VI. Transistor Amplifiers

VI. Transstor Amplfers 6. Introducton In ths secton we wll use the transstor small-sgnal model to analyze and desgn transstor amplfers. There are two ssues that we need to dscuss frst: ) What are the mportant propertes of an amplfer? and 2) How can we add a sgnal to the bas n a real crcut? 6.. Amplfer Parameters In Secton, we showed that the response of a two-port network s completely determned f we sole the smple crcut shown. Three parameters defne the propertes of a two-port network. sg sg 2port Network ) The rato of / s called the oltage transfer functon of the crcut. For oltage amplfers, / = const and s called the amplfer oltage gan (or gan for short): o L Voltage Gan: A = In general A depends on the load, L. A specal case of the gan s of partcular nterest: Open-loop Gan: A o = c = L 2) As the combnaton of the two-port network(amplfer) and the load s a two-termnal network, t can be modeled by ts Theenn equalent. Furthermore, as ths combnaton does not contan an ndependent source, t reduces to a resstor, called the nput resstance: sg sg 2port Network o L Input esstance: = In general depends on the load, L. 3) The combnaton of the amplfer and the nput ( and sg ) s also a two-termnal network and can be modeled by ts Theenn equalent. We denote the Theenn resstance of ths combnaton as o. sg sg 2port Network o L ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-

In order to calculate o we need to set the ndependent oltage source = 0 and compute the equalent resstance seen between the output termnals,.e., Output esstance: o = o sg =0 (Theenn esstance) In general o depends on the load, sg. The Theenn oltage source of the amplfer/nput combnaton, V T, s the open-crcut oltage, c, and s related to the open-loop gan of the amplfer: V T = c = L = A o Therefore, the load sees a Theenn equalent crcut wth V T = A o and T = o. Combnng models of the nput and output ports, we arre at a model for an amplfer whch conssts of three crcut elements as s shown below (left). Voltage Amplfer Model o A o o sg sg o o A o L The amplfer crcut model allows us to sole any amplfer confguraton f we know alues of A o, and o (smlar to usng Theenn Theorem to label any two-termnal network wth T and V T ). For example, we can fnd the oerall oltage gan of the amplfer as: = L o L A o = sg = = sg A = sg A o L o L We see that the open-loop gan A o s the maxmum alue for the amplfer gan A. In addton, to maxmze /, we need and o 0. A practcal oltage amplfer, thus, s desgned to hae a large and a small o (.e., sg and o L ). A oltage-controlled oltage source s an deal oltage amplfer as and o = 0. We wll use ths amplfer model later to fnd parameters of mult-stage amplfers. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-2

6..2 Capactor Couplng As dscussed before, a constant bas oltage should be added to the sgnal to ensure that MOS s always n saturaton (or BJT s acte). There are practcal ways to accomplsh ths task. ) Drect Couplng: For a mult-stage amplfer, the basng scheme can be set up such that the bas oltage of each stage matches the bas oltage of the followng stage (see Problem 28, for example). Usually, bas wth two oltage supples s used such that for the frst amplfer stage, V G = 0 (or V B = 0 for BJT). Integrated crcut chps use ths scheme n order to aod usng capactors whch takes a lot of space on the chp. Because of the drect couplng between stages, basng becomes a dffcult desgn problem. 2) Capacte couplng. Snce a capactor becomes an open crcut for bas, t can be used to couple the sgnal to the crcut as s shown below for a MOS amplfer. V DD V DD D D D gs C c G V GS GS _ D DS ds gs I D C c G V DS 0 V GS V GS _ gs C c G gs _ d ds ds eal Crcut Bas Crcut Sgnal Crcut The frst ssue wth ths method s that DC sgnals cannot be amplfed. Secondly, the capactor mpedance, Z C = /(ωc), depends on the frequency. As a result / wll also depend on frequency and the amplfer would not behae lnearly for an arbtrary sgnal (whch ncludes many frequences). We note, howeer, that at hgh enough frequences, the mpedance of a capactor becomes ery small and the capactor effectely becomes a sort crcut. For frequences hgher than ths alue, / wll be ndependent of frequency and the lnear behaor of the amplfer s recoered. The frequency at whch we can gnore the mpedance of the couplng capactors s called the lower cut-off frequency of the amplfer. The frequency range aboe the lower cut-off frequency s called the mdband where the amplfer should be operated. (You wll see n ECE 02 that the gan of a transstor amplfer drops at hgh frequences and there s also a upper cut-off frequency). As the capacte couplng scheme confnes bas oltages n each stage, basng s smpler wth ths scheme. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-3

