EC1QUEBANK

COURSE/ LESSON PLAN SCHEDULESTAFF NAME: S.SANKAR CLASS: II YEAR -ECESUBJECT: ELECTRONIC CIRCUITS- I / EC 2205A) TEXT BOOKS

1. Millman J and Halkias .C., Integrated Electronics, TMH, 2007.2. S. Salivahanan, N. Suresh Kumar and A. Vallavaraj, Electronic Devices and Circuits,2nd Edition, TMH, 2007.

B) REFERENCES1. Robert L. Boylestad and Louis Nashelsky, Electronic Devices and Circuit Theory, 9th Edition,

Pearson Education / PHI, 2007.2. David A. Bell, Electronic Devices & Circuits, 4th Ediion, PHI, 20073. Floyd, Electronic Devices, Sixth Edition, Pearson Education, 2002.4. I.J. Nagrath, Electronic Devices and Circuits, PHI, 2007.5. Anwar A. Khan and Kanchan K. Dey, A First Course on Electronics, PHI, 2006.6. B.P. Singh and Rekha Singh, Electronic Devices and Integrated Circuits, Pearson Education, 2006.7. Rashid M, Microelectronics Circuits, Thomson Learning, 2007.

C) LEGEND:L: Lecture BB: Black Board Tx : Text Book Rx: Reference Book pp: Pages

Sl.NoLecture

hourTopic to be covered Teaching

aid required

Book No /Page No

UNIT I TRANSISTOR BIAS STABILITY

01 L1 BJT BB Tx1 pp (282-287) ,Tx2 pp (149) Rx1 pp (131-146 )

02 L2 Need for biasing – Stability factor BB Tx1 pp (282-287),Tx2 pp (174) Rx1 pp ( 212) ,Rx2 pp(124-129)

03 L3 Fixed bias circuit, Load line and quiescent point

BB Tx1 pp (283-285) ,Tx2pp (181) Rx1 pp (163-172)

04 L4 Variation of quiescent point due to hFE variation within manufacturers tolerance, Stability factors

BB Tx1 pp (284-286),Tx2 pp (180)

05 L5, L6 Different types of biasing circuits, Method of stabilizing the Q point

BB Tx1 pp ( 287-294),Tx2pp (181-193 ) ,Rx1 pp ( 177-179,186-189) ,Rx2 pp (130-141)

06 L7 Advantage of Self bias (voltage divider bias) over other types of biasing

BB Tx1 pp (290-294 ) Tx2 pp (185)

07 L8, L9 Bias compensation – Diode, Thermister and Sensistor compensations

BB Tx1 pp (299-303) Tx2 pp (193-194)

08 L10, L11 Biasing the FET and MOSFET BB Tx1 pp (335-339 ),Tx2 pp (213), Rx1 pp ( 290-303),Rx2 pp (291-298 )

09 L12 TRIAC BB Tx2 pp(235-238)

UNIT II MIDBAND ANALYSIS OF SMALL SIGNAL AMPLIFIERS

10 L13 CE, CB and CC amplifiers BB Tx1 pp (251-255 ),Tx2pp (279) Rx1 pp ( 201-202)

11 L14 Method of drawing small-signal equivalent circuit ,

BB Tx1 pp (263-274),Tx2 pp (284),Rx2 pp (185-200 )

VECW ELECTRONIC CIRCUITS-I EC 2205

12 L15, L16 Midband analysis of various types of single stage amplifiers to obtain gain, input impedance and output impedance

BB Tx1 pp (263-274 ) Tx2 pp (284)

13 L17 Miller’s theorem, Comparison of CB, CE and CC amplifiers and their uses

BB Tx1 pp (255-258),Tx2 pp (251) Rx1pp (600-601),Tx1pp (253-255)

14 L18 Methods of increasing input impedance using Darlington connection and bootstrapping

BB Tx1 pp ( 274-279),Tx2pp (308) Rx1 pp ( 633-634)

15 L19CS, CG and CD (FET) amplifiers

BB Tx1 pp (313-320 ),Tx2pp (313) Rx1 pp (627-630),Rx2 pp (339-352 )

