Samacheer Kalvi 12th Physics Guide Chapter 9 Semiconductor Electronics

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 9 Semiconductor Electronics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 9 Semiconductor Electronics

12th Physics Guide Semiconductor Electronics Text Book Back Questions and Answers

Part I:

Textbook Evaluation:

I. Multiple choice questions:

Question 1.
The barrier potential of a silicon diode is approximately
a) 0.7 V
b) 0.3V
c) 2.0 V
d) 2.2V
Answer:
a) 0.7 V

Question 2.
Doping a semiconductor results in
a) The decrease in mobile charge carriers
b) The change in chemical properties
c) The change in the crystal structure
d) The breaking of the covalent bond
Answer:
c) The change in the crystal structure

Question 3.
A forward-biased diode is treated as
a) An open switch with infinite resistance
b) A closed switch with a voltage drop of 0V
c) A closed switch in series with a battery voltage of 0.7V
d) A closed switch in series with a small resistance and a battery.
Answer:
d) A closed switch in series with a small resistance and a battery.

Question 4.
If a half -wave rectified voltage is fed to a load resistor, which part of a cycle the load current will flow?
a) 0° – 90°
b) 90° – 180°
c) 0° – 180°
d) 0° – 360°
Answer:
c) 0°-180°

Question 5.
The primary use of a zener diode is
a) Rectifier
b) Amplifier
c) Oscillator
d) Voltage regulator
Answer:
d) Voltage regulator

Question 6.
The principle in which a solar cell operates
a) Diffusion
b) Recombination
c) Photovoltaic action
d) Carrier flow
Answer:
c) Photovoltaic action

Question 7.
The light emitted in an LED is due to
a) Recombination of charge carriers
b) Reflection of light due to lens action
c) Amplification of light falling at the junction
d) Large current capacity.
Answer:
a) Recombination of charge carriers

Question 8.
When a transistor is fully switched on, it is said to be
a) Shorted
b) Saturated
c) Cut-off
d) Open
Answer:
b) Saturated

Question 9.
The specific characteristic of a common emitter amplifier is
a) High input resistance
b) Low power gain
c) Signal phase reversal
d) Low current gain
Answer:
c) Signal phase reversal

Question 10.
To obtain sustained oscillation in an oscillator,
a) Feedback should be positive
b) Feedback factor must be unity
c) Phase shift must be 0 or 2π
d) All the above
Answer:
d) All the above

Question 11.
If the input to the NOT gate is A = 1011, its output is
a) 0100
b) 1000
c) 1100
d) 0011
Answer:
a) 0100

Question 12.
The electrical series circuit in digital form is
a) AND
b) OR
c) NOR
d) NAND
Answer:
a) AND

Question 13.
Which one of the following represents a forward bias diode? (NEET)

Answer:
a) A node must have a high potential

Question 14.
The given electrical network is equivalent to

a) AND gate
b) OR gate
c) NOR gate
d) NOT gate
Answer:
c) NOR gate

Question 15.
The output of the following circuit is 1 when the input ABC is

a) 101
b) 100
c) 110
d) 010
Answer:
a) 101

II. Short Answer Questions:

Question 1.
Define electron motion in a semiconductor.
Answer:
To move the hole in a given direction, the valence electrons move in the opposite direction. Electron flow in an N-type semiconductor is similar to electrons moving in a metallic wire. The N-type dopant atoms will yield electrons available for conduction.

Question 2.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:

Intrinsic Semiconductor	Extrinsic Semiconductor
1. It is pure form of semiconductor	Small amount of impurity is added
2. No doping takes place in intrinsic semiconductor	Doping takes place
3. Number of free electrons in conduction is equal to number of holes in valence band.	Number of free electrons and holes are not equal
4. It has bad electrical characteristics	It has good electrical conductivity

Question 3.
What do you mean by doping?
Answer:
The process of adding impurities to the intrinsic semiconductor is called doping.

Question 4.
How electron-hole pairs are created in a semiconductor material?
Answer:
When external voltage is applied in semiconductor free electrons from valence band moves to conduction band and recombined with holes. So electron-hole pairs are created.

Question 5.
A diode is called a unidirectional device. Explain
Answer:
A diode is called a unidirectional device, i.e., current flows in only one direction (anode to cathode internally) when a forward voltage is applied, the diode conducts and when a reverse voltage is applied, there is no conduction. A mechanical analogy is a rat chat, which allows motion in one direction only.

Question 6.
What do you mean by leakage current in a diode?
Answer:

1. Under biasing very small current (μA) flows across the junction in a diode.
2. It is due to minority charge carriers.
3. This current is called leakage current or reverse saturation current.

Question 7.
Draw the output waveform of a full-wave rectifier.
Answer:

Input and output waveforms

Question 8.
Distinguish between avalanche and Zener breakdown.
Answer:

Avalanche breakdown	Zener breakdown
1. The phenomenon of increasing the free electrons or current in the semiconductor by applying a higher voltage	The process in which electrons are moving across the barrier from valence band of P-type material to conduction band of lightly filled n -material
2. Depletion layer is thick	Depletion layer is thin
3. Electric field is weak	Electric field is strong
4. Produces pairs of electrons and holes	Produces electrons
5. Doping is low	Doping is heavy
6. Breakdown voltage vary	Voltage remains constant

Question 9.
Discuss the biasing polarities in NPN and PNP transistors.
Answer:
In a PNP transistor, base and collector will be negative with respect to emitter indicated by the middle letter N whereas base and collector will be positive in an NPN transistor [indicated by the middle letter P.

Question 10.
Explain the current flow in an NPN transistor.
Answer:

1. In NPN transistor electron flow from Emitter to collector. So conventional current flow from collector to emitter.
2. Electrons from emitter region flow towards base region constitute emitter current (I_E). Electrons after reaching base region recombine with holes – Most of electrons reach collector region. This constitute collector current (I_c). After recombination of holes in base region by bias voltage V_EE constitute base current I_B.
   I_E = I_B + I_C [Since base current is very low I_c ≈ I_C]

Question 11.
What is the phase relationship between the AC input and output voltages in a common emitter amplifier? What is the reason for the phase reversal?
Answer:
In a common emitter amplifier, the input and output voltages are 180° out of phase or in, opposite phases. The reason for this can be seen from the fact that as the input voltage rises, so the current increases through the base circuit.

Question 12.
Explain the need for a feedback circuit in a transistor oscillator.
Answer:
If the portion of the output fed to the input is in phase with the input, then the magnitude of the input increases. It is necessary for sustained oscillations, so feed back circuit is needed for transistor oscillator.

Question 13.
Give circuit symbol, logical operation, truth table, and Boolean expression of AND, OR, NOT, NAND, NOR, and EX-OR gates.
Answer:

Question 14.
State De Morgan’s first and second theorems.
Answer:
First Theorem:
The complement of the sum is equal to the product of its complements
\(\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)
Second Theorem:
The complement of the product is equal to sum of its complements
\(\overline{A \cdot B}=\bar{A}+\bar{B}\)

III. Long Answer Questions:

Question 1.
Elucidate the formation of a N – type and P – type semiconductors.
Answer:
1. N – type semiconductor:
N – type semiconductor is obtained by doping a pure Germanium or silicon with dopant from group V pentavalent elements like phosphorous, Arsenic, Antimony.

n-type extrinsic semiconductor:
(a) Free electron which is loosely attached to the lattice
(b) Representation of donar energy level.