6..3 Analyss of Transstor Amplfer Crcuts Analyss of a transstor amplfer crcut follows three steps as we need to address seeral ssues: bas, lnear response (to small sgnals), and the mpact of couplng capactors. Bas: Zero out the sgnal and replace capactors wth open crcuts. Compute transstor bas pont parameters. Md-band Small Sgnal esponse: ) Compute g m, r o (and r π for BJT) from bas pont parameters 2) Draw the sgnal crcut (e.g., ground bas oltage sources) 3) Assume capactors are short crcut. 4) Inspect the crcut. If you dentfy the crcut as a prototype crcut, you can drectly use the formulas for that crcut. Otherwse, replace the transstor wth ts small sgnal model and sole for A, A o, and o. Frequency-response: Couplng and bypass capactors ntroduce poles at low frequences. In Secton 6.7, We wll ntroduce a method to compute the low-frequency poles. ECE02 nclude a more thorough reew of the amplfer frequency response. It turns out that there are four basc amplfer confguratons (4 for BJT and 4 for MOS). Furthermore, the nput and output resstances can be found usng elementary forms. These formulas allows one to compute propertes of any amplfer readly. These bass form are dscussed n the Appendx. Snce we are only focusng on dscrete and smple amplfer confguratons n ths course, amplfer parameters are computed drectly from the crcut n the followng sectons. Notes: ) Small-sgnal models of PNP and NPN transstors (or PMOS and NMOS transstors) are smlar. Thus, the formulas dered below can be used for ether case. 2) The small-sgnal model of a BJT s smlar to that of a MOS wth the excepton of the addtonal resstor r π (the nput termnals n a MOS s open crcut). As such, we expect that formulas for MOS amplfers would be the same as those of BJT amplfers f we set r π. 3) For MOS crcuts, we use the common approxmaton g m r o as g m = 2I D V OV, r o V A I D g m r o = 2V A V OV typcally g m r o s 50 or more. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-4

4) For BJT crcuts, we also use the common approxmaton g m r o as g m = I C V T, r o = V A V CE I C V A I C g m r o = V A V T typcally g m r o s seeral thousands. In addton, g m r π = β 5) In many text books (e.g., Sedra & Smth), the formulas for BJT amplfers are gen n terms of β & r e (nstead of g m & r π ) where r e = = r π g m β wth r e typcally n 0s or 00 Ω range. Here we keep the g m form so we that can see the comparson to MOS amplfers. 6) Some manufacturer spec sheet for BJTs(e.g., spec sheet for 3N3904) use the older notaton (hybrd π model) for BJT whch are h fe β, h re r π, and h oe /r o 6.2 Common-Dran and Common-Collector Amplfers Common-Dran or Source Follower Confguraton Crcut shown s the generc sgnal crcut of a common-dran amplfer (.e., we hae zeroed out all Bas sources). sg C c G S o L Note that the nput s appled at the gate and the output s taken at the source. As the dran s grounded (for sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the common-dran amplfer. It s mportant to realze that as a transstor can be based n many ways, seeral complete crcuts (.e., ncludng bas elements) wll reduce to the aboe sgnal crcut form of a common-dran amplfer. Some examples are gen below ( and sg n the nput are not shown for smplcty). ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-5

