CHAPTER 3 Frequency Response of Basic BJT and MOSFET Amplifiers

Transcription

1 HAPT 3 Frequency espnse f asc JT and MOFT Aplfers eew aterals n Appendces III and V In ths chapter yu wll learn abut the eneral fr f the frequency dan transfer functn f an aplfer. Yu wll learn t analyze the aplfer equalent crcut and deterne the crtcal frequences that lt the respnse at lw and hh frequences. Yu wll learn se specal technques t deterne these frequences. JT and MOFT aplfers wll be cnsdered. Yu wll als learn the cncepts that are pursued t den a wde band wdth aplfer. Fllwn tpcs wll be cnsdered. eew f de plt technque. Ways t wrte the transfer.e., an functns t shw frequency dependence. and-wdth ltn at lw frequences.e., D t f L. Deternatn f lwer band cut-ff frequency fr a snle-stae aplfer shrt crcut te cnstant technque. and-wdth ltn at hh frequences fr a snle-stae aplfer. Deternatn f upper band cut-ff frequency- seeral alternate technques. Frequency respnse f a snle dece JT, MOFT. ncepts related t wde-band aplfer den JT and MOFT exaples. 3. A shrt reew n de plt technque xaple: Prduce the de plts fr the antude and phase f the transfer functn Ts s 0s s 5 /0 /0, fr frequences between rad/sec t 0 6 rad/sec. We frst bsere that the functn has zers and ples n the nuercal sequence 0 zer, 0 ple, and 0 5 ple. Further at ω= rad/sec.e., lt less than the frst ple at ω=0 rad/sec, T s 0s. Hence the frst prtn f the plt wll fllw the asypttc lne rsn at 6 d/ctae, r 0 d/decade, n the nehbrhd f ω= rad/sec. The antude f Ts n decbels wll be apprxately 0 d at ω= rad/sec. abn aut, Ph.D. Pae 3. 3/6/03

2 The secnd asypttc lne wll cence at the ple f ω=0 rad/sec, runnn at -6 d/ctae slpe relate t the preus asyptte. Thus the erall asyptte wll be a lne f slpe zer,.e., a lne parallel t the ω- axs. The thrd asyptte wll cence at the ple ω=0 5 rad/sec, runnn at -6 d/octae slpe relate t the preus asyptte. The erall asyptte wll be a lne drppn ff at -6 d/ctae bennn fr ω=0 5 rad/sec. nce we hae cered all the ples and zers, we need nt wrk n sketchn any further asypttes. The three asypttc lnes are nw sketched as shwn n fure 3.. Asyptte lnes Fure 3.: The asypttc lne plts fr the Ts. The actual plt wll fllw the asypttc lnes ben 3 d belw the frst crner pnt.e.,at ω=00.e., 57 d,and 3 d belw the secnd crner pnt.e., ω=0^5,.e. 57 d. In between the tw crner pnt the plt wll apprach the asypttc lne f cnstant alue 60 d. The antude plt s shwn n fure 3.. abn aut, Ph.D. Pae 3. 3/6/03

3 Mantude plt heaer lne Fure 3.:de antude plt fr Ts Fr phase plt, we nte that the s n the nueratr wll e a cnstant phase shft f +90 derees snce s j 0 j, anle: tan / 0 tan 90, whle the ters n the dennatr wll prduce anles f respectely. The ttal phase anle wll then be: tan /0, and 5 tan / tan /0 tan /0 3. Thus at lw frequency << 00 rad/sec, the phase anle wll be clse t 90. Near the ple frequency ω=00, a -45 wll be added due t the ple at akn the phase anle t be clse t +45. The phase anle wll prressely decrease, because f the frst tw ters n φω. Near the secnd ple ω=0 5, the phase anle wll apprach tan 0 /0 tan 0 / e., -45 derees. The student n encuraed t draw the cure 3. plfed fr f the an functn f an aplfer reealn the frequency respnse ltatn abn aut, Ph.D. Pae 3.3 3/6/03

4 3.. Gan functn at lw frequences lectrnc aplfers are lted n frequency respnse n that the respnse antude falls ff fr a cnstant d-band alue t lwer alues bth at frequences belw and abe an nteredate rane the d-band f frequences. A typcal frequency respnse cure f an aplfer syste appears as n fure3.3. Fure 3.3: Typcal frequency respnse functn antude plt fr an electrnc aplfer Usn the cncepts f de antude plt technque, we can apprxate the lwfrequency prtn f the sketch abe by an expressn f the fr T L Ks s, r s a K T L s. In ths K and a are cnstants and s=jω, where ω s the physcal,.e., a / s easurable anular frequency n rad/sec. In ether case, when the nal frequency s ery uch saller than the ple frequency a, the respnse T L s takes the fr Ths functn ncreases prressely wth the frequency Ks / a. s j, fllwn the asypttc lne wth a slpe f +6 d per ctae. At the ple frequency a, the respnse wll be 3 d belw the preus asypttc lne, and hencefrth fllw an asypttc lne f slpe -6+6=0 f zer d/ ctae. Thus T L s wll rean cnstant wth frequency, assun the d-band alue. Nte that T L s s a frst rder functn n s a snle te-cnstant functn. abn aut, Ph.D. Pae 3.4 3/6/03

5 The frequency at whch the antude plt reaches 3 d belw the d-band.e., the flat prtn f the antude respnse cure an alue s knwn as the -3 d frequency f the an functn. Fr the lw-frequency seent.e., T L s f the antude plt ths wll be denated by f L r ω L =π f L. In a practcal case the functn T L s ay hae seeral ples and zers at lw frequences. The ple whch s clsest t the flat d-band alue s knwn as the lw frequency dnant ple f the syste. Thus t s the ple f hhest antude an all the ples and zers at lw frequences. Nuercally the dnant ple dffers fr the -3 d frequency. ut fr splcty, ne can apprxate dnant ple t be f sae alue as the -3d frequency. The -3d frequency at lw frequences s als setes referred t as the lwer cut-ff frequency f the aplfer syste. The frequency respnse ltatn at lw frequency ccurs because f cupln and bypass capactrs used n the aplfer crcut. Fr snle-stae aplfers,.e.,,..,g aplfers these capactrs ce n seres wth the nal path.e., they fr a lp n the nal path, and hence pedes the flw f nal cupled t the nternal ndes.e., ndes f the JT, G ndes f the MOFT f the acte dece. The students can cnnce theseles by cnsdern the sple llustratn presented n fure 3.4. r r s s r T s Ks s a Fure 3.4: Illustratn the fratn f a zer n the ltae transfer functn because f a capactr n the nal lp. The cntrlln ltae π fr the V has a zer because f the presence f. abn aut, Ph.D. Pae 3.5 3/6/03

