S.E. Sem. III [ETRX] Electronic Circuits and Design I

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1 S.E. Sem. [ETRX] Electronic ircuits and Design Time : 3 Hrs.] Prelim Paper Solution [Marks : 80 Q.1(a) What happens when diode is operated at high frequency? [5] Ans.: Diode High Frequency Model : This purely resistive A model for the diode works well when the frequency of the A signals is sufficiently low. At high frequencies, we need to include the effects that arise due to these time varying signals and the charge separation that exists in the depletion region and in the ulk p and n regions of the diode under forward ias conditions. Within the device and the depletion region there exists an electric field. For A signals, this electric field is varying with time. As you ve learned in electromagnetics, a time varying electric field is a displacement current. The effects of a displacement current are modeled y equivalent circuit capacitances : We won t do anything with this effect now. This is presented primarily as an FY. Q.1() Explain why configuration is referred as Voltage follower. [5] Ans.: As amplifier provides Av = 1 = Vo. mplies that Vo = Vi, i.e. output follows input, Vi hence the name voltage follower Q.1(c) Draw transfer and drain characteristics of EMOSFET. [5] Ans.: The transfer and drain characteristics of EMOSFET is as shown D (ma) +5 V +3 V +2 V V GS = +1V V DS (V) Q.1(d) Explain construction of Schottky diode and draw its characteristics. [5] Ans.: t is mainly used as a rectifier at signal frequency exceeding 300 MHz due to it s quick response time and lower noise Figure. t has more uniform junction region and is more rugged than PN diode. 1 2 D (ma) DSS 1 V DS constant Drain characteristics V DS (V) V T Transfer characteristics

2 : S.E. ED - onstruction : t is a metalsemiconductor junction diode with no depletion layer. t uses a metal (like gold, silver, platinum, tungsten etc.) on one side of junction and usually an N type doped silicon semiconductor on the other side. The diode construct & its symol are shown in aove Figure. Gold leaf metal contact A node (+) (+) N () Silicon dioxide screen ntype silicon metal Metal semiconductor Metal metal athode () Fig. 1: Schottky Barrier diode (a) asic structure () symol Applications : 1) Because of these quality schottky diode can easily rectify signals of frequencies exceeding 300 MH 3, it can produce an almost perfect half wave rectified output. 2) t is commonly used in switching power supplied that operates at frequencies of 20 GHz. 3) Due to low noise Figure it is extremely important in communication receivers and radar units etc. Fig. 2 : Application of Schottky Barrier diode as rectifier. Q.1(e) What is the Voltage Regulator explain simple zener shunt voltage Regulator. [5] Ans.: A regulator is an electronic circuit that provides constant and stailized output independent of variation of applied input voltage. The Zener Shunt Regulator is as shown. R S The circuit diagram of zener shunt regulator is z as shown. The zener is connected in shunt with the load, and zener is kept reverse iased. The unregulated input V i must e greater than V 0, atleast y 5 to 10 V. The resistor R S will ensure a minimum current through zener when V i = V i min, keeping the zener in ON state. Also it limits the maximum value of zener current when V i = V i max. n the ON state, zener maintains a constant voltage across its terminals. Therefore V 0 = V z. Since V z is a stale zener reference source, the output voltage is stale and hence the load gets a constant voltage. V 0 V t V i V z V 0 R 2

3 Prelim Paper Solution Q.2(a) Draw Energy and diagram of pin junction diode under : (i) Zero Bias, (ii) Forward ias and (iii) Reverse Bias Ans.: (a) zero ias, () reverse ias, and (c) forward ias [10] Q.2() Ans.: For the given circuit find Steady State D Parameters cq and V ceq Given = 100 and V BE = 0.7 V, also state in which region the circuit is working. 6 K 1.5 K 1 K 5 V 100 ohm [10] 3

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5 Prelim Paper Solution Q.3(a) Ans.: For the given MOSFET amplifier, Determine Dq, V GSq and V DS. R 1 K 22 M R 2 18 M V dd 3 K 40 V R D R s 820 V s D s V GS (Th) = 5 V D (ON) = 3 ma V GS (ON) = 10 V [10] 5