16 L20 Multistage amplifiers. BB Tx1 pp (349-387)

17 L21 Basic emitter coupled differential amplifier circuit - Bisection theorem

BB Tx1 pp ( 507-510),Tx2pp (323)

18 L22 Differential gain – CMRR BB Tx1 pp ( 508,523-524),Rx1 pp ( 649-658),Rx2 pp (399-403 )

19 L23 Use of constant current circuit to improve CMRR, Derivation of transfer characteristic

BB Tx1 pp ( 508-510),Tx2pp (325) Tx1 pp ( 510-512)

20 L24 Feedback Amplifiers BB Tx2 pp(486-500)

UNIT III FREQUENCY RESPONSE OF AMPLIFIERS

23 L25 General shape of frequency response of amplifiers - Definition of cutoff frequencies and bandwidth

BB Tx1 pp (389-391),Tx2 pp (389) Rx2 pp ( 232-233)

24 L26, L27 Low frequency analysis of amplifiers to obtain lower cutoff frequency Hybrid, π equivalent circuit of BJTs

BB Tx1 pp ( 397-401),Tx2pp (397) Rx1 pp (586-588) ,Rx2 pp ( 234-237)

25 L28, L29 High frequency analysis of BJT amplifiers to obtain upper cutoff frequency – Gain Bandwidth Product

BB Tx1 pp ( 403-407) Tx2 pp (403)

26 L30, L31 High frequency equivalent circuit of FETs

BB Tx1 pp ( 412-416),Tx2pp (412) Rx1 pp ( 602-606)

27 L32 High frequency analysis of FET amplifiers - Gain-bandwidth product of FETs

BB Tx1 pp ( 412-416),Tx2pp (417) Rx1 pp (609-611 )

28 L33 General expression for frequency response of multistage amplifiers

BB Tx1 pp (147-419 ),Tx2pp (349) Rx1 pp (613-614 )

29 L34 Calculation of overall upper and lower cutoff frequencies of multistage amplifiers

BB Tx1 pp ( 417-419),Tx2pp (351)

30 L35 Amplifier rise time and sag and their relation to cutoff frequencies

BB Tx2 pp ( 351)

31 L36 Oscillators BB Tx2 pp(511-518)

UNIT IV LARGE SIGNAL AMPLIFIERS

32 L37 Classification of amplifiers, Class A large signal amplifiers

BB Tx2 pp (427-429),Rx1 pp ( 747-752)

33 L38 second harmonic distortion, higher order harmonic distortion

BB Tx2 pp (429-432) Rx1 pp (773-774 )


34 L39, L40 transformer-coupled class A audio power amplifier – efficiency of Class A amplifiers

BB Tx2 pp (432-435) Rx1 pp ( 754-760) Rx2 pp (606-611 )

35 L41, L42 Class B amplifier – efficiencypush-pull amplifier - distortion in amplifiers

BB Tx2 pp (436-442) Rx1 pp ( 761-768) Rx2 pp ( 612-615)

36 L43, L44 complementary-symmetry (Class B) push-pull amplifier

BB Tx2 pp (442-443)

37 L45 Class C, Class D amplifier – Class S amplifier

BB Tx2 pp (443-445) Rx1 pp (780-782 )

38 L46 MOSFET power amplifier, BB Tx2 pp (445-446),Rx2 pp (635-636 )

39 L47 Thermal stability and heat sink BB Tx2 pp (446-448)

40 L48 Operational Amplifiers BB Tx2 pp(707-714)

UNIT V RECTIFIERS AND POWER SUPPLIES

41 L49 Classification of power supplies BB Tx2 pp (617-618)

42 L50, L51 Rectifiers - Half-wave, full-wave and bridge rectifiers with resistive load

BBTx2 pp (619-628),Rx1 pp(74-80),Rx2 pp ( 52-67)

43 L52, L53 Analysis for V dc and ripple voltage

with C, L, LC and CLC filtersBB

Tx2 pp (629-635)Rx2 pp (80-82,95-102 )

44 L54 Voltage multipliers BBTx2 pp (637-638),Rx1 pp (98-100),Rx2 pp (86-89)

45 L55 Voltage regulators ,Zener diode regulator

BBTx2 pp (638-643)Rx2 pp (7073 )