V DD G V SS V DD C c S G2 S L L L V DD V SS G = G G2 G, G, S, C c, C c, We now proceed wth the sgnal analyss by replacng the MOS wth ts small-sgnal model. Usng node-oltage method (there s one node, ): gs = sg G gs _ g m gs S r o D Node S r o L g m gs = 0 r o S L g m ( ) = 0 G S o L = g m(r o S L ) g m (r o S L ) As can be seen, the gan of ths amplfer s less than. For large S L, t achees a gan of (snce g m r o ). Thus, ths confguraton s also called the Source Follower. Fndng s easy as = and = G (see crcut). Therefore, = G. Note that f G were not present (see example complete crcuts aboe).. To fnd o we need to zero out and compute the Theenn Equalent resstance seen at the output termnals. Because of the presence of the controlled source, the standard method s to attach a x oltage source to the crcut and compute x. sg G G gs _ g m gs S S r o D o x x x We see from the crcut that g = 0 and gs = x. Thus: ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-6

x = x x g m gs = x x x r o S r o S /g m o = x = S r o S x g m g m In summary, the general propertes of the common-dran amplfer (source follower) nclude a oltage gan, a large nput resstance (whch can be made nfnte n some basng schemes) and a small output resstance. Ths type of crcut s called a buffer and often used when there s a msmatch between nput resstance of one stage and the output resstance of the preous stage. Addtonally, L = o as A but o. As such, ths crcut can be used to amplfy the sgnal current (and power) and dre a load (used typcally as the last stage of an amplfer crcut). Common-Collector or Emtter Follower Confguraton Crcut shown s the generc sgnal crcut of a common-collector amplfer (.e., we hae zeroed out all Bas sources). sg C c B E o L Note that the nput s appled at the base and the output s taken at the emtter. As the collector s grounded (for sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the common-collector amplfer. As can be seen ths confguraton s analogous to MOS common-dran. Smlarly to the MOS case, the BJT can be based many ways. Seeral complete crcuts (.e., ncludng the bas elements) wll reduce to the aboe sgnal form of a common-collector amplfer. Some examples are gen below. V CC B V EE V CC C c E C c2 B2 E L L L V CC V EE B = B B2 B, B, E, C c, C c, ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-7

We now proceed wth the sgnal analyss by replacng the BJT wth ts small-sgnal model. Usng node-oltage method (there s one node, ): π = sg π B C π _ r π gm π E r o Node E r o L r π g m π = 0 B E o L The last two terms n the aboe equaton can be smplfed by notng /r π = g m /β: r π g m π = ( ) g m β g m( ) = g m ( ) β β g m ( ) Substtutng back n the node equaton, we get: E r o L g m ( ) = 0 r o E L g m ( ) = 0 = g m(r o E L ) g m (r o E L ) Smlar to the MOS common-dran amplfer, the gan of the common collector amplfer s less than, but usually close to because of the large BJT g m. Ths confguraton s also called the Emtter Follower. To fnd = /, we note that by KCL = π and π = r π = ( ) o r π = g m(r o E L ) g m (r o E L ) = π = r π g m r π (r o E L ) = π = B r π β(r o E L ) = = B [r π β(r o E L )] g m (r o E L ) B [r π β(r o E L )] ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-8

Note that when emtter degeneraton basng s used, we need to hae B (β) E. In ths case, B (smlar to the common-dran amplfer n whch = G ). If B s not present, the nput resstance s would be ery large (MΩ leel). To fnd o we need to zero out and compute the Theenn Equalent resstance seen at the output termnals. We attach x oltage source to the crcut and compute x. The calculaton s straght forward and s left as an exercse (Hnt: π = [r π /(r π B sg )]. o = E r o r π B sg β In summary, the general propertes of the common-collector amplfer (emtter follower) nclude a oltage gan close to unty, a large nput resstance and a small output resstance (smlar to the common-dran amplfer). Thus, emtter follower s also used as a buffer or for amplfyng the sgnal current (and power) and dre a load. 6.3 Common-Source and Common-Emtter Amplfers Common-Source Confguraton Crcut shown s the generc sgnal crcut of a common-source amplfer (.e., we hae zeroed out all Bas sources). Note that the nput s appled at the gate and the output s taken at the dran. As the source s grounded (for sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the commonsource amplfer. sg C c G D o L If source degeneraton basng s used for the common-source confguraton, a resstor S should be present. A by-pass capactor s typcally used so that the sgnal by-passes S,effectelymakngthesource grounded for the sgnal as s shown. sg C c G D S o C S L We replace the MOS wth ts small-sgnal model. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-9

