Learn/ 12th Physics/ Unit 18 /Long Questions

Unit 18: Electronics — Long Questions

12th Class Physics · Unit 18: Electronics

1.What is p-n junction? Also describe forward biased and reverse biased p-n junction.

p-n JUNCTION

A p-n junction is formed when a crystal of Germanium or silicon is grown in such a way that its one half is doped with trivalent impurity and one half with pentavalent impurity. p-n junction is an important building block of electronics.

Characteristics of p-n Junction

(i) n-region of p-n junction contain free electrons as majority charge carriers.

(ii) p-region contains holes as majority charge carriers.

(iii) Depletion Region

Just after the formation of p-n junction. The free electrons in n-region due to their random motion, diffuse into the p-region. As result of diffusion, a chargeless region is formed around the junction in which no charge carriers present which is called depletion region.

In the figure, black dots represents the free electron and small circles show the holes whereas the circles with +ve and -ve ions which make the depletion region.

(iv) Potential Barrier

Due to the diffusion of electrons from n to p region, a potential difference develops across the depletion region called potential barrier. Its value is 0.7 volt for silicon and 0.3 volt for Germanium. This potential difference stops further diffusion of electrons into the p-region.

Forward Biased p-n Junction

If a battery is connected to a p-n junction such that the p side is positive and n side is negative, then the external potential of battery supplies energy to free electron to n-region and holes in p-region. A current of the order of few milliampere flows across the p-n junction. In this state the p-n junction is said to forward biased.

Variation of Current with Biasing Voltage

Variation of current through junction with biased voltage is studied by the circuit shown in figure. The values of biasing voltage is recorded. The graph is drawn between current and biased voltage which is as shown. It is observed from graph that as forward biased voltage is increased, the current also increases. If ΔVf is the increase in biased voltage and ΔIf is the increase in current then the forward resistance of p-n junction is given by

fr = ΔVf / ΔIf

It is the resistance offered by the p-n junction when it is conducting. The value of this resistance is only a few ohm.

Reverse Biased p-n Junction

When a battery is connected to p-n junction such that p-side is -ve and n-side is +ve then the p-n junction is reverse biased. In this situation no current flows due to majority charge carriers. However, a very small current due to minority charge carriers flows across the junction which is of the order of few micro amperes. It is known as reverse current or leakage current.

Variation of Current with Biasing Voltage

Variation of reverse current with applying biasing voltage can be studied in circuit shown in a figure. A graph is drawn between reverse current and biased voltage is as shown. It is seen that reverse voltage increase from zero, the reverse current quickly rises. As the reverse voltage is further increased, the reverse current remains constant. The resistance offered by p-n junction is very high of the order of several mega ohm.

As the reverse voltage is increased, the kinetic energy of the minority charge carriers with which they cross the depletion region also increases till it is sufficient to break a covalent bond. When the covalent bond breaks, more electron hole pairs are created. Thus minority charge carriers begins to multiply due to which the reverse current begins to increase till a point is reached when the junction breaks down and reverse current rises. After break down, the reverse curve will rise to very high value which will damage the junction. p-n junction is also known as semi-conductor diode whose symbolic representation is as shown. The arrow head represents the p-n region and known as anode. The vertical line shows the n-region and is known as cathode.

2.What is meant by rectification? Explain the action of semi-conductor diode as half wave rectifier.

RECTIFICATION

Conversion of alternating current into direct current is called rectification. Semi-conductor diode (p-n junction) is used for rectification. There are two types of rectification (1) Half wave rectification (2) Full wave rectification

(1) Half Wave Rectification

A half wave rectification is shown in figure. When an alternating voltage of period T called input – (input signed) voltage applied to diode D which is connected in series with resistance R. Half of alternating current cycle are converted into direct current. During +ve half cycle of input alternating voltage. That is 0 → T/2, the diode is forward biased. So, it offers a very low resistance and current flows through R. The flow of current through R causes a potential drops across it. During -ve half cycle T/2 → T the diode is reverse biased and it offers a very high resistance. So, practically no current flows through R and potential drop across R is almost zero. The current through resistance flows in only one direction which means that it is direct current. However this current flows in pulses as shown.

(2) Full Wave Rectification

When both half of the input voltage can be used to send unidirectional current in output circuit, then such rectification is called full wave rectification. Its circuit consist of4 diodes connected in bridge type arrangement as shown in figure. During the +ve half cycle i.e., 0 → T/2, the terminal A of the bridge is +ve with respect to its other terminal B. Now, diode D₁ and D₃ become forward biased and conduct current. Current flows through the circuit as shown in figure.

