RBSE Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

Rajasthan Board RBSE Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits Textbook Exercise Questions and Answers.

RBSE Class 12 Physics Solutions Chapter 14 Semiconductor Electronics: Materials, Devices and Simple Circuits

RBSE Class 12 Physics Semiconductor Electronics: Materials, Devices and Simple Circuits Textbook Questions and Answers

Question 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.
Answer:
(c) Holes are minority carriers and pentavalent atoms are the dopants.

Question 14.2.
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:
(d) Holes are majority carriers and trivalent atoms are the dopants.

Question 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 sfatments 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:
(c) (Eg)_C > (Eg)_Si > (Eg)_Ge
 
Question 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.
Answer:
(c) hole concentration in p-region is more as compared to n-region

Question 14.5.
When a forward bias is applied to a p-n junction, it
(а) raises the potential, barrier
(b) reduces the majority carrier current to zero
(c) lowers the potential barrier
(d) None of the above.
Answer:
(c) lowers the potential barrier

Question 14.6.
For transistor action, which of the following statements are correct?
(а) 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 collectpr junction are forward biased
Answer:
(b) The base region must be very thin and lightly doped
(c) The emitter junction is forward biased and collector junction is reverse biased.

Question 14.7.
For transistor amplifier, the voltage gain.
(a) remains constant for all frequencies
(b) is high at high and low frequencies
(c) is low at high and low frequencies and constant in the middle frequency range
(d) None of the above.
Answer:
(c) is low at high and low frequencies and constant in the middle frequency range

Question 14.8.
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:
Given input frequency = 50 Hz.
Output frequency
For Half wave rectifier = 50 Hz
For full wave rectifier = 50 x 2
= 100 Hz.

Question 14.9. For a C.E. transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2V. Suppose the current amplification factor of the transistor is 100, find the input signal voltage and base current, if the base resistance is 1 kΩ.
Answer:
Given ß = 100, R_i = 1 kΩ,
R₀ = 2 kΩ
Voltage gain,
A = Current gain x Resistance gain
= 100 x \(\frac{\mathrm{R}_0}{\mathrm{R}_i}\) = 100 x \(\frac{2}{1}\) = 200

∴ I_B = \(\frac{\mathrm{I}_{\mathrm{C}}}{\beta}=\frac{1 m \mathrm{~A}}{100}\) = 0.01 mA = 10 µA.

Question 14.10.
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 a.c. signal.
Answer:
Given A_V1 =10 and A_V2 = 20
V_i1 = 0.01V
So voltage gain of 1st amplifier
A_V1 = \(\frac{\mathrm{V}_{01}}{\mathrm{~V}_{i 1}} \)
∴ V₀₁ = V_i1 A_V1
or V₀₁ = 0.01 X 10 = 0.1 V
And voltage gain of 2nd amplifer
A_V2 = \(\frac{V_{02}}{V_{01}}\)
0r V₀₂ = V₀₁ A_V2
= 0.1 x 20
or V₀₂ = 2V.

Question 14.11.
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Answer:
Band gap E_g = 2.8 eV
Energy band gap corresponding to wavelength 6000 nm is given by

Since E_g' < E_g, so it cannot detect a wavelength of 6000 nm.

ADDITIONAL EXERCISES

Question 14.12.
The number of silicon atoms per m³ in 5 x 10²². This is doped simultaneously with 5 x 10²² atoms per m³ of Arsenic and 5 x 10²⁰ per m³ atoms of Indium. Calculate the number of electrons and holes. Given that n_i = 1.5 x 10¹⁶ m^-3. Is the material n-type or p-type?
Answer:
Given n = 5 x 10²² m^-3
N_D = 5 x 10²² m^-3
N_A = 5 x 10²⁰ m^-3
= 0.05 x 10²² m^-3
n_i = 1.5 x 10¹⁶ m^-3.
Since n_e - n_h = N_D - N_A .....................(1)
And n_e - n_h = n_i² ....................(2)
So n_e + n_h = \(\sqrt{\left(n_e-n_h\right)^2+4 n_e n_h}\)
Using Eq. (1) and (2), we get
n_e + n_h = \(\sqrt{\left(\mathrm{N}_{\mathrm{D}}-\mathrm{N}_{\mathrm{A}}\right)^2+4 n_i^2}\) .........................(3)
Adding (1) and (3), we get
n_e = \(\frac{1}{2}\left[\left(\mathrm{~N}_{\mathrm{D}}-\mathrm{N}_{\mathrm{A}}\right)+\sqrt{\left(\mathrm{N}_{\mathrm{D}}-\mathrm{N}_{\mathrm{A}}\right)^2+4 n_i^2}\right]\)

