Semiconductor Electronics: Materials, Devices And Simple Circuits Class 12th Physics Part Ii CBSE Solution

Class 12th Physics Part Ii CBSE Solution
Exercises
  1. In an n-type silicon, which of the following statement is true:A. Electrons are majority…
  2. Which of the statements given in Exercise 14.1 is true for p-type semiconductors.A.…
  3. Carbon, silicon and germanium have four valence electrons each. These are characterised by…
  4. In an unbiased p-n junction, holes diffuse from the p-region to n-region becauseA. free…
  5. When a forward bias is applied to a p-n junction, itA. raises the potential barrier. B.…
  6. For transistor action, which of the following statements are correct:A. Base, emitter and…
  7. For a transistor amplifier, the voltage gainA. remains constant for all frequencies. B. is…
  8. In half-wave rectification, what is the output frequency if the Input frequency is 50 Hz.…
  9. For a CE-transistor amplifier, the audio signal voltage across the collected resistance of…
  10. Two amplifiers are connected one after the other in series (cascaded). The first amplifier…
  11. A p-n photodiode is fabricated from a semiconductor with the band gap of 2.8 eV. Can it…
Additional Exercises
  1. The number of silicon atoms per m^3 is 5 × 10^28 . This is doped simultaneously with 5 ×…
  2. In an intrinsic semiconductor, the energy gap Eg is 1.2eV. Its hole mobility is much…
  3. In a p-n junction diode, the current I can be expressed as Where I0 is called the reverse…
  4. You are given the two circuits as shown in Fig. 14.44. Show that circuit (a) acts as OR…
  5. Write the truth table for a NAND gate connected as given in Fig. 14.45. Hence identify the…
  6. You are given two circuits as shown in Fig. 14.46, which consist of NAND gates. Identify…
  7. Write the truth table for circuit given in Fig. 14.47 below consisting of NOR gates and…
  8. Write the truth table for the circuits given in Fig. 14.48 consisting of NOR gates only.…

Exercises
Question 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.


Answer:

In an n-type silicon holes are minority carriers and pentavalent


atoms are the dopants as it is formed by doping of a pentavalent atom in a tetravalent crystal.


Pentavalent atoms have an excess electron than a tetravalent atom, hence a number of holes are less than the number of electrons. So, holes are the minority carriers.



Question 2.

Which of the statements given in Exercise 14.1 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 the dopants.


Answer:

In a p-type semiconductor holes are majority carriers and trivalent atoms are dopants as it is formed by doping of a trivalent atom in a tetravalent crystal.


Trivalent atoms have an electron less than a tetravalent atom, hence a number of holes are greater than the number of electrons (As Absence of electron represents hole).So, holes are the majority carriers.



Question 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


Answer:

For an insulator, the band gap energy is always higher, (Eg≈ 6ev). And in case of silicon and germanium, silicon has higher energy gap than germanium. (This is experimental data.)


Question 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.


Answer:

In an unbiased p-n junction diode, the holes diffuse from the p-region to the n-region because of the difference in their concentration in the regions. Holes concentration is more in p-region, hence to achieve equilibrium they diffuse towards less concentrated n-region.


Question 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.


Answer:

When a forward bias voltage is applied, The N-region is attached to the negative terminal of the battery and the P-region is connected to the positive terminal of the battery. Negative-negative and positive-positive repulsion occur on both sides, so it opposes the junction barrier voltage, hence the barrier voltage is reduced/lowered. The diagram of a diode in forward bias is shown as below:



Question 6.

For transistor action, which of the following statements are correct:
A. Base, emitter and collector regions should have similar size and doping concentrations.

B. The base region must be very thin and lightly doped.

C. The emitter junction is forward biased and collector junction is reverse biased.

D. Both the emitter junction as well as the collector junction are forward biased.


Answer:

The base region is made very thin and lightly doped, as it’s the main function is to control the flow of electron through the transistor. It’s lightly doped hence less number of majority carriers will be there resulting in a lesser percentage of emitter current flow through the base than the collector. That’s why base current is negligibly small.


Also, emitter junction is always forward biased and collector junction is reverse biased. This setup results in the transistor being in the active region. Otherwise, no current flows through the transistor. (Cut-off or saturation region)


Hence, both B) and C) are correct.


Question 7.

For a transistor amplifier, the voltage gain
A. remains constant for all frequencies.

B. is high at high and low frequencies and constant in the middle frequency range.

C. is low at high and low frequencies and constant at mid frequencies.

D. None of the above


Answer:

The gain is maximum and constant in mid frequency ranges and is low at both high and low-frequency ranges. The figure gives the Distribution of a transistor-amplifier gain over a frequency range.



Question 8.

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.


