NCERT Solutions Class 12 Physics Chapter 14

Q.1 In n-type silicon, which of the following statement is true:
(a) Electrons are 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 dopants.

Ans.

Correct answer: (c) Holes are minority carriers and pentavalent atoms are the dopants.

Explanation: In n-type semiconductors, n_e >> n_h (number of electrons are greater than number of holes) i.e., holes are minority carriers and this type of semiconductors (Si or Ge) is obtained by doping with pentavalent atoms which are also called as donors.

Q.2 Which of the statement given is true for p-type semiconductors?
(a) Electrons are 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 dopants.

Ans.

Correct answer: (d) Holes are majority carriers and trivalent atoms are dopants.

Explanation: In p-type semiconductors, n_{e <<} n_h (number of holes are greater than number of electrons) i.e., holes are majority carriers and this type of semiconductors (Si or Ge) is obtained by doping with trivalent atoms which are also called as acceptors.

Q.3 Carbon, silicon and germanium have four valence electrons each. These are characterized by valence and conduction bands separated by energy band gap respectively equal to (E_g)_C, (E_g)_Si and (E_g)_Ge.

Which of the following statements is true?

\begin{array}{l}{\text{(a) (E}}_{\text{g}}{\text{)}}_{\text{Si}}{\text{< (E}}_{\text{g}}{\text{)}}_{\text{Ge}}{\text{< (E}}_{\text{g}}{\text{)}}_{\text{C}}\\ {\text{(b) (E}}_{\text{g}}{\text{)}}_{\text{C}}{\text{< (E}}_{\text{g}}{\text{)}}_{\text{Ge}}{\text{> (Eg)}}_{\text{Si}}\\ {\text{(c) (E}}_{\text{g}}{\text{)}}_{\text{C}}{\text{> (E}}_{\text{g}}{\text{)}}_{\text{Si}}{\text{> (E}}_{\text{g}}{\text{)}}_{\text{Ge}}\\ {\text{(d) (E}}_{\text{g}}{\text{)}}_{\text{C}}{\text{= (E}}_{\text{g}}{\text{)}}_{\text{Si}}{\text{= (E}}_{\text{g}}{\text{)}}_{\text{Ge}}\end{array}

Ans.

Correct answer: (c) (E_g)_C > (E_g)_Si > (E_g)_Ge

Explanation: Energy band gap for C (diamond), Si and Ge are 5.4 eV, 1.1 eV and 0.7 eV respectively.

Therefore, energy band gap is maximum in carbon and least in germanium (E_g)_C > (E_g)_Si > (E_g)_Ge among the given elements.

Q.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.

Ans.

Correct answer: (c) hole concentration in p-region is more as compared to n-region.

Explanation: In an unbiased p-n junction the, holes diffuse from the p-region to n-region because in p-region concentration of holes are greater than concentration of holes in n-region, therefore hole diffusion takes place from higher concentration to lower concentration i.e., from p-region to n-region.

Q.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.

Ans.

Correct answer: (c) lowers the potential barrier.

Explanation: When a forward bias is applied to p-n junction, it lowers the potential barrier by reducing the depletion layer due to cancellation of charges present in the depletion layer.

Q.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?

Ans.

In half wave rectifier the frequency of output signal is same as that of input signal and in full wave rectifier the frequency of output signal is double than that of input signal. The reason is, a half wave rectifier rectifies only one half cycle of input A.C.

Therefore, frequency of the output A.C.is equal to frequency of input A.C which is 50 Hz. A full wave rectifier rectifies both halve cycles of the A.C. input. Thus, the frequency of output A.C. is twice of frequency of input A.C. which is 100 Hz.

Q.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?