46 L56 principles of obtaining a regulated power supply,

BB Tx2 pp (644-646)

47 L57 regulator with current limiting, Over voltage protection

BB Tx2 pp (652-653)

48 L58 Switched mode power supply (SMPS) BB Tx2 pp (654-662)

49 L59 Power control using SCR BB Tx1 pp (715-719)

50 L60 IC voltage regulators BB Rx2 pp(582-583)

Prepared by HOD -ECE(S. KARTHIKEYAN)

UNIT-I TRANSISTOR BIASING STABILITY

1. What is meant by Biasing?It is the process of maintaining proper flow of zero signals IC and VCE during the passage of signal. Biasing keeps EB junction forward bias and CB junction reverse bias.

2. What is the need for biasing?


To operate transistor in desired region, apply external DC voltage of correct polarity and magnitude to the two junctions of the transistor.

3. What is Q point?It is the point on load line which represents the DC current and voltage, when no input signal is applied.

4. How to select the Q point of amplifierQ point should be located at the center of load line which ensure that the amplified signal will be an exact replica of the input signal

5. What are the factors affecting the Q pointChanges in temperature, 2) Changes in the value of β, 3) Change of parameters from one transistor to other

6. Define Stability factor. What is its ideal value (MAY 2005)

S=∆ IC∆ICO

∕ constant V BE∧β S'=∆ IC∆V BE

∕ constant ICO∧β S' '=∆ IC∆ β

∕ constant Ico∧V BE

It’s ideal value =17. What are the advantages and disadvantages of different biasing methods?

Biasing method Advantages DisadvantagesFixed Bias Good flexibility

Require less componentPoor thermal stabilityQ point shift due to β

Collector – Base Bias

Provides better thermal stability than fixed bias

Requires only one resistor

stability factor is highReduce the gain of the amplifier

Voltage divider or Self Bias

Good stability Possible to avoid signal loss

Require more componentsstability factor depends on RB & RE

8. What is Stabilization Technique?It refers to the use of resistive biasing circuits which permits IB to vary so as to keep IC constant.

9. What is Bias Compensation?Compensation technique is used to stabilize the Q point instead of DC biasing circuits where the reduction of gain crosses the tolerable limit. It uses the temperature sensitive devices such as diodes, transistor, thermistors, sensistor to compensate for the variation in currents.

10. How the compensation achieved in Diode compensation for VBE

Change in VBE is compensated by change in voltage across diode11. How the compensation achieved in Diode compensation for ICO

In order to compensate for ICO, diode saturation current is equal to transistor leakage current which gives IC= β IB

12. How the compensation achieved by using Thermistor?Thermistor resistance decreases exponentially with increase in temperature. As resistance decreases, VBE decreases so that IB decreases.

13. How the compensation achieved by using Sensistor?Sensistor resistance increases with increase in temperature. As resistance increases, VR2 decreases so that forward emitter bias decreases. As a result IC decreases.

14. What is meant by Thermal Runaway?The self destruction of an unbiased transistor is known as thermal runaway. To avoid it; the operating point is to be stabilized.

15. Define Thermal resistanceThe steady state temperature rise at the collector junction of a transistor is directly proportional to power dissipated at the junction. ΔT=Tj-TA=QPD

16. What is the condition of Thermal Stability?Rate at which heat is released at collector junction must not exceed the rate at which heat can be dissipated under steady state condition. ΔPC ΔTJ < 1 θ

17. Why thermal runaway is not there in FET(Dec09)FET has positive temperature coefficient. As temperature increases, resistance increase, cureent flow through it decreases, so power dissipation also decreases .This reduces its junction temperature and prevents thermal runaway.

18. Mention the applications of FETUsed in chopper amplifier, used as constant current source, used in Filp-Flop, used in attenuator

19. Compare BJT and JFFT


BJT JFETBipolar device [ current conduction by both electron and Hole]

Unipolar device[ current conduction is by only one type of carrier either by electron or by hole]

Low input impedence due to forward bias High input impedence due to reverse bias

Current driven device Voltage driven device

High noise level Low noise level

20. Compare JFET and MOSFETJFET MOSFEToperate in Depletion mode only operate in either Depletion or

Enhancement modeHigh input impedence Very High input impedencegate current is larger than MOSFET Less gate current

PART-B

1. What is d.c load line? How will you select the operating point? Explain it using common emitter amplifier characteristics as an example?