Answer:

During a cycle, half wave rectifier conducts once and full wave rectifier conducts twice, hence the output frequency for the half-wave rectifier will be same i.e. 50 Hz, whereas the output frequency for the full wave rectifier will be double, i.e. 100 Hz.



Output frequency for full wave rectifier = 100Hz, for half-wave rectifier it’s 50 Hz.


(Actually, this is application of p-n junction diode)



Question 9.

For a CE-transistor amplifier, the audio signal voltage across the collected resistance of 2k is 2 V. Suppose the current the amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1k.


Answer:

Given that,


Rc = 2K-Ω = 2000 Ω (Collector Resistance)


V = 2 Volts (Audio signal voltage across the collector resistance)


β = 100 (Current amplification factor of the transistor)


RB = 1K-Ω (Base resistance)


Let, the Input signal voltage be Vi


Base current = IB


We have the relation for voltage amplification as:



⇒ 


⇒ 


⇒Vi = 0.01 V


Therefore, the input signal voltage of the amplifier is 0.01 V.


We know that Base Resistance:


RB = Vi/IB


⇒ IB = Vi/RB = 0.01 V/1000Ω = 10μA


Therefore, the base current of the amplifier is 10μA



Question 10.

Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.


Answer:

Let A1 = 10 (Voltage gain of the first amplifier)


and A2 = 20 (Voltage gain of the second amplifier)


Vi = 0.01 V (Input signal voltage)


Let, Output AC signal voltage = Vo


We have the relation:


V0 = A × Vi


Output after the first stage VO1 = A1 × Vi


Then, output after the 2nd stage VO2 = VO1 × A2 = Vi × A1 × A2


Hence, the final output = VO = 20 × 10 × 0.01 = 2 Volts


Therefore, the output AC signal of the given amplifier is 2 V.



Question 11.

A p-n photodiode is fabricated from a semiconductor with the band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?


Answer:

Given that the Energy band gap of photodiode, Eg = 2.8 eV


Wavelength, λ = 6000 nm = 6000 × 10–9 m


The energy of a signal is given by the relation:


 …. (I)


Where,


h = Planck’s constant = 6.626 × 10–34J-s


c = Speed of light = 3 × 108m/s


So, E = 3.3 × 10–20 J [From equation (I)]


We know that, 1.6 × 10–19J = 1 eV


Therefore,


∴ E = 0.21 eV


The energy of any particular signal of wavelength 6000 nm is 0.21 eV, which is less than given 2.8 eV − the energy band gap of the photodiode. So, the photodiode cannot detect the signal.


(Actually, this is example of application of photodiode)




Additional Exercises
Question 1.

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?


Answer:

Given,

Number of Silicon atoms per m3 = 5 × 1028


Number of Arsenic atoms per m3 (nAs) = 5 × 1022


Number of Indium atoms per m3 (nIn) = 5 × 1020


Intrinsic charge carrier concentration (ni) = 1.5 × 1016 m-3


So, number of electrons per m-3 (ne) = nAs – ni


i.e., ne = 5 × 1022 – 1.5 × 1016 = 4.99 × 1022 m-3


If nh is the number of holes per m-3, then according to Mass-action law,


ni2 = nenh


So, nh = ni2/ne


i.e., nh = 


So, nh= 4.5 × 109 m-3


The number of electrons is higher than the number of holes. So, the given material is n-type.


NOTE: Arsenic is pentavalent and Indium is trivalent. In an n-type semiconductor, electrons are the majority charge carriers i.e., nenh. In a p-type semiconductor, holes are the majority charge carriers i.e., nhne.



Question 2.

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.


Answer:

Given,

Energy gap (Eg) = 1.2eV


The temperature dependence of intrinsic carrier concentration is given as



Where T is the temperature


Eg is the forbidden energy gap


KB is the Boltzmann’s constant (8.62 × 10-5 eV/K)


no is a constant.


At temperature T1 = 600K,


 ……….(1)


At temperature T2 = 300K,


 …..(2)


Dividing eqn. (1) by eqn. (2), we get



⇒ 


⇒ 


⇒ 


⇒ 


So, 


Hence, the ratio of conductivity at 600K to that at 300K is 1.09 × 105.


NOTE: 1. Eg is the forbidden energy gap which is the energy gap between the valence band and conduction bands.


2. Intrinsic semiconductors are pure semiconductors with fewer conductivities. Trivalent and pentavalent impurities are added to intrinsic semiconductors to form p-type and n-type semiconductors respectively. This process is called doping.



Question 3.

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?