Ans.

\begin{array}{l}{\text{Since energy gap E}}_{\text{g}}\text{of p-n diode greater than energy E of the photon of wavelength 6000 nm,}\\ \text{thus the given wavelength cannot be detected}\text{.}\\ \text{Here}\\ {\text{E}}_{\text{g}}\text{= 2}\text{.8 eV = 2}\text{.8×1}{\text{.6×10}}^{\text{-19}}\text{J = 4}{\text{.48×10}}^{\text{-19}}\text{J}\\ {\text{λ = 6000 nm = 6000×10}}^{\text{-9}}\text{m}\\ \text{As energy}\text{is E=}\frac{\text{hc}}{\text{λ}}\\ \text{E =}\frac{\text{6}{\text{.62×10}}^{\text{-34}}{\text{Js×3×10}}^{\text{8}}\text{ms-1}}{{\text{6000 x 10}}^{\text{-9}}\text{m}}\text{= 3}{\text{.31×10}}^{\text{-20}}\text{J}\\ \text{In}\text{terms}\text{of}\text{eV}\\ \text{1}{\text{.6×10}}^{\text{-19}}\text{J = 1 eV}\\ \therefore \text{3}{\text{.31×10}}^{\text{-20}}\text{J = 0}\text{.207}\text{eV}\end{array}

Since energy gap E_g of p-n diode is greater than energy E of the photon of wavelength 6000 nm, thus the given wavelength cannot be detected.

Q.8 The number of silicon atoms per m³ is 5 × 10²⁸. This is doped simultaneously with 5 × 10²² atoms per m³ of Arsenic and 5 × 10²⁰ per m³ atoms of Indium. Calculate the number of electrons and holes. Given that n_i = 1.5 × 10¹⁶ m^-3. Is the material n-type or p-type?

Ans.

$\begin{array}{l}\text{As Arsenic is a pentavalent impurity and Indium is a tetravalent impurity, thus}\\ {\text{n}}_{\text{e}}\text{=}\left({\text{5×10}}^{\text{22}}{\text{m}}^{\text{-3}}{\text{– \hspace{0.17em}5×10}}^{\text{20}}{\text{m}}^{\text{-3}}\right)\text{=}\left(\text{5 – 0.05}\right){\text{×10}}^{\text{22}}{\text{m}}^{\text{-3}}{\text{= 4.95 ×10}}^{\text{22}}{\text{\hspace{0.17em}m}}^{\text{-3}}\\ \text{As,\hspace{0.17em}\hspace{0.17em}}\mathrm{m}\text{ass-action\hspace{0.17em}law\hspace{0.17em}\hspace{0.17em}given\hspace{0.17em}\hspace{0.17em}by}\\ {\text{n}}_{\text{e}}{\text{\hspace{0.17em}n}}_{\text{h}}{{\text{= n}}_{\text{i}}}^{\text{2}}\\ \text{from\hspace{0.17em}this,\hspace{0.17em}we\hspace{0.17em}get}\\ {\text{n}}_{\text{h}}\text{=}\frac{{{\text{n}}_{\text{i}}}^{\text{2}}}{{\text{n}}_{\text{e}}}\text{=}\frac{{\left({\text{1.5×10}}^{\text{16}}{\text{m}}^{\text{-3}}\right)}^{\text{2}}}{{\text{4.95×10}}^{\text{22}}{\text{m}}^{\text{-3}}}\\ {\text{n}}_{\text{h}}{\text{= 4.55×10}}^{\text{9}}{\text{m}}^{\text{-3}}\\ {\text{As n}}_{\text{e}}{\text{\hspace{0.17em}\hspace{0.17em}>\hspace{0.17em}\hspace{0.17em}n}}_{\text{h}}\text{,\hspace{0.17em}so\hspace{0.17em}it\hspace{0.17em}is\hspace{0.17em}a\hspace{0.17em}\hspace{0.17em}n-type semiconductor.}\end{array}$

Q.9 In an intrinsic semiconductor the energy gap E_g is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration n_i is given by

{\text{n}}_{\text{i}}={\text{n}}_0\text{exp}\left(-\frac{{\text{E}}_{\text{g}}}{{\text{2k}}_{\text{B}}\text{T}}\right)

where n₀ is a constant.

Ans.