A current flows through the circuit in the direction shown by arrows in figure (a). During the negative half cycle i.e., T/2 → T, the terminal A is negative with respect to terminal B, now the diode D₂ and D₁ become forward bias and conduct current. A current flows through the circuit in the direction shown by arrows in figure (b). It can be seen that the direction of current flow through the resistance R is the same in both the halves of the cycle. Thus both halves of the alternating input voltage send a unidirectional current. However the output voltage is not smooth but pulsating. It can be made smooth by using a circuit known as filter.

3.What are the specially designed p-n junction? (OR) What are the application of p-n junction?

SPECIALLY DESIGNED p-n JUNCTION

In addition to a use of p-n junction, as rectifier, many types of p-n junction have made for special purposes. The following p-n junctions are most commonly used

(i) Light Emitting Diode (LED)

(ii) Photo-diode

(iii) Photovoltaic cell

Light Emitting Diode

LED are made from semi-conductors. Such as Galium Arsenide Phosphide and galium Arsenide in which potential barrier in n and p side is such that when an electron combine with the hole during forward biased, a photon of visible light is emitted. These diodes are commonly used as small light sources e.g., inductors etc. A specially formed arra of seven LED's is used for displaying digits in electronic devices.

Photo Diode

Photo diode is used for detection of light both visible and invisible. It is operated in reverse biased condition. When no light is incident on the junction, the reverse current is almost zero. But when p-n junction is exposed to light, the reverse current increases with the intensity of light as shown. A photo diode can turn its current ON and OFF in nanoseconds. Thus, it is the fastest photo detection device. Applications of photo diode are

(i) Detection both visible and invisible light.

(ii) Automatic switching.

(iii) Optical communication systems.

(iv) Logic circuits.

Photo-Voltaic Cell

There are p-n junctions in which potential barriers between p and n regions is used to drive a current through external circuit when light is incident on junctions. The current is directly proportional to the intensity of light. A single silicon photo-voltaic cell produces a small voltage of 0.6 volt and a current of the order of few milli ampere. In order to obtain greater power, a series-parallel thousands of small cells are used. They are called photo-voltaic panels and are commonly used in satellites and space stations.

4.What is transistor?

TRANSISTORS

A transistor consist of a single crystal of germanium or silicon which is grown in such a way

n-p-n and p-n-p Transistors

If a thin layer of p-type substance is sandwiched between two n-type substances, the transistor is called n-p-n transistor. If a thin n-type substance is sandwiched between two p-type substance, the transistor is called p-n-p transistor.

Emitter Base and Collector

The central region of transistor is known as base which is very thin of the order of 10-6 meter. The thicker regions on either side of the base are called emitter and collector. The emitter and collector have greater strength of impurity. The collector is comparitively larger than emitter. The emitter has greater concentration of impurity as compare to collector.

Operation of Transistor

For normal operation of transistor batteries VBB and VCC are connected in such away that its emitter-base junction is forward biased and collector-base junction is reverse biased. The battery VCC is of much higher value than VBB. Figure shows biasing arrangement for p-n-p transistor. In actual partice n-p-n transistors is generally used.

Current Flow in a n-p-n Transistor

Figure (a) shows a n-p-n transistor at the instant when the biasing voltage is applied. Electrons in the emitter have not yet entered the base region. After the application of the biasing voltage, emitter base junction is forward biased, so emitter injects a large number of electrons in base region (Figure b). These free electrons in the base can flow in either of two directions. They can either flow out of the base to the positive terminal of VBB or they can be attracted towards the collector because of battery VCC. Since the base is extremely thin, very few electrons manage to recombine with holes and escape out of the base. Almost all of the free electrons injected from the emitter into the base are attracted into the collector by the large positive VCC (Figure c). Thus, in a normally biased transistor due to above mentioned flow of electrons, we can say, that an electronic current IE flows from the emitter into the base. A very small part of it, current IB, flows out of the base, the rest of IC flows out of the collector (Figure).

The flow of conventional current is shown in figure. In future we will use conventional current only. From the figure, it can be seen that

IE = IC + IB ........ (i)

As very few electrons flow out of base, so IB is very small as compared to IC.

It is also found that for a given transistor the ratio of collector current IC to base current IB is nearly constant i.e.,

β = IC / IB ........ (ii)

The ratio β is called current gain of transistor. Its value is quite large of the order of hundreds. Eq. (i) and (ii) are fundamental equations of all transistors.

