Text Book Back Questions and Answers - Chapter 9 Semiconductor Electronics 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Semiconductor Electronics
Multiple Choice Questions
Question 1.
The barrier potential of a silicon diode is approximately,
(a) $0.7 \mathrm{~V}$
(b) $0.3 \mathrm{~V}$
(c) $2.0 \mathrm{~V}$
(d) $2.2 \mathrm{~V}$
Answer:
(a) $0.7 \mathrm{~V}$
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 $0 \mathrm{~V}$
(c) A closed switch in series with a battery voltage of $0.7 \mathrm{~V}$
(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) $0^{\circ}-90^{\circ}$
(b) $90^{\circ}-180^{\circ}$
(c) $0^{\circ}-180^{\circ}$
(d) $0^{\circ}-360^{\circ}$
Answer:
(c) $0^{\circ}-180^{\circ}$

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.
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 $2 \pi$
(d) All the above.
Answer:
(d) All the above.
Question 11.
If the input to the NOT gate is $\mathrm{A}=1011$, 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 forward bias diode?

Answer:

Question 14.
The given electrical network is equivalent to 

(a) AND gate
(b) OR gate
(c) NOR gate 1
(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
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 $\mathrm{N}$ - type semiconductor is similar to electrons moving in a metallic wire. The $\mathrm{N}$ - type dopant atoms will yield electron available for conduction.
Question 2.
Distinguish between intrinsic and extrinsic semiconductors.
Answer:
1. Intrinsic:
- These are pure semiconducting tetravalent crystals.
- Their electrical conductivity is low.
- There is no permitted energy state between valence and conduction band.

- Their electrical conductivity depends on temperature.
2. Extrinsic:
- These are semiconducting tetravalent crystals doped with impurity atoms group III (or) V.
- Their electrical conductivity is high.
- There is no permitted energy state of the impurity atom between valence and conduction band.
- Their electrical conductivity depends on temperature as well as dopant concentration.
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:
The free electrons from electron hole pairs, enable current to flow in the semiconductor when an external voltage is applied. The holes in the valence band also allow electron movement within the valence band itself and this also contributes to current flow. This process is called electron - hole pair generation.
Question 5.
A diode is called as a unidirectional device. Explain?
Answer:
Diode is called as 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 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:
The leakage current in a diode is the current that the diode will leak when a reverse voltage is applied to it. Under the reverse bias, a very small current in $\mu \mathrm{A}$, flows across the junction. This is due to the flow of the minority charge carriers called the leakage current or reverse saturation current.
Question 7.
Draw the output waveform of a full wave rectifier.
Answer :

Question 8.
Distinguish between avalanche and zener breakdown.
Answer:
1. Avalanche Breakdown:
- It occurs in junctions which are lightly and have wide depletion widths.
- It occurs at higher reverse voltages when thermally generated electrons get enough kinetic energy to produce more electrons by collision.
- At reverse voltage above $6 \mathrm{~V}$ breakdown is due to avalanche effect.
- Electric field produced is weak in nature.
- Charge carriers obtain energy from the applied potential.
2. Zener Breakdown:
- It occurs in junctions which are heavily doped and have narrow depletion widths.
- It occurs due to rupture of covalent bonds by strong electric fields set up in depletion region by the reverse voltage.
- At reverse voltage below $6 \mathrm{~V}$ breakdown is due to zener effect.
- A strong electric field is produced
- Zener current is independent of applied voltage.
Question 9.
Discuss the biasing polarities in an 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 $\mathrm{P}]$
Question 10.
Explain the current flow in a NPN transistor?
Answer:
1. The conventional flow of current is based on the direction of the motion of holes
2. In NPN transistor, current enters from the base into the emitter.
Question 11.
What is the phase relationship between the $\mathrm{AC}$ 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 $180^{\circ}$ 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:

The circuit used to feedback a portion of the output to the input is called the feedback network. If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases. It is necessary for sustained oscillations.

Question 13.
Give circuit symbol, logical operation, truth table, and Boolean expression of-
1. AND
2. OR
3. NOT
4. NAND
5. NOR
6. EX - OR gates.
Answer:
1. AND gate Circuit symbol:
$A$ and $B$ are inputs and $Y$ is the output.

It is a logic gate and hence $\mathrm{A}, \mathrm{B}$, and $\mathrm{Y}$ can have the (a) Two input AND gate value of either 1 or 0 .