1. Dopants has 5 valence electrons, Germanium has 4 valence electrons.
2. During the process of doping few Ge atoms’ are replaced by group V dopants.
3. Four of the five valence electrons of impurities bound with 4 valence electrons of neighbouring Ge atoms.
4. The 5th valence electron of impurity atom is loosely attached, it has not formed covalent bond.
5. Energy level loosely attached electron is just below the conduction band edge which is called donor energy level.
6. At room temperature electrons easily moves to conduction band with absorption of thermal energy.
7. These thermally generated electrons leave holes in valance band.
8. Such a semiconductor doped with a pentavalent impurity is called n-type semiconductor.

2. P-type semiconductor.
1. Trivalent atom from group III elements such as Boron, Aluminium, Gallium, Indium is added with Germanium or silicon is a p-type semiconductor.
2. The dopants with 3 valence electrons bound with neighbouring Germanium atom one electron position of the dopant in Ge lattice will remain vacant.
3. The missing electron position in covalent band is denoted as a hole.
4. To make complete covalent bond with all 4 neighbouring atom, dopant needs one more electrons.
5. Dopants accept electron from neighbouring atoms. This impurity is called acceptor impurity.
6. Energy level of hole created by each impurity atom is just above valence band, which is called acceptor energy level.
7. In such extrinsic semiconductor holes are majority carriers and thermally generated electrons are minority carriers. This semiconductor is called a p-type semiconductor.

P-type extrinsic semiconductor
(a) Hole generated by the dopant
(b) Representation of acceptor energy level.

Question 2.
Explain the formation of PN junction diode. Discuss its V-I characteristics.
Answer:

p-n junction diode
(a) Schematic representation
(b) Circuit symbol

i. A p-n junction diode is formed when p-type semiconductor is fused with N-type semiconductor.
ii. Depending on polarity of external source to the p-n junction there are two types of biasing
1. Forward bias
2. Reverse bias

1. Forward bias:
i. If positive terminal of external voltage source is connected to p-side and negative terminal to n-side forward bias takes place.
ii. Electron moves to n-side holes move to p side Recombination takes place near junction and reduce depletion region.
iii. Electron from n-side accelerate towards p side it experience reduced potential barrier at junction.
iv. Applied voltage is increased, width of depletion region and barrier potential further reduced.
v. So large number of electrons pass through junction.

2. Reverse bias:
i. If positive terminal of external voltage source is connected to p-side and negative terminal to n-side reverse bias takes place.

ii. Depletion region is increased potential barrier is also increased.
iii. Majority charge carriers from both sides experience a great barrier to cross the junction. So diffusion current reduces.
iv. The current flows under reverse bias is called reverse saturation current I_S

V-I Characteristics:

1. Forward characteristics:
i. A graph is plotted by taking forward bias voltage in X axis and current i Y axis.
ii. Current flow is negligible when applied voltage is less than threshold voltage beyond that increase in current is significant even for small increase in voltage.
iii. Graph shows clearly current flow is non linear. It doesnot obey ohm’s law.
iv. Forward resistance r_f = \(\frac{\Delta \mathrm{V}}{\Delta \mathrm{I}}\)
v. Diode behaves as conductor when it is forward bias.

p-n junction diode
(a) diode under forward bias
(b) forward characteristics

2. Reverse characteristics:
i. In reverse bias p-region is connected to negative terminal and n-region to positive terminal, Graph is drawn for reverse bias characteristics.
ii. Very small current in pA flows across junction. This is due to leakage current or reverse saturation current.
iii. The reverse bias voltage can be increased only up to rated value otherwise diode will enter into breakdown region.

p-n junction diode
(a) diode under reverse bias
(b) reverse characteristics

Forward and reverse characteristics of a diode

Question 3.
Draw the circuit diagram of a half wave rectifier and explain its working.
Answer:
i. The circuit consist of transformer, p-n junction diode and resistor.
ii. In half wave rectifier either a positive half or the negative half of AC input is passed through it, other half is blocked.

(a) Input signal
(b) half wave rectifier circuit
(c) input and output waveforms

During positive half cycle:
1. When positive half cycle of input signal passes. A becomes positive with respect to B.
2. Diode is forward bias, current flows through R_L output voltage V₀ is developed.
3. Waveform is shown in figure.

During negative half cycle:
1. When negative half cycle of input signal passes A is negative with respect to B.
2. Diode is reverse bias, does not conduct No current is passed through R_L. No voltage drop across R_L.
3. Negative half cycle of ac supply is suppressed at output waveform is shown in figure.
4. Efficiency of half wave rectifier is 40.67%.

Question 4.
Explain the construction and working of a full wave rectifier.
Answer:
i. Full wave rectifier consist two p-n junction diodes a center tapped transformer and load resistor (R_L)
ii. Due to centre tap transformer output voltage rectified by each diode is only one half of total secondary voltage.

(a) Full wave rectifier circuit
(b) Input and output waveforms

During positive half cycle:
1. Input signal passes through the circuit M is positive, G is zero, N is negative potential.
2. D₁ forward bias, D₂ reverse bias.
3. Current flows through MD₁ AGC
4. Positive half cycle of voltage appears across R_L in direction G to C.

During negative half cycle:
1. Terminal N is positive, G is at zero, M is negative potential.
2. D₂ is forward bias, D₁ is reverse bias.
3. Current flows through ND₂BGC
4. Negative half cycle of voltage across R_L from G to C.
5. Hence in full wave rectifiers both postive and negative half cycle of input signals pass through the circuit in same direction.
6. Efficiency of full wave rectifier is 31.2%.

Question 5.
What is an LED? Give the principle of operation with a diagrajm?
Answer:
1. LED is a p-n junction diode which emit visible or invisible light when it is forward bias.
2. It converts electrical energy to light energy.
3. It consists of p-layer, n-layer and substrate. External resistance in series with biasing source is required to limit forward current through LED. It has anode and cathode.

(a) Circuit symbol of LED
(b) Inside view of LED
(c) Schematic diagram to explain recombination process

4. When p-n junction is forward bias, conduction band electron on n-side and valence band hole on p side diffuse across the junction.
5. When they cross junction, they become excess minortiy carriers. These excess minority carriers recombine with opposite charged majority carriers in respective region.
6. During recombination process energy is released is the form of light or heat.
7. The colour of the light is determined by energy band gap of the material.
8. LED’s with wide range colours such as blue (Sic), green (AlGaP), red(GaAsP), white(GaInN) is also available.

Question 6.
Write notes on Photodiode.
Answer:
                     
(a) Circuit symbol
(b) Schematic view of photodiode

1. Photodiode is a p-n junction diode which converts an optical signal into electric current.
2. It works in reverse bias.
3. It consists of F p-n junction semiconductor made of photosensitive material kept safely inside a plastic case.
4. It has small transparent window to allow light.
5. When photon (hυ) strikes depletion region of diode some valence band electrons elevated to conduction band. In turns holes are developed in valence band. This creates electron hole pair.
6. Amount of electron – hole pair generated depends on intensity of light.
7. These electrons and hole swept across the junction by electric field. Thus holes moves to -n side, electrons to p side.
8. When external circuit is made, electrons flow through external circuit and constitute the photo current.

Applications:
Alarm system, count items in conveyer belt, photo conductors, CD players, smoke detectors, detectors for computed tomography etc.