6 3.. Gan functn at hh frequences A slar scenar exsts fr the respnse at hh frequences. y cnsdern the raph n F.3.3 at frequences beynd.e., hher than the d-band seent, we can prpse K the fr f the respnse functn as: T H s. K and b are cnstants. Other s b Kb K alternate frs are: TH s, r TH s. Nte that n all cases, fr s b s / b frequences << the ple frequency b, the respnse functn assues a cnstant alue.e., the d-band respnse. Fr T H s, whch s a frst-rder functn, the frequency b beces the -3db frequency fr hh frequency respnse, r the upper cut-ff frequency. When there are seeral ples and zers n the hh frequency rane, the ple wth the sallest antude and hence clsest t the d-band respnse zne s referred t as the hh frequency dnant ple. Aan, nuercally the hh frequency dnant ple wll be dfferent fr the upper cut-ff frequency. ut n st practcal cases, the dfference s sall. In case the hh frequency respnse has seeral ples and zers, ne can frulate the functn as s/ z s/ z.. TH s 3. s/ s/.. p p In an nterated crcut scenar cupln r by-pass capactrs are absent. The frequency dependent an functn.e., transfer functn s prduced because f the ntrnsc capactances parastc capactances f the deces. As a cnsequence the zers ccur at ery hh frequences and nly ne f the ples fall n the nal frequency rane f nterest, wth the ther ples at substantally hher frequences. Thus f p s the ple f sallest antude, the aplfer wll hae p as the dnant ple. In such case T H s p, and p s p wll als be the -3 d r upper cut-ff frequency f the syste. Otherwse, the -3 d frequency H can be calculated usn the frula edra and th, Mcrelectrnc rcuts, 6 th edn., ch.9, p.7, Oxfrd Unersty Press, 00. abn aut, Ph.D. Pae 3.6 3/6/03

7 H [.....] p p z z / plfed frst rder fr f the aplfer an functn nsdern the dscussns n sectns 3..- we can frulate the splfed fr f the aplfer an functn can then be cnsdered as : As=A M F L s F H s 3.4 In 3.4, A M s ndependent f frequency, F L has a frequency dependence f the fr s/s+w L, whle F H has a frequency dependence f the fr w H /s+w H. Thus fr frequences hher than w L and fr frequences lwer than w H the an s clse t A M. Ths s a cnstant an and the frequency band w H - w L s called the d-band frequences. n the d-band frequences the an s cnstant.e., A M. At frequences << w L, F L s ncreases wth frequency re: de plt by rtue f the s n the nueratr, at 6d/ctae. As the frequency ncreases, the rate f ncrease slws dwn and the de plt eres wth the cnstant alue A M shrtly after w=w L. At w=w L the respnse falls 3 d belw the ntal asypttc lne f slpe 6d/ctae. larly, as frequency ncreases past w H, the respnse As tends t fall ff, passn thruh 3d belw A M n d at w=w H and then fllwn the asypttc lne wth slpe nus 6d/ctae drawn at w=w H. It s f nterest t be able t fnd ut these tw crtcal frequences fr basc snle stae aplfers pleented usn JT r MOFT. 3.3 plfed hh-frequency ac equalent crcuts fr JT and MOFT deces It can be nted that fr aplfers pleented n nterated crcut technly nly the upper cut-ff frequency w H s f nterest. T nestate ths we ust be falar wth the ac equalent crcut f the transstr at hh frequences. The eleents that affect the hh frequency behar are the parastc capactrs that exst n a transstr. These arse because a transstr s ade by layn dwn seeral secnductr layers f dfferent cnductty.e., p-type and n-type aterals. At the junctn f each par f dsslar layers, a capactance s enerated. We wll cnsder the splfed hh-frequency equalent crcuts fr the JT and MOFT as shwn n Fs In these dels abn aut, Ph.D. Pae 3.7 3/6/03

8 each transstr s asned wth nly tw parastc capactance asscated wth ts nternal ndes. These arse ut f the secnductr junctns that are nled n buldn the transstr. Fr the JT, the base ateral prduces a sall resstance r x, whch assues prtance fr hh nal frequency applcatns nal prcessn. The dels fr N-type.e., NPN, NMOFT and P-type.e., PNP, PMOFT transstrs are cnsdered sae. In re adanced dels used n ndustres re nuber f parastc capactances and resstances are eplyed Hh frequency respnse characterstcs f a JT r X r r Fure 3.5: plfed ac equalent crcut fr a JT dece fr hh nal frequency stuatn. An prtant perfrance paraeter f a JT dece s the sall nal shrt crcut current an f the dece under de f peratn. Thus n F.3.5, f we nsert an ac current surce at ternal and seek the ac shrt-crcut utput current at nde, we can cnstruct the ac equalent crcut as n F.3.6. The shrt-crcut current an / f the dece can be dered fr the KL equatns returnn ternal t ac rund at the ndes and. Wrtn =/r n eneral, we et 3.5 x, 0 x x s s abn aut, Ph.D. Pae 3.8 3/6/03

9 abn aut, Ph.D. Pae 3.9 3/6/03

10 r X r r Fure 3.6: nfurn the JT dece fr shrt-crcut current an calculatn. ln fr π and ntn that at nde whch s shrt crcuted fr ac = - π +s µ π, we can fnally dere the shrt-crcut current an f the JT under de f peratn as: h fe s x s s 3.6 x s s r q.3.6 represents a transfer functn wth a lw-frequency.e., 0 alue f h fe lwfrequency = h fe s s=jω=0 = - r π =-β, the falar sybl fr the current an f a JT n peratn. ecause µ s ery sall, the zer f h fe jω.e., / µ les at ery hh frequences. Usn the sybl h fe 0 fr lw-frequency 0 alue f h fe, and fr frequences << z, the de antude plt f h fe appears as n F h fe 0 j h fe h fe 0 T Fure 3.7: The de antude plt f h fe jω. abn aut, Ph.D. Pae 3.0 3/6/03