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7 Prelim Paper Solution Q.3() Explain working principle, characteristics and applications of Tunnel diode. [10] Ans.: Tunnel diode is a specially made p-n junction device which exhiits negative resistance over part of the forward ias characteristic. t has extremely heavy doping on oth sides of the junction and an arupt transition from the p-side to the n-side. The tunneling effect is a majority carrier effect and is consequently very fast. The tunnel diode is useful for oscillation or amplification purposes. Because of the thin junction and short transit time, it is also useful for microwave applications in fast switching circuits. Volt-amp haracteristics of a Tunnel Diode The volt-ampere characteristics of a tunnel diode are shown in Figure elow. (a) Fig. 1 : (a) Sum of tunneling and injection, () V- characteristic of tunnel diode. The tunnel effect controls the current at very low values of forward ias where the normal or the injection current is very small as shown in Figure aove. The mechanism of tunneling is purely a quantum mechanical phenomenon. An electron on one side of the arrier will have a certain proaility of leaking through the arrier if arrier is very thin. f oth p and n type materials of a junction are heavily doped, the depletion region ecomes very narrow; as narrow as of the order of 100 A. Fig. 2: Symol for tunnel diode Q.4(a) Derive expression for ripple factor for section filter. [10] Ans.:.. Filter : n case of filter the ripple factor is directly proportional to the R, whereas in c Filter, the ripple factor is inversely proportional to R, it means in oth the circuit depends upon the R, to overcome the aove prolem, we use Filter, which is comination of & filter, and in Filter the ripple factor is independent of load resistance R. () 7

8 : S.E. ED - As shown in the figure, the rectifier output is given to the inductor. The inductor provides more opposition to the A component and less opposition to the D component. This results that, most of the A component appears across the inductor and whole of the D component passes to the load. f at all any part part of A component manages to reach the load, then it is removed y the capacitor. Derivation for Ripple Factor : Assumption Mode : i) While otaining the expression for the ripple factor, we will consider only the second harmonics component and neglect all other higher order harmonics. ii) nductive Reactance X > apacitive Reactance X. The Full wave Rectifier output is pulsating in nature, hence its Fourier Series is given as : 2Vm 4Vm 4Vm cos2wt cos4wt 3 15 Neglecting higher order harmonics. 2Vm 4Vm cos2wt 3 2V n the aove equation m represents the d.c. component availale at the output of 4Vm Rectifier and cos2wt represents the A component availale at the Rectifier 3 output. The peak value of the A current is given as : 4Vm cos2wt 3 peak X Assuming cos 2 wt = 1, 4Vm peak 3X RMS D D A D D peak 4Vm 2 3 2X 2 RMS V 3 2X dc 2 2V 3 2X m But 2V m R V dc 8

9 2 V RMS 3 X dc Now the RMS voltage developed across the load is given as : V X RMS RMS 2 Vdc VRMS X 3 X VRMS 2 X V 3 X dc 2 X 3 X where X 1 2w & X 9 2w Prelim Paper Solution Q.4() Derive expression for nput resistance, Voltage gain, current gain and output [10] resistance for E amplifier with R E unypassed. Ans.: ommon Emitter BJT Amplifier with R E unypassed (a) ircuit diagram : V i +V () ac Equivalent circuit : Short V and short all the capacitors (c) Replace transistor y its h parameter model : RB V i R B i 0 h ie Q V i R B R V 0 R R E (1 + h fe ) R E V 0 h fe R i Vi.h ie (1h fe)r R E 0 Q R E1 h (1h )R ie fe E R V 0