n_h = 4.55 x 10⁹
Since n_e >> n_h, so the material is n-type.

Question 14.13.
In an intrinsic semiconductor the energy gap Eg is 1.2 eV. Its hole mobility is independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration ni is expressed as:
n_i = n₀ exp\([\frac{-\mathrm{E}_{\mathrm{g}}}{2 \mathrm{k}_{\mathrm{B}} \mathrm{T}}]\)
when n₀ is a constant.
Answer:
Since µ_e > µ_h and for the intrinsic semiconductor n_e = n_h = n_i can be written as
σ_e = µ_en_e
Using the equation for ni given in the problem
σ = eµ_en₀ exp.\(\left[\frac{-\mathrm{E}_g}{k_{\mathrm{B}} \mathrm{T}}\right]\)
All the pre-exponential terms are assumed independence of temperature. We can put a constant.
σ₀ = eµ_en₀
and express the conductivity.

Takin gk_B = 8.62 x 10^-5 eVK^-1 and solving, we get
\(\frac{\sigma(600 \mathrm{~K})}{\sigma(300 \mathrm{~K})}\) ≈ 1 x 10⁵
This shows that the conductivity of an intrinsic semiconductor is greatly influenced by temperature.

Question 14.14.
In a p-n junction diode the current I can be expressed.
I = I₀ \(\left[\exp \cdot\left(\frac{e \mathrm{~V}}{\mathrm{k}_{\mathrm{B}} \mathrm{T}}\right)-\mathbf{i}\right]\)
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, K_B is the Boltzmann’s constant (8.6 x 10^-5 eV/K) and T is the absolute temperature. If for a diode I₀ = 5 x 10^-12 A and T = 300 K, then
(а) 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 current if reverse bias voltage changes from IV to 2 V?
Answer:
Given that
I = I₀ \(\left[\exp \cdot\left(\frac{e \mathrm{~V}}{\mathrm{k}_{\mathrm{B}} \mathrm{T}}\right)-\mathbf{i}\right]\)
I₀ = 5 x 10^-12 A; T = 300 K and k_B = 8.6 x 10^-5 e V/K

(a) For V = 0.6 V
\(\frac{e \mathrm{~V}}{k_{\mathrm{B}} \mathrm{T}}=\frac{e \times 0.6 \mathrm{~V}}{8.6 \times 10^{-5}}(e \mathrm{~V} / \mathrm{K})(300 \mathrm{~K})=\frac{100}{4.3}\)
∴ exp. \(\left(\frac{100}{4.3}\right)\) = 1.2585 x 10¹⁰
∴ I = 5 x 10^-12 x 1.2585 x 10¹⁰ = 0.0629 A

(b) V = 0.7 V

I = 5 x 10^-12 x 6.069 x 10¹¹ = 3.0349 A
Increase in current = 3.0349 - 0.0629 = 2.927 A

(c) Dynamic Resistance 
= \(\frac{\delta \mathrm{V}}{\delta \mathrm{I}}=\frac{0.1}{2.927}\) = 0.0336 Ω

(d) For reverse bias V = -1 volt
I = I₀ [e (-38.76) - 1] = -5 x 10-12 A
For V = -2V
I = 5 x 10^-12 A
Change in current is zero.

Question 14.15.
You are given two circuits as shown in Fig. TBQ 14.15. Show that circuit (a) acts as OR gate while the circuit
(b) acts as AND gate.