Answer:

Given,

Reverse saturation current (Io) = 5 × 10-12 A


Temperature (T) = 300K


The diode current (I) is given by the relation,



Where Io is the reverse saturation current


V is the voltage across the diode


kB is the Boltzmann’s constant


T is the absolute temperature


e is the electronic charge (1.6 × 10-19 C)


Note: The value of kB (in electron volts) is 8.6 × 10–5 eV/K and also 1.38 × 10-23 kg m2s-2K-1)


(a) For V = 0.6V,



⇒ 


⇒ 


So, I = 5.42 × 10-7 A


(b) For V = 0.7V,



⇒ 


⇒ 


So, I = 3.74 × 10-6 A


Hence, increase in current = 3.74 × 10-6 - 5.42 × 10-7 = 3.198 × 10-6 A


(c) 


So, dynamic resistance = 


i.e., Dynamic resistance = 31269.5 Ω = 31.26kΩ


(d) If the reverse bias voltage changes from 1V to 2V, then the diode current remains equal to reverse saturation current. So, dynamic resistance will be infinite.


NOTE: A diode is said to be in forward bias if the p-side and n-side are connected to the positive and negative terminals of the battery respectively. Electric current flowing through the diode in this condition is known as forward diode current. If the p-side is connected to the negative terminal and n-side is connected to the positive terminal of the battery, then the diode is said to be in reverse bias.



Question 4.

You are given the two circuits as shown in Fig. 14.44. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate.



Answer:

The circuit (a) is as follows:


STEPS:


1. The output of the NOR gate is the complement of A + B i.e., (A + B)’.


2. The NOT gate gives the complement of this output.


3. So, the output Y = A + B.


The truth table for this circuit is as follows:



The circuit (b) is as follows:



STEPS:


1. The NOT gates give the complements of the inputs A and B i.e., A’ and B’.


2. The NOR gate gives the output as (A’ + B’)’.


3. Using De Morgan’s law, (A’ + B’)’ = ((A∙B)’)’ = A∙B


4. So, the output Y = A∙B.




Question 5.

Write the truth table for a NAND gate connected as given in Fig. 14.45.



Hence identify the exact logic operation carried out by this circuit.


Answer:

The output is obtained as (A∙A)’ = A’. (Since, A∙A = A)

Hence, the given gate acts as a NOT gate.


The truth table is given as follows:



NOTE: NAND gate is a universal gate because it can be used to implement any Boolean function without using any other gates.



Question 6.

You are given two circuits as shown in Fig. 14.46, which consist of NAND gates. Identify the logic operation carried out by the two circuits.



Answer:

The circuit (a) is as follows:


The first NAND gate gives the output (A∙B)’. The second NAND gate gives the output ((A∙B)’∙(A∙B)’)’.


Using De Morgan’s law,


((A∙B)’∙(A∙B)’)’ = ((A∙B)’)’ + ((A∙B)’)’ = (A∙B) + (A∙B) = A∙B (Since, A + A = A)


So, the output is Y = A∙B. The circuit acts as an AND gate.


The truth table for the circuit is as follows:



The circuit (b) is as follows:



The first NAND gate gives (A∙A)’ = A’ as the output. The second NAND gate gives (B∙B)’ = B’ as the output. The last NAND gate gives (A’∙B’)’ as the output.


Using De Morgan’s law,


(A’∙B’)’ = (A’)’ + (B’)’ = A + B


So, the output is A + B. The circuit acts as an OR gate.


The truth table for the circuit is as follows:



NOTE: NAND gate is a universal gate because it can be used to implement any Boolean function without using any other gates.



Question 7.

Write the truth table for circuit given in Fig. 14.47 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.)


Answer:

The given circuit is as follows:


The first NOR gate gives the output (A + B)’. The second NOR gate gives the output ((A + B)’ + (A + B)’)’.


Using De Morgan’s law,


((A + B)’ + (A + B)’)’ = ((A + B)’)’∙((A + B)’)’ = (A + B)∙(A + B) = A + B (Since, A∙A = A)


So, the output is Y = A + B and the circuit acts as an OR gate.


The truth table for the given circuit is as follows:



NOTE: NOR gate is a universal gate because it can be used to implement any Boolean function without using any other gates.



Question 8.

Write the truth table for the circuits given in Fig. 14.48 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.



Answer:

The circuit (a) is as follows:


The circuit gives the output (A + A)’ = A’.


So, the output is Y = A’ and the circuit acts as a NOT gate.


The truth table for the circuit is as follows:



The circuit (b) is as follows:



Similar to the circuit (a), the first two gates give inverted output as A’ and B’. The third NOR gate gives the output (A’ + B’)’.


Using De Morgan’s law,


(A’ + B’)’ = (A’)’∙(B’)’ = A∙B (Since, (A’)’ = A)


Hence, the output is Y = A∙B and the circuit acts as an AND gate.


The truth table is as follows:



NOTE: NOR gate is a universal gate because it can be used to implement any Boolean function without using any other gates.


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