NCERT solutions for class 12 physics chapter 14 Semiconductor Electronics Materials Devices And Simple Circuits: This chapter is mainly divided into two parts. Analog electronics and digital electronics. The first 10 questions of solutions of NCERT class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits deal with semiconductor devices, their properties, characteristics, and applications. The next five questions of CBSE NCERT solutions for class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits is related to logic gates. As far as the current scenario of electronics is considered this chapter is very important and is the base of electronics. NCERT solutions for class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits are important for competitive exams like NEET and JEE Mains also. Solutions of NCERT helps students in the preparation of the CBSE board exam as well.
Q. 14.1 In an ntype 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.
An Ntype semiconductor has electron as majority carriers and holes as minority carriers. It is formed when we dope pentavalent impurity in Silicon atom. Some pentavalent dopants are phosphorus, arsenic, and bismuth.
Hence the correct option is C.
Q. 14.2 Which of the statements given in Exercise 14.1 is true for ptype 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 the dopants
In a ptype semiconductor, holes are the majority carrier and electrons are the minority carrier. It is formed when a trivalent atomlike aluminium is doped in a silicon atom. Hence correct option for ptype conductor would be (d).
(a)
(b)
(c)
(d)
Since carbon is a nonmetal, its energy band gap would be highest and energy band gap of Ge would be least as it is a metalloid.
Hence correct option would be (c)
(a) free electrons in the nregion attract them
(b) they move across the junction by the potential difference.
(c) hole concentration in pregion is more as compared to nregion.
Answer:
Charge flows from higher concentration to the lower concentration in a junction. In this case, holes are diffusing from the pregion to nregion and hence the concentration of hole is greater in p region.
and hence correct option would be (c)
Q. 14.5 When a forward bias is applied to a pn junction, it
(a) raises the potential barrier
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier.
When a pn junction is forward biased, the negative voltage repels the electron toward junction and give them the energy to cross the junction and combine with the hole which is also being pushed by a positive voltage. This leads to a reduction in the depletion layer which means a reduction in potential barrier across the junction.
Hence correct option would be (c)
As we know :
output frequency for halfwave rectifier = input frequency, and hence output frequency in halfwave rectifier will be 50Hz.
also, output frequency for fullwave rectifier = 2*(input frequency) and Hence output frequency in fullwave rectifier will be 2*50 = 100 Hz.
Q. 14.7 A pn photodiode is fabricated from a semiconductor with bandgap of Can it detect a wavelength of
Given
the energy band gap of photodiode is 2.8eV.
wavelength = = 6000nm =
The energy of signal will be
where c is speed of light(300000000m/s) , h is planks constant ( = )
putting the corresponding value
The energy of signal =
=
=
The energy of the signal is 0.207eV which is less than 2.8eV ( the energy and gap of photodiode). Hence signal can not be detected by the photodiode.
Given:
number of Silicon atoms per =
number of Arsenic atoms per =
number of Indium atoms per =
number of thermally generated electrons
Now,
Number of electrons
=
number of holes is
in thermal equilibrium
Now, since the number of electrons is higher than number of holes, it is an ntype semiconductor.
Energy gap of given intrinsic semiconductor = E_{g} = 1.2eV
temperature dependence of intrinsic carrier concentration is given by
Where is constant, is Boltzmann constant = ,
T is temperature
Initial temperature = T1 = 300K
the intrinsic carrier concentration at this temperature :
Final temperature = T2 = 600K
the intrinsic carrier concentration at this temperature :
the ratio between the conductivities at 300K and at 600K is equal to the ratio of their intrinsic carrier concentration at these temperatures
Therefore the ratio between the conductivities is .
Q. 14.10 (a) In a pn junction diode, the current I can be expressed as
(a) What will be the forward current at a forward voltage of
As we have
Here, , and , = Boltzmann constant =
When the forward voltage is 0.6V:
Hence forward current is 0.0625A
Q.14.10 (b) In a pn junction diode, the current I can be expressed as
(b) What will be the increase in the current if the voltage across the diode is increased to
As we have
Here, , and , = Boltzmann constant =
When the forward voltage is 0.7V:
When the forward voltage is 0.6V:
Hence the increase in the forward current is
Q. 14.10 (c) In a pn junction diode, the current I can be expressed as
(c) What is the dynamic resistance?
Dynamic Resistance =
Resistance change = 0.7  0.6 = 0.1
Current change = 2.967(calculated in prev question)
Therefore
,
Q.14.10 (d) In a pn junction diode, the current I can be expressed as
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
As we have
Here, , and , = Boltzmann constant =
When reverse voltage is 1V, V= 1
When the reverse voltage is 2V:
In both case current is very small and approximately equal to the reverse saturation current, hence their difference is negligible which causes dynamic resistance of infinity.
Q. 14.11 (a) You are given the two circuits as shown in Fig. 14.36. Show that circuit
(a) acts as OR gate while the circuit
Fig. 14.36
Here, THE Input = A and B
Output = Y
The left part of the figure acts as a NOR and right part acts as NOT Gate.
The output of NOR gate =
the output of the NOR gate would be the input of NOT Gate and hence
Y = =
Hence the figure functions like an OR Gate.
or compare the truth table by giving different input and observing the output
INPUTS