Question 7.
Explain the working principle of a solar cell. Mention its applications.
Answer:
11. Solar cell converts light energy directly into electricity or electric potential difference by photovoltaic effect.

Cross-sectional view of a solar cell

2. It generates emf when radiations falls on p-n junction. A solar cell is of two type p-type and n-type.
3. Both types use a combinations of p-type and n-type silicon which together forms the p-n junction.
4. In solar cell, electron – hole pairs are generated due to absorption of light near junction.
5. Electrons move towards n-type silicon, holes moves towards p-type silicon layer.
6. Electrons reaching n-side are collected by front contact and holes reaching p-side are collected by back electrical contact.
7. Thus potential difference is developed across solar cell. When external load is connected, photocurrent flows through it.
8. Many solar cells are connected in series or parallel to form solar panel or module.

Applications:
1. Widely used in calculators, watches, toys, portable power supplies etc.
2. Used in satellites and space stations.
3. Solar panels are used to generate electricity.

Question 8.
Sketch the static characteristics of a common emitter transistor and bring out the essence of input and output characteristics.
Answer:

Static characteristics of a NPN transistor in common emitter configuration

i. The circuit to study the static characteristics of NPN transistor is given in figure
ii. Bias supply voltages V_BB and V_CC bias, base – emitter junction and collector – emitter junction
iii. Junction potentials are V_BE and Y_CE
iv. R₁ and R₂ are used to vary base and collector currents respectively.

1. Input Characteristics:
i. Input Characteristics curve gives relationship between I_B and V_BE at constant V_CE
ii. For constant collector emitter voltage V_CE, Base emitter voltage V_BE increases for corresponding Base current I_B which is recorded and graph is ploted.
iii. The curve looks like forward characteristics

Input characteristics of a NPN transistor in common emitter configuration

iv. Beyond knee voltage base current increases with increase in base emitter voltage for silicon 0.7 V & for Germanium 0.3 V
v. Increase is V_CE decreases I_B. This shift the curves outward.
vi. Input resistance R_i = \(\left(\frac{\Delta V_{B E}}{\Delta I_{B}}\right)_{V_{C B}}\)

2. Output characteristics:

Output characteristics of a NPN transistor in common emitter configuration

i. The output characteristics gives the relation between ∆I_C with respect to ∆V_CE at constant I_B
ii. Initially base current I_B is set to particular value. Increasing collector emitter voltage V_CE corresponding collector current I_C increases. A graph is plotted.

iii. Saturation region:
When V_CE increased above 0V, I_C increases rapidly almost independent of I_B called knee voltage Transistor operated above this knee voltage

iv. Cut off region:
A small I_C exist even after I_B is reduced to zero. This current is due to presence of minority carriers across collector – base junction and the surface leakage current (I_CEO). This region is called cut-off region.

v. Active region:
In this region emitter – base junction is forward bias, collector – base junction is reverse bias. Transistor in this region can be used for voltage, current and power amplification.

vi. Breakdown region:
If V_CE is increased beyond rated value, given I_c increases enormously leading to junction breakdown of transistor. This avalanche breakdown can damages the transistor.
vii. Output resistance R_o = \(\left(\frac{\Delta V_{C E}}{\Delta I_{C}}\right)_{I_{B}}\)

Question 9.
Describe the function of a transistor as an amplifier with the neat circuit diagram. Sketch the input and output wave form.
Answer:
i. Amplification is the process of increasing the signal strength (increase in the amplitude)
ii. NPN transistor is connected in CE configuration
iii. To start with, Q point is fixed to get maximum signal swing at the output.
iv. R_C – to measure output voltage
C₁ allows only AC signals to pass,
Bypass capacitor C_E provides a low resistance path
Coupling capacitor C_C is used to couples next stage amplifier.
V_s input source signal
I_C = BI_B (∴ B = \(\frac{I_{C}}{I_{B}}\))
V_CE = V_CC – I_CR_C

                       
(a)Transistor as an amplifier
(b) Input and output waveform showing 180° phase reversal.

Working of the amplifier:

During positive half cycle:
1. Input signal V_s increases the forward voltage across emitter base. I_B increases. also increases I_C also increases β times.
2. This increase the voltage drop across R_c which decreaseV_CE.
3. Therefore input signal in the positive direction produces an amplified signal in negative direction at the output. Hence output signal is reversed by 180°

During negative half cycle:
1. Input signal V_s decreases the forwarded voltage across emitter – base. As a result I_B decreases, I_C increases.
2. Increase in I_C decreases potential drop across R_c and increases V_CE.
3. Input signal in negative direction produces amplified signal in the positive direction at the output.
4. Therefore 180° phase reversal is observed during negative half signal.

Question 10.
Transistor functions as a switch. Explain.
Answer:
1. In saturation and cut-off region transistor works as an electronic switch that helps to turn ON or OFF a circuit.
2. presence of dc source at the input (saturation region):

Transistor as a switch

3. When high input voltage (V_in = +5V) is applied, I_B increases and in turn increases I_C.
4. Transistor will move into saturation region (turned ON)
5. Increase in I_C increases voltage drop across R_c lowering the output voltage close to zero.
6. Transistor acts like a closed switch (ON condition)

Absence of dc source at the input (cut-off region) :
1. At low input voltage (V_in = 0V), decreases the I_B and in turn decreases I_C.
2. Transistor moves to cut off region (turned ON)
3. Decrease in I_C decreases voltage drop across R_c, increasing output voltage +5V transistor acts as open switch (OFF condition)
4. It is manifested that high input gives low output and low input gives high output.

Question 11.
State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example.
Answer:
1. Complement law:
\(\overline{\mathrm{A}}\) = A
A Y = \(\overline{\mathrm{A}}\)
O Y = \(\overline{\mathrm{O}}\) = 1
I Y = \(\overline{\mathrm{I}}\) = \(\overline{\mathrm{O}}\)

2. OR laws:

3. AND laws:

4. Commutative laws:
A + B = B + A
A.B = B.A

5. Associative laws
A + (B + C) = (A + B) + C
A.(B.C) = (A.B).C

6. Distributive laws
A (B + C) = AB + AC
A + BC = (A + B) (A + C)

Question 12.
State and prove De Morgan’s First and Second theorems.
Answer:
First Theorem:
The complement of the sum is equal to the product of its complements
\(\overline{A+B}=\bar{A} \cdot \bar{B}\)

Second Theorem:
The complement of the product of two inputs is equal to the sum of its complements.
\(\overline{\mathrm{A} \cdot \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\)

IV. Numerical Problems:

Question 1.
The given circuit has two ideal diodes connected as shown in figure below. Calculate the current flowing through the resistance R₁

Solution:
D₁ will act as Reverse bias so current will not pass through it
Total Resistance = 4Ω
V = 10 V
Current flow through R₁I = \(\frac{\mathrm{V}}{\mathrm{R}}\)
= \(\frac{10}{4}\) = 2.5 A

Question 2.
Four silicon diodes and a 10Ω resistor are connected as shown in figure below. Each diode has a resistance of 1Ω. Find the current flows through the 18Ω resistor.

Solution:
Current flow through D₁ and D₃ = 2Ω
Current flow through D₂ and D₄ = 2Ω
These two are in parallel so
Net resistants = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1Ω
Total Resistance = 1Ω+ 18Ω = 19Ω
V = 2.5Ω
I = \(\frac{\mathrm{V}}{\mathrm{R}}\) = \(\frac{2.5}{19}\)
I = 0.13 A

Question 3.
Assuming V_CESat = 0.2 V and β = 50, find the minimum base current (I_B) required to drive the transistor given in the figure to saturation.

