Lecture 14: More MOS Circuits and the Differential Amplifier

Transcription

1 Lecture 4: More MOS rcuts an the Dfferental Aplfer Gu-Yeon We Dson of nneern an Apple Scences Harar Unersty We

2 Oerew Rean S&S: hapter 5.0, 6.~, 6.6 ackroun Han seen soe of the basc aplfer crcuts we can bul wth MOSFs, conclue ateral n hapter 5 of S&S by lookn at the hhfrequency oel for MOSFs. We wll use ths hh-frequency oel when analyzn the hh-frequency operaton of MOSFs n aplfer crcuts. he perforance of aplfer crcuts can be proe by usn a fferental par topoloy. Dfferental par aplfers hae two nputs poste an neate ternals. he fferental par aplfer s what we assue for the eal aplfer when we learne about op ap crcuts. We wll now nestate how to bul these aplfers. he rean n Sectons 6.~ proes portant an necessary backroun nforaton for fferental aplfers. hen, we wll skp to 6.6 whch nestates buln fferental aplfer wth MOSFs. We S54 - Lecture 4

3 MOSF nternal apactances Fro our stuy of the physcal operaton of MOSFs, we can see that there are nternal capactances Gate capactance fro ate oxe (parallel plate an frnn capactors) s,, b Juncton capactances fro source-boy an ran-boy epleton layer capactances (reerse base PN junctons) sb, b s,, b b We S54 - Lecture 4 3

4 MOS Gate ap he three ate capactances ( s,, b ) epen on the transstor s oe (reon) of operaton n troe (lnear) reon ( DS sall), channel has unfor epth s WL ox n saturaton reon, channel s tapere an pnche off near the ran. We can approxate the capactances as follows: s WL ox 3 0 n cut off, no channel but oel capactance between bulk an ate b WL here s also an oerlap capactance that shoul be ae to s an s WL o o ox 0 ox We S54 - Lecture 4 4

5 MOS Juncton ap he epleton-layer capactances of the two reerse-base pn junctons are oerne by the follown equaton: sb sb 0 S 0 b b 0 D 0 where 0 s the bult-n potental of the pn junctons (approx. 0.6~0.8, a functon of the N A an N D opn concentratons) We S54 - Lecture 4 5

6 Hh-Frequency Moel We ust auent our (low-frequency) sall-snal oel of the MOS transstor wth these capactors n orer to accurately oel ts operaton at hh frequences coplete splfe when source s connecte to the boy, oel s splfe (reoe sb ) n saturaton, b ~ 0 further splfy oel by reon b for han calculatons We S54 - Lecture 4 6

7 MOSF f We can aan fn the f as a fure of ert for the transstor s hh-frequency operaton (unty current an frequency) Sole for the short crcut output current w.r.t an nput current assues s sall an rops out n aboe eq. o s s s s s s ( ) s o s ( ) s ω s note that f s a functon of We S54 - Lecture 4 7

8 J Dfferental Par Dfferental par crcuts are one of the ost wely use crcut buln blocks. he nput stae of eery op ap s a fferental aplfer asc haracterstcs wo atche transstors wth etters shorte toether an connecte to a current source Deces ust always be n acte oe Aplfes the fference between the two nput oltaes, but there s also a coon oe aplfcaton n the non-eal case Let s frst qualtately unerstan how ths crcut works. NO: hs qualtate analyss also apples for MOSF fferental par crcuts We S54 - Lecture 4 8

9 ase Assue the nputs are shorte toether to a coon oltae, M, calle the coon oe oltae equal currents flow throuh Q an Q etter oltaes equal an at M -0.7 n orer for the eces to be n acte oe collector currents are equal an so collector oltaes are also equal for equal loa resstors fference between collector oltaes 0 What happens when we ary M? As lon as eces n acte oe, equal currents flow throuh Q an Q Note: current throuh Q an Q always a up to, current throuh the current source So, collector oltaes o not chane an fference s stll zero. Dfferental par crcuts thus reject coon oe snals We S54 - Lecture 4 9

