6. Frequency Response

Transcription

1 6. Frequency esnse eading: Sedra & Sith: hater.6, hater 3.6 and hater 9 (MOS rtins, EE 0, Winter 0, F. Najabadi

2 Tyical Frequency resnse an liier U t nw we have ignred the caacitrs. T include the caacitrs, we need t slve the circuit in the requency dain (r use Phasrs. wer cut- requency: Uer cut- requency: H Band-width: B H

3 lassiicatin aliiers based n the requency resnse aliier (caacitively-culed aliier (directly-culed 0 Tuned r Band-ass aliier (High Q

4 Hw t ind which caacitrs cntribute t the lwer cut- requency nsider each caacitr individually. et 0 (caacitr is en circuit: I v (r M des nt change, caacitr des NOT cntribute t I v (r M 0 r reduced substantially, caacitr cntributes t Exale: c v i 0 v 0 ntributes t N change in v es NOT cntribute t

5 Hw t ind which caacitrs cntribute t the higher cut- requency nsider each caacitr individually. et (caacitr is shrt circuit: I v (r M des nt change, caacitr des NOT cntribute t H I v (r M 0 r reduced substantially, caacitr cntributes t H Exale: c N change in v es NOT cntribute t H v 0 ntributes t H

6 Hw t ind id-requency circuit ll caacitrs that cntribute t lw-requency resnse shuld be shrt circuit. ll caacitrs that cntribute t high-requency resnse shuld be en circuit. Exale: c cntributes t cntributes t H shrt circuit en circuit

7 w-frequency esnse

8 w-requency resnse a S aliier Each caacitrs gives a le. ll les cntribute t (exact value r siulatin I ne le is at least tw ctave (actr 4 higher than thers (e.g., in the abve igure, is arxiately equal t that le (e.g., in abve gd arxiatin r design & hand calculatins: 3

9 w-requency resnse a S aliier c en: v i 0 v 0 c en: v 0 s en: Gain is reduced substantially (r S a. T S a. With S ( x x x 3 sig G G M M sig r g s s s s s s ω ω ω See S&S r detailed calculatins (S&S assues r and S, ] / (/ [ (, ( 3 S s c sig G c g r g r ω ω ω ll caacitrs cntribute t (v is reduced when 0 r cas en circuit

10 Finding les by insectin. Set v sig 0. nsider each caacitr searately (assue thers are shrt circuit!, e.g., n 3. Find the ttal resistance seen between the terinals the caacitr, e.g., n (treat grund as a regular nde. 4. The le assciated with that caacitr is 5. wer-cut- requency can be und r n π 3 n n * lthugh we are calculating requency resnse in requency dain, we will use tie-dain ntatin instead hasr r (i.e., v sig instead sig t avid cnusin with the bias values.

11 Exale: w-requency resnse a S aliier Exainatin circuit shws that caacitrs cntribute t the lw-requency resnse. In the llwing slides with cute les intrduced by each caacitr (care with the detailed calculatins and nte that we exactly get the sae les. Then 3

12 Exale: w-requency resnse a S aliier π c ( G sig. nsider c : Terinals c. Find resistance between aacitr terinals

13 Exale: w-requency resnse a S aliier π S [ S (/ g / r g ] / g ( / r g. nsider S : // g ( / / rrgg / g ( / r g Terinals S. Find resistance between aacitr terinals

14 Exale: w-requency resnse a S aliier 3 π c ( r. nsider c : Terinals c. Find resistance between aacitr terinals

15 High-Frequency esnse liier gain alls due t the internal caacitive eects transistrs

16 aacitive Eects in n Junctin Majrity arriers harge stred is a unctin alied vltage. We can deine a sall-signal caacitance, j j dq d Q In reverse-bias regin, analysis shw (see S&S 54-56: j0 j / 0 : Junctin built-in vltage j0 : aacitance at zer reversed-bias vltage. : grading ceicient (/ t /3. Fr rward-bias regin: j j0 J ( 0

17 aacitive Eects in n Junctin Minrity arriers Excess inrity carriers are stred in and n sides the junctin. The charge deends n the inrity carrier lie-tie (i.e., hw lng it wuld take r the t diuse thrugh the junctin and recbine. Gives iusin aacitance, d d is rrtinal t current ( d 0 r reverse-bias d T I τ T

18 Sall Signal Mdel r a dide j d everse Bias Frward Bias d j j0 I T τ d 0 T j j0 ( / 0 r Junctin caacitances are sall and are given in et-farad (F F 0 - F

19 aacitive Eects in MOS. aacitance between Gate and channel (Parallel-late caacitr. aacitance between Gate & Surce and Gate & rain due t the verla gate electrde (Parallel-late caacitr 3. Junctin caacitance between Surce and Bdy (everse-bias junctin 4. Junctin caacitance between rain and Bdy (everse-bias junctin

20 aacitive Eects in MOS Parallel-Plate caacitances (deends n the channel shae eine: W gate x v W v x gs Tride gate v gs Saturatin 3 gate v ut- gs gd v gd gate v gd v gb gate Pinched hannel N hannel Junctin caacitances sb sb0 ( SB / 0 db db0 ( B / 0