11 It s bsered that at the frequency r h, h fe drps t fe 0,.e., -3db belw h fe 0. Ths frequency s knwn as the β cut-ff frequency fr the JT under de f peratn. At frequences uch hher than, h fe jω chanes as see eq.3.6 Ths reaches a antude f unty.e. =, at a frequency. j T 3.7 Ths s knwn as the transtn frequency f the JT fr peratn as aplfer. The transtn frequency T T f s a ery prtant paraeter f the JT fr hhfrequency applcatns. Fr a en JT, the hh-frequency peratnal lt f the dece can be ncreased by ncreasn ω T a an ncrease n, the ac transcnductance f the JT. Ths, hweer, ples an ncrease n the D bas current snce = I/V T and hence an ncrease n the D pwer cnsuptn f the syste. ecalln the relatn r π =β+, we can deduce that T h fe0 3.8 In real JT deces, and. Hence, the zer frequency z wll be >> the transtn frequency ω T. nce h fe jω beces < beynd ω T, the zer frequency bears n practcal nterest Hh frequency respnse characterstcs f a MOFT abn aut, Ph.D. Pae 3. 3/6/03

12 D G G D G D s s d s r Fure 3.8: plfed ac equalent crcut fr a MOFT dece fr hh nal frequency stuatn. A splfed ac equalent crcut fr the MOFT s shwn n fure 3.8. The bdy ternal fr the MOFT, and the asscated parastc capactances as well as the bdy transcnductance b hae nt been shwn. y fllwn a prcedure slar t that f a JT, t can be shwn that the shrt crcut current an f the MOFT cnfured as a aplfer s en by sd whch can be apprxated s s d as fr frequences well belw the zer frequency /s d. s s d Under the abe assuptn the frequency at whch the antude f the current an beces unty.e., the transtn frequency, beces: T 3.9 s d The transtn frequency f a MOFT s a ery prtant paraeter fr hh frequency peratn. Ths can be ncreased a an ncrease n wth the attendant ncrease n the D bas current and hence ncrease n D pwer dsspatn. abn aut, Ph.D. Pae 3. 3/6/03

13 3.4 alculatn f ω L the lwer cut-ff frequency hrt rcut Te nstant ethd Fure 3.9a depcts a typcal -JT aplfer wth cupln capactrs, 3, and the by-pass capactr. ach f these capactrs fall n the nal path fr the peratn f the aplfer and hence nfluences the ltae an functn n ters f ntrducn seeral ples and zers n the an transfer functn. A splfed ethd t deterne the ples s t cnsder nly ne f the capactrs effecte at a te and assue that the ther capactrs behae apprxately as shrt crcuts. ecause nly ne capactr s present n the syste, t s easy t deterne the te cnstant paraeter f the asscated ac equalent crcut. Hence the ethd s knwn as shrt crcut te cnstant ethd T. Fures 3.9b-d shw the three ac equalent crcuts under the assuptn f nly ne f,, r present n the crcut. The lcatns t be used fr the calculatn f the equalent Theenn resstance fr each f the capactrs,, 3 are shwn n blue lnes n the daras. The nternal capactances f the JT ffer ery hh pedance at lw frequences and hence they are cnsdered as pen crcuts s these are nt shwn. V L The r X r r L a b r X r r L r X r r 3 3 The L The c d abn aut, Ph.D. Pae 3.3 3/6/03

14 abn aut, Ph.D. Pae 3.4 3/6/03 Fure 3.9: a cheatc f a aplfer wth fur resstr basn; b the ac equalent crcut wth, 3 as shrt crcuts; c the ac equalent crcut wth, 3 as shrt crcuts, and d the ac equalent crcut wth, as shrt crcuts. Analyss f the equalent crcut n F.3.9b s strahtfrward. y nspectn, the Theenn resstance asscated wth s r r x Th, where the ntatn ples n parallel wth. The asscated te-cnstant s Th, and the crrespndn ple-frequency s L = / Th. larly, the Theenn resstance fr 3 s L Th r 3 see F.3.9d. The crrespndn ple-frequency s L3 = / 3 Th3. The calculatn f the Theenn resstance asscated wth can be splfed cnsderably by assun r as nfnty. Then by nspectn see F.3.9c, fe x Th h r r. The crrespndn ple-frequency s L = / Th. A re adenturus student ay dscard the assuptn f r nfnty and prceed t set up a 3 by 3 ndal adttance atrx NAM see Appendx III by usn the substtutns L c cp sp x p r r r,,, and by nsertn a duy current surce x at the nde labeled as n F. 3.9c. The NAM wll appear as V V V x cp sp p p p p In the abe =/,, πp =/r πp, =/r, and s n, hae been used. Wth the further assuptn t s ery d f r x s << r π f V V, the atrx equatn 3.0, beces, after rearraneent.e., chann sdes: x cp sp p p p p V V V 3. Then Th s en by V / x. The result s usn Maple prra cde: cp sp cp sp cp p cp sp p sp p cp sp sp cp p p cp sp p Th Nw ntrducn the assuptn 0.e., r nfnty, ne wll et

15 Th p cp sp p sp cp p cp sp cp sp cp 3. ubsttutn back n ters f the resstance ntatns,.e., =/,, πp =/r πp, =/r, and s n, ne can et Th sp r p sp r p r p 3.3 Usn r πp =h fe, and splfyn, ne arres at Th sp sp r r p p / h / h fe fe,.e., Th sp r p h r r x h fe fe ******** The erall lwer -3 d frequency s calculated apprxately by the frula L L L3 L. If ut f the seeral ples f the lw-frequency transfer functn F L s, ne s ery lare cpared t all ther ples and zers, the erall lwer -3 d frequency L beces dnant ple.e., larest f L r L r L3. If the nuercal alues f the arus ple frequences are knwn by exact crcut analyss fllwed by nuercal cputatn, the lwer 3-d frequency can be calculated apprxately by a frula f the fr... where, ω, ω,.. are the L. 3 nddual ple frequences and the zer-frequences are ery sall cpared wth the ple frequences. xaple 3.4.: nsder the fllwn alues n a JT aplfer. =50, = = 0 k, r =500, r x =5, h fe =00 and =k, =.5k, L =3.3 k, V A =0 lts, I A. Further, =uf and = 0uF and 3 = uf. What s L? Accrdn t abe frulas, Th =.05k, Th =5.5 and Th3 =.39k+3.3k = 4.69k. Then L = rad/s, L = rad/sec and L3 = 3. rad/sec. Then, L L L L rad/s, whch s pretty clse t L. abn aut, Ph.D. Pae 3.5 3/6/03

