BEC101/BEC201; Fundamentals of Electronics Engineering UNIT-1 : SEMICONDUCTOR DIODE (1-1 J to 1-48 J) Semiconductor Diode: Depletion layer, V-I characteristics, ideal and practical Diodes, Diode Equivalent Circuits, Zener Diodes breakdown mechanism (Zener and avalanche), Diode Application: Diode Configuration, Half and Full Wave Wetification, Clippers, Clampers, Zener diode as shunt regulator, Voltage-Multiplier Circuits, Special Purpose two terminal Devices: Light-Emitting Diodes, Photo Diodes, Varactor Diodes, Tunnel Diodes. UNIT-2 : BJT AND FET Bipolar Junction Transistor: Transistor Construction, Amplification action. Common Base, Collector Configuration, Field Effect Transistor: Construction and Characteristic of JFETs. Transfer Characteristic, MOSFET (MOS) (Depletion and Enhancement) Type, Transfer Characteristic. INIT : OPERATIONAL AMPLIFIERS (8-1 J to 3-25 J) Introduction, Op-Amp basic, Practical Op-Amp Circuits inverting Amplifier, Non-inverting Amplifier, Unit Follower, Summing Amplifier, Integrator, Differentiator). Differential and Common-Mode Operation, Comparators. UNIT-4 : DIGITAL ELECTRONICS Number system & representation, of Basic and Universal Gates, of Boolean function. K Map (2-15 to 2-40 3) Operation, Common Emitter, Common (4-1 to 4-204) Binary arithmetic, introduction using Boolean algebra simplification Minimization upto 6 Variables. UNIT-5 : FUNDAMENTALS OF COMM. ENGG. (5-1 J to 5-33 J) Fundamentals of Communication Engineering: Basics of signal representation and analysis, Electromagnetic spectrum Elements of a Communication System, Need of modulation and typical applications, Fundamentals of amplitude modulation and demodulation techniques. Introduction to Wireless Communication: Overview of wireless communication, cellular communication, different generations and standards in cellular communication systems, Fundamentals of Satellite & Radar Communication, SHORT QUESTIONS (SQ-1 J to SQ-26 J) SOLVED PAPERS (2020-21 TO 2021-22) (SP-1 J to SP-13 J)—————E—— ee ltttt—t— Semiconductor, Diode Semiconductor Diode 1-2 (Sem-1& 2) PART-1 ‘Semiconductor Diode : Depletion Layer, ‘eal and Practical Diodes, Diode Equi V-I Characteristics, ivalent Circuits. CONTENTS Part-l + Semiconductor Died Layer, Vi Characteistien. ioe Diede Equivalent Create Part2 + Zener Diodes Break renkaewn 1-60 to Mechanism (Zener and Avalanche) yee Part-3. + Diode Application : Diode 18d to 1-20) Configuration, Half and Full Wave Rectification Part-4 : Clippers, Clampers 1-205 to 1-825 Part-5 : Zener Diode as Shunt Regulator, 1-924 to 1-403 Part-6 : Special Purpose Two Terminal ea... I-40J to 1-435 Devices : Light Emitting Diodes, Photodiodes Part-7: Varactor Diodes, Tunnel Diode: 148d to 1-470 1-1 (Sem-1 & 2) GasTA | Explain semiconductor diode and also draw its equivalent cireuit. “Answer 1. Adiode is an electrical device allowing (forward bias) with far greater east (reverse bias). 2 ‘The most common type of diode in modern cireuit design is the ‘semiconductor diode (p-n junction). ‘Adiode is a two-layer somiconductar consisting ofp type semeonduston A diode and n-igpe semiconductor material the equivalent cireuit of ‘diode is shown in Fig. 1-11. surrentto low only inone direction fe than in the other direction switch ty ee Diode symbol Equivalent circuit Pie 4 tman equilibrium, pa junto the fee lctrons rom the n-type rein Inan oglu janctin 0 the ptype ide where hey wl wail aifase ros fan ols into aye material: Seman recombine wih omen juntion inthe oppeite iestion and recombine. eatin of roo electrons and hates in the 5 The soommbinaion row rgin on either side ofthe junction hay Suneton leaves mie. Tha arrow roqion which bas ees. dtd ona age neale the depletion ave. 7 aaa cr aetrans iat te prep wear patel cree nt Chana) in the mean 17. Similarly, diffusion of hl Teaves fixed ions (acceptors) tes near the pn interface in the n-type region ) with negative charge:incering 184 omg» Fundamentals of Blectronies Eng! tho pn interfaces lose their neutrality and becgn, 8 The regions ney ape eharge region oF depletion layer, charged, forming the space a pe oe tefeb ee ef Migated | oe leet ——— Miratad from n-type A Hi from p-type. Potential barrier /Contact a potential s o Space charge” | region or depletion layer Fig. 14.2, junction semiconductor, 9. This separation of charges causes an electric field to extend aeross the depletion layer, A potential difference must therefore exist across the depletion layer. Que 12. | Explain the VA characteristic of p-n junction diode, ‘Draw well labelled characteristic. 1. Forthe forward bias fap junet : apnunetion,pypois connected to the poste terminal white the ype to negative terminal obttery. 2 The potential canbe varied ith potent potential ca arid wth tential vider, Aone forward vl (05 tors 3 Vir te ptental hae eit nd current starts owing This voltae iv known as Uvesbol oF kee voltage (Vj). — a A the rwanda pl neroseeyand enol ols he forward current exponentially ws shown in Fig. 12.1 4. Beyond certain ae val ‘ala, reduce un eriemely larg cat which may destroy the junction due to overheating. vie Revert ban 1. ‘Tha pgp in connected to th connected to the positive terminal of a battery. a Sahin in ona 7 bigh and pracy 1-45 (Sem-1& 2) ‘Semiconductor Diode ‘the order of wA flows in the cireuit due to 3, Inpractical, a small current of The reverse current minority earriers. Ths is known as reverse current is shown in Fig. 1.2.1. Reverse bins region Volt-ampere charactoristies of p-njunetion. A. Asthe reverse bias is increased from zero, the reverse current quickly ‘iso to its maximum or saturation value. ‘The slight increase is due to {Impurities on the surface, which behaves asa resistor and hence obeys chm's lave. Ths gives rise to a current called surface leakage current. Ifthe reverse voltage is further inereased, the kinetic energy of electrons becomes so high that they knock out from the semiconductor atoms. At this stage, breakdown ofjunetion gecurs and there is a sudden rise of ‘reverse current, Now the junction is destroyed completely. 6 ‘Thus, p-n junction diode is one-way device which offers alow resistance swhon forward biased and beliaves like an insulator when reverse biased. "Thus, it ean be used as areetifier i, for eonverting alternating current, into direct current, Quets. |] Sketch and explain ideal and practical V-1 characteristics of a p-n junction diode. Fig. 1.24 | ‘Answer A. Ideal V-I character 1. Anideal diode is perfeet eonductor in forward bi in reverse bias, 1d « perfect resistor \go characteristic of the ideal diode is shown in ‘Tho current-volta Fig. 1341Fundamentals of Electronics Engineering 1-55 Sem-1& 2) Ip Ve Ve Ik Fig. 1.2.1. Vi characteristics ofan ideal diode, reverse biased, it acts as an open circuit. oz rae et saa the voltage across the diode is zero and in the reverse direction, the current through the diode is zero as shown in Fig. 1.3.1. * ae Pr uy nen ee Sis enaan oie inte wr se ee Fig. 132. ‘The resiator approximates he semicond ON/OPPertrwrdtialeereba Aron leon leakage current. a ‘Semiconductor Diode a 1-64 (Sem-1&2) we “The ability of diode to withstand reverse-bias voltage is limited, Ifthe 4 ‘opliedreverse-bins voltage becomes to large diode will experience a jzavy conduction known as breakdown, which is usually destructive, Tho current versus voltage curve for areal diode looks ike as shown in Fig 132, PART-2 ‘Zener Diodes Breakdoun Mechanism (Zener and Avalanche). ener es (Questions-Answers ‘Long Answer Type and Medium Answer Type Questions es_e—oas aa Gaeta | Explain the V-l characteristi ofp-n junction diode, How iis differfrom zenerdiode? [ARTO 2016-17, Gems Marks 07 OR Explain input and output characteristics of zener diode, (ATU 2016-16, Sem-I) Marks 05] == 1 junction diode t Refer Q. 1.2, Page 1-3, 1. Zener diode is areverse-biased heavily oped p-n-junetion diode which Peperated in the breakdown region. Fig. 14.1 shows the symbol of zener diode, o——_f aa Cathode Anode ig. LA.1. Zener diode 2. When a zener diode is forward biased, its characteristics are just same ts the ordinary diode and itisshown in Fig. 1.4.2 8. When ener diode is reverse biased then it gives constant current up eenstain voltage, When the reverse bias voltage is increased beyond that voltage, the current increased rapidly as shown in Fig. 1.42Fundamentals of Electronics Et ronies Engineering 1-75 (Be mel & 9) Ima) Fig. 14.2. VI characteristic of zener diode, ‘The cut-off value of voltage beyond which zener diode tft reverse current ly ical srr vologe Vor reo “ ot reakdown vgs Tebrehdon orsnr hag doped pon he amt zener diode can be used as a voltage sic voltage toa load. Difference: S.No| regulator to provide a constant ‘P-n junction diode 1. | The electricity flows in one direction, 2 [the Teverse as The reverse bias mak permanently damages the | elect Footn‘the Eipledon regions "| Sieon 7 10" 1 both the Zener diode ‘The cletrity ows nb direction. s 3. |The width of depletion ‘region is large because the p and n region is lightly doped, 4. | Thisis used for retification ‘The width of depletion region ia narrow becaute the p and Fesion sesvily doped This is used for voltage ‘regulation, a ‘Que 15. ] Explain reverse breakdown of a diode. =a Te mah ince 1-84 (Sem-1& 2) ‘Semiconductor Diode 185 Gente) CC 44. Tener current is independent of the applied voltage. It depends only on ‘the external resistance. ‘This breakdown is called as zener breakdown as shown in Fig. 1.5.1 ‘This breakdown occurs at low voltage. ‘ener breakdown valanche breakdown | a Fig. 15:1. The LV charncterstis comparison ‘between zener and avalanche breakdown. fi, Avalanche breakdown : 1 Avalanche breakdown takes place in slightly thick junetion than the Zener breakdown case, Itmeans both sides of junetion are lightly doped. 2, Inthis case, the eleetriefeld across the depletion region layer) isnot so trong to produce zener breakdown for the same applied voltage of ‘zener breakdown case. 4, Here, the minority carriers accelerated by the field collide with the semiconductor atoms inthe depletion region. During collision the kinetie energy of electrons is transferred to other ‘covalent bonds, thus the energy transferred to covalent bonds increases {he band energy, hence covalent bonds are broken and electron-hole pairs are generated, 15, The newly generated carriers transfer their energy to other covalent ondeand break more bonds and thus extremely large numbers ferries tre generated due to cumulative process of avalanche multiplication 6 This breakdown is called avalanche breakdown as shown in Fig. 5.1 ‘This breakdown occurs at higher voltages. PART-3. Diode Application : Diode Configuration, Half and Full ‘Wave Rectification.Fundamentals of Electronics Engineering 1-95 Gemigs, 7 ————— ‘Questions-Answers ‘Long Answer Type and Medium Answer Type Questions Que 1.6. | Explain the series diode configuration. =a 1 The series circuit of diode is shown in Fig. 1.6.1 2 The state of the diode is first determined by replacing the diode with a resistive element as shown in Fig. 1.6.2). The resulting direction offic ‘2 match with the arrow in the diode symbol and since Z > V,, the diode isin the “ON” state. The network is then redrawn as shown in Fig. 1.6.2(5) with the ‘sppropriate equivalent model for the forward biased silicon diode, Si E REV, Fig. 16.1. Series diode configuration. 4. The resulting voltage and current level are the folowing Vp= Vg Vaz EB Vg Yq Fig. 162. 5. In Pig. 1.6.3 the diode is reversed. Replacing the diode with a resistor clement as shown in Fig. 16.4 (a) will reveal that the resulting current direction does not match the arrow in the diode symbol. 6 The diode is in the “OFF” state, resulting in the equivalent circuit of Fig. 16.41, 4-105 Gem-1 & 2) Semiconductor Diode ee iar RE +—_ Fig. 1.6.3. Reversing the diode 4, Due to open circuit, the diode current is 0 and the voltage across the for Ris the following. sitar Fe TE Rap R=(0AR=0V 4 Vy-E 1S 1B Fo Fig. 164. Gael. | Sketch Vo for given cireuit configuration (Fig. 1.7.0. av yo} Vo 22k0 si Fig U7, ‘ARTU 2017-18 (Gem-I), Marks 35, Answer : 1. Assume input waveform, svt -sy| - Fig. 17.2. 2. For positive half yelo. Diode isin ON state. 2 For posit ve Det re ive half eyele Diode is in OFF state, For negative alee, Dade isin OFF 9vFundamentals of Electronics Engineering 4. Output waveform as shown in Fig. L7.3. Vo mv ot -9V Fig. 173. IB | For the circuit shown in Fig. 18.1, determine I, 1, av tov an V,=07V oltage across kA, aplying KVL in oop Ly, -¥,+10V-07V-07Ve0 ‘ | We have, TALI oma gy ly lyly y-12J (Gem-1& 2) Semiconductor Diode ppl KCL at nde Ty = ly 12dabmk GATT bapain the working of half wave and full wave bridge ” OR Explain the bridge rectifier with clear diagram. ARTO TOTTI Gen, aria 3 on . raw the circuit and discuss the working of full wave bridge rectifier Pith suitable input-output waveform. (AKTU 2016-17(Sem-D, Marks 05) oR With help of neat circuit diagrams, explain the working of a full ‘ARTO 2020-21 (Sem-D, Marks 05 1.075 mA~0.1167mA wave bridge rectifier. “Aas [A Half wave rectifier :Its shown in Fig. 1.9.1. As the name signifies only half portion of input is rectified either positive halfor negative hal). Only ‘ingle diode and a step down transformer are required for this cieuit, mov Fig. 1.9.1, Half wave rectifier cireuit, ame output voltage ‘Wo ak Fig. 1.9.2, (a) Input voltage (6) Output voltage.Fundamentals of Electronics Engineering 1185 Gem a, DB RulsovreiiersTscritmt start oege taken enei Tee me i. Centertap rectifier: Input o— AC voltage 220 50H: De Fig. 1.9.3. Full wave rectifier, 1 fncentr-tap fll wave rectifier drei diode D, will be ON for positive halfeyce (2,00) and the diode D, willbe OFF during this a’ 2 Fhe diode D, will be ON for negative halfeyee (x to 2x) while the Aiode D, willbe OFF during this eycle, 3. The input and output waveforms are shown in Fig. 1.944 Bean ay Be Yig: 1.94. (a) Input wavetorm (6) Output wavelorm ih, Bridge reetifie 8 fhe i Pig. 18.50) only diodes D, and D, are ve half eyele while diodes D, and D, afeastics for total outer ela shown in Fig. 19.500 and Fig 19'S # active diodes for aparticular eycle, Semiconductor Diode 1-144 (Sem-1 & 2) ectifier cteuit Fig. 19.5 (a) Basi bridge ret 1) Outpat of bg eter fr postive half eye (6) Output of bridge rectifier for negative hal eye (4) Output fr fall eye Explain the following terms in context with half wav reétifier and full wave rectifier: i. DC voltage and DC current HRMS value of current 1 derive an expression for ripple factor. (ARTU 2015-16(Sem-I0, Marks 75 oR For a half wave recti1:15 (Sem-1 & 2) vse with the help of sn the operation of fall wave Did ithe Esp mei aca tr e real diagretrive ite riple factor and ree (ARTO 2017-18(Sem-I), Marks 07] on " ing of full wave bridge rectifi sw the circuit and discuss the working . liter rectifier? ARTO 2016-17(Sem-I), Marks 07] er [ase OTST, Maver 7] oR 1c. What is the value of the ripple factor Define the term ripple factor. What isthe voice fora half wave rectifier and a ARTO 202021 Gem aa A. Half wave rectifier: ‘Operation : Refer Q. 19, Page 1-12J, Unit-1 , Marks 08 i i PIV: 1. Aswe know during negative half eyele, diode willbe reverse biased and there will be no output voltage. 2 Inthe negative half eycle the voltage across diode will be maximum vihen inpat reaches tits maximum valve V,,. This maximum voltage ‘uring augative half eyeleis called as the peak inverse voltage (PIV). 2. PIV or half wave rectifier is given by PIV-V,, DC voltage and DC current : ‘Suppose currentin load i, =I, sinat, OSotsn (1102) sots2n Here 11,= Peak value of current i, V, ee DC voltage for half wave rectifier, Ving" Ine «Ry, 1.102) (1.103) iv, RMS value of current: ‘The rms value of the current flowing through the loa¢ given as 1-165 Sem-1 & 2) zen ZR aeo 42 Replacing the value of i, from ea. (1.10.1) - Es a0 ae 2 = [AE Joy sin 2a © YaexaP 2 3. Ba (1.104) (AC and DO). {Thus instantaneous value of AC in output is, ge too. Game Yeh =o? aot) (Efe rtie-2itio.do = (+T-2le seme v. Ripple factor: to the DC component inthe rectifier output. ‘The ripple factor is given as Since, Togs =Iy[2 0 Tog If 80, 7 = vi. Rectification efficiency :The efficiency is given as Output DC power Pao. Tnpyt AC power ~ Pye where, R, -+ load resistance, > diode resistance G13 -R 12? (7+) a Semiconductor Diode Yo [[f-sin® o aeons "0 ata] (1.104) (1.105) 7s define as the ratio ofr.m.s value of AC component io F, Rae c nyeeR)__ Refer Q.1.9, Page 1-123, Unit-l ii, RMS value of current i, PIV PeakcInverse Voltage) Is same asthe hal wa thevaloe of BV is difrent fr ceneraptypefulwave eter In centre-tap full wave rectifiers) PIV = 2V,, In bridge type full wave rectifiers| PIV = iv. DC voltage and DC current : 2p, atao = 2f7, sina dan 21, Ine= Seay v. Ripplefactor mw - (Tm) fray, Gz) 4" yar) “ @t,)n R, TW, + BR * ng 4.7V,Vo=4V40.7V 24.7 for V, <4.7V, diode “off” and therefore Wry, v 9 |_47v. ° Fig. 1232, PART-5| Zener Diode as Shunt Regulator, Voltage-Multiplier Circuits. Questions-Answers ‘Long Answer Type and Medium Answer Type Questions ‘Que 124, | How zener diode is used as shunt regulator ? Explain it. ‘Answer 1. Thecircuit diagram for zener voltagelshunt regulator circuit is shown in Fig. 124.1. 2, The zener diode is selected with V, equal to the voltage desired across the load. Under reverse biased condition, voltage across zener diode practically remains constant, even if the current through it changes by a large extent. 4. Under normal conditions, the input current /, resistor R. The input voltage V, can be written as Vie LR+V,=U,+1)R+V, ‘When the input voltage ¥, increases, asthe voltage across zener diode remains constant, the drop across 2 will increase with a corresponding inerease in I, +1, f, +1, lows throughFundamentals of Electronies Engineering 188 bemg = Lee 4-34d (Sem-1 & 2) Semiconductor Diode +h Tt Tt Ts y, vA’ y, 3R, 5ka ; * ° cover Ry = 10kQ \ omy ig. 1241 Zener valtag glate. eae & AsV bsacontenthevoligeacitholndwlaloremainscouig, & PIRI oy and henee, J, will be a constant. ‘omstant Trin Rs = Vimin 2. Therefore an ineresein 1 wl renin nnerese jn, which totale tb voliage soso Ths, eer dots oed conse 8 oe ‘Similarly, ‘To operate zener diode as voltage regulator, the reverse voltage appli tozener dode never enceeds PIV ofthe diode andat the sameting ye, 4. Fromen- (1.25.0, applied input voltage must be greater than the breakdown voltage et the zener diode, Que 125. | For the circuit shown below, determine the value of maximum and Teis) I, Similarly, Tpimani= Fan H, = 14-5 =| yum zener diode current. BE enpcarjunl omen eet : s. 1.26.1). Wit Se saageringotthesenr dite ie) for given circuit. : * 1k Tp i Vir50V Ry, Fig 1.36.1. TARTU 2017-16(Sem-I0, Marks 07| 1. Apply KCL at node A, Given isk+h, Iau 32mA, V,= 10 ‘To Find : Range of, and R,, maximum wattage. From the cireuit, (1.25.1) 2 Wehave, where, ‘Therefore,3. Hence, 25k and 4 Let Ipgig= 8:08, Therefore, 1, = 40mA-5 mA = 35 mA Yo. Bina 7 OF 6. Hence, the range of is between 8 mA and 35 mA w between 285.72 0 and 1.25 ko. llerangeote, Maximum wattage of zener diode is, Tons Ve SE mA % 10 =320 x 10 = 0.92 Ww Que127.] Determine the range of V,for the Fig. 1.27.1, that iy ‘maintain the zener diode in “ON” state, Fig. 1271. AKTU 2017-18 em-D, Marks 35 “Answer Given :Ry=2200, V,=20V, lige = 60 mA, Ri, ‘To Find : Range of, Lo Wetme eM R T2097 2 6, Vina = 2 RM, inn = ek Vina = 28679 3. Now cures rough, ie 1 lang’, #60 + 16.67 = 7661 mA Rg Vz= 16.67 x 10-9 «220 + 20 Vigan = 9687 Hence, th range of the range ofV, es between 23.67 V and 36.87 V 4-86 (Sem-1 &2) Semiconductor Diode Guo IGB. | Draw the V-I characteristics of zener diode. Determine the network of figure given below, determine the range of V,, that tre maintain V, at 8V andnot exceed the maximum power rating of Thezener diode. Fig. 1281. Taewer | A. Vi characteristics of zener diode : Refer Q. 1.4, Page 1-6), Usit-1 2.0, V,= 8V, Prous 400 mW, Ry = 0:20%0, ¥, =8V. 1 (m+ R).V, @2KO» 0.22%00.8V ago Vina = ORF 8028 . j098V 720 3 Wo know, 400 1 = Tog BV 400 mW 50 mA av 4. Now, current through Rs 1 Tomas SmA-+ 9636 mA = 86.964 = IR + V, = 86.36 mA «8248 08+ 8V=1508Va Fundamentals of Electronics Engineering 1879 (em, & 2) Describe with the help of cireuit dis = 1-98 (Sem-t & 2) Semiconductor Diode sala ten Gerona © ENT cr Enplan with suitable cut that how diode ats as «yay Were ubleeaiea dy OR ‘ge During second negative halfeyele, the diode D, and diode D, are ON, The ‘Draw and discuss voltage tripler circuit, rata eT) + Veg + Veg Vag~2V q+ Vg Vg |AKTU 2016-17(Sem.1 Mm _RRRReTTT we Define Voltage Multiplier. Draw the cireuitand e Yaltage across C,-> Voy = Vu of Voltage Tripler and Quadrupler circuit. Voltage across Cy» Ven" 2¥ Veltage across C > Vea=2V5, ABTU 2021-22 (Sem-D, Marka To] voltage across C, > Ve,= 2Vy pln the working “Anaw 4 Voltage doubler : Taking output across C | 1 Fig. 1201 sh of Veu=2V, 1 shows a circuit of general multipier this eire one, obs euler, ler and udroplr uit cane g-Yoltage tripler: Taking output across C, and C, ireuit-is a Doubler (2V,) Vee Vex Vea" Vu Vy * Vy Cireuit-2is a Tripler 3V,) 10, Voltage quadrupler Taking output across C, and C, Gireuit-is a Quadrupler(4V,) Vine Vou Ver=2¥q + 2Vq=4Vq 3. During positive half yee the i ‘i : gor, Dering postve inf, hediotsD is ONanditchargescapactrc, ‘ae SH] What is voltage multiplier using p-n junction diode ? 4. Inthe first nogative half ye plain the operation of voltage doubler the diode Dis ON and a Wy Weg = Vy + Voy = 2V,)- In , is ON and it charges C, to : is 2 a Vs Ven= 2) inthis cyl th charg on capaci’, Balin full wave voltage doubler with eons ner TARTU 2017 16(Sem-1), Marks 3.5) oR With help of a neat diagram, explain the working of a voltage doubler cireuit. ARTU 202021 GemD, Marks 05 Taswer 1ethe circuit which produces a greater DC output voltage than ACinput rae ing a eciner eu alld as volnge ltl. 2. Thooutputooltage doubler is twice the peak input voltage. 4. There are two types of voltage doubler ciruits Hatt wave voltage doubler 1. Thecirit gram is showninFg. 1302 2. Dring postive hal-yte ofthe inp voltage diode Dy One Gide , iorevert based (OFF). Capacitor Soak value ofsocdndary voltage V, wit plants forward biased eharges to the M¥e- 1.20.1. General circuit diagram of multiplier.B i Input Fig. 1.80.1, Half wave voltage doubler, During negative half-cycle, diode D, is ON while diode Capacor Cy is charged up fo vole Le, th sum voltage and the vatage across C, e During positive halfeyele: Voy = Vy Ya Ve=Ve Cy sia Ld o mah Semiconductor Diode | ig. 1.30.4, Full wave voltage doubler. PART-6 ‘special Purpose Two Terminal Devices : Light Emitting Diodes, Photodiodes. ‘Long Answer Type and Medium Answer Type Questions Que 131, | What is LED? Giveits principle of working, construction ‘and applications. Fig. 130.2, During negative halfeyele: =Vy— Vert Veg 0 ~Vi- Vi Veg=0 fea Vm Vou Gg Dy | Vm Dy *F Yo Fig. 1308, Full wave voltage double ‘There dagram shown nig. 1304 ring postive half eel, Dg ON ane Simpelic yen a # ON and, is OFF. Capacivar wit : Vor" Vy, herp aaa Ds i OFF and Dy is ON, Capacitor C, charge iu Vi Thus output voltage sith no load connect aie age with no load connected Aaewer | [A LED: LED is specil type of semiconductor that under Coed iad mits external radiations in llravilt,vsble and infrared Stelons of electromagnetic spectrum. B. Construction of LED : 1 LED is just not an ordinary pon junction diode where silicon is used, Tie a tse compound heving elements Ike gallium, arsenic and Tabephorus which ave oemitransparent unlike silicon which is opaque 4 Inallsemicondvctor pn junctions ome ofits energy willbe given offs heat and some inthe form of photons 4. Inthe materials sucha gallium arsenide phosphide GaAsP)or glia ietehide (GaP), the numberof photons of ight energy emitted is aclent to create avistble light sourceFundamentals of Electronics Engineering Recombization epee ne 1-415 (Sem-1& 2, Principle of LED: ‘The process involves Generation of electron-hole pair (EHP) by excitation of semiconductor, Recombination of EHP. Extraction of photons from the semiconductor. ‘The characteristic for LED isgiven in Fig. 1.31.2. sited ght) Tae it Fig. 131.2. Characteristics. Working: ‘When LED isin forward bas condition, the electrons from.n-type material ‘ross the p- junction and recombines with holes inthe p-type material ‘When recombination takes place, the recombining electrons release ‘energy in the form of heat and light. ‘The emission depends upon the type of material, 2, GaAs - infrared radiation invisible) GaP + red or green light (visible) GaAsP -» red or yellow light (visible) ‘Applications: Display LEDs like calculator, digital locks et. Light soureein optical fibre communication. Light source ina ours 1 source detector package ike smoke detectors, tachometers, proximity detectors ete. pee ee detectors, og Semi &2) Semicondetor Diode Gueis.| What is photodiode ? What are its different types ? seneribe the construction of« photodiode with its operation. Rewer ‘A. Photodiodes: 1) Tuoteminl eve devel reed to hotnshrtion ar 2 Photodiode na emiconustor junction device whove operations Fimited to reverse bias region. 44. The typesof photodiode are i. pondiode i, pin diode fi, Avalanche diode xt The output current ofa reverse bias p-n junction changes when device jsexposed to illumination. 15, The ariation in the output current flux. The construction and symbol is shown in Fi LPs linear with respect to luminous 32.1 16. This diode is designed in such amanner that the rays are aliowed to fall “only onone surface across the junction, The remaining sdesare restricted forthe light to penetrate. 7. As the temperature due to illumination increases, more and more ‘lectron-hole pairs are generated and results in increasing the reverse saturation current, When light rays fallon depletion width ‘Wit creates electron-hole pair and electrons are swept into n-region and holes into p-region very rapidly. ‘This gives rise toa photo current. Thisis the basic principle of eperation of photodiode. B. Photodiode characteristics : 1. The Fig, 1.32.2 shows the V-I charueteristis of pn junetion photodiode with different illumination level143JSem1&2 | yy Sem 2 oe | “Tis is also called as varicap, Fundamentals of Electronics Engineering SemicandectorDinde ‘WC (voltage variable capacitance) or L tuning diode js constructed from semiconductor material, ts 2 Trendent variable capacitor. als Ris simply a voltage ++] eool- - + +10 e@ol- - 2 When no light ray is incident, the diode has a small reverse current |, known as dark current 3. The dark curren is that current which exists only with no applied slminstion, 4. The increase in reverse voltage doesnot inerease the reverse current significantly because al available charge carriers have already being gua ‘swept scross the junction. en - 15. The photodiode can also be used as a variable resistor controlled by Lb aerer id raractor diode in reverse-bias condition and symbol light intensity. 7 V+ larger, then W-> wider 6. Photodiode operates in quadrants i, in third quadrant, both and V V— smaller, then W-> narrower Rarer ead parce s being delivered tothe device from external 4 The depletion region W inthis case acts like an insulator preventing the reat. conduction between then and prregion ofthe diode just lke a dielectric 7. Infourth quadrant, Vis positive and Fis negative and power i delivered which separates the two plates of capacitor. from the junction to the eternal circit. In applications, usually third 5. Let areaof plates = ‘quadrant operation is preferred. stance between two capacitor plates=W then capacitance, C ae where ¢ = Permittvity ofthe semiconductor materials Varactor Diodes, Tunnel Diodes. I (Questions-Answers Ik - ———— so ‘Long Answer Type and Medium Answer Type Questions We ee Fig. 1.382. Varieap characteristics : GQueiSs.| Explain input and output characteristics of varactor + ily 188.3 hows the characteristics of a typical commercially available — ETT DOTS Teo Ty Marin 06) * We se tha there th nil sharp deine in wth erent odes is limited to about 20 V. a The normal range of V_ for varactor di sven Write a short note on varactor diode. ETO. 4 Interms ofthe applied reverse bias, the transition capacitances [AREU S696-81:ese-D, Marks 06 | approximately byFundamentals of Electronics Engineering 3-485 (Sem-1 & y ASS eeTINE ABT Bemt a, k Wr Var where k= Constant determined by the semiconductor material any construction technique V;= Knee potential ‘Vg = Magnitude of the applied reverse bias potential = 2 for alloy junction = V8 for diffused junction. CytpF) Cr 100) (cw) 80 60; 40 va) at ote Fig. 1.33.3. 5. _Interms ofthe capacitance atthe zero-bias condition C10), the capacitance ‘a5 a function of Vi is given by om __ a, ‘QueT34] Explain working and characteristics of Tunnel diode with the help of neat diagram. [AKTU201616(Semd =10 -12 Discuss the construction and working of tunnel diode, Also sketch its I-V characteristics and explain. [ARTU 2017-18(Sem-I, Marks 07] OR Explain principle of operation and construction of tunnel diode. Draw its VI characteristic, [ATU 2016:17(Sem-I), Marks 525) OR Explain the V-I characteristic of tunnel diode, [AKTU 2016-17(Sem-D, Marks 065 oR Explain tunnel diode. [AKTU 2017-18(Sem-D), Marks 3.5) Principle of Operation of tunn A Torandbiny conte 28D; : z serra cnn Sethe i site dd arte eens horn By os Fig 1341, ‘The thin barrier across the junction would permit tunneling of electrons Increased, more such fife states onthe msde wold come opponite to eee SoSSace eee Sonietcmean iremareonons shoeing meas aenetanae = a a ‘component of current would dominate. This results in normal p-Fundamentals of Electronics Engineering pas Cuma ii In reverse bias condition + L ‘When negative wotag sapped tothe psde wih respect ton-side, this would push Bp above Be ait thet copration being equal the applied voltage. ied filled states of the p-side are directly opposite to the empty states on then-side crocs avery narrow baie, ‘Thus, the electrons from p-side would easly tunnel through this thin Gel rd the process of quantum mechanical tunneling and appear on these. Since electrons travel from psie to n-side, the actual direction of current is fom nto pide and is nopative, ‘As the amount of revere bias is increased, more uber of fled states {nthe valence band ofthe p-side would appear opposite tothe mamber ot empty states inthe contin band of tesa the vere Coren ‘would keep on increasing wth verse vltage. B. Characteristics : Due to heavy doping, the cursent in negative resistance region is suddenly increased witha smal applied voltage Due tothis reduced depletion region the sudden ineeasemewreat (at small applied voltage) sealed “earrier punching through fet. 2 —-y Negative resistance region 1_//___ Superimposed semiconductor 7 diode characteristic ye N, PE Oe poses te 2 Vp W. Fig. 1.342. VERY IMPORTANT QUESTIONS: Following questions are very important. These questions ‘may be asked in your SESSIONALS as well as UNIVERSITY EXAMINATION. 1-48 (Sem-1& 2) “qu: Beplain the va carsaena so well labelled charaeterisgiot® fn j ame Refer 12, : 2 Explain the wor rectifier. aw Refer @ 13, Q3. Explain the followis | rectifier and full lunetion diode, Draw ‘king of halt ‘wave and full wave brid; ridge ‘wa Refer Q. 1.10, Q4. Sketch Vo: ch Vo for each network of Fig. 1, ; c for the input shown, 120 Fig ame Refer Q. 1.16. 7 @5. What do mean by clipper? Draw the oa the given circuit. eieretern of ov y, ° ‘Ang Refer Q.1.22, @6. Describe with the help of circuit diagram working of voltage tripler. ‘Aus Refer Q. 1.29. Q.7. Whatis voltage multiplier using pn junction diode ? Explain the operation of voltage doublers. ‘Aum Refer Q. 1.30 Q.8. Explain working and characteristics of Tunnel diode with the help of neat diagram. ‘Anu Refer Q. 1.34 900UNIT Bipolar Junction Transistor : ‘Transistor Construction, Operation, Amplification Action Common Base, Common Emitter, Common Collector Configuration Field Effect Transistor Constructions and Characteristice of JFBTs, Transfer Characteristics MOSFET (MOS) (Depletion and. Enhoncement) Type, Transfer Characteristic 21d Bem-1 & 2) gp Sem-1 &2) CC AANA BsTand FET Bipolar Junction Transistor: Transistor Construction, Operation, Amplification Action, 1 2-25 to 2-75 2 215 to 217 * b 3 2185 to 2260 4 6 6 1 E Questions Answers ‘Long Answer Type and Medium Answer Type Questions Fake [Bevin sr, mel ‘A janetion transistor isa three-layer semiconductor device, sandwich “of one type of semiconductor material between two layers ofthe other type, ‘There are two types of transistors npn transistor and pnp transistor. ‘When a layer of p-type material is sandwiched between two layers of notype material, the transistor is known as npn as shown in Fig. 2.1.1, ‘Similarly, when a layer of n-type materials sandwiched between two layers of p-type material, the transistor is known as pnp transistor as shown in Fig. 2.1.10. ‘The transistors are made either from silicon or germanium crystal, ‘The symbols for npr and pnp transistors are also shown in Fig. 2.1.1, Collector miter 2 : ° ake (eset ctopm tani caller : saitee—[> BD 5 he BiSatare fpr 8 Pig 2h [A transistor has three doped regionee Fundamentals of Electronics Engineering Ne“ Emitter i. This isthe left hand section ofthe transistor. 4 The main function ofthis region is to supply majority charge carriey, (either holes or electrons) tothe base and hence itis more heaviy oped in comparison to other regions. he Base: i. Themidte section ofthe transistor is known as base This is very lightly doped and is very thin (10 ~® m) as compared ty cither emitter or collector so that it may pass most of the injecteg charge carriers tothe collector © Collector i. Theright hand section of transistor is known as collector. 4% The main fonction ofthe colletoris to collet majority charge carrie ‘through the base. This is moderately doped. =] 28d Sem-1gy 8. The arrow head direction indicates the conventional direction of current flow 1c. in ease of npn itis from base to emitter while in case of pnp is from emitter to base. 8 ‘In most ofthe cases the collector region is made large due to the fact ‘uatcollector has to dissipate power, med 10, The junction between emitter and bese ma; Junction and junction between collector-base junction, 3 becalled as emitter-base Base and coletor maybe tlleas 1. The emitter-base junction is ways forward biased w collector-base junetion is reversed biased. ores 12. Theresistanee of emi talons unc ter-base junction is very small as compared to 18. So, forward bias applied to emit ter-base junction is gene smal and reverse Mason collclon bone a ees T-base junction is mueh higher: Qe22. | Describe the operation of pnp transistor, aa io transistor with miter ba base junction ed junction a revered need 2 Thebes tp pregon sre repel ‘towards the base. en % The potential barrier “ hat doles crn haan nb Bon is edaced a forward Ue miter current 08 ad penetrate inton:regen ema ‘ A 'gion, constitute The width of bs ive seis ery thin and only 6 4 {ss letronn freon whch enter ie Feombine tutes the current the ae gas emt 8D ByTand FET Te remaining holes areable to drift across the bare and enter into the Mieetor region. seliecure swept up by negative collector voltage Veg. This constitutes Pe acre 2 oe = 6. Rig 22.1 Operation of pnp transor. 1, As holes reach the collet, electrons are emitted from the negative ferminal of battery and neutralige these holes Now a covalent bond near the emitter electrode breaks down. The Tiberated electron enters the positive terminal of battery Vey. This process is repeated again and again. 4, Asthe width ofthe base region is very smal, the rato of hole current to electron current is very large 20 the elestron eurrent may be neglected 10, Thusonly the hole current plays the important role in the operation of pnp transistor Que23. | Explain the operation of npn transistor. oR Explain various current components in npn transistor with help of suitable diagram, ‘ARTO 2016-17(Sem-D, Marks 05 a 1. The biasing of mpn transistor is shown in Fig. 2.3.1. 2 ‘The emitter base junction is forward biased because electrons are repelled from the negative terminal of battery Vg towards the junetion. 8. The collector base junction is reversed biased because electrons are flowing away from the collector junetion towards the positive collector battery terminal Veg. 4 ‘The electrons in emitter region are repelled from the negative terminal of the battery towards the emitter junction,