sg G D G gs _ g m gs S r o D o L Inspecton of the crcut shows that = gs. Furthermore, r o, D, and L are n parallel and by KCL a current of g m gs flows n r o D L (from the ground to ). Ohm Law: = g m gs (r o D L ) = g m (r o D L ) The negate sgn n the gan s ndcate of a 80 phase shft n the output sgnal. Inspectng the crcut, we fnd = / = G. To fnd o we need to zero out and compute the Theenn Equalent resstance seen at the output termnals. Snce n ths crcut gs = 0 when = 0, the controlled source g m gs becomes an open crcut and the output resstance can be found by nspecton to be o = D r o. In summary, the general propertes of the common-source amplfer nclude a large oltage gan, a large nput resstance (and can be made nfnte wth some basng schemes) but a medum output resstance. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-0

Common-Emtter Confguraton Crcut shown s the generc sgnal crcut of a common-emtter amplfer (.e., we hae zeroed out all Bas sources). Note that the nput s appled at the base and the output s taken at the collector. As the emtter s grounded (for sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the commonemtter amplfer. sg C c B C o L If emtter degeneraton basng s used for ths confguraton, a resstor E should be present. A by-pass capactor s typcally used so that the sgnal by-passes E, effectely makng the emtter grounded for the sgnal as s shown. sg C c B C E o C E L We replace the BJT wth ts small-sgnal model. sg B C gm π π r π r B o C _ E o L Inspecton of the crcut shows that = π. Furthermore, r o, C, and L are n parallel and by KCL a current of g m π flows n r o C L (from the ground to ). Ohm Law: = g m π (r o C L ) = g m (r o C L ) The negate sgn n the gan s ndcate of a 80 phase shft n the output sgnal. Inspectng the crcut, we fnd = / = B r π. To fnd o we need to zero out and compute the Theenn Equalent resstance seen at the output termnals. Snce n ths crcut π = 0 when = 0, the controlled source g m π becomes an open crcut and the output resstance can be found by nspecton to be o = C r o. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-

In summary, the general propertes of the common-emtter amplfer nclude a large openloop oltage, a medum nput resstance and a medum output resstance. 6.4 Common-Source and Common-Emtter Amplfers wth Degeneraton Common-Source Confguraton wth a Source esstor Crcut shown s the generc sgnal crcut of a common-source amplfer wth degeneraton (.e., wth a source resstor). Note that the nput s appled at the gate and the output s taken at the dran smlar to a common-emtter amplfer sg C c G D S o L We replace the MOS wth ts small-sgnal model. Usng node-oltage method (there are two nodes, and s ): Node s : s s g m gs = 0 S r o sg G G gs _ g m gs S r o D D o L Node : s g m gs = 0 D L r o S s S D L = 0 where the last equaton s found by summng the frst two. Substtutng for gs = s n the frst equaton, computng s, and substtutng n the thrd equaton, we get: g m r o ( D L ) = D L r o S (r o g m ) g m r o ( D L ) D L r o g m r o S = g m ( D L ) g m S ( D L )/r o Compared to a CS amplfer wth no S, the amplfer gan s substantally reduced wth the presence of S. Howeer, the gan has become much less senste to changes n g m. Inspectng the crcut we fnd = / = G, smlar to a common-source amplfer. To fnd o, we set = 0 and compute the Theenn Equalent resstance seen at the output termnals. We attach x oltage source to the crcut and compute x. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-2

sg G D x x By KCL, a current of g m gs should flow n r o and should flow n S. Snce gs = S : G gs _ g m gs S S r o D o x KVL : x = r o ( g m gs ) S KCL: x = r o ( g m S ) S = (r o g m r o S S ) x = r o S g m r o S r o (g m S ) x = x D x = x D x r o (g m S ) o = x X = D [r o (g m S )] In summary, source degeneraton has led to an amplfer wth a lower gan whch s less senste to transstor parameters. Common-Emtter Confguraton wth an Emtter esstor Crcut shown s the generc sgnal crcut of a common-emtter amplfer wth degeneraton (.e., wth a emtter resstor). Note that the nput s appled at the base and the output s taken at the collector smlar to a common-emtter amplfer. sg C c B C E o L We replace the BJT wth ts small-sgnal model. Usng node-oltage method (there are two nodes, and e ): sg π _ B r π gm π E r o C C o L Node e : 0 = e E e r π e r o g m π B E Node : 0 = C L e r o g m π = 0 e e = 0 E C L r π ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-3

