Fundamentals of Microelectronics Chapter 7 CMOS ... - DECOM

Fundamentals of MicroelectronicsChapter 7 CMOS Amplifiers CH1 Why Microelectronics? CH2 Basic Physics of Semiconductors CH3 Diode Circuits CH4 Physics of Bipolar Transistors CH5 Bipolar Amplifiers CH6 Physics of MOS Transistors CH7 CMOS Amplifiers CH8 Operational Amplifier As A Black Box 7.1 General Considerations 7.2 Common-Source Stage 7.3 Common-Gate Stage 7.4 Source Follower 7.5 Summary and Additional Examples1CH7 CMOS Amplifiers 2Chapter OutlineMOS BiasingV = −(V −V) +GSTH1V1=WµnCoxRSL12 ⎛ RV2 DD⎞V1+ 2V1⎜−VTH⎟⎝R1+ R2⎠ Voltage at X is determined by V DD , R 1 , and R 2 . V GS can be found using the equation above, and I D can befound by using the NMOS current equation.CH7 CMOS Amplifiers 3CH7 CMOS Amplifiers 41

Self-Biased MOS StageCurrent SourcesI R + V + R I = VDDGSSDDD The circuit above is analyzed by noting M 1 is in saturationand no potential drop appears across R G .CH7 CMOS Amplifiers 5 When in saturation region, a MOSFET behaves as a currentsource. NMOS draws current from a point to ground (sinks current),whereas PMOS draws current from V DD to a point (sourcescurrent).CH7 CMOS Amplifiers 6Common-Source StageOperation in SaturationR I < VD DDD−( V −V)GSTHλ = 0A = −gRvA = −vmD2µCnCH7 CMOS Amplifiers 7oxWILDRD In order to maintain operation in saturation, V out cannot fallbelow V in by more than one threshold voltage. The condition above ensures operation in saturation.CH7 CMOS Amplifiers 82

CS Stage with λ=0CS Stage with λ ≠ 0A = −gRRRvinout= ∞= RmLLA = −gRRvinout= ∞( R || r ) However, Early effect and channel length modulation affectCE and CS stages in a similar manner.mLL= R || rOO9CH7 CMOS Amplifiers 10CS Gain Variation with Channel LengthCS Stage with Current-Source LoadA =vW2µnCoxL ∝λ I Since λ is inversely proportional to L, the voltage gainactually becomes proportional to the square root of L.D2µnCoxWLICH7 CMOS Amplifiers 11DA = −gRvout= rm1O1( r || r )|| rO1O2O2 To alleviate the headroom problem, an active currentsourceload is used. This is advantageous because a current-source has a highoutput resistance and can tolerate a small voltage dropacross it.CH7 CMOS Amplifiers 123

PMOS CS Stage with NMOS as LoadCS Stage with Diode-Connected LoadA = −gv( rO1||r2)m2 O Similarly, with PMOS as input stage and NMOS as the load,the voltage gain is the same as before.CH7 CMOS Amplifiers 13A =−gvA =−gvm1m11⋅gm2( W/L)1=−( W/L)⎛ 1⎜ || rO2|| r⎝gm2 Lower gain, but less dependent on process parameters.CH7 CMOS Amplifiers 14O1⎞⎟⎠2CS Stage with Diode-Connected PMOS DeviceCS Stage with Degeneration⎛ 1 ⎞A = −g⎜ || r || r ⎟v m2 o1o2⎝ gm1⎠ Note that PMOS circuit symbol is usually drawn with thesource on top of the drain.15RDAv=−1+ RSgλ = 0m Similar to bipolar counterpart, when a CS stage isdegenerated, its gain, I/O impedances, and linearity change.CH7 CMOS Amplifiers 164

Example of CS Stage with DegenerationCS Stage with Gate ResistanceVRG= 0RDAv= −1 1+g gm1m2 A diode-connected device degenerates a CS stage.CH7 CMOS Amplifiers 17 Since at low frequencies, the gate conducts no current,gate resistance does not affect the gain or I/O impedances.CH7 CMOS Amplifiers 18Output Impedance of CS Stage with DegenerationOutput Impedance Example (I)r ≈ g r R + routm OSO⎛ 1 ⎞ 1Rout = rO1⎜1+ g ⎟+⎝ g ⎠ gm1m2m2 Similar to the bipolar counterpart, degeneration boostsoutput impedance.CH7 CMOS Amplifiers 19 When 1/g m2 is parallel with r O2 , we often just consider 1/g m2 .CH7 CMOS Amplifiers 205

