Semiconductor Electronics - Mohit Nain · CHAPTER 14 : Semiconductor Electronics 1.4 p-n Junction A...

- CHAPTER 14

SemiconductorElectronics

Chapter Analysis w.r.t, Last 3 Year's Board ExamsThe analysis given here gives you an analytical picture of this chapter and will helpyou toidentify the concepts of the chapter that are to befocussed more from exam point of view.

Number of Questions asked in last 3 years

2015 2016 2017Delhi All India Delhi All India Delhi All India

-Very Short Answer (I mark)

f--Short Type I Answer (2 marks) IQShort Type II Answer (3 marks) 2 Qs IQ IQ 2 Qs 2 Qs

Long Answer (5 marks)Value Based Questions (4 marks)---

• In 2015, InDelhi set, two question of 3 marks each based on Fabrication andCE-Transistors were asked. In All India set, one question of 2 marks based on Intrinsicand Extrinsic Semiconductors and one questions of 3 marks based on Logic Gateswere asked.

• In 2016, only one question of3 marks based on p-n hmction Diode in All India set.• In 2017, in Delhi set, 2 question of 3 marks each based on p -n Junction and other is

numerical question based CE-Transistor were asked. In All India set, two question of3 marks based on Forward and Reverse Biased, Solar Cell and Circuit basednumerical were asked.

On the basis of above analysis, it can be said that from exam point of view Fabrication,Intrinsic and Extrinsic Semiconductors, Logic Gates, p-n Junction and Solar C-ellare mostimportant concepts of the chapter.

[TOPIC 1] Semiconductor, Diode and ItsApplications

1.1Classification of Metals, Semiconductors and InsulatorsOn the basis of the relative value of electrical conductivity (cr)or resistivity (p = ~ J the solids are broadly

classify as· Metals They possess very low resistivity or high conductivity.

p _1O-2_1O-8Qm, cr-102_108Sm-1

. Semiconductors They have resistivity or conductivity intermediate to metals and insulators.p_1O-5_1O-6Qm, cr-1O-5to 100Sm-1

· Insulators They have high resistivity or low conductivity.p _1011_1019Qm, o _10-11_10-19 Sm-1

Types of Semiconductors on the basis of their chemical composition are given below(i) Elemental Semiconductors These semiconductors are available in natural form, e.g. silicon

and germanium.(ii) Compound Semiconductors These semiconductors are made by compounding the metals,

e.g. CdS, GaAs, CdSe, Inp, anthracene, polyaniline, ete.

1.2 Energy BandIn a crystal due to interatomic interaction, valence electrons of one atom are shared by more than oneatom. Now, splitting of energy level takes place. The collection of these closely spaced energy levels arecalled an energy band.· Valence Band Valence band is the highest energy band which includes the energy levels of the

valence electrons.· Conduction Band Conduction band is the energy band above the valence band. The lowest unfilled

allowed energy band next to valence band is called conduction band.· Energy Band Gap The minimum energy required for shifting electrons from ~B~:'

valence band to conduction band is called energy band gap (E 9)' !lEg

· Forbidden Energy Gap (LVJ 9) Energy gap between conduction band and valence VB max.band, LVJ 9 = (CB)min- (VB)max min.

· Fermi Energy The highest energy level in the conduction band filled up with electrons at absolutezero is called Fermi level and the energy corresponding to the Fermi level is called Fermi Energy. Itis the maximum possible energy possessed by free electrons of a material at absolute zero temperature(i.e. 0 K).

CHAPTER 14 : Semiconductor Electronics 425

Differences between conductor, insulator and semiconductoron the basis of energy bands

Conductor (Metal) Insulator Semiconductor

In conductor, either there is noenergy gap between theconduction band which is partiallyfilled with electrons and valenceband or the conduction band andvalence band overlap each other.

In insulator, the valence bandis completely filled, theconduction band is completelyempty and energy gap is quitelarge that small energy fromany other source cannotovercome it.

In semiconductor also, like insulatorsthe valence band is totally filled and theconduction band is empty but theenergy gap between conduction bandand valence band, unlike insulators isvery small.

Thus, many electrons from belowthe fermi level can shift to higherenergy levels above the fermi levelin the conduction band andbehave as free electrons byacquiring a little more energyfrom any other sources.

Thus, electrons are bound tovalence band and are not freeto move and hence, electricconduction is not possible inthis type of material.

Thus, at room temperature, someelectrons in the valence band acquirethermal energy greater than energyband gap and jump over to theconduction band where they are freeto move under the influence of even asmall electric field and acquire smallconductivity.

Conduction ConductionElectron ,( band b~denergyEc···················

Ev JiJiiiffi{UjiiJJii dValenceband

For metals

Empty conduction band

!>-EC~e>

! lEg >3 eVe~ Ev ValenceUJ band

1.3 SemiconductorsSemiconductors are the materials whose conductivity lies between metals and insulators. They arecharacterised by narrow energy gap (- 1eV) between the valence band and conduction band.

Classification of Semiconductor on the Basis of PurityIntrinsic SemiconductorsIt is a pure semiconductor without any significant dopant species present.

ne = nh = ni

where, ne and nh are number densities of electrons and holes respectively and ni is called intrinsic carrierconcentration. An intrinsic semiconductor is also called an undoped semiconductor or i-typesemiconductor.The total current I is the sum of the electron current I, and hole current Ih.

I=Ie+Ih·

where, L, = electron current, Ih = hole current

Extrinsic SemiconductorsPure semiconductor when doped with the impurity, it is known as extrinsic semiconductor.

426

Extrinsic semiconductors are basically of two types:

n-type SemiconductorIn this type of extrinsic semiconductormajority charge carriers are electrons andminority charge carriers are holes, i.e. ne > nh'Here, we dope Si or Ge with a pentavalentelement, such as As, P or Sb of group V, thenfour of its electrons bond with the foursilicon neighbours, while fifth remains veryweakly bound to its parent atom.Formation of n-type semiconductor is shownbelow:

.: ~. :,...···0····..···0···....··0· ..·..······...•V·.......•.V·.......•·V· ··i ". ~ ': ~ ....

. • \ ~~e-unbondedfree······8·..·····•.e...··..···8··..··..··· electron donated....... :+: .•........ :+5•........•• +~..•......... by pentavalent

i) : \ i) (+5 valency) atom~ i ~ i ....~:::::~e:~:::::~e:~::::~:e::::~:::::i ~ ~'. ~ \. . ... ., ..Pe;:tavalent danor atam (As. sb, P. etc.) daped fortetravalent Si or Ge giving n-type semiconductor

Donor energy level lies just below theconduction band

Conduction band- 1.

Eg= 1 evL~~~~;·I:v·:s ...• · T Ed= 0.01 eV

Valence band

n-type

p-type Semiconductor• In this semiconductor, majority charge

carriers are holes and minority chargecarriers are electrons i.e. nh > ne'

• In a p-type semiconductor, doping is donewith trivalent impurity atoms. i.e. Thoseatoms which have three valence electronsin their valence shell.

o Chopterwise CBSE Solved Popers PHYSICS

• Formation of p-type semiconductor is shown below:

:~:::8:~:::::~e~:::::~:8~::::::::~.... ~.... i ~.. .. .... .. .,.. .. ... .. .'.. :. .:· ·8· ··@· ··8·..•..···....• ( ~.......•. r ·k·~:-·<~!t~ .

.. .. i>CI, • .

::::~e:~:::::~e:~:~~~:(3::::~::::~.... ~. :.... ".. . .... " " :

Trivalent acceptor atom (In. AI. 8. etc) doped intetravalent Si or Ge lattice giving p -type semiconductor

Acceptor energy level lies just above the valence band

Conduction band

Eg= 1 evl A t I I 1.~:::~~~~~7~~'~-::' T e,= 0.01 eV

Valence band

p-type

• At equilibrium condition, nenh = n;

• Minimum energy required to create a hole-electronpair, hv e E9

where, Egis energy band gap.. h hc1.e.Eg= vrrtin =--

Amax

• Electric current, I = eA (neve + nhvh)where, A is area of cross-section andve and vh are speed of electron and hole respectively.Mobility of charge carriers.u = 2:. , where E is applied

Eelectric field.Hence, ve = !leE and vh = !lhEElectrical conductivity, o = .!. = e (Jlene + nhllh)

pwhere, ne and nh are concentration of electron andhole respectively and u, and u , are mobilities ofelectron and hole, respectively, applying the formula

V Ex! EA1= R = p! = P= !(neve + nhvh)

A

CHAPTER 14 : Semiconductor Electronics

1.4 p-n JunctionA p-n junction is an arrangement made by a closecontact of n-type semiconductor and p-typesemiconductor.

Formatian of Depletion Region in p-n.JunctionDuring formation of p-n junction, due to theconcentration gradient across p and n-sides, holesdiffuse from p-side to n-side(p ~ n) and electronsdiffuse from n-side to p-side (n ~ pl.This space charge region on either side of thejunction topr .•ier is known as depletion region.

E d ift - Electrondiffusionlectron rI -eeEllEll~~:?%~

~~:: v/. . "illeeEllEll '~~;'~7.eeEllEll ~t"--'-::-,--=-~'-'"

.M<'II10< oc

1\6{ XI00

Holediffusiorl--Hole drift

Depletion region is the small region in thevicinity of the junction which is depleted of freecharge carriers. Width of depletion region is of theorder oflO-6m. The potential difference develo; jacross the depletion region is called the poterr' Ibarrier.

Semiconductor Diodelp-n Junction Diode• A semiconductor diode is basically a p-n

junction with metallic contacts provided at theends for the application of an external voltage.

• A p-n junction diode is represented as

~Anode Cathode .

• The direction of arrow indicates theconventional direction of current (when thediode is under forward bias).

• The graphical relations between voltage appliedacross p-n junction and current flowing throughthe junction are called I-V.

427

I-V (Current-Voltage) Characteristicof p-n Junction DiodeForward Biased Characteristic• Junction diode is said to be forward bias when

the positive terminal of the external battery isconnected to the p -side and negative terminal tothe n-side of the diode.Similarly, if the positive terminal of a battery isconnected to n-side and negative terminal to thep-side, then the p-n junction is said to be reversebiased. The circuit diagram and I-Vcharacteristics of a forward biased diode isshown below:

8

I:n :~ 3

Li: 21~

Battery o 0.1 0.2 0.3 0.4 0.5Forw~sM

(b)(a)

Forward bias

(i) In forward biasing width of depletion layerdecreases.

(ii) In forward biasing resistance offeredRForward '" Ion -25n

Reverse Biased CharacteristicIn reverse biased, the applied voltage supports theflow of minority charge carriers across thejunction. So, very small current flows across thejunction due to minority charge carriers.

428

The circuit diagram and 1- V characteristics of areverse biased diode is shown below.

p n--- Reverse bias M

-10 -8 -6 -4 -2 0--~~~~~~O~ ~

2 .s

4 ~l6 138 3l10~

<lla:

Breakdownvoltage

Battery'------I ~~f--:-+_.......J

(a) (b)

(i) In reverse biasing width of depletion layerincreases.

(ii) In reverse biasing resistance offeredRReverse "" 1050

(iii) Reverse bias supports the potential barrierand no current flows across the junctiondue to the diffusion of the majority carriers.

• The DC resistance of a junction diode, rDC = ~I

• The dynamic resistance or AC-resistance of. . di d LlVjunction 10 e, rAC = -

M

1.5 Diode as a RectifierThe process of converting alternatingvoltage/current into direct voltage/current iscalled rectification. Diode is used as a rectifierfor converting alternating current/Voltage intodirect current/voltage. There are two ways ofusing a diode as a rectifier i.e.

Diode as a Half-Wave RectifierDiode conducts corresponding to positive halfcycle and does not conduct during negative halfcycle. Hence, AC is converted by diode intounidirectional pulsating DC. This action is knownas half-wave rectification.

o Chapterwise CBSE Solved Papers PHYSICS

Transformer

+p-n

51 :A

::;c.::;o1

·8" .. ----_._.

Circuit diagram for Diode as a half-wove rectifier

The input and output waveforms have beengiven below:

(a)InputAC

..J

a:(/)(/)eo<tI<ll

Fg(b)

Input and output waveforms

Diode as a Full-Wave RectifierIn the full-wave rectifier, two p-n junction diodes,Dj and D2 are used. Its working is based on theprinciple that junction diode offer very lowresistance in forward bias and very high resistancein reverse bias .

Transformer

01, • -.-: •• --r:.,...-. -:-.• ---j .;.t------,:P1 51:

Circuit diagram of full-wove rectifier


The input and output waveforms havebeen given below:

<:(

CilE.2

~coCilE

jDue tolDue to Due tOiDue to

I I I I

01 : O2 : 01 : 02 :: : : I:I I I II I I II I I I

• The average value or DC value obtained from ahalf-wave rectifier,

10IDC =-1t

• The average value or DC value obtained from afull-wave rectifier,

210IDC=-1t

• The pulse ~uency of a half-wave rectifier isequal to frequency of AC.

• The pulse frequency of a full-wave rectifier isdouble to that of AC.

1.6 OptoelectronicJunction Devices

Semiconductor diodes in which carriers aregenerated by photons. i.e. photo-excitation, suchdevices are known as optoelectronic devices.These are as follows:

Light Emitting Diode (LED)It is a heavily doped forward biased p-n junctiondiode which spontaneously converts electricalenergy into light energy, like infrared and visiblelight.

429

Its symbol is~hv

~

p

Metaliisedcontact

A forward biased LED

V-I characteristics of LED are shown below:

J(mA)30 Silicon

-10V

2015

10

--+-----t--"""'-'--.......-...,..,...- V (volt)

LEDs has the following advantages overconventional incandescent low power lamps.

(a) Fast action and no warm up time required(b) It is nearly monochromatic(c) Low operational voltage and less power

consumed, long life, ruggedness(d) Fast ON-OFF switching capability in

nanoseconds.

PhotodiodeA photo diode is a special type of junction diodeused for detecting optical signals. It is a reversebiased p-n junction made from a photosensitivematerial. Such a way that light can fall on itsjunction.

430

V-I characteristics of photodiode are shownbelow:

Reverse bias

J(mA)

v

[(mAl

We observe from the figure that current inphotodiode changes with the change inlight intensity (I), when reverse bias isapplied. In light operated switches.

Solar CellSolar cell is a p-n junction diode which convertssolar energy into electrical energy. It is based onthe photovoltaic effect. Its symbol is

~~,..

pPhoto current through an illuminated p-n junction

V-I characteristics of solar cell are shown below:/

Open circuit voltage(Voc)-,---+--_--V

Iec

o Chopterwise CBSE Solved Papers PHYSICS

The materials used for solar cell are Si and GaAs.

Zener DiodeZener diode is a reverse biased heavily doped p-njunction diode. It is operated in breakdownregion.

Its symbol is -cf--V-I characteristics of Zener diode are shownbelow:

J(mA)

Reverse biasVz

-v

Forward bias

/ (JJ.A)

. Zener Diode as a Voltage Regulator Whenthe applied reverse voltage (V) reaches thebreakdown voltage (V,) of the Zener diode thereis a large change in the current. So, after thebreakdown voltage V" a large change in thecurrent can be produced by almost insignificantchange in the reverse bias voltage i.e. Zenervoltage remains constant even though thecurrent through the Zener diode varies over awide range. The circuital arrangement is shownhere .

• This breakdown in a diode due to the band toband tunneling is called Zener breakdown.

Unregulatedvoltage

LoadRL

Regulatedvoltage, Vz

Circuit diagram af Zener diode as voltage regulator

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 1o 1 Mark Questions1. The graph shown in the figure represents a

plot of current versus voltage for a givensemiconductor. Identify the region, if anyover which the semiconductor has anegative resistance. All India 2013

B

Voltage01)

2. What is the difference between an n-typeand a p- type extrinsic semiconductor?Oelhl2012C

3. What happens to the width of depletionlayer of a p- n junction when it is

(i) forward biased?(ii) reverse biased? All India 2011

4. Why cannot we take one slab of frtypesemiconductor and physically join it toanother slab of n-type semiconductor to getp-n junction? All India 2010C

01

-2V~

5. What is the most common use ofphotodiode? All India 2009

6. State the relation between the frequency vof radiation emitted by LED and the bandgap energy E of the semiconductor used tofabricate it. All India 2009C

7. Figure shows the I- V characteristics of agiven device. Name the device and writewhere it is used.

Reverse biasVz

/(mA)

Forward bias

V(volt)

/(mA)Oelhl2009

o 2 Marks Questions8. The V-I characteristic of a silicon diode is

as shown in the figure. Calculate theresistance of the diode at (i) I = 15 mAand (ii) V = -10V

(1 mAl

30

-10V

20 -------- Silicon r-:\--

15

10

o0.70.8 V

Foreign 2015

9. Distinguish between 'intrinsic' and'extrinsic' semiconductors? All India 2015

10. Explain, with the help of a circuitdiagram, the working of a p-n. junctiondiode as a half-wave rectifier. All India 2014

11. Draw energy band diagram of n-type andp-type semiconductor at temperatureT> OK.Mark the donar and acceptorenergy level with their energies. Foreign 2014

12. Distinguish between a metal and aninsulator on the basis of energy banddiagram. Foreign 2014

13. Explain, with the help of a circuitdiagram, the working of a photodiode.Write briefly how it is used to detect theoptical signals. Oelhl 2013

432

14. Assuming that the two diodes Dl and D2used in the electric circuit shown in thefigure are ideal, find out the value of thecurrent flowing through 1n resistor.