OUTPUT 

A B  Y 
0 0  0 
0 1  1 
1 0  1 
1 1  1 
Q. 14.11 (b) You are given the two circuits as shown in Fig. 14.36. Show that circuit
Answer:
The output of NOT gate ( left part of the circuit) is the input of the NOR gate
Hence the output of total circuit Y =
=
=
Hence the circuit functions as AND gate.
or give the inputs 00,01,10,11 and observe the truth table
INPUTS

OUTPUT 

A B  Y 
0 0  0 
0 1  0 
1 0  0 
1 1  1 
The truth table is the same as that of AND gate
Q. 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.
Here A is both input of the NAND gate and hence Output Y will be
Hence circuit functions as a NOT gate.
The truth table for the given figure:
Input  Output 
A  Y 
0  1 
1  0 
a)
A and B are inputs of a NAND gate and output of this gate is the input of another NAND gate so,
Y =
Y=
Y=
Hence this circuit functions as AND gate.
b)
A is input to the NAND gate output of whose goes to the rightmost NAND gate. Also, B is input to the NAND gate whose output goes to the rightmost NAND gate.
Y =
Y = +
Y = A + B
Hence the circuit functions as an OR gate .
Alternative method
fig. a
construct the truth table by giving various input and observe the output
INPUT 
INTERMEDIATE OUTPUT 
OUTPUT 
00  1  0 
01  1  0 
10  1  0 
11  0  1 
The above truth table is the same as that of an AND gate
fig. b
INPUTS  OUTPUT 
00  0 
01  1 
10  1 
11  1 
The above truth table is the same as that of an OR gate
(Hint: then and inputs of second NOR gate will be and hence Similarly work out the values of for other combinations of and Compare with the truth table of OR, AND, NOT gates and find the correct one.)
A and B are the input od a NOR gate and Output of this NOR gate is the Input of Another NOR gate whose Output is Y. Hence,
Y =
Y = .
Y = A + B
Hence Circuit behaves as OR gate.
Truth table
INPUTS  OUTPUT 
00  0 
01  1 
10  1 
11  1 
Figure 14.40
a)
A is the two input of the NOR gate and Hence Output Y is:
Y =
Y =
Hence circuit functions as a NOT gate.
TRUTH TABLE:
INPUT  OUTPUT 
0  1 
1  0 
b) A is the two input of a NOR gate whose output(which is ) is the one input of another NOR gate. B is the two input of NOR gate whose output (which is ) is the input of another NOR gate. Hence,
Y =
Y = .
Y = A.B
Hence it functions as AND gate.
TRUTH TABLE:
INPUTS  OUTPUT 
00  0 
01  0 
10  0 
11  1 
As far as CBSE board exam is considered you can expect more theoretical questions from NCERT physics chapter 14. With the help of NCERT solutions for class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits, you can prepare well and it is easy to score in this chapter. CBSE NCERT solutions for class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits help in preparations of exams like NEET and JEE Mains. For NEET exams one or two questions and for JEE Mains one question is expected from this chapter.
Q.14.10 In a pn junction diode, the current I can be expressed as
where is called the reverse saturation current, is the voltage across the diode and is positive for forward bias and negative for reverse bias, and is the current through the diode, is the Boltzmann constant and is the absolute temperature. If for a given diode and then
(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?
Q. 14.5 When a forward bias is applied to a pn 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.
Q. 14.10 In a pn junction diode, the current I can be expressed as
where 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 is the current through the diode, is the Boltzmann constant and is the absolute temperature. If for a given diode and then
(a) What will be the forward current at a forward voltage of