5.How is it used as an amplifier? Give its circuit diagram. Deduce the relation for its voltage gain.

TRANSISTOR AS AN AMPLIFIER

Conversion of low voltage input into a large voltage output is called amplification.

In majority of electronic circuits, transistors are basically used as amplifiers. An amplifier is thus the building block of every complex electronic circuit. It is for this reason that study of transistor amplifier is important.

The circuit in figure is a transistor voltage amplifier. The battery VBB forward biases the base-emitter junction and VCC reverse biases the collector-base junction. VBE and VCE are the input and output voltages respectively. The base current is

IB = VBE / rie

where rie is base emitter resistance of the transistor. The transistor amplifies it β times. So

IC = βIB = β VBE / rie

The output voltage Vo = VCE is determined by applying KVL equation in the output loop which gives

VCC = IC RC + VCE or VCE = VCC - IC RC

Substituting the value of IC and replacing VCE by Vo.

Vo = VCC - β (VBE RC) / rie ........ (a)

When the input signal Vin changes from VBE to VBE + ΔVin. This causes a little change in base current from IB to (IB + ΔIB) due to which the collector current changes from IC to (IC + ΔIC). As the collector current changes, the voltage drop across RC i.e., (IC RC) also changes due to which the output voltage Vo changes by ΔVo. Substituting the changed values in equation (a).

Vo + ΔVo = VCC - β (VBE + ΔVin) (RC / rie) ........ (b)

Subtracting eq. (a) from eq. (b)

ΔVo = -β ΔVin (RC / rie)

ΔVo / ΔVin = -β RC / rie

Therefore the gain of the amplifier is

A = ΔVo / ΔVin = -β RC / rie

The value of the factor β RC / rie is of the order of hundreds, so the input voltage is amplified. The negative sign shows that there is a phase shift of 180° between the input and the output signals. This is also called common emitter amplifier.

6.How a transistor is used as a switch?

TRANSISTOR AS A SWITCH

Fig. (i) shows a circuit in which transistor is used as a switch. The collector (C) and emitter (E) behaves as terminals of switch. The circuit in which current is to be turned OFF and ON is connected across these terminals. The base (B) and emitter (E) act as control terminals which decide the state of the switch.

Working

In order to turn on the switch a large P.D VB is applied between control terminals B.E. This inject a large current IB into base circuit due to which very large current IC began to flow in C.E circuit. This value of large current is possible only when resistance between C and E drops to such a small value that the potential drops across C.E is nearly 0.1 volts. In figure (i) ammeter is at ground potential. So, we can suppose that collector is also at ground potential and C.E circuit can be drawn as shown in figure (ii). C.E switch is closed and bulb glows due to large value of collector current.

To turn the switch OFF, the base current IB is set zero by removing the base circuit as shown in figure (iii). As I become zero and C.E circuit becomes open. Now the resistance between C and E nearly becomes infinite which open the C.E switch as shown in figure (iv).

7.Explain operational amplifier.

OPERATIONAL AMPLIFIER

It is an important electronic circuit that is used in almost every electronic devices. So, instead of making amplifier circuit by discrete components, the whole amplifier is integrated on a small silicon chip and enclosed in a capsule. Pins connected with working terminals such as input, output and power supply project outside the capsule. The enclosed circuit of amplifier is used by making required connection with the (three) pins. Such an integrated amplifier is known as operational amplifier.

The operational amplifier in symbolic form is shown in figure. It has 2 input terminals. One is known as inverting input (-) and the other is known as non-inverting input (+). A signal that is applied at inverting (-) input, appears after amplification, at the output terminal with a phase shifted to 180°.

A signal that is applied at non-inverting (+) input is amplified at output without any phase change. An operational amplifier has a large no of characteristics.

Input Resistance

It is the resistance between +ve and -ve inputs of the amplifier as shown in figure. Its value is very high of the order of several mega ohm. Due to high input resistance Rin, practically no current flows between two input terminals. It is very important feature of operational amplifiers.

Output Resistance

It is the resistance between the output terminals and ground. Its values is only a few ohms.

Open Loop Gain

It is the ratio of output voltage Vo to the voltage difference between inverting and non-inverting inputs when there is no external connection between the output and inputs that is

AOL = Vo / (V_+ - V_-) = Vo / Vi

The open loop gain of the amplifier is very high. It is of the order of 105.

8.Draw the circuit diagram of inverting amplifier and label it. Evaluate a relation for its gain.