Answer:
The truth table of circuit (a) is:

A	B	A’	Y
0	0	1	0
0	1	0	0
1	0	0	0
1	1	0	1

Thus we find that the circuit (a) operation of OR gate is performing the operation of OR gate.

A	B	A’	B’	Y
0	0	1	1	0
0	1	1	0	0
1	0	0	1	0
1	1	0	0	1

Thus we find that the circuit (b) is performing the operation of AND gate.

Question 14.16.
Write the truth table for a NAND gate connected as given in Fig. TBQ 14.16.

Answer:
The truth table of the given NAND gate connected as shown in Fig. TBQ 14.16 (b) is

Question 14.17.
You are given two circuits as shown in Fig. TBQ 14.17 which consist of NAND gates. Identify the logic operation carried out by the two circuits.

Answer:
(a) The circuit (a) is AND gate, because the circuit can be represented as:

So the truth table is :

A	B	Y’ =\( \overline{\mathrm{A} \cdot \mathrm{B}}\)	Y = \(\overline{\mathrm{A} \cdot \mathrm{B}}\)
0	0	1	0
0	1	1	0
1	0	1	0
1	1	0	1

Hence the circuit (c) behaves like an AND gate
(b) The circuit (d) is OR gate, because the circuit can be represented as:

So the truth table is :

A	B	\(\overline{\mathrm{A}}\)	\(\overline{\mathrm{B}}\)	Y = \(\overline{\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}}\) = A + B
0	0	1	1	0
0	1	1	0	1
1	0	0	1	1
1	1	0	0	1

Hence the circuit (b) behaves like an OR gate.

Question 14.18.
Write the truth table for circuit given in the figure TBQ 14.18 (a) 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 truth of OR, AND, NOT gates and find the correct one.

Answer:
The truth table is shown in Fig. TBQ 14.18 (b)

Thus we find that the circuit is performing the operation of OR gate.

Question 14.19.
Write the truth table for the circuit given in the following figures consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.

Answer:
In Fig. TBQ 14.19 (a), two inputs A and B are given to a NOR gate and its output is given to the input of a NOT gate.
So the truth table is:

A	B	Y = \(\overline{\mathrm{A}+\mathrm{B}}\)
0	0	1
1	1	0

This is the truth table of OR gate, Hence Fig. TBQ 14.19 (a) acts as an OR gate.
In Fig. TBQ 14.19 (b), inputs A and B are given to two NOT gates and their output is given to a NOR gate.
So the truth table is:

A	B	\(\overline{\mathrm{A}}\)	\(\overline{\mathrm{B}}\)	Y = \(\overline{\overline{\mathrm{A}}+\overline{\mathrm{B}}} \overline{\bar{A}}+\overline{\bar{B}}\) = A.B
0	0	1	1	0
0	1	1	0	0
1	0	0	1	0
1	1	0	0	1

This is the truth table of an AND gate, hence Fig. TBQ 14.19 (b) acts as an AND gate. 

SELECTED EXEMPLAR PROBLEMS

MCQ I. (with one correct option)

Question 14.1.
The conductivity of a semiconductor increases with increase in temperature because
(a) number density of free current carriers increases.
(b) relaxation time increases.
(c) both number density of carriers and relaxation time increase.
(d) number density of current carriers increases, relaxation time decreases but effect of decrease in relaxation time is much less than increase in number density.
Answer:
(d) n increases, r decreases but decrease in r is much less than increase in n.

Question 14.2.
In Fig. EP 14.2 V₀ is the potential barrier across a p-n junction, when no battery is connected across the junction
(a) 1 and 3 both correspond to forward bias of junction
(b) 3 corresponds to forward bias of junction and 1 corresponds to reverse bias of junction.
(c) 1 corresponds to forward bias and 3 corresponds to reverse bias of junction.
(d) 3 and 1 both correspond to reverse bias of junction.

Answer:
(b) A p-n junction during forward bias opposes the potential barrier and during reverse bias, it supports the potential barrier.