10 ase & 3 Q base roune an Q base at All current flows throuh Q No current flows throuh Q tter oltae at 0.3 an Q s J not F -αr Q base roune an Q base at - All current flows throuh Q No current flows throuh Q tter oltae at -0.7 an Q s J not F -αr We S54 - Lecture 4 0

11 ase 4 Apply a sall snal auses a sall poste to flow n Q Requres sall neate n Q snce an be use as a lnear aplfer for sall snals ( s a functon of ) Dfferental par respons to fferences n the nput oltae an entrely steer current fro one se of the ff par to the other wth a relately sall oltae Let s now take a quanttate look at the lare-snal operaton of the fferental par We S54 - Lecture 4

12 S54 - Lecture 4 We J Dff Par Lare-Snal Operaton Frst look at the etter currents when the etters are te toether Soe anpulatons can lea to the follown equatons an there s the constrant: Gen the exponental relatonshp, sall fferences n, can cause all of the current to flow throuh one se S e α S e α e e e e e

13 Notce - ~ 4 enouh to swtch all of current fro one se to the other For sall-snal analyss, we are ntereste n the reon we can approxate to be lnear sall-snal conton: - < / We S54 - Lecture 4 3

14 S54 - Lecture 4 We 4 J Dff Par Sall-Snal Operaton Look at the sall-snal operaton: sall fferental snal s apple expan the exponental an keep the frst two ters e α e e e α ( ) ( ) ( ) α α α α α c α α c ultply top an botto by e

15 S54 - Lecture 4 We 5 Dfferental oltae Gan For sall fferental nput snals, <<, the collector currents are We can now fn the fferental an to be ( ) R R ( ) R R c c R A

16 J Dff Par Dfferental Half rcut We can break apart the fferental par crcut nto two half crcuts whch then looks lke two coon etter crcuts ren by / an / We S54 - Lecture 4 6

17 We can then analyze the sall-snal operaton wth the half crcut, but ust reeber paraeters r π,, an r o are base at / nput snal to the fferental half crcut s / c / r π π π r ο R oltae an of the fferental aplfer (output taken fferentally) s equal to the oltae an of the half crcut c A ( R ro ) We S54 - Lecture 4 7

18 oon-moe Gan When we re the fferental par wth a coon-oe snal, M, the ncreental resstance of the bas current effects crcut operaton an results n soe an (assue to be 0 when R was nfnte) M αr R r e M αr R M αr R We S54 - Lecture 4 8

19 f the output s taken fferentally, the output s zero snce both ses oe toether. Howeer, f taken snle-enely, the coon-oe an s fnte αr Ac R f we look at the fferental an snle-enely, we et A R hen, the coon rejecton rato (MRR) wll be MRR A A c R whch s often expresse n MRR 0lo0 A A c We S54 - Lecture 4 9

20 M an Dfferental Gan quaton nput snals to a fferental par usually conssts of two coponents: coon oe ( M ) an fferental( ) M hus, the fferental output snal wll be n eneral o A ( ) A c We S54 - Lecture 4 0

21 MOS Dff Par he sae basc analyss can be apple to a MOS fferental par W ( ) D, µ nox GS, t L W D, µ nox ( GS, t) L an the fferental nput oltae s GS GS W D D µ nox L Wth soe alebra (etale n S&S p.59) D, ± GS t D D D ( ) ( ) ( ) ( ) GS t GS t GS t We S54 - Lecture 4

22 We et full swtchn of the current when ax ( t) GS We S54 - Lecture 4

23 Next e We wll fnsh off S&S 6.6 by lookn at the sources of offset oltaes n MOS fferental par crcuts. hen, we wll rest current sources yet aan an ntrouce schees that proe ther perforance (hher output resstance). hen, we wll fnsh off MOS fferental par crcuts wth an acte-loae fferental aplfer. hen, we wll shft ears a lttle an look at a crcut technque calle cascon that helps to proe the perforance of aplfer crcuts. he book coers ths technque wth Js. We wll look at the n the context of MOSFs. We S54 - Lecture 4 3