16 xaple 3.4.: What f, L = 800 rad/sec s t be dened? We can assue, fr exaple, L = 0.8 L,, ω L = L = 0. L and L3 =0. L. Then, den the alues f the capactrs, and 3. The student can try ther relate allcatns t. 3.5: alculatn f ω H the hher cut-ff frequency eeral alternate ethds exst n the lterature. The fllwn are presented. 3.5.: Open crcut te-cnstant OT ethd Ths s slar t the case as wth lw frequency respnse. Fr hh frequency peratn, we are nterested n the capactr whch wll hae lwer reactance alue snce ths capactance wll start t derade the hh frequency respnse sner than the ther. Thus, we can cnsder ne capactr at a te and assue that the ther capactrs are t sall and hae reasnably hh reactance alues fr a, the reactance s / s that they culd be cnsdered as pen crcuts. We then calculate the asscated te cnstant. Thus the ethd s naed as pen crcut te cnstant OT ethd. We shall llustrate the ethd usn the case f a JT aplfer. nsder fure 3.0a whch shws the ac equalent crcut f the -JT aplfer f F.3.9a. The hh frequency equalent crcut fr the JT has been ncluded. The cupln and by-pass capactrs are assued as shrt crcuts fr hh frequency stuatn. abn aut, Ph.D. Pae 3.6 3/6/03

17 Th + - r X r r L r X r r L a b r X r Th r L c Fure 3.0: a hh frequency equalent crcut f the aplfer n F.3.9a; b the equalent crcut wth µ pen; c the equalent crcut wth π pen. ase : pen The ac equalent crcut t deterne the Theenn equalent resstance Thπ acrss π s shwn n F.3.0b. y sple nspectn Th r rx The hh frequency ple due t ths stuatn s H =/ π Thπ. ase : π pen We nw need t deterne the Theenn equalent resstance Thµ acrss µ. The asscated equalent crcut s shwn n F.3.0c. We can use a duy nal current surce x and carry ut few steps f basc crcut analyss see F.3.. abn aut, Ph.D. Pae 3.7 3/6/03

18 x r X r V V V r L Fure 3.: qualent crcut fr calculatn Thµ KL at V nde es: V G x G 0, r r x 3.3a KL at V nde es: x V V GL GL 0, r L 3.3b ln 3.3a,b fr V and V we can fnd Thµ =V -V / x = s ++ s L L The hh frequency ple fr s H = / Thµ. When the tw ple frequences are cparable n alues, the upper -3 d frequency s en apprxately by H Th Th 3.4 If, hweer, the tw alues are wdely apart say, by a factr f 5 r re, the upper -3 d frequency wll be called as the dnant hh frequency ple and wll be equal t the lesser f H and H. 3.5.: Applcatn f Mller s there Ths there helps splfyn the ac equalent crcut f the JT aplfer by ren the capactr, whch runs between tw flatn ndes.e., between the base sde t the cllectr sde. In prncple, f an adttance Y 3 runs between ndes and wth Y at nde t rund and Y at nde and rund and f K s the ltae an V /V between ndes and, then Y 3 can be splt nt tw parts ne ben n parallel abn aut, Ph.D. Pae 3.8 3/6/03

19 wth Y wth a alue Y 3 -K and anther becn n parallel wth Y wth a alue - /KY 3. The there can be appled t all cases f flatn eleents cnnected between tw ndes n a syste. As a result f ths prncple, the hh frequency equalent crcut f the JT aplfer see F.3.5 splfes t fure 3.. r X V K V V V r r Fure 3.: Hh frequency equalent crcut f a -JT dece after applcatn f Mller s there In the abe K,. Fure 3. s a sple tw nde K crcut and can be cnenently analyzed. Mller s there s ery effecte when the adttances Y and Y hae ne f ther ends runded fr ac nals. After the equalent crcut s splfed as abe, ne can apply the OT ethd t deterne the hh frequency ples. xaple 3.5..: nsder F.3.3a whch shws the hh frequency equalent crcut fr the aplfer n F.3.9a. Fure 3.3b s a reduced fr f F.3.3a, sutable fr analyss by ndal atrx frulatn. In F.3.3b, the fllwn expressns hld: r X, /,, a alculate the lw frequency ltae an between ndes labeled and n F.3.3b. Ths aunts t nrn the presence f π and µ fr ths calculatn. Let ths an be K. abn aut, Ph.D. Pae 3.9 3/6/03

20 r X r r L r r L x r r c Fure 3.3: a hh frequency equalent crcut f the aplfer n F.3.9a, b the equalent crcut adjusted fr ndal adttance atrx NAM analyss, c Transfred ac equalent crcut after applcatn f Mller s there. b Use Mller s there t fnd the new crcut cnfuratn n the fr f F.3.. c Apply OT ethd t dere the ple frequences fr hh frequency respnse f the aplfer. d Gen that =50Ω, =8 kω, =47 kω, r X =0Ω, =40 hs, r =50 kω, =.7 kω, L =4.7 kω, =70 Ω, h fe 0=h F = 49, π =. pf, µ =0. pf, fnd the hh frequency ples by usn The OT ethd dscussed n sectn The OT ethd after applyn Mller s there ntrduced n sectn lutn : y nspectn f F.3.3b, the ltae an K= / = Further, recalln r π =h fe 0 =49, we et r π =5 Ω. r L =-66.4 abn aut, Ph.D. Pae 3.0 3/6/03

21 9 Part c : Th r rx =57. Ω. H =/ π Thπ = rad/s Thµ = s ++ s L see deratns n 3.5.= Ω, H = / Thµ = rad/sec. The abe s the result by OT ethd wthut takn recurse t Mller s there. Part b: Usn Mller s there the equalent crcut f F.3.3b transfr t fure 3.3c. Usn K=-66.4, the new capactance alues bece: 3 K 7.930,.00 K Nw, we need t recalculate the Theenn resstances Th r Ω, r 658Ω. Th L 9 Then, 9.0 rad/sec, and rad/sec. H Th H Th Part d: nce H s << H, H s the dnant hh-frequency ple fr the 9 aplfer. Hence the -3d frequency s apprxately.87 0 rad/sec,.e., the dnant hh frequency ple f the syste. Part d: nce the H, H alues are nt wdely apart.e., dffer by a factr f 5 r re, we wll estate the hher -3d frequency by adptn the frula H Th Th rad/sec. Part d resed: If we had used the sae frula wth the alues fund n part c, we wuld et H Th Th rad/sec. nclusn: In practce the re cnserate alue shuld be chsen,.e., the upper -3 9 d frequency wll be.680 rad/sec Transfer functn analyss ethd The te-cnstant ethds dscussed abe lead t deternatn f the ples and zers f the transfer functn. We need yet t deterne the d-band an functn and cbne ths wth the ples and zers t fr the erall transfer functn fr hh r lw frequency respnse f the aplfer. Ths nles a tw-step prcess. Alternately, abn aut, Ph.D. Pae 3. 3/6/03