Question 14.3.
In Fig. EP 14.3. assuming the diodes to be ideal.
(a) D₁ is forward biased and D₂ is reverse biased and hence current flows from A to B
(b) D₂ is forward biased and D₁ is reverse biased and hence no current flows from B to A and vice versa.
(c) D₁ and D₂ are both forward biased and hence current flows from A to B.
(d) D₁ and D₂ are both reverse biased and hence no current flows from A to B and vice versa

Answer:
(d) Here both the diodes are reverse biased as it is clear from the figure.

Question 14.4.
A 220 V A.C. supply is connected between points A and B (Fig. EP 14.4). What will be the potential difference V across the capacitor?
(a) 220V.
(b) 110 V
(c) 0V.
(d) 220 \(\sqrt{2}\) V.

Answer:
(d) A p-n junction conducts only during positive half, and the potential across the capacitor is equal to the peak voltage of a.c. 
i.e. E₀ = E_v \(\sqrt{2}\) = 220 \(\sqrt{2}\) V

Question 14.5.
Hole is
(a) an anti-particle of electron.
(b) a vacancy created when an electron leaves a covalent bond.
(c) absence of free electrons.
(d) an artificially created particle.
Answer:
(b) (Knowledge based question.)

Question 14.6.
The output of the given circuit in Fig. EP 14.6.
(а) would be zero at all times.
(b) would be like a half wave rectifier with positive cycles in output.
(c) would be like a half wave rectifier with negative cycles in output.
(d) would be like that of a full wave rectifier.

Answer:
(c)
The p-n junction is forward biased during +ve half of input ac and its resistance is low, and large potential difference will appear across the resistance connected in series. And voltage across p-n junction is zero.
The p-n junction is reverse biased during -ve half of input a.c. and its resistance is high, and will be more than resistance connected in series and there will be voltage across the p-n junction with negative cycle in output. Hence the circuit would be half wave rectifier with negative cycle in output.

Question 14.7.
In the circuit shown in Fig. EP 14.7, if the diode forward voltage drop is 0.3 V, the voltage difference between A and B is
(а) 1.3 V
(b) 2.3 V
(c) 0
(d) 0.5 V

Answer:
(b) V - 0.3 = RI
V - 0.3 = [(5 + 5) x 10³] x 0.2 x 10^-3
= 10 x 10³ x 0.2 x 10^-3
= 2
or V = 2 + 0.3 = 2.3 volt.

Question 14.8.
Truth table for the given circuit (Fig. EP 14.8.) is

Answer:
(c) Here C = A B & D = \(\overline{\mathrm{A}}\). B 
And E = C + D = A.B + \(\overline{\mathrm{A}}\) B
Truth table is

A	B	\(\overline{\mathrm{A}}\)	C = A.B	D = \(\overline{\mathrm{A}}\) B	E = C + D
0	0	1	0	0	0
0	1	1	0	1	1
1	0	0	0	0	0
1	1	0	1	0	1

MCQ II. (with more than one correct options)

Question 14.9.
When an electric field is applied across a semiconductor
(а) electrons move from lower energy level to higher energy level in the conduction band.
(b) electrons move from higher energy level to lower energy level in the conduction band.
(c) holes in the valence band move from higher energy level to lower energy level.
(d) holes in the valence band move from lower energy level to higher energy level.
Answer:
(a), (c) Knowledge based question.

Question 14.10.
Consider a npn transistor with its base-emitter junction forward biased and collector base junction reverse biased. Which of the following statements are true?
(a) Electrons cross over from emitter to collector.
(b) Holes move from base to collector.
(c) Electrons move from emitter to base.
(d) Electrons from emitter move out of base without going to the collector.
Answer:
(a), (c) Knowledge based question.

Question 14.11.
Figure EP 14.11 shows the transfer characteristics of a base-biased CE transistor. Which of the following statements are true?

(a) At V_i = 0.4 V, transistor is in active state.
(b) At V_i = 1V, it can be used as an amplifier.
(c) At V_i = 0.5 V, it can be used as a switch turned off.
(d) At V_i = 2.5 V, it can be used as a switch turned on.
Answer:
(b), (c), (d)
At V_i = 0.4 V, there is no current and transistor is in cut-off state.

