter is very high as compared to the junction resistance. Potential of potential barrier for \(G e\) is \(V_B=0.3 \mathrm{~V}\) and for silicon is \(V_B=0.7 \mathrm{~V}\). On the average, the potential barrier in P-N junction is \(\sim 0.5 \mathrm{~V}\).

Q14.20: Draw the output waveform across the resistor (Fig. 14.8).

Answer: The diode acts as a half-wave rectifier; it offers low resistance when forward biased and high resistance when reverse biased. So the output is obtained only when a positive input is given, so the output waveform is

SA

Q14.23: 

(i) Name the type of a diode whose characteristics are shown in Fig. 14.9 (A) and Fig. 14.9(B).
(ii) What does the point P in Fig. (A) represent?
(iii) What does the points P and Q in Fig. (B) represent?

Answer: (i) Fig. (a) represents the characteristics of Zener diode and curve (b) is of solar cell.
(ii) In Fig. (a), point P represents Zener breakdown voltage.
(iii) In Fig. (b), the point Q represents zero voltage and negative current. Which means the light falling on the solar cell with at least minimum threshold frequency gives the current in the opposite direction to that due to a battery connected to the solar cell. But for the point Q the battery is short-circuited. Hence, it represents the short-circuit current.
And the point P in Fig. (b) represents some open circuit, voltage on the solar cell with zero current through the solar cell.
It means there is a battery connected to a solar cell, which gives rise to the equal and opposite current to that in the solar cell by virtue of light falling on it.

Q14.24: Three photo diodes D1, D2, and D3 are made of semiconductors having band gaps of \(2.5 \mathrm{eV}, 2 \mathrm{eV}\) and 3 eV, respectively. Which ones will be able to detect light of wavelength 6000Å?

Answer: According to the problem,
Wavelength of light \(\lambda=6000 Å=6000 \times 10^{-10} \mathrm{~m}\)
Energy of the light photon
\(
E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}=2.06 \mathrm{eV}
\)
For the incident radiation to be detected by the photodiode, the energy of the incident radiation photon should be greater than the band gap. This is true only for D2. Therefore, only D2 will detect this radiation.

Q14.25: Two-car garages have a common gate that needs to open automatically when a car enters either of the garages or cars enter both. Devise a circuit that resembles this situation using diodes for this situation.

Answer: 

\(
\begin{aligned}
&\text { OR gate glves output according to the truth table. }\\
&\begin{array}{|c|c|c|}
\hline A & B & C \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 1 \\
\hline
\end{array}
\end{aligned}
\)

Q14.26: Explain why elemental semiconductor cannot be used to make visible LEDs.

Answer: Specially designed diodes, which give out light radiation when forward-biased. LED’s are made of GaAsp, Gap, etc.
These are forward-biased \(P-N\) junctions which emits spontaneous radiation.
In elemental semiconductor, the band gap is such that the emission are in the infrared region and not in the visible region.
\(
\begin{aligned}
& \lambda=\frac{h c}{E_g}=\frac{1242 \mathrm{eVnm}}{E_{\mathrm{g}}} \\
& \text { for } \mathrm{Si} ; E_g=1.1 \mathrm{eV}, \lambda=\frac{1242}{1.1}=1129 \mathrm{~nm} \\
& \text { for } \mathrm{Ge} ; E_g=0.7 \mathrm{eV}, \lambda=\frac{1242}{0.7}=1725 \mathrm{~nm}
\end{aligned}
\)

Q14.27: Write the truth table for the circuit shown in Fig. 14.11. Name the gate that the circuit resembles.

Answer: Truth Table
\(
\begin{array}{|c|c|c|}
\hline A & B & Y \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 0 \\
\hline 1 & 0 & 0 \\
\hline 1 & 1 & 1 \\
\hline
\end{array}
\)
This is an AND gate.

Q14.28: A Zener of power rating 1 W is to be used as a voltage regulator. If Zener has a breakdown of 5 V and it has to regulate voltage which fluctuates between 3 V and 7 V, what should be the value of Rs for safe operation (Fig.14.12)?

Answer: According to the problem, power \(=1 \mathrm{~W}\)
Zener breakdown voltage, \(V_z=5 \mathrm{~V}\)
Minimum voltage, \(V_{\min }=3 \mathrm{~V}\)
Maximum voltage, \(V_{\max }=7 \mathrm{~V}\)
We know, \(P=V I\)
So, current \(I_{Z_{\max }}=\frac{P}{V_Z}=\frac{1}{5}=0.2 \mathrm{~A}\)
For safe operation \(R_s\) will be equal to
\(
R_s=\frac{V_{\max }-V_Z}{I_{Z_{\max }}}=\frac{7-5}{0.2}=\frac{2}{0.2}=10 \Omega
\)

LA

Q14.29: If each diode in Fig. 14.13 has a forward bias resistance of \(25 \Omega\) and infinite resistance in reverse bias, what will be the values of the current \(I_1, I_2, I_3\) and \(I_4\)?

Answer: \(I_3\) is zero as the diode in that branch is reverse biased. Resistance in the branch AB and EF are each \((125+25) \Omega=150 \Omega\).
As AB and EF are identical parallel branches, their effective resistance is \(\frac{150}{2}=75 \Omega\)
\(\therefore\) Net resistance in the circuit \(=(75+25) \Omega=100 \Omega\).
\(\therefore\) Current \(I_1=\frac{5}{100}=0.05 \mathrm{~A}\).
As resistances of AB and EF are equal, and \(I_1=I_2+I_3+I_4, I_3=0\)
\(
\therefore I_2=I_4=\frac{0.05}{2}=0.025 \mathrm{~A}
\)

Q14.30: Draw the output signals \(C_1\) and \(C_2\) in the given combination of gates (Fig. 14.15).