Boolean equation:
$
\mathrm{Y}=\mathrm{A} . \mathrm{B}
$
It performs logical multiplication and is different from arithmetic multiplication.
Logic operation:
The output of AND gate is high (1) only when all the inputs are high (1). The rest of the cases the output is low. Hence the output of AND gate is high (1) only when all the inputs are high.
2. OR gate:
Circuit Symbol
$A$ and $B$ are inputs and $Y$ is the output.
 

Boolean equation:
$
\mathrm{A}+\mathrm{B}=\mathrm{Y}
$
It performs logical addition and is different from arithmetic addition.

Logic operation:
The output of OR gate is high (logic 1 state) when either of the inputs or both are high.
3. NOT gate:
Circuit symbol
$A$ is the input and $Y$ is the output.

Boolean equation:
$
\mathrm{Y}=\bar{A}
$
Logic operation:
The output is the complement of the input.

(b) Truth Table
It is represented with an overbar. It is also called as inverter. The truth table infers that the output $\mathrm{Y}$ is 1 when input $\mathrm{A}$ is 0 and vice versa. The truth table of NOT.
4. NAND gate:
$A$ and $B$ are inputs and $Y$ is the output.

Boolean equation:
$
\mathrm{Y}=\mathrm{Y}=\overline{A B}
$
Logic operation:
The output Y equals the complement of AND operation. The circuit is an AND gate followed by a NOT gate. Therefore, it is summarized as NAND. The output is at logic zero only when all the inputs are high. The rest of the cases, the output is high (Logic 1 state).
5. NOR gate:
Circuit symbol: A and B are inputs and $\mathrm{Y}$ is the output.

Boolean equation:
$
\mathrm{Y}=\mathrm{A}+\mathrm{B}
$
Logic operation:
$\mathrm{Y}$ equals the complement of OR operation (A OR B). The circuit is an OR gate followed by a NOT gate and is summarized as NOR. The output is high when all the inputs are low. The output is low for all other combinations of inputs.
6. $\mathrm{Ex}$ - OR gate:
Circuit symbol
$\mathrm{A}$ and $\mathrm{B}$ are inputs and $\mathrm{Y}$ is the output. The Ex-OR operation is denoted as $\bigoplus$.
Boolean equation:
$
\begin{aligned}
& \mathrm{Y}=\mathrm{A} \cdot \bar{B}+\bar{A} \cdot \mathrm{B} \\
& \mathrm{Y}=\mathrm{A} \oplus \mathrm{B}
\end{aligned}
$
Logic operation:
The output is high only when either of the two inputs is high. In the case of an Ex-OR gate with more
than two inputs, the output will be high when odd number of inputs are high.

Question 14.
State De Morgan's first and second theorems De Morgan's First Theorem?
Answer:
The first theorem states that the complement of the sum of two logical inputs is equal to the product of its complements.
$
\overline{A+B}=\bar{A} \cdot \bar{B}
$
De Morgan's Second Theorem:
The second theorem states that the complement of the product of two inputs is equal to the sum of its complements.
$
\overline{A \cdot B}=\overline{A+B}
$
Long Answer Questions
Question 1.
Elucidate the formation of a $\mathrm{N}$ - type and $\mathrm{P}$ - type semiconductors.
Answer:
1. $\mathrm{N}$ - type semiconductor:
A n-type semiconductor is obtained by doping a pure Germanium (or Silicon) crystal with a dopant from group V pentavalent elements like Phosphorus, Arsenic, and Antimony. The dopant has five valence electrons while the Germanium atom has four valence electrons.
During the process of doping, a few of the Germanium atoms are replaced by the group V dopants. Four of the five valence electrons of the impurity atom are bound with the 4 valence electrons of the neighbouring replaced Germanium atom. The fifth valence electron of the impurity atom will be loosely attached with the nucleus as it has not formed the covalent bond.

The energy level of the loosely attached fifth electron from the dopant is found just below the conduction band edge and is called the donor energy level. At room temperature, these electrons can easily move to the conduction band with the absorption of thermal energy. It is shown in the figure (c). Besides, an external electric field also can set free the loosely bound electrons and lead to conduction.
It is important to note that the energy required for an electron to jump from the valence band to the conduction band $\left(\mathrm{E}_{\mathrm{g}}\right)$ in an intrinsic semiconductor is $0.7 \mathrm{eV}$ for $\mathrm{Ge}$ and $1.1 \mathrm{eV}$ for $\mathrm{Si}$, while the energy required to set free a donor electron is only $0.01 \mathrm{eV}$ for $\mathrm{Ge}$ and $0.05 \mathrm{eV}$ for Si.