Question 14.12.
In a npn transistor circuit, the collector current is 10 mA. If 95 percent of the electrons emitted reach the collector, which of the following statements are true?
(a) The emitter current will be 8 mA.
(b) The emitter current will be 10.53 mA.
(c) The base current will be 0.53 mA.
(d) The base current will be 2 mA.
Answer:
(b), (c)
Given I_c = 10 mA
And I_c = \(\frac{95}{100}\) I_e
or I_e = \(\frac{100}{95}\) I_c = \(\frac{10 \times 100}{95}\) = 10.53 mA
I_b = I_e - I_c = 10.53 - 10 = 0.53 mA

Question 14.13.
In the depletion region of a diode
(a) there are no mobile charges
(b) equal number of holes and electrons exist, making the region neutral.
(c) recombination of holes and electrons has taken place.
(d) immobile charged ions exist.
Answer:
(a), (b), (d)
Electrons and holes are not present in depletion layer hence their recombination cannot take place.

Question 14.14.
What happens during regulation action of a Zener diode?
(a) The current and voltage across the Zener remain fixed.
(b) The current through the series resistance (Rg.) changes.
(c) The Zener resistance is constant.
(d) The resistance offered by the Zener changes.
Answer:
(b), (d)
During regulation action of a Zener diode, the current through resistance changes and the resistance offered by the Zener changes.

Question 14.15.
To reduce the ripples in a rectifier circuit with capacitor filter
(a) R_L should be increased. 
(b) input frequency should be decreased.
(c) input frequency should be increased.
(d) capacitors with high capacitance should be used.
Answer:
(a), (c), (d)
Since ripple factor with capacitor filter is:

So to reduce γ, Increase R_L, C and v.

Question 14.16.
The breakdown in a reverse biased p-n junction diode is more likely to occur due to
(a) large velocity of the minority charge carriers if the doping concentration is small.
(b) large velocity of the minority charge carriers if the doping concentration is large.
(c) strong electric field in a depletion region if the doping concentration is small.
(d) strong electric field in a depletion region if the doping concentration is large.
Answer:
(a), (d)

Very Short Answer Type Questions

Question 14.17.
Wliy are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?
Answer:
The dopant should be such that the size of the dopant atom should not distort the pure semi-conductor lattice structure and it should easily contributes a charge carrier, which are provided by group XIII or group XV elements.

Question 14.18.
Sn, C, and Si, Ge are all group XIV elements. Yet, Sn is a conductor, C is an insulator while Si and Ge are semiconductors. Why?
Answer:
The energy gap for Sn is 0 eV, for C is 5.4 eV and Si is 1.1 eV and for Ge is 0.7 eV. So Sn is a conductor, C is an insulator, Si and Ge are semi-conductors.

Question 14.19.
Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction?
Answer:
No, because the resistance of the voltmeter should be very high as compared to the junction resistance and resistance of the junction is nearly infinite.

Question 14.20.
Draw the output waveform across the resistor (Fig. EP 14.20).

Question 14.21.
The amplifiers X, Y and Z are connected in series. If the voltage gains of X, Y and Z are 10, 20 and 30, respectively and the input signal is 1 mV peak value, then what is the output signal voltage (peak value)
(i) if d.c. supply voltage is 10V?
(ii) if d.c. supply voltage is 5V?
Answer:
(i) If d.c. supply voltage is 10V, the output signal voltage
∆V₀ = 10 x 20 x 30 x 10^-3 = 6V
(ii) If d.c. supply voltage is 5 V, then output peak will not exceed 5 V. So V₀ = 5 V.

Question 14.22.
In a CE transistor amplifier there is a current and voltage gain associated with the circuit. In other words there is a power gain. Considering power a meausre of energy, does the circuit violate conservation of energy?
Answer:
There is no violation of conservation of energy because the extra power required for amplified output is obtained from d.c. source.

Short Answer Type Questions

Question 14.23.

(i) Name the type of a diode whose characteristics are shown in Fig. EP 14.23 (a) and Fig. EP 14.23 (b).
(ii) What does the point P in Fig. EP 14.23 (a) represent?
(iii) What does the points P and Q in Fig. EP 14.23 (b) represent?
Answer:
(i) Zener junction diode and solar cell
(ii) Zener breakdown voltage
(iii) Q-Short circuit current and P-open circuit voltage.

