Lecture 22ANNOUNCEMENTS•Midterm #2:Th11/15 3:30‐5PM in Sibley Aud. (Bechtel Bldg.)OUTLINEMidterm #2: Th11/15 3:305PM in Sibley Aud. (Bechtel Bldg.)• HW#11: Clarifications/revisions to Problems 1, 3, 4 were madeOUTLINE• Differential Amplifiers–General considerations– BJT differential pair• Qualitative analysis•Largesignal analysis•Large‐signal analysis• Small‐signal analysis• Frequency responseEE105 Fall 2007 Lecture 22, Slide 1Prof. Liu, UC BerkeleyReading: Chapter 10.1‐10.2“Humming” Noise in Audio Amplifier• Consider the amplifier below which amplifies an audio signal from a microphone.p• If the power supply (VCC) is time‐varying, it will result in an additional (undesirable) voltage signal at the output, id “h i ” i b thperceived as a “humming” noise by the user.EE105 Fall 2007 Lecture 22, Slide 2Prof. Liu, UC BerkeleySupply Ripple Rejection• Since node X and Y each see the voltage ripple, their voltage difference will be free of ripple.pprinvXvvAv+=YXrYvAvvvv=−=invYXvAvv=EE105 Fall 2007 Lecture 22, Slide 3Prof. Liu, UC BerkeleyRipple‐Free Differential Output• If the input signal is to be a voltage difference between two nodes, an amplifier that senses a differential signal is needed.EE105 Fall 2007 Lecture 22, Slide 4Prof. Liu, UC BerkeleyCommon Inputs to Differential Amp.• The voltage signals applied to the input nodes of a differential amplifier cannot be in phase; otherwise, the differential p p; ,output signal will be zero.+=rinvXvvAv0=−+=YXrinvYvvvvAv0YXvvEE105 Fall 2007 Lecture 22, Slide 5Prof. Liu, UC BerkeleyDifferential Inputs to Differential Amp.• When the input voltage signals are 180° out of phase, the resultant output node voltages are 180° out of phase, so that p gp,their difference is enhanced.ArinvYrinvXvvAvvvAv+−=+=invYXrinvYvAvv 2=−EE105 Fall 2007 Lecture 22, Slide 6Prof. Liu, UC BerkeleyDifferential Signals• Differential signals share the same average DC value and are equal in magnitude but opposite in phase.q g pp p• A pair of differential signals can be generated, among other ways, by a transformer.EE105 Fall 2007 Lecture 22, Slide 7Prof. Liu, UC BerkeleySingle‐Ended vs. Differential SignalsEE105 Fall 2007 Lecture 22, Slide 8Prof. Liu, UC BerkeleyBJT Differential Pair• With the addition of a “tail current,” an elegant and robust differential pair is achieveddifferential pair is achieved.EE105 Fall 2007 Lecture 22, Slide 9Prof. Liu, UC BerkeleyCommon‐Mode Response• Due to the fixed tail current, the input common‐mode value can vary without changing the output common‐mode value.y gg p21BEBEVV=22121EECCBEBEIII ==2EECCCYXIRVVV −==EE105 Fall 2007 Lecture 22, Slide 10 Prof. Liu, UC BerkeleyDifferential Re sponseEECIII==01EECCCXCVVIRVVI−==02EE105 Fall 2007 Lecture 22, Slide 11 Prof. Liu, UC BerkeleyCCYVV=Differential Re sponse (cont’d)EECIII==02EECCCYCVVIRVVI−==01EE105 Fall 2007 Lecture 22, Slide 12 Prof. Liu, UC BerkeleyCCXVV=Differential Pair Characteristics• A differential input signal results in variations in the output currents and voltages, whereas a common‐mode input signal g,p gdoes not result in any output current/voltage variations.EE105 Fall 2007 Lecture 22, Slide 13 Prof. Liu, UC BerkeleyVirtual Ground• For small input voltages (+∆V and ‐∆V), the gmvalues are ~equal, so the increase in IC1and decrease in IC2are ~equal in q,C1C2qmagnitude. Thus, the voltage at node P is constant and can be considered as AC ground. IIIEEC∆+=1IIIIIEECC∆−=∆+=221IIC∆22V∆0VgIVmCP∆=∆=∆10EE105 Fall 2007 Lecture 22, Slide 14 Prof. Liu, UC BerkeleyVgImC∆−=∆2Extension of Virtual Ground• It can be shown that if R1 = R2, and the voltage at node A goes up by the same amount that the voltage at nodeBgoes downup by the same amount that the voltage at node Bgoes down, then the voltage at node X does not change.00=XvEE105 Fall 2007 Lecture 22, Slide 15 Prof. Liu, UC BerkeleySmall‐Signal Differential Gain• Since the output signal changes by ‐2gm∆VRCwhen the input signal changes by 2∆V, the small‐signal voltage gain is –gmRC. • Note that the voltage gain is the same as for a CE stage, but that the power dissipation is doubled. CmCmvRgVRgA−=∆−=2CmvgV∆2EE105 Fall 2007 Lecture 22, Slide 16 Prof. Liu, UC BerkeleyLarge‐Signal AnalysisTininEEVVVI21exp−TininTCVVVVI211exp1−+=ininEECTVVII212exp1−+=TVexp1+EE105 Fall 2007 Lecture 22, Slide 17 Prof. Liu, UC BerkeleyInput/Output CharacteristicsVVEE105 Fall 2007 Lecture 22, Slide 18 Prof. Liu, UC BerkeleyTininEECoutoutVVVIRVV2tanh2121−−=−Linear/Nonlinear Regions of OperationAmplifier operating in linear region Amplifier operating in non-linear regionEE105 Fall 2007 Lecture 22, Slide 19 Prof. Liu, UC BerkeleySmall‐Signal AnalysisEE105 Fall 2007 Lecture 22, Slide 20 Prof. Liu, UC BerkeleyHalf Circuits• Since node P is AC ground, we can treat the differential pair as two CE “half circuits.”CmininoutoutRgvvvv−=−−2121inin21EE105 Fall 2007 Lecture 22, Slide 21 Prof. Liu, UC BerkeleyHalf Circuit Example 1Omininoutoutrgvvvv−=−−2121EE105 Fall 2007 Lecture 22, Slide 22 Prof. Liu, UC BerkeleyHalf Circuit Example 2()1311|||| RrrgAOOmv−=EE105 Fall 2007 Lecture 22, Slide 23 Prof. Liu, UC BerkeleyHalf Circuit Example 3()1311|||| RrrgAOOmv−=EE105 Fall 2007 Lecture 22, Slide 24 Prof. Liu, UC BerkeleyHalf Circuit Example 4CRA−=EmvRgA+1EE105 Fall 2007 Lecture 22, Slide 25 Prof. Liu, UC BerkeleyDifferential Pair Frequency Response• Since the differential pair can be analyzed using its half circuit, its transfer function, I/O impedances, locations of poles/zeros are the same as that of its half circuit.EE105 Fall 2007 Lecture 22, Slide 26 Prof. Liu, UC
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