The group V pentavalent impurity atoms donate electrons to the conduction band and are called donor impurities. Therefore, each impurity atom provides one extra electron to the conduction band in addition to the thermally generated electrons. These thermally generated electrons leave holes in valence band. Hence, the majority carriers of current in an n-type Semiconductor are electrons and the minority carriers are holes. Such a semiconductor doped with a pentavalent impurity is called an n-type semiconductor.

2. $\mathrm{P}$ - type semiconductor:
Here, a trivalent atom from group III elements such as Boron, Aluminium, Gallium and Indium is added to the Germanium or Silicon substrate. The dopant with three valence electrons are bound with the neighbouring Germanium atom as shown in Figure (a). As Germanium atom has four valence electrons, one electron position of the dopant in the Germanium crystal lattice will remain vacant. The missing electron position in the covalent bond is denoted as a hole.

To make complete covalent bonding with all four neighbouring atoms, the dopant is in need of one more electron. These dopants can accept electrons from the neighbouring atoms. Therefore, this impurity is called an acceptor impurity. The energy level of the hole created by each impurity atom is just above the valence band and is called the acceptor-energy level, as shown in Figure (b).

For each acceptor atom, there will be a hole in the valence band in addition to the thermally generated holes. In such an extrinsic semiconductor, holes are the majority carriers and thermally generated electrons are minority carriers. The semiconductor thus formed is called a $\mathrm{p}$-type semiconductor.
Question 2.
Explain the formation of PN junction diode. Discuss its V-I characteristics.
Answer:
Formation of depletion layer:
A $\mathrm{p}-\mathrm{n}$ junction is formed by joining $\mathrm{n}$-type and $\mathrm{p}$-type semiconductor materials as shown in figure.
(a) Since the $\mathrm{n}$-region has a high electron concentration and the p-region a high hole concentration, electrons diffuse from the $\mathrm{n}$-side to the $\mathrm{p}$-side. This causes diffusion current which exists due to the concentration difference of electrons. The electrons diffusing into the p-region may occupy holes in that region and make it negative.

The holes left behind by these electrons in the $\mathrm{n}$-side are equivalent to the diffusion of holes from the $\mathrm{p}$ side to the $\mathrm{n}$-side. If the electrons and holes were not charged, this diffusion process would continue until the concentration of electrons and holes on the two sides were the same, as happens if two gasses come into contact with each other.

But, in a $\mathrm{p}-\mathrm{n}$ junction, when the electrons and holes move to the other side of the junction, they leave behind exposed charges on dopant atom sites, which are fixed in the crystal lattice and are unable to move. On the $\mathrm{n}$-side, positive ion cores are exposed and on the $\mathrm{p}$-side, negative ion cores are exposed as shown in Figure
(b) An electric field $\mathrm{E}$ forms between the positive ion cores in the $\mathrm{n}$ - type material and negative ion cores in the p-type material. The electric field sweeps free carriers out of this region and hence it is called depletion region as it is depleted of free carriers. A barrier potential $\mathrm{V}_{\mathrm{b}}$ due to the electric field $\mathrm{E}$ is formed at the junction as shown in Figure.
(c) As this diffusion of charge carriers from both sides continues, the negative ions form a layer of negative space charge region along the $\mathrm{p}$-side. Similarly, a positive space charge region is formed by positive ions on the $\mathrm{n}$-side. The positive space charge region attracts electrons from $\mathrm{p}$-side to $\mathrm{n}$-side and the negative space charge region attracts holes from $\mathrm{n}$-side to $\mathrm{p}$-side.

This moment of earners happen in this region due to the formed electric field and it constitutes a current called drift current. The diffusion current and drift current flow in the opposite direction and at one instant they both become equal. Thus, a $\mathrm{p}-\mathrm{n}$ junction is formed.
V-I characteristics of a junction diode:
Forward characteristics:
It is the study of the variation in current through the diode with respect to the applied voltage across the diode when it is forward biased. An external resistance $(\mathrm{R})$ is used to limit the flow of current through the diode. The voltage across the diode is varied by varying the biasing voltage across the dc power supply.