Question 14.24.
Three photodiodes D₁, D₂ and D₃ are made of semiconductors having band gaps of 2.5 eV, 2 eV and 3eV, respectively. Which ones will be able to detect light of wavelength 6000 Å?
Answer:
Energy of incident light
E = hv = \(\frac{h c}{\lambda}=\frac{6.62 \times 10^{-34} \times 3 \times 10^8}{6000 \times 10^{-10}}\) = 2 eV
So only diode D₂ will detect this radiation, because the incident radiation photon should be greater than the band gap.

Question 14.25.
If the resistance, R₁ is increased (Fig. EP 14.25), how will the readings of the ammeter and voltmeter change?

Answer:
From Fig. EP 14.25 (a) we find that

If R₁ is increased, I_B will decrease.
Also I_C = ßI_B
As I_B decreased, I_C also decreased. So there is decrease in ammeter and voltmeter readings.

Question 14.26.
Two car garages have a common gate which needs to open automatically when a car enters either of the garages or cars enter both. Devise a circuit that resembles this situation using diodes for this situation.
Answer:
The circuit is an OR gate as shown in the figure.

The gate output is given by the truth table.

A	B	Y = A + B
0	0	0
0	1	1
1	0	1
1	1	1

Question 14.27.
How would you set up a circuit to obtain NOT gate using a transistor?
Answer:
The circuit diagram using a transistor to obtain NOT gate is as shown in the figure

Truth table 

A	Y
0	1
1	0

Question 14.28.
Explain why elemental semi-conductor cannot be used to make visible LEDs.
Answer:
The band gap of elemental semiconductors is such that the emissions are in IR region and not in visible region as required in LEDs.

Question 14.29.
Write the truth table for the circuit shown in Fig. EP 14.29. Name the gate that the circuit resembles.

Answer:

A	B	Y = V0
0	0	0
0	1	0
1	0	0
1	1	1

The circuit resembles AND gate

Question 14.30.
A Zener of power rating 1 W is to be used as a voltage regulator. If Zener has a breakdown of 5V and it has to regulate voltage which fluctuated between 3V and 7 V, what should be the value of Rs for safe operation (Fig. EP 14.30)?

Answer:
I_{z max} = \(\frac{\mathrm{P}}{\mathrm{V}_{\mathrm{z}}}=\frac{1}{5}\) = 0.2 A = 200 mA
R_s = \(\frac{\mathrm{V}_{\mathrm{s}}-\mathrm{V}_{\mathrm{z}}}{\mathrm{I}_{\mathrm{z} \max }}=\frac{2}{0.2}\) = 10 Ω

Long Answer Type Questions

Question 14.31.
If each diode in Fig. EP 14.31 has a forward bias resistance of 25 Ω and infinite resistance in reverse bias, what will be the values of the currents I₁, I₂, I₃ and I₄?

Answer:
Here diode in branch AB and EF are forward biased and diode in branch CD is reverse biased (infinite resistance).
∴ I₃ = 0
Resistance in branch AB and EF all each = 125 + 25 = 150 Ω
These two resistances all in parallel.
So their equivalent resistance is \(\frac{150}{2}\) = 75 Ω
∴ Total resistance in the circuit
= 75 + 25 = 100 Ω
∴ Current I₁ = \(\frac{5}{100}\) = 0.05 A
Since I₁ = I₂ + I₃ + I₄
or I₁ = I₂ + I₄ ..............(i) [∵ I₃ = 0]
Since resistance in branch AB and EF is the same 
So I₄ = I₂
So eq. (i) becomes
I₁ = 2I₂
or I₂ = \(\frac{\mathrm{I}_1}{2}=\frac{0.05}{2}\) = 0.025 A
And I₄ = 0.025 A

Question 14.34.
Consider the circuit arrangement shown in Fig. EP 14.34 (a) for studying input and output characteristics of npn transistor in CE configuration.  