Solution:
V_CC = 3V
β = 50
I_B =?
Collector current
V_CC = 3 – 0.2 = 2.8 V
I_C = \(\frac{\mathrm{V}_{\mathrm{CC}}}{\mathrm{R}_{\mathrm{c}}}\)
= \(\frac{2.8}{1 \times 10^{3}}\) = 2.8 × 10^-3 A

β = \(\frac{\mathrm{I}_{\mathrm{c}}}{\mathrm{I}_{\mathrm{B}}}\)
I_B = \(\frac{\mathrm{I}_{\mathrm{c}}}{\beta}\) = \(\frac{2.8 \times 10^{-3}}{50}\)
= 56 × 10^-6 A
I_B = 56 µA

Question 4.
A transistor having α = 0.99 and V_BE = 0.7 V, is given in the circuit. Find the value of the collector current.

Solution:
V_cc = 12 V
α = 0.99
V_BE = 0.7 V
I_C =?
Applying Kirchoff’s voltage law,
V_BE + (I_C + I_B) 1k + 10k.I_B + (I_C + I_B) 1k = 12 ………………(1)
I_B = \(\frac{\mathrm{I}_{\mathrm{C}}}{\beta}\)

β = \(\frac{\alpha}{1-\alpha}\) = \(\frac{0.99}{1-0.99}\) = 0.99

I_B = \(\frac{\mathrm{I}_{\mathrm{C}}}{99}\)

I_C = β I_B
I_C = 99 I_B
I_C = \(\frac{\mathrm{I}_{\mathrm{C}}}{99}\)
V_BE = 0.7 V
Substitute in equation (1)
\(\frac{0.7}{10^{3}}\) + (99 I_B + I_B) + 10 I_B + (99 I_B + I_B) = \(\frac{12}{10^{3}}\)
100 I_B + 10 I_B + 100 I_B = \(\frac{12-0.7}{10^{3}}\)
210 I_B = \(\frac{11.3}{10^{3}}\) ;
I_B = 0.054 × 10^-3 A
I_C = 99 I_B = 99 × 0.054 × 10^-3 A
I_C = 5.34 × 10^-3 A
I_C = 5.34 mA

Question 5.
In the circuit shown in the figure, the BJT has a current gain (β) of 50. For an emitter-base voltage V_EB = 600 mV, calculate the emitter-collector voltage V_EC (in volts)

Solution:
β = 50
V_EB = 600 × 10^-3V
R_B = 60 kΩ
R_C = 500 kΩ
V_B = V_E – V_EB
= 3 – 0.6 = 2.4 V
I_B = \(\frac{\mathrm{V}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}\)
= \(\frac{2.4}{60 \times 10^{3}}\)
= 0.04 × 10^-3 A
I_B = 40 µA
I_C = β I_B
= 50 × 40 × 10^-6
I_C = 2mA
V_C = I_C R_C = 500I_C
= 500 × 2 × 10^-3
V_EC = V_E – V_C
= 3 – 1 = 2V

Part II:

12th Physics Guide Semiconductor Electronics Additional Questions and Answers

I. Choose the correct answer:

Question 1.
A transistor has α = 0.95, it has change in emitter current of 100 milliampere then the change in collector current is
a) 95 mA
b) 99.05 mA
c) 100.95 mA
d) 100 mA
Answer:
a) 95 mA
Solution:
α = \(\frac{\Delta \mathrm{I}_{\mathrm{c}}}{\Delta \mathrm{I}_{\mathrm{e}}}\)
∆I_C = ∆I_e × α
= 100 × 0.95
= 95 mA

Question 2.
Diode can works as
a) Demodulator
b) Modulator
c) Amplifier
d) Rectifier
Answer:
a) Demodulator

Question 3.
Three amplifiers of each with gain 10 volt are connected in series then the overall amplification is ______.
a) 10/3
b) 13
c) 1000
d) none of these
Answer:
c) 1000
Solution:
10 × 10 × 10 = 1000

Question 4.
A Transistor has an α = 0.95 then β is
a) 1/19
b) 19
c) 1.5
d) 0.95
Answer:
b) 19
Solution:
β = \(\frac{\alpha}{1-\alpha}\) = \(\frac{0.95}{1-0.95}\)

Question 5.
In p – type semiconductor germanium is doped with _______.
a) aluminium
b) boron
c) gallium
d) all of these
Answer:
d) all of these

Question 6.
For an npn transistor, the collector current is 10 mA. If 90% of electron emitted reach the collector then
a) base current will be 1 mA
b) base current will be 10 mA
c) emitter current will be 11 mA
d) emitter current will be 9 mA
Answer:
a) & c)
Solution:
0.9 I_e10 = I_C
I_e = \(\frac{10}{0.9}\) = 11 mA
I_B = 11 – 10 = 1 mA

Question 7.
The depletion layer in p-n junction region is caused by _______.
a) drift of holes
b) diffusion of charge carriers
c) drift
d) drift of electrons
Answer:
b) diffusion of charge carriers

Question 8.
When – n – type semiconductor is heated ______.
a) number of electrons only increase
b) number of holes increase
c) number of electrons and holes remains same
d) number of electrons and holes increase equally
Answer:
d) number of electrons and holes increase equally

Question 9.
The device that can act as complete electronic circuit is _______.
a) junction diode
b) IC
c) transistor
d) diode
Answer:
b) IC

Question 10.
In following circuit the output Y for all possible input A and B is expressed by truth table

a)

b)

c)

d)

Answer:
c)

Y’ = \(\overline{A+B}\)
Y = \(\overline{Y^{\prime}}\) = \(\overline{\overline{A+B}}\)
Y = A + B

Question 11.
Which one is correct for reverse bias is applied in junction diode
a) Increase minority career
b) lower potential barrier
c) raise in potential barrier
d) increase in majority carrier
Answer:
c) raise in potential barrier

Question 12.
In common emitter configuration V_c = 1.5 V_c change in base current from 100 µA to 150 µA produce change in collector current from 5 mA to 10 mA. The current gain (β) is
a) 75
b) 100
c) 50
d) 67
Answer:
b) 100
Solution:
∆I_B = 50 µA
∆I_C = 5 mA
β = \(\frac{\Delta \mathrm{I}_{\mathrm{B}}}{\Delta \mathrm{I}_{C}}\)
= \(\frac{5 \mathrm{~mA}}{50 \mu \mathrm{A}}\)
= 100

Question 13.
To use a transistor as an amplifier
a) the emitter base junction is forward biased and base collector junction is reverse biased
b) no bias voltage is required
c) both junction are forward bias
d) both junctions are reverse bias
Answer:
a) the emitter base junction is forward biased and base collector junction is reverse biased

Question 14.
The circuit shown in figure contains two diodes each with forward resistance of 50Ω and with infinite backward resistance. If battery voltage is 6V the current through 100Ω resistance is

a) Zero
b) 0.02 mA
c) 0.03 mA
d) 0.36 mA
Answer:
b) 0.02 mA
Solution:
Current will not pass through D₂
Total resisatnce = 50 + 150 + 100 = 300Ω
Current = \(\frac{6}{300}\) = 0.02 V