The forward bias voltage and the corresponding forward bias current are noted. A graph is plotted by taking the forward bias voltage (V) along the $\mathrm{X}$ - axis and the current (I) through the diode along the $\mathrm{Y}$ - axis. This graph is called the forward V-I characteristics of the $\mathrm{p}-\mathrm{n}$ junction diode. Three inferences
can be brought out from the graph:

(i) At room temperature, a potential difference equal to the barrier potential is required before a reasonable forward current starts flowing across the diode. This voltage is known as threshold voltage or cut-in voltage or knee voltage $\left(\mathrm{V}_{\text {th }}\right)$. It is approximately $0.3 \mathrm{~V}$ for Germanium and $0.7 \mathrm{~V}$ for Silicon. The current flow is negligible when the applied voltage is less than the threshold voltage. Beyond the threshold voltage, increase in current is significant even for a small increase in voltage.
(ii) The graph clearly infers that the current flow is not linear and is exponential. Hence it does not obey Ohm's law.
(iii) The forward resistance $\left(\mathrm{r}_{\mathrm{f}}\right)$ of the diode is the ratio of the small change in voltage $(\Delta V)$ to the small change in current $(\Delta \mathrm{I}), \mathrm{r}_{\mathrm{f}}=\frac{\Delta V}{\Delta I}$

However, if the applied voltage is increased beyond a rated value, it will produce an extremely large current which may destroy the junction due to overheating. This is called as the breakdown of the diode and the voltage at which the diode breaks down is called the breakdown voltage. Thus, it is safe to operate a diode well within the threshold voltage and the breakdown voltage.
Reverse characteristics:
In the reverse bias, the p-region of the diode is connected to the negative terminal and n-region to the positive terminal of the dc power supply. A graph is drawn between the reverse bias voltage and the current across the junction, which is called the reverse characteristics of a p-n junction diode.
 

Under this bias, a very small current in $\mu \mathrm{A}$, flows across the junction. This is due to the flow of the minority charge carriers called the leakage current or reverse saturation current. Besides, the current is almost independent of the voltage. The reverse bias voltage can be increased only up to the rated value otherwise the diode will enter into the breakdown region.
Question 3.
Draw the circuit diagram of a half wave rectifier and explain its working Half wave rectifier circuit:
Answer:
The half wave rectifier circuit. The circuit consists of a transformer, a p-n junction diode and a resistor. In a half wave rectifier circuit, either a positive half or the negative half of the AC input is passed through while the other half is blocked. Only one half of the input wave reaches the output. Therefore, it  is called half wave rectifier. Here, a p-n junction diode acts as a rectifying diode. 

During the positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal A becomes positive with respect to terminal $B$. The diode is forward biased and hence it conducts. The current flows through the load resistor $\mathrm{R}_{\mathrm{L}}$ and the $\mathrm{AC}$ voltage developed across $\mathrm{R}_{\mathrm{L}}$ constitutes the output voltage $\mathrm{V}_0$ and the waveform of the diode current.
During the negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal A is negative with respect to terminal $\mathrm{B}$. Now the diode is reverse biased and does not conduct and hence no current passes through $\mathrm{R}_{\mathrm{L}}$. The reverse saturation current in a diode is negligible. Since there is no voltage drop across $\mathrm{R}_{\mathrm{L}}$, the negative half cycle of ac supply is suppressed at the output.
The output of the half wave rectifier is not a steady dc voltage but a pulsating wave. This pulsating voltage is not sufficient for electronic equipments. A constant or a steady voltage is required which can be obtained with the help of filter circuits and voltage regulator circuits. Efficiency ( $\eta)$ is the ratio of the output de power to the ac input power supplied to the circuit. Its value for half wave rectifier is $40.6 \%$.
Question 4.
Explain the construction and working of a full w ave rectifier.
Answer:
Full wave rectifier:
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier. It consists of two $\mathrm{p}$-n junction diodes, a center tapped transformer, and a load resistor $\left(\mathrm{R}_{\mathrm{L}}\right)$. The centre is usually taken as the ground or zero voltage reference point. Due to the centre tap transformer, the output voltage rectified by each diode is only one half of the
total secondary voltage.

During positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal $M$ is positive, $G$ is at zero potential and $\mathrm{N}$ is at negative potential. This forward biases diode $\mathrm{D}_1$ and reverse biases diode $\mathrm{D}_2$. Hence, being forward biased, diode $\mathrm{D}_1$ conducts and current flows along the path $\mathrm{MD}_1 \mathrm{AGC} . \mathrm{As}$ a resul t, positive half cycle of the voltage appears across $R_L$ in the direction $G$ to $C$.
During negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal $N$ is positive, $G$ is at zero potential and $\mathrm{M}$ is at negative potential. This forward biases diode $\mathrm{D}_2$ and reverse biases diode $\mathrm{D}_1$. Hence, being forward biased, diode $\mathrm{D}_2$ conducts and current flows along the path $\mathrm{ND}_2 \mathrm{BGC}$. As a result, negative half cycle of the voltage appears across $R_L$ in the same direction from $G$ to $C$.