2. Draw the circuit diagram of Fixed -bias circuit using CE configuration and explain how it stabilizes operating point.

3. Draw the circuit diagram of Collector – Base bias circuit using CE configuration and explain how it stabilizes operating point.

4. Draw the circuit diagram of self-bias circuit using CE configuration and explain how it stabilizes operating point.

5. Derive the Stability factor for the different biasing method6. Explain the bias compensation techniques by using i)Diode ii) Themistor iii) Sensistor 7. Draw the circuit of a common source FET amplifier & explain its operation? 8. With the help of neat diagram explain the voltage divider biasing method for JFET9. Explain how field effect transistor can be biased10. Draw the biasing circuit for MOSFET using common source configuration and explain(May04)11. Draw and explain voltage divider bias using FET (May03)

UNIT II MIDBAND ANALYSIS OF SMALL SIGNAL AMPLIFIERS 1. Compare CB,CE,CC amplifiers

CB CE CC

input impedance very low (100Ω) moderate(750Ω) very high(100Ω)output impedance very high(450Ω) moderate(45KΩ) very low (25Ω)current gain 1 high highvoltage gain high(150) high(150) 1application for High Frequency for Audio Frequency for Impedance matching

2. What are the effects of unbypassed RE? Increase in the Input Impedance Reduction in voltage amplification Improved stability of voltage gain

3. Explain the function of bypass capacitor in an amplifier circuitsIt offers low reactance to AC signal which increases the voltage gain of amplifier

4. Why coupling capacitor is used to connect a signal source to an amplifierIt blocks DC voltage but passes AC signal, because of this biasing conditions are maintained constant.

5. Define the various h parameters (May03)Input impedance=h11=Vi/Ii reverse voltage gain=h12= Vi / Vo

forward current gain =h21= Io/Ii output admittance =h22=Io/Vo

6. Draw the hybrid model for transistor


7. State Miller theorem (May04)It states that, if the gain ratio of two node is 1:K then an impedance of Z connecting the two nodes can be replaced with Z/K impedance between the first node and ground and a KZ/(K-1) impedance between the second node and ground.

8. What does Bootstrapping Mean?(May03)If the gain of the amplifier is 1, then if the voltage at one end of resistor changes, then there is same changes at the other end of resistor. It is as if resistor were pulling itself up by its bootstraps.

9. Why the Darlington connection is not possible for more number of stages.When number of stages increases, the leakage current also increases and gets multiplied by the current gain. Voltage gain will also reduce.

10. What are the features of Darlington Emitter Follower?High current gain, Low voltage gain, High input resistance, Low output resistance

11. Draw the LF model of JFET (May03)

12. Draw the HF model of JFET

13. Mention the advantages of h parameters Easy to measure Real number up to RF Easily correctable from one configuration to other.

14. Write the equation form which the small signal LF equivalent of JFET is formed(May05)Id=gmVgs+Vds/rd

15. What is the need for muli stage amplifier?In single stage amplifier, the parameters input impedance, Voltage gain, Bandwidth and Output impedance are not fulfilled. So the multistage amplifier is needed for these requirements.

16. What are the features of Cascode Amplifier? Input and current gain are equal to the corresponding value of single stage CE amplifier Output resistance is equal to that of CB amplifier Bandwidth is very large

17. What is cascade amplifier?It consists of CE and CB configurations.CB provide a good HF operation

18. What is the voltage gain of cascade amplifier?It is the product of voltage gains of the various stagesAV=AV1.AV2.AV3………AVN

19. Define CMRRIt is the ratio of differential gain to common mode gain [ Ad/ Ac]

20. What are the features of Differential Amplifier? High input impedance Low output impedance Large Bandwidth High CMRR

21. State Bisection theoremIn a Bisected network, if all the connecting wires are open circuit, then the impedance at input and output is Z1/2OC.If the connecting wires are short circuit, then the impedance at input and output is Z1/2SC

22. How does constant current improve CMRR?In the circuit, RE will ideally be ∾ and Ac will be zero making CMRR ∾.RE replaced with constant current circuit will improve CMRR.


