ECE315 / ECE515 Lecture 5 Date:

Transcription

1 Lecture 5 Date: Common Source Amplfer MOSFET Amplfer Dstorton

2 Example 1 One Realstc CS Amplfer Crcut: C c1 : Couplng Capactor serves as perfect short crcut at all sgnal frequences whle blockng any dc C c2 : Couplng Capactor serves as perfect short crcut at all sgnal frequences whle blockng any dc ths makes v o = v d R L : Load Resstor could be actual resstor to whch the output s requred or could be nput resstance of another amplfer stage where more than one amplfcaton stage s needed C S : Bypass Capactor (μf) range provdes small mpedances (deally perfect short crcut) at all sgnal frequences

3 Example 1 (contd.) g 0 Rn RG v v sg R n Rn R sg Usually, R G s very hgh (of the order of MΩ) and therefore: Open Loop Voltage Gan (e, when there v v sg Now, vgs v s no feedback loop from o/p to the /p): Avo gmro RD The overall voltage gan from the sgnal-source to the load s: v v g v r R R o m gs o D L v v A g r R R o o v m o D L vn vgs v sg R G RG R R G g r R R For the determnaton of R out, the sgnal v sg has to be set to zero (replace G v m o D L RG Rsg the sgnal generator wth a short crcut) smple nspecton gves: Rout ro RD sg

4 The total output voltage s the sum of the small-sgnal output voltage and the DC output voltage! MOSFET Amplfer Dstorton D t = I D + d (t) v (t) 4. 0 V V K = = 5K ECE315 / ECE515 v O t = V DS + v o (t) ma V 2 V T = 2. 0 V Lets look at the last example. You needed to perform a small-sgnal analyss to determne the small-sgnal open-crcut v voltage gan A v = o (t) We found that the small-sgnal voltage gan s: Say the nput voltage to ths amplfer s: A v (t) vo v o ( t) v ( t) v ( t ) V cosωt Q: What s the largest value that V can take wthout producng a dstorted output? A: Well, we know that the small-sgnal output s: vo( t ) Avo v ( t ) BUT, ths s not the output voltage! 5.0V cosωt 5.0

5 MOSFET Amplfer Dstorton (contd.) Note for ths example, the DC output voltage s the DC dran voltage, and that ts value s: V O V D 10 V Thus, the total output voltage s : vo ( t ) VD vo( t ) V cosωt It s very mportant that you realze there s a lmt on both how hgh and how low the total output voltage v O (t) can go. That s rght! If the total output voltage v O (t) tres to exceed these lmts even for a moment the MOSFET wll leave saturaton mode. And leavng saturaton mode results n sgnal dstorton!

6 MOSFET Amplfer Dstorton (contd.) Let s break the problem down nto two separate problems: 1) If total output voltage v O (t) becomes too small, the MOSFET wll enter the trode mode 2) If total output voltage v O (t) becomes too large, the MOSFET wll enter the cutoff mode We ll frst consder problem 1. For a MOSFET to reman n saturaton, v DS (t) must reman greater than the excess gate voltage V GS V T all the tme t. Snce the source termnal of the MOSFET n ths crcut s connected to ground, we know that V S = 0. Therefore: And so the MOSFET wll reman n saturaton only f the total output voltage remans larger than V GS V T = V G V T. v t V V DS GS T vds( t) vd( t) vo( t) V GS V G v t V V O GS T Thus, we conclude for ths amplfer that the output floor s: v t DS VGS VT

7 MOSFET Amplfer Dstorton (contd.) Here, V GS = 4.0 V and V T = 2.0 V. Therefore: G t L V V V Thus, to reman n saturaton, the total output voltage must reman larger than the floor voltage at all tme t. v t L 2.0 V O Snce ths total voltage s: v t O ( ) 0 V cosωt we can determne the maxmum value of small-sgnal nput magntude: V cosωt V cosωt V cosωt 1.6 Snce cosωt can be as large as 1.0, we fnd that the magntude of the nput voltage can be no larger than 1.6 V,.e., V 1.6 V If the nput magntude exceeds ths value, the MOSFET wll (momentarly) leave the saturaton regon and enter the dreaded trode mode!