Hence in a full wave rectifier both positive and negative half cycles of the input signal pass through the circuit in the same direction as shown in figure (b). Though both positive and negative half cycles of ac input are rectified, the output is still pulsating in nature. The efficiency ( $\eta$ ) of full wave rectifier is twice that of a half wave rectifier and is found to be $81.2 \%$. It is because both the positive and negative half cycles of the ac input source are rectified.
Question 5.
What is an LED? Give the principle of operation with a diagram.
Answer:
Light Emitting Diode (LED):
LED is a $p-n$ junction diode which emits visible or invisible light when it is forward biased. Since, electrical energy is converted into light energy, this process is also called electroluminescence. The cross-sectional view of a commercial LED is shown in figure (b). it consists of a p-layer, n-layer and a substrate. A transparent window is used to allow light to travel in the desired direction. An external resistance in series with the biasing source is required to limit the forward current through the LED. In addition, it has two leads; anode and cathode:

When the $\mathrm{p}-\mathrm{n}$ junction is forward biased, the conduction band electrons on $\mathrm{n}$-side and valence band holes on p-side diffuse across the junction. When they cross the junction, they become excess minority carriers (electrons in $\mathrm{p}$-side and holes in $\mathrm{n}$-side).

These excess minority carriers recombine with oppositely charged majority carriers in the respective regions, i.e. the electrons in the conduction band recombine with holes in the valence band as shown in the figure (c). During recombination process, energy is released in the form of light (radiative) or heat (non-radiative). For radiative recombination, a photon of energy ho is emitted.
For non-radiative recombination, energy is liberated in the form of heat. The colour of the light is determined by the energy band gap of the material. Therefore, LEDs are available in a wide range of colours such as blue ( $\mathrm{SiC})$, green (AlGaP) and red (GaAsP). Now a days, LED which emits white light (GalnN) is also available.
Question 6.
Write notes on Photodiode?
Answer:
Photodiodes:
A p-n junction diode which converts an optical signal into electric current is known as photodiode. Thus, the operation of photodiode is exactly opposite to that of an LED. Photo diode words in reverse bias. Its circuit symbol is shown in figure (a). The direction of arrows indicates that the light is incident on the photo diode.

The device consists of a $\mathrm{p}-\mathrm{n}$ junction semiconductor made of photosensitive material kept safely inside a plastic case. It has a small transparent window that allows light to be incident on the $\mathrm{p}-\mathrm{n}$ junction. Photodiodes can generate current when the $\mathrm{p}-\mathrm{n}$ junction is exposed to light and hence are called as light sensors.

When a photon of sufficient energy (hv) strikes the depletion region of the diode, some of the valence band electrons are elevated into conduction band, in turn holes are developed in the valence band. This creates electron - hole pairs. The amount of electron - hole pairs generated depends on the intensity of light incident on the $p$-n junction. These electrons and holes are swept across the $p$-n junction by the electric field created by reverse voltage before recombination takes place. Thus, holes move towards the $\mathrm{n}$-side and electrons towards the $\mathrm{p}$-side.
When the external circuit is made, the electrons flow through the external circuit and constitute the photocurrent. When the incident light is zero, there exists a reverse current which is negligible. This reverse current in the absence of any incident light is called dark current and is due to the thermally generated minority carriers.
Question 7.
Explain the working principle of a solar cell. Mention its applications.
Answer:
Solar cell:
A solar cell, also known as photovoltaic cell, converts light energy directly into electricity or electric potential difference by photovoltaic effect. It is basically a $\mathrm{p}-\mathrm{n}$ junction which generates emf when solar radiation falls on the $\mathrm{p}-\mathrm{n}$ junction. A solar cell is of two types: $\mathrm{p}$-type and $n$-type. Both types use a combination of p-type and n-type silicon which together forms the p-n junction of the solar cell.
The difference is that p-type solar cells use p-type Silicon as the base with an ultra-thin layer of n-type Silicon as shown in Figure, while n-type solar cell uses the opposite combination. The other side of the p-Silicon is coated with metal which forms the back electrical contact. On top of the n-type Silicon, metal grid is deposited which acts as the front electrical contact. The top of the solar cell is coated with anti-reflection coating and toughened glass.

