Lecture 24Common-Mode (CM) ResponseEquilibrium Overdrive VoltageMinimum CM Output VoltageDifferential ResponseSmall-Signal ResponseSmall-Signal Differential GainLarge-Signal AnalysisMaximum Differential Input VoltageMOSFET vs. BJT Differential PairsEffect of Doubling the Tail CurrentEffect of Doubling W/LSmall-Signal AnalysisVirtual Ground and Half CircuitMOSFET Diff. Pair Frequency ResponseExampleHalf Circuit Example 1Half Circuit Example 2MOSFET Cascode Differential PairMOSFET Telescopic Cascode AmplifierCM to DM Conversion Gain, ACM-DMMOS Diff. Pair with Active LoadAsymmetric Differential PairThevenin Equivalent of the Input PairSimplified Diff. Pair w/ Active LoadEE105 Fall 2007 Lecture 24, Slide 1 Prof. Liu, UC BerkeleyLecture 24OUTLINE•MOSFET Differential AmplifiersReading: Chapter 10.3-10.6ANNOUNCEMENTS•The last HW assignment (HW#12) will be due 12/6•Prof. Liu will be away on Tuesday 12/4 (no office hour that day)EE105 Fall 2007 Lecture 24, Slide 2 Prof. Liu, UC BerkeleyCommon-Mode (CM) Response•Similarly to its BJT counterpart, a MOSFET differential pair produces zero differential output as VCM changes.2SSDDDYXIRVVV EE105 Fall 2007 Lecture 24, Slide 3 Prof. Liu, UC BerkeleyEquilibrium Overdrive Voltage•The equilibrium overdrive voltage is defined as VGS-VTH when M1 and M2 each carry a current of ISS/2. LWCIVVoxnSSequilTHGSEE105 Fall 2007 Lecture 24, Slide 4 Prof. Liu, UC BerkeleyMinimum CM Output Voltage•In order to maintain M1 and M2 in saturation, the common-mode output voltage cannot fall below VCM-VTH. •This value usually limits voltage gain.THCMSSDDDVVIRV 2EE105 Fall 2007 Lecture 24, Slide 5 Prof. Liu, UC BerkeleyDifferential ResponseEE105 Fall 2007 Lecture 24, Slide 6 Prof. Liu, UC BerkeleySmall-Signal Response•For small input voltages (+V and -V), the gm values are ~equal, so the increase in ID1 and decrease in ID2 are ~equal in magnitude. Thus, the voltage at node P is constant and can be considered as AC ground. IIIEED22VgIVgIVmDmDP21 ;0IIIEED21EE105 Fall 2007 Lecture 24, Slide 7 Prof. Liu, UC BerkeleySmall-Signal Differential Gain•Since the output signal changes by -2gmVRD when the input signal changes by 2V, the small-signal voltage gain is –gmRD. •Note that the voltage gain is the same as for a CS stage, but that the power dissipation is doubled.EE105 Fall 2007 Lecture 24, Slide 8 Prof. Liu, UC BerkeleyLarge-Signal Analysis 2211214212ininoxnSSinoxnDDVVLWCIVVLWCIIinEE105 Fall 2007 Lecture 24, Slide 9 Prof. Liu, UC BerkeleyMaximum Differential Input Voltage•There exists a finite differential input voltage that completely steers the tail current from one transistor to the other. This value is known as the maximum differential input voltage. equilTHGSininVVVV 2max21EE105 Fall 2007 Lecture 24, Slide 10 Prof. Liu, UC BerkeleyMOSFET vs. BJT Differential Pairs•In a MOSFET differential pair, there exists a finite differential input voltage to completely switch the current from one transistor to the other, whereas in a BJT differential pair that voltage is infinite. MOSFET Differential Pair BJT Differential PairEE105 Fall 2007 Lecture 24, Slide 11 Prof. Liu, UC BerkeleyEffect of Doubling the Tail Current•If ISS is doubled, the equilibrium overdrive voltage for each transistor increases by , thus Vin,max increases by as well. Moreover, the differential output swing will double. 22EE105 Fall 2007 Lecture 24, Slide 12 Prof. Liu, UC BerkeleyEffect of Doubling W/L•If W/L is doubled, the equilibrium overdrive voltage is lowered by , thus Vin,max will be lowered by as well. The differential output swing will be unchanged.22EE105 Fall 2007 Lecture 24, Slide 13 Prof. Liu, UC BerkeleySmall-Signal Analysis•When the input differential signal is small compared to 4ISS/nCox(W/L), the output differential current is ~linearly proportional to it:•We can use the small-signal model to prove that the change in tail node voltage (vP) is zero: 212121421ininSSoxnoxnSSininoxnDDVVILWCLWCIVVLWCII 2122110vvvgvgmm PPininvvvvvv2121EE105 Fall 2007 Lecture 24, Slide 14 Prof. Liu, UC BerkeleyVirtual Ground and Half Circuit•Since the voltage at node P does not change for small input signals, the half circuit can be used to calculate the voltage gain. DmvPRgAv0EE105 Fall 2007 Lecture 24, Slide 15 Prof. Liu, UC BerkeleyMOSFET Diff. Pair Frequency Response•Since the MOSFET 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 the half circuit’s.EE105 Fall 2007 Lecture 24, Slide 16 Prof. Liu, UC BerkeleyExample 33,3113133,1311,1111])/1([1DBGDDoutpSBDBmmGDGSmYpGDmmGSSXpCCRCCggCCgCggCREE105 Fall 2007 Lecture 24, Slide 17 Prof. Liu, UC BerkeleyHalf Circuit Example 11331||||1OOmmvrrggA0Half circuit for small-signal analysisEE105 Fall 2007 Lecture 24, Slide 18 Prof. Liu, UC BerkeleyHalf Circuit Example 2 212SSmDDvRgRAHalf circuit for small-signal analysis0EE105 Fall 2007 Lecture 24, Slide 19 Prof. Liu, UC BerkeleyMOSFET Cascode Differential PairHalf circuit for small-signal analysis1331 OmOmvrgrgA EE105 Fall 2007 Lecture 24, Slide 20 Prof. Liu, UC BerkeleyMOSFET Telescopic Cascode Amplifier )(||7551331 OOmOOmmvrrgrrggA Half circuit for small-signal analysisEE105 Fall 2007 Lecture 24, Slide 21 Prof. Liu, UC BerkeleyCM to DM Conversion Gain, ACM-DM•If finite tail impedance and asymmetry are both present, then the differential output signal will contain a portion of the input common-mode signal.SSmCMDSSDmDSSDGSCMRgVIRIgIRIVV2122 2121DDoutoutoutDDDoutDDoutRIVVVRRIVRIV SSmDCMoutRgRVV2/1 EE105 Fall 2007 Lecture 24, Slide 22 Prof. Liu, UC BerkeleyMOS Diff. Pair with Active Load•Similarly to its BJT counterpart, a MOSFET differential pair can use an active load to enhance its single-ended output.EE105 Fall 2007 Lecture 24, Slide 23 Prof. Liu, UC BerkeleyAsymmetric Differential Pair•Because of the vast difference in
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