NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

Q. In an n-type silicon, which of the following statement is true:

  • (i) Electrons are majority carriers and trivalent atoms are the dopants.
  • (ii) Electrons are majority carriers and pentavalent atoms are the dopants.
  • (iii) Holes are minority carries and pentavalent atoms are the depants.
  • (iv) Holes are majority carriers and trivalent atoms are the dopants.
  • Ans. (iii) The electrons are the majority carriers in n-type silicon while the holes are the minority carriers. An n-type semiconductor is obtained when pentavalent atoms, such as phosphorus, are doped in silicon atoms.

Q. Which of the statements given in previous question is true for p-type semiconductors?

Ans. (iv) The holes are the majority carriers in p-type semiconductor while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium are doped in silicon atoms.

Q. In an unbiased p-n junction, holes diffuse from the p-region to n-region because:

  • (i) free electrons in the n-region attract them.
  • (ii) they move across the junction by the potential difference.
  • (iii) hole concentration in p-region is more as compared to n-region.
  • (iv) All the above.
Ans (iii) The diffusion of charge carriers across a junction occurs from the region of higher concentration to the region of lower concentration. Thus, the p-region has 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.

Q. Can a slab of p-type semiconductor be physically joined to another n-type semiconductor slab to from p-n junction ? Justify your answer.

Ans. No, any slab howsoever flat will have roughness much larger than the interatomic crystal from spacing (2Å to 3Å) and hence continuous contact at the atomic junction will not be possible. The junction will behave as discontinuous for the following charger carriers.

Q. When a forward bias is applied to a p-n junction, it

  • (i) raises the potential barrier.
  • (ii) reduces the majority carrier current to zero.
  • (iii) lowers the potential barrier.
  • (iv) None of the above.

Ans. (iii) As soon as forward bias is applied to a p-n junction, it lowers the values of 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.

Q. The number of silicon atoms per m3 is 5 × 1023. This is doped simultaneously with 5 × 1023 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni = 1.5 × 1016 m– 3. Is the material n-type or p-type.

Ans. Given,

Number of silicon atoms (N) = 5 × 1023 atoms/m3

Number of arsenic atoms (nAs) = 5 × 1022 atoms/m3

Number of indium atoms (nIn) = 5 × 1020 atoms/m3

Number of thermally-generated electrons (ni) = 1.5 × 1016 electrons/m3

Number of electrons (ne)= 5 × 1022 – 1.5 × 1016 ≈ 4.99 × 1022

Let number of holes = nh

In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as

$$n_en_h=n_i^{2}\\\therefore n_h=\frac{n_i^{2}}{n_e}=\frac{(1.5×10^{16})^{2}}{4.99×10^{22}}\approx4.51×10^{9}\\\text{The number of electrons is approximately}\space4.99×10^{22}\text{and the number of holes are}\space 4.51×10^{9}.\\\text{Since, the number of electrons is more than the number of holes, the material is an n-type semiconductor.}$$

Q. 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. Given, energy band gap of the given photodiode (Eb) = 2.8 eV.

Wavelength (λ) = 6000 nm = 6000 × 10– 9 m

The energy of a signal is given by

$$\text{E}=\frac{hc}{\lambda}\\\text{Here, h = Planck’s constant = 6.626×10}^{-34}\text{Js}\\c=\text{Speed of light =}3×10^{8}\text{m/s}\\\text{E}=\frac{6.626×10^{-34}×3×10^{8x}}{6000×10^{-9}}\\=3.313×10^{-20}\text{J}\\\text{But}\space1.6×10^{-19}\text{J}=1e\text{V}\\\therefore\text{E}=3.313×10^{-20}\text{J}\\=\frac{3.313×10^{-20}}{1.6×10^{-19}}=0.207\text{eV}\\\text{Energy of a signal of wavelength 6000 nm is 0.207 eV,}\\\text{ which is less than 2.8 eV the energy band gap of a photodiode.}\text{ Therefore, the photodiode cannot detect the signal.}$$

Q. 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 Given, input frequency = 50 Hz

For a half-wave rectifier, the output frequency is equal to the input frequency.

∴ Output frequency = 50 Hz

For a full-wave rectifier, the output frequency is twice the input frequency.

∴ Output frequency = 2 × 50 = 100 Hz

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

$$\textbf{I}=\textbf{I}_0\textbf{exp}.\bigg(\frac{\textbf{eV}}{\textbf{2k}_\textbf{B}\textbf{T}}-1\bigg)$$

 where T0 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, kB is the Boltzmann constant (8.6 × 10– 5 eV/K) and T is the absolute temperature. If for a given diode I0 = 5 × 10– 12 and T = 300 K, then

  • (i) What will be the forward current at a forward voltage of 0.6 V ?
  • (ii) What will be the increase in the current it the voltage across the diode is increased to 0.7 V ?
  • (iii) What is the dynamic resistance ?
  • (iv) What will be the current if reverse bias voltage changes from 1 V to 2 V ?