In a solar cell, electron-hole pairs are generated due to the absorption of light near the junction. Then the charge carriers are separated due to the electric field of the depletion region. Electrons move towards $n-$ type Silicon and holes move towards p-type Silicon layer. The electrons reaching the $n$-side are collected by the front contact and holes reaching p-side are collected by the back electrical contact.
Thus a potential difference is developed across solar cell. When an external load is connected to the solar cell, photocurrent flows through the load. Many solar cells are connected together either in series or in parallel combination to form solar panel or module. Many solar panels are connected with each other to form solar arrays. For high power applications, solar panels and solar arrays are used. Applications:
- Solar cells are widely used in calculators, watches, toys, portable power supplies, etc.
- Solar cells are used in satellites and space applications
- 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 Transistor in Common Emitter Mode:
The know-how of certain parameters like the input resistance, output resistance, and current gain of a transistor are very important for the effective use of transistors in circuits. The circuit to study the static characteristics of an NPN transistor in the common emitter mode is given in figure.
The bias supply voltages $\mathrm{V}_{\mathrm{BB}}$ and $\mathrm{V}_{\mathrm{CC}}$ bias the base-emitter junction and collector- emitter junction respectively. The junction potential at the base-emitter is represented as $\mathrm{V}_{\mathrm{BE}}$ and the collector-emitter as $\mathrm{V}_{\mathrm{CE}}$. The rheostats $\mathrm{R}_1$ and $\mathrm{R}_2$, are used to vary the base and collector currents respectively.
The static characteristics of the BJT are:
1. Input characteristics
2. Output characteristics
3. Transfer characteristics

1. Input Characteristics:
Input Characteristics curves give the relationship between the base current $\left(\mathrm{I}_{\mathrm{B}}\right)$ and base to emitter voltage $\left(\mathrm{V}_{\mathrm{BE}}\right)$ at constant collector to emitter voltage $\left(\mathrm{V}_{\mathrm{CE}}\right)$ and are shown in figure. Initially, the collector to emitter voltage $\left(\mathrm{V}_{\mathrm{CE}}\right)$ is set to a particular voltage (above $0.7 \mathrm{~V}$ to reverse bias the junction). Then the base-emitter voltage $\left(\mathrm{V}_{\mathrm{BE}}\right)$ is increased in suitable steps and the corresponding base-current $\left(I_B\right)$ is recorded. A graph is plotted with $V_{B E}$ along the $x$-axis and $I_B$ along the $y$-axis. The procedure is repeated for different values of $\mathrm{V}_{\mathrm{CE}}$.

The following observations are made from the graph:
1. The curve looks like the forward characteristics of an ordinary $p-n$ junction diode.
2. There exists a threshold voltage or knee voltage $\left(\mathrm{V}_{\mathrm{k}}\right)$ below which the base current is very small. The value is $0.7 \mathrm{~V}$ for Silicon and $0.3 \mathrm{~V}$ for Germanium transistors. Beyond the knee voltage, the base current increases with the increase in base-emitter voltage.
3. It is also noted that the increase in the collector-emitter voltage decreases the base current. This shifts the curve outward. This is because the increase in collector-emitter voltage increases the width of the depletion region in turn, reduces the effective base width and thereby the base current.
Input resistance:
The ratio of the change in base-emitter voltage $\left(\Delta \mathrm{V}_{\mathrm{BE}}\right)$ to the change in base current $\left(\Delta \mathrm{I}_{\mathrm{B}}\right)$ at a constant collector-emitter voltage $\left(\mathrm{V}_{\mathrm{CE}}\right)$ is called the input resistance $\left(\mathrm{R}_{\mathrm{j}}\right)$. The input resistance is not linear in the lower region of the curve.
The input resistance is high for a transistor in common emitter configuration. Output Characteristics: The output characteristics give the relationship between the variation in the collector current $\left(\Delta \mathrm{I}_{\mathrm{C}}\right)$ with respect to the variation in collector- emitter voltage $\left(\Delta \mathrm{V}_{\mathrm{CE}}\right)$ at constant input current $\left(\mathrm{I}_{\mathrm{B}}\right)$ as shown in figure.