22 we can apply ndal atrx analyss see Appendx III technque t deterne the erall transfer functn as the frst peratn. nce the hh frequency equalent crcut f the transstr has tw capactrs, the transfer functn wll be f rder tw n s.e., secnd deree n s. Fr such a transfer functn there exsts a sple rule t deterne the dnant ple f the transfer functn. Further, by nrn the frequency dependent ters.e., ceffcents f s n the transfer functn, we can dere the d-band an f the syste. Hence the transfer functn analyss pens up aenues fr dern seeral prtant netwrk functns fr the aplfer n hand. Fllwn exaples llustrate seeral cases. xaple : Deratn f the ltae an transfer functn f a -JT aplfer nsder the ac equalent crcut f the JT aplfer f F.3.9a whch s redrawn n F.3.4a fr cnenence. Applyn Theenn-Nrtn equalence prncple t the left f the arrw-head n blue, we can cnert the crcut n F.3.4a t r X r r L r r L Fure 3.4: a hh frequency equalent crcut f the cplete aplfer, b the equalent crcut adjusted fr ndal adttance atrx NAM analyss. F.3.4b, whch s cnenent fr ndal analyss.e., t has fewer nuber f ndes and s dren by a current surce. In F. 3.3b nte that r 3.5 X, /,, Usn the ntatns = /, π =/r π, = L, =/,and s n, we can frulate the adttance atrx fr the tw ndes labeled as, syste n F.3.3b. Thus, s s s s V V 3.6 abn aut, Ph.D. Pae 3. 3/6/03

23 abn aut, Ph.D. Pae 3.3 3/6/03 Ntn that π = V, and brnn - π = V n the left sde nsde the atrx, 0 V V s s s s 3.7 The utput nal ltae f the syste s =V, en by V s s s s s s On carryn ut the tasks f ealuatn f the deternants, ne can fnd p s s s Wrtn Ds = p s s, and substtutn fr we fnd X r s D s 3.9 The ltae an transfer functn s: X r s D s s T 3.0 We can ake tw prtant deratns fr the result n 3.0 Lw-frequency ltae an A M : Ths s btaned fr 3.0 n apprxatn ω0,.e., s0. Thus X M r A 3. Assun the cpnent and dece paraeter alues as n xaple 3.5.., we et A M =-63. Hh frequency dnant ple ω HD When the dennatr Ds f the transfer functn s f secnd rder, ne can estate the dnant ple fr the least alued rt f the dennatr plynal. The technque s explaned as fllws. We can wrte Ds n the fr : s +bs + c. Then the hh frequency dnant ple.e., the ple wth the least antude f all the hh frequency ples s en by pd = c/b

24 .e., the rat f the cnstant ter n Ds t the ceffcent f the s ter n Ds. Ths fllws easly by wrtn. Then, s bs c s s s s b, f α<< β. Further, fr αβ=c, we deduce c / c / b, as the dnant ple f the ltae an transfer functn. Hence, ω HD = / 3. nsder the ltae an functn f the JT aplfer n xaple as an llustratn. We can fnd ω HD apprxately by substtutn the alues fr the pertnent paraeters,, and s n. Thus, the upper cut-ff frequency.e., upper -3d 9 frequency s calculated as.6580 rad/sec. The student ay cpare ths alue wth 9 the alues btaned preusly,.e., rad/sec usn OT ethd, and rad/sec usn Mller s there fllwed by OT ethd, respectely. xaple : Deratn f the ltae an transfer functn VTF f a -JT aplfer V 3 r X r r L L Fure 3.5: a cheatc f the -JT aplfer stae, b the hh-frequency equalent crcut. Fures 3.5a-b depct respectely the scheatc and hh frequency equalent crcuts f the aplfer stae. All cupln and by-pass capactrs behae as shrt crcuts and hence they d nt appear n F.3.5b. abn aut, Ph.D. Pae 3.4 3/6/03

25 abn aut, Ph.D. Pae 3.5 3/6/03 Fr ndal adttance atrx NAM frulatn we need t transfr the ltae surce wth ts nternal resstance t ts Nrtn equalent,.e., a nal current surce = n parallel wth the student s encuraed t cplete ths part t dfy F.3.5b. The NAM by nspectn wll be: s s s s s s s x ut π =. Hence 3.3 reduces t 0 0 x s s s s s s s s 3.4 In rder t cpare the perfrance f the and JT aplfers, t wll be cnenent t assue r X as nelble. Then nde wll be at zer ac ptental and 3.4 wll apprxate t by dscardn the rw and clun asscated wth the nde 0 s s 3.5 The VTF s en by / = / = ] [ s s Usn the VTF we can deduce The d-band an.e., s jω, ω0 fr de f peratn s: A M 3.6 The dnant hh frequency ple fr de f peratn s ω HD = 3.7

