Power amplifiers


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Power amplifiers

  1. 1. Power AmplifiersText Book:[1] Jacob Millman, Cristos C. Halkias, “IntegratedElectronics: Analog and Digital Circuits and Systems”,34th Reprint 2004, Tata-McGraw-Hill. Chapter: 18[2] Robert Boylestad and Louis Nashelsky, “ElectronicDevices and Circuit theory”, Fifth Edition, Prentice-hallof India Private Limited 1994. Chapter:16[3] Albert Paul Malvino, “Electronic Principles”, TataMcGraw-Hill, 1999. Chapter: 11
  2. 2. Amplifier:An amplifier is an electronic device that increases voltage, current or power of a signal.According to the class of operation, the amplifiers can be classified as:• Class A, (ii) Class B, (iii) Class AB, and (iv) Class C.Class A: A class Aamplifier is one in whichthe operating point and theinput signal are such thatthe current in the output The output signal of classcircuit flows at full times. A amplifiers varies for a full 360o of the cycle.
  3. 3. Class B: A class B amplifier is one in which the operating point is atan extreme end of its characteristic, so that the quiescent power isvery small. If the signal voltage is sinusoidal, amplification takesplace for only one-half a cycle.A class B circuit provides an output signal varying over one-half theinput signal cycle, of for 180o of input signal.Class AB: A class AB amplifier is one operating between the twoextremes defined for class A and class B. Hence the output signal iszero for part but less than one-half of an input sinusoidal signal.For class AB operation the output signal swing occurs between 180oand 360o and is neither class A nor class B operation.
  4. 4. Class C: A class C amplifier is one in which theoperating point is chosen so that the output current (orvoltage) is zero for more than one-half of the inputsinusoidal signal cycle.The output of a class C amplifier is biased for operation atless than 180o of the cycle and will operate only with atuned (resonant) circuit which provides a full cycle ofoperation for the tuned or resonant frequency.
  5. 5. Series-Fed Class A Amplifier The fixed-bias circuit connection shown in Fig. 16.1 can be usedto discuss the main features of a class A amplifier. The beta (β) of power transistor is generally less than 100, theoverall amplifier circuit using power transistors being capable ofhandle large power or current not providing much voltage gain.DC BiasThe dc bias set by VCC and RB fixes the dc base-bias current at VCC − VBE IB = (16.1) RBwith the collector current then being IC = βI B (16.2)with the collector-emitter voltage then VCE = VCC − I C RC (16.3)
  6. 6. To appreciate the importance of the dc bias the operation of the poweramplifier, consider the collector characteristic shown in Fig. 16.3.A dc load line is drawn using the values of VCC and RC.The intersection of the dc bias value of IB with the dc load line thendetermines the operating point (Q-point) for the circuit.The quiescent-point values are those calculated using Eqs. (16.1)through (16.3).If the dc bias collector current is set atone-half the possible signal swing(between 0 and VCC/RC) the largestcollector current swing will be possible.Additionally, if the quiescent collector-emitter voltage is set at one-half thesupply voltage, the largest voltageswing (between 0 and VCC) will bepossible.
  7. 7. AC OperationWhen an input ac signal is applied to the amplifier of Fig. 16.1, theoutput will vary from its dc bias operating voltage and current.A small input signal, as shown in Fig. 16.4(a), will cause the basecurrent to vary above and below the dc bias point, which will thencause the collector current (output) to vary from the dc bias point set aswell as the collector-emitter voltage to vary around its dc bias value.Power ConsiderationsThe input power with no inputsignal drawn from the supply is Pi (dc) = VCC I CQ (16.4)Even with an ac signal applied, theaverage current drawn from thesupply remains the same.So Pi(dc) represent the input power.
  8. 8. Output PowerThe output voltage and current varying around the bias point provideac power to the load.The ac power is delivered to the load RC, in the circuit of Fig. 16.1 isgiven byUsing rms value :Po (ac) = VCE (rms) I C (rms) (16.5a) 2 Using peak value :Po (ac) = I C (rms) RC (16.5b) VCE (p) I C (p) 2 Po (ac) = (16.5a ) VCE (rms) 2Po (ac) = (16.5c) 2 RC I C (p) RC Po (ac) = (16.5b) 2 2 VCE (p) Po (ac) = (16.5c) 2 RC
  9. 9. Using peak - to - peak value : V (p - p) I C (p - p) Po (ac) = CE (16.5a ) 8 2 I C (p - p) RC Po (ac) = (16.5b) 8 2 VCE (p - p) Po (ac) = (16.5c)Efficiency 8 RCThe efficiency of an amplifier represents the amount of ac powerdelivered (transferred) from the dc source.The efficiency of the amplifier is calculated using Po (ac) %η = × 100% (16.8a ) Pi (dc) 2 VCE (p - p) %η = × 100% (16.8b) 8RC × VCC I CQ
  10. 10. Maximum EfficiencyFor the class A series amplifier the maximum efficiency can bedetermined using the maximum voltage and current swings.For the voltage swing it is Maximum VCE (p − p) = VCC VCC Maximum I C ( p − p) = RCUsing the maximum voltage swing in Eq. (16.7c) yields 2 VCC Maximum Po (ac) = 8 RCThe maximum power input can be calculated using the dc biascurrent set to half the maximum value. 2 VCCThus V /2 Maximum Pi (dc) = VCC CC = RC 2 RC
  11. 11. We can then use Eq. (16.8) to calculate the maximum efficiency: 2 VCC 2 RC maximum Po (ac) maximum %η = × 100% = × × 100% maximum Pi (dc) 8 RC V 2 CC 2 = × 100% = 25% 8 The maximum efficiency of a class A amplifier is thus seen to be 25%. A form of class A amplifier having maximum efficiency of 50% uses a transformer to couple the output signal to the load as shown in Fig. 16.6.