23. How does current mirror method proves CMRR?In this method, output current is forced to be equal to input current i.e. output current is the mirror images of input current.PART B

1. Derive the expressions for current gain, voltage gain, input impedance and output impedance for an emitter follower circuit.(Dec02)

2. Explain with circuit diagram the bootstrapped Darlington emitter follower(Dec02)3. Derive the expressions for the voltage gain of common collector amplifier4. Derive the expressions for the current gain, input impedance, voltage gain and output admittance of a small

signal amplifier in terms of the h- parameters5. Explain in detail with neat sketch operation of common mode differential amplifier6. Draw and explain midband analysis of CE configuration in BJT amplifiers To obtain Ai,Av,Zi and Yo7. How does constant current source increase the gain and hence CMRR in differential amplifier8. Write and discuss in detail the improving methods of CMRR and its Measurement9. Derive the expressions for CMRR for a small signal model with a common mode input voltage10. Draw the transistor hybrid model in CE configuration and determine its h-parameter from its characteristics11. Explain briefly multistage amplifiers12. Derive the expressions for the current gain, input impedance, voltage gain and output impedance for a

voltage divider common source FET amplifier

UNIT III FREQUENCY RESPONSE OF AMPLIFIER1. Define Bandwidth

It is defines as the difference between the half power frequency.BW= f2-f1 Hz2. Definition of Midband gain

The midband of an amplifier is the band of frequencies between 10 f1 and 0.1 f2.3. How coupling capacitor effect the bandwidth of an amplifier

It offers a large reactance at LF, due to this, voltage drop across them increases which in turn reduce the gain of the amplifier

4. What is the effect of bypass capacitor?At LF, reactance is not equal to zero, but it has finite value. The parallel combination of RE and CE will offer a finite impedance. So RE is not properly bypassed and the voltage gain will reduced.

5. What is the effect of internal transistor capacitancesAt high frequencies, the coupling and bypass capacitors act as short and do not affect the amplifier response.However, at high frequencies, the internal capacitances, commonly known as junction capacitances do come into play, reducing the current gain.

6. Why it is not possible to use the h parameters at HF?At HF,h parameter become complex and its values are frequency dependent.

7. Define alpha cutoff frequencyIt is the frequency at which the short circuit CB current gain of the transistor drops by 3db from its value

at LF. fα=1+hfe /2 πrb 'e [ce+cc ] 8. Define beta cutoff frequency

It is the frequency at which the short circuit CE current gain of the transistor drops by 3db from its value at LF. fβ=1 ∕ 2 πrb ' e [ce+cc]9. Define fTIt is the frequency at which the short circuit CE current gain becomes unity. fT=hfe fβ10. What is rise time?The time difference between the t1 and t2 which corresponds to 10% and 90% value of the final value is called rise time. tr=2.2 R2C211. Give the relation between bandwidth and rise timeBW=fH=0.35 ∕ tr12. What is tilt or sag?Tilt=P=(V−V ') ∕ V

13. Give the significance of two capacitors in hybrid Π modelDiffusion capacitance is offered by the forward biased BE junction and represents the excess minority carrier storage at the base emitter junction. Its value is 100Pf.Transition capacitance is offered by reverse biased CB junction. Its value is 3pF.14. Give the expression for lower cutoff frequency of multistage amplifier


fL(n)=fL/√21/n-1 fL(n) – Lower 3dB frequency of identical cascaded stages, fL- Lower 3dB frequency of single stage, n – Number of stages

15. Give the expression for Higher cutoff frequency of multistage amplifier

fH(n)=fH.√21/n-1 fH(n) – Higher 3dB frequency of identical cascaded stages,fH - Higher 3dB frequency of single stage, n – Number of stages