26 xaple : Deterne the d-band ans and dnant hh frequency ples f the JT aplfer perated as see F.3.9a and see F.3.5a des f peratn. The deces, the D basn cndtns, and the crcut cpnents are assued dentcal fr bth the cnfuratns. [Gen =50Ω, =8 kω, =47 kω, =70 Ω, =.7 kω, L =4.7 kω =40 hs, r =50 kω, r X =0Ω,h fe 0=h F = 49 π =. pf, µ =0. pf] On substtutn nt the expressns 3.6 and 3.7, we et: A M =57.7 /, 9 ω HD = rad/sec. Nte I: We bsered that the -JT aplfer has sewhat lwer ltae an.e., 57.7 cpared wth that f -JT.e., 63. n antude aplfer, but has a 9 hher hh-frequency bandwdth ~ dnant hh-frequency ple= rad/sec 9 cpared wth that f -JT.e., rad/sec aplfer. Nte II: It s knwn that the -JT aplfer has derate t hh kω t tens f kω nput resstance, as well as derately hh kω utput resstance. In cparsn, a - JT aplfer has lw Ω t tens f Ω nput resstance whle a derately hh kω utput resstance. Nte III: A -JT s preferred er a -JT aplfer fr st rad-frequency MHz and abe applcatns because t has a hher hh-frequency bandwdth fr dentcal dece and basn cndtns, and affrds t prde better pedance atchn at the nput wth the rad-frequency F surce n the rane f 50Ω t 75Ω. xaple : Deratn f the ltae an transfer functn VTF f a - MOFT aplfer wth acte lad nsder fures 3.6a-b whch depct respectely the scheatc f a -MOFT aplfer and the asscated hh frequency equalent crcut. abn aut, Ph.D. Pae 3.6 3/6/03

27 V DD V G r d s d s s r V a b Fure 3.6: a cheatc f a -MOFT aplfer wth acte lad, b asscated hh-frequency equalent crcut. Usn a duy nput current surce x when nt shwn explctly, the student shuld adpt ths prncple, the adttance atrx s s s d sd x 3.8 sd s d d s After re-arrann the student s suested t wrk ut the detals, we can fnd the VTF en by sd s x 0 s s d d s d x d xercse : an yu sketch the de antude plt fr the abe VTF? Assue, =50 µa/v, r, r = MΩ each, d, d = 0 ff each f fet.e., 0-5, s = 00 ff. xaple : Deratn f the ltae an transfer functn VTF f a G- MOFT aplfer. Fures 3.7a-b shw the scheatc daras f a G-MOFT aplfer M under deal ladn and practcal ladn cndtns respectely. Fure 3.7c depct the hh frequency equalent crcut. Nte that nw we need t nclude the bdy- abn aut, Ph.D. Pae 3.7 3/6/03

28 abn aut, Ph.D. Pae 3.8 3/6/03 d 3 bd 3 Fure 3.7: a cheatc f the G-MOFT aplfer wth deal lad and bas surce, b practcal basn and acte lad arraneent, c hh-frequency equalent crcut del fr the G-MOFT stae, d re cplete hh-frequency del fr a MOFT. transcnductance b f the MOFT, snce the tply s such.e., the aplfyn dece s n a tte ple cnnectn that the surce and bdy ternals f the aplfyn transstr M cannt be cnnected tether! It ay be nted that despte the cplcated lk f F.3.7c, there are nly tw nal ndes n the equalent crcut. Thus the ndal adttance atrx wll appear as: 3 bs b s bs b s x d bd d bd bs s s s Wrtn = s + bs, = d + bd + d + bd, and bsern that the ate and bdy ternals f the MOFTs are at zer ac ptentals s that s =- s =-, bs =- s =-, we can re-wrte the adttance atrx equatn as b b x s s e-arrann, we et 0 3 x b b s s 3.3 The VTF s en by:

29 3 b 3.3 x 0 b s s x 0 xercse I: Fnd the explct expressn fr the VTF usn 3.3. xercse II: Usn 3.3, and the knwlede that the drn pnt pedance DPI at the nput s en by Z dp, fnd the expressn fr the Z dp. x xercse III: Gen that r r M, r M, 00 h, 0., 3 b 00 ff, bs 0 ff, d 5 ff, bd 0 ff, d 0 ff, bd ff fnd the s 5 expressns fr the VTF and the Z dp xercse IV: Fnd the expressn fr the trans-pedance functn TIF = x fr the aplfer stae. xaple : Deratn f the ltae an transfer functns VTF fr a -JT and D-MOFT aplfers Fures 3.8a-d shw respectely the scheatc and hh frequency equalent crcuts f a -JT and D-MOFT aplfers. The basn/ladn s arraned by d abn aut, Ph.D. Pae 3.9 3/6/03

30 Fure 3.8: a cheatc f a -JT aplfer, b hh frequency equalent crcut f the aplfer, c cheatc f a D-MOFT aplfer, d hh frequency equalent crcut f the D aplfer eplyn acte resstrs Q n F.3.8a, and M n F.3.8c. It ay be nted that snce ne ternal f the capactr µ and f d s runded fr ac, nether f these capactrs are subjected t the Mller effect anfcatn, as are n the cases wth JT and MOFT aplfers. The student s encuraed t cplete the analyss and dere the expressns fr the VTF / s f the and the D aplfers.eeber that fr the MOFT the bdy ternal s cnnected t a D ltae. 3.6: Wde band ult-stae aplfers 3.6.: - ascde JT aplfer It has been knwn that a stae has hh ltae an and hh nput resstance whle a stae has hh ltae an wth lw nput resstance. Fr hh frequency peratn aplfer wll hae a saller band wdth than the aplfer snce n the frer the capactance s anfed due t Mller effect. Ths has a crrespndn effect n reducn the band wdth. In the stae, snce the base s at ac rund, the capactance has already ne ternal t rund and s nt subjected t Mller effect anfcatn. the stae affrds t hher peratn band wdth. It s nterestn t cnsder a cpste aplfer cntann the and staes s that adantaes f bth the cnfuratns culd be shared. That s what happens n a cascde aplfer where the stae recees the nput nal, whle the stae acts as a lw abn aut, Ph.D. Pae /6/03

31 resstance lad fr the stae. ecause f the lw resstance lad, the Mller effect anfcatn f the capactr n the stae s drastcally reduced. ecause f the lw resstance lad, hweer, the ltae an n the stae drps. ut t s adequately cpensated by the lare ltae an f the stae whch wrks fr a lw nput resstance t a hh utput resstance : Analyss fr the bandwdth and d-band an The scheatc cnnectn f a cascde aplfer and the asscated ac equalent crcut are shwn n Fures 3.9a-b. The equalent crcut has sx ndes. nertn the nal surce n seres wth and r X t ts Nrtn equalent wll brn the nuber f ndes t fur. Fr the cn base transstr Q we can nre r X as ery sall ~ zer. Thus the nuber f ndes reduces t three. These are labeled as ndes,,3 n F.3.9b. Fr quck hand analyss we can adpt the fllwn splfcatn prcedure. We can nw ntrduce the assuptn that r s ery lare nfnty, and ree t. Ths separates nde# and #3 wth the cntrlled current surce π splt nt tw parts- ne runnn fr nde#3 t rund, and the ther fr rund t nde#. V s + - V I Q x Q s + - r X r X r r ' ' r r 3 V a abn aut, Ph.D. Pae 3.3 3/6/03