$\begin{array}{l}{\text{Initial\hspace{0.17em}temperature,\hspace{0.17em}T}}_{\text{1}}\text{= 300 K}\\ \text{For\hspace{0.17em}this\hspace{0.17em}intrinsic\hspace{0.17em}carrier\hspace{0.17em}concentration\hspace{0.17em}can\hspace{0.17em}be\hspace{0.17em}written\hspace{0.17em}as}\\ {\text{n}}_{\text{1}}{\text{=\hspace{0.17em} n}}_{\text{0}}{\text{e}}^{\frac{{\text{-E}}_{\text{g}}}{{\text{2\hspace{0.17em}k}}_{\text{B\hspace{0.17em}}}{\text{T}}_{\text{1}}}}\\ {\text{Similarly\hspace{0.17em}for\hspace{0.17em}final\hspace{0.17em}temperature,\hspace{0.17em}T}}_{\text{2}}\text{= 600 K\hspace{0.17em}}\\ \text{Intrinsic\hspace{0.17em}carrier\hspace{0.17em}concentration\hspace{0.17em}can\hspace{0.17em}be\hspace{0.17em}written\hspace{0.17em}as}\\ {\text{n}}_{\text{2}}{\text{= \hspace{0.17em}n}}_{\text{0}}{\text{e}}^{\frac{{\text{-E}}_{\text{g}}}{{\text{2\hspace{0.17em}k}}_{\text{B\hspace{0.17em}}}{\text{T}}_{\text{2}}}}\\ \text{The\hspace{0.17em}ratio\hspace{0.17em}between\hspace{0.17em}the\hspace{0.17em}conductivity\hspace{0.17em}at\hspace{0.17em}600 K\hspace{0.17em}and\hspace{0.17em}at\hspace{0.17em}300 K\hspace{0.17em}is equal\hspace{0.17em}to}\\ \frac{{\text{n}}_{\text{2}}}{{\text{n}}_{\text{1}}}\text{=}\frac{{\text{n}}_{\text{0}}{\text{e}}^{\frac{{\text{-E}}_{\text{g}}}{{\text{2\hspace{0.17em}k}}_{\text{B\hspace{0.17em}}}{\text{T}}_{\text{1}}}}}{{\text{n}}_{\text{0}}{\text{e}}^{\frac{{\text{-E}}_{\text{g}}}{{\text{2\hspace{0.17em}k}}_{\text{B}}{\text{\hspace{0.17em}T}}_{\text{2}}}}}\\ \frac{{\text{n}}_{\text{2}}}{{\text{n}}_{\text{1}}}{{\text{= \hspace{0.17em}e}}^{\frac{{\text{E}}_{\text{g}}}{{\text{2\hspace{0.17em}k}}_{\text{B}}}}}^{\left[\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{\hspace{0.17em}-\hspace{0.17em}}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right]}\\ \frac{{\text{n}}_{\text{2}}}{{\text{n}}_{\text{1}}}{\text{=\hspace{0.17em} e}}^{\frac{\text{1.2 eV}}{{\text{2\hspace{0.17em}×\hspace{0.17em}8.6\hspace{0.17em}×\hspace{0.17em}10}}^{\text{-5}}\text{eV/K}}\left[\frac{\text{1}}{\text{300 K}}\text{\hspace{0.17em}\hspace{0.17em}-\hspace{0.17em}\hspace{0.17em}}\frac{\text{1}}{\text{600 K}}\right]}\\ \frac{{\text{n}}_{\text{2}}}{{\text{n}}_{\text{1}}}{\text{=\hspace{0.17em} 1.12×10}}^{\text{5}}{\text{= 1×10}}^{\text{5}}\end{array}$

Q.10 In a p-n junction diode, the current I can be expressed as

\text{I}={\text{I}}_0\text{exp}\left(\frac{\text{eV}}{{\text{k}}_{\text{B}}\text{T}}-\text{1}\right),

where I₀ is called 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 8.62 × 10^-5 eV/K and T is the absolute temperature. If for a given diode I₀ = 5 × 10^-12 A and T = 300 K, then
(a) What will be the forward current at forward voltage of 0.6 V?
(b) What will be the increase in the current if voltage across the diode is increased to 0.7 V?
(c) What is the dynamic resistance?
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

Ans.

