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CALVIN ENGR 332 - Chapter 9

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PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Chapter 9 Output Stages And Power AmplifiersLow Output Resistance – no loss of gainSmall-Signal Not applicableTotal-Harmonic Distortion (fraction of %)EfficiencyTemperature RequirementsFig. 9.1 Collector current waveforms for transistors operating in (a) class A, (b) class B, (c) class AB, and (d) class C amplifier stages.Fig. 9.2 An emitter follower (Q1) biased with a constant current I supplied by transistor Q2.Class ATransfer CharacteristicsIE1I iLThe bias current I must ve greater the largest negative current valueOtherwise Q cutts offThe transfer characyteristic of the emitter follower for this figure isvOvIvBE1Where vBE1 depends on the emitter current iE1 and thus on the loadcurrent iL.If we neglect the relative small changes in vBE1 (60mV for every factot of10 change in iE) the transfer curve results Fig. 9.3 Transfer characteristic of the emitter follower in Fig. 9.2. This linear characteristic is obtained by neglecting the change in vBE1 with iL. The maximum positive output is determined by the saturation of Q1. In the negative direction, the limit of the linear region is determined either by Q1 turning off or by Q2 saturating, depending on the values of I and RL.Class ATransfer CharacteristicsFrom figure 9.3 we can see thatvomaxVCCVCE1satIn the negative direction, the limite of the linear region is determined either by Q1 turning offvOminI RLor by Q2 saturatingvOminVCC VCE2satDepending on the values of I and RL. The absolutely lowest output voltage is that given by the previous equation and is achieved provided that the bias current I is greater than the magnitude of the corresponding load currentIVCC VCE2satRLClass ATransfer CharacteristicsExercises D9.1 and D9.2Class ASignal Waveforms0 5 10101vo t( )t0 5 10012vcE1 t( )t0 5 10012ic1 t( )t0 5 1000.51pD1 t( )tClass APower DissipationP VCCILargest Power Dissipation When vo = 0Q1 must be able to withsatnd a continuous dissipation of VCC*IThe power dissipation of Q1 depends on the value of RL.If RL is infinite, iC1 = I and the dissipation in Q1 depends on vo.Maximum power dissipation will occur when vo = -VCC since vCE1 will be 2VCC.pD1 = 2VCC*I. This condition would not normally persist for a prolonged interval, sothe design need not be that conservative. The average pD1 = VCC*IWhen RL is zero a positive voltage would result in a theoretically infinite current (large practical value) would flow through Q1. Short-circuit protection is necessary.Class APower Conversion Efficiencyload_power PL supply_power PS Voaverage voltagePL12Vo2RLPS2 VCC I14Vo2I RL VCC14VoI RLVoVCCVoVCC VoI RLmaximum efficiency is obtained when VoVCCI RLClass AExercise 9.4Vopeak 8 I 100 103 RL100 VCC10PLVopeak22100 PL0.32Pplus VCCI Pplus 1Pminus VCCI Pminus 1 PSPplus PminusPLPS  0.16CLASS AMany class A amplifiers use the same transistor(s) for both halves of the audio waveform. In this configuration, the output transistor(s) always has current flowing through it, even if it has no audio signal (the output transistors never 'turn off'). The current flowing through it is D.C. A pure class 'A' amplifier is very inefficient and generally runs very hot even when there is no audio output. The current flowing through the output transistor(s) (with no audio signal) may be as much as the current which will be driven through the speaker load at FULL audio output power. Many people believe class 'A' amps to sound better than other configurations (and this may have been true at some point in time) but a well designed amplifier won't have any 'sound' and even the most critical 'ear' would be hard-pressed to tell one design from another. NOTE: Some class A amplifiers use complimentary (separate transistors for positive and negative halves of the waveform) transistors for their output stage. Class APower Conversion EfficiencyFig. 9.5 Class B output stage. Class BCircuit OperationCLASS 'B' A class 'B' amplifier uses complimentary transistors for each half of the waveform. A true class 'B' amplifier is NOT generally used for audio. In a class 'B' amplifier, there is a small part of the waveform which will be distorted. You should remember that it takes approximately .6 volts (measured from base to emitter) to get a bipolar transistor to start conducting. In a pure class 'B' amplifier, the output transistors are not "biased" to an 'on' state of operation. This means that the the part of the waveform which falls within this .6 volt window will not be reproduced accurately. The output transistors for each half of the waveform (positive and negative) will each have a .6 volt area in which they will not be conducting. The distorted part of the waveform is called 'crossover' or 'notch' distortion. Remember that distortion is any unwanted variation in a signal (compared to the original signal). The diagram below shows what crossover distortion looks like.Fig. 9.6 Transfer characteristic for the class B output stage in Fig. 9.5.Fig. 9.7 Illustrating how the dead band in the class B transfer characteristic results in crossover distortion.Class ABCircuit OperationClass ABOutput ResistanceClass ABExercise 9.6Calvin College - ENGR 332Class AB Output Stage Amplifier Consider the class AB circuit (illustrated below) with Vcc=15 V, IQ=2 mA, RL=100 ohms. Determine VBB. Determine the values of iL, iN, iP, vBEN, vEBP, vI, vO/vI, Rout, and vo/vi versus vO for vO varying from -10 to 10V. Note that vO/vI is the large signal voltage gain and vo/vi is the incremental gain obtained as RL/(RL+Rout). The incremental gain is equal to the slope of the transfer curve. Assume QN and QP to be matched, with IS=10E-13.Class ABExercise 9.6under quiescent conditions iN=iP=IQ vO=vI=0Solving for VBBVBB 1 IS 1013 VT 0.025 IQ 2 103 RL 100GivenIQ IS eVBB2VTVBB Find VBB( ) i 0 100 VBB 1.186Class ABExercise 9.6vOi10i5 iLivOiRL10 0 100iLivOiClass ABExercise 9.6vBENiVT lniNiIS10 5


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CALVIN ENGR 332 - Chapter 9

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