Question 15.
Type of material which emits white light in LED
a) GaInN
b) Sic
c) AlGaP
d) GaInP
Answer:
a) GalnN

Question 16.
Blue colour LED is made up of
a) Sic
b) AIGaP
c) GaAsP
d) GaInP
Answer:
a) Sic

Question 17.
The energy band gap is maximum in
a) metals
b) super conductors
c) insulators
d) semiconductors
Answer:
c) insulators

Question 18.
The part of transistor which is mostly heavily doped to produce large majority carriers
a) emitter
b) base
c) collector
d) None of these
Answer:
a) emitter

Question 19.
In the middle of depletion layer of reverse – biased p-n junction
a) electric field is zero
b) potential is maximum
c) electric field is maximum
d) potential is zero
Answer:
c) electric field is maximum

Question 20.
What is the current flowing in the circuit?

a) 1.71 A
b) 2.0 A
c) 2.31 A
d) 1.33 A
Answer:
b) 2.0 A
Solution:
D₁ is reverse bias will not conduct = \(\frac{12}{6}\) = 2 A

Question 21.
A diode as a rectifier converts _______.
a) a.c to d.c
b) d.c to a.c
c) AC only
d) dc only
Answer:
a) ac to d.c

Question 22.
An oscillator is nothing but an amplifier with
a) positive feedback
b) large gain
c) no feed back
d) negative feedback
Answer:
a) positive feedback

Question 23.
In Intrinsic semiconductor at room temperature number of electrons and holes are
a) equal
b) zero
c) unequal
d) infinite
Answer:
a) equal

Question 24.
The level formed due to impurity atom in forbidden energy gap, very near to valence band is p-type semiconductor is called
a) An acceptor level
b) A donor level
c) Conduction level
d) A forbidden level
Answer:
a) An acceptor level

Question 25.
The symbol represents _______ gate.

a) NAND gate
b) OR gate y
c) AND gate
d) NOT gate
Answer:
d) NOT gate

Question 26.
In p-n junction, avalanche current flows is circuit when biasing
a) Forward
b) Reverse
c) Zero
d) Excess
Answer:
b) Reverse

Question 27.
For detecting light, which is correct?
a) Photodiode has to be forward biased
b) Photodiode has to be reversed biased
c) LED has to be connected in forwarded bias
d) LED in reverse bias
Answer:
b) Photodiode has to be reversed biased

Question 28.
What is the output Y in the above circuit, when all the three inputs A, B and C are 0?

a) 0
b) 1
c) 10
d) 11
Answer:
a) 0

Question 29.
Which type of semiconductor device does not need any bias voltage
a) Photodiode
b) Zero diode
c) Solar cell
d) Transistor
Answer:
c) Solar cell

Question 30.
The frequency of oscillator is_____ 2 1
a) f = \(\frac{1}{2 \pi \mathrm{LC}}\)

b) ω² = \(\frac{1}{\mathrm{LC}}\)

c) ω = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

d) f = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
Answer:
d) \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

Question 31.
The logic circuit shown in figure represents ______ gate.

a) OR
b) AND
c) NOR
d) NAND
Answer:
d) NAND

Question 32.
Maximum efficiency of full wave rectifier
a) 100%
b) 81.2%
c) 40.6%
d) 95%
Answer:
b) 81.2%

Question 33.
The diffusion current in P-N junction is from
a) p side to n side
b) n side top side
c) Both (a) and (b)
d) None
Answer:
a) p side to n side

Question 34.
The current through an ideal PN junction shown in figure

a) 5 mA
b) 10 mA
c) 70 mA
d) 100 mA
Answer:
b) 10 mA
Solution:
V = 5-(-2)
V = 7V
I = \(\frac{7}{100}\)
I = 10 MA

Question 35.
Efficiency of half wave rectifier is
a) 81.2%
b) 100%
c) 40.6%
d) 95%
Answer:
c) 40.6%

Question 36.
In common base amplifier, phase difference between input voltage and output voltage
a) 0
b) π/4
c) π/2
d) π
Answer:
a) 0

Question 37.
What will be the input of A and B for Boolean expression \(\overline{(A+B)}\) \(\overline{(\mathrm{A} \cdot \mathrm{B})}\) = 1
a) 0,0
b) 0,1
c) 1,0
d) 1,1
Answer:
a) 0,0
Solution:
\(\overline{0} \cdot \overline{0}\) = 1
1.1 = 1

Question 38.
The value of current flowing through AB in this circuit

a) 10 mA
b) 20 mA
c) 15 mA
d) 11 mA
Solution:
PD = 3 -(-7) = 10V
I = \(\frac{10}{1000}\)
= 10^-2 A = 10 mA

Question 39.
To get output, 1 for following circuit. The correct choice for input is ________.

a) A = 0, B = 1, C = 0
b) A = 1, B = 0, C = 0
c) A = 1, B = 1, C = 0
d) A = 1, B = 0, C = 1
Answer:
c) A = 1, B = 1, C = 0
Solution:
Y’ = A+B
Y = (A + B) C = 1
(1 + 0).1 = 1

Question 40.
The given electrical network is equal to

a) AND gate
b) OR gate
c) NOR gate
d) Ex-OR gate
Answer:
c) NOR gate
Y₁ = \(\overline{A+B}\) = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\)
Y₂ = \(\overline{\mathrm{Y}}_{1}\) = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\) = A + B
Y = \(\overline{A+B}\)

Question 41.
Which gate will give high input when odd numbers of input are high
a) NAND gate
b) OR gate
c) NOR gate
d) EX-OR gate
Answer:
d) EX-OR gate

Question 42.
The current gain (β) of a transistor in common emitter mode is 40. To change the collector current by 160 mA at constant V_CE, The necessary change in the base current is
a) 0.2 mA
b) 4 µA
c) 4 mA
d) 40 mA
Answer:
c) 4 mA

Question 43.
Which one is Barkhausen condition for sustained Oscillation
a) phase shift 0°
b) loop gain is unity
c) \(|A B|\) = 0
d) Both (a) and (b)
Answer:
d) Both (a) and (b)

Question 44.
Tank circuit consists of ________.
a) Capacitor
b) Inductor
c) Both (a) and (b)
d) Diode
Answer:
c) Both (a) and (b)

Question 45.
The given electrical network is equivalent to:

(a) NAND gate
(b) OR gate
(c) NOT gate
(d) Ex-OR gate
Answer:
b) OR gate

Question 46.
The maximum reverse bias that can be applied before entering into zener region is referred as
a) PIV rating
b) PAV rating
c) RIM rating
d) None
Answer:
a) PIV rating

Question 47.
For reverse voltage between 4 and 6V ___________ are present
a) Zener effect
b) Avalanche effect
c) Both effects
d) None
Answer:
c) Both effects

Question 48.
Solar cell works on _______ effect.
a) Photo emissive
b) Photo electric
c) Photo conducting
d) Photo voltaic
Answer:
d) Photo voltaic

Question 49.
Which one is the symbol for photodiode
a)

b)

c)

d)
Answer:
c)

II. Choose the Correct Statement:

Question 1.
Which of the statement is not true?
a) The resistance of intrinsic semiconductor decreases with increase of temperature
b)Doping pure Si with trivalent impurities give p-type semiconductor.
c) The majority carriers in n type semiconductors are holes
d) A p-n type junction can act as a semiconductor diode
Answer:
c) The majority carriers in n type semiconductors are holes