$\begin{array}{l}\text{(a)}\\ {\text{Given:\hspace{0.17em}I}}_{\text{0}}{\text{= 5×10}}^{\text{-12}}\text{A,\hspace{0.17em}\hspace{0.17em}V = 0.6 V,\hspace{0.17em}\hspace{0.17em}T = 300 K,}\\ {\text{k}}_{\text{B}}{\text{=8.6×10}}^{\text{-5}}{\text{eV\hspace{0.17em}K}}^{\text{-1}}{\text{=8.6×10}}^{\text{-5}}{\text{×1.6×10}}^{\text{-19}}{\text{J\hspace{0.17em}K}}^{\text{-1}}\\ {\text{Current,\hspace{0.17em}I = I}}_{\text{0}}\text{exp}\left(\frac{\text{eV}}{{\text{2k}}_{\text{B}}\text{T}}\text{– 1}\right)\text{,}\\ \text{If\hspace{0.17em}V = 0.6 V,}\\ {\text{I = 5×10}}^{\text{-12}}\text{A}\times \text{exp}\left(\left(\frac{{\text{1.6×10}}^{\text{-19}}\text{C×0.6 V}}{{\text{8.6×10}}^{\text{-5}}{\text{eV/K×1.6×10}}^{\text{-19}}\text{C×300 K}}\right)\text{– 1}\right)\\ {\text{I = 5×10}}^{\text{-12}}\text{A}\times \text{exp}\left(\text{22.3}\right)\text{= 0.024 \hspace{0.17em}A}\end{array}$

$\begin{array}{l}\text{(b)}\\ \text{Let,\hspace{0.17em}V’\hspace{0.17em}and\hspace{0.17em}I’\hspace{0.17em}are\hspace{0.17em}voltage\hspace{0.17em}and\hspace{0.17em}current\hspace{0.17em}across\hspace{0.17em}the\hspace{0.17em}diode\hspace{0.17em}respectively.}\\ \text{If\hspace{0.17em}the\hspace{0.17em}voltage\hspace{0.17em}across\hspace{0.17em}the\hspace{0.17em}diode\hspace{0.17em}is\hspace{0.17em}increased\hspace{0.17em}to\hspace{0.17em}0.7 \hspace{0.17em}V,\hspace{0.17em}then}\\ {\text{Current,\hspace{0.17em}I’ = I}}_{\text{0}}\text{exp}\left(\frac{\text{eV’}}{{\text{2k}}_{\text{B}}\text{T}}\text{\hspace{0.17em} – \hspace{0.17em}1}\right)\text{\hspace{0.17em}will\hspace{0.17em}become}\\ {\text{I = 5×10}}^{\text{-12}}\text{A}\times \text{exp}\left(\left(\frac{{\text{1.6×10}}^{\text{-19}}\text{C×0.7 V}}{{\text{8.6×10}}^{\text{-5}}{\text{eV/K×1.6×10}}^{\text{-19}}\text{C×300 K}}\right)\text{– 1}\right)\\ {\text{I = 5×10}}^{\text{-12}}\text{A}\times \text{exp}\left(\text{26.13}\right)\text{= 1.12\hspace{0.17em}A}\\ \text{Hence,\hspace{0.17em}the\hspace{0.17em}increase\hspace{0.17em}in\hspace{0.17em}current,ΔI = \hspace{0.17em}I’ – I}\\ \text{ΔI = 1.12 A – 0.024\hspace{0.17em}A =1.096\hspace{0.17em}A}\approx \text{1.1\hspace{0.17em}A}\end{array}$

$\begin{array}{l}{\text{(c) Dynamic resistance of diode, R}}_{\text{d}}\text{=}\frac{\text{ΔV}}{\text{ΔI}}\\ {\text{R}}_{\text{d}}\text{=}\frac{\left(\text{0.7 V – 0.6 V}\right)}{\text{1.1 A}}\text{= 0.090 Ω}\end{array}$

(d) In reverse bias, V becomes negative in the relation so for 1 V and 2 V reverse voltage the current I will be equal to I₀ thereby giving i.e infinite dynamic resistance.

Q.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.

Ans.

\begin{array}{l}\text{If fig}\left(\text{a}\right)\text{, the output}\\ {\text{y}}_{\text{1}}\text{=}\overline{\text{A+B}}\text{and Y =}\overline{\overline{{\text{y}}_{\text{1}}}}\\ \text{Y =}\overline{\overline{\text{A + B}}}\text{= A + B}\\ \text{Thus as per Boolean algebra fig}\left(\text{a}\right)\text{represents an OR gate}\text{.}\\ \text{In fig}\left(\text{b}\right)\\ \text{Y =}\overline{{\text{y}}_{\text{1}}{\text{+ y}}_{\text{2}}}\text{=}\overline{{\text{y}}_{\text{1}}}\cdot \overline{{\text{y}}_{\text{2}}}\\ \text{But,}\overline{{\text{y}}_{\text{1}}}\text{= A}\text{and}\overline{{\text{y}}_{\text{2}}}\text{= B}\\ \text{Thus,}\text{Y =}\overline{{\text{y}}_{\text{1}}}\cdot \overline{{\text{y}}_{\text{2}}}\text{= A}\cdot \text{B,}\text{which is an AND operation}\text{.}\end{array}

Q.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

Ans.