Select the values of RB and RC for a transistor whose V_BE = 0.7 V, so that the transistor is operating at point Q as shown in the characteristics graph in Fig. EP 14.34 (b).
Given that the input impedance of the transistor is very small and V_CC = V_BB = 16 V, also find the voltage gain and power gain of circuit making appropriate assumptions.
Answer:
Here V_BE = 0.7 V, V_CC = V_BB = 16 V
(From the graph)
V_CE = 8 V
I_C = 4 mA = 4 x 10^-3 A
and I_B = 30 µA = 30 x 10^-6 A
∴ V_CC = I_C R_C + V_CE
or R = \(\frac{\mathrm{V}_{\mathrm{CC}}-\mathrm{V}_{\mathrm{CE}}}{\mathrm{I}_{\mathrm{C}}}\)
= \(\frac{16-8}{4 \times 10^{-3}}\) = 2 kΩ
As V_BB = I_BR_B + V_BE

Question 14.35.
Assuming the ideal diode, draw the output waveform for the circuit given in Fig. EP 14.35. Explain the waveform.

Answer:
When input voltage is equal or less than 5V, the diode will be reverse biased offering high resistance and diode appears in open circuit. And input waveform is passed to output terminal. So the sine wave input is to dip off all positive going portion above 5V. When input is more than 5V, the diode is conducting and there will be no voltage beyond 5V as the voltage beyond 5V will appear across R. When input voltage is -ve there will be opposition to 5 V battery in p-n junction. Due to it, reverse bias voltage decreases and appear across output. When input voltage is more than -5V, the diode will be reverse biased and junction diode appears in open circuit as shown in the given figure.

Question 14.36.
Suppose a ‘n’-type wafer is created by doping Si crystal having 5 x 10²⁸ atoms/m³ with 1 ppm concentration of As. On the surface 200 ppm Boron is added to create ‘P' region in this wafer. Considering n_i = 1.5 x 10¹⁶ m^-3
(i) Calculate the densities of the charge carriers in the n and p regions
(ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.
Answer:
(i) In ‘n' region: number of e^- is due to As
∴ n_e = N_D = 1 x 10^-6 x 5 x 10²⁸ = 5 x 10²² m^-3.
The minority carriers (hole) in n type water is
n_h = \(\frac{n_i^2}{n_e}=\frac{\left(1.5 \times 10^{16}\right)^2}{5 \times 10^{22}}=\frac{2.25 \times 10^{32}}{5 \times 10^{22}}\)
or n_h = 0.45 x 10¹⁰ m^-3.
Similarly, when boron is implanted a p-type is created with holes
n_h = N_A = 200 x 10^-6 x 5 x 10²⁸ = 1 x 10²⁵ m^-3
Minority carriers created in p-type wafer is
n_e = \(\frac{n_i^2}{n_h}=\frac{2.25 \times 10^{32}}{1 \times 10^{25}}\) = 2.25 x 10⁷ m^-3
(ii) When p-n junction is reverse biased, the minority carrier holes of n-region wafer (n_h = 0.45 x 10¹⁰ m^-3) would contribute more to the reverse saturation current than minority electrons (n_g = 2.25 x 10⁷ m^-3) of p-region water.

Question 14.39.
For the transistor circuit shown in Fig. EP 14.39 evaluate V_E, R_B, R_E given I_c = 1 mA, V_CE = 3V, V_BE = 0.5 V and V_CC = 12 V. ß = 100.

Answer:
Collector current is almost equal to emitter current
i.e. I_c = I_E and R_c = 7.8 kΩ
R_E = 9 - R_c = 1.2 kΩ
∴ V_E = 1.2 V
V_B = V_E + V_BE = 1.7 V

Question 14.40.
In the circuit shown in Fig. EP 14.40, find the value of Rc.

Answer:
I_E = I_C + I_B and I_C = ßI_B ...................(i)
I_CR_C + V_CE + I_ER_E = V_CC ....................(ii)
R I_B + V_BE + I_ER_E = V_CC ..........................(iii)
From eq. (iii), I_E = I_C = ßI_B
From eq. (iii), I_B (R + ßR_E) = V_CC - V_BE