21 Sall signal r MOS in high-requencies Fr surce cnnected t bdy gs Saturatin 3 gate v db gd v db0 ( B / 0

22 High-requency resnse a S aliier MOS internal caacitrs are shwn utside the transistr t see their iact. ll MOS caacitrs cntribute t H (v is reduced when r cas shrt circuit 3 Fr, all culing ( c and c and by-ass caacitrs are shrt circuit gd shrt: Inut is cnnected t utut Gain is reduced t db shrt: v 0 Fr gs shrt: v i 0 v 0

23 High-requency resnse a S aliier In General: One internal caacitrs shrts inut t the grund ( gs here One internal caacitrs shrts utut t the grund ( db here 3 One internal caacitrs shrts inut t utut ( gd here gd aears in arallel t db See S&S 7-74 r detailed calculatins (S&S assues db 0 sig M M G x s / ω G sig g ( r ω in, ω ( in gs G gd sig ( g r ( gd db ( r gd aears in arallel t gs (with a uch larger value,

24 High-requency-relevant caacitrs High-requency-relevant caacitrs aear between inut & grund, utut & grund, and inut & utut. aacitrs that are cnnected between inut & utut rvide eedback. In the case S aliier, we saw that they aeared in the transer unctin as caacitrs in arallel t inut & grund and utut & grund caacitrs. We can use Miller s There t relace caacitrs cnnected between inut & utut and siliy the analysis.

25 Miller s There I ( (, /( I I /, /( I ( ( nsider an aliier with a gain with an iedance attached between inut and utut and eel the iedance nly thrugh I and I We can relace with any circuit as lng as a current I lws ut and a current I lws ut.

26 Miller s There I an iedance is attached between inut and utut an aliier with a gain, it can be relaced with tw iedances between inut & grund and utut & grund Other arts the circuit

27 Exale Miller s There: Inverting aliier n n v v v v 0 0 ( v v i ecall r EE 00, i 0 is large Slutin using Miller s there: 0 / 0 0 i n v v / ( / ( / ( v v v v i n i /

28 Finding H by insectin. Set v sig 0. Use Miller s There t relace caacitr between inut & utut with tw caacitrs at the inut and utut. 3. nsider each caacitr searately (assue thers are en circuit!, e.g., n 4. Find the ttal resistance seen between the terinals the caacitr, e.g., n (treat grund as a regular nde. 5. ute the 6. Uer cut- requency can be und r: H n π n n 3...

29 autin: Methd in revius slide is called the tie-cnstant arxiatin t H (see S&S age 74. Since, the abve rula give This is the crrect rula t ind H Hwever, S&S gives a dierent rula in age 7 (cntradicting rulas 74. Ignre this rula (S&S Eq n n n π... 3 n Σ n n H π n n H n Σ π (S&S Eq H

30 lying Miller s There t aacitrs j / ( / ( ω

31 Exale: High-requency resnse a S aliier ircuit includes which is ten used t set the dinate le. The irst ste is t identiy which caacitrs are relevant t highrequency resnse ( gs, db, gd, and. The ther caacitrs, c and c are relevant t lw-requency resnse. t high requency, c and c will be shrt.

32 Exale: High-requency resnse a S aliier Use Miller s There t relace caacitr between inut & utut ( gd with tw caacitrs at the inut and utut. d g( r g vg ( ( g g gd, i v gd gd gd gd, ( / ( / g gd gd gd * ssuing g >> in gs gd, i db gd,

33 Exale: High-requency resnse a S aliier π in ( G sig. nsider in : Terinals in. Find resistance between aacitr terinals

34 Exale: High-requency resnse a S aliier π ( r. nsider : Terinals. Find resistance between aacitr terinals

35 High-requency resnse a S aliier / / / ( ( /, ( / ( H db gd gd gs in sig G in sig G G M r g r r g π π

36 Miller s There vs Miller s rxiatin Fr Miller There t wrk, rati / (aliier gain shuld be indeendent eedback iedance. This was crrect r O exale where the gain the chi, 0, reains cnstant when is attached (utut resistance the chi is sall. Hwever, the caacitr that cnnect the inut and utut changes the requency resnse the aliier (i.e., its gain and s we cannt strictly aly Miller s There. In ur analysis, we used id-band gain the transistr and ignred changes in the requency resnse due t the eedback caacitr. This is called Miller s rxiatin. Miller s rxiatin nly gives arxiate values the les and the higher cut- requency. Mre irtantly, Miller s rxiatin isses the zer intrduced by the eedback resistr (which can cause unstable eratin.

37 Exale: High-requency resnse a G aliier ( gd db db is ignred in the abve. Including bdy eect, ne sees db actually aears between drain and grund (arallel t in the abve circuit and is absrbed in. Nte that gd is als between the drain and the grund and is in arallel t.

38 Exale: High-requency resnse a G aliier π gs [ sig ( r / g r ] Terinals gs. nsider gs : r g r r g r r g r. Find resistance between aacitr terinals

39 Exale: High-requency resnse a G aliier π [ r ( g sig ]. nsider : r ( gsig r ( gsig. Find resistance between aacitr terinals

40 High-requency resnse a G aliier / / / ( ] ( [ /, ( / / ( ( H db gd sig i sig gs i sig i i M g r r g r r g π π