10 : S.E. ED - Q.5(a) Ans.: (d) To find input resistance R i = V i / i : Vi Vi.h ie (1h fe)re RB R i =.. R h (1h )R i i B ie fe E = h (1 h )R.R ie fe E B R h (1h )R B ie fe E (e) To find output resistance R 0 : R 0 = R = R h (1h )R B ie fe E (f) To find voltage gain A v = V 0 / V i : Output Voltage (V 0).R 0 A v = = nput Voltage (V ).h (1h )R hfe R = h (1h )R ie fe E i ie fe E hfe R = h (1h )R ie fe E The ve sign indicates that input and output voltages are 180 out of phase. (g) To find current gain A = 0 / i output current A i = input current = 0 i = 0. i hfe RB =. R h (1h )R B ie fe E Design Single Stage E amplifier for the given specifications [10] A v 100, S = 10, V o = 3 V, f = 20 Hz and Ri 3K, also calculate Av, Ri and Ro for the designed circuit. Step 1: Selection of ias We select voltage divider ias with R E partially y-passed. Step 2: Selection of transistor We select B having following parameters. P Dmax = 250 mw cmax = 0.1 A V EO 45 V h fe = 330 h ie = 4.5 K V E sat = 0.25 V Step 3: Design of R et R 2 = 10 K and A V = 22 h fe.r A V = h (1h )R ie fe E1 et R E1 = 100 ; ¼ W R = K R = K R sat = 1.5 K; ¼ W R.R E R R = k E Step 4: To determine D Q point, V EQ = 1.5 [V E sat + V op ] = 1.5 [ ] = V 10

11 Prelim Paper Solution V cc et V EQ = (mid point iasing) 2 V cc = 9.75 V et V cc = 10 V et V RE = 10% V cc = 2V V VE VRE = R = 2.75 ma Step 5: Design of R E VRE R E = = 363 R E = R E1 + R E2 R E2 = 263 R E2 sat = 240 ; ¼ W Step 6: Design of R 1 & R 2 R3 Given S = 10 = 1 + R E (R E = RE1 R E2 ) = 340 R 3 R.R 1 2 = R1 R2 = 9.R E = 3060 V B = V B + V RE = = 1.7V R2 also, V B =.V R1 R2 R2 R1 R = V3 2 V = 0.17 R 1 = 18 K R 2 = 3.68 K R 1 sat = 17 K; ¼ W R 2 sat = 3.6 K; ¼ W To maintain R 1 = 312, let R 2 = 4.3 K; ¼ W Step 7: Design of capacitor (a) R i = R 1 R 2 (h ie + (1 + h fe ) R E1 ) = 3.14 K 1 f 1 = 2R, given f = 20 H E, 1 = 2.5 f i 1 1 sat = 10 f; 5 V () R 0 = R + R 2 = 11.5 K 1 f 2 = 2R sat = 1f; 15 V 10 (c) f = 2R E. E 2 = 0.69 f E = 796 f E = 1000 f; 5 11

12 : S.E. ED - Step 8: Designed ircuit 17K R +V = 10V 1.5K 2 Step 9: alculation for derived circuit 1. R 0 = R = 1.2 K 2. R i = R R 2 [h ie + (1 + h fe ) R 1 ] = K 3. h fe.r A 0 = h (1h )R = ie fe E1 Q.5() What is lamping circuit, explain with neat nput and output waveforms for [10] negative lamping circuit. Ans.: Sometimes it is necessary to add a dc level to the ac signal. The circuits which are used to add dc level as per the requirement into the signal ac are known as lamper ircuits. The capacitor, diode and the resistor are the 3 asic elements of the clamper circuits. The clamper circuit is also known as dc restorer (or) dc insertor circuit. Depending upon whether the positive d.c. (or) negative dc shift is to e introduced, clampers are classified into two types : 1. Positive lamper 2. Negative lamper 1. The simple positive clamper circuit using capacitor, diode D and load resistor R is as shown in figure. The circuit is used to add positive level to the ac output voltage V o. V i The operation of the circuit can e explained as follows : 1 10f R 1 V in B 147 B R E f V 0 onsider the first negative half cycle, during which diode D ecomes forward iased and starts conducting. Hence, capacitor starts charging through diode D and it gets charged almost equal to V m 0.7V with the polarity as shown in figure. Just after the negative peak, the diode ecomes reverse iased and capacitor remains charged to a value V m 0.7 with the same polarity. Now capacitor can discharge through load resistor R. But the value of the R is selected on the higher side such that the discharging time constant R is so large that capacitor discharges very little which can D R 10K 4.3 K R 2 RE2 100 t 1000f R V O 12