Answer: \(C_1=\overline{A+B}\)
\(C_2=A \cdot B\)

Q14.31: Assuming the ideal diode, draw the output waveform for the circuit given in Fig. 14.17. Explain the waveform.

Answer: When the input voltage is greater than 5 V, the diode is conducting (output will be 5V). When input is less than 5 V, the diode is an open circuit (output is same as the input).

Q14.32: Suppose a ‘ \(n\) ‘-type wafer is created by doping Si crystal having \(5 \times 10^{28}\) atoms \(/ \mathrm{m}^3\) with 1 ppm concentration of As. On the surface 200 ppm Boron is added to create the ‘ P ‘ region in this wafer. Considering \(n_i=1.5 \times 10^{16} \mathrm{~m}^{-3}\), (i) Calculate the densities of the charge carriers in the \(n \& p\) regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when the diode is reverse-biased.

Answer: (i) In ‘ \(n\) ‘ reglon; number of \(e^{-}\) is due to As:
\(
\begin{aligned}
& n_e=N_D=1 \times 10^{-6} \times 5 \times 10^{28} \text { atoms } / \mathrm{m}^3 \\
& n_e=5 \times 10^{22} / \mathrm{m}^3
\end{aligned}
\)
The minority carriers (hole) is
\(
n_h=\frac{n_i^2}{n_e}=\frac{\left(1.5 \times 10^{16}\right)^2}{5 \times 10^{22}}=\frac{2.25 \times 10^{32}}{5 \times 10^{22}}
\)
\(
n_h=0.45 \times 10 / \mathrm{m}^3
\)
Similarly, when Boron is implanted a ‘p’ type is created with holes
\(
\begin{aligned}
& n_h=\mathrm{N}_{\mathrm{A}}=200 \times 10^{-6} \times 5 \times 10^{28} \\
& =1 \times 10^{25} / \mathrm{m}^3
\end{aligned}
\)
This is far greater than \(e^{-}\) that existed in ‘ n ‘ type wafer on which Boron was diffused.
Therefore, minority carriers in created ‘p’ region
\(
\begin{aligned}
n_e= & \frac{n_i^2}{n_h}=\frac{2.25 \times 10^{32}}{1 \times 10^{25}} \\
& =2.25 \times 10^7 / \mathrm{m}^3
\end{aligned}
\)
(i1) Thus, when reverse blased \(0.45 \times 10^{10} / \mathrm{m}^3\), holes of ‘ \(n\) ‘ region would contribute more to the reverse saturation current than 2.25 \(\times 10^7 / \mathrm{m}^3\) minority \(e^{-}\) of p type region.

Q14.33: An X-OR gate has the following truth table:
\(
\begin{array}{|c|c|c|}
\hline \mathrm{A} & \mathrm{~B} & \mathrm{Y} \\
\hline 0 & 0 & 0 \\
\hline 0 & 1 & 1 \\
\hline 1 & 0 & 1 \\
\hline 1 & 1 & 0 \\
\hline
\end{array}
\)
It is represented by following logic relation
\(
\mathrm{Y}=\overline{\mathrm{A}} \cdot \mathrm{~B}+\mathrm{A} \cdot \overline{\mathrm{~B}}
\)
Build this gate using AND, OR and NOT gates.

Answer: 

Q14.34: Consider a box with three terminals on top of it as shown in Fig.14.18 (a):

Three components, namely, two germanium diodes and one resistor, are connected across these three terminals in some arrangement. A student performs an experiment in which any two of these three terminals are connected in the circuit shown in Fig. 14.18 (b). The student obtains graphs of current-voltage characteristics for the unknown combination of components between the two terminals connected in the circuit. The graphs are
(i) when A is positive and B is negative

(ii) when A is negative and B is positive

(iii) When B is negative and C is positive

(iv) When B is positive and C is negative

(v) When A is positive and C is negative

(vi) When A is negative and C is positive

From these graphs of current-voltage characteristic shown in Fig. 14.18 (c) to (h), determine the arrangement of components between A, B, and C.

Answer: The V-I characteristics of these graphs are discussed in points:
(a) In the V-I graph of condition (i), reverse characteristics are shown in Figure (c). Here, A is connected to the n-side of the p-n junction \(I\), and \(B\) is connected top side of the p-n junction \(I\) with a resistance in series.
(b) In the V-I graph of condition (ii), a forward characteristic is shown in figure (d), where 0.7 V is the knee voltage of p-n junction I. 1/slope \(=(1 / 1000) \Omega\).
It means \(A\) is connected to the \(n\)-side of \(p-n\) junction \(I\) and \(B\) is connected to the \(p\)-side of \(p-n\) junction \(I\) and resistance \(R\) is in a series of p-n junction I between \(A\) and \(B\).
(c) In the V-I graph of condition (iii), a forward characteristic is shown in figure (e), where 0.7 V is the knee voltage. In this case, the p-side of p-n junction II is connected to C and the n-side of \(\mathrm{p}-\mathrm{n}\) junction II to B.
(d) In V-I graphs of conditions (iv), (v), (vi), also concludes the above connection of p-n junctions I and II along with a resistance \(R\).

Thus, the arrangement of p-n I, p-n II, and resistance \(R\) between \(A, B\) and \(C\) will be as shown in the figure.