The thrd equaton s the sum of the frst two. Fndng e from the thrd equaton and substtutng n the 2nd equaton, we get: g m ( C L ) = g m E (( C L )/r o )( E /r π ) = g m ( C L ) g m E E /r π ( E /r π )( C L )/r o g m ( C L ) g m E ( E /r π )( C L )/r o Where we hae used E /r π = g m E /β g m E. If ( C L )/r o β (a ery good approxmaton), we can drop the last term to fnd: g m( C L ) = C L g m E E /g m whch s the expresson often used. Note that the amplfer gan s reduced wth the presence of E but t has become substantally less senste to any change n β (only through /g m whch s usually E ). From the crcut we fnd = / = B ( / b ). The exact formulaton for / b s cumbersome. A good approxmaton whch leads to a smple expresson s r o E. In ths case, r o can be remoed from the crcut and = b r π ( b g m π ) E = b r π ( b β b ) E = b [r π (β) E ] = B [r π (β) E ] To fnd o, we set = 0 and compute the Theenn Equalent resstance seen at the output termnals. Because of the presence of the controlled source, we need to attach x oltage source to the crcut and compute x. Calculatons are left as an exercse. The output resstance s gen by ( )] β E o = C [r o C r π E B sg In summary, emtter degeneraton has led to an amplfer wth a lower gan whch s much less senste to transstor parameters and a substantally larger nput resstance. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-4

6.5 Common-Gate and Common-Base Amplfers Common-Gate Confguraton Crcut shown s the generc sgnal crcut of a common-gate amplfer. Note that the nput s appled at the source and the output s taken at the dran. As the gate s grounded (for sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the common-gate amplfer. sg C c D S o L In some cases, the gate has to based to a DC alue (for example usng oltage dder crcut shown). Howeer, snce G = 0, the gate remans grounded for the sgnal. We replace the MOS wth ts small-sgnal model. Usng node-oltage method (there s one node, ): sg C c G V G2 DD D S o L gs = 0 = G D g m ( ) = 0 L D r o = g mr o r o D L r o r o D L g mr o r o sg _ = gs _ gm gs S S r o D o L = g m (r o D L ) To fnd, t s easer to wrte = S (seecrcut). ByKCLatnodeS,current g m gs wll flow n r o and current wll flow n D L. Thus: = ( g m gs )r o ( D L ) _ G = gs _ gm gs S S r o D D L = r o g m r o ( D L ) = r o ( D L ) g m r o ( D L )/r o g m = S ( D L )/r o g m ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-5

To fnd o, we set = 0 and compute the Theenn Equalent resstance seen at the output termnals. Because of the presence of the controlled source, we need to attach x oltage source to the crcut and compute x. G D x x ByKCL,acurrentof g m gs shouldflow nr o and shouldflown S sg. Then, gs = s = ( S sg ) and sg _ = gs _ gm gs S sg S r o D o x KVL : x = r o ( g m gs ) S KCL: x = r o g m r o ( S sg ) ( S sg ) x r o (g m ( S sg )) x = x D x = x D x r o (g m ( S sg )) o = x X = D [r o (g m ( S sg ))] In summary, the general propertes of the common-gate amplfer nclude a large oltage gan, a small nput resstance and a medum output resstance (t has the same gan and output resstance alues as that of a common-source confguraton but has a much lower nput resstance). ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-6