01 2Q•...v

2Q

1 Q

6V Delhi 2013C

15. Mention the important considerationsrequired while fabricating a p-ri junctiondiode to be used as a Light EmittingDiode (LED). What should be the order of

, band gap of an LED, if it is required toemit light in the visible range? Delhi 2013

16. Write two characteristics featuresto distinguish between n-type and p-typesemiconductors. All Indio 2012

17. Give two advantages of LED's over theconventional incandescent lamps.Foreign 2012

18. The current in the forward bias is knownto be more (- mA) than the current in thereverse bias (- J.lA). What is the reason, tooperate the photodiode in reverse bias?Delhi 2D12

19. How does a light emitting diode (LED)work? Give two advantages of LED's overthe conventional incandescent lamps.Foreign 2D12

20. (i) Why are Si and GaAs preferredmaterials for fabrication in solarcells?

(ii) Draw V-I characteristic of solar celland mention its significance.All Indio 2012C

21. Name the semiconductor device that canbe used to regulate an unregulated DCpower supply. With the help of I-Vcharacteristics of this device, explain itsworking principle. Delhi 2011

22. How is forward biasing different fromreverse biasing in a p- n junction diode?Delhi 2D11


23. Explain, how a depletion region is formedin a junction diode? Delhi 2011

24. Draw the circuit diagram showing how ap-n junction diode is

(i) forward biased (ii) reverse biasedHow is the width of depletion layeraffected in the two cases? All Indio 2011C

25. Carbon and silicon both have four valenceelectrons each, then how are theydistinguished? Delhi 2DllC

26. Name the device, D which is used as avoltage regulator in the given circuit andgive its symbol.

+

I R 121

uatingR~O

Consttage volta

1 1

Fluetvol

antge

Delhi 2011C

27. Draw the circuit diagram of anilluminated photodiode in reverse bias.How is photodiode used to measure lightintensity? Delhi 201D

28. Write the main use of the(i) photodiode

(ii) Zener diode. All Indio 2010C

o 3 Marks Questions29. (i) In the following diagram, which bulb

out of ~ and ~ will glow and why?01 O2

~tJrB2

(ii) Draw a diagram of an illuminatedp-ri junction solar cell.

(iii) Explain briefly the three processesdue to which generation of emf takesplace in a solar cell. All Indio 2017


30. (i) In the following diagram, is thejunction diode forward biased orreverse biased?

~v

(ii) Draw the circuit diagram of a fullwave rectifier and state how itworks? All Indio 2017

31. Write the two processes that take place inthe formation of a p-n. junction. Explainwith the help of a diagram, the formationof depletion region and barrier potentialin a p-ri junction. Delhi 2017

32. A Zener diode is fabricated by heavilydoping both p- and n- sides of thejunction. Explain, why?Briefl~plain the use of Zener diode as aDC voltage regulator with the help of acircuit diagram. Delhi 2017

33. (i) Explain with the help of a diagramthe formation of depletion region andbarrier potential in a p -n junction.

(ii) Draw the circuit diagram of ahalf-wave rectifier and explain itsworking. All Indio 2016

34. (i) Describe the working principle of asolar cell. Mention three basicprocesses involved in the generationof emf.

(ii) Why are Si and GaAs preferredmaterials for solar cells? Foreign 2016

35. With what considerations in view, aphotodiode is fabricated? State itsworking with the help of a suitablediagram.Even though the current in the forwardbias is known to be more than in thereverse bias, yet the photo diode works inreverse bias. What is the reason? Deihl 2015

36. (i) Distinguish between n-type andp- type semiconductors on the basisof energy band diagrams.

(ii) Compare their conductivities atabsolute zero temperature and atroom temperature. Delhi 2015C

433

37. Draw the energy band diagrams of(i) n -type and

(ii) p-type semiconductor attemperature, T> OK.

In the case n-type Si semiconductor, thedonor energy level is slightly below thebottom of conduction band whereas inp-type semiconductor, the acceptor energylevel is slightly above the top of thevalence band. Explain, what role do theseenergy levels play in conduction andvalence bands. All Indio 2015C

38. Write any two distinguishing featuresbetween conductors, semiconductors andinsulators on the basis of energy banddiagrams. All Indio 2014

39. (i) How is a photo diode fabricated?(ii) Briefly explain its working. Draw its

V-I characteristics for two differentintensities of illumination. Foreign 2014

40. Draw the circuit diagram of a full-waverectifier using p-ti junction diode. Explainits working and show the output inputwaveforms. Delhi 2012

41. Draw V-I characteristics of a p-ri junctiondiode. Answer the following questions,giving reasons.

(i) Why is the current under reversebias almost independent of theapplied potential upto a criticalvoltage?

(ii) Why does the reverse current show asudden increase at the criticalvoltage?

Name any semiconductor device whichoperates under the reverse bias in thebreakdown region. All Indio 2012

42. Draw a labelled diagram of a full-waverectifier circuit. State .Its working principle. Show theinput-output waveforms. All Indio 2011

43. Name the important processes that occursduring the formation of a p- n junction.Explain briefly, with the help of a suitablediagram, how a p- n junction is formed.Define the term 'barrier potential'?Foreign 2011

434

44. (i) Why is a photo diode operated inreverse bias mode?

(ii) For what purpose is a photo diodeused?

(iii) Draw its I-V characteristicsfor different intensities ofillumination. HOTS; All India 2011C

45. (i) Why are Si and GaAs preferredmaterials for solar cells?

(ii) Describe briefly with the help of anecessary circuit diagram, theworking principle of a solar cell.All India 2011C

46. (i) Describe the working of LightEmitting Diodes (LEDs).

(ii) Which semiconductors are preferredto make LEDs and why?

(iii) Give two advantages of using LEDsover conventional incandescent lowpower lamps. All India 2011

47. An AC signal is fed into two circuits 'X' and'Y' and the corresponding output in the twocases have the waveforms as shown.

(i) Identify the circuits 'X and 'Y'. Drawtheir labelled circuit diagrams.

(ii) Briefly explain the working ofcircuit Y.

(iii) How does the output waveform fromcircuit Y get modified when a capacitoris connected across the outputterminals parallel to the load resistor?

~~ @-o, 0,

~[}-~All India 2009

48. With the help of a suitable diagram, explainthe formation of depletion region in a p-tijunction. How does its width change whenthe junction is

(i) forward biased and(ii) reverse biased? All India 2009

49. (i) With the help of circuit diagrams,distinguish between forward biasingand reverse biasing of p-ti junctiondiode.


(ii) Draw V-I characteristics of ap-n junction diode in(a) forward bias. (b) reverse bias.All India 2009

50. Explain with the help of a circuit diagramhow a Zener diode works as a DC voltageregulator? Draw its V-I characteristics.All India 2009

51. How is a Zener diode fabricated so as tomake it a special purpose semiconductordiode? Draw the circuit diagram of aZener diode as a voltage regulator andexplain its working. All India 2009C

o 5 Marks Questions52. (i) State briefly the processes involved

in the formation of p-ri junction,explaining clearly how the depletionregion is formed.

(ii) Using the necessary circuitdiagrams, show how the V-Icharacteristics of a p-ri junction areobtained in (a) forward biasing(b) reverse biasing

How are these characteristics made use ofin rectification? Delhi 2014

53. (i) Draw the circuit arrangement forstudying the V-I characteristics of ap-n junction diode in (a) forward and(b) reverse bias. Briefly explain howthe typical V-I characteristics of adiode are obtained and draw thesecharacteristics.

(ii) With the help of necessary circuitdiagram, explain the working of aphotodiode used for detecting opticalsignals. All India 2014C

54. (i) Explain with the help of diagram,how a depletion layer and barrierpotential are formed in a junctiondiode.

(ii) Draw a circuit diagram of afull-wave rectifier. Explain itsworking and draw input and outputwaveforms.Delhi 2014C


55. (i) How is a depletion region formed inp-n junction?

(ii) With the help of a labelled circuitdiagram. Explain how a junctiondiode is used as a full-wave rectifier.Draw its input, output waveforms.

(iii) How do you obtain steady DC outputfrom the pulsating voltage? Delhi 2D13C

56. Why is a Zener diode considered as aspecial purpose semiconductor diode?Draw the I-V characteristics of Zenerdiode and explain briefly, how reversecurrent suddenly increase at thebreakdown voltage?Describe briefly with the help of a circuitdiagram, how a Zener diode works toobtain a constant DC voltage from theunregulated DC output of a rectifier.Delhi 2009C; Foreign 2012

57. (i) Describe briefly, with the help of adiagram, the role of the twoimportant processes involved in theformation of a p-ri junction.

(ii) Name the device which is used as avoltage regulator. Draw thenecessary circuit diagram andexplain its working. HOTS; All Indio 2012

58. (i) Draw the circuit diagram of a p-njunction diode in(a) forward bias.(b) reverse bias.How are these circuits used to studythe V-I characteristics of a silicondiode? Draw the typical V-Icharacteristics.

(ii) What is a Light Emitting Diode(LED)? Mention two importantadvantages of LEDs overconventional lamps.

Delhi 2010C; All Indio 2010

59. (i) Draw I-V characteristics of a Zenerdiode.

(ii) Explain with the help of a circuitdiagram, the use of a Zener diode asa voltage regulator.

435

(iii) A photodiode is operated underreverse bias although in the forwardbias, the current is known to be morethan the current in the reverse bias.Explain, giving reason. HOTS; Foreign 2010

60. (i) Draw a circuit arrangement forstudying V-I characteristics of a p-njunction diode in(a) forward bias and(b) reverse bias.Show typical V-I characteristics of asilicon diode.

(ii) State the main practical applicationof LED. Explain, giving reason, whythe semiconductor used for fabricationof visible light LEDs must have aband gap of at least (nearly) 1.8 eV.Delhi 2010C

61. (i) How is a Zener diode fabricated so asto make it a special purpose diode?Draw I-V characteristics of Zenerdiode and explain the significance ofbreakdown voltage.

(ii) Explain briefly, with the help of acircuit diagram, how a p-n junctiondiode works as a half-wave rectifier.Delhi 2009C

62. (i) Draw the typical shape of the V-Icharacteristics of a p-ti junction diodeboth in (a) forward (b) reverse bias

J r L 'r:configuration. How do we infer, fromthese characteristics that a diode canbe used to rectify alternatingvoltages?

(ii) Draw the circuit diagram of afull-wave rectifier using a centre-taptransformer and two p-ti junctiondiodes. Give a brief description of theworking of this circuit. Delhi 2009C

o Explanations1. Resistance of a material can be found out by the

slope of the curve V versus I. Part Be of the curveshows the negative resistance as with the increasein current and uecrease in voltage. (1)

436 [21 Chapterwise CBSE Solved Papers PHYSICS

2. p-typeSemiconductor

n-typeSemiconductor

(i) It is formed bydopingpentavalentimpurities.

It is formed bydoping trivalentimpurities.(112)

(ii) The electrons aremajority carriersand holes areminority carriers.(n,»nh)

The holes aremajority carriersand electrons areminority carriers(nh> >n,)(112)

3. (i) Width of depletion layer decreases inforward bias.

(ii) Width of depletion layer increases in reversebias. (1)

4. In this way;-c6ntinuous contact cannot beproduced at atomic level and junction willbehave as a discontinuity for the flowing chargecarrier. (1)

5. The photodiode can be used as a photodetectorto detect optical signals. In light operatedswitches. (1)

6. In LED, energy of the photon should be equalto or less than the band gap energy i.e.

hv s; Eswhere, Es = band gap energy,

v = frequency of emitted photon. (1)

7. Zener diode, which is used as a DC voltageregulator. (1)

8. (i) From the given curve, we havevoltage, V = 0.8 volt for current,

I = 20 mA for voltage,V = 0.7volt for current,I=10mA

=> LV = (20-10) mA = 10x 10-3A (1)

=> ~V= (0.8-0.7) =0.1 V. ~V:. Resistance, R = -

LV

=> R = 0.110x 10 3

=> R=10n

(ii) For V = -10V, we have1= -1 J.LA= -1 Xl0-6A

=> R=_1_0_=1.OXI07n1X10-6 (1)

9. Intrinsicsemiconductor

Extrinsicsemiconductor

1. It is a puresemiconductormaterial with noimpurity atomsin it.

It is prepared by dopinga small quantity ofimpurity atoms to thepure semiconductor.

2. The number offree electrons inthe conductionband and thenumber of holesin valence band isexactly equal.n, = nh = nj

The number of freeelectrons and holes isnever equal. There is anexcess of electrons inn-type n, > nj

semiconductors andexcess of holes in p-typenh > nj semiconductors.

10.

(2)

p-n Junction Diode as a Half-Wave Rectifier ACvoltage to be rectified is connected to the primarycoil of a step-down transformer. Secondary coil isconnected to the diode through resistors Ru acrosswhich output is obtained.

Transformer A x

B YWorking During positive half cycle of the inputAC, the p-n junction is forward biased. Thus, theresistance in p-n junction becomes low and currentflows. Hence, we get output in the load. Duringnegative half cycle of the input AC, the p-n junctionis reverse biased. Thus, the resistance of p-n junctionis high and current does not flow. Hence, no outputin the load. So, for complete cycle of AC, currentflows through the load resistance in the samedirection. (2)

(a)InputAC

IIIII

(b) tInput and output waveforms


11. The required energy band diagram is shownbelow: I Conduction band I

Acceptor energy level~

I."",,.':,., ',.:':.':'.,,',..:'110.04 eVV~I~D9.~ba.Dd

(a) p-type

I Conduction band 1

10.045 eV••_.'\ ...••........•0-- ••_.-..

Donor energy level

IHV~I~r,9~6~g81(b) n-type

12. (i) Metal For metals, the valence band iscompletely filled and the conduction band canhave two possibilities either it is partially filledwith an extremely small energy gap betweenthe valence and conduction bands or it isempty, with two bands overlapping each otheras shown below:

Partiallyfilled { Conduction bandconductionband

Filledvalence band{I.;X~f~~~~:~~~P~~~

Overlapping ~ .conduction band Conduction band

Filledvalence band~ .;!yaW6.1.f§~ri§;.On applying small even electric field, metalscan conduct electricity. (1)

(ii) Insulators For insulator, the energy gapbetween the conduction and valence bands arevery large, also the conduction band ispractically empty, as shown below:

When an electric field is applied across such asolid, the electrons find it difficult to acquire.So, a large amount of energy is required toreach the conduction band. Thus, theconduction band continues to be empty. Thatis why no current flows through insulators. (1)

437

(1)13. Working of photodiode A junction diode made

from light sensitive semiconductor is called aphotodiode. A photo diode is a p-n junction diodearranged in reverse biasing.

hv

R (1)

(1)

The number of charge carriers increases whenlight of suitable frequency is made to fall on thep-n junction, because new electron holes pairs arecreated by absorbing the photons of suitablefrequency. Intensity of light controls the numberof charge carriers. Due to this propertyphotodiodes are used to detect optical signals. (1)

According to the question

01 z n14.

A B

O2 2n0 C

+ - 1 nE F

6V

RAB = 2+1;= 3n1 1 1-=-+-R' 2 3

= 3+2=2n-1;

6 6R'=~n

SV 6

IEF =-=-= SAR 6/ S (2)

15. For LEDs, the threshold voltages are much higherand slightly different for different colours. Thereverse breakdown voltages of LEDs are lowgenerally around 5V. It is due to this reason, thecare is taken that high reverse voltages do notappear across LEDs. There is very little resistanceto limit the current in LED. Therefore, a resistormust be used in series with the LED to avoid anydamage to it. (1)

438 o Chapterwise eSSE Solved Papers PHYSICS

u Q5 17. When we apply sufficient voltage to LED, electron~ 3: c 2~ u .Q o Ql Ql

E = Ql ::::l :.c move across the junction into p-region and get~ Ql~ 000 ~a: <{

attracted to the holes there holes are sent from50 p region to n region (where they are minority

<{ carriers). Thus, electrons and holes recombine ..§.40 During each recombination, the electric potential-C energy is converted into the electromagnetic

~ 30 energy and a photon of light with a characteristic::::lo frequency is emitted, this is how, LED works.12<Il 20 Li9h~~0 pu,

10RL

n0 2 3 4 5 V

16.