Initially, the base current $\left(\mathrm{I}_{\mathrm{B}}\right)$ is set to a particular value. Then collector- emitter voltage $\left(\mathrm{V}_{\mathrm{CE}}\right)$ is increased in suitable steps and the corresponding collector current $\left(\mathrm{I}_{\mathrm{C}}\right)$ is recorded. $\mathrm{A}$ graph is plotted with the $\mathrm{V}_{\mathrm{CE}}$ along the $\mathrm{x}$-axis and $\mathrm{I}_{\mathrm{C}}$ along the $\mathrm{y}$-axis.
This procedure is repeated for different values of $\mathrm{I}_{\mathrm{B}}$, The four regions in are:
(i) Saturation region:
When $\mathrm{V}_{\mathrm{CE}}$ is increased above $0 \mathrm{~V}$, the Ic increases rapidly to a saturation value almost independent of $\mathrm{I}_{\mathrm{B}}$ (Ohmic region, OA) called knee voltage. Transistors are always operated above this knee voltage.
(ii) Cut-off region:
A small collector current $\left(\mathrm{I}_{\mathrm{C}}\right)$ exists even after the base current $\left(\mathrm{I}_{\mathrm{B}}\right)$ is reduced to zero. This current is due to the presence of minority carriers across the collector-base junction and the surface leakage current ( $\mathrm{I}_{\mathrm{CEO}}$ ). This region is called as the cut-off region, because the main collector current is cut-off.
(iii) Active region:
In this region, the emitter-base junction is forward biased and the collector-base junction is reverse biased. The transistor in this region can be used for voltage, current and power amplification.
(iv) Breakdown region:
If the collector-emitter voltage $\left(\mathrm{V}_{\mathrm{CE}}\right)$ is increased beyond the rated value given by the manufacturer, the collector current $\left(\mathrm{I}_{\mathrm{C}}\right)$ increases enormously leading to the junction breakdown of the transistor. This avalanche breakdown can damage the transistor.
Output Resistance:
The ratio of the change in the collector-emitter voltage $\left(\Delta \mathrm{V}_{\mathrm{CE}}\right)$ to the corresponding change in the collector current $\left(\Delta \mathrm{I}_{\mathrm{C}}\right)$ at constant base current $\left(\mathrm{I}_{\mathrm{B}}\right)$ is called output resistance $\left(\mathrm{R}_{\mathrm{O}}\right)$.
$
\mathrm{R}_0=\left(\frac{\Delta \mathrm{V}_{\mathrm{CB}}}{\Delta \mathrm{I}_{\mathrm{C}}}\right)_{\mathrm{I}_{\mathrm{B}}}
$
The output resistance for transistor in common emitter configuration is very low.

Question 9.
Describe the function of a transistor as an amplifier with the neat circuit diagram. Sketch the input and output wave form.
Answer:
Transistor as an amplifier:
A transistor operating in the active region has the capability to amplify weak signals. Amplification is the process of increasing the signal strength (increase in the amplitude). If a large amplification is required, the transistors are cascaded with coupling elements like resistors, capacitors, and transformers which is called as multistage amplifiers.
Here, the amplification of an electrical signal is explained with a single stage transistor amplifier as shown in figure (a). Single stage indicates that the circuit consists of one transistor with the allied components. An NPN transistor is connected in the common emitter Configuration.

To start with, the $\mathrm{Q}$ point or the operating point of the transistor is fixed so as to get the maximum signal swing at the output (neither towards saturation point nor towards cut-off). A load resistance, $R_C$ is connected in series with the collector circuit to measure the output voltage. The capacitor $\mathrm{C}_1$ allows only the ac signal to pass through. The emitter bypass capacitor $C_E$ provides a low reactance path to the amplified ac signal. The coupling capacitor $C_C$ is used to couple one stage of the amplifier with the next stage while constructing multistage amplifiers. $V_S$ is the sinusoidal input signal source applied across the base-emitter.
Collector currrent , $\mathrm{I}_{\mathrm{C}}=\mathrm{I} \beta_{\mathrm{B}}=\left[\therefore \beta=\frac{I_C}{I_B}\right]$
Applying Kirchhoff's voltage law in the output loop, the collector-emitter voltage is given by $\mathrm{V}_{\mathrm{CE}}=\mathrm{V}_{\mathrm{CC}}-\mathrm{I}_{\mathrm{C}} \mathrm{R}_{\mathrm{C}}$
Working of the amplifier:
1. During the positive half cycle:
Input signal $\left(\mathrm{V}_{\mathrm{S}}\right)$ increases the forward voltage across the emitter-base. As a result, the base current $\left(\mathrm{I}_{\mathrm{B}}\right)$ increases. Consequently, the collector current $\left(\mathrm{I}_{\mathrm{C}}\right)$ increases $\beta$ times. This increases the voltage drop
across $\mathrm{R}_{\mathrm{C}}$ which in turn decreases the collector-emitter voltage $\left(\mathrm{V}_{\mathrm{CE}}\right)$. Therefore, the input signal in the positive direction produces an amplified signal in the negative direction at the output. Hence, the output signal is reversed by $180^{\circ}$ as shown in figure (b).
2. During the negative half cycle:
Input signal $\left(\mathrm{V}_{\mathrm{S}}\right)$ decreases the forward voltage across the emitter-base. As a result, base current $\left(\mathrm{I}_{\mathrm{B}}\right)$ decreases and in turn increases the collector current $\left(\mathrm{I}_{\mathrm{C}}\right)$. The increase in collector current $\left(\mathrm{I}_{\mathrm{C}}\right.$ ) decreases the potential drop across $\mathrm{Rc}$ and increases the collector-emitter voltage $\left(\mathrm{V}_{\mathrm{CE}}\right)$. Thus, the input signal in the negative direction produces an amplified signal in the positive direction at the output. Therefore, $180^{\circ}$ phase reversal is observed during the negative half cycle of the input signal.
Question 10.
Transistor functions as a switch. Explain.
Answer:
The transistor in saturation and cut - off regions functions like an electronic switch that helps to turn $\mathrm{ON}$ or OFF a given circuit by a small control signal.