Question 2.
Which one is correct statement A — p type semiconductor is
i) A silicon crystal is doped with arsenic impurity
ii) Si doped with aluminium impurity
iii) Ge doped with phosphorous impurity
iv) Ge doped with boron impurity
a) (i) and (ii) are correct
b) (ii) and (iii) are correct
c) (i) and (iv) are correct
d) (i) only correct
Answer:
b) (ii) and (iii) are correct

Question 3.
Read the following statement carefully
Y: The resistivity of semiconductor decreases with increase of temperature
Z: The resistivity value of semiconductor is 10^-5 to 10⁶ Ωm
a) Y is true but Z is false
b) Y is false but Z is true
c) Both Y and Z are true
d) Both Y and Z are false
Answer:
c) Both Y and Z are true

Question 4.
Which one of the following statement is correct?
a) The depletion region of P-N junction diode increase with forward bias
b) The depletion region of P-N junction diode decrease with reverse bias
c) The depletion region of PN junction diode does not change with biasing
d) The depletion region of PN junction diode decreases with forward biasing.
Answer:
d) The depletion region of PN junction diode decreases with forward biasing.

Question 5.
Assertion:
The size, capacity of chips are progressed enormously with advancement of technology
Reason:
Computers, mobile phones are possible in small size and low cost of ICs
a) Assertion and Reason are true and Reason is correct explanation of Assertion
b) Assertion and Reason are true but Reason is not correct explanation of Assertion
c) Assertion is true but Reason is false
d) Assertion is false but Reason is true
Answer:
a) Assertion and Reason are true and Reason is correct explanation of Assertion

III. Fill in the blanks:

Question 1.
The reverse saturation current of silicon diode doubles for every _______ rise in temperature.
Answer:
10°C

Question 2.
The external voltage applied to p-n junction is called _______.
Answer:
bias voltage

Question 3.
A semiconductor has ________ resistance coefficient.
Answer:
negative

Question 4.
The most commonly used semiconductor is ____________.
Answer:
silicon

Question 5.
Find the odd one out. Trivalent atoms used in doping
a) Boron
b) Aluminium
c) Indium
d) Arsenic
Answer:
d) Arsenic

Question 6.
Find the odd one out Pentavalent dopant used in n-type semiconductor
a) Phosphorus
b) Antimony
c) Galium
d) Arsenic
Answer:
c) Galium

IV. Match the following:

Question 1.

a) 1 – d 2 – b 3 – a 4 – c
b) 1 – d 2 – a 3 – b 4 – c
c) 1 – d 2 – a 3 – c 4 – b
d) 1- c 2 – a 3 – b 4 – d
Answer:
b) 1 – d 2 – a 3 – b 4 – c

Question 2.

1. Silicon	a. 6 ev
2. Germanium	b. holes
3. Insulator	c. 1.1 ev
4. Deficiency of electron	d. 0.7 ev

a) 1 – c 2 – d 3 – b 4 – a
b) 1 – a 2 – b 3 – c 4 – d
c) 1 – c 2 – d 3 – a 4 – b
d) 1 – b 2 – c 3 – a 4 – d
Answer:
c) 1 – c 2 – d 3 – a 4 – b

V. Two Mark Questions:

Question 1.
Define Energy band.
Answer:
Band of very large number of closely spaced energy levels in a very small energy range is known as energy band.

Question 2.
What is valence band?
Answer:
The energy band formed due to the valence orbits is called valence band.

Question 3.
What is called conduction band?
Answer:
The energy band formed by unoccupied orbits is called conduction band.

Question 4.
What is forbidden energy gap?
Answer:
The energy gap between valence band and conduction band is called forbidden energy gap.

Question 5.
What are passive and active components?
Answer:
Passive components:
Components that cannot generate power in a circuit.
Active components:
components that can generate power in a circuit.

Question 6.
Give example for Pentavalent elements.
Answer:
Phosphorus, Arsenic and Antimony

Question 7.
Write – the examples of trivalent impurities.
Answer:
Boron, Aluminum, Gallium and Indium

Question 8.
Differntiate Donor impurities, Acceptor impurities.
Answer:

Donor Impurity	Acceptor Impurity
Impurity atoms which donate electrons to the conduction band are called Donor impurities. Ex: Pentavalent elements like Phosphorous, Arsenic and Antimony	Dopants (or) Impurities which accept electron from neighbouring atoms are called acceptor impurity. Ex: Trivalent elements like Boron, Aluminium, Gallium, Indium

Question 9.
What is barrier potential?
Answer:
The difference in potential across the depletion layer is called barrier potential.

Question 10.
What are extrinsic semiconductors?
Answer:
An extrinsic semiconductor is a semiconductor doped by a specific impurity which is able to deeply modify its electrical properties, making it suitable for electronic applications (diodes, transistors etc.) or optoelectronic applications (light emitters and detectors).

Question 11.
What is called bias voltage?
Answer:
The external voltage applied to the p-n junction is called bias voltage.

Question 12.
What is called Forward bias and reverse bias in p-n junction diode?
Answer:

1. If positive terminal of external voltage source is connected to p-side and negative terminal to n-side is called forward bias.
2. If positive terminal of battery is connected to n-side and negative terminal to p-side. The junction is said
   to be reverse bias.

Question 13.
What is reverse saturation current?
Answer:
The current that flows under a reverse bias is called the reverse saturation current. It is represented as I_s.

Question 14.
Write down the applications of Zener diode.
Answer:
The Zener diode can be used as:

1. Voltage regulators
2. Peak clippers
3. Calibrating voltages
4. Provide fixed reference voltage in a network for biasing
5. Meter protection against damage from accidental application of excessive voltage.

Question 15.
In the combination of the following gates, write the Boolean equation for output Y in terms of inputs A and B

Answer:
The output at the 1st AND gate : \(\mathrm{A} \overline{\mathrm{B}}\)
The output at the 2nd AND gate : \(\overline{\mathrm{A}} \mathrm{B}\)
The output at the OR gate : Y = \(\mathrm{A.} \overline{\mathrm{B}}+\overline{\mathrm{A}} \cdot \mathrm{B}\)

Question 16.
What are the characteristics of ideal diode?
Answer:

1. It acts like conductor when it is forward biased.
2. When it is reverse bias it acts like an insulator
3. The barrier potential is assumed to be zero and hence it behaves like a resistor.

Question 17.
What are Optoelectronic devices?
Answer:
Optoelectronics deals with devices which convert electrical energy into light and light into electrical energy through semiconductors.

Question 18.
Define efficiency of rectifier.
Answer:
Efficiency (η) is the ratio of the output DC power to the ac input power supplied to the circuit.

Question 19.
What is Zener diode. Draw its symbol
Answer:
Zener diode is a reverse biased heavily doped silicon diode. It is specially designed to operate in breakdown region

Question 20.
What is called optoelectronics?
Answer:

1. 1. Optoelectronics deals with devices which convert electrical energy into light and light into electrical energy through semiconductors.
2. Optoelectronic devices are LED, photodiode and solar cells.

Question 21.
What is light emitting diode?
Answer:
Light Emitting Diode (LED) is a p-n junction diode which emits visible or invisible light when it is forward biased. Symbol is

Question 22.
What is photodiode? Draw the circuit symbol.
Answer:
A p-n junction – diode which converts an optical signal into electric current is known as photodiode. Circuit symbol is

Question 23.
What is solar cell?
Answer:
A solar cell is photo voltaic cell converts light energy directly into electricity or electric potential difference by photovoltaic effect.