  12. 12. Example 16.1 Calculate the (i) base current, collector current andcollector to emitter voltage for the operating (or Q) point (ii) inputpower, output power, and efficiency of the amplifier circuit in thefollowing figure for an input voltage in a base current of 10 mA peak.Solution:(i) the Q-point can be determined by DCanalysis as: V − 0.7V 20V − 0.7V I = CC = =19.3 mA B R 1 KΩ Q B I = βI = 25(19.3mA) = 0.48 A C B V = 20V − (0.48A)(20Ω) =10.4 V CE Q
  13. 13. (ii) When the input ac base current increases from its dcbias level, the collector current rises by I (p) = βI (p) = 25(10mA) = 250 mA peak C B I 2 (p) (250×10− 3)2 Po(ac) = C Ro = 20 = 0.625 W 2 2 P (dc) =V I = (20V)(0.48A) = 9.6W i CC C Q Po (ac) 0.625W %η = ×100% = = 6.5% P (dc) 9.6W i
  14. 14. Class B AmplifierClass B operation is provided when the dc bias transistor biased justoff, the transistor turning on when the ac signal is applied.This is essentially no bias and the transistor conducts current for onlyone-half of the signal cycle.To obtain output for the full cycle of signal, it is necessary to use twotransistors and have each conduct on opposite half-cycles, thecombined operation providing a full cycle of output signal.Since one part of the circuit pushesthe signal high during one-halfcycle and the other part pulls thesignal low during the other half-cycle, the circuit is referred to as apush-pull circuit. Figure 16.12 shows a diagram for push-pull operation.
  15. 15. The arrangement of a push-pull circuit is shown in Fig. 18-8.When the signal on transistor Q1 is positive, the signal on Q2 isnegative by an equal amount.During the first half-cycle of operation, transistor Q1 is driven intoconduction, whereas transistor Q2 is driven off. The current i1 throughthe transformer results in the first half-cycle of signal to the load.During the second half-cycle of operation, transistor Q2 is driven intoconduction, whereas transistor Q1 is driven off. The current i2 throughthe transformer results in the second half-cycle of signal to the load.The overall signal developed across the load then varies over the fullcycle of signal operation.
  16. 16. Consider an input signal (base current) of the form ib1=Ibmcosωt appliedto Q1.The output current of this transistor is given by Eq. (18.33): i = I + B + B cosωt + B cos 2ωt + B cos3ωt + ........... (18.33) 1 C 0 1 2 3The corresponding input signal to Q2 is i = −i = I cos(ωt + π ) b2 b1 bm The output current of this transistor is obtained by replacing ωt by ωt+π in the expression for ib1. That is, i = i (ωt + π ) (18.34) 2 1 Hence, i = I + B + B cos(ωt + π ) + B cos 2(ωt + π ) + B cos3(ωt + π ) + ........... 2 C 0 1 2 3 which is i = I + B − B cosωt + B cos 2ωt − B cos3ωt + ........... (18.35) 2 C 0 1 2 3
  17. 17. The total output current is then proportional to thedifference between the collector current in the tworesistors. That is, i = k (i − i ) = 2k (B cosωt + B cos3ωt + ...........) (18.36) 1 2 1 3This expression shows that a push-pull circuit will balanceout all even harmonics in the output and will leave thethird-harmonic tram is the principle source of distortion.This conclusion was reached on the assumption that thetwo transistors are identical. If their characteristics differappreciably, the appearance of even harmonics must beexpected.The fact that the output current contains no even-harmonic.
  18. 18. Advantages of Push-Pull System1) Push-pull circuit give more output per active device: Because noeven harmonics are present in the output of a push-pull amplifier, sucha circuit will give more output per active device for a given amount ofdistortion. For the same reason, a push-pull arrangement may be usedto obtain less distortion for a given power output per transistor.2) Eliminates any tendency toward core saturation and nonlineardistortion of transformer: The dc components of the collectorcurrent oppose each other magnetically in the transformer core. Thiseliminates any tendency toward core saturation and consequentnonlinear distortion that might arise from the curvature of thetransformer-magnetizing curve.3) Ripple voltage will not appear in the load: The effects of ripplevoltages that may be contained in the power supply because ofinadequate filtering will be balance out. This cancellation resultsbecause the currents produced by this ripple voltage are in opposedirections in the transformer winding, and so will not appear in theload.