The NAND gate shown in the truth table has only one input. Therefore, the truth table is

\text{Since}\text{Y}=\overline{\text{A}}\text{in this case},\text{the circuit is actually a NOT gate with the truth table}.

Q.13 You are given two circuits as shown in figure, which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Ans.

\begin{array}{l}\text{In the figure}\left(\text{a}\right)\text{,}\\ \text{Y =}\overline{{\text{y}}_{\text{1}}\cdot {\text{y}}_{\text{1}}}\text{=}\overline{{\text{y}}_{\text{1}}}\\ \text{As,}\left({\text{y}}_{\text{1}}\cdot {\text{y}}_{\text{1}}{\text{= y}}_{\text{1}}\right)\\ \text{But}{\text{y}}_{\text{1}}\text{=}\overline{\text{A}\cdot \text{B}}\text{thus}\text{Y =}\overline{\overline{\text{A}\cdot \text{B}}}\text{= A}\cdot \text{B}\\ \text{Therefore the figure}\left(\text{a}\right)\text{represents AND gate}\text{.}\\ \text{In figure}\left(\text{b}\right)\text{,Y=}\overline{{\text{y}}_{\text{1}}\cdot {\text{y}}_{\text{2}}}\\ \text{As}{\text{y}}_{\text{1}}\text{=}\overline{\text{A}\cdot \text{A}}\text{=}\overline{\text{A}}\text{and}{\text{y}}_{\text{2}}\text{=}\overline{\text{B}\cdot \text{B}}\text{=}\overline{\text{B}}\\ \text{Thus,}\text{Y =}\overline{{\text{y}}_{\text{1}}\cdot {\text{y}}_{\text{2}}}\text{=}\overline{\overline{\text{A}}\text{.}\overline{\text{B}}}\text{= A+B,}\\ \text{Therefore, figure}\left(\text{b}\right)\text{represents OR gate}\text{.}\end{array}

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

(Hint: A = 0, B = 1 then A and B inputs of second NOR gate will be 0 and hence Y = 1. Similarly work out the values of Y for other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)

Ans.

\begin{array}{l}{\text{Let y}}_{\text{1}}\text{be the output, which appears at the first operation of NOR gate}\text{.}\\ {\text{So that, y}}_{\text{1}}\text{=}\overline{\text{A + B}}\mathrm{\dots }\left(\text{i}\right)\\ \text{As the output is fed into an other NOR gate, thus}\\ \text{Y =}\overline{{\text{y}}_{\text{1}}{\text{+ y}}_{\text{1}}}\text{=}\overline{{\text{y}}_{\text{1}}}\\ \text{Y =}\overline{\overline{\text{A + B}}}\text{Using}\left(\text{i}\right)\\ \text{Y= A+B which is an OR operation}\text{.}\\ \text{The circuit is an OR gate and its truth table is given below}\end{array}

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

Ans.

\begin{array}{l}\text{In figure}\left(\text{i}\right)\text{, Y =}\overline{\text{A+A}}\text{therefore as per Boolean algebra}\\ \text{Y =}\overline{\text{A+A}}\text{=}\overline{\text{A}}\\ \text{As, this is a NOT operation therefore, figure}\left(\text{i}\right)\text{represents a NOT gate}\text{.}\\ \text{Its truth table is given below}\end{array}

\begin{array}{l}\text{In}\text{figure}\text{(ii)}\\ \text{Y =}\overline{\overline{\text{A}}\text{+}\overline{\text{B}}}\text{=}\overline{\overline{\text{A}}}\cdot \overline{\overline{\text{B}}}\text{= A}\cdot \text{B}\\ \text{Thus, figure}\left(\text{ii}\right)\text{represents AND gate}\text{. Its truth table is given below}\end{array}