Question 21.
Write down the applications of Oscillators.
Answer:
Applications of oscillators:

• to generate periodic sinusoidal or nonsinusoidal waveforms.
• to generate RF carriers.
• to generate audio tones
• to generate clock signals in digital circuits.
• as sweep circuits in TV sets and CRO.

Question 25.
What is Input resistance?
Answer:
The ratio of the change in base emitter voltage (∆V_BE) to change in base current (∆I_B) at constant collector emitter voltage (V_CE) is called input resistance (R_i)
R_i = \(\left(\frac{\Delta V_{B E}}{\Delta I_{B}}\right)_{V_{C E}}\)

Question 26.
What is output resistance?
Answer:
The ratio of change in collector – emitter voltage (∆V_CE) to the corresponding change is collector current (∆I_C) at constant base current.
R₀ = \(\left(\frac{\Delta \mathrm{V}_{\mathrm{CE}}}{\Delta \mathrm{I}_{\mathrm{C}}}\right)_{\mathrm{I}_{\mathrm{B}}}\)

Question 27.
What is forward current gain in common emifter mode?
Answer:
The ratio of change in collector current (∆I_C) to change in base current (I_B) at constant collector – emitter voltage (V_CE) is called forward current gain (B)
B = \(\left(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right)_{\mathrm{V}_{\mathrm{CE}}}\)
The value is very high and range from 50 to 200.

Question 28.
What is amplification?
Answer:

1.  Amplification is the process of increasing the signal strength (increase the amplitude)
2. If large amplification is required, the transistors are cascaded.

Question 29.
Write the Barkhausen conditions for sustained oscillations?
Answer:

1. The loop phase shift is 0° or integral multiple of 2n.
2. The loop again must be unity \(|A B|\) = 1
   A – Voltage gain,
   B – Feedback ratio.

Question 30.
Draw the circuit diagrams of transistor in CB and CC modes.
Answer:

CB Mode	CC Mode

Question 31.
Differentiate damped and undamped Oscillations.
Answer:

Damped Oscillation	Undamped Oscillation
1. If amplitude of electrical oscillations decreases with time due to energy loss is damped oscillation.	1. The amplitude of electrical oscillation remains constant with time is damped oscillation.

VI. Three Mark Questions:

Question 1.
Differentiate Intrinsic semiconductor and Extrinsic semiconductors.
Answer:

Intrinsic Semiconductor	Extrinsic Semiconductor
1. It is in pure form	It formed by adding trivalent or pentavalent impurity.
2. Holes and electrons are equal	No. of holes are more in p-type. No of electrons are more in n-type.
3. Conductivity increases with raise in temperature	Conductivity depends on the amount of impurity added
4. Conductivity is slightly increased	Conductivity is greatly increased.

Question 2.
Find the differences between N-type and p-type semiconductors.
Answer:

N – type	P – type
1. Pentavalent impurities are added	Trivalent impurities are added.
2. Majority carriers are electrons.	Majority carriers are holes.
3. Minority carriers are holes	Minority carriers are electrons
4. They are negatively charged materials	They are positively charged materials

Question 3.
Write briefly about common — base configuration. Draw the circuit symbol for it.
Answer:
NPN transistor in common base configuration

(a) Schematic circuit diagram

(b) Circuit Symbol

1. Base is common to both input and output
2. Emitter current I_E is input collector current is output current I_C.
3. Input signal is applied between base and emitter output is measured between collector and base.

Question 4.
What is Zener diode? Write the applications.
Answer:
Zener diode is reverse biased with heavily doped silicon diode specially works in breakdown region. Zener diode can be used as

1. Voltage regulators
2. Peak Clippers
3. Calibrating Voltages
4. Provide fixed reference voltage in-network for biasing
5. Meter protection against damage from accidental application of excessive

Question 5.
What are the applications of LED?
Answer:

1. Indicator lamps on front panel of scientific and lab equipments
2. Seven – segment displays
3. Traffic – signals, exit signals in emergency vehicle lighting.
4. Industrial process control, position encoders, bar graph readers.

Question 6.
List the applications of photodiodes
Answer:

1. Alarm system
2. Count items on a conveyer belt
3. Photo conductors
4. CD players smoke detectors
5. Detectors for computed demography

Question 7.
Explain the different modes of transistor biasing.
Answer:
1. Forward Active:

• In this bias emitter – base junction is forward bias, collector base junction is reverse bias. .
• Transistor is in active mode of operation.
• Transistor functions as amplifier.

2. Saturation:

• Emitter base junction and collector-base junction are forward biased.
• Very large flow of current across junction.
• Transistor is used as closed switch.

3. Cut-Off:

• Emitter – base and collector-base junction are reverse biased.
• Transistor is acts as an open switch.

Question 8.
Write a note on common – Emitter configuration Draw the circuit diagram.
Answer:
NPN transistor in common emitter configuration

(a) Schematic circuit diagram

(b) Circuit Symbol
1. Emitter is common to both input and output.
2. Base current I_B is input current I_C is output current.
3. Input signal is applied between base and emitter. Output is measured between collector and emitter.

Question 9.
Write a note on common – collector configuration.
Answer:
1. Collector is common to both the inputs and outputs.
2. In is input current and Ig is output current.
3. Input signal is applied between base and collector. Output is measured between emitter and collector.
NPN transistor in common collector configuration

(a) schematic circuit diagram

(b) Circuit diagram

Question 10.
Write the applications of Oscillators.
Answer:

1. To generate a periodic sinusoidal or non-sinusoidal wave forms.
2. To generate RF carriers.
3. To generate audio tones .
4. To generate clock signals in digital circuits.
5. As sweep circuits in TV sets and CRO.

Question 11.
Write a short note on NAND gate.
Answer:
The circuit symbol of NAND gate is

Boolean Equation:
Y = \(\overline{A \cdot B}\)

Logic operation:
1. The output Y equals the complement of AND
2. The output is logic zero only when all inputs are high

Question 12.
Write a not on NOR gate.
Answer:
Circuit symbol:

A, B inputs, Y – output

Boolean equation:
Y = \(\overline{A+B}\)

Logic operation:
1. Y equals to complement of OR
2. The output is high when all inputs are low.

Question 13.
Write a note on Ex – OR gate.
Answer:
The circuit symbol is

A, B inputs, Y – output

Boolean equation is
Y = \(A \cdot \bar{B}+\bar{A} \cdot B\)
Y = \(A \oplus B\)
Logic operation:
The output is high only when either of two inputs is high.

Question 14.
Write down the concept in details of Integrated Chips (IC’s) Integrated Chips
Answer:
An integrated circuit is also referred as an IC or a chip or a microchip. It consists of thousands to millions of transistors, resistors, capacitors, etc. integrated on a small flat piece of semiconductor material that is normally Silicon. Integrated circuits (ICs) are the keystone of modem electronics. With the advancement in technology and the emergence of Very Large Scale Integration (VLSI) era it is possible to fit more aind more transistors on chips of same piece.

ICs have two main advantages over ordinary circuits: cost and performance. The size, speed, and capacity of chips have progressed enormously with the advancement in technology. Computers, mobile phones, and other digital home appliances are now made possible by the small size and low cost of ICs. ICs can function as an amplifier, oscillator, timer, microprocessor and computer memory.