8 MOSFET Amplfer Dstorton (contd.) Now let s consder problem 2 For the MOSFET to reman n saturaton, the dran current must be greater than zero D > 0. Otherwse, the MOSFET wll enter cutoff mode. Applyng Ohm s Law to the dran resstor, we fnd the dran current s: t s evdent that dran current s postve only f: v O 15 V In other words, the upper lmt (.e., the celng ) on the total output voltage s: Snce ths total voltage s: v ( t) V cosωt we can conclude that n order for the MOSFET to reman n saturaton mode: Therefore, we fnd: 5.0 O V cosωt D V v 15 v DD O O R C 5 L V 15.0V DD V cosωt 15.0 Snce cosωt can be as large as 1.0, we fnd that the magntude of the nput voltage can be no larger than: V 1.0 V

9 MOSFET Amplfer Dstorton (contd.) If the nput magntude exceeds 1.0 V, the MOSFET wll (momentarly) leave the saturaton and enter the cutoff regon! In summary: 1) If, V > 1.6 V, the MOSFET wll at tmes enter trode, and dstorton wll occur! 2) If, V > 1.0 V, the MOSFET wll at tmes enter cutoff, and even more dstorton wll occur! To demonstrate ths, let s consder three examples: 1. V < 1. 0 V The output sgnal n ths case remans between V DD = 15.0V and V G V T = 2.0 V for all tme t. Therefore, the output sgnal s not dstorted. L V DD 15 G V O L V V T 10 2 v ( t ) O t

10 MOSFET Amplfer Dstorton (contd.) V > V > 1. 0 V The output sgnal n ths case remans greater than 2 G T for all tme t. However, the small-sgnal output s now large enough so that the total output voltage at tmes tres to exceed L V DD 15. For these tmes, the MOSFET wll enter cutoff, and the output sgnal wll be dstorted. L V V

11 MOSFET Amplfer Dstorton (contd.) 3. V > 1. 6 V In ths case, the small-sgnal nput sgnal s suffcently large so that the total output wll attempt to exceed both lmts (.e., V DD = 15.0 V and V G V T = 2.0 V). Therefore, there are perods of tme when the MOSFET wll be n cutoff, and perods when the MOSFET wll be n saturaton.

12 Effect of Input / Output Loadng ECE315 / ECE515 Sgnal swng: Upswng lmted by resstve dvder: Downswng not affected by loadng Voltage gan: nput loadng (R S ): no effect because gate does not draw current output loadng (R L ): It detracts from voltage gan because t draws current.

13 Common Source (CS) Amplfer Major Lmtatons: ECE315 / ECE515 Increase n A v by ncreasng the R D leads to smaller V D.e, the voltage V DS essentally lmts the voltage swng R D dffcult to fabrcate n smaller chp area a major constrant for ICs Problems wth the precson of R D Actve loads overcome these problems Dode-connected load (FETs n whch dran and gate are ted to work as resstors) Current source (such as a FET operatng n saturaton mode) FET operatng n trode mode We can make a two termnal devce from a MOSFET by connectng the gate and the dran!

14 Dode-Connected Load or Enhancement Load Q: How does ths enhancement load resemble a resstor? A: For ths we need to consder the -v curves for both. For a Resstor Now consder the same curve for an enhancement load. Snce the gate s ted to the dran, we fnd v G = v D, and thus v GS = v DS. As a result, we fnd that v DS > v GS V T always. Therefore, we fnd that f v GS > V T, the MOSFET wll be n saturaton ( D = K(v GS V T ) 2 ), whereas f v GS < V T, the MOSFET s n cutoff ( D = 0).

15 Dode-Connected Load (contd.) ECE315 / ECE515 Snce for enhancement load = D and v = v GS, we can descrbe the enhancement load as: 0 for v V 2 K v VT for v VT T So, resstors and enhancement loads are far from exactly the same, but: 1) They both have = 0 when v = 0. 2) They both have ncreasng current wth ncreasng voltage v.

16 Dode-Connected Load (contd.) Therefore, we can buld a common source amplfer wth ether a resstor, or n the case of an ntegrated crcut, an enhancement load. For the dode-connected load amplfer, the load lne s replaced wth a load curve (v = V DD v DS )! And the transfer functon of the CS Amplfer s:

17 Dode-Connected Load - Analyss Step 1 - DC Analyss If, V > V T then I = K(V V T ) 2 I or: V V T Step 3 Determne the small-sgnal crcut gm 2K V GS VT 2K V VT Step 2 Determne g m and r o r o K I I K V V D T 2 You need to replace all enhancement loads wth ths small-sgnal model whenever you are attemptng to fnd the small-sgnal crcut of any MOSFET amplfer.