The semiconductor used for fabrication of visibleLEDs must at least have a band gap of 1.8eV(spectral range of visible light is from aboutO.41lm to 0.71lm i.e. from about 3 eV to 1.8 eV). [1)

(i) In n-type semiconductor, the semiconductor isdoped with pentavalent impurity. In it theelectrons are majority carriers and holes areminority carriers or n, > > nil (n,= numberdensity of electrons, nh= number density ofholes). In energy band diagram of n-typesemiconductor, the donor energy level ED isslightly below the bottom of Ee conductionband and thus, the electron can move toconduction band, even with small supply ofenergy.

Ec· • • • .}__+ .ED 0.01 eV

Eg

jn-type [1)

(ii) In p-type semiconductor, the semiconductor isdoped with trivalent impurity. In thissemiconductor, the holes are the majority carriersand electrons are the minority carriers i.e.nil » n,.

--:"j---'--- Ec

Eg

0.01 eV {--i---------------· EAo 0 0 0 0 0 0 0 ~ Ev

p-type

In energy band diagram of p-lype, the acceptorenergy level is slightly above the top of valenceband Ev' Thus, even with small supply of energyelectron from valence band can jump to level, E A

and ionise the acceptor, negatively. [1)

[1)

18.

Advantages of LEDs over incandescent lamps(i) Since, LEDs do not have a filament that can

burn out, hence, they last longer.(ii) They do not get hot during use, hence fast

action, no warm up time required.Hand calculators, cash registers, digital clocks,ete. use seven-segment red or green displays. Eachsegment is an LED and depending on whichsegment is energised, the display lights up thenumbers 0 to 9, as shown in figure.

~

LED segment

I-I n I:I:IU. U "__1 I

,-, CC -100- _11_1IU_I

When photodiode is illuminated with light due tobreaking of covalent bonds, equal number ofadditional electrons and holes come intoexistence whereas fractional change in minoritycharge carrier is much higher than fractionalchange in majority charge carrier. Since, thefractional change of minority carrier current ismeasurable significantly in reverse bias than thatof forward bias. Therefore, photodiode areconnected in reverse bias. [2)

[1)

19. A light emit ting diode is simply a forward biasedp-n junction which emits spontaneous lightradiation. At the junction, energy is released inthe form of photons due to the recombination ofthe excess minority charge carrier with themajority charge carrier. (1)

Advantages:(i) Low operational voltage and less power.

(ii) Fast action and no warm up time required. [1)


20. (i) The energy for the maximum intensity of thesolar radiation is nearly 1.5 eV. In order tohave photo excitation, the energy of radiation(hv) must be greater than energy band gap(Eg), i.e., hv » Eg• Therefore, thesemiconductor with energy band gap about1.5 eV or lower and with higher absorptioncoefficient, is likely to give better solarconversion efficiency. (1)

The energy band gap for Si is about 1.1 eV,while for GaAs, it is about 1.43 eV. The gasGaAs is better inspite of its higher band gapthan Si because it absorbs relatively more energyfrom the incident solar radiations being ofrelatively higher absorption coefficient. (1)

It

(ii)

(Vod -V

Open circuit voltage (VAc)A

BIsc

Short circuit currentcharacteristic of a solar cell

21.

(i) V-I curve is drawn in the forth quadrant,because a solar cell does not drawscurrent but supply current to the load.

(ii) In V-I curve, the point A indicates themaximum voltage Voc being supplied bythe given solar cell when no current isbeing drawn from it. Voc is called theopen circuit voltage.

(iii) In V-I curve, the point B indicates themaximum current lsc which can beobtained by short circuiting the solar cellwithout any load resistance. lsc is calledthe short circuit current. (1)

Zener diode is used as voltage regulator. (112)

Principle Zener diode is operated in the reversebreakdown region. The voltage across it remainsconstant, equal to the breakdown voltage forlarge charge in reverse current. (1)

mA

Forwardbias

JlA

439

22. Differences between forward and reverse biasesare given below:

Forward bias Reverse bias

Positive terminal ofbattery is connected top-type and negativeterminal to n- typesemiconductor.

Positive terminal ofbattery is connectedto n-type and negativeterminal to p- typesemiconductor.

Depletion layer is verythin.

Depletion layer isthick.

p-n junction offers verylow resistance.

p-njunction offersvery high resistance.

An ideal diode havezero resistance.

An ideal diode haveinfinite resistance.

(1/2 x 4 = 2)23. With the formation of p-njunction, the holes

from p-region diffuse into the n-region andelectrons from n-region diffuse into p-region andelectron-hole pair combine and get annihilated.This in turn, produces potential barrier, VB acrossthe junction which opposes the further diffusionthrough the junction. Thus, small region forms inthe vicinity of the junction which is depleted offree charge carrier and has only immotile ions iscalled the depletion region. (2)

24. Circuit diagram of forward biased and reversebiased p-n junction diode is shown below: (1)

BThe width of depletion layer(i) decreases in forward bias.

(ii) increases in reverse bias. (1/2 • 1 =1)

25. The four valence electrons of carbon are presentin second orbit while that of silicon in third orbit.So, energy required to extricate an electron fromsilicon is much smaller than carbon.Therefore, the number of free electrons forconduction in silicon is significant on contrary tothe carbon. This makes silicons is conductivitymuch higher than carbon. This is the maindistinguishable property. (2)(1/2)

440

26. Device, D is a Zener diode. (1)

Symbol of Zener diode (1)

-t>f--27. Circuit diagram of illuminated photodiode in

reverse bias is shown below: (1)

Reverse bias

1(~)

v11----- .•...11112----- .•..

h-----74-----14> h > 12> 11 I (mA)

Reverse bias currents through a photodiode

Hence, frequency of light v such that hv » E 9'

where E 9 i/i band gap of increasing intensityI] ')2,13, ete. The value ofreverse saturation currentincreases with the increase of intensity of light.Thus, the measurement of charge in the reversesaturation current can give the intensity ofincident light.

28. (i) Main use of photo diode In demodulation ofoptical signal and detection of optical signal. (1)

In light operated switches, electronic counters.

(ii) Main use of Zener diode As DC voltageregulator. (1)

(i) D] diode is forward biased, hence current willflow in B] bulb and D2 is reverse biased, sothere will be no current in B2. Hence, B] willglow. (1)

(ii) The diagram of illuminated p-n junction solarcell is given below

29.

p n

(iii) Processes due to generation of emf takes placein a solar cell are given below(a) When light photon reach the junction, the

excited electrons from the valence bandto conduction band creating equalnumber of holes and electrons.


(b) These electron hole pair move in oppositedirection due to junction field. Theirmovement in opposite direction createspotential difference (photo-voltage).

(c) When load is connected in the externalcircuit, current starts flowing through itdue to photo-voltage. (1)

30. (i) The given diagram shown below.

~v

The circuit above can be redrawn as follows.---111------,

R

(1)

As the p-section is connected to negativeterminal of the battery, the diode shown isreverse biased. (1)

(ii) During the first half of input cycle, the upperend of the coil is at positive potential andlower end at negative potential. The functiondiode D] is forward biased and D2 in reversebiased. Current flows in output load in thedirection shown in figure. During the secondhalf of input cycle, D2 is forward biased. Inthis way, current flows in the load in thesingle direction as shown in figure. (2)

01

(1)

31. Two processes that takes place during theformation of p-njunction are diffusion and driftof charge carriers.In an n-type semiconductor, the concentration ofelectrons is more than that of holes. Similarly, in ap-type semiconductor, the concentration of holes ismore than that of electrons. Formation ofdepletion region during formation of p-n junctionand due to the concentration gradient across p andn-sides, holes diffuse from p-side to n-side


(p -t n) and electrons diffuse from n-side top-side (n -t p). The diffused charge carrierscombine with their counterparts in theimmediate vicinity of the junction and neutraliseeach other. (1)

Thus, near the junction, positive charge is builton n-side and negative charge on p-side.

- ElectrondiffusionElectrondrift -

D·ii::DeeE9E9eeE9E9

: : - Depletion regionHole diffusion-

-Hole driftp-n junction formation process

32.

This sets up potential difference across thejunction and an internal electric field E, directedfrom n-side to p-side. The equilibrium isestablished when the field E, becomes strongenough to stop further diffusion of the majoritycharge carriers (however, it helps the minoritycharge carriers to drift across the junction). (1)

The region on either side of the junction whichbecomes depleted (free) from the mobile chargecarriers is called depletion region or depletionlayer. The potential difference developed acrossthe depletion region is called the potentialbarrier. (1)

In Zener diode, both p and n-side of the functionare heavily doped. Heavy doping ensures highjunction field and low breakdown voltage. (1)

The circuit diagram of a voltage regulator usinga Zener diode is shown in figure. Theunregulated DC voltage is connected to theZener diode through a series resistance R, inreverse biased. Thus, any charge in the inputvoltage result is charge voltage drop across R,without any change in voltage across the Zenerdiode. Therefore, Zener diode acts as a voltageregulator. (1)

r---~~VV~~-r-----'---O+LoadOutput

(constantvoltage)

(Vz)

Unregulated-=- voltage

Excess current bypass when V.Xl ~ V,

441

33. (i)

(1)

Hole

p-type ~ n-typeDepletion layer

Electron drift~Electron diffusionplij - - : ::f-Depletion region

Hole diffusion'~ :.--Hole drift

~

v : .: :: : ve

X1: : :X2: : :

The small region in the vicinity of the junctionwhich is depleted of free charge carriers andhas only immobile ions is called depletionregion. The accumulation of negative charges inthe p-region and positive charge in the n-regionsets up a potential difference across thejunction. This acts as a barrier and is calledbarrier potential VB'

V'~

'-----\,...-~----.l.- 0 T/2 T t(1)

0110'Step down Transformer

(1)

Working(a) During positive half cycle of input

alternating voltage, the diode is forwardbiased and a current flows through theload resistor RL and we get an outputvoltage.

(b) During other negative half cycle of theinput alternating voltage, the diode isreverse biased and it does not conduct(under break down region). (1)

Hence, AC voltage can be rectified in thepulsating and unidirectional voltage.

442 o Chapterwise CSSE Solved Papers PHYSICS

34. (i) Principle A solar cell works on the principle ofphoto voltaic effect according to which whenlight photons of energy greater than energyband gap of a semiconductor are incident onp-n junction of that semiconductor,electron-hole pairs are generated which giverise to an emf. Thus, working principle of asolar cell is same as that of a photodiode.However, no bias is applied in a solar cell andthe junction area is kept much larger so thatmore solar radiation may be incident.cc;;J

Depletion layer

Generation of emf: Three basic processes areinvolved in the generation of emf by a solar cellwhen solar radiations are incident on it. Theseare

,(a) t¥ generation of electron-hole pairs closeto the junction due to incidence of lightwith photo energy hv ~ E b' (1'12)

(b) the separation of electrons and holes dueto the electric field of the depletion region.So, electrons are swept to n-side and holesto p-side.

(c) the electrons reaching the n-side arecollected by the front contact and holesreaching p -side fIe ~llected by the backcontact. Thl!sL.p"''.,,~!<!ebtcomes positive andn-side become negative giving rise to aphotovoltage. When a4 external load R L isconnected as shown in figure, aphotocurrent IL begins.cnflcw through theload. ' "'.' y/ I

(ii) Refer to Ans. 45 (i). (1'/2)

35. The photodiode is a special purpose silicon diode.It is fabricated with a transparent window toexpose its junction to light radiations. It alwaysworks on the reverse bias condition below thebreakdown voltage. (1)

hv>Eg

'--v---'p-side- +

'--v---'n-side

R

When visible light of energy (hv > E 9) enters itsdepletion region, the electron-hole pairs aregenerated. These charge carriers are separated bythe junction's electric field and are made to flowacross the junction and causes reverse saturationcurrent. The value of the reverse saturationcurrent depends on the intensity of incidentradiation and is independent of reverse bias. (1)

The photo diode is operated in reverse biascondition because in the reverse bias condition,the change in the reverse saturation current isdirectly proportional to the change in incidentlight intensity. Thus, photodiode can be used todetect the optical signals. It cannot be done, ifthe photodiode is forward biased. (1)

36. (i) • •• •• •~Q)cQ)

cet5Q)

m

Ec • •~:oo;·~vnn.n.n..m.~[EO-----<>--<>------<>--<>---

• •• •(a)

Energy band diagram ofn-type semiconductor

at T>O K(1%)

>-~Q)cQ)

cet5Q)

m• •

• •

(b)Energy band diagram ofp-type semiconductor

at T>O K

In n-type extrinsic semiconductors, thenumber of free electrons in conduction bandis much more than the number of holes invalence band. The donor energy level lies justbelow the conduction band. In p-typeextrinsic semiconductor, the number of theholes in valence band is much more than thenumber of free electrons in conduction band.The acceptor energy level lies just above thevalence band.

CHAPTER 14 : Semiconductor Electronics 443

(ii):>.!?

EelQlcQlC

Eg

e Ev~ • • • •tn • • • •• •

Energy band diagram ofsemiconductor at T =0 K

At absolute zero temperature (0 K) conductionband of semiconductor is completely empty,i.e., 0=0.

Hence, the semiconductor behaves as aninsulator. At room temperature, some valenceelectrons acquire enough thermal energy andjump to the conduction band where they arefree to conduct electricity. Thus, thesemiconductor acquires a small conductivity atroom temperature. (1'10)

37. The required energy band diagrams are given inAns.36. (2)

The donor energy level ED is just below thebottom of the conduction band. At roomtemperature this small energy gap is easilyconverted by the thermally excited electrons. Theconduction band has more electrons as they havebeen contributed both by thermal excitation anddonor impurities. Whereas the acceptor energylevel E A lies slightly above the top of the valenceband. At room temperature, many electrons ofthe valence band get excited to these acceptorenergy levels, leaving behind equal number ofholes in the valence band. These holes canconduct current. Thus, the valence band has moreholes than the electrons in the conduction bandrn

38. Refer to text. (Pg-425) (3)

39. (a) A photodiode is fabricated by allowing light tofall on a diode through a transparent window.It is fabricated such that the generation of e-hpairs take place near the depletion region. (1)

(b) Refer to Ans. 13.V-I characteristics:

currentt mA

Reverse bias

Voltage'1-----~1'2-----'

40. In these type of questions, we have to mind that infull-wave rectifier, full cycle of the input will be used.

The circuit diagram of full-wave rectifier is shownbelow:

+

+The input and output waveforms have been givenbelow:

~~~~Or----+----t----t----+--£~II

I I I I

Due to:Due to: Due to'Due to:01 : 02 : 01 : O2 :

I I I I,I I I II I I I~Ql

~ Ol Or---'--+.;....;.--.f----f-'---+--"S ~ : Timeo~ I

I (2)

41.

, Its working.based on the principle that junctiondiode offer very low resistance in forward biasand very high resistance in reverse bias.

V-I characteristic of p-n junction diode(1)

100 80 60

100

80

60

4020

(2)(1)

444

42.43.

(i) Under the reverse bias condition, the holes ofp-side are attracted towards the negativeterminal of the battery and the electrons ofthe n-side are attracted towards the positiveterminal of the battery. This increases thedepletion layer and the potential barrier.However, the minority charge carriers aredrifted across the junction producing a smallcurrent.At any temperature, the number of minoritycarriers \5 constant, so there is the smallcurrent at any applied potential. This is thereason for the current under reverse bias tobe almost independent of applied potential.At the critical voltage, avalanche breakdowntakes pl~ce which results in a sudden flow oflarge current. (1]

(ii) At the critical voltage, the holes in the n-sideand conduction electrons in the p-side areaccelerated due to the reverse bias voltage.These minority carriers acquire sufficientkinetic energy from the electric field andcollide with a valence electron.Thus, the bond is finally broken and thevalence electrons move into the conductionband resulting in enormous flow of electronsand thus, formation of hole-electron pairs.Thus, there is a sudden increase in thecurrent at the critical voltage.Zener diode is a semiconductor device whichoperates under the reverse bias in thebreakdown region. . (1]

Refer to Ans. 40. (3]

... 11-- ./' ve "

I \p n_ 0 --<> Ieee (±)(±)(±),__ 0

0-.0 0 Ieee (±)(±)(±)I_ 0 _

rI o 0 o-Ieee (±)(±)(±)I--<> • • l-• --<> 0 Ieee (±)(±)(±)I__ 0

_ 0 --<> Ieee (±)(±)(±)I_0 •I Depletion I

layer

- IIII +B

During the formation of p-n junction, diffusionof charge takes place. As, soon as p-typesemiconductor is joined with n-typesemiconductor, diffusion of free charges acrossthe junction starts. (2]

o Chnpterwise eBSE Solved Papers PHYSICS

For explanation of formation of p-n junctionRefer to Ans. 23.Potential barrier The accumulation of =ve'charges in the p-region and +ve charges in then-region sets up a potential difference across thejunction (p-n) is called potential barrier (VB) whichopposes the further diffusion of electrons andholes. (1]

44. In these type of questions, we should mind that thediode is connected reverse biased or forward biased.

(i) Photodiode is connected in reverse bias andfeeble reverse current flows due to thermally.generated electron-hole pair, known as darkcurrent. When light of suitable frequency (v)such that hv » E 9' where E 9 is band gap is .incident on diode, additional electron-hole pairgenerated and current grows in the circuit. ( 1)

(ii) Refer to Ans. 28 (i) (2]

(iii) Refer to Ans. 27. (1]

45. (i) The energy for the maximum intensity of the solarradiation is nearly1.5 eV. In order to havephotoexcitation, the energy of radiation (hv) mustbe greater than energy band gap (E 9)' Therefore,the semiconductor with energy band gap about1.5 eV or lower than it and with higher absorptioncoefficient is likely to give better solar conversionefficiency. The energy band gap for Si is about1.1 eV, while for GaAs, it is about 1.53 eV. TheGaAs is better inspite of its higher band gap thanSi because it absorbs relatively more energy fromthe incident solar radiations being of relativelyhigher absorption coefficient. (1]

(ii) When light of frequency, v such that hv » E 9

(band gap) is incident on junction, thenelectron-hole pair liberated in the depletionregion drifts under the influence of potentialbarrier. The gathering of these charge carriersmake p-type as positive electrode and n-type asnegative electrode and hence, generatingphoto-voltage across solar cell.

RL

It

~v

(2]


46. (i) Working of LED LED is a forward biasedp-njunction which converts electrical energy intooptical energy of infrared and visible light region.Being in forward bias, thin depletion layer and lowpotential barrier facilitate diffusion of electron andhole through the junction when high energyelectron of conduction band combines with thelow energy holes in valence band, then energy isreleased in the form of photon, may be seen in theform of light. (1)

(ii) Semiconductors with appropriate band gap (E 9)close to 1.5 eV are preferred to make LED sizeGaAs, CdTe, ete.The other reasons to select these materials arehigh optical absorption, availability of rawmaterial and low cost.

(iii) Uses of LEOs(a) LED can operate at very low voltage and

consumes less power in comparison toincandescent lamps.

(b) Unlike the lamps, they take very lessoperational time and have long life.

47. (a) X = Half-wave rectifierY = Full-wave rectifier

Transformer o

y

x

RL OutputPrimary

B(Half-wave rectifier)

Centre-TapTransformerr----------, A Diode (0,), r-~'----~~.~----,

CentreTap

B, L--+,~~--~>r----~,-----------, Diode (02)

II x

RL Output

y

(Full-wave rectifier)

(b) Refer to Ans. 40. (1)

(c) A capacitor oflarge capacitance is connected inparallel to the load resistor RL . When thepulsating voltage supplied by the rectifier isrising, the capacitor C gets charged. If there isno external load, the capacitor would have

445

remained charged to the peak voltage of therectified output. (1)

However, when there is no load and therectified voltage starts falling, the capacitorgets discharged through the load and thevoltage across capacitor begins to fall slowly.

(1)

Filtered outputI

-I

48. Refer to Ans. 23 and 24. (3)

49. (i) Refer to Ans. 24 for circuit diagram. (1)

(ii) (a) Refer to Ans. 22 for difference betweenforward and reverse bias.

(1) (b)Reverse bia~ (V)

-8 -6 -4

10

(1) 15Reverse bias characteristic curve (1)

V-I characteristic of a p-n junction in reversebias is shown above.For forward bias V-I characteristic curveRefer to Ans. 41 (ii). (1)

Explanation refer to Ans. 21. (2)50.

Regulatedvoltage (Vo)

!

Unregulatedvoltage (Vz)

Iz

(1)

446

51. Zener diode fabrication Zener diode is made byheavily doping of both p and n-typesemiconductors and hence, the width of depletionlayer becomes thin which lead to produce largeelectric field to increase the current even onapplying reverse voltage of 4 or 5 V. (2)

For circuit diagram Refer to Ans. 59. (1)

52. (i) p-n Junction A p-n junction is anarrangement made by a close contact of n-typesemiconductor and p-type semiconductor.There are various methods of forming p-njunction diode. In one method, an n-typegermanium crystal is cut into thin slices calledwafers. An aluminium film is laid on an n-typewafer which is then heated in an oven at atemperature of about 600°C. Aluminium thendiffuses into the surface of wafer. In this way,a p-type semiconductor is formed on n-typesemiconductor. (1)

Formation of Depletion Region in p-nJunction In an n-type semiconductor, theconcentration of electrons is more thanconcentration of holes. Similarly, in a p-typesemiconductor, the concentration of holes ismore than that of concentration of electrons.During formation of p-n junction and due tothe concentration gradient across p andn-sides, holes diffuse from p-side to n-side(p ~ n) and electrons diffuse from n-side top-side (n ~ pl.

- ElectrondiffusionElectrondrift -

Dp' iiiiDn'eeee .eeee

: : - Depletion regionHole diffusion-

-Holednft

The diffused charge carriers combine withtheir counterparts in the immediate vicinity ofthe junction and neutralise each other. (1)

Thus, near the junction, positive charge is builton n-side and negative charge on p-side. Thissets up potential difference across the junctionand an internal electric field E; directed fromn-side to p-side. The equilibrium is establishedwhen the field E; becomes strong enough tostop further diffusion of the majority chargecarriers (however, it helps the minority chargecarriers to diffuse across the junction). Theregion on either side of the junction which

o Chapterwise CSSE Solved Papers PHYSICS

becomes depleted (free) from the mobilecharge carriers is called depletion region ordepletion layer. The width of depletionregion is of the order oflO-6 mThe potential difference developed across thedepletion region is called the potential barrier.Potential barrier depends on dopantconcentration in the semiconductor andtemperature of the junction. (1)

(ii) (a) Forward Biased CharacteristicsThe circuit diagram for studying forwardbiased characteristics is shown in the figure.Starting from a low value, forward bias voltageis increased step by step (measured byvoltmeter) and forward current is noted (byammeter). A graph is plotted between voltageand current. The curve so obtained is theforward characteristic of the diode.

P n

8

~7

It:o 4'E~ 3~2

Battery(a)

Ge

oo 0.1 0.2

FO~M(b)

At the start when applied voltage is low, thecurrent through the diode is almost zero. It isbecause of the potential barrier, which opposesthe applied voltage. Till the applied voltageexceeds the potential barrier, the currentincreases very slowly with increase in appliedvoltage (OA portion of the graph).With further


increase in applied voltage, the current increasesvery rapidly (AB portion of the graph), in thissituation, the diode behaves like a conductor. Theforward voltage beyond which the currentthrough the junction starts increasing rapidly withvoltage is called knee voltage. If line AB isextended back, it cuts the voltage axis at potentialbarrier voltage. (1)

(b) Reverse Biased CharacteristicsThe circuit diagram for studying reverse biasedcharacteristics is shown in the figure.

P n

BatteryL-----1- IIII j.:...+ __ -l

(a)

Reverse bias (\I)-10 -8 -8 -4 -2 0

ac 2 -c

4.§.C

6 ~8

8 ~Q)

105ia:

Breakdownvoltage

o(b)

In reverse biased, the applied voltage supports theflow of minority charge carriers across thejunction. So, a very small current flows across thejunction due to minority charge carriers.Motion of minority charge carriers is alsosupported by internal potential barrier, so all theminority carriers cross over the junction.Therefore, the small reverse current remainsalmost constant over a sufficiently long range ofreverse bias, increasing very little with increasingvoltage (OC portion of the graph). This reverse

447

current is voltage independent up to certainvoltage known as breakdown voltage and thisvoltage independent current is called reversesaturation current. (1)

Use of p-n Junction Characteristics inRectificationFrom forward and reverse characteristics, it isclear that current flows through the junctiondiode only in forward bias not in reverse bias i.e.current flows only in one direction.

53. (i) Refer to Ans. 52 (ii). (3)

(ii) Refer to Ans. 27. (2)

54. (i) Refer to Ans. 43. (3)(ii) Refer to Ans. 40. (2)

55. (i) Refer to Ans. 52 (i). (1)(ii) Refer to Ans. 42. (1)

(iii) A full-wave bridge rectifier using four diodes(full-wave bridge rectifier) gives acontinuous, unidirectional but pulsatingoutput voltage or current. (1)

The rectified output is passed through a filtercircuit which removes the ripple and analmost steady DC voltage (or current) isobtained. (2)

Zener diode works only in reverse breakdownregion that is why it is considered as a specialpurpose semiconductor. (1)

I - V characteristics of Zener diode is given below:(1)

56.

I (mA)

Forward bias

-VM

I(IlA)

Reverse current is due to the flow of electronsfrom n -+ p and holes from p -+ 11. As, the reversebiased voltage increase the electric field acrossthe junction, increases significantly and whenreverse bias voltage V = Vz' then the electric fieldstrength is high enough to pull the electronsfrom p-side and accelerated it to n-side. (1)

448

These electrons are responsible for the highcurrent at the breakdown.

Unregulated I-_ •.h_-. ---evoltage

Regulatedvoltage, Vz

Voltage regulator converts an unregulated DCoutput of rectifier into a constant regulated DCvoltage, using Zener diode. The unregulatedvoltage is connected to the Zener diode through aseries resistance Rs such that the Zener diode isreverse biased. If the input voltage increases, thencurrent through Rs and Zener diode increases.Thus, the voltage drop across Rs increases withoutany change in the voltage drop across Zenerdiode. This is because of the breakdown region,Zener voltage remain constant even though thecurrent through Zener diode changes. Similarly, ifthe input voltage decreases, the current throughRs and Zener diode decreases. The voltage dropacross Rs' decreases without any change in thevoltage across the Zener diode. (1)

Now, any change in input voltage results thechange in voltage drop across Rs, without anychange in voltage across the Zener diode.Thus,Zener diode acts as a voltage regulator. (1)

57. When we are dealing with depletion layerformation we have to keep in mind the majoritycharge carriers, diffusion will always happens fromhigh concentration to low concentration.

(i)Fictitious battery

Junction VaAcceptor ion

Hole e e e e:e:0:00 0 0 o I I I •I I Ie e e e:e:0:00 0 0 0 I I I •

I I Ie e e e:e:0:00 0 0 0 I I I •I I Ie e e e:e:0:00 00 0 0 o I I I • • •

p-type 1.----1 n-type


The two process involved in the formation ofp-n junction.(a) Diffusion (b) Drift. (112 x 2 = 1)

Holes and electrons diffuse from p to n and n top respectively.The majority charge carrier drifts under theinfluence of applied electric field such that(a) holes along applied Eand(b) electron opposite to E (1/2 x 2 = 1)

(ii) Zener diode is used as voltage regulator forexplanation Refer Ans. 21 and Ans. 50for circuit. (3)

58. (i) Refer to Ans. 52. (2)

(ii) Refer to Ans. 46 (i) and (ill). (3)59. (i) Refer characteristics curve of Ans. 21. (1)

(ii) From the figure, it is clear that the device, X isa full-wave rectifier. Circuit diagram as shownin figure below: (1)

R1+

Unregulatedvoltage

1

Iz

f(1)

RL Regulatedvoltage (Vo)

1Zener diode connected with unregulated DCvoltage in reverse bias. When the input voltageincreases, then current through R\ increase andhence, voltage drop across R\, increases whilevoltage across the Zener diode remainsconstant. The voltage across Zener dioderemains constant beyond Zener voltage andhence, same/constant regulated voltage isobtained across RL. (1)

(ill) In n-type semiconductor,n, > nh ... (i)

On incidence of light of suitable frequency,there is equal rise in number of electrons andholes [i.e. ill! (say) 1

1 1 tin ill!~ -<-or-<-», nh n, nh

where, ill! = change in electron or hole chargecarrier. Thus, fractional change in minoritycharge carrier (hole) is much higher thanfraction change in majority charge carrier(electron). Also, minority charge carriercontribute in drift current in reverse bias.Thus, with incidence of light, fractional changein minority charge carrier is significant.Therefore, photodiode should be connected inreverse bias for measuring light intensity. (1)


60. (i) Refer to Am. 52.(ii) Refer to Am. 46.

61. (i) Refer to Am. 56.(ii) Circuit diagram of p-n junction diode as

half-wave rectifier is shown below:p-n

+ A +T:50-:5o1

R

B

Diode conducts corresponding to positive halfcycle and does not conduct during negative halfcycle, hence AC is converted by diode intounidirectional pulsating DC.

449.

(3)

(2)

(3)

This action is known as half-wave rectification.

~~l°r----+----r----+----+--E -g Time

~~IO~---+----+----+----T--c3 -g Time

(1)

(1) 62 (i) Refer to Am. 52. (3)From these two graphs we see that thejunction diodes operates mainly in forwardbias, this characteristic of junction diode canbe used to make it a rectifier.

(ii) Refer to Ans. 40. (2)

[TOPIC 2] Transistors and Its Applicationsand Logic Gates

2.1 Junction TransistorA junction transistor is three terminalsemiconductor device consisting of two p-njunctions formed by placing a thin layer ofdoped semiconductor (p-type or n-type) betweentwo thick similar layers of opposite type.There are two types of transistor:

p-n-p transistorHere, two thicker segments of p-type (termed asemitter and collector) are separated by asegment of n-type semiconductor (base).

Emitter Base Collector

E C

n-p-n transistorHere, two thicker segments of n-typesemiconductor (emitter and collector) are separatedby a segment of p-type semiconductor (base).

Emitter Base Collector

E Cn P n

Bn-p-n transistor

Transistors SchematicRepresentation

n- p- n transistor p -n- p transistorE C E P P C

B B

450

Transistor Action orWorking of Transistorp-n-p TransistorFrom given figure, we can see that, theemitter-base junction is forward biased.Collector-base junction is reverse biased.

n-Basep-Emitter p-Colleetor. ..----'---.. .p:::: ~ :::: p

Ie 0-- I 0--+

E :::: J.. :::: c0--: 0--0---+ I 0---+

lcVES

+ - + 11-VEE Ie Ie Vcc

Flow of charge carriers in p n p tronststor

Bie

L-I f--~+ II -Vca

Action of p-n-p transistorand its biasing