  19. 19. Class B amplifier operationThe transistor circuit of Fig. 18-8 operates class B if R2=0 because a silicon transistor is essentially at cutoff if the base is shorted to the emitter.Advantages of Class B as compared with class A operation2. It is possible to obtain greater power output,3. The efficiency is higher, and4. There is negligible power loss at no signal.Disadvantages of Class B as compared with class A operation2. The harmonic distortion is higher,3. Self-bias cannot be used, and4. The supply voltage must have good regulation.
  20. 20. The graphical construction from which to determine the output currentand voltage waveshapes for a single transistor operating as a class Bstage is indicated in Fig. 18-9.
  21. 21. Input power: The power supplied to the load by an amplifier isdrawn from the power supply which provides the input or dc power.The amount of this power can be calculated using Pi (dc) = VCC I dc (16.17)In class B operation the current drawn from a single power supplyhas the same form of a full-wave rectified signal. So, 2 2 I = I ( p ) (16.18) and Pi (dc) = VCC I (p) (16.19) dc π πOutput Power: The output across the load is given by 2 VL (rms) Po (ac) = (16.20) RL 2 VL ( p ) 2 VL ( p - p ) Po (ac) = = (16.21) 2 RL 8 RL
  22. 22. Efficiency: The efficiency of the class B amplifier can be calculatedusing the basic equation Po (ac) %η = × 100% Pi (dc) 2 VL ( p ) π %η = ×100% 2 RL 2VCC I ( p ) V ( p) using, I (p) = L RL 2 VL ( p ) π RL %η = ×100% 2 RL 2VCC VL (p) π VL ( p ) %η = ×100% (16.22) 4 VCC
  23. 23. Maximum Efficiency: Equation (16.22) shows that the larger thepeak voltage, the higher the circuit efficiency, up to a maximum valuewhen the , this maximum efficiency then being VCC 2 VCC Maximum, I (p) = Maximum, I dc = RL π RL 2 2VCC 2 VCC 2 VCCMaximum, Pi (dc) = VCC = Maximum, Po (ac) = π RL πRL 2 RL 2 VCC πRL π Maximum, %η = × = ×100% = 78.5% 2 RL 2V 2 4 CC The maximum efficiency of a class B amplifier is thus seen to be 78.5%.
  24. 24. Crossover distortionThe distortion caused by the nonlinear transistor input characteristic isindicated in Fig. 18-12. The iB-vB curve for each transistor is drawn,and the construction used to obtain the output current (assumedproportional to the base current) is shown.In the region of small currents (for vB<Vγ) the output is much smallerthan it would be if the response were linear. This effect is calledcrossover distortion.In order to overcome the problemof crossover distortion, thetransistor must operate in a classAB mode.
  25. 25. (16.10) Calculate (i) the input power, (ii) the output power, (iii) thepower handled by each transistor, and (iv) the efficiency for aninput of 12 V (rms) of a push-pull Class-B amplifier circuit havingbiasing voltage, VCC= 25 V and load resistance, RL= 4 Ω.Solution: Given, Vi (rms)=12V, VCC= 25 V, load resistance, RL= 4 Ω. Vi ( p ) = 2Vi (rms ) = 2 × 12V ≈ 17 VSince, the operating point is selected in cutoff region, thusVi(p)=VL(p)=17V VL ( p ) 17 V I L ( p) = = = 4.25A RL 4Ω 2 2 I dc = I L ( p ) = × 4.25A = 2.71A π π Pi (dc) = VCC I dc = 25V × 4.25A = 67.75W
  26. 26. 2 ( p) VL (17 V )2 Po (ac) = = = 36.125W 2 RL 2 × 4ΩPower dissipated by each output transistor is: P2Q Pi − Po 67.75 − 36.125 PQ = = = = 15.8 W 2 2 2 Po 36.125 Efficiency, %η = × 100% = × 100% = 53.3% Pi 67.75
  27. 27. Class C OperationWith class B, we need to use a push-pull arrangement. That’swhy almost all class B amplifiers are push-pull amplifiers.With class C, we need to use a resonant circuit for the load.This is why almost all class C amplifiers are tuned amplifiers.With class C operation, the collector current flows for lessthan half a cycle.A parallel resonant circuit can filter the pulses of collectorcurrent and produce a pure sine wave of output voltage.The main application for class C is with tuned RF amplifiers.The maximum efficiency of a tuned class C amplifier is 100percent.
  28. 28. Figure 11-13a shows a tuned RF amplifier.The ac input voltage drives the base, and an amplifiedoutput voltage appears at the collector.The amplified and inverted signal is then capacitivelycoupled to the load resistance.Because of the parallel resonant circuit, the output voltage ismaximum at the resonant frequency, given by: 1 fr = (11.14) 2π LC
  29. 29. On either side of the resonant frequency fr, the voltage gaindrops as shown in Fig. 11-13b.For this reason, a tuned class C amplifier is always intendedto amplify a narrow band of frequencies.This makes it ideal for amplifying radio and televisionsignals because each station or channel is assigned a narrowband of frequencies on both sides of a center frequency.