OP-AMP AS INVERTING AMPLIFIER

Figure shows the circuit of an op-amps when used as an inverting amplifier. The input signal Vi which is to be amplified, is applied at inverting terminal (-) through a resistance R1. Vo is its output. The non-inverting terminal (+) is grounded, i.e., its potential is zero. We know that AOL is very high, of the order of 105 for any value of Vo, V_+ - V_- ≈ 0 or V_+ ≈ V_-. Since V_+ is at ground so V_- is virtually at ground potential i.e., V_- = 0. Referring to figure.

Current through R1 = I1 = (Vi - V_-) / R1 = (Vi - 0) / R1 = Vi / R1

Current through R2 = I2 = (V_- - Vo) / R2 = (0 - Vo) / R2 = -Vo / R2

As practically no current flows between (-) and (+) terminals, so according to Kirchhoff's current rule I1 = I2

or Vi / R1 = -Vo / R2 or Vo / Vi = -R2 / R1

As Vo/Vi is defined as gain G of the inverting amplifier, so

G = -R2 / R1 ........ (i)

The negative sign indicates that the output signal is 180° out of phase with respect to input signal. It is interesting to note that the closed loop gain depends upon the two externally connected resistances R1 and R2. The gain is independent of what is happening inside the amplifier.

If R1 = 10 kΩ and R2 = 100 kΩ, the gain of the amplifier is

G = Vo / Vi = -R2 / R1 = -100 Ω / 10 kΩ = -10

9.Draw the circuit diagram of non-inverting amplifier and label it. Evaluate a relation for its gain.

OP-AMP AS NON-INVERTING AMPLIFIER

The circuit diagram of op-amp as non-inverting amplified is shown in figure. In this case the input signal Vi is applied at the non-inverting terminal (+). As explained earlier, due to high open loop gain of amplifier, the inverting (-) and non-inverting (+) inputs are virtually at the same potential. That is,

V_- ≈ V_+ = Vi

Also, from figure

Current through R1 = I1 = (0 - V_-) / R1 = (0 - Vi) / R1 = -Vi / R1

Current through R2 = I2 = (Vi - Vo) / R2

As practically no current flows between (-) and (+) terminals, so by Kirchhoff's current rule I1 = I2.

Hence -Vi / R1 = (Vi - Vo) / R2

or Vi (1/R1 + 1/R2) = Vo / R2

or Gain = Vo / Vi = 1 + R2 / R1 ........ (ii)

Again the gain of the amplifier is independent of the internal structure of the op-amp. It just depends upon the two externally connected resistances R1 and R2. The positive sign of gain indicates that the input and output signals are in phase.

10.Write a note on op-amp as a comparator.

OP-AMP AS A COMPARATOR

Op-amp usually requires two power supplies of equal voltage but of opposite polarity. Most op-amp operate with VCC = ±12V supply.

As the open loop gain of the op-amp is very high (105), even a very small potential difference between the inverting and non-inverting inputs is amplified to such a large extent that the amplifier gets saturated, i.e., its output either becomes equal to +VCC or -VCC. This feature of op-amp is used to compare tw voltages. Figure shows the circuit of an op-amp used as comparator. VR is reference voltage which is connected with (+) terminal and V is the voltage which is to be compared with the reference VR. It is connected with (-) terminal.

When V_- > V_+ or V_- > VR, then Vo = -VCC

and if V_- < V_+ or V_- < VR, then Vo = +VCC

11.Explain comparator as a night switch.

COMPARATOR AS A NIGHT SWITCH

Suppose it is required that when intensity of light falls below a certain level, the street light is automatically switched on. This can be accomplished by using op-amp as a comparator. In figure resistances R1 and R2 form a potential divider. The potential drop across R2 provides the reference voltage VR to the (+) input of the op-amp. Thus

VR = R2 / (R1 + R2) × VCC ........ (i)

LDR is a light dependent resistance. The value of its resistance RL depends upon the intensity of light falling upon it. RL and R3 form another potential divider. The potential drop across R3 is V' which is given by

V' = R3 / (RL + R3) × VCC ........ (ii)

V' provides the voltage to (-) input of the op-amp. V' will not be a constant voltage but it will vary with the intensity of light. During day time, when light is falling upon LDR, RL is small. According to eq. (ii), V' will be large such that V' > VR so that Vo = -VCC. The output of the op is connected with a relay system which energizes only when Vo = +VCC and then it turn on the street lights. Thus when Vo = -VCC, the light will not be switched ON.

As it gets darker, RL becomes larger and V' decreases. When V' becomes just less than VR, the output of op-amp switches to +VCC which energizes the relay system and the street lights are turned ON.