The resistance of emitter-base junction is verylow. So, the voltage of VEE(VEB) is quite small(i.e., 1.5 'V).The current in p-n-p transistor is carried byholes and at the same time their concentrationis mMtroned.,.But in external circuit, the currentis due tothe flow of electrons.In this case,

I, = Ib + l , [using Kirchhoff's law]where, Ie = emitter current

Ib = base currentand I, = collector current

o ehopterwise eBSE Solved Papers PHYSICS

n-p-n TransistorIn the base, Ie and I c flow in opposite directions. Inthis transistor, the emitter-base junction is forwardbiased and its resistance is very low.So, the voltage of VEE is quite small.

p-Basen-Emitter n-Collector. ....---'--0. '--:::: n C---

E n

-VCB-

Ia- + -IIII~+---'VEE h Ie vcc

Flow of charge carriers in n-p-n transistorn-p-n

E C

t-----<~- II +

VEa VcsAction of n-p-n transistor and its biasing

The collector base junction is reverse biased. Theresistance of this junction is very high. So, the voltageof Vcc (VCB) is quite large (45 V). In n-p-n transistor,the current is carried inside as well as in externalcircuit by the electrons. Thus, in this [I b « Ie] casealso, IE = I B + I c [Kirchhoff's first law]In the base, IE and Ie flow in opposite direction.

Transistors Configuration(i) Common Base (CB) mode

(ii) Common Emitter (CE) mode(iii) Common Collector (CC) mode

2.2 Characteristics of aTransistor

The graphical representation of the variations amongthe various current and voltage variables of atransistor are called transistor characteristics.


Common Base TransistorCharacteristicsHere emitter-base circuit is forward biased withbattery VEEand collector-base circuit is reversebiased with battery Vcc.

I p-n-pe CI'

The common base characteristics of a transistor areof two types:

(i) Emitter or Input Characteristics A graphicalrelation between the emitter voltage and emittercurrent at constant collector voltage, is calledemitter or input characteristics. The graph isplotted between emitter current andcorresponding emitter voltage.

Ie (mA) VEB= - 4 V, VCB= -2 V. . . . .10 --t--i-- - ~---~--

""""-IQ.)t I I I a' I__ 8 - - + - - of - - :- -:- - -:- --

~ 6 : : P: : ::J --i--i- ai--i--o "JC\",Ci3 4 --1"-- - -1--- t- --

t: : v,C::'E 2 -- •.- - - -1- - -1- - -'- --W I I I I I

, '0.5' 0.7' ,'----L--,.L..:...-':-'-:-'---::-'-:,.......,...".-..... VEBM0.2 0.4 0.6 0.8 1.0

Emitter-base voltage (VEB)-

(ii) Collector or Output Characteristics Agraphical relation between the collector voltageand collector current at constant emittercurrent, is called collector or outputcharacteristics. The graph is plotted betweencollector current and corresponding collectorvoltage.

Ie (mA)

, , , , ,, , , , ,Ie, , , , ,, , , ,

'Ie, , , , ,, , , ,'I, , , , , e, , , ,'I, , , , , e, , , ' __ 'IA =-- - --

(/8) Base current

~20 =20mAC~15 =15mA:Jo010 = 10 mA

~ 5 =5mA8 OmA

5V10V15V20V25V VcBMCollector voltage (VCB)-

45t

Common Emitter TransistorCharacteristicsHere, base-emitter circuit is forward biased withbattery VBE and emitter- collector circuit is '"'reverse biased with battery Vcc.

These two characteristics can be studiedas shown below:

(i) Emitter or Input Characteristics Agraphical relation between the emittervoltage and the emitter current by keepingcollector voltage constant is called inputcharacteristics of the transistor.

Ib(llA)

100

80

60

40

20

VCE=10V

B

0.2 0.4 0.6 0.8 1.0 VSEM ~

Input resistance It is defined as theratio of change in base-emitter voltage 1

(AVBE)to the resulting change in the base, 21

current (AIb) at constant collector-emittervoltage (VCE)'It is reciprocal of slope of

I b - VBE cu:e. ~n(p:~B:e)sistance,

I AI b VCE = constant

(ii) Collector or Output Characteristics Agraphical relation between the collectorvoltage and collector current by keepingbase current constant is called output.characteristics of the transistor.

452

lb60 J.IA

r: 50 J.IA

"40 J.IA30 J.IA20 J.IA10 J.IA

~ 10.s~ 8C~ 6:;o.9 4~"0o 2

o 2 4 6 8 10 12 14 16Collector to emitter voltage (VcE!in volts

NOTE From the output characteristics, we defineoutput resistance of transistor as the ratio ofchange in collector-emitter voltage to theresulting change in collector current at constantbase current. Thus, output resistance,

ra =(~VCE)~/c Ib = constant

= Reciprocal of slope of Ie - VCEcurve.The current amplification factor (~) of atransistor in CE configuration is defined as theratio of change in collector current to the changein base current at a constant collector-emittervoltage when the transistor is in active state.

.. ~AC=(~;:tE=constantIts value is very large (~AC> > 1)

Region of Operation of Junctions

Region Collectorjunction Emitter junction

Cut-off Reverse biased Reverse biased

Active Reverse biased Forward biased

Saturation Forward biased Forward biased

2.3 Transistor as an Amplifier(CE configuration)

An amplifier is a device which is used forincreasing the amplitude of input signal. Thecircuit diagram for p-n-p transistor as anamplifier is shown in the figure given below:


/~P 1 RL

C\ CE -.!c cPV L Ie 1 :vccInput OutputAC + AC

signal Ib Ie signalVSE

Circuit diagram of transistor as on amplifier

When no AC voltage is applied to the input circuit,we have

Ie=Ib+I, ... (i)Due to collector current I" the voltage drop acrossload resistance (RL) is I,RL. Therefore, thecollector-emitter voltage VCE is given by

VCE = Vcc - IeRL ... (ii)

Gains in Common-EmitterAmplifiera and ~ parameters of a transistor are defined asa =!.£ and ~ = !.£.a is about 0.95 to 0.99 and ~ is

Ie Ibabout 20 to 100. The various gains in acommon-emitter amplifier are as follows:

(i) DC Current Gain It is defined as the ratio ofthe collector-current to the base current andis denoted by ~DC- Thus,

~DC =!.£ = _1_, _ = I,/Iei, Ie-I, I-I,/Ie

[':Ie=Ib+I,l

[.:a = I,IIel~=~I-a

(ii) AC Current Gain It is defined as the ratioof the change in the collector-current tothe change in the base-current at a constantcollector to emitter voltage and is denoted

by ~AC' ()Thus, ~AC = ~'

b VCE

The value of~ is from 15 to 100, for atransistor.

(iii) AC Voltage Gain It is defined as the ratio ofthe change in the output voltage to the changein the input voltage and is denoted by Av.


Suppose, on applying an AC input voltagesignal, the input base-current changes byM b and correspondingly the outputcollector-current changes by Me' If Rin andRout be the resistances of the input and theoutput circuits respectively, then

A_Me X Rout Me Routv- x--si, xRin si, Rin

Now, Me/ si, is the AC current gain ~AC'

(iv) AC Power Gain It is defined as the ratioof the change in the output power to thechange in the input power.Since, power = current x voltage, we have

AC . Change in output powerpower gam = --"'-----"----'--Change in input power

= AC current gain x AC voltage gain= ~AC xAv

. Relationship between a and ~

~=~I-a

and a=-~-I+~

Voltage gain = Current gain x Resistance gain

FeedbackWhen a portion of the output power is returnedback to the input in phase, this is termed aspositive feedback.

Input fTrairiSiErtOr'l--------,.r-=-i TransistorII II amplifier 1f-----------r~Output

'----=------11 FeedbackUI network ~

2.4 Analog and Digital CircuitsAnalog CircuitsCircuits use signals (current or voltage) in theform of continuous, time-varying voltage orcurrent.

453

A sinusoidally varying alternating voltage as shownin figure is the simplest analog signal. Theelectronic circuits which process analog signals arecalled analog circuits.

Vb /\ /"\J\J

Digital CircuitsCircuits use two discrete level of current or voltagewhich are termed as binary signals. These twovalues are represented as 0 and 1 (low or high i.e.ON or OFF)The electric circuits which process digital signalsare called digital circuits.

V

Level1

2.5 Logic GateA gate is a digital circuit that follows certainrelationship between input and output voltage.Logic gates are building blocks of electroniccircuits. In logic gates, there exist a logicalrelationship between output and input(s).It has one output but one or more inputs .· Truth Table It is a table that shows all possible

input combinations and the correspondingoutput combinations for a logic gate. .~· Boolean Operators Just as in ordinary algebra,

mathematical operators like addition, subtractionand multiplication are used, similarly, in Booleanalgebra three basic operators like OR, AND andNOT are used.· Boolean Expression The expression shows the

combination of two Boolean variables thatresults into a new Boolean variable is known asBoolean expression.

454

Basic Logic GatesThere are three basic logic gates:

OR GateBoolean expression of OR gate is given as

Y=A+B.(a) It has two or more inputs and one output.(b) In this gate, if anyone of the input or all the

inputs are 1, then output is 1.Symbol

Truth TableA yB

o o oo

o

..~~-------- .........•...•..A \ ,,\ ,\ 01

,,, , y, C ,,r: r/, R /

~

, /, -= .../I

-;'..J:...5V B ~I --_L?2..- ~'

,

o~'-- --,-

Realisatlon of ORgate

AND GateBoolean expression of AND gate is given as

Y=A·B(a) It has two or more inputs and one output.(b) It has output 1, only when all inputs are 1.

BA~ySymbol ~

Truth TableA B y

o o oo o

o 0

o Chopterwise CBSE Solved Papers : PHYSICS

A~----------------- .•..,, \

\

0, \ yr

L__~~ ~~~~~_// -L~r5V.:r...

-1 Bo~

Realisatian of AND gate

NOT GateBoolean expression of NOT gate is given as Y = it .

(a) It has one input and one output.(b) It gives an inverted version of its input i.e. if

input is 1, then output is 0 and vice-versa.Symbol ~ .

Truth TableA y

oo

~

Reallsatlan of NOTgate

Combination of GatesVarious combinations of three basic gates can beused to produce complicated digital circuits,which are also called gates. Differentcombinations of basic gates are given below:

NOR GateBoolean expression of NOR gate is given asY=A+B.


Here NOT operation is applied after OR gate.If all its inputs are 0, then its output will be l.

A~-Symbol B~Y=A+B

Truth Table----------------------Input Output

A B Y=A+B0 0 1

0 0

0 0

1 0

NAND Gate (NOT AND)Boolean expression of NAND gate is given asY = A-B.

Here, AND gate followed by a NOT gate.If all the inputs are 1, then output will be O.

Symbol ~~Y

Truth Table----------------~=---A B Y=A·B

o oo

oo

NAND and NOR gates are called universal gates.

XDR GateBoolean expression of XOR gate is given asY=A-B+A·B=AE9B

Symbol

Truth Table

A B Y=AEaB

o ooo

oo

455

XNDR GateBoolean expression of XNOR gate is given asY = A· B = AB + AB.

Here, XOR gate is followed by a NOT gate.

Symbol

Truth Table

A B Y=A·B

o 0o 0

o o

Some Useful Laws of Boolean Algebra(i) de-Morgan's Theorem It states that the

complement of the whole sum is equal tothe product of individual complements andvice-versa.

(a) A+B=A·B

(b) A·B=A+Bde-Morgan's theorem also states that

(a) A+B = A·B = A·B

(b) A·B = A·B = A+B(ii) Commutative laws

(a) A+B=B+A(b) A·B =B·A

(iii) Associative laws(a) A+(B+C) = (A+B)+C(b) A·(B·C) = (A·B)·C

(iv) Distributive laws(a) A·(B+C) = A·B+A·C(b) (A+B)·(A+C) = A+B·C

(v) Absorption laws(a) A+A·B = A(b) A·(A+A) = A(c) A.(A+B) = A·B

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 2o 1 Mark Questions1. In a transistor, doping level in base is

increased slightly. How will it affect(i) collector current and

(ii) base current? Delhi 2011

2. Draw the logic circuit of a NAND gateand write its truth table.Foreign 2011

3. Draw the logic circuit of AND gate andwrite its truth table. Foreign 2011

4. Draw the logic circuit of NOT gate andwrite its truth table. Foreign 2011

5. Write the truth table for the followingcircuit. Name the equivalent gate thatthis circuit represents.

;~yForeign 2010

6. A given logic gate inverts the inputapplied to it. Name this gate and giveits symbol. Delhi 2010C

7. The truth table of a logic gate has theform given here. Name this gate anddraw its symbol.

A B y

o oo o

o oo

All Indio 2010C

8. The truth table of a logic gate has the formgiven here. Name this gate and draw itssymbol.-------------------------

A B Y

0 0 10 1 11 0 11 1 0

All India 2010C

9. Give the logic symbol of NOR gate. All India 2009

10. Give the logic symbol of NAND gate.All India 2009

11. Give the logic symbol of AND gate. All India 2009

12. Define current amplification factor incommon-emitter mode of transistor.All India 2010C, Delhi 2009C

o 2 Marks Questions13. Draw a circuit diagram of n-p-ri transistor

amplifier in CE configuration. Under whatcondition does the transistor act as anamplifier? All India 2014

14. Write the truthtable for the Acombination of thegates shown.Name the gatesused. All India 2014

15. Identify the logic A~gates marked P and B Q XQ in the given circuit.Write the truth tablefor the combination. Delhi 2014

16. The outputs of two NOT gates are fed to aNOR gate. Draw the logic circuit of thecombination of gates. Give its truth table.Identify the gate represented by thiscombination. Delhi 2014C

17. The input waveforms A and B and the outputwaveform Yof a gate are shown below.


Name the gate it represents, write itstruth table and draw the logic symbol ofthis gate.

A' Iri.L.IIII

B

IIIljr----;-----;

I II II II II I

uY I II II II II I

t = 0 2 4 5

All Indio 2014C

18. Identify the equivalent gate representedby the circuit shown in the figure. Drawits logic symbol and write the truthtable.

A

B

Foreign 2014

19. In the given circuit diagram, a voltmeterV is connected across a lamp L. Howwould

(i) the brightness of the lamp and(ii) voltmeter reading V be affected, if

the value of resistance R isdecreased? Justify your answer.

L

+~i 9VI-'-

R

457

20. Draw a typical output characteristics of ann-p-ri transistor in CE configuration. Showhow these characteristics can be used todetermine output resistance? All Indio 2013

21. In the circuit shown in the figure, identifythe equivalent gate of the circuit and makeits truth table.

A

B

Y

All Indio 2013


A

7

BAll Indio 2013

Y


A

BAll Indio 2013

24. Describe briefly with the help of a circuitdiagram, how the flow of current carriersin a p-n- p transistor is regulated withemitter-base junction in forward biasedand base-collector junction in reversebiased. All Indio 2012

25. Distinguish between analog signal anddigital signal. All Indio 2012

26. Draw the output waveform at X using thegiven inputs, A and B for the logic circuitshown below. Also, identify the logicoperation performed by this circuit.

A

Bx

458 o ehapterwise eSSE Solved Papers PHYSICS

1 11 1

U 11111

1 11 11 1 11 n111i1 <I 1

t1 t21

t3 t4 ts t6 . t71 1

! ! !Delhi 2012; 2011

27. In the given circuit, a voltmeter V isconnected across lamp L. What changeswould you observe in the lamp L and thevoltmeter V if the value of resistor R isreduced?

Delhi +2DllC

28. Draw the transfer characteristic curve of abase-biased transistor in CE configuration.Explain clearly how the active region of theVo versus Vi curve, in a transistor is usedas an amplifier? Delhi 2011

29. Draw the output waveform at X using thegiven inputs, A and B for the logic circuitshown below. Also, identify the logicoperation performed by this circuit.Delhi 2011; 2DDB .

A

Bx

1 1 1AU i 34.1111

1 1 11 1 11 1 11 1 I1 11 1

BI I1 1 1

! ! !t1 tz t3 t4 ts t6 t7

30. Write the truth table for the logic circuitshown below and identify the logicoperation performed by this circuit.Delhi 2011

A

y

B

31. Identify the logic gates X and Y in thefigure. Write down the truth table ofoutput Z for all possible inputs A and B.All India 2011C

A

Bz

32. (i) For the digital circuit givenbelow, write the truth table showingoutputs Y1 and Y2 for all possibleinputs of A and B.

A Y1

B

(ii) Show output waveform for allpossible inputs of A and B.All India 2011C

33. (i) Identify the logic gates marked P andQ in the given logic circuit.

(ii) Write down the output at X for theinputs,A = 0, B = 0 and A = I, B = 1 .All India 2010

~~x

(i) Identify the logic gates marked P andQ in the given logic circuit.

A~ Q XB


(ii) Write down the output at Xfor the inputs A = 0, B = 0and A = 1 , B = 1 .

All Indio 2010

35. Define the following terms.(i) Input resistance ri'

(ii) Current amplification factor 13 of atransistor used in its CEconfiguration. All Indio 2010C

36. The following figure shows the inputwaveforms A, B and the outputwaveform Y of a gate. Identify the gate,write it, .ruth table and draw its logicsymbol. Deihl 2009Ah

....,---II

8

I I I II I I II I I I IyU I I UI I

I II II I

I I I I I II I I I I I

0 1 2 3 4 5 6

37. The output of a 2-input AND gate isfed to a NOT gate. Give the name ofthe combination and its logic symbol.Write down its truth table.Deihl 2009, Foreign 2008

38. (i) Sketch the output waveform froman AND gate for the inputs, A andB shown in the figure.I I I II I I I

LJ LJA, I II I II I II I II I nI II II I

8' II I II I I! ! !0 1 3 4 5 6 7

(ii) If the output of the above ANDgate is fed to a NOT gate, namethe gate of the combination, soformed. Delhi 2009

459

39. Draw the logic symbol of the gate whosetruth table is given as below:

Inputs Output

A B y

0 0 10 1 01 0 01 1 0

If this, logic gate is connected to NOT gate,what will be the output when

(i) A = 0, B = 0 and (ii) A = 1 , B = 1?Draw the logic symbol of the combination.Foreign 2009

40. A logic gate is obtained by applying output ofOR gate to a NOT gate. Name the gate soformed. Write the symbol and truth table ofthis gate. Foreign 2009

41. A logic gate is obtained by applying output ofAND gate to a NOT gate. Name the gate soformed. Write the symbol and truth table ofthis gate. Foreign 2009

42. The two circuits shown here are acombination... r-___...

A

8y

A

8

(i) Three NAND gates.(ii) Three NOR gates.

Write truth tables for each of thesecombinations. Delhi 2009C

o 3 Marks Questions43. For a CE transistor amplifier, the audio

signal voltage across the collector resistanceof 2 kn is 2V. Given, the current amplificationfactor of the transistor is 100, find the inputsignal voltage and base current, if the baseresistance is 1 kn. Deihl 2017

460

44. (i) Write the functions of the threesegments of a transistor.

(ii) The figure shows the inputwaveforms A and B for 'AND' gate.Draw the output waveform and writethe truth table for this logic gate.

t, t2 t3 t4 t5 t6 t7 t8, I II I II I I

A

(Input)

III~nIII

All India 2017

45. (i) Write the functions of threesegments of a transistor.

(ii) Draw the circuit diagram forstudying the input and outputcharacteristics of n-p-n transistor incommon emitter configuration.Using the circuit, explain how input,output characteristics are obtained.Delhi 2016

46. Identify the gates P and Q shown in thefigure. Write the truth table for thecombination of gates shown in figurebelow:

A ----1;\.-~ YB~

Name the equivalent gate representingthis circuit and write its logic symbol.Deihl 2015

47. Draw a circuit diagram of a CE transistoramplifier. Briefly explain its working andwrite the expression for (i) current gain,(ii) voltage gain of the amplifier. Deihl 2015

48. Draw a circuit diagram of a transistoramplifier in CE configuration.Define the terms

(i) input resistance and(ii) current amplification factor. How are

these determined using typical inputand output characteristics?All India 2015


49. Output characteristics of an n-p-ritransistor in CE configuration is shown inthe figure. Determine

50mA

a 2 4 6 8 10 12 14 16 18VCE(V) -

(i) dynamic output resistance(ii) DC current gain and

(iii) AC current gain at an operating pointVCE = 10 V, when IB = 30 ~A.Delhi 2D13

50. Draw the transfer characteristic of abase-biased transistor in CE configuration.Mark the regions where the transistor canbe used as switch. Explain briefly itsworking.

51. You are given a circuit below. Write itstruth table. Hence, identify the logicoperation carried out by this circuit. Drawthe logic symbol of the gate whichcorresponds to All India 2011

XA

z

B

52. You are given a circuit below. Write itstruth table. Hence, identify the logicoperation carried out by this circuit. Drawthe logic symbol of the gate whichcorresponds to All India 2011

A

zB


53. Draw transfer characteristics of acommon-emitter n- p-n transistor. Pointout the region in which the transistoroperates as an amplifier.Define the following terms used intransistor amplifiers:

(i) Input resistance(ii) Output resistance

(iii) Current amplification factor.Foreign 2011

54. Draw the general shape of the transfercharacteristics of a transistor in itsCE configuration. Which regions of thischaracteristic of a transistor are usedwhen it works as an amplifier? All Indio 2010C

55. Give the circuit diagram of acommon-emitter amplifier using an n-p-ritransistor. Draw the input and outputwaveforms of the signal. Write theexpression for its voltage gain.HOTS. All Indio 2010

56. The inputs A and B shown here are usedas the inputs for three different gates G1,

G2 and G3• The outputs obtained in thethree cases have the forms shown.Identify the three gates and write theirtruth tables. All India 2009C

Lil!1JJ1j',I I I I• I I I I I

I I I I IA I I ----4--- I I

I I I I I I I I I

8' I I I I I I I I

CllllUlJ' ,I I I II I I IL___ I I I

I I I I I I I I II I I I I I I I IG I I I I I I I I I

1ITIlrTllJI I I I I I I II I I I I I I I, , ,, , ,, , ,, , ,

G I I I

2, I I I I II I I I I I I I

LLLf1JJlj',I I I II I I IG3 I I I I _I I I I I I I II I I I I I I I

01234567

57. The inputs A and B shown here are usedas the inputs for three different gates G1,

G2 and G3 one by one. The outputsobtained in the three cases have the formsshown. Identify the three gates and writetheir symbols.All India 2009C

461

tt1JJlljjA ~--+--I : '---I : : :

LlJTI1Itj!! I : : I I : :

I I I I IB I --i---{--- I I l. 1

I I I I I, , ,, , ,, , ,G1' '+--I'-+--I,,,G2: I---+-+--+--+---1

I I I I I

GJ---~--~--LJ--U---~---;I I I I I I I I I

012345678

58. Draw the labelled circuit diagram of acommon-emitter transistor amplifier.Explain clearly, how the input and outputsignals are in opposite phase? All India 2009

59. The inputs A and B are inverted by usingtwo NOT gates and their outputs are fedto the NOR gate as shown below:

A

y

Analyse the action of the gates (1) and (2)and identify the logic gate of the completecircuit so obtained. Give its symbol andthe truth table. All India 2009

o 5 Marks Questions60. (i) Differentiate between three segments

of a transistor on the basis of theirsize and level of doping.

(ii) How is a transistor biased to be inactive state?

(iii) With the help of necessary circuitdiagram, describe briefly, how n-p-ritransistor in CE configurationamplifies a small sinusoidal inputvoltage. Write the expression for theAC current gain. Delhi 2014

61. (i) Explain briefly with the help of acircuit diagram, how an n- p-ntransistor in CE configuration is usedto study input and outputcharacteristics.

462

(ii) Describe briefly the underlyingprinciple of a transistor amplifierworking as an oscillator. Hence, usethe necessary circuit diagram toexplain how self sustainedoscillations are achieved in theoscillator. Delhi 2D14C

62. (i) Draw the circuit diagram of ann-p-ri transistor with emitter-basejunction forward biased andcollector-base junction reversebiased. Describe briefly, how themotion of charge carriers in thetransistor constitutes the emittercurrent IE' the base current IB andthe collector current Ie. Hence,deduce the relation, IE = IB + Ie·

(ii) Explain with the help of a circuitdiagram, how a transistor works asan amplifier? All Indio 2D14C

63. (i) Why is the base region of atransistor thin and lightly doped?

(ii) Draw the circuit diagram forstudying the characteristics of ann-p-n. transistor in common-emitterconfiguration.Sketch the typical (a) input and(b) output characteristics in thisconfiguration.

(iii) Describe briefly, how the outputcharacteristics can be used toobtain the current gain in thetransistor? Deihl 2D13C

64. The set-up shown below can produce anAC output without any external inputsignal. Identify the components 'X' and'Y' of this set up. Draw the circuitdiagram for this set-up. Describe brieflyits working. Ail Indio 2D12C

Input Output~ ~

o ehopterwise eSSE Solved Papers PHYSICS

65. Draw a simple circuit of a CE transistoramplifier. Explain its working. Show thatthe voltage gain Av of the amplifier is givenby A v = 13AcR L / rj' where 13AC is the currentgain, RL is the load resistance and rj is theinput resistance of the transistor. What isthe significance of the negative sign in theexpression for the voltage gain. Deihl 2012

66. (i) Draw the circuit for studying the inputand output characteristics of atransistor in CE configuration. Showhow from the output characteristics theinformation about the currentamplification factor (13 AC) can beobtained.

(ii) Draw a plot of the transfercharacteristics (V, versus Vi) for abase-biased transistor in CEconfiguration. Ail Indio 2010; Foreign 2012

67. (i) Using the necessary circuit diagram,draw the transfer characteristics of abase-biased transistor in CEconfiguration. With the help of thesecharacteristics, explain briefly how thetransistor can be used as an amplifier?

(ii) Why are NAND gate called universalgates? Identify the logical operationscarried out by the circuit given asbelow: Foreign 2011

A

B ~--,'---"y

68. (i) Draw the circuit diagram of abase-biased n-p-n transistor inCE configuration. Explain, how thiscircuit is used to obtain the transfercharacteristic Vo - Vi characteristic.

(ii) The typical output characteristicsIe versus VCEof an n -;r n transistor inCE configuration is shown in the figure.Calculate(a) the output resistance ro and(b) the current amplification factor I3AC'Forelan 2010


60mAI'..50mAV

/ 40mA30mA

20mA

/ 10mA

~ 10c 8.5.- 8;g

6

4

2o 2 4 6 8 10 12 14 16

Collector to emitter voltage (VC8 in volts

69. (i) Draw the circuit diagram used forstudying the input and outputcharacteristics of an n- p-re transistorin the CE configuration.Show the typical shapes of these twocharacteristics.

(ii) How are the(a) input resistance(b) current amplification factor of the

transistor determined from thesecharacteristics? Delhl2010C

70. (i) Draw a circuit diagram to study theinput and output characteristics ofann -p -n transistor in its common-emitter configuration. Draw thetypical input and outputcharacteristics.

(ii) Explain with the help of a circuitdiagram, the working of an n -p- ntransistor as a common-emitteramplifier. Delhl2009C

71. Draw a circuit diagram of an n- p-ritransistor with its emitter base junctionforward biased and base- collectorjunction reverse biased. Describe brieflyits working. Explain, how a transistor inactive state exhibits a low resistance atits emitter-base junction and highresistance at its base-collector junction?Foreign 2009

72. Draw a labelled circuit diagram of abase-biased transistor in common-emitterconfiguration. Plot the transfercharacteristics of this base biasedtransistor indicating the different regionsof its operation. Delhl2009C

463

o Explanations1. (i) Collector current decreases.

(ii) Base current increases.2. Logic circuit of a NAND gate

Truth table

(1/2 x 2 = 1)

A B Y=A·B

0 0 10 1 11 0 11 1 0

(1/2 x 2 = 1)

3. Logic circuit of a AND gate:=O-yTruth table

A B Y=A·B

0 0 0

0 0

0 0

(1/2 x 2 = 1)4. Logic circuit of a OT gateA--t>----y

Output is high when input is low and vice-versa.Truth table

Input

o

Output

A Y=A

o

(112x 2 = 1)

5. The given combination consists of NOR gate andNOT gate, so equivalent gate is OR gate.Truth table

A B Y=A+B

0 0 00 1 11 0 11 1 1 (112x 2 = 1)

464

From the truth table, it is clear that the output is1 only when at least one of the inputs is at thehigh state i.e. 1

6. This logic gate is NOT gate and its symbolsA ----{:>o----- y = Ii

7. Logic gate is NOR gate.Symbol :=I>-y

NOR gate (112" 2 = 1)

8. Logic gate is NAND gate.Refer to Ans. 2. (1/2 " 2 = 1)

9. Logic gate is NOR gate.Symbol Refer to Ans. 7. (1)

10. Refer to Ans. 2. (1)

11. Refer to Ans. 3. (1)

12. Current amplification factor in common emittermode,

13.

PAC = I tilc Iti I B VCE = constant (1)

Circuit diagram of n-p- n transistor amplifier in CEconfiguration is given below:

Ie

Ie Ie

14.

The condition for the amplifier to work is thatthe base-emitter junction should be forwardbiased and collector-base junction should bereversed biased. (2)

The truth table is given as shown below:

A B Y'=A+ B Y = A·(A + B)

0 0 0 00 1 1 01 0 1 11 1 1 1

The logic gates used are R is OR gateand S is AND gate.

o Chopterwise C8SE Solved Papers PHYSICS

(1)

15. The logic gates are P is NAND gate and Q is OR gate. (1)

The truth table is given as shown below:

A B A·B A·B X=B+A-B

0 0 0 10 I 0 11 0 0 11 1 1 0

16. Refer to Ans. 21.(1)

(2)

A0---1 )<0---'-'--\:o---y

Bo----I ")(}---'"'---{

17. Refer to Ans. 2.18. Refer to Ans. 23.

and logic symbol is

:~Y=A+B

(2)

(1)

(1)

19. The given figure in question is common-emitter (CE)configuration of an n-p-n transistor. The input circuitemitter is forward biased and collector circuit isreverse biased. (1)

As, the base resistance R decreases, the input circuitwill become more forward biased thus, decreasingthe base current (IB) and increasing the emittercurrent (IE). This will increase the collector current(Icl as IE =~B + Ic· .When Ic increases which flows through the lamp,the voltage across the bulb will also increase thusmaking the lamp brighter and as the voltmeter isconnected in parallel with the lamp, the reading inthe voltmeter will also increases. (1)

Output characteristics is the plot betweencollector-emitter voltage (VCE) and the collectorcurrent (Ie) at different constant values of basecurrent (18).

Vec

20.

Base current (Is)

60 jlAI:..

50 jlA

/ 40 jlA30 jlA20 jlA, 10 jlA

~ 10.s~ 8c~ 6:soo 4U~8 2

o 2 4 6 8 10 12 14 16Collector to emitter voltage {VcE! in volts

(1)

(1) (1)


Output resistance is defined as the ratio ofvariation of collector-emitter voltage (dId andcorresponding change in collector current (dI dwhen base current remains constant. Initiallywith the increase in VeE the collector currentincreases almost linearly. this is because thejunction is not reverse biased. When the supplyis more than required to reverse bias. thebase-collector junction. Ie increases very littlewith VeE. The reciprocal of slope of the linearpart of the curve gives the value of output

resistance. i.e. ro = (d VeE )IB = constantsi; (1)

21. A

Y

B

A'= A. B'=B~ Y'= A'B'

Y =y= A' + B'

~ =A·B=A·B

The equivalent gate of the given circuit is ANDgate.Truth table

A B A' B' Y' Y

I I 0 0 0 I

I 0 0 I 0

0 I I 0 I 0

0 0 I I I 0

22. Y1 = A and Y2 = 13. . Y = Y1 + Y2 = Y1 . Y2

=A·B=A·B

The equivalent gate of the given circuit is ANDgate. (1)Truth table

A B II Yz Y

I I 0 0 I

I 0 0 I 0

0 I I 0 0

0 0 I I 0

465

23. Y=A'·B'=A+B=A+B

A

(1)

Y=A'·B'

B

Truth table

A B A'=A·A B' = B· B Y = A· B=A+B

001 I 0o I I o I100 I 1I I 0 o I

(1)

Thus. the equivalent gate is OR gate.24. Emitter-base Collector-base

jUnctio~ \n

( ~unction

()--+ ()--+

()--+ ()--+

()--+ ()--+

IE Forward Reverse Icbiasing 18 biasing

+ IIVEB VCB (1)

(1)

(1)

Heavily doped emitter is subjected to electric fieldby emitter-base battery and consequently. holesgets drifted towards collector through thin andlightly doped base region. Nearly 5% hole. whichdrifted from emitter combined with electron in baseregion and remaining nearly 95% hole reaches tocollector under the influence VeE. (1)

25. A signal in which current or voltage changescontinuously with the time is called analogsignals. A signal in which current or voltage cantake only two discrete values is called a digitalsignal. Example of analog circuit amplifier. radio.oscillator etc. Examples of digital circuit VCR.electronic watches. robots. modern computers.

(lx2=2)

Eorl

o~~--~--,---+------+------~--

(1)Analogsignal(Asinusoidallyvaryingalternatingvoltage)

466

Pulse duration1l

forI 1Pulse Pulserise\ amplitude

O~--~O----~--~O----~--~--~~Time-

Digital signal

26. Equivalent gate is OR gate. If input A or B or bothare 1. then the output of OR gate is 1. Booleanexpression of OR gate is given as A + B = X

Logic symbol of OR gate and the outputwaveform as shown below:::=[>---0 X =A + B

I I 1 I 1 I I I 1 I

A~I I I I I I I

: : I 1 I 1 I : I

B~I I I I I I II I I I I I II I I I I I I

X: t-----:---t---i ~~ii}---1i.......

Truth table

Inputs Output

A B X=A+B

0 0 00 I 11 0 11 I I (2)

27. Lamp glows brighter and voltmeter readingincreases with the decrease of R. Input currentincreases in turn by transistor action which willlead to increase in collector current. This makeslamp brighter and hence. voltmeter reading goesup. (2)

The transfer characteristic curve of base biasedtransistor in CE configuration as shown below:

28.

I

Cut-off Active 1 Saturationregion region: region

--- ----+"IIIIIIIIIIIIIIIIIIII

Vj-

IZI ehapterwise eBSE Solved Papers PHYSICS

Rs B

VBB

(1)

Applying Kirchhoff's rule to the input and outputcircuits separately. we get

VBB = VBE + [BRB

VCC= VCE + IeRe

VCE = Vee - IeRe

VBB = DC input - voltage (V;)

VCE = DC output - voltage (Vo)

As. V; increases slightly above 0.6 V. a current Ieflows in the output circuit and the transistorarrives in active state.

.. Vo=Vcc-IeRe

with the growth of Ie. Ve decrease linearly.Also. voltage gain in active state is given by

Av = - ~Vo (':~Vo > ~V;)~V;

There is voltage gain and hence amplification ofvoltage takes place. Thus. transistor used as anamplifier. (1)

29. Logic circuit behaves like an AND gateFor truth table (1)Refer to Ans. 3.

t, tz t3 t4 ts t6 t7 t8r I I I I I I I:. : M : r-r--:~ l--...:.-.l : :

A I I I I I I :

(Inputs): : I I I : I,"1: : r---i1 : r--i~ : l--...:.-.l :B I I I : I I I

I I I I I 1 II I n I I II I I I II I I I IX " , 1 '

(Output) ! I I ! I I ! (1)

Logic operation of the graphic circuit

X=A·B=AB


30. The truth table of given system is as follows: (1/2)

A B C=A·A D=A·B Y=C·D

0 0 1 1 0

0 0 1

1 0 0 1 1

0 1 1 0 1

(l1/a)

The correct performs the logic operation of ORgate.

31. X: AND gateY: NOT gateZ: Y=A·B (112 x 2)

A B zo oo

oo

(1)

32. (i) As, Y1 = A + B, Y2 = A + B

A B Y1 Y2

0 0 0 1

0 1 1 0

1 0 1 0

1 1 1 0(1)

(U) t1 t2 ts t4 ts te t7I I I I I I I

(Inp:t:l

I

II

II I

I I I II I I II I I II I I I

I I I I I II I I I I I II I I I I I I

B

Y2(Output) (1)

33. (i) P: NAND gate; Q: OR gate(U) X=A·B+ B (1I2x2= 1)

For A = 0, B = 0, X = 0.0 + 0 = 1 + 0 = 1For A =1, B = 1, X = 1.1 + 1 = 0 + 1 = 1. (1)

467

34. (i) P: NOT gate and Q: OR gate(U) X = (A + B)

For A = B = 0, X = 0 + 0 = 1 + 0 = 1For A = 1, B = 1, X = T + 1 = 0 + 1 = 1

35. (i) The input resistance, rj of transistor in CEconfiguration is defined as the ratio of smallchange in base-emitter voltage to thecorresponding small change in the base current,when the collector emitter voltage is keptconstant, i.e.

r. = ( ~:BLE= constant

[1/2x2= 1)

(1)

(1)

(U) The current amplification factor of a transistor inCE configuration is equal to the ratio of the smallchange in the collector current (Md to the smallchange in base current when collector-emittervoltage is kept constant, i.e.

~ = (Me)M B vCE ::::constant (1)

36. Gate From the given output waveform, it is clearthat output is zero only when both inputs are 1, sothe gate is NAND gate. (112)

Truth table

A B Y

o 0o

oo

(1)

Logic symbol

~:=[J-YY= AB (1/2)

37. When output of a two inputs AND gate is fed to aNOT gate, then the combination is called NANDgateLogic symbol~=cr-Y (1)

(1)For truth table, refer Ans. 36.

468

38. (i) Output of AND gate is Y = A·B. In this case,output will be I only when both inputsare 1.

iHJI I II I I

i H8 -~--+---!I I I

I I II I Inil

I II II I

X I I ::(Output) 0 1 ; ; 4 5 ~ ~

Input and output waveforms for an AND gate

A--+-(Inputs)

(ii) NAND gate will be formed.

39. NOR gateSymbol

On connecting the given gate with NOT gate,the output

Y=A+B=A+B(i) If A=O,B=O~Y=A+B=O+O=O

(ii) If A=l,B=1 44.Y=I+I=1. (1)

40. Logic gate~=D---v-YEquivalent so formed gate is NOR gate.Symbol ~=D-Y (1)

Truth table Y = A + B

A A B B Y=A·B

0 0

0 1 0 0

0 0 0

0 0 0

(1)

41. NAND gate will be formed for logic symbol andtruth table, refer to Ans. 36. (2)

o ehopterwise eBSE Solved Papers PHYSICS

42. (i)

A A B B Y=A+B=A+B

0 1 0 1 00 I I 0 I1 0 0 1 11 0 I 0 1

(1)

(ii)A A B B Y=A+B=A·B

0 I 0 1 00 1 I 0 01 0 0 I 01 0 1 0 1

(1)(1)

(1)43. Given, amplification factor = iL = 100

ibi = collector voltage = __ 2_ = 10-3 Ac collector resistance 2 x 103

. _ i, _ 10-3 -10-5 AIb------

100 100 (1)(1)

Also given, base resistance = 103 Q

Input signal voltage = ib x R = 10-5 X 103y = 10-2 v (1)

B input signal voltage 10-2 10-5 Aase current = = -- =base resistance 103 (1)

(i) The functions of all the three segments oftransistor are given below.The emitter supplies the majority carriers forcurrent flow. The collector collects them. Thebase acts as an accelerator for charge carriersand send them to collector. It also regulates theflow of majority carriers in the circuit. (1)

(ii) Output of an AND gate is at high potential whenboth inputs A and B are supplied with highpotential. So, waveform is

t1 t2 t3, ,, ,, ,,,,A:,

(Input) l,,8'i-, --;.-,,,,,,

(Output) (Y) :;-----;.-,,,,,,

(1)

t8,,,,,,,,,, ,~n, ,, ,, ,, ,, ,, ,, ,, ,


Truth table for AND gate is given below.

Inputs Output

A B y

o o oo o

o o

45.

(1]

(i) Functions of three segments of a transistor areas follows:Emitter It supplies a large number of majoritycharge carriers for the flow of current throughthe transistor.Base It controls the flow of majority chargecarriers from emitter to collector.Collector It collects a major portion of themajority carriers supplied by emitter for thecircuit operation.

R8 Vcc

VBB

Circuit arrangement for studying the input and outputcharacteristics of transistor in CE configuration (1]

(a) Input characteristics Input characteristicmeans we have to plot the graphicalrepresentation between IBand VBE. VBE is theemitter to base voltage or the forward biasvoltage and IB is the base current. In thisforward biasing, E is at lower potential than B.We will be plotting IB versus VBE because baseis at higher potential than emitter, so that willbe reflected here. Now go on varying VBE. Forsilicon diode we have knee voltage around0.7 V. After overcoming the knee voltage,current will rise sharply.The input characteristic will be different if wego on increasing the VCE. It will be shiftingright, means for the same VBEwe will begetting lower input current lB.

469

100

80

60

40

20

VCE = 1 V

VCE = 5 V

VcE=10V

0.2 0.4 0.6 0.8 1.0 VSE(V) (1]

Input resistance (r;) This is defined as theratio of change in base-emitter voltage (6VBE)to the resulting change in base current (MB)

at constant collector-emitter voltage (VCE). Thisis dynamic (AC resistance) and as its valuevaries with the operating current in thetransistor.

r; = (~BE)B VCE

(b) Output characteristics A graph showingthe variation of collector current Ie withcollector emitter voltage VCE at constant basecurrent IBis called the output characteristic ofthe transistor.A study of these curves reveals the followingfeatures.(i) When the voltage VCE increases from 0 to

about 0.2 V, the collector current Ieincreases rapidly.

(ii) Once the voltage VCE exceeds the kneevoltage the output current Ie varies veryslowly but linearly with VCE for a givenbase current lB. (1)

Base current (/a)60 J.IA

/' 50 J.IA40 J.IA30 J.IA

V 20 J.IA10 J.IA

~12.5.~~ 10

C 7.5~::::lo

j(5u

5

2.5

(ill)

o 0.5 1 1.5 2 2.5 3 3.5 4Collector to emitter voltage (VCE) in volts

Output resistance (ro) This is defined asthe ratio of change in collector-emittervoltage (6VCE) to the change in collectorcurrent (Mcl at a constant base current IB:

470

46. Consider the given figure,

A -----l;'\~ y

B~

P is AND gate and Q is a NOT gate.Truth Table

A B X X=y0 0 0 1

0 1 0 1

1 0 0 1

1 1 1 0

The equivalent gate representing this circuit isNAND gate. Its logic symbol is

A~ -B~Y=A.B

47. Circuit diagram of a CEtransistor amplifierIe

48.p-n-p RL

0VCE ~

V IE OutputACInputAC L ::=:: Vcc signal

signal -=-+

IB IeVBE (1)

WorkingThe emitter-base circuit is forward biased by alow voltage battery VSE' that means the resistanceof input circuit is small. The collector-emittercircuit is reversed biased by a high voltage batteryVco that means the resistance of the outputcircuit is high.RL is a load resistance connected incollector-emitter circuit.The weak input AC signal is applied across thebase-emitter circuit and the amplified output isobtained across the collector emitter circuit.When no AC voltage is supplied to the inputcircuit, we have

... (i)

121 Chapterwise CSSE Solved Papers PHYSICS

(1)

Due to collector current Ie' the voltage dropacross the load resistance (Rd is IeRL. Therefore,the collector emitter voltage VCE is given by

VCE '"' Vcc - IeRL ... (ii)when the input AC voltage signal is applied acrossthe base-emitter circuit, it changes base-emittervoltage and hence, emitter-current IE changeswhich in turn changes the collector current Ic.So, the collector-emitter voltage VCE varies in theaccordance with Eq. (ii).This variation in VCE appears as an amplifiedoutput. (1)

Current gain and voltage gain of amplifier

(i) DC current gain, ~DC = IeIs

AC current gain, ~AC = (!Hc )VCEIlls

(ii) ACvoltage gain, A v = II Ie X RII t,X R

= llIe x~st, R (1)

(1)

(1)

(1)

(i) Input resistance may be defined as the ratio ofthe small change in the base - emitter voltage(IlVse) to the resulting change in base current(Is) at constant collector-emitter voltage (VCE) .

Input resistance, R; = IIVBE (1)si,

(ii) Current amplification factor (r) is defined asthe ratio of change in emitter current (ME) tothe change in base current (MB).

. r=MEu. msi,49. (i) Dynamic output resistance is given as

R" t = (IlVCE) 12-8u llIc IB~ constant (3.6- 3.4)X 10 3

4 = 20kn0.2 x 10-3 (1)


(ii) DC current gain,

~DC = Ie = 3.5 mAI8 30~A

= 350 = 116.673

(iii) AC current gain,

~ = Me = (4.7-15)mA = 1.2x 10-3 =120DC M8 (40-30)~A 10xlO-6

In the linear region of output characteristics ~AC

is close to ~DC' (1)

Transfer characteristics The graph between Vi)and Vi is called the transfer characteristics of thebase-biased transistor, is shown in figure.

Cut-off ActiveV region i region

a III

3.5 X 10-3

30 X 10-6"

50.

Saturation__ region

Transfer characteristic

When the transistor is used in the cut-off orsaturation state. It acts as a switch.

Vo = Vcc -IcRc ~ Vi = l8R8 + VBETransistor as a switch The circuit diagram oftransistor as a switch is shown in figure.

Ie

Rc+

Va

vcc

Sase-biased transistor in CEconfiguration

As long as Vi is low and unable to forward bias thetransistor, Vo is high (at Vd. if Vi is high enoughto drive the transistor into saturation, then Vo islow, very near to zero.When the transistor is not conducting it is said to beswitched off and when it is driven into saturation itis said to be switched on. This shows that if wedefme low and high states as below and abovecertain voltage levels corresponding to cut-off andsaturation of the transistor, then we can say that alow input switches the transistor off and a highinput switches it on. (1)

471

51. Truth table of given circuit is as shown below:

(1)

A B X=A Y=B Z=X+Y

0 0 0

0 0 00 1 0 0

0 0

This circuit carries out by the logic operationof AND gate which can also be verified byde-Morgan's theorem

Z=X+Y

=A+B

=A+B=A·B

So, the circuit corresponds to AND gate.

Symbol AB=D--Z=A.a

(2)

(1)

52. Z = A + B = A· B = AB

Refer to Ans. 51. (3)

(1) 53.

Cut-off Activeregion region Iv, : :

I : Saturation: __ re-"9'-io_n _IIIIIII

Transfer characteristic ofbose-biased transistor (W2)

(1)The active region of transfer characteristic curveoperates as an amplifier. (1/2)(i) Refer to Ans. 35 (i). (1/2)

(ii) Output resistance The ratio of variation ofcollector emitter voltage (VCE) andcorresponding change in collector current (Me)when base current remains constant is calledoutput characteristic curve.

R - (~VCE)out - --Me 18 =conSlant

(ill) Refer to Ans. 35 (i), (ill). (112)

412

54. For curve, refer Ans. 53. (1)

The active region of transfer characteristic curve

(

'Sfused as an amplifier. (2)

, Whenever CE circuit is used as an amplifier the

j~~~~;f~~;~li~~i~f;r~i;~~~:hr:~~~~f_55.

Circuit diagram of a common-emitter amplifier(1'12)

Ie

VeE Ie Output

- Vcc 1VBB =

~doutput

(1'1.)

Voltage gain It is equal to the ratio of smallch~rige'in 'output voltage at the collector to that of

(tl change in input voltage, i.e.Av = Output voltage

Input voltage : '= ,iVCE = (MdRout = ~AC X 'Rout

,iVEB ME Rm Rm

=> Voltage gain = ~AC X RoutBm

C\V

InputAC voltage

where, ~AC is AC current gain.56. For G1

Gate NAND gateTruth table

A B

(112 x 2 = 1)

For G2

Gate NOR gate


Truth table

A B

o oo o

o oo

(112 x 2= 1)

ForG3

Gate AND gateTruth table

A B G3 = A·B

0 0 0

0 0

0 0

(1'1.)

(112 x 2 = 1)

57. (i) G1 : AND gate (ii) G2: NOR gate

(iii) G3: OR gate. (112 x 3)

Truth table

A B G1 G2 G3

0 0 0 1 0

0 1 0 0 1

1 0 0 0 1

1 1 1 0 1

(1'12)

58. For figure refer to Ans. 55. (2)

Relationship between input and output signals ofn -p- n transistor amplifier. When positive halfcycle is fed into input circuit, forward bias ofemitter base circuit decreases. This lead todecrease IE and by transistor action, collectorcurrent decreases. Since, output voltage,Vo = VCE - Ie Ru therefore, decrease in collectorcurrent, irIcreases the output voltage. As, thecollector is connected with the negative terminalof battery Vee, the irIcrease in collector voltageimply that negatively of collector increases.Thus, correspondirIg to positive half cycle of inputAC, a negative amplified cycle is obtained atcollector and vice-versa. This shows that outputand input signals are in opposite phase. (1)


59. The gates 1 and 2 are NOR gates acting as NOTgate.For I and 2,

When A = 0, A = 0, Y = °+ °= 0 = I

Similarly, B = 0, B = 0, Y = °+ 0= 0 = I (1)

Logic gate of complete circuit is AND gate. (1)

Truth table (1)Refer to Ans. 3. (2)

60. (i) The base region of the transistor is physicallylocated between the emitter and the collectorregion and is made from lightly doped highresistivity material. The emitter and collectorregions are heavily doped. But the doping levelin emitter is slightly greater than that ofcollector and the area of collector region isslightly more than that of emitter. (1)

In term of doping level,Emitter region> collector region> baseregionIn term of area of the region,Collector region> emitter region> baseregion. (1)

The area of the collector region is greater thanthat of emitter. This is because the collectorregion has to handle more power than theemitter and also it has to collect more numberof charge carriers to constituent the current.Emitter is heavily doped to provide largenumber of majority charge carriers, while baseand collector are lightly doped to accept thesecharge carriers from emitter. (1)

(ii) The conditions of a transistor for to be inactive state are below:(a) The input circuit should be forward biased

by using a low voltage battery.(b) The output circuit should be reverse

biased by using a high voltage battery. (1)

(ill) CE configuration refer to Ans. 65. (1)

61. (i) Common-emitter TransistorCharacteristics To study the characteristics ofan n-p-n transistor in common-emitter mode,required circuit is shown in the figure. Here,base-emitter circuit is forward biased withbattery VBEand emitter- collector circuit isreverse biased with battery Vce.

From circuit diagram, we come across to knowthat it is made up of two sections, i.e. input andoutput. (1)

These two characteristics can be studied as shownbelow:

(a) Emitter or Input CharacteristicsA graphical relation between the emittervoltage and the emitter current by keepingcollector voltage constant is called inputcharacteristics of the transistor.Adjust collector-emitter voltage at a suitablehigh value VCE (say = + 10 V). It is necessary soas to make the base-collector junction reversebiased.

Now, with the help of rheostat graduallyincreases, the value of base-emitter voltage VBEinsmall steps and note the corresponding values ofbase current lB' (1)

'B~)100

lc

VeE=10V80604020

VBEM0.2

473

Vee

Input resistance It is defined as the ratio ofchange in base-emitter voltage (L~.vBE)to theresulting change in the base current (!VB) atconstant collector-emitter voltage (VeE)' It isreciprocal of slope of IB-VBE curve.

Input resistance, R i = (~BE) = constantB VCE

474 o ehapterwise eBSE Solved Papers PHYSICS

(b) Collector or Output CharacteristicsA graphical relation between the collector voltageand collector current by keeping base currentconstant is called output characteristics of thetransistor. To study output characteristics oftransistor we keep value of base current IB fixed(say at 10 J.1A) with the help of VBE. Now, graduallychange the value of VCE and note the values ofcollector current t.. (1)

Plot Ie - VCE graph. Repeat the process for differentconstant values of lB'

The output characteristics are as shown below:

Base current (18)

60~/ 50~

40~30~20~10~

~ 10.5:;g 8C~ 6'So

~ 4.!l1(5o 2

o 2 4 6 8 10 12 14 16Collector to emitter voltage (VcEl in volts

(ii) Feedback When a portion of the outputpower is returned back to the input in phasethis is termed as positive feedback. (1)

Input'r=--i Transistor II r-i amplifier II---------.-~ Output

'----------11 Feedback UI network ~

Feedback network The phenomenon of mutualinductance is used to take a part of output in coil(L') back into input coil (L).

When the switch K is closed, collector currentbegin to flow through L', which in turn increasesthe magnetic flux linked with L' and hence with L.This leads to produce an induce emf in L, whichincreases the forward bias. This also increases thebase current and hence collector current along withthe charging of capacitor takes place with upperplate as positive.This phenomenon is repeated again and again tillthe collector current reaches to its maximum value.

Inductivecoupling n-p-n +

'----l~KJ-VBB

Circuit diagram of tronsistor as on oscillator

At maximum value of Ie' current through L'does not change and therefore flux remainsunchanged and emf in L' and L reduces to zero .Now, the discharging of capacitor beginsthrough L. The positivity of upper platedecreases and forward bias decrease, whichresults in the form of decrease in base currentand hence, decrease in collector current. Thisphenomenon repeats till collector currentreduces to zero and emf in the coil L alsoreduces to zero.Thus, the time duration in which collectorcurrent grows from zero to maximum, thecurrent in coil L of tank circuit complete its halfcycle. The duration in which collector currentreduces from maximum to zero, the current in L'completes its next half cycle.

Ie ICollectorl'---+--J(--.......;--+-currentin L'

Current in coil f---+--+---I-,---t--L of input

circuit t-Rise and fall of currents

The frequency of oscillation is given by1

V=---2rc.JLCThus, the AC of desired frequency andamplitude can be obtained by taking appropriatevalue of inductance, capacitance and strength ofbattery B. (1)


62. (i) In this transistor, the emitter-base junction isforward biased and its resistance is very low. So,the voltage of VEE is quite small. (1)

The collector-base junction is reverse biased. Theresistance of this junction is very high. So, thevoltage 'of Vee (VeB) is quite large ('" 45 V).Electrons in emitter are repelled towards base bynegative potential of VEE on emitter, resultingemitter current 1£. (1)

The base being thin and lightly doped has lowdensity of holes, thus when electrons enter thebase region, then only a few holes getneutralised by electron hole combination,resulting in base current (Is). The remainingelectrons pass over to the collector, due to highpositive potential of collector, resulting incollector current (I cl .As, one electron reaches to collector. it getsneutralised by the flow of one electron from thenegative terminal of the battery Vee to collectorthrough connecting wire. Then, one electronflow from negative terminal of battcry Vce topositive terminal of battery VEE and one electronflow from negative terminal of VEE to emitter. (1)

When the electron coming from emittercombines with the holes in base, then deficiencyof hole in the base is compensated by thebreaking of covalent bond there. The electron, soreleased flows to the positive terminal of batteryVEE, through connecting wire.Thus, in n-p-n transistor, the current is carriedinside as well as in external circuit by theelectrons. Thus, in this case also,

1£ = Is + I e [Kirchhoff's first law]In the base, IE and Ie flow in opposite direction. (1)

p-basen-emitter region n-collector, , .,-----'--, , ,--- :--- ---:::::.e ---

~n - --- nL--- :--- ---___ --.-0 ------ i:=- ------ - -----VEB

BVCB-.--

18-.1 + -1111:+1

ic

IE lcVEE vcc

Flow of charge carriers in n-p-n transistor

(ii) Refer to Ans. 61.

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63. (i) Thin and lightly doped base region contains asmaller number of majority charge carriers,reducing rate of recombination resulting smallIs and collector current almost equal to 1£resulting large voltage and power gain. (1)

(ii) Refer to Am. 61 (i). (1)

(iii) Refer to Ans. 68 (i). (3)

64. X ~ Amplifier, Y ~ Feedback networkTransistor as an oscillator: In an oscillator, theoutput at a desired frequency is obtained withoutapplying any external input voltage. The commonemitter n -p - n transistor as an oscillator is shown inthe following figure. A variable capacitor C ofsuitable range is connected in parallel to coil L togive the variation in frequency. (1)

(1)

f- .--:f-:::Output

(1)

(1)

Oscillator actionAs in an amplifier, the base-emitter junction isforward biased while the base collector junction isreverse biased. When the switch S is put on, a surgeof collector, current flows in the coil T2. Theinductive coupling between coil T2 and Ii cause acurrent to flow in the emitter circuit i.e., feedbackfrom input to output. As a result of positivefeedback, the collector current reaches atmaximum. When there will be no further feedbackfrom T2 to Ii, the emitter current begins to fall andcollector current decreases. Therefore, the transistorhas reverted back to its original state. The wholeprocess now repeats itself.The resonance frequency(f) of the oscillator is given by

1=_1_21tfLC (2)

The tank of tuned circuit is connected in theoscillator side. Hence, it is known as tuned collectoroscillator.

476

65. While finding gain for CE configuration weshould mind that it will depend upon the loadresistance, input resistance as well as output willbe inverted.For circuit diagram refer to Ans. 55. (1)

Working In the circuit, emitter is forwardbiased and collector is reversed biased. Thismakes input resistance (R;n) very low and outputresistance (Rout) high. During the positive halfcycle of input AC decrease the forward bias.Hence, emitter current, IE and by transistoraction collector current decreases. This tend toincrease the collector voltage which is givenby Vo = VCE = Vcc - IeRLThe high value of RL produces large change ill Vocorresponding to low change in Vi' Thus,amplified pulse is obtained at collector. (1)

Voltage gain It is equal to the ratio of changein output voltage (VeE) corresponding to thechange in input voltage (~VBE)' i.e.

Voltage gain, Av = ~VCE ,;, (Md RL~VBE (M8) Vi

where, RL and R, are output resistances (loadresistance) and input resistance of transistorrespectively.

AV=(Me)RL =~AeRLM8 ri ri

where, ~Ae is AC current gain = MeM8

The output voltage of CE amplifier differ in phasefrom the input voltage by 1800 or n rad. Theopposite phase is represented by negative sign. (1)

.. Voltage gain = - ~Ae RLri (112)

66. (i) The circuit is as shown below:

Ra cB

teVBB

(112)

Rc

+/' Vo

vcc


Current amplification factor (J3Ad is the ratioof change in collector current (Md to the changeis based current (M B) at constant collectorvoltage, i.e.

.• ~AC = Me IM B VCE = constant (1)

Output characteristics represent the variation oft c with Ve' keeping 18 constant.

--------~_r_--- la = 400 mA

.-...&.---- la = 300 mA

__-r--- le= 200 mA

la=100mA

Vc(volt) - (1)

(1)

From above graph at Ve = V, the value of collectorcurrent increases with the increase in the basecurrent, lB' Thus,

~AC = Me I [AC current gain]M B vCE = constant

(U) Transfer characteristics curve for a base-biasedtransistor in CE configuration.

Cut-offregion Active

region Saturationregion

'------'-------'----_ Vi (2)

67.

Hence, low input give high output and high inputgives low output.(i) For curve refer to Ans. 53, CE amplifier circuit

Ans.55. (2)

The active region of a transfer characteristicscurve can be used to explain the transistor asan amplifier.

(1)


The resistance of output circuit is large beingin reverse bias and resistance of input circuitis low being in forward bias.When input voltage, VBE comes in activeregion, Ie flows in output and Vo variessignificantly as Vo = VCE = Vcc - IeRLThis change in output voltage is obtained asamplified form. (1)

(ii) NAND gates are termed as universal gatesbecause all three basic gates namely AND, ORand NOT can be made using NAND gate. (1)

The given circuit perform the logic operations

of AND gate as Y = (A·B) = A·B

(i) For n-p- n transistor in CE configuration circuitdiagram.Refer to Ans. 66 (ii). (1)

For transfer characteristic curve Refer toAns.28. (2)

(ii) (a)

68.

(b)

The output resistance (ro)

_ (L'l.VCE)

Me Is = constant

From the given graph, at IB = 60~,

VCE = 2V, VCE = 16 VCollector current changes from 8 mA to8.5 mA,i.e. L'l.VCE =16 - 2=14 V

Me = 8.5 - 8 = 0.5 mA = 5 x 10-4 A

.. ro =(:~t=60~ sx1:0-4

ro = 2.8 x 104 nro = 28 kn

The current amplification factor,

~AC= (Me)M B VCE = constant

At VCE= 2 V, IB = 10 ~ to 60 ~.. MB=(60-1O)=50~

Ie changes from 1.5 mA to 8 mA.. Me=8-1·.5=6.5mA

=> ~ =(Me) 6.5x 10-3 AAC M 50x 10 6 A

B VCE

6.5 X 103 2~AC= =1.3xlO=>~Ac=13050 (1)

477

(1)

69. (i) Refer to Ans. 66 (ii), 28. (3)

(ii) Refer to Ans. 35 (i), (ii). (2)

70. (i) Refer to Ans. 57. (3)

(ii) For circuit refer to Ans. 51. (1)

Working of n-p-n transistor as CEamplifier In the circuit, output resistance isvery high whereas input resistance is very lowbeing reverse and forward bias, respectively.When current, Ie grows in output circuit,potential difference across the collectordecreases significantly as per relation

Vo ;= VCE = Vcc - IeRLWhen input voltage is fed into input circuit, VEBchanges, which in turn change IB and IE' Bytransistor action, Ie change and thus, outputvoltage changes in amplified manner. (1)

71. n -p- n transistor in CB configuration Since,the base is common in input and output circuits,therefore transistor is connected in CBconfiguration.

ho--..,---{/}---,-=-T

Input =} r B ,=} Output

VSE

1Is =vccVSS

(2%)

(1)

Working When input voltage, VBE is sufficientto make flow of emitter current, collector currentflows in output circuit. In this condition, thecircuit is said to be in active state.The small change in VEB, produces sufficientchange in emitter current and hence, in collectorcurrent. The input circuit offers very smallresistance as ample change in emitter currentoccurs corresponding to small change in inputvoltage.This lead to produce large change in outputvoltage in spite of smaller change in collectorcurrent (IE < Id. This shows that output circuitoffer high resistance. (2%)

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72. Refer to (first part)


VBB

VInput

ACvoltage

RL IIe Vo

1=-

~~

(H) Second part 66 (ii).

n-p-n transistor as a common emitter amplifier

(5)

Value Based Questions (From Complete Chapter)o 4 Marks Questions1. Meeta's father was driving her to the

school. At the traffic signal she noticedthat each traffic light was made of manytiny lights instead of a single bulb. WhenMeeta asked this question to her father,he explained the reason for this.Answer the following questions based onabove information:

(i) What were the values displayed byMeeta and her father? '

(ii) What answer did Meeta's fathergive?

(iii) What are the tiny lights in trafficsignals called and how do theseoperate? Deihl 2016

Ans. (i) Values displayed by Meeta, are curiosity tolearn and good observation. Values displayedby her father are patience andknowledgeable. (1)

(ii) Meeta's father most probably explained herthe benefits of using tiny bulbs (LEOs) over asingle bulb.(a) Tiny lights are semiconductor devices

which consume very less power than asingle bulb.

(b) Tiny lights are very cheap.

(c) If some of these tiny lights are notworking, then traffic system will not beaffected. But if a single bulb is fused,traffic system will be disturbed. (2)

(Hi) Tiny lights in traffic signals are called LEOs.LEOs are operated in forward biasing and emitsspontaneous radiation. (1)

2. Ameen had been getting huge electricity billfor the past few months. He was upsetabout this. One day his friend Rohit, anelectrical engineer by profession, visited hishouse. When he pointed out his anxietyabout this to Rohit, his friend found thatAmeen was using traditional incandescentlamps and using old fashioned airconditioner. In addition there was no properearthing in the house. Rohit advised him touse CFL bulbs of 28 W instead oflOOOW-200Vand also advised him to getproper earthing in the house. He madesome useful suggestion and asked him tospread this message to his friends also.

(i) What qualities/values, in your opiniondid Rohit possess?

(ii) Why CFLs and LEDs are better thantraditional incandescent lamps?

(iii) In what way earthing reduceselectricity bill? oelhl2015C


Ans. (i) Being an engineer, Rohit was well awaredabout energy saving and use of moderntechnology. (1)

(ii) CFLs and LEOs consumes less power incomparison of traditional incandescent lampand also give more light and it can save upto85% on energy bill. (1)

(iii) The earth wire acts as negative terminal. Theflowing current from positive cable to earthgrounding will not be counted by electricmeter because it does not pass the negativecable. In this way, you can reduce theelectricity bill and save your money. (2)

3. Kritika Singh was enjoying TVprogramme at her home with heryounger brother Sorya at night.Suddenly, the light went off causingdarkness all over. Sorya asked her tobring candle along with matchstick toput the TV switch off. Kritika at oncepicked the mobile phone and pressed thebutton lighting up the surrounding. Heryounger brother was surprised andasked, where the light was coming.Kritika proudly showed her mobile.Read the above passage and answer thefollowing questions.

(i) Which value is displayed byKritika?

(ii) Name the material used in theformation of LED.

(iii) LED works, in which biasing?Ans. (i) Kritika is a creative and intelligent girl. She

has good knowledge of Physics.(ii) The semiconductor material is used in the

formation of LED e.g. Ga, As, P.(iii) LED works in forward biasing i.e. n-type is

connected with negative terminal of thebattery and p-type is connected with positiveterminal of the battery.

Light Emitting Diode (LED) Speciallydesigned diodes, which give out lightradiations when forward biased. LED's aremade of GaAsP, GaP etc.

479

4. Ashwin was given 3 semiconductors A, Band C with respective band gaps of 3 eV,2 eV and 1 eV for use in a photodetector todetect A = 1400nm. He found that thephotodetector was not working with thesesemiconductors and did not know why? Hisfriend Akash found out the reason for itand explained it to him.Read the above passage and answer thefollowing questions.

(i) Why did the photodetector not work?(ii) What according to you are the values

shown by Akash?Ans. Energy related to wavelength A is

E = he = 6.634x 10-34 x 3x 108 =1.42 x 1O.-19JA 1400x 10 9

E _1.42 X 10-19

= 0.88 eV [:. 1 eV = 1.6 x 10-19J]1.6x 10 19 .

(i) The photodetector does not work because theenergy related to wavelength, 0.88 eV is lessthan band gaps of semiconductors A, Band C.

(2)

(4)

(ii) The values of Akash are(a) helping attitude.(b) presence of mind.(c) high degree of awareness.(d) concern for his friend, kindness. (2)

5. Garima and Gaurav want to purchase anew TV set. They visited electronic shops tolook for some branded TV. The dealershowed them LCD and LED TV's. Now,they were confused which set to buy.Finally, after discussing with friends,reading relevant literature and searchingthe internet, they decided to purchase LED.

(i) Which value is being highlighted byGarima and Gaurav?

(ii) What is the difference between LEDand LCD?

(i) Interpretation skill application skill. analyticalthinking is being highlighted by both. (1)

(ii) LED stands for Light Emitting Diode. LED is asemiconductor diode. Whereas, LCD stands forLiquid Crystal Displays. LCDs came as areplacement technology for large and heavycathode ray tubes.

Ans.

480 o Chopterwise CBSE Solved Papers PHYSICS

LCD is a devices, which consists of several parts,whereas an LED is a single component device. LCDis only used as a display device, whereas LED areused in various other applications such as flashlightsand indicators. LEDs are capable of producing lightwhereas liquid crystals cannot produce light. LEDdisplays consume less power in general than thesame sized LCDs.LED displays can produce more brightness andcontrast than the counter part LCD displays. [31

6. Gautam went for a vacation to the villagewhere his grandmother lived. Hisgrandmother took him to watch 'nautanki' oneevening. They noticed a blackbox connected tothe mike lying nearby. Gautam's grandmotherdid not know what that box was. When sheasked this question to Gautam, he explainedto her that it was an amplifier.

(i) Which values were displayed by thegrandmother? How can inculcation ofthese values in students be promoted?

(ii) What is the function of an amplifier?(iii) Which basic electronic device is used in

the amplifier?Ans. (i) The values displayed by Gautam's grandmother are

(a) Curiosity (b) Awareness [2)

The inculcation of these values in students can bepromoted by positive mental state, improvement inmotives and healthy supportive environment.

(H) The function of an amplifier to increase theamplitude of variation of alternating voltage orcurrent or power. [1)

(Hi) Transistor is the basic electronic device used in theamplifier. [1)

7. (i) Figure shows the input waveform whichis converted by a device 'X' into an outputwaveform. Name the device and explainits working using the proper circuit.Derive the expression for its voltage gainand power gain.

(ii) Draw the transfer characteristic of basebiased transistor in CE configuration.Explain clearly which region of the curveis used in an amplifier. Delhl2015C

Ans. (i) Device X ~ CE AmplifierPower gain Ap of the transistor may beexpressed as:AC power gain (Ap) = Current gain (J3Acl xVoltage gain (A,,)

Ap = ~;,.C"RLR [2)

As power is always positive, hence power gainAp is always positive.Refer to Ans. 48.(H) Refer to Ans 50. [2)

8. Sorya usually enjoyed loud music.One day his music system was notproducing very loud sound. Sorya getdisappointed. He discussed hisproblem with his big brother Kamal.Kamal advised him to connect anamplifier in series with the amplifierof sterosystem. This increased thesound of Sorya music systemconsiderably.Read the above passage and answerthe following questions.

(i) Which values is displayed byKamal?

(ii) Why the loudness of musicsystem get increased onconnecting amplifier?

(iii) Give the equation for currentgain in the transistor as anamplifier.

Ans. (i) Kamal is caring for his youngerbrother, good subject knowledge helpshim to develop application skill. [1)

(ii) On connecting two amplifiers in series, itsamplification gains get multiplied. Henceloudness of music system increases. [21

(iii) Current gain in the transistor as anamplifier is given by ~AC = M C

MBwhere, Mc = change in collector current

M B = change in base current [1)

Semiconductor Electronics - Mohit Nain · CHAPTER 14 : Semiconductor Electronics 1.4 p-n Junction A...

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