# Ncert Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuits

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.

## NCERT Solutions for class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits exercises:

(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 N-type 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.

(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 p-type semiconductor, holes are the majority carrier and electrons are the minority carrier. It is formed when a trivalent atom-like aluminium is doped in a silicon atom. Hence correct option for p-type conductor would be (d).

(a) $\inline (E_{g})_{Si} < (E_{g})_{Ge}< (E_{g})_{C}$

(b) $\inline (E_{g})_{C} < (E_{g})_{Ge}> (E_{g})_{Si}$

(c) $\inline (E_{g})_{C} > (E_{g})_{Si}> (E_{g})_{Ge}$

(d) $\inline (E_{g})_{C} = (E_{g})_{Si}= (E_{g})_{Ge}$

Since carbon is a non-metal, its energy band gap would be highest and energy band gap of Ge would be least as it is a metalloid.

$\inline (E_{g})_{C} > (E_{g})_{Si}> (E_{g})_{Ge}$

Hence correct option would be (c)

## Q14.4  In an unbiased p-n junction, holes diffuse from the p-region to n-region beca

(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

Charge flows from higher concentration to the lower concentration in a junction. In this case, holes are diffusing from the p-region to n-region and hence the concentration of hole is greater in p region.

and hence correct option would be (c)

(a) raises the potential barrier

(b) reduces the majority carrier current to zero.

(c) lowers the potential barrier.

(d) none of the above.

When a p-n 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 half-wave rectifier = input frequency, and hence output frequency in half-wave rectifier will be 50Hz.

also, output frequency for full-wave rectifier = 2*(input frequency) and Hence output frequency in full-wave rectifier will be 2*50 = 100 Hz.

Given

the energy band gap of photodiode is 2.8eV.

wavelength = $\lambda$ = 6000nm = $6000*10^{-9}$

The energy of signal will be          $\frac{hc}{\lambda }$

where c is speed of light(300000000m/s) , h is planks constant ( = $6.626 * 10^{-34}Js$ )

putting the corresponding value

The energy of signal =    $\frac{(6.626 * 10^{-34} * 3*10^8)}{6000*10^{-9}}$

=   $3.313*10^{-20}J$

=   $0.207eV (since 1.6*10^{-20}= 1eV)$

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  $\inline m^{3}$   =  $\inline 5\times 10^{28}.$

number of Arsenic atoms per  $\inline m^{3}$ =  $\inline 5\times 10^{22}.$

number of Indium atoms per  $\inline m^{3}$  =  $\inline 5\times 10^{20}$

number of thermally generated electrons $\inline n_{i}=1.5\times 10^{16}\; m^{-3}.$

Now,

Number of electrons

$n_e =$ $5 * 10 ^{22}-1.5*10^{16}$ = $4.99*10^{22}(approx)$

number of holes is $n_h$

in thermal equilibrium

$n_h*n_e=n_i^2$

$n_h=n_i^2/n_e$

$n_h= (1.5*10^{16})^2/4.99*10^{22}$

$n_h= 4.51 * 10^9$

Now, since the number of electrons is higher than number of holes, it is an n-type semiconductor.

## Q. 14.9  In an intrinsic semiconductor the energy gap $\inline E_{g}$  is  $\inline 1.2\; eV.$  Its hole mobility is much smaller than electron mobility and independent of  temperature. What is the ratio between conductivity at $\inline 600K$  and that at $\inline 300K$ Assume that the temperature dependence of intrinsic carrier concentration $\inline n_{i}$ is given by

$\inline n_{i}=n_{0}\; exp\left [ -\frac{E_{g}}{2K_{B}T} \right ]$

Where, $\inline n_{0}$  is constant.

Energy gap of given intrinsic semiconductor = Eg = 1.2eV

temperature dependence of intrinsic carrier  concentration $\inline n_{i}$ is given by

$\inline n_{i}=n_{0}\; exp\left [ -\frac{E_{g}}{2K_{B}T} \right ]$

Where is constant, $K_B$ is Boltzmann constant = $8.862 * 10^{-5}eV/K$

T is temperature

Initial temperature = T1 = 300K

the intrinsic carrier concentration at this temperature :

$n_{i1} = n_0exp[\frac{-E_g}{2K_B*300}]$

Final temperature = T2 = 600K

the intrinsic carrier concentration at this temperature :

$n_{i2} = n_0exp[\frac{-E_g}{2K_B*600}]$

the ratio between the conductivities at 300K and at 600K  is equal to the ratio of their intrinsic carrier concentration at these temperatures

$\frac{n_{i2}}{n_{i2}} = \frac{n_0exp[\frac{-E_g}{2K_B*600}]}{n_0exp[\frac{-E_g}{2K_B*300}]}$

$= exp\frac{E_g}{2K_B}[\frac{1}{300}-\frac{1}{600}]=exp[\frac{1.2}{2*8.62*10^{-5}}* \frac{2-1}{600}]$

$= exp[11.6] = 1.09 * 10^{5}$

Therefore the ratio between the conductivities is $1.09 * 10^{5}$.