PART B1. Explain the analysis of BJT and FET at low frequencies.2. Explain the analysis of BJT and FET at high frequencies.3. Obtain the expression for the various parameters for amplifier at high frequency.4. Derive the expression for the CE short circuit current gain of transistor at high frequency.5. Derive an expression for voltage and current gain, cutoff frequencies including source resistance and gain

bandwidth product.6. Explain the concept of rise time and sag and derive the relationship to cutoff frequencies.7. Derive the expression for the CE short circuit gain of transistor at high frequency8. Draw the high frequency hybrid π model for a transistor in the CE configuration and explain the

significance of each component9. Derive expressions for the short circuit current gain of common emitter amplifier at high frequency. Define

beta cut off frequency and transition frequency and derive their values in terms of circuit parameters10. Explain high frequency analysis of FET and gain bandwidth product.11. Derive the upper and lower cut off frequency of RC coupled BJT amplifier12. Derive the expressions for transistor conductance for hybrid π common emitter transistor model.13. What specific capacitance has the greater effect on the high frequency response of a cascade FET

amplifier?

UNIT IV LARGE SIGNAL AMPLIFIER

1. What is power amplifier?The final stage in multistage amplifiers, such as audio amplifiers and radio transmitters, designed to deliver maximum power to the load, rather than maximum voltage gain, for a given percent of distortion.

2. What are the features of power amplifier?Power transistors are required, Impedance matching is necessary

3. Compare the different classes of power amplifiers

Class A Class B Class C Class ABposition of Q point at the center of load line at the cut off Below cut off just below cut offEfficiency 25 to 50% 78.5% above 95% between 50 & 78.5%Conduction angle 360° 180° less than 180° between 180°& 360°

4. What are the advantages and disadvantages of different types of power amplifier?Types of power amplifier

Advantages Disadvantages

Class A simple circuit distortion less output

very low efficiency large power dissipation

Class B high efficiency compared to Class A zero power dissipation Impedance matching is possible

efficiency is not so high cross over distortion occur

Class C very high efficiency low power loss

distorted output

Class AB elimination of cross over distortion preferred to Audio system

low efficiency cannot be used as audio amplifiers

Class D High efficiency amplify Digital signals

complicated design

5. What is cross over distortion? How it can be eliminated?For making transistor ON, it is necessary that VBE voltage must exceed 0.7 v.Due to this, in class B amplifier while crossing over from one half cycle to other, as long as input is below 0.7v, none of the


transistor is ON and output is zero. Due to this, there is distortion in the output, which is called the cross over distortion.To overcome this distortion, a small forward bias is kept applied to the transistors so that when input is zero, this additional forward bias can make the transistor ON immediately, eliminating cross-over distortion.

6. What is meant by Harmonic distortion?It states that the presence of those frequency component in the amplifier output which are absent in the input side of amplifier. The frequency component which has the same frequency of the input is known as the fundamental frequency component and others are known as Harmonics.

7. Define conversion efficiency of power amplifierIt is a measure of the ability of an active device to convert the DC power of the supply into an AC power delivered to the load

8. Define thermal resistance.The temperature rise of a junction is proportional to the power dissipation. The constant of proportionality between the two is called thermal resistance. It is defined as the temperature rise per unit watt of heat dissipation.

θ = (T2-T1/Pd) °C/W

9. Define Amplitude distortionThe dynamic characteristics of a transistor is non linear. Due to this, the output waveform will be slightly different from the AC input signal. This type of distortion is known as non linear or Amplitude distortion.

10. Define Frequency DistortionThe change in gain of the amplifier with change in the frequency of input AC signal is called frequency distortion. It takes place when various frequency component in the input signal are amplified differently

11. Define Phase DistortionPhase Distortion or Delay Distortion occurs in a non-linear transistor amplifier when there is a time delay between the input signal and its appearance at the output. This time delay will increase progressively with frequency within the bandwidth of the amplifier.

12. Why class A amplifier must not be operated under no signal conditions.Under no signal condition, the entire d.c. power input PDC = VccICQ, is dissipated as the heat. Thus power dissipation is maximum under no signal condition. This may increase the transistor junction temperature beyond safe value, which may lead to transistor damage. To avoid this, class A amplifier must not be operated under no signal condition.

13. What are the advantages and disadvantages of transformer coupled class A amplifierAdvantages :

The efficiency of the operation is higher than directly coupled amplifier. The d.c bias current that flows through the load in case of directly coupled amplifier is stopped in

case of transformer coupled. The impedance matching required for maximum power transfer is possible.

