Small-Signal Analysis of BJT Differential Pairs

Transcription

1 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 1/1 SallSgnal Analyss of BJT Dfferental Pars Now lets conser the case where each nput of the fferental par conssts of an entcal D bas ter B, an also an A sallsgnal coponent (.e., 1 an ) ( ) O 1 t ( ) O t B ( ) Q Q 1 t 1 E 1 E ( ) t B BE 1 BE EE As a result, the opencrcut output oltages wll lkewse hae a D an sallsgnal coponent.

2 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par /1 ecall that we can alternately express these two sallsgnal coponents n ters of ther aerage (coonoe): c ( t) ( t) 1 an ther fferental oe: Such that: ( t) ( t) ( t) 1 = c = c 1.E.: ( ) O 1 t ( ) O t ( ) ( ) Q Q t 1 B c t c E 1 E B BE 1 BE EE

3 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 3/1 Now, let s eterne the sallsgnal oltage gan of ths aplfer! Q: What o you ean by gan? s t: A o o1 o o1 o or Ao or Ao or Ao??? 1 1 A: Actually, none of those efntons! Ths s a fferental aplfer, so we typcally efne gan n ters of ts coonoe ( A c ) an fferental ( A ) gans: o1 o o1 o c = an A = c c A So that: ( ) ( ) ( ) 1 t = A t A t o c c ( ) ( ) ( ) t = A t A t o c c Q: So how o we eterne the fferental an coonoe gans? A: The frst step of course s to accoplsh a D analyss; turn off the sallsgnal sources!

4 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 4/1 O 1 O B Q Q 1 B Ths D analyss s qute sple! EE 1. Snce the D base oltage B s the sae for each transstor, we know the two etter currents wll each : = = E 1 E We know one current, we know e all! = =α 1 = = B1 B 1 1 ( β )

5 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 5/1 Lkewse, for the BJTs to n acte oe, we know that: > 0 >. B B Fro KL, the collectoltage s: = = α Therefore, n orer for the BJTs to n the acte oe: B α > < B α. Now, we eterne the sallsgnal paraeters of each transstor: g = g = = α 1 T T r r B = = = π1 π T 1 ( β ) 1 T r = r = A o1 o α

6 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 6/1 3. Turnng off the D sources: o1 o ( ) t ( ( ) Q Q t ) c t 1 c e1 e An now nsertng the hybrp BJT oel: ( ) o1 t ( ) o t ( ) c t ( ) c t ( t) 1 g 1 g ( t) Now, tyng ths scheatc up a bt:

7 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 7/1 ( ) o1 t ( ) o t 1 g g 1 ( ) c t ( ) c t ( t) ( t) Q: Ykes! How o we analyze ths ess? A: n a wor, superposton! Q: see, we turn off three sources an analyze the crcut wth the one reanng source on. We then oe to the next source, untl we hae four separate analyss then we a the results together, rght? A: t s actually uch easer than that!

8 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 8/1 We frst turn off the two fferentaloe sources, an analyze the crcut wth only the two reanng (equal alue) coonoe sources. ( ) o1 t ( ) o t 1 g g 1 ( ) c t ( ) c t Fro ths analyss, we can eterne thngs lke the coonoe gan an nput resstance!

9 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 9/1 We then turn off the two coonoe sources, an analyze the crcut wth only the two (equal but opposte alue) fferentaloe sources. ( ) o1 t ( ) o t 1 g g 1 Fro ths analyss, we can eterne thngs lke the fferental oe gan an nput resstance! Q: Ths stll looks ery ffcult! How o we analyze these fferental an coonoe crcuts? A: The key s crcut syetry. We notce that the coonoe crcut has a perfect plane of reflecton (.e., blateral) syetry:

10 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 10/1 ( ) o1 t c1 c ( ) o t ( ) e1 t ce1 g g 1 e 1 e ce ( ) e t 1 n1 ( ) c t n ( ) c t The left an rght se of the crcut aboe are rror ages of each other (nclung the sources wth equal alue c ). The two ses of the crcut a perfectly an precsely equalent, an so the currents an oltages on each se of the crcut ust lkewse perfectly an precsely equal! For exaple: = 1 = o1 o = ce1 ce = e1 e an g = n1 n g 1 = c1 c = = e1 e

11 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 11/1 Q: Wat! You say that cause of crcut syetry that: =. e1 e But, just look at the crcut; fro KL t s eent that: = e1 e How can both stateents correct? A: Both stateents are correct! n fact, the stateents (taken together) tell us what the sallsgnal etter currents ust (for ths coonoe crcut). There s only one possble soluton that satsfes the two equatons the coonoe, sallsgnal etter currents ust equal to zero! = = = e1 e e 0 Hopefully ths result s a bt obous to you. f a crcut possess a plane of perfect reflecton (.e., blateral) syetry, then no current wll flow across the syetrc plane. f t, then the syetry woul estroye! Thus, a plane of reflecton syetry n a crcut s known as rtual open no current can flow across t!