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

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, $\inline k_{B}$  is the Boltzmann constant $\inline (8.6\times 10^{-5}eV/K)$  and $\inline T$ is the absolute temperature. If for a given diode $\inline I_{0}=5\times 10^{-12}A$  and $\inline T=300\; K,$ then

(a)  What will be the forward current at a forward voltage of $\inline 0.6\; V\; ?$

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $\inline I_{0}=5\times 10^{-12}A$ ,  $\inline T=300\; K,$ and , $\inline k_{B}$  = Boltzmann constant =  $\inline (8.6\times 10^{-5}eV/K)$     $=(1.376*10^{-23}J/K)$

When the forward voltage is 0.6V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.6}{1.376*10^{-23}*300}-1 ]=0.0625A$

Hence forward current is 0.0625A

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

where I0 is called the reverse saturation current, $\inline V$ is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $\inline I$ is the current through the diode, $\inline k_{B}$  is the Boltzmann constant $\inline (8.6\times 10^{-5}\; eV/K)$ and $\inline T$ is the absolute temperature. If for a given diode $\inline I_{0}=5\times 10^{12}A$  and  $\inline T=300\; K,$  then

(b) What will be the increase in the current if the voltage across the diode is increased to $\inline 0.7 \; V?$

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $\inline I_{0}=5\times 10^{-12}A$ ,  $\inline T=300\; K,$ and , $\inline k_{B}$  = Boltzmann constant =  $\inline (8.6\times 10^{-5}eV/K)$     $=(1.376*10^{-23}J/K)$

When the forward voltage is 0.7V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.7}{1.376*10^{-23}*300}-1 ]=3.029A$

When the forward voltage is 0.6V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*0.6}{1.376*10^{-23}*300}-1 ]=0.0625A$

Hence the increase in the forward current is

$I(whenv=0.7) - I(whenv=.6)$   $= 3.029- 0.0625 = 2.967A$

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

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 $(8.6\times 10^{-5}V/K)$ and  $T$  is the absolute temperature. If for a given diode$I_{0}=5\times 10^{-12}A$   and  $T=300\; K,$  then

(d) What will be the current if reverse bias voltage changes from 1 V to 2 V?

As we have

$I=I_{0}\; [exp \frac{eV}{K_{B}T}-1 ]$

Here, $\inline I_{0}=5\times 10^{-12}A$ ,  $\inline T=300\; K,$ and , $\inline k_{B}$  = Boltzmann constant =  $\inline (8.6\times 10^{-5}eV/K)$     $=(1.376*10^{-23}J/K)$

When reverse voltage is 1V, V= -1

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-1)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$

When the reverse voltage is -2V:

$I=5*10^{-12}\;[ exp \frac{1.6*10^{-19}*(-2)}{1.376*10^{-23}*300}-1 ]\approx5\times10^{-12}$

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.

(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 =  $\overline{A+B}$

the output  of the NOR gate would be the input of NOT Gate and hence

Y = $\over\overline{A+B}$ = $\overline\overline{A+B}$

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

(b) acts as AND gate.

The output of NOT gate ( left part of the circuit) is the input of the NOR gate

Hence the output of total circuit Y = $\over(\overline A + \overline B)$

=  $\overline{\overline A}.\overline{\overline B}$                                  $\overline{A+B}=\overline A. \overline B$

= $A*B$

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

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

$Y = \overline {A*A}$

$Y = \overline {A} + \overline A$

$Y = \overline {A}$

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 = $\over(\overline {A.B})(\overline {A.B})$

Y=   $\over(\overline {A.B})$$+$$\over(\overline {A.B})$

Y= $AB$

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 = $\over \overline A .\overline B$

Y = $\over\overline A .$ + $\over\overline B.$

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:  $A=0,B=1$  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.)

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 =    $\over(\overline{A+B} + \overline{A+B})$

Y =     $\over\overline {A+B}$ .  $\over\overline {A+B}$

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 =   $\overline {A+A}$

Y = $\overline {A}$

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 $\overline {A}$)  is the one input of another NOR gate. B is the two input of NOR gate whose output (which is $\overline {B}$)  is the input of another NOR gate. Hence,

Y = $\over\overline {A} + \overline {B}$

Y = $\over\overline {A}$.$\over\overline {B}$

Y = A.B

Hence it functions as AND gate.

TRUTH TABLE:

 INPUTS OUTPUT 00 0 01 0 10 0 11 1

## NCERT solutions for class 12 physics- chapter wise

 NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields Solutions of NCERT class 12 physics chapter 2 Electrostatic Potential and Capacitance CBSE NCERT solutions for class 12 physics chapter 3 Current Electricity NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism Solutions of NCERT class 12 physics chapter 5 Magnetism and Matter CBSE NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction NCERT solutions for class 12 physics chapter 7 Alternating Current Solutions of NCERT class 12 physics chapter 8 Electromagnetic Waves CBSE NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions Solutions of NCERT class 12 physics chapter 11 Dual nature of radiation and matter CBSE NCERT solutions for class 12 physics chapter 12 Atoms NCERT solutions for class 12 physics chapter 13 Nuclei NCERT solutions for class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

## Significance of solutions of NCERT class 12 physics chapter 14 semiconductor electronics materials devices and simple circuits:

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 circuitsyou 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.

Exams
Articles
Questions