Lecture 17. Differential Amplifiers II Current Mirror Load and Single-Ended Output

Transcription

1 Lecture 17 Differential mplifiers II Current Mirror Load and Single-Ended Output In this lecture you will learn: Differential mplifiers Use of Current Mirrors in Differential mplifiers Small Signal and Large Signal Models with Current Mirrors ECE 315 Spring 007 Farhan Rana Cornell Uniersity Differential mplifier: Reiew Difference-Mode Gain: od d gm1ro1 R Common-Mode Gain: c oc ic o1 gm1 ro1 R r 1 oc 1 g g r r R m1 mb1 o1 Common-Mode Rejection Ratio (CMRR): CMRR d r 1 oc 1 gm1 gmb1ro1 c ro1 R ~ gm1 roc Large if r oc is large ECE 315 Spring 007 Farhan Rana Cornell Uniersity 1

2 Differential mplifier: Reiew DD B DD Difference-Mode Gain: Just copy from the preious sle: IC O1 o1 i d O o i d M IC od d gmn ron rop ECE 315 Spring 007 Farhan Rana Cornell Uniersity FET CS mplifier Operation: Reiew DD DD IN I D OUT IN ID in OUT out I I i D D d gs ds GS DS g g g m gs o ds g m in o out ssume a NFET CS amp with an eal current source bias ID Since: 0 gmin goout 0 out gm gmo r in go ECE 315 Spring 007 Farhan Rana Cornell Uniersity

3 Differential mplifier: Reiew DD B DD Difference-Mode Gain: od d gmn ron rop O1 o1 O o IC i d i d M IC Lets get the aboe results GIN by another way. ECE 315 Spring 007 Farhan Rana Cornell Uniersity DD Differential mplifier: Double-Ended Output B O1 o1 DD For : i g g d1 mn gs1 on ds1 g g mn on o1 1 IC M IC ECE 315 Spring 007 Farhan Rana Cornell Uniersity 3

4 I IC BIS i DD 1 d1 Differential mplifier: Double-Ended Output B O1 o1 DD M IC For : i g g ECE 315 Spring 007 Farhan Rana Cornell Uniersity d1 mn gs1 on ds1 g g mn on o1 For : 1 3 g g Equating: mp gs3 op ds3 g 0 g mp op o1 g g g g mn on o1 op o1 mn o1 gon gop g r r mn on op Differential mplifier: Reiew O1 o1 DD B DD O o o1 gmn ron rop o o1 gmn ron rop od o1o gmn ron rop IC M IC Difference-Mode Gain: od d gmn ron rop ECE 315 Spring 007 Farhan Rana Cornell Uniersity 4

5 Differential mplifier: Double-Ended Output The FET differential amplifiers consered had a double-ended output i1 o1 o1 d d i1 i c ic c i1 i i - o o d d i1 i c ic c i1 i Difference-Mode Output: od o1 o d Common-Mode Output: o o oc 1 cic ECE 315 Spring 007 Farhan Rana Cornell Uniersity Differential mplifier: Conersion to Single-Ended Output Suppose one tries to connect a load to one of the outputs: i1 i - o1 o R L out - out o1 d c ic d We hae lost half of the oltage We can do better. Try another scheme: i1 i out o1 - o od o1 o d ECE 315 Spring 007 Farhan Rana Cornell Uniersity RL out We hae recoered the full signal but this scheme is not always practical - 5

6 Differential mplifier: Conersion to Single-Ended Output DD DD R I1 I GS1 - - GS R O1 O I I D1 D R L This is an awkward way to connect the load at the output! Most of the time what one would really like to hae is a single-ended output (without loosing the factor of ): i1 i - o i1 i out o d c ic d d RL out - ECE 315 Spring 007 Farhan Rana Cornell Uniersity DD FET Differential mplifier: Current Mirror Load DD O1 OUT out M I1 I The current mirror load will let us recoer the factor of two with a single-ended output! ECE 315 Spring 007 Farhan Rana Cornell Uniersity 6

7 DD O1 o1 1 IC FET Differential mplifier: Current Mirror Load ~ S DD M Now a small signal difference-mode input signal is also applied OUT IC out i g g d1 mn gs1 on ds1 g g mn on o1 ssume approximate small signal ground ECE 315 Spring 007 Farhan Rana Cornell Uniersity I BIS IC i DD d1 O1 o1 1 ssume approximate small signal ground FET Differential mplifier: Current Mirror Load ~ S DD M Now a small signal difference-mode input signal is also applied OUT IC out ECE 315 Spring 007 Farhan Rana Cornell Uniersity 1 gmngs1gonds1 gmn gono1 1 3 gmpgs3 gopds3 gmpo1gopo1 gmpo1 Equating: gmn gono1 gmpo1 mn o1 gon gmp g g g mn mp 7