Common-Base Confguraton Crcut shown s the generc sgnal crcut of a common-base amplfer. Note that the nput s appled at the emtter and the output s taken at the collector. As the base s grounded (for sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the common-base amplfer. sg C c V CC C E o L In some cases, the base has to based to a DC alue (for example usng oltage dder crcut shown). In ths case, a bypass capactor s needed to keep the base grounded for the sgnal. We replace the BJT wth ts small-sgnal model. Comparng the small sgnal crcut of the common base amplfer wth that of a common-gate amplfer of the preous page, we see that the two crcuts are dentcal f we replace S wth E r π (and D wth C ). As such we can use the results from the common-gate amplfer analyss to get: sg _ = π _ sg B r π C b C c gm π E E B B2 r o C C C E o o L L = g m (r o C L ) = E r π ( C L )/r o g m o = C [r o (g m ( E r π sg ))] In summary, the common-base confguraton has a large open-loop oltage, a small nput resstance and a medum output resstance (t has the same gan and output resstance alues as that of a common-emtter confguraton but a much lower nput resstance). ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-7

6.6 Summary of Amplfer Confguratons The common-source (CS) and common-emtter (CE) amplfers hae a hgh gan and arethemanconfguratonnapractcalamplfer. Ignorngbasresstors G or B,the CS confguraton has an nfnte nput resstance whle the CE amplfer has a modest nput resstance. Both CS and CE amplfer hae a rather hgh output resstance r o and a lmted hgh-frequency response (you wll see ths n 02). Addton of source or emtter resstor (degenerated CS or CE) leads to seeral benefts: a gan whch s less senste to temperature, a much larger nput resstance for CE confguraton, a better control of amplfer saturaton, and a much mproed hghfrequency response. Howeer, these are realzed at the expense of a lower gan. The common-gate (CG) and commons-base (CB) amplfers hae a hgh gan (smlar to CS and CE) but a low nput resstance. As such, they are only used for specalzed applcatons. CG and CB amplfers hae an excellent hgh-frequency response. They are typcally used n combnaton wth a CS or CE stage (such as cascode amplfers) The source-follower and emtter-follower confguratons hae a hgh nput resstance, a gan close to unty, and a low output resstance. They are employed as a oltage buffer and/or as the output stage to ncrease the current and power to the load. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-8

6.7 Low Frequency esponse of Transstor Amplfers Up to now, we hae neglected the mpact of the couplng and by-pass capactors (assumed they were short crcut). Each of these capactors ntroduce a pole n the response of the crcut. For example, let s consder the couplng capactor at the nput to the amplfer (C c n amplfer confguratons that we examned before). We need to perform the analyss n the frequency doman (oltage are represented by captal letter as they are n phasor form): A = V o V = V V sg = V V sg = V o V sg = L o L A o sg /(sc c ) sg s sω p, sg A s sω p sg sg 2πf p = ω p C c C c ( sg ) o A o o L Where A s the md-frequency gan that we hae calculated for all transstor confguratons (.e., wth capactors short). As can be seen, the couplng capactor C c has ntroduced a low-frequency pole and the amplfer gan falls at low frequences. One can compute the mpact of the couplng capactor at the output n a smlar manner. t s straghtforward to show (left as an exercse): sg sg o A o L V o V sg = sg A s sω p2 2πf p2 = ω p2 ( L o ) Smlarly, f a by-pass capactor s present (see page 5-8), t wll ntroduce yet another pole, f p3. The followng method allows one to compute the poles by nspecton.. Zero out V sg. 2. Consder each capactor separately (.e., assume all other capactors are short) 3. Compute,, the total resstance between the termnals of the capactor. The pole ntroduced by that capactor s gen by f p 2πC ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-9

Wth all poles assocated wth by-pass and couplng capactors n hand, we can fnd the oerall frequency response of the amplfer as V o V sg = A s sω p s sω p2 s sω p3 and the lower cut-off frequency s located at 3dB below maxmum alue, A (the mdfrequency gan). If poles are suffcently separated (such as the fgure aboe), the lower cut-off frequency of the amplfer s gen by the hghest-frequency pole. Otherwse, fndng the lower cut-off frequency would be cumbersome. A smple approxmaton for hand calculatons (whch s surprsngly ery good) s to set f l f p f p2 f p3 The next two pages nclude a summary of formulas for dscrete transstor amplfers. These formulas are correct wthn approxmaton of g m r o and β both of whch are always ald. Many of these formulas can be smplfed (before pluggng n numbers) as they nclude resstances that are n parallel and typcally one s much smaller (at least by a factor of ten) than the others. For example, n a common emtter amplfer, we often fnd that C L and C r o. Then, r o C L C and the gan formula can be smplfed to A = C /r e. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-20