12 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 1/1 ( ) o t c c ( ) o t ( ) e t ce ce g g 0 0 ( ) e t n ( ) c t The rtual Open n ( ) c t Thus, we can take paf scssors an cut ths crcut nto two entcal halfcrcuts, wthout affectng any of the currents oltages the two crcuts on ether se of the rtual open are copletely nepenent! ( ) o t g The oonmoe Half rcut n ( ) c t

13 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 13/1 Now, snce an, we can splfy the crcut by approxatng t as an open crcut: ( ) o t g n ( ) c t Now, let s analyze ths halfcrcut! Fro Oh s Law: = r π n An fro KL: n = g Thus cobnng: ( g r ) = = β π

14 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 14/1 Q: Ykes! How can = β?? The alue β s not equal to 1!! A: You are rght ( β 1)! nstea, we ust conclue fro the equaton: = β that the sallsgnal oltage ust equal to zero! = Q: No way! f = 0, then g = 0. No current s flowng, an so the output oltage o ust lkewse equal to zero! A: That s precsely correct! The output oltage s approxately zero: 0 o 0 Q: Why you say approxately zero?? A: eer, we neglecte the output resstance o r n our crcut analyss. f we ha explctly nclue t, we woul fn that the output oltage woul ery sall, but not exactly zero.

15 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 15/1 Q: So what oes ths all ean? A: t eans that the coonoe gan of a BJT fferental par s ery sall (alost zero!). A c o = c 0 Lkewse, we fn that: n Such that the coonoe nput resstance s really bg: 0 c n!!! The coonoe coponent of nputs 1 an hae rtually no effect on a BJT fferental par! Q: So what about the fferental oe? A: Let s coplete our superposton an fn out!

16 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 16/1 ( ) o1 t ( ) o t 1 g g 1 Q: Hey, t looks lke we hae the sae syetrc crcut as fore won t we get the sae answers? ( ) o1 t c1 c ( ) o t ( ) e1 t g g 1 ce1 e 1 e ce ( ) e t 1 n1 n

17 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 17/1 A: Not so fast! Look at the twosall sgnal sources they are equal but opposte. The fact that the two sources hae opposte sgn changes the syetry of the crcut. nstea of each current an oltage on ether se of the syetrc plane ng equal to the other, we fn that each current an oltage ust equal but opposte! For exaple: = 1 = o1 o = ce1 ce = e1 e an g = n1 n =g 1 = c1 c = e1 e Ths type of crcut syetry s referre to as o syetry; the coonoe crcut, n contrast, possesse een syetry. Q: Wat! You say that cause of crcut syetry that: =. e1 e But, just look at the crcut; fro KL t s eent that: = e1 e

18 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 18/1 How can both stateents correct? A: Both stateents are correct! n fact, the stateents (taken together) tell us what the sallsgnal etteltages ust (for ths fferentaloe crcut). There s only one possble soluton that satsfes the two equatons the fferentaloe, sallsgnal etter oltages ust equal to zero! e1 = e = e = 0 More generally, the electrc potental at eery locaton along a plane of o reflecton syetry s zero olts. Thus, the plane of o crcut syetry s known as rtual groun! ( ) o t c c ( ) o t e = 0 g g ce1 e e ce e= 0 n The rtual Groun n

19 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 19/1 Agan, the crcut has two solate an nepenent hales. We can take our scssors an cut t nto two separate halfcrcuts : ( ) o t g n The DfferentalMoe Half rcut Note the only fference (ase fro the sallsgnal source) tween the fferental halfcrcut an ts coonoe counterpart s that the etter s connecte to groun t s a coonetter aplfer! Let s reraw ths halfcrcut an see f you recognze t: n ( ) o t g

20 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 0/1 Q: Hey, we e seen ths crcut (about a llon tes) fore! We know that: g g ( ) = t o o t ( ) ( r ) ( ) t An also: n ( ) = t 1 r π ght? A: Exactly! Fro ths we can conclue that the fferentaloe sallsgnal gan s: t ( ) o 1 = g ( ) t A An the fferental oenput resstance s: n n ( ) t ( ) t = r π n aton, t s eent (fro past analyss) that the output resstance s: out = ro

21 5/11/011 Dfferental Moe Sall Sgnal Analyss of BJT Dff Par 1/1 Now, puttng the two peces of our superposton together, we can conclue that, gen sallsgnal nputs: = c = c 1 The sallsgnal outputs are: ( ) ( ) ( ) ( ) 1 t = A t A t A t o c c ( ) ( ) ( ) ( ) t = A t A t A t o c c Moreoer, f we efne a fferental output oltage: o o1 o t t t ( ) ( ) ( ) Then we fn t s relate to the fferental nput as: ( t) = A( t) o Thus, the fferental par akes a ery goo fference aplfer the kn of gan stage that s requre n eery operatonalaplfer crcut!