18 CS Amplfer wth Dode-connected load Q: What s the small-sgnal open-crcut voltage gan, nput resstance, and output resstance of ths amplfer? A: The values that we wll determne when we follow precsely the same steps as before!! Step 1 DC Analyss Let s ASSUME that both M 1 and M 2 are n saturaton. Then we ENFORCE: I K V V K V V 2 D1 1 GS1 T 1 Contnung wth the ANALYSIS, we can fnd the dran current through the enhancement load (I D2 ), I K V V D2 2 GS2 T G T1 2

19 CS Amplfer wth Dode-connected Load (contd.) I I D1 D2 2 2 K V V K V V 1 G T 1 2 GS2 T 2 K1 V V V V K GS2 G T 1 T 2 2 Snce V DS2 = V GS2 and V DS1 = V DD V DS2, we can lkewse state that: K1 V V V V K DS2 G T 1 T 2 2 K1 V V V V V K DS1 DD T 2 G T 1 2 Now, we must CHECK to see f our assumpton s correct. The saturaton assumpton wll be correct f: K1 V V V V K DS2 G T 1 T 2 2 V V f V V GS1 T 1 G T 1 Step 2 Calculate small-sgnal parameters g 2 K V V and g 2K V V m1 1 G T 1 m2 2 GS2 T 2 r 1 1 and r o I I o1 2 1 D 2 D

20 CS Amplfer wth Dode-connected Load (contd.) Step 3 Determne the small-sgnal crcut Frst, let s turn off the DC sources: We now replace MOSFET M 1 wth ts equvalent smallsgnal model, and replace the dode-connected load wth ts equvalent small-sgnal model. M 2 M 1 v O (t) v (t) + _

21 CS Amplfer wth Dode-Connected Load (contd.) We fnd that: gs1 v v v v v ( g v g v ) ( r r ) o m1 gs1 m2 gs2 o1 o2 ( g v g v ) ( r r ) m1 m2 o o1 o2 Rearrangng, we fnd: A vo v ( r r ) g g o v 1 ( r r ) g g o1 o2 m1 m1 o1 o2 m2 m2 gs2 o W g 2 K I 1 D K m Therefore: 1 L A vo gm2 2 K 2 D K I W L 2 Strong man devce and weak load devce gves hgher gan In other words, we adjust the MOSFET channel geometry to set the small-sgnal gan of ths amplfer!

22 CS Amplfer wth Dode-Connected Load (contd.) Now let s determne the small-sgnal nput and output resstances of ths amplfer! It s evdent that: R v Now for the output resstance, we know that the open-crcut output voltage s: oc v ( g v g v ) ( r r ) o m1 gs1 m2 gs2 o1 o2 Lkewse, the short-crcut output current s: ( g v g v ) os m1 gs1 m2 gs2 Thus, the small-sgnal output resstance of ths amplfer s equal to: oc v ( g v g v ) ( r r ) o m1 gs1 m2 gs2 o1 o2 R ( r r ) o sc o1 o2 ( g v g v ) o m1 gs1 m2 gs2

23 CS Amplfer wth Dode-Connected Load (contd.) Dode-connected load Man devce A g R g v m D m1 A v g 1 g m2 mb2 R D A g v 1 g m2 mb2 g g n ox 2 D2 m1 m ncox ( W / L) 1I D1 1 2 C ( W / L) I 1 g g mb2 m2 If varaton of η wth the output voltage s neglected the gan s ndependent of bas currents and voltages However, for ths to happen the devce M 1 has to reman n saturaton ths ensures that current I D1 s constant ensures constant g m1 In other words, the gan remans relatvely constant for the varaton n nput and output sgnals ensures that nput-output relatonshp s lnear

24 CS Amplfer wth Constant Current Source Now consder ths NMOS amplfer usng a current source. Note no resstors or capactors are present! Ths s a common source amplfer. I D stablty s not a problem! Q: I don t understand! Wouldn t the small-sgnal crcut be: A: Remember, every real current source (as wth every voltage source) has a source resstance r o. A more accurate current source model s therefore:

25 CS Amplfer wth Constant Current Source (contd.) Ideally, r o =. However, for good current sources, ths output resstance s large (e.g., r o = 100 kω). Thus, we mostly gnore ths value (.e., approxmate t as r o = ), but there are some crcuts where ths resstance makes qute a dfference. Ths s one of those crcuts! Therefore, a more accurate amplfer crcut schematc s: And so the small-sgnal crcut becomes the famlar:

26 CS Amplfer wth Constant Current Source (contd.) Therefore, the small sgnal model s: Now go ahead and do the analyss Constant Current Source NFET deal current source PFET deal current source As long as a MOS transstor s n saturaton regon and λ=0, the current s ndependent of the dran voltage and t behaves as an deal current source seen from the dran termnal.

27 Constant Current Source (contd.) Example of poor current source ECE315 / ECE515 Snce the varaton of the source voltage drectly affects the current of a MOS transstor, t does not operate as a good current source f seen from the source termnal