12.What is digital system?

DIGITAL SYSTEM

Digital system is that one which deals quantities or variables which have two discreet values or states.

For example

(i) A switch can be either open or closed.

(ii) The answer of a question can either be yes or no.

(iii) A certain statement can either true or false.

(iv) A bulb can be either off or on.

Various designs are used to represent the two quantized states of such quantities. The most common of these are

| | 1 | 2 | 3 | 4 | 5 | 6 |
| --- | --- | --- | --- | --- | --- | --- |
| One of the states | True | High | 1 | Yes | On | Closed |
| The other state | False | Low | 0 | No | Off | Open |

Mathematical calculations of these quantities can be best carried if they are represented by binary regions 1 and 0. In describing functions of digital systems, a closed switch will be shown as 1 and open switch will be shown as 0. Similarly a lighter bulb will be described as 1 and off bulb will be described as 0.

Boolean Algebra

For mathematical calculations, we need two basic mathematical operations addition and subtraction. Similarly, in digital system, we require special algebra known as Boolean algebra for the calculation of quantities which have values 1 and 0. 0 and 0. Boolean algebra is based on three basic operations.

• AND operation

• OR operation

• NOT operation

These operations are implemented in the study of logic states.

13.Describe the fundamental logic gates.

FUNDAMENTAL LOGIC GATES

The electronic circuit which implement the various logic operation are known as logic gates. In these gates, high and low states, 1 and 0 states are supposed by certain voltage levels. One particular voltage level represent a high (1) and another voltage level represent a low (0). For example figure show the range 1 and zero level for a certain type of digital gates. Thus, if voltage of 3.5 volt is applied to a gate. It will accept it as high or 1 and if a voltage of 0.5 volt is applied to a gate will accept it as low or 0.

OR Gate

It is symbolically represent as shown

It has two or more inputs and a signal output X. The output have value 1 when at least one of its inputs A and B is at one. Thus output will be 0 only when both the inputs are zero. Thus it implements the truth table of OR operation as shown. The mathematical notation for OR operation is

X = A + B

AND Gate

The AND gate is symbolically written as shown

It has two or more inputs and single output. It is designed in such that it implements of truth table of AND operation i.e., its output is only one when both of its inputs are at 1 and for all other combination of the values of inputs, and the output is zero. The mathematical expression for AND gate is

X = A . B

NOT Gate

It performs the operation of inversion or complementation that is why it is also known as inverter. It changes a logic level to its opposite level i.e., it changes 1 to zero (0) and zero (0) to 1. The symbolic representation of NOT gate is shown in figure.

Whenever a bar is placed on any variable. This shows that value of variable have been inverted. For example,

1̄ = 0 or 0̄ = 1

The bubble (0) indicates the operation of inversion. Mathematical symbol for NOT gate is

X = Ā

The truth table for NOT gate are as shown.

14.Draw the symbols of logic gates for the following Boolean functions. Write their respective truth labels X = A + B, X = A . B.

OTHER LOGIC GATES

NOR Gate

In NOR gate, the output of OR gate is inverted. Its symbol is shown in figure. And its truth table is as given, the mathematical symbol for NOR operation is

X = A̅ + B̅

NAND Gate

In NAND gate the output of AND gate is inverted. Its symbol is shown in figure. The bubble shown in figure tells that output of AND gate is inverted. The truth table implemented by it shown. The mathematical symbol of NAND operation is that

X = A̅ . B̅

Exclusive OR Gate

Consider, a boolean function X of a variable A and B such that

X = A̅B + AB̅

The first term of function X is A•NOTing the variable A with NOT of B. The second term is NOT of A and AND of B. The function X is obtained by ORing these two term. It can be constructed by combining OR, AND and NOT gate according to figure. The value of function X is obtained by drawing truth table which gives the value of X for value of variable A and B.

The value of X is zero then the output have the same values and it is 1 when input have different values. It can be verified that the above circuit implement the truth table.

Exclusive-NOR Gate (XNOR)

The exclusive NOR gate is obtained by inverting the output of a XOR gate. Its symbol is shown in figure. The bubble shown at the output in this figure shows that the output of XOR gate has been inverted. So its Boolean expression is given by

X = A̅B̅ + AB̅

The truth table of XNOR gate is given in the table. Its output is 1 when its two inputs are identical and 0 when the two inputs are different. Like XOR gate, it is also constructed by a combination of NOT, AND and NOR gates by the so shown in figure.