32 Fure 3.9: a cheatc f a - cascde aplfer, b hh frequency equalent crcut. The splfed equalent crcut appears n fure 3.0. Fure 3.0: plfed hh frequency equalent crcut fr the - cascde aplfer. In F.3.0 we hae used rx r, r r. A careful scrutny f the cntrlled current surce π between rund t nde # and the fact that the cntrlln ltae π s als effecte acrss the sae par f ndes wth the pste ternal at rund nde labeled reeals that the cntrlled current surce π can be equalently replaced by a cnductance f.e., resstance / cnnected acrss nde# and rund. Ths bseratn leads t the equalent crcut shwn n fure 3.. abn aut, Ph.D. Pae 3.3 3/6/03

33 Fure 3.: nersn f F.3.0 after relcatn f the cntrlled current surce seents π each. Nw we can apply the Mller effect anfcatn cnsderatn t the flatn capactr µ runnn between ndes # and #. The ltae an between these ndes s nrn the feed-frward current thruh µ whch s a ery sall capactance apprxately en by the prduct f - and. nsdern the fact that = r r r, and that r, we can apprxate as. Hence the Mller anfcatn turns ut t be K, whch s ery clse t when the transstrs Q and Q are atched t each ther. y the prncple f Mller effect the flatn capactr µ can nw be replaced by tw runded capactr f alue µ -K,.e., µ at nde #, and µ -/K,.e., µ at nde#. Ths leads t the fnal splfed fr f the equalent crcut as n fure 3.. abn aut, Ph.D. Pae /6/03

34 Fure 3.: Fnal splfed hh frequency equalent crcut fr the cascde aplfer. Fure 3. present three dsjnt sub-crcuts wth three dstnct te cnstants. Fr F.3.a, the te cnstant s a. Fr F.3.b and c the te cnstants are respectely, b, and c. The hh frequency cut ff.e., upper -3d frequency s bandwdth f the - cascde aplfer. H a b c, whch can be cnsdered as the The d-band an f the cascde aplfer can be btaned by nrn all the capactrs as pen crcuts and cnsdern the crcut n fure 3.3. Fure 3.3: qualent crcut fr the - cascde aplfer fr d-band.e., lw frequency an calculatn. y nspectn, s = s r r r r 3.33 X X The d-band an s then r AM 3.34 r r s X abn aut, Ph.D. Pae /6/03

35 xercse :nsder an NPN JT dece wth h F =00, f T =6000 MHz, π = 5 pf based t perate at I = A. Gen that r X =0 Ω, V A =50 V, and the nal surce resstance =00 Ω. The lad resstance s.7 kω. athe JT s used as a aplfer wth abe en paraeters. Fnd the d-band an A M, the upper cut-ff frequency f -3d, and hence the Gan-andwdth GW f the aplfer nte: GW=A M tes f -3d btw atched JT deces wth abe specfcatns are used t cnstruct a - cascde aplfer. Fnd the d-band an A M, the upper cut-ff frequency f -3d, and hence the Gan-andwdth GW f the aplfer nte: GW=A M tes f -3d. chw des the GW cpare wth the GW ascde? Hnt: the student need t frst deterne, r π, r, and µ fr the JT fr the en nfratn xaple : On usn a typcal JT deces fr PI sulatn lbrary, we et the fllwn results: Aplfer de - cascde Gan and wdth 80.MHz 793.5MHz 66.MHz G MOFT cascde aplfer It s easy t cnstruct a cascde f -G aplfer staes usn MOFT deces. The scheatc and the hh frequency equalent crcut are shwn n fures 3.4a-b respectely. I shuld be nted that the surce ternals f transstrs M,M cannt be cnnected t the respecte bdy substrate ternals and hence the asscated parastc abn aut, Ph.D. Pae /6/03

36 V DD M3 VG3 V G M VDD V bs s bs G s s d b s D r bd 3 G3 d 3 D3 r 3 s + - M I V s + - b G s s d s r D 3 bd 3 bd V a b Fure 3.4: a cheatc f a -G MOFT cascde aplfer wth acte lad and current surce/rrr basn, b hh frequency equalent crcut. capactances need be cnsdered. Fr M, hweer, snce the and ternals are bth at ac rund, the capactance bs s shrted ut. The equalent crcut n F.3.4b has three unrunded ndes shwn n blue shaded crcles. The ltae an / s can be calculated usn standard ndal atrx frulatn. Alternately, the crcut can be splfed by adptn apprxatn prcedures as used n the - cascde aplfer. Ths s left as an exercse t the nterested students. Nte that Mller anfcatn effect wll be applcable t d : Wde band dfferental aplfer wth JT deces The cascde aplfer s snle ended,.e., has ne ternal fr nal nput. It s f nterest t nestate the pssblty fr a wde band dfferental aplfer. Twards ths we can arue that a cn dfferental aplfer wth tw staes acceptn the dfferental nput nals wll nt be desrable, because each stae wll suffer deradatn n hh frequency perfrance because f Mller effect n. In the fllwn we dscuss seeral pssble cnfuratns fr wde-band dfferental aplfer. abn aut, Ph.D. Pae /6/03

37 3.6.3.: - cascde dublet as wde band dfferental aplfer The - cascde aplfer that we dscussed already can be cnsdered as a half crcut fr cnstructn a dfferental aplfer. Fure 3.5 presents the scheatc V V 0 0 V V 4 x x d d I V Fure 3.5: - cascde dublet as wde band dfferental aplfer cnfuratn. The analyss f the an bandwdth GW can be carred ut as n the - cascde aplfer by cnsdern nly ne half crcut f the cnfuratn. The dfferental aplfer wll presere all the erts f the parent - cascde n addtn t prdn the specal features f a dfferental aplfer.e., cn de rejectn, ren een harnc cpnents : - cascade dublet as wde band dfferental aplfer The Mller anfcatn n a stae culd be entrely reed f were zer. ut that ples a stae, whch has a lw ltae an less but clse t. The lss n ltae an can be cpensated partally, by cascadn the stae wth a stae. ut then we are ettn a ltae an nly fr ne stae. y usn the - cascade abn aut, Ph.D. Pae /6/03