Disadvantages : Due to transformer, the circuit becomes bulkier, heavier and costlier compared to directly coupled

circuit. The circuit is complicated to design and implement compared to directly coupled circuit. The frequency response of the circuit is poor.

14. What are the advantages and disadvantages of complementary symmetry class B amplifier.Advantages:

As the circuit is trasformerless, its weight, size and cost are less. Due to common collector configuration, impedance matching is possible. The frequency response improves due to transformer less class B amplifier circuit.

disadvantages : The circuit needs two separate voltage supplies. The output is distorted to cross-over distortion.

15. Compare class B Push-pull and Complementary Symmetry amplifiers


parameters Push-pull Complementary Symmetrytypes of transistors both should be either p-n-p and n-p-n one should be p-n-p and other

should be n-p-nuse of transformer driver and output transformers are used transformers are not neededconfiguration both the transistors operate in CE

configurationboth the transistors operate in CC configuration

Efficiency low high

16. What is the use of Heat Sink?Power transistors are temperature dependent devices. As they handle large currents they can be heated which leads to self destruction. To avoid this, the transistor is fixed on a metal sheet preferably Aluminium to dissipate heat from the transistor.

PART B

1. Draw and explain the operation of series fed directly coupled class A power amplifier and derive its efficiency.

2. For the transformer coupled class A power amplifier circuit, explain and derive the expressions for its efficiency

3. Draw a neat diagram of push pull class B amplifier. Explain its working and derive its efficiency4. Draw and explain the operation of complementary-symmetry class B power amplifier and derive its

efficiency.5. With circuit diagram, explain the operation of Class C power amplifier and derive its efficiency.6. With neat diagram, explain the MOSFET power amplifier.7. Give the design procedure for Heat sink8. Draw a Quasi complementary-symmetry power amplifier and explain its merits.

UNIT V RECTIFIERS AND POWER SUPPLIESPART A

1. Define RectifiersA rectifier is defined as electronic device used for converting a.c. voltage into unidirectional voltage. A rectifier utilizes unidirectional conduction device like PN junction diode.

2. Define voltage regulation Voltage regulation = VNL-VFL

VFL

3. Define ripple factor ripple factor = rms value of a.c. component of output D.C. or average value of the component4. Define-efficiency of a half-wave rectifier along with its maximum value

The efficiency of a half wave rectifier is defined as the ratio of d.c. output power to a.c. input powerEfficiency = dc output power/ ac input power

5. Define peak inverse voltage (PIV)Peak inverse voltage is defined as the maximum reverse voltage that a diode can withstand without destroying the junction.

6. Define Transformer utilization factor (TUF)Transformer utilization factor is defined as the ratio of d.c power to that of a.c rating of the transformer secondary. TUF = Pdc / Pac rated

7. Give the expressions of form factor and peak factor for a half wave rectifierForm factor = rms value / average valuePeak factor = peak value / rms

8. Compare the different types of rectifier circuitsparameters HWR FWR Bridge rectifieraverage value Im / π 2Im / π 2Im / πRms value Im / 2 Im / √2 Im / √2efficiency 40% 81.2% 81.2%TUF 28.7% 69.3% 81.2%ripple factor 1.21 0.48 0.48ripple frequency 50Hz 100Hz 100HzPIV Vm 2Vm Vmcenter tap transformer

not required required not required

9. What are advantages and disadvantages of different types of rectifier circuits?Types of rectifiers Advantages Disadvantages


HWR Simple circuit High ripple factor Low efficiency Low TUF

FWR Low ripple factor Better efficiency Better TUF

Cost of center tap transformer high

Need large size of diodesBridge rectifier Center tap transformer is not

required Efficiency high TUF is high

Four diodes are needed As two diodes conduct

simultaneously, the voltage drop increases, output decreases

10. Compare the different types of filtersparameters Capacitor filter Inductor filter L Section filter π Section filterripple factor 1

4 √3 fCRR

3√2ωLR

6√2ω2LC

1

4 √3(ω2LC1C2RL)Useful in reducing ripple in

load voltagereducing ripple in load current

reducing ripple in load current

reducing ripple in load voltage

suitable for light load application

heavy load application

light & heavy load application

all loads

surge current through

very high low low low

11.What is the need of filter circuit?

The output of rectifier circuit consists of DC and Ripple components. To remove the ripples,the filter circuits are used.

12. What is Bleeder resistor?This resistance is connected across the output of the filter, to place minimum load

13. What are advantages and disadvantages of filters?Types of filters Advantages Disadvantagescapacitor easy to design

reduction in the ripple content increase in the average load voltage

ripple factor is dependent on the load

regulation is poor diodes have to handle large peak

currentsInductor filter low ripple factor at heavy load currents

reduce ripple in output ripple factor is poor bulky and costly

L Section filter good load regulation low ripple factor and not dependent on

load

power loss takes place in inductor

bulky and costly

π Section filter low ripple factor high DC voltage

power loss takes place in inductor

bulky and costly high peak diode current

14. Why is a simple capacitor filter is not suitable for heavy loads.As the load current increases, for the same d.c output voltage the load resistance decreases. This increases ripple content in the output for heavy loads. Practically for heavy loads, the d.c output voltage decreases and shows very poor regulation. Hence the simple capacitor is not suitable for heavy loads.

15. State the three factors that determine the stability of the voltage regulators.The output d.c. voltage V0 depends upon the input unregulated dc voltage Vin, load current IL and temperature T. The three factors that determine the stability of voltage regulator are

1. Input regulation factor, Sv = ∆Vo/∆Vin when ∆IL=0,∆T = 0.2. Output resistance, Ro = ∆Vo/∆ IL when ∆Vin =0, ∆T =0 3. Temperature coefficient ST= ∆Vo/∆T when ∆Vin=0, ∆IL = 04.16. Define Line regulation


Line regulation is defined as the change in output voltage for a change in regulated load voltage due to change in line voltage, keeping the load current and temperature constant.

Line regulation =VLH-VLL LH=load voltage with high line voltage ,LL= load voltage with low line voltage17. What are the factors affecting the output voltage of a regulated power supply?

Load current, Input voltage, Temperature18. Compare shunt regulator and series regulator

Shunt regulator Series regulator The control element is in parallel with the load. The control element is in series with the load.Any change in output voltage is compensated by changing the current Ish through the control element as per the control signal

Any change in output voltage is compensated by adjusting the voltage across the control element as per the control signal.

Regulation is poor Regulation is good19. Compare Rectifier and regulator

Rectifier Regulator1. Rectifier converts pure sinusoidal input into pulsating d.c. output.

Regulator converts pulsating d.c. input into constant d.c. output.

2. The output contains ripples. The output is ripple free.3. Output voltage changes with respect to load current, input voltage and temperature.

Output voltage changes with respect to load current, input voltage and temperature

4.Not provided with over load protection, short circuit protection, thermal shutdown etc.

Provided with all sorts of protection circuits.

20 .What are the advantages and disadvantages of SMPS. advantages disadvantages

Efficiency is high due to less heat dissipation Protection against excessive output voltage Higher power handling capacity Reduced harmonic feedback into the supply

main

Load regulation is poor No isolation between input and output Radio frequency interference to the neighbouring

circuits Transient response is slow

21. What are the disadvantages of the linear voltage regulators? Low efficiency Need large value of capacitors Input transformer is bulky and costly.

22. What is the basic concept of SMPS? In SMPS, series pass transistor operates as a switch. Pulses from generator are applied to switch, when it is ON

it connects the input as it is to the input of the filter. When it is OFF, filter input is disconnected and gives rectangular waveform. Then filter converts the rectangular into smooth dc voltage by removing the ripple contents.

PART B

1. Draw the circuit of a) Half waveb) Full wavec) Bridge rectifiers and explain its working .Obtain its ripple factor and Efficiency.

2. With the circuit diagram, explain the operation ofa) Capacitorb) Inductor c) LC d) CLC filters and derive its ripple factor.

3. Explain the working of different types of voltage multipliers4. With a neat diagram, explain the

a) Series transistorized voltage regulatorsb) Shunt transistorized voltage regulators.

5. With necessary sketches, explain the operation of SMPS.


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