Output Impedance Example (II)CS Core with Biasingout( g m 1rO 21) r O 1R ≈ + In this example, the impedance that degenerates the CSstage is r O , instead of 1/g m in the previous example.CH7 CMOS Amplifiers 21R1|| R2−RDR1|| R2Av= ⋅ , Av= − gmRRG+ R1|| R 12+ RRG+ R1|| R2Sgm Degeneration is used to stabilize bias point, and a bypasscapacitor can be used to obtain a larger small-signalvoltage gain at the frequency of interest.CH7 CMOS Amplifiers 22DCommon-Gate StageSignal Levels in CG StageA = g RvmD Common-gate stage is similar to common-base stage: arise in input causes a rise in output. So the gain is positive.CH7 CMOS Amplifiers 23 In order to maintain M 1 in saturation, the signal swing atV out cannot fall below V b -V TH .CH7 CMOS Amplifiers 246

I/O Impedances of CG StageCG Stage with Source Resistance1Rin=gλ = 0mRout= R Dvoutvout( 1/ gm)RD= gmRD⇒ =gmRD=1/ gMvvin( 1/ gm) + R 1Sin+ RS1/ g + RgmSm The input and output impedances of CG stage are similarto those of CB stage.CH7 CMOS Amplifiers 25 When a source resistance is present, the voltage gain isequal to that of a CS stage with degeneration, only positive.CH7 CMOS Amplifiers 26Generalized CG BehaviorExample of CG Stage( )ORout= 1 + gmrORS+ r When a gate resistance is present it does not affect the gainand I/O impedances since there is no potential drop acrossit ( at low frequencies). The output impedance of a CG stage with source resistanceis identical to that of CS stage with degeneration.CH7 CMOS Amplifiers 27vvoutinVXVin=gm11//g m11//gm21+ RSgm2gm1R=1 + ( g m 1+ gDm 2) R S11=RS( gm1 + gm2) + 1⎡ ⎛ 1 ⎞ ⎤Rout≈ ⎢gm1rO1⎜|| RS⎟+ rO1|| Rg⎥⎣ ⎝ m2⎠ ⎦ Diode-connected M 2 acts as a resistor to provide the biascurrent.CH7 CMOS Amplifiers 28D7

CG Stage with BiasingSource Follower StagevoutR=v R ||in33||( 1/ gm)( 1/ g ) +m⋅gRRGm DAv

Output Resistance of Source FollowerSource Follower with Biasing1 WI( ) 2D= µnCoxVDD−IDRS−VTH2 L1 1Rout= || rO|| RL≈ || RLg gm The output impedance of a source follower is relatively low,whereas the input impedance is infinite ( at lowfrequencies); thus, a good candidate as a buffer.CH7 CMOS Amplifiers 33m R G sets the gate voltage to V DD , whereas R S sets the draincurrent. The quadratic equation above can be solved for I D .CH7 CMOS Amplifiers 34Supply-Independent BiasingExample of a CS Stage (I) If R s is replaced by a current source, drain current I Dbecomes independent of supply voltage.CH7 CMOS Amplifiers 35⎛ 1⎞Av= − gm 1⎜|| rO1|| rO2|| rO3⎟⎝ gm 3⎠1Rout= || rO1|| rO2|| rO3gm 3 M 1 acts as the input device and M 2 , M 3 as the load.CH7 CMOS Amplifiers 369

Example of a CS Stage (II)Examples of CS and CG StagesrO2Av=−1 1+ || rg gm1 M 1 acts as the input device, M 3 as the source resistance,and M 2 as the load.CH7 CMOS Amplifiers 37m3O3[( 1+gm2rO2)RSrO2] rO1A +v_ CS= −gm1||r // rO2=+ RSgO1v_ CG1m2 With the input connected to different locations, the twocircuits, although identical in other aspects, behavedifferently.CH7 CMOS Amplifiers 38AExample of a Composite Stage (I)Example of a Composite Stage (II)RDAv=1 1+g gm1m2 By replacing the left side with a Thevenin equivalent, andrecognizing the right side is actually a CG stage, thevoltage gain can be easily obtained.CH7 CMOS Amplifiers 39vvout21|| rgm3= −1|| rgm2O3O2|| r This example shows that by probing different places in acircuit, different types of output can be obtained. V out1 is a result of M 1 acting as a source follower whereasV out2 is a result of M 1 acting as a CS stage withdegeneration.CH7 CMOS Amplifiers 40inO41+gm110