These extremely small ICs can perform calculations and store data using either digital or analog technology. Digital ICs use logic gates, which work only with values of ones and zeros. A low signal sent to a component on a digital IC will result in a value of 0, while a high signal creates a value of 1.

Question 15.
In the following diagrams, indicate which of the diodes are forward biased and which are reverse biased?

Answer:
a) Forward biased
b) Reverse biased
c) Reverse biased

Question 16.
Find the current through the Zener diode when the load resistance is 1 kΩ. Use diode approximation.

Answer:
Voltage across AB, V_z = 9V
Voltage drop across R = 15 – 9 = 6V
Current through the resistor
I = \(\frac{\mathrm{V}}{\mathrm{R}}\)
= \(\frac{6}{1 \times 10^{3}}\) = 6 mA
Voltage across load resistor,
V_AB = 9V
Current through load resistor,
I_L = \(\frac{\mathrm{V}_{\mathrm{AB}}}{\mathrm{R}_{\mathrm{L}}}=\frac{9}{2 \times 10^{3}}\)
= 4.5 mA
Current through the zener diode,
I_z = I – I_p = 6 mA = 1.5 mA

VII. Five Mark Questions:

Question 1.
Explain the forward and reverse bias of Zener diode. Discuss its V-I Characterestics.
Answer:
1. The forward characteristics of Zener diode is similar to ordinary p-n junction diode.
2. It starts conducting around 0.7 H. However reverse characteristics is highly significant.
3. The increase in reverse voltage normally generate reverse current. While in Zener diode when reverse voltage (V₂) increase in current is very sharp the voltage remains constant.
4. If reverse current is increased further diode will be damaged V_z – Zener breakdown voltage
5. I_z(min) minimum current to sustain breakdown
6. I_z(max) maximum current limited by maximum power dissipation.
7. Zener diode is operated is reverse bias having voltage greater than V_z and current less than I_z(max)
8. The reverse characteristics is not exactly vertical so diode posses zener dynamic impedence.
9. Zener resistance in inverse of slope in breakdown region.
10. Increase in the Zener current produces only a very small increase in reverse voltage.
11. Voltage of ideal Zener diode doesn’t change. V_z remains almost constant even when I_z increases.

(a) forward bias

(b) reverse bias

(c) V-I characteristics

Question 2.
How zener diode works as a voltage regulator? Explain.
Answer:

Circuit to study voltage regulation by Zener diode

1. A Zener diode working in breakdown region can serve as a voltage regulator.
2. It maintains constant output voltage V₀ even when input voltage V_i or load current I_L varies.
3. In this circuit, V_i is regulated to constant voltage Zener voltage V_z at output represented as V₀ using Zener diode.
4. The output is maintained constant as long as V_i. does not fall below V₂.
5. When potential developed across the diode is greater than V_z diode moves into Zener breakdown region. It conducts, large current through R_i.
6. The total current is always less than the maximum Zener diode current. Under all conditions V₀ = V_z Thus output voltage is regulated.

Question 3.
Transistor works as oscillator Explain.
Answer:

Block diagram of an Oscillator

1. An electronic oscillator converts dc energy into ac energy of high frequency
2. An oscillator circuit consist of tank circuit, an amplifier and feed back circuit.
3. Tank circuit generates electrical oscillations and acts as ac input source to transistor amplifier.
4. Amplifier amplifies input ac signal Feed back circuit produces Block diagram of an oscillator a portion of output to tank circuit to sustain the oscillations without energy loss.
5. So an oscillator does not require an external output signal. It is self – sustained.
6. Feedback Network Portion of output is fed to input is in phase with input the magnitude of input increases. It is necessary for sustained oscillations.
Tank Circuit:

tank circuit

1. The LC tank circuit consist of inductance and capacitor connected in parallel
2. Whenever energy is supplied to tank circuit from DC source, energy is stored in inductor and capacitor alternatively This produces oscillations of definite frequency
3. In order to produce undamped oscillations, positive feedback is provided from output to frequency f = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)

VIII. Additional Problems:

Question 2.
In PNP transistor circuit, the collector current is 10 mA. If 90% of the holes reach the collector, find emitter and base currents.
Solution:
Here, I_E = 10 mA
As 90% of the holes reach the collector, so the collector current.
I_C = 90% of I_E
I_C = \(\frac { 90 }{ 100 }\) I_E
I_E = \(\frac { 100 }{ 90 }\) I_C = \(\frac { 100 }{ 90 }\) x 10
I_E ≃ 11mA
Base current, I_B = I_E – I_C = 11 – 10
I_B = 1 mA.

Question 2.
An NPN BJT having reverse saturation current I_S = 10^-15 A is biased in the forward active region with V_BE = 700 mV and the current gain (β) may vary from 50 to 150 due to manufacturing variations. What is the maximum emitter current (in μA)
Solution:
I_S = 10^-15 A
V_BE = 700
V_T = 25 mV
β range from 50 to 150
I_C = I₀ e^(V_BE/V_T)
I_E = \(\frac { β+1 }{ β }\) I_C
I_E = \(\frac { β+1 }{ β }\) I_S e^V_BE/V_T
I_E will be maximum when β is 50
= 1.02 × 10^-15 × e^{700 × 10-3/25 × 10-3}
I_E = 1475 μA

Question 3.
The current gain β of silicon transistor used in circuit shown in figure is 50. (Barrier potential of silicon is 0.69 V) find I_B, I_E and I.

Solution:
V_BB = 2V
V_CC = 10V
β = 50
R_B = 10 kΩ
Barrier potential of Si V_BE = 0.69 V
R_C = 1 kΩ
V_BB = I_B R_B + V_BE
I_B = \(\frac{\mathrm{V}_{\mathrm{BB}}-\mathrm{V}_{\mathrm{BE}}}{\mathrm{R}_{\mathrm{B}}}\)
= \(\frac{2-0.69}{10 \times 10^{3}}\) = 131 µA
β = \(\frac{\mathrm{I}_{\mathrm{c}}}{\mathrm{I}_{\mathrm{B}}}\)
I_C = I_Bβ
= 131 × 10^-6 × 50 = 6.55 mA
∴ I_E = I_C + I_B
I_E = 6.55 mA + 131 µA
= 6.55 mA + 0.131 mA
I_E = 6.681 mA

Question 4.
You are given the two circuits shown in figure show that circuit
(a) acts as OR gate while the circuit
(b) acts as AND gate
Solution:
(a) A, B input Y output

(a) A, B inputs Y output

output of NOT gate = \(\overline{A+B}\)
This will be input for NOT gate.
Its output will be
\(\overline{\overline{A+B}}\) = A + B
∴ Y = A + B
Hence the circuit functions as OR gate.

(b) A, B are inputs Y – output

Y = \(\overline{\mathrm{A}}+\overline{\mathrm{B}}\)
= \(\overline{\overline{\bar{A}} \cdot \overline{\bar{B}}}\)
Y = A.B
Hence this circuit functions as AND gate.

Question 5.
In the circuit shown in the figure, the input voltage V_i is 20V, V_BE = 0V and V_CE = 0V, what are the values of I_B I_C and β?

Solution:

Question 6.
Name the logic gate equivalent of the output Y. If the same waveform I / P is given to both the inputs A and B, draw the corresponding output waveform.

Solution:

b) The output waveform will be the same as that of input waveform as it is OR (Addition) operation and same input is given to both A and B.