13 Prelim Paper Solution e neglected. For a good clamper circuit R Time constant should e ten times the time period of input signal. Hence the output voltage is given as : V o = V i + V c But V c = V m 0.7 V o = V i + V m 0.7 et V i = V m, then the V o gets, V o = 2V m 0.7 et V i = 0, then V o = V m 0.7 et V i = V m, then V o = 0.7 This indicates that output voltage is equal to sum of V i and dc level of capacitor. The figure shows the input and output waveform for positive clamper. t can e oserved that, peak to peak value of the output is same as peak to peak value of the input signal. Only difference is that a positive dc level is added into the output signal. Waveforms : V = V P(in) 0.7 V P(in) V P(in) V o V [2V P(in) 0.7] V PP = 2V P (in) V P (in) = +V m V P (in) = V m The figure shows the circuit diagram of negative clamper. t is a circuit which adds negative dc level to the ac output voltage. The operation of the circuit can e explained as follows: During the positive half cycle, diode ecomes forward iased and strats conducting. Therefore, current flows V in D R V O through the circuit and charges the capacitor upto the value V m 0.7 with the polarity as shown in figure. t t 13

14 : S.E. ED - Once the capacitor charges to V m 0.7 then diode ecomes reverse iased and acts as an open circuit. Now capacitor can discharge through R, ut value of R is selected so high that the time constant R is very large and capacitor discharges very little which can e neglected. Now when diode is acting as a reverse iased then output voltage is given as : V o = V i V c where V c = V m 0.7 V o = V i V m et V i = 0, then V o = V m et V i = V m, then V o = et V i = V m, then V o = 2 V m Waveform : V in V P(in) V P(in) V o 0.7V 0 0 V PP = 2V P (in) 2V P(in) +0.7 V = [V P(in) 0.7 ] This indicates that output voltage is V m 0.7 dc level of ac. The circuit adds negative dc level in the signal. Hence, it is called as Negative lamper. n this circuit also, peak to peak voltage of the output is same as the input. Q.6(a) For the voltage divider iased E MOSFET circuit derive equation of nput Resistance, [10] Voltage gain and output resistance for S amplifier. Ans.: (a) ircuit diagram : V i 1 R 1 +V DD V P (in) = +V m V P (in) = V m R D 2 V 0 t t R 2 R s s 14

15 Prelim Paper Solution () a.c. equivalent circuit : Short V DD and short all capacitor. Replace FET y its equivalent : (d) To find voltage gain A V = V 0 /V i : V 0 = g m. V gs. (r d R D ) V i = V gs V gmv gs(r d R 0 D) A V = V V i f R is connected at the output, then A g (r R R ) gs v m d D (e) To find R i : R i = R G (f) To Find output impedance R 0 : R 0 = r d R D Q.6() Derive equation of nput resistance, urrent gain and Voltage gain for [10] amplifier. Ans.: (a) ircuit diagram : V i V i R G V gs g m V gs 1 R B Since output is taken from emitter, emitter voltage follows the ase voltage and hence the name emitter follower. () ac Equivalent circuit : Short V, short all capacitors. +V Q R E r d R D V 0 Q 2 V 0 V 0 V i R B R E 15

16 : S.E. ED - (c) Replace transistor y its equivalent : i (d) To find input resistance R i = V i / i : h = ie (1 h fe)r E.RB R h (1 h )R R h (1 h )R = B ie fe E B ie fe E (e) To find output resistance R 0 : hie 1 R 0 = R E R 1 h g (f) To find voltage gain A v = V 0 / V i : A v = R i fe m Output Voltage (V ) nput Voltage (V ) (g) To find current gain A = 0 / i : output current A i = input current h ie V i R B Vi.h (1h ) R ie fe E h (1h )R i 0 E (1 h fe).re =.h (1h )R = 0 i ie fe E = 0. (1 h fe) RB =. R h (1h )R = (1 h fe).rb R h (1h )R B ie fe E ie fe E h fe (1+h fe ) R E B ie fe E V0 i (1 h fe)re = h (1h )R ie fe E R 0 16