by NCERT Solved Exercise Questions – Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices And Simple Circuits
14.1 In an 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 the dopants.
Ans – The correct statement is (c).
The majority carriers in an n-type silicon are electrons, while the minority carriers are holes. Pentavalent atoms, such as phosphorus, are doped into silicon atoms to produce an n-type semiconductor.
14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
Ans – The correct statement is (d).
The majority carriers in a p-type semiconductor are holes, while the minority carriers are electrons. Trivalent atoms, such as aluminium, can be doped into silicon atoms to produce a p-type semiconductor.
14.3 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 (Eg) C, (Eg) Si and (Eg) Ge. Which of the following statements is true? (a) (Eg) Si < (Eg) Ge < (Eg) C (b) (Eg) C < (Eg) Ge > (Eg) Si (c) (Eg ) C > (Eg ) Si > (Eg) Ge (d) (Eg) C = (Eg) Si = (Eg) Ge
Ans – The correct statement is (c).
Carbon has the largest energy band gap while germanium has the smallest among the three elements mentioned.
These elements’ energy band gaps are related as follows: (Eg)C > (Eg)Si > (Eg)Ge
14.4 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 – The correct statement is (c).
From the area of higher concentration to the area of lower concentration, charge carriers diffuse across a junction. The p-region in this instance has a higher concentration of holes than the n-region. Therefore, holes diffuse from the p-region to the n-region in an unbiased p-n junction.
14.5 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 – The proper answer is (c).
An increase in forward bias reduces the potential barrier at a p-n junction. The potential barrier opposes the applied voltage in a forward bias situation. As a result, the potential obstacle at the junction is lessened.
14.6 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 – A half wave rectifier simply adjusts the output frequency by half of the input A.C. cycle.
A.C. = input frequency A.C. = 50 Hz
A full wave rectifier corrects the output frequency’s two half cycles of the AC input.
A.C. = 2 x the input frequency A.C. = 2 x 50 = 100 Hz
14.7 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 – Energy band gap of the given photodiode, Eg = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9 m
To respond to incident light, the photo diode requires 2.8 eV of energy. The provided photo diode is unable to detect photons with a wavelength of 6000 nm because E < Eg.
14.8 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 ni = 1.5 × 1016 m–3. Is the material n-type or p-type?
Ans – We are aware that one free electron is received for every atom doped with arsenic. Similar to this, a vacancy is produced for every indium atom. Consequently, the quantity of free electrons added as a result of the pentavalent impurity
Number of silicon atoms, N = 5 × 1028 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
14.9 In an intrinsic semiconductor the energy gap Eg is 1.2eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration ni is given by where n0 is a constant.
Ans – For an intrinsic semiconductor, ne = nh = ni, and conductivity is given by = e (ne e +nh h). Additionally, the mobility of holes (p,,) and electrons (pj) Conductivity thus equals enepe. Dependence of intrinsic carrier concentration on temperature
kB = Boltzmann constant = 8.62 × 10−5 eV/K
T = Temperature
n0 = Constant
14.10 In a p-n junction diode, the current I can be expressed as where I0 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 A and T = 300 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 reverse bias voltage changes from 1 V to 2 V?
Ans – I0 = Reverse saturation current = 5 × 10−12 A
T = Absolute temperature = 300 K
kB = Boltzmann constant = 8.6 × 10−5 eV/K = 1.376 × 10−23 J K−1
V = Voltage across the diode
(a) When V = 0.6 V
(b) When V = 0.7 V,
(c)
(d) The current (I) will almost always stay equal to I0 in both circumstances if the reverse bias voltage is increased from 1 V to 2 V. The dynamic resistance in the reverse bias will therefore be unbounded.
14.11 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 – (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.
14.12 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 –A acts as the two inputs of the NAND gate and Y is the output, as shown in the following figure.
| A | A | y = A.A¯ |
| 0 | 0 | 1 |
| 1 | 1 | 0 |
Since Y = A¯ in this case, the circuit is actually a NOT gate with the truth table
| A | Y |
| 0 | 1 |
| 1 | 0 |
14.13 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.
Ans –A and B are the inputs and Y is the output in both of the illustrated circuits.
14.14 Write the truth table for circuit given in Fig. 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 Yfor other combinations of A and B. Compare with the truth table of OR, AND, NOT gates and find the correct one.)
Ans – A and B are the inputs of the given circuit. The output of the first NOR gate is 
. It can be observed from the following figure that the inputs of the second NOR gate become the out put of the first one.
14.15 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.
Ans – A acts as the two inputs of the NOR gate and Y is the output, as shown in the following figure.