1. Presence of dc source at the input (saturation region):
When a high input voltage $\left(V_{\text {in }}=+5 V\right)$ is applied, the base current $\left(I_B\right)$ increases and in turn increases the collector current. The transistor will move into the saturation region (turned ON). The increase in collector current $\left(\mathrm{I}_{\mathrm{C}}\right)$ increases the voltage drop across Rc, thereby lowering the output voltage, close to zero. The transistor acts like a closed switch and is equivalent to ON condition.
2. Absence of dc source at the input (cut-off region):
A low input voltage $\left(\mathrm{V}_{\text {in }}=0 \mathrm{~V}\right)$, decreases the base current $\left(\mathrm{I}_{\mathrm{B}}\right)$ and in turn decreases the collector current $\left(I_{\mathrm{C}}\right)$. The transistor will move into the cut-off region (turned OFF). The decrease in collector current (Ic) decreases the drop across Rc, thereby increasing the output voltage, close to $+5 \mathrm{~V}$.

The transistor acts as an open switch which is considered as the OFF condition. It is manifested that, a high input gives a low output and a low input gives a high output. In addition, we can say that the output voltage is opposite to the applied input voltage. Therefore, a transistor can be used as an inverter in computer logic circuitry.
Question 11.
State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example. Answer:
Law's of Boolean Algebra: The NOT, OR and AND operations are $\bar{A}, \mathrm{~A}+\mathrm{B}, \mathrm{A}$. B are the Boolean operations. The results of these operations can be summarised as:
Complement law

The complement law can be realised as $\overline{\overline{\mathrm{A}}}=\mathrm{A}$ OR laws:

AND laws:

The Boolean operations obey the following laws.
Commutative laws
$
\mathrm{A}+\mathrm{B}=\mathrm{B}+\mathrm{A}
$
A. $\mathrm{B}=\mathrm{B} \cdot \mathrm{A}$
Associative laws
$
A+(B+C)=(A+B)+C
$
A. $(B \cdot C)=(A \cdot B) \cdot C$
Distributive laws
$
\begin{aligned}
& \mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC} \\
& \mathrm{A}+\mathrm{BC}=(\mathrm{A}+\mathrm{B})(\mathrm{A}+\mathrm{C})
\end{aligned}
$
The above laws are used to simplify complicated expressions and to simplify the logic circuitry.
Question 12.
State and prove De Morgan's First and Second theorems.
Answer:
De Morgan's First Theorem:
The first theorem states that the complement of the sum of two logical inputs is equal to the product of
its complements.
Proof:
The Boolean equation for NOR gate is $Y=\overline{A+B}$. The Boolean equation for a bubbled AND gate is $Y$ $=\bar{A} \cdot \bar{B}$. Both cases generate same outputs for same inputs. It can be verified using the following truth table.

From the above truth table, we can conclude $\overline{A+B}=\bar{A} \cdot \bar{B}$. Thus De Morgan's First Theorem is proved. It also says that a NOR gate is equal to a bubbled AND gate. The corresponding logic circuit diagram is shown in figure.

De Morgan's Second Theorem:
The second theorem states that the complement of the product of two inputs is equal to the sum of its complements.
Proof:
The Boolean equation for NAND gate is $\mathrm{Y}=\overline{A B}$
The Boolean equation for bubbled OR gate is $\mathrm{Y}=\bar{A}+\bar{B}$. A and $\mathrm{B}$ are the inputs and $\mathrm{Y}$ is the output. The above two equations produces the same output for the same inputs. It can be verified by using the truth table.

From the above truth table we can conclude $\overline{A . B}=\bar{A}+\bar{B}$. Thus De Morgan's Second Theorem is proved. It also says, a NAND gate is equal to a bubbled OR gate. The corresponding logic circuit
diagram is shown in figure.