8 DD FET Differential mplifier: Current Mirror Load DD We hae: 1 O1 o1 1 IC ssume approximate small signal ground ~ S OUT out M IC o1 g g mn mp What we hae now is the following situation: is trying to push down extra current because its GS4 has become more negatie nd M is trying to draw less current because its GS has become less positie The output node oltage OUT must increase by a lot so that the current pushed down by is decreased and the current pulled in by M is increased until these currents are both equal ECE 315 Spring 007 Farhan Rana Cornell Uniersity IC DD 1 O1 o1 1 ssume approximate small signal ground FET Differential mplifier: Current Mirror Load DD So far we hae: gmn o1 g ~ S OUT out M IC ECE 315 Spring 007 Farhan Rana Cornell Uniersity mp Lets look at the right arm now: gmngs gonds g g 4 g g mn on out mp gs4 op ds4 g g mp o1 op out Equating: gmn gonout gmpo1 gopout g g g g out gmn g r r g g mn on out mn op out d mn on op on op 8

9 FET Differential mplifier with Current Mirror: Small Signal nalysis DD DD Need to make and analyze a small circuit model and calculate: M O o I1 i1 i i d o (calculated under a pure difference-mode input) Difference-Mode Gain o c ic Common-Mode Gain (calculated under a pure common-mode input) ECE 315 Spring 007 Farhan Rana Cornell Uniersity FET Differential mplifier with Current Mirror: Small Signal nalysis ECE 315 Spring 007 Farhan Rana Cornell Uniersity 9

10 FET Differential mplifier with Current Mirror: Small Signal nalysis ECE 315 Spring 007 Farhan Rana Cornell Uniersity FET Differential mplifier with Current Mirror: Small Signal nalysis Small signal circuit model Note the lack of symmetry!! ECE 315 Spring 007 Farhan Rana Cornell Uniersity 10

11 Small Signal nalysis: Difference-Mode Input There are three unknown internal oltage ariables: We therefore need three equations Try KCL at three different nodes o1 o s ECE 315 Spring 007 Farhan Rana Cornell Uniersity Small Signal nalysis: Difference-Mode Input (1) (1) gmn s gon o1s gmbs gmpo1 0 small mn s on mb s on mp o1 g g g g g 0 gmn s gon gmb s gmpo1 0 ECE 315 Spring 007 Farhan Rana Cornell Uniersity 11

12 Small Signal nalysis: Difference-Mode Input () () gmn s on o s mb s mp o1 op o 0 g g g g gmn s gon gop o gon gmb s gmpo1 0 ECE 315 Spring 007 Farhan Rana Cornell Uniersity Small Signal nalysis: Difference-Mode Input Now subtract () from (1): g g g 0 mn on op o g g r r mn o o mn on op gon gop o g r r d mn on op ECE 315 Spring 007 Farhan Rana Cornell Uniersity 1

13 Small Signal nalysis: Difference-Mode Input () One can also get the correct results by assuming a-priori that the in the differencemode the oltage s is approximately zero (i.e. the node is a small signal ground) If s is assumed to be approximately zero then: gmn gmpo1 gmn g o1 mn o o gmp gon gop gon gop g r r mn on op ECE 315 Spring 007 Farhan Rana Cornell Uniersity Small Signal nalysis: Common-Mode Input g mp o1 g mp o 1 One good way to think about the amplifier in common-mode operation: If the output resistance of is assumed to be ery large, then the current mirror, as the name suggests will ensure that the drain currents of and M are entical (both equal to g mp o1 ) as you can see aboe as well. ECE 315 Spring 007 Farhan Rana Cornell Uniersity 13

14 Small Signal nalysis: Common-Mode Input g mp o1 g mp o 1 One good way to think about the amplifier in common-mode operation: If the output resistance of is assumed to be ery large, then the current mirror, as the name suggests will ensure that the drain currents of and M are entical (both equal to g mp o1 ) as you can see aboe as well. Since the input oltages, ic, on the two ses are also the same, the circuit (at least the bottom portion) has complete left-right symmetry in common-mode operation ECE 315 Spring 007 Farhan Rana Cornell Uniersity Small Signal nalysis: Common-Mode Input So one can assume that the in the common-mode, een with the current mirror, no current flows in the bottom most horizontal wire (i.e. it is a small signal open) ECE 315 Spring 007 Farhan Rana Cornell Uniersity 14

15 Small Signal nalysis: Common-Mode Input (1) (3) Doing KCL at (1) and (3) gies: gmn ron gmp o1 1 ron roc mn gmp g g r r mb on oc ic ECE 315 Spring 007 Farhan Rana Cornell Uniersity Small Signal nalysis: Common-Mode Input () Using the left-right symmetry gies: c o ic r on o o o1 1 g mp r oc g g mn mp on g g r r mn mb ECE 315 Spring 007 Farhan Rana Cornell Uniersity r on oc 15

16 Small Signal nalysis: Output Resistance i t R r r out on op Left for homework ECE 315 Spring 007 Farhan Rana Cornell Uniersity Two-Port Model of a FET Differential mplifier with Current Mirror i1 R out i R in - i1 i d i1i c o g r r d c mn o ic on R out ron rop op gmn r g on mp 1 r r g g r r g on oc mn mb on oc mp Rin ECE 315 Spring 007 Farhan Rana Cornell Uniersity 16

17 FET Cascode Differential mplifier with Cascode Current Mirror DD M7 M8 DD What if one needs a large gain and a large output resistance from a differential (transconductance) amplifier? M5 M6 O o g r r g r r R out mn on on mp op op g g r r g r r d mn mn on on mp op op BIS M I1 i1 i i ECE 315 Spring 007 Farhan Rana Cornell Uniersity ECE 315 Spring 007 Farhan Rana Cornell Uniersity 17