1EE105 - Fall 2006Microelectronic Devices and CircuitsProf. Jan M. Rabaey (jan@eecs)Lecture 13: MOS Single Stage Amplifiers2Overview Last lecture– The Amplifier as a Two-Port This lecture– Common Source (Read § 8.3-8.5)– Common Gate (Read § 8.8.2)– Common Drain (Read § 8.9.2)23Common-Source Amplifier (again)How to isolate DC level?4DC BiasNeglect all AC signals5 V2.5 VChoose IBIAS, W/L35Load-Line Analysis to find QQDDDoutRDVVIR−=110kslope =0V10kDI =5V10kDI =6Input – Output Characteristics+VDDVTnMOSFET is saturated Æhigh slopeMOSFET is triode Ælow slopevIN= vGS0(negative supply, VSS= 0)vO= vDSHigher value of RL means higher gain47Small-Signal AnalysisinR =∞8svsRinRoutRLRminGvinv+−Two-Port Parameters:Find Rin, Rout, GminR =∞mmGg=||out o DRrR=Generic Transconductance Amp59Two-Port CS ModelReattach source and load one-ports:L10Maximize Gain of CS Amp Increase gm(more current) Increase RD(free? Don’t need to dissipate extra power) Limit: Must keep the device in saturation For a fixed current, the load resistor can only be chosen so large – To have good swing we’d also like to avoid getting too close to edge-of-saturation(||)vmDoAgR r=−,DSDDDDDSsatVVIRV=− >How to enable l arge curr ent a nd lar ge r esi sta nce at the sam e time?611Answer: Current Source Load Current independent of voltage for ideal source 12CS Amp with Current Source Supply713Load Line for DC BiasingBoth the I-source and the transistor are idealized for DC bias analysis14Two-Port ParametersFrom currentsource supplyinR =∞||out o ocRrr=mmGg=815P-Channel CS AmplifierDC bias: VGS= VBIAS – VDD sets drain current –IDp= ISUP16Two-Port Model ParametersSmall-signal model for PMOS and for rest of circuit917Common Gate Amplifier18CG as a Current Amplifier: Find Aiout dtiii==−1iA =−1019CG Input ResistanceAt input:Output voltage: touttmgsmbtovvigvgvr⎛⎞−=− + +⎜⎟⎝⎠(|| ) ( ||)out docL tocLvir R ir R=− =gstvv=−()||toc L ttmtmbtovrRiigvgvr⎛⎞−=+ +⎜⎟⎝⎠20Approximations… We have this messy result But we don’t need that much precision. Let’s start approximating:11||1mmbtooc Lin toggirrRRvr++==+1mmboggr+>>||ocLLrRR≈0LoRr≈1inmmbRgg=+1121CG Output Resistance()0sstmgs mbsSovvvgv g vRr−−−− + =11tsmmbSoovvggRrr⎛⎞+++=⎜⎟⎝⎠22CG Output ResistanceSubstituting vs= itRS11ttS m mbSooviR g gRrr⎛⎞++ + =⎜⎟⎝⎠The output resistance is (vt/ it)|| roc|| 1oout oc S m o mb oSrRrR grgrR⎛⎞⎛⎞=+++⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠1223Approximating the CG RoutThe exact result is complicated, so let’s try tomake it simpler:Sgmμ500≈ Sgmbμ50≈Ω≈ kro200][||SSombSomoocoutRRrgRrgrrR +++=][||SSomoocoutRRrgrrR ++≅Assuming the source resistance is less than ro,)]1([||][||SmoocSomoocoutRgrrRrgrrR +=+≈24CG Two-Port ModelFunction: a current buffer• Low Input Impedance• High Output Impedance1325Common-Drain Amplifier21()2DS ox GS TWICVVLμ=−2DSGS ToxIVVWCLμ=+Weak IDSdependence26CD Voltage GainNote vgs= vt–vout||outmgs mb outoc ovgvgvrr=−()||outmt out mb outoc ovgvv gvrr=−−1427CD Voltage Gain (Cont.)KCL at source node:Voltage gain (for vSBnot zero):()||outmt out mb outoc ovgvv gvrr=−−1||mb m out mtoc oggv gvrr⎛⎞++ =⎜⎟⎝⎠1||out minmbmoc ovgvggrr=++1out min mb mvgvgg≈≈+28CD Output ResistanceSum currents at output (source) node:|| ||tout o octvRrri=tmt mbtigv gv=+1outmmbRgg≈+1529CD Output Resistance (Cont.)ro|| roc is much larger than the inverses of the transconductances Æ ignore1outmmbRgg≈+Function: a voltage buffer• High Input Impedance• Low Output
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