$$\textbf{Ans}.\space\text{In}\space \text{a p-n junction diode, the expression for current is given by}\\\text{I}=\text{I}_0\text{exp.} \bigg(\frac{\text{eV}}{2\text{k}_\text{B}\text{T}}-1\bigg)\\\text{Here},\\\text{I}_0=\text{Reverse saturation current =}\space5×10^{-12}\text{A}\\\text{T}=\text{Absolute temperature}=330k\\k_B=\text{Boltzmann constant}=8.6×10^{-5}\text{eV/K}=1.376×10^{-23}\text{J k}^{-1}\\\text{V = Voltage across the diode}\\\text{(i) Forward voltage (V) = 0.6 V}\\\therefore\space\text{Current}(\text{I})=5×10^{-12}\\\bigg[\text{exp}\bigg(\frac{1.6×10^{-19}×0.6}{1.376×10^{-23}×300}\bigg)-1\bigg]\\=5×10^{-12}×\text{exp}[22.36]\\=0.0256\text{A}\\\text{Thus, the forward current is about 0.0256 A.}\\\text{(ii) For forward voltage (V) =}0.7\text{V},\text{we can write :}\\\text{I}’=5×10^{-12}\\\bigg[\text{exp}\bigg(\frac{1.6×10^{\normalsize-19}×0.6}{1.376×10^{\normalsize-23}×300}\bigg)-1\bigg]\\=5×10^{-12}×\text{exp}[26.25]\\=1.257\text{A}\\\text{Therefore, the increases in current,}\\\Delta\text{I}=\text{I’}-\text{I}\\=1.257-0.0256=1.23\text{A}\\\text{(iii)}\space\text{Dynamic resistance}=\frac{\text{Change in voltage}}{\text{Change in current}}\\=\frac{0.7-0.6}{1.23}=\frac{0.1}{1.23}=0.081Ω$$

(iv) If the reverse bias voltage changes from 1 V to 2 V, then the current (I) will almost remain equal to I0 in both cases. Therefore, the dynamic resistance in the reverse bias will be infinite.

Q. C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors ?

Ans. The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out an electron from these atoms (i.e., ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.
Q. 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 (Eg) C, (Eg)Si and (Eg)Ge. Which of the following statements is true ?
  • (i) (Eg)Si < (Eg)Ge < (Eg)C
  • (ii) (Eg)C < (Eg)Ge > (Eg)Si
  • (iii) (Eg)C > (Eg)Si > (Eg)Ge
  • (iv) (Eg)C = (Eg)Si = (Eg)Ge

Ans. (iii) The energy band gap of carbon is the maximum among given elements and that of germanium is the least.

Thus, energy band gap of these elements are related as : (Eg)C > (Eg)Si > (Eg)Ge

Q. The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). What is the reason then to operate the
photodiodes in reverse bias ?

Ans. Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n >>p). On illumination, let the excess electrons and holes generated be Δn and Δp, respectively.

n′ = n + Δn

p′ = p + Δp

Here n′ and p′ are the electron and hole concentrations at any particular illumination and n and p are carriers concentration when there is no illumination. Remember Δn = Δp and n >> p. Hence, the fractional change in the majority carriers (i.e., Δn/n) would be much less than that in the minority carriers (i.e., Δp/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes are preferably used in the reverse bias condition for measuring light intensity.

Q. In an intrinsic semiconductor the energy gap Eg 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 300 K ? Assume that the temperature dependence of intrinsic carrier concentration ni is given by

$$\textbf{n}_{\textbf{i}}=\textbf{n}_\textbf{0}\space\textbf{exp}\bigg[-\frac{\text{E}_g}{2\text{k}_\text{B}\text{T}}\bigg]\\\textbf{Where}\space \textbf{n}_\textbf{0}\textbf{is a constant.}$$

Ans. Given, energy gap of the intrinsic semiconductor
(Eg) = 1.2 eV.

The temperature dependence of the intrinsic carrier concentration is given by

$$\text{n}_i=\text{n}_0\space\text{exp}\bigg[-\frac{\text{E}_\text{g}}{2\text{k}_\text{B}\text{T}}\bigg]\\\text{Here},k_B=\text{Boltzmann constant}=8.62×10^{-5}\text{eV/K}\\(\text{T}=\text{Temperature}, n_0 = \text{Constant})\\\text{Initial temperature}(\text{T}_1)=300\text{k}.\\\text{The intrinsic carrier-concentration at this temperature can be given by}\\ n_{i1}=n_{0}\space\text{exp}\bigg[-\frac{E_g}{2k_B×300}\bigg]\space…(\text{i})\\\text{Final temperature}(\text{T}_2)=600\text{k}.\\\text{The intrinsic carrier-concentration at this temperature can be given as}\\n_{i2}=n_0\space\text{exp}\bigg[\frac{\text{E}_g}{2k_B×300}\bigg]\space\text{…(ii)}$$

The ratio between the conductivity at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperature.