Unit 3 important topics, Summaries of Analog Electronics

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js not possible to discuss a) all i it p yieations e of the . are use to perform on following wavel tection k of jgnal at a desired dy tive © tive PC ’ ari 5: the ott pand we can say that a clampin adds Ce one: to the sign@! divided into two groups namely ut a sine wal ; ° t and pushe Fig. 2-27- pe of original signal has not changed : only ther wile a positive clamper- ne neat, s {a e that the sha uch a clamper is ¢ yawards sO he signal dov that positive pea clamper circuit. peak-to-peak or rms yalue of i rhe per about the ped output has does not change the tt ig: 2,27, the input wave ov in this case. ri value her 2 ranges the peak and average value © ef 3 clear that input wavel fa waveform F OY P orm. has a peak value 2 J form and clam Ce Scanned with OKEN Scanner visio eatin (PPLICATIO NS - 109 and average value over a cycle is zero. The clam ; . ped outpu Fl petween 20V and OV. Therefore the peak value of litrnpmeel Meee is 204 18 + erage Vi .c, value © _—* i and average value (or cc, value 2 10V) ig 10V, Mence we conclude that a clumper changes the peak value as well as the average value of a waveform. 7.10.1. Positive Clamper ; fig. 2.28 shows a circuit of a positive clamper. Vad Cc i i . oT — © ° m2 T ' Yo ia} z A, Ves =-V 5 ; Fig. 2.28. Positive clamper. Here the input signal is assumed to be a square wave with time period 7. The clamped output is obtained across R,. One things should be remember that for the Proper operation of the circuit the charging time (1, = R, x C) where R, is the forward resistance of the diode and C is the capacitor connected. Between the diode and input supply is small as compared to the discharging time (ty = R, x Cj}. Where R, is the value of the load resistance, this condition is based on the fact that voltage across the Cpacitar will not discharge significantly during the interval of diode in non-conducting. Unstone the discharging time is deliberately made much greater than the charging ime. Operation (] During the negative half cycle of the input signal, the diode is forward biased. Therefore the diode behaves as a short circuit (S.C.} as shown in Fig. 2.29. Under this situation the capagitor will charge to V volts very quickly. From figure it is clear that during this interval, the output voltage is directly across the short circuit. Therefore V,,, = OV. : ee c [ alk —o+ f 4k T O+ ney [se AL Vou VeeV" [~~ Os. AL Vos ——O- o- Fig. 2.29, Fig. 2.30, C® scanned with OKEN Scanner g Fig. 2.33. 7 Fig. 2.34. , npring the negative half cycle the diode is reverse biased and behaves as an @ qircuit a6 shown in Fig. 2.34. Since the discharging time (c= R, « Cis in od of input signal, the capacitor almost remains : than the time peri ing the off time of the diode. Applying KVL io out Fig. 2.35. Vos’ “4 - hat are the main application areas © if Mas i the need of rectifiers ? ; ("Deine the verm spectification” , ouit™ > os OWN the diodes are used in rectifier ani fiers 2 — ecilié cave! . a ‘ : res scanned with UM ie oe ys oN —— as SLAC PROS AYV geet! eye ANE 0.04 mA a. “ Fp vA = fh ye 404 Sol. whee Conrent ampli - coy the wattie HP artes 1A wes ors M Exagmypte tt ie ace, + curren . pind the Base f an in Fig. _ gumtion js shawl in rif Sol. The circuit for CR cont iid . 5K S$ 4-9y 49V 2,98 mA . le= FQ = + 7 ies =| We know, OF et | ; 2 mA Fig. 11.13. Ea (0.98 i sict ts) Ip=lg- tc (since JE Ic ® =1-0.98= 0.02 mA (Ans. . ctions, the emitter current is 1 mA. If collector current is ne . Example 11.4. In a common base conn eollector current’ vhen 0. = 0.95, Also calculate 26 WA when the emitter circuit is open. Determine the ota the base current. Sol. Total collector current, Ic = ole + Icgo where Ip=1 mA; a= 0.95 and Icgo = 20, pA = 0.02 mA Ice = 0.95 x 1+ 0.02 = 0.97 mA (Ans.) Ig = Ig — Ic (since Ip = Ic + Ip) = 1-097 = 0.03 mA (Ans.) Example 11.5. The emitter current Ig in a transistor is 3 mA. If the leakage current Icgo is 3 pA and o. = 0.98, calculate the collector and base current. Sol. We know that, Ie alg + lego wie a= 0.98 ; Ip=3 mA and Iego = 5 LA = 0,005 mA . Ic = 0.98 x 3 + 0.005 = 2.945 mA Now, IgsIo+Iy ie : Ip = In . : ~1e=3- 2.945 = 0.055 MA = 55 WA (Ans) 11.13, TRANSISTOR CHARACTERISTICS Oniy @ (current amplification factor) of a transistor . . A i wan does scribe i . ae electsical behaviour of a transistor can be described by studying the describe its behavious. The complet voliages of (hat tansistor. These selations can be displayed convenie tipo between various currents and are known as the characteristics of a transistor, Althongh mai AY on Sraph and the curves thus obtain ca Pea ; ai VMavy sets of for a given configuration but the most important are fput charact As of characteristic curves can be piotle these two sets of characteristics completely describe the bet eristics and output characteristics. 10 fact, ° hay , Vviou . i T Of a Wansistor, =| Gs Scanned with OKEN Scanner SMI: DEVIC AND CIRCUDTS men chat acteristics + NCB configur Se yy vo I Se vo lane Ven at CONStANE emitter ci Characteristic Curves Aiffereny Stttings of 16 shows We Pe ee ime SIS of 4 lypieal PRP Aransistor a8 Con ‘uration, The following Points from these characteristics ing : 2-0 ® In We active regi 3 BON, where Latstorbase imetion is Teversc. Sa ETN NASed the Collector Cuirent Ie is almost ' 4 ation, the curve plotted between collector CUrreny \ irrent Ip is called output characteristic, CQ, 1) ye \ Matted c Tp. Fi (ind) ‘ : output 21H feree eee trates eee ate. 1-0 05 Ty large 1e=0-SmA le=0 mA eee 5 -6-7 5 _ cur oF neg (ON——___} its value is ve; Ows in the collector circuit, and its value ranges c OWA ili i _ Output resista, i collector current i? Output resistance CB gs Tt may be seen al constant J Alc Ee from the Output chara Aguration is very i Put resistance of a pnp ansistor in CB Configuration from the in Fig. 11.15 at aq Point Where fp = 1.9 mA and Vep = — 10 V. Sol. Let a small change in 7p around the given point Jo Ale= 125-075-905 mA From the curve for Vos =~ 10 V, the Cortesponding chap , AVep =0 ige in emitter-base voltage, ie, 1-065 = 0.95 Vv AV, Input resistance, \ EB T= ——* at constant y =-10V Rly ce =20V 00s Ro 100°Q O5 mA Os x 103 ‘This value of 7; is somewhat higher than an amplifier, the operating point lies at the par i i i ted What is expected” In fact, when a transistor is bara ule Uof the curve where it is more Steeper, At that po value of rj will be low (nearly 20 Q) as expected, co @® Scanned with OKEN Scanner dhs MMVPMVLOR TF MITTER Eg NNEC The common Milter gj Uit ar Ber (OR cE, CONE as and 11.17 (b) spectively, ty this a ene TOC nn tranesc MGURATION, : is taken from the Collec » Angemen the inn. a iStor an tr ; : an T and cite : Ne in UCis con, ip ANSiStOr is shown In Fig. 11.17 circuits and hence the DAME Cony LS, the Smiter of ’eled between by mend vi AT @ MON emit Hon ANSistor js common to both lanieend output OF Common emitter Configuration, epee Io Ip INPUT ~ Ae (SIGNAL) NFU OUTPUT (SIGNaty i (2) Common emitter ¢; ‘eg Nec” Mutter circuit H Smit of npn Uansistor () Comon emitter Circuit of pap transistor Fig. 11.17, Base current amplification factor «B) The ratio of output current to input Current in common emitter confi amplification factor, In thi ‘guration is known as base current S arrangement, the outp ut Current is collector Current Je and the input Current is base current th. Thus, the ratio of Be in base current is knawwn as base current amplification factor. It is generally represen ted by Greak letter § (beta), Ale ‘e ar Relation between B and a The relation between B and o can be derived as under : peti - We know, iy ; aoe i Alg Ig=Ic¢+ Ip inl Alg= Alc + Alp A Aly = Alg- Alc or : s in equ. (1), we get, Substituting the value of A/g in eq rt B= aig Alc BlelMg B= AypfAly -AlciAlg 1 ~ or C® Scanned with OKEN Scanner RS 7 Example 11.9. In cr Configuration 175 iret is 5 V. Find the value of base ee pee drop dross 540 j B= 59 “ resistor connected in the collector * _ Voltage drop Resistance Of Re Sol, Collector current 5 lox : C Sho =i mA Now ! Ig= £ (sine B =| B Ip a = ao 0.02 mA (Ans.) Example 11.10. In@ CE configuration, th F J 1 th i is connected in the collector circuit, the volte fiona, supe ee rine eolecor: age d it icy emitter voltage (Vcg) and (ii) base Curent fy "OP across itis 0.5 V. For e = 0.98, determine (i) collector- Sol. (1) Collector-emitter voltage, Ver = Voc ~ drop in Re 210-05 =9.5 V (it) Collector current, Ie 05 V., 0:5 mA 1kQ Current amplification factor, BS __ 0.98 Las Ta” 1-098. Io. 05 1 | Base current, Ine . = 7 = 00102 mA (Ans) | 7 ! Example 11.11. Jn an apn silicon-transistor a 20.995, Tp Ico) = 0.5 WA. Determine Ic, Ip, B andilego. ‘ Sol. We know, le = ale + Icro =, “59,995 x 10 mA + 0.5 pA 95 + 0,0005 = 9.9505 mA (Ans.) = 10 miA; leakage current Icpo (or 4 Ip =le-Ic ' oan = 10 - 9.9505 = 0.0495 mA = 49.5 HA {Ans.) * a 0.995 a . G = = 199, (Ans, Baa T0595 - 2 Ane) Now, Io = Bla + Iceo es or “ego = Ic - Bla = 9.9505 - 199 x 0.0495 = 0.1 mA = 100-HA (Ans) 11.16. CHARACTERISTICS OF COMMON-EMITTER (CE) ieee ke ee To determine the characteristics of a wansistor in CE configuration, the circuit is me . = in Fig 1.19, The tase emir oa Tae sug pono or ona ngs i tage can be varied by adjusting the ; tee | ie oun ci, cons are read from the miliammeters and ao On the basis of these reading the input and output characteristic, curves are plotted on the graph. ‘ceo = (B+ 1) ego = (299 + 1) 05 = 200 x 0.5 = 100 HA . ‘nals ; i yer, for pnp wénsistor the termina ‘ ircui in Fig, 11.19 is for **The circuit arrangement shown : a anes of all the batteries, miliammeter and voltmeters will hav ‘ Cc Scanned with OKEN Scanner BLEC TRON Ee DEMICHS ARD CULCUETS AT ( Fy. (1.19, ; 1. Input characteristics : In CE configuration, the curve blotted between base CUFTERL Ty ap b emitter Voltage Vag at constant collector-emitter voltage Vor is called input characteristic, Ae. To draw ; the input characteristic, note down the Teadings of ammeter, (1p) connected in the base !p (A) Circuit for various yal ucs of Var at constant Vor. 4 Veg=2V Plot the curve on the graph taking Ty along y-axis and Vpg along x-axis as shown in Fig. 11.20, Draw r 8 family of curves for different Settings of Veg, The 79 following points are worth noting from these ie cteristics : a () These Curves are similar to those Obtained gq b.. } : for CB configuration ie, like a forward diode 7 characteristic, The only difference is that in thiscase ‘"" Tp increases less Tapidly with increase in Ver. 30 Hence, the input resistance of CE configuration is comparatively higher than that of CB configuration, 20 (i) The change in Veg does not result in a 10}. large deviation of the curves and hence, the effect Of change in Vcg on the input characteristics js 0 ignored for all practical Purposes, Input resistance. The ratio of change in Fig. 11.20. base-emitter voltage (A Vpz) to thé resulting change in base Current (Ap) at constant collector-emitter vollage (Vee) is known as input resistance, i.e, 7 AVog Input resistance, ye th al constant Veg Jn CE configuration, the typical value of input ‘Tesistance is of the order of a few hundred a 2. Output characteristics : In CE configuration, the curve Dlotted between collector current fc@ collector-emitter voltage Vog at constant base current Tp is called output characteristic. csi A family of characteristic curves can be plotted for different Settings of fp. Fig, 11.21 shows the mists characteristics of a typical *npn transistor in CE co nfiguration. ‘The following poins [rom these characte may be noted : *Por a pnp transistor all the quantides e.g, Veg, Ic and fy will be negative, C® Scanned with OKEN Scanner a eT DENT CTs AND CMMs Te sistance, Anund thy sy 7 determing the Let the small operating point t % dynamic Output change in Ver Ne from 7.5 Vito be (a) BO Vor s 12.5 — 7S 8 Vv The Somesponding change in fe at constant Ips 40) he, Al Ce 40 02 mA AVer = at Constant Ip = 40 A Ale sv uu 1c are to take the Value of Ic To = 40 LA at - From gtaph, when Ip = 40 HA: ’ To=48 mA at Veg =10V lo 48 mA” “Bac = = 5 aa = 120 (Ans. Ip 40 WA fans) In order to determine ac, Current gain (B) ie, Be Alc/Alp, Voe=10V. draw a vertical line Corresponding to From the given characteristics it is clear that when base Current changes from 30 HA to 49 BA, Correspondingly the collector current changes from 3.6 MA to 4.8 mA, A fol : fs a constant Veg = 10 Vv Alp _ 48-36) ma 12x 10-3 | © 4) lant or eee 120 (Ans,) (40 ~ 30) HA 10x 19-6 : 11.17. COMMON COLL) CONFIGURATION) transistor is shown in Fig. 11.23 and 11.24 iF INPUT — INPUT { (SIGNAL) ; (SIGNAL) | c ¥ [-—=—-—_] laren iors Common collector circuit of ApH lennsistor \ Cammon collector circuit of pip transistor Fig. 11.23. ; Fig. 11.24, Q® Scanned with OKEN Scanner — _fgelvely. In (his arrangement, the jy 41088 the cmitler and Collector, Thue Mpu Vis voy Mected pf 179 : ce the name “sy ME Collee EUW LOH hase © jgcnils and aie the tame Common Colleen Celor Of the itannie ANd Colfcetoe While output j Current amplification factor (y) Clor COnnECtiog ; “OMIM LG both inn Eat ES taken The ratio of output curr T common COL LO iy UTeNt | au CHiran, « S - - {put Current js CMitlor ¢ known 4S curreny atin, Thus the ratio of chan T CURA Jey Mput and outpur Collector Configuration, collector comnection, the ; wlification . ¢ actor, ie stor. Tie Lo? Ut emitter ¢ "lt: whereas the input geen, 12 2 common lification factor, Tt ig '0 the o P Ont is base current j amplifies Scutcrally fepresentey by G ae ute in base current js kn es § reak letter ° KNOWN as current a Aly Y (Gamma), Relation between y ang a p We know, ys Alg ; Aly : iy and a= fe . Al; fii) J, Now, Ihe le+ Ip or Alg= Alc + Aly or Alp = Alp ~ Alc Substituting the value of Alp in eqn, @, we get, : y= Ale Alg}/Alp 1 = Alpe Ae oe “Ale~ Ale ~ Algfalg ~ AldAlp ~ te 1: oe ‘The above relation shows that the value of y is nearly equal to 8. However, this circuit arrangement is seldom used for amplification because in this arrangement input resistance is high (about 750 kQ) and Output resistance is very low (abaut 25 Q). D ue to this reason, the voltage gain is very low (less than 1). This circuit arrangement is Primarily used for impedance matching. Collector current We have seen that Ic = Op + Icso and Ig = Ic + Ip = (lg + Icpo) + Ip or Ig— lg = Ip + Icgo or Ig (l-a)=Ip+lepo. 1 1 ) * te= ta (2) + leno (pPy = *(B +1) Ig + (B+ 1) Iezo RATIONS = Fiitaeape alee oe eae eee three configurations ie. CB, CE cud eee e T ation in which a and Fe ed nica a ueadon is different in different jeep aes oie Heine ae lansist ie 0 be com ses that depends upon the requifement €8- ted for impedance matching. aoe but a sal f voltage gain, ie . ee | tion its characteristics must be ae eee i i tf 1 the application of a particular configural Hence, itis concluded that fo a A a Ls les =——— opt _ ta ear be C® Scanned with OKEN Scanner “stSTORS Bs Operation. When . *signal is applied in the emitter by, sg this junction increases, This increases the 1 oe eases the collector current, The increased coll ; ec Junction, durin i 1B DOSILi ve hah cy che, tee Sox went, 5 0: A af clectrons fj iS fron i A emitter to cflentiee Ve Sahel incr and thus i ie { Joad resistor Re. Howe . . OF curren - ie as pase fiiaen decease tee negative half cycle of ibe Oduces mone vektaap fg ati or-bas ses, Thi le signs é Paden ascs. This decre, Signal, the forward tine, Smeg ases the collector curre Ch Comtcnesdiy Cone: . Ollector cu DL which cone MARES Dy. yg a across emil c- Hence, an ampli fi , ified signal appears 2 ; Pllied signal appears acrons Ge clear the yoliage joad resistor. Collector current analysis. The praphi ; When no signal is applied, uhe input circuit ie emniter hn of collector current is shorwn in Fig, 1224. js forward biased by the battery Vga. Therefore, a d.c. collector curten Io Ic flows in the collector circuit. This current is called 0 ead collector current. When the signal voltage is applied ducing 0 signal - half cycle, the forward bias on emitter-base junction siases Positive jncrease in total collector current ic. Whereas, during ne causing cycle, the forward bias-on emitter-base junction decreases which cau decrease in total collector current ic. Thus, total collector canon consists of two components, namely : ” () The d.c. collector current Jc when no signal is applied. This js due to forward bias created at the emitter-base junction by the bias drop across the collector load resistor R pattery Vgp. (ii) The a.c. collector current i,, due to signal applied in the Graph for cofecer coat emitter-base junction. , : Fe. Lae Total collector-cusrent, ic = tes + Ic. - The useful output is the voltage drop across collector load resistor Re cae & fe 2c copes of which flows through it because of applied signal. The purpose of 2270 Soul ois Se current ics ie. dc. component J¢ is only to ensure that emitter-base junction is forward Dass S&S SS 1121. SYMBOLS USUALLY EMPLOYED FOR CURRENTS AND VOLTAGES IN TRASSEEROR APPLICATIONS Usually, the symbols shown in the following table.are employed for currents and Volluges 3 SST applications : ; ; S.No. Particulars Instantaneous ac. de, valses } Tra gr sete values { beau 1. | Emitter current fe : = i 2. | Collector current fe is | 5 i 3. | Base current ‘ b VCE 5 ciey { 4. | Collector-emitter voltage ; Yee : vs 5 VgB BS _| | Emitter-base voltage Veb "122. TRANSISTOR LOAD LINE ANALYSIS wy § ally required to be analysed ie, for & given. Before applying a transistor, the transistor circuit is generally can be obtained by plotting be ifector current. This * Oleg “See the value of co ifiers. However, 10 actual emblance to & sine wave. f sine waves of different emitter voltage we must know ause it is convenient for testing the ampl hich is a complex wove having little rest “He ey USEE Sine : : ; Vere sine wave is considered as a signal bec Jex waves are Ue sum of a number 0} Nice, th My oN . Hl ingoagye ott may be a speech on music ctc., W Hering ~ fourier series analysis, all these comp! “Nee, sine wave in considered as a signal. (CE Scanned with OKEN Scanner