38 as a half crcut, t s pssble t deelp a wde band - dublet dfferental aplfer. The scheatc s shwn n fure 3.6. V d d I V Fure 3.6: A - cascade dublet wde band dfferental aplfer. The aplfer n F.3.6 wll prde wde band peratn wth d ltae an. ut there are fur transstrs whch need be suppled wth D bas current. The D pwer cnsuptn wll be hh. A tte-ple cnfuratn where tw transstrs are stacked n ne clun wll result n a lwer D pwer cnsuptn. Ths s pssble by usn cpleentary.e., NPN and PNP transstrs t fr the - cascade. Fure 3.7 presents the cnfuratn : - cascade dublet dfferental aplfer wth cpleentary transstrs In the cnfuratn f F.3.7 D bas current s t be suppler nly t tw cluns f transstrs. cpared wth F.3.6, the cpleentary - cascade dublet dfferental aplfer F.3.7 wll cnsue less D pwer. Analyss fr an bandwdth A splfed analyss fr the band wdth f the dfferental aplfer can be carred ut n the assuptn that the NPN and the PNP transstrs are atched pars t s seld true n practce and then wrkn n ne half crcut f the syste. Thus cnsdern the par abn aut, Ph.D. Pae /6/03

39 V d d I V V Fure 3.7: cheatc f cpleentary - cascade dublet dfferental aplfer Q,Q3 we can cnstruct the hh frequency equalent crcut as shwn n fure 3.8. We hae nred r x and r t further splfy the analyss. We can nw rearrane the drectn f the cntrlled current surce n Q3 by reersn the drectn f the cntrlln ltae π. Ths leads t fure 3.9a. Fure 3.9b reeals further splfcatn by cbnn the tw dentcal parallel π, r π crcuts whch ccur n seres cnnectn. The current π cn tward and lean fr the nde arked, can be lfted ff fr ths junctn and has been shwn as a snle current π eetn the nde. A rule f- thub reardn such splfcatns s: a resstance n seres cnnectn can be nelected as shrt crcut f t s sall cpared wth ther resstances. larly, a resstance n shunt cnnectn can be nelected as an pen crcut f t s hh cpared wth ther resstances. abn aut, Ph.D. Pae /6/03

40 d r r Fure 3.8: Apprxate hh frequency equalent crcut fr the scheatc n F.3.7. In F.3.9b we can see tw dsjnt crcuts shwn by blue and le reen lnes wth asscated dstnct te cnstants. The te cnstants are: r, and. The dnant hh frequency ple s: H, whch can be rearded as the band wdth f the aplfer. + d - r r, + d - r /, a b Fure 3.9: Further splfcatn and cpactn f the crcut n F.3.8. The student s suested t cnstruct the lw-frequency fr f the equalent crcut n F.3.9b and fnd the expressn fr the d-band an A M. The an bandwdth s then A M ω H. abn aut, Ph.D. Pae /6/03

41 3.6.4:Wde band aplfers wth MOFT transstrs nhanceent de MOFT can be cnnected n the sae way as JT staes t dere cascde and wde band dfferental aplfers fr hh frequency applcatns. The analyss fllws slarly. The student s encuraed t draw the scheatcs wth MOFT deces by fllwn the crrespndn JT cnfuratns n sectns : Practce xercses 3.7.: hw by apprprate analyss KL/KVL/Ndal atrx that the ltae nal cupled acrss the nternal base-etter junctn f a JT.e., π has a a zer at ω=0, when a ltae surce nal s fed t the base f the JT a a seres capactr, and b has a zer at a fnte frequency when parallel, netwrk s n seres wth the ltae surce. Use the lw-frequency equalent crcut fr the transstr. nsder the representate cases as shwn belw. 3.7.: Fr the JT aplfer belw, en =600 hs, = k hs, h fe =99, I = A, =.5 k hs, =. k hs, L = k hs, = μf, =5 μf, 3 = 0 μf. What wll be the lwer -3d frequency fr the aplfer? abn aut, Ph.D. Pae 3.4 3/6/03

42 3.7.3: In a JT, aplfer the netwrk paraeters are: = 00 hs, = k hs, r x = 50 hs, I = A, h fe = 99, π =. pf, μ =0. pf, V A =50 V, =.5 k hs. Fnd, a The te cnstants asscated wth the capactrs usn pen-crcut te cnstant ethd. b What s the apprxate upper cut-ff frequency? c In the equalent crcut f the aplfer use Mller s there assun a an f between the nternal cllectr and base ternals f the JT, and re-draw the equalent crcut. d Deterne the ple frequences n the equalent crcut dered n step c abe. e What wll be the apprxate upper cut-ff frequency usn the results n d? f Use the full transfer functn deternatn ethd t the equalent crcut f the aplfer and deterne the ple frequences usn exact slutn f Ds=0, where the ltae an functn s : Ns/Ds. Deterne the dnant ple fr the Ds dered n step f abe. h What are yur estates abut the upper cut-ff frequences f yu use the results n f and? Tabulate the upper cut-ff frequency alues btaned n steps b, e, f, and. abn aut, Ph.D. Pae 3.4 3/6/03

43 3.7.4: In a MOFT aplfer, yu are en the fllwn: =00 hs, s =0. pf, d =0 ff, =50 μ h, I D =50 μa, V A =0 V, and L = 5 k hs. The MOFT aplfer s cnfured t perate as aplfer. Fnd the dnant hh-frequency ple f the aplfer usn: a Mller s there b Full ndal analyss c Open-crcut te cnstant ethd 3.7.5: nsder a basc MOFT current rrr crcut. Fnd an expressn fr the hh frequency current transfer functn s/ n s. Use the hh frequency ac equalent crcut del fr the transstrs : Fr the JT aplfer shwn belw, deterne the hh frequency ltae an H transfer functn n the fr: As AM. Gen μ =0.5 pf, f T =600 MHz, h fe =49. s H abn aut, Ph.D. Pae /6/03

44 3.7.7: Fr the D- cascade MOFT cpste aplfer, t s en that, V A =0 V, s =0.5 pf, d = 0. pf, I D = 50V G - V TH μa, V TH = V. state f H fr ths aplfer syste. The utput resstance f each current surce s Mea hs : Fnd f H fr the -G cascade MOFT aplfer syste shwn belw. Use necessary data fr prble # abe. abn aut, Ph.D. Pae /6/03

45 abn aut, Ph.D. Pae /6/03