Common Dran (Source Follower): Summary of Dscrete MOS Amplfers A = g m(r o S L ) g m (r o S L ) = G o = g m S sg C c G S o L Common Source: A = g m (r o D L ) sg C c D = G o = D r o f p3 = 2πC s [ S (/g m )] G S C S o L Common Source wth Source esstance: g m ( D L ) A = g m S ( D L )/r o D = G o = D [r o (g m S )] sg C c G S o L Common Gate: A = g m (r o D L ) D = S ( D L )/r o g m o = D [r o (g m ( S sg ))] sg C c S o L f l = Σ j f pj and f p = /[2πC c ( sg )] and f p2 = /[2π ( L o )] ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-2

Summary of Dscrete BJT Amplfers Common Collector (Emtter Follower): A = g m(r o E L ) g m (r o E L ) = B [r π β(r o E L )] sg C c B E o L o = E r o r π B sg β Common Emtter: A = g m (r o C L ) C = B r π sg C c o = C r o f p3 = 2πC e [ E (/g m ( B sg )/β)] B E C E o L Common Emtter wth Emtter esstor: A g m( C L ) g m E = B [r π (β) E ] ( )] β E o = C [r o r π E B sg sg C c B C E o L o C Common Base: C = g m (r o C L ) = E r π ( C L )/r o g m o = C [r o (g m ( E r π sg ))] sg C b C c B E o L f p3 = /[2πC b CB ] CB B [r π (β)( sg E )] f l = Σ j f pj and f p = /[2πC c ( sg )] and f p2 = /[2π ( L o )] ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-22

6.8 Exercse Problems Problem to 3: Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). 9 V 9 V 9 V 8k 8k 4 V 22k k 22k k 22k k k 8k 5 V Problem Problem 2 Problem 3 Problem 4 5 V 4 V 4 V 34k k 00 nf 4.3mA 4.3mA 00 4.7 µ F 5.9k 50 47 µ F V EE V EE Problem 5 Problem 6 Problem 7 5 V 5 V 00 5.9k 4.7 µ F 50 47 µ F 00 4.7 µ F 34k k 00 nf 34k k 00 nf 5.9k 50 Problem 8 Problem 9 ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-23

5 V 5 V 00 5.9k 4.7 µ F 50 47 µ F 00 4.7 µ F 34k k 00 nf 34k k 00 nf 5.9k 50 Problem 8 Problem 9 5 V 00 4.7 µ F 34k k 00 nf 3V 2.3k 47 µ F ma 3V 47 µ F 5.9k 270 00 nf 00 nf 240 47 µ F 2.3k 2.3k 3V Problem 0 Problem Problem 2 3V Problem 4-7. Fnd the bas pont and amplfer parameters of ths crcut (V tn = 4 V, V tp = 4 V, µ p C ox (W/L) = µ n C ox (W/L) = 0.4 ma/v 2, and λ = 0.0 V. Ignore the channel-wdth modulaton effect n basng calculatons.) 2.5 V 8 V 0nF k 3k 00nF 00.3M 0k 000nF 400 2k 500k Problem 3 Problem 4 ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-24

3V V SS 5 V 0k 0.7 ma 00 5 V 5 V 0.7 ma Problem 5 Problem 6 Problem 7 V SS Problem 8-24. Fnd the bas pont and amplfer parameters of ths crcut (V tn = V, V tp = V, µ p C ox (W/L) = µ n C ox (W/L) = 0.8 ma/v 2, and λ = 0.0 V. Ignore the channel-wdth modulaton effect n basng calculatons. 5 V 5 V 00.8M 00nF 0k 00 nf 00.2M 00nF 0k µ F.2M 0k µ F.8M 0k 00 nf Problem 8 Problem 9 9 V 5 V 0k 00 nf 00 00.2M 00nF 0k 0k µ F.8M 0k 00 nf 6 V Problem 20 Problem 2 ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-25

5 V 6 V 5 V 00.8M 00nF 0k 00 nf 00 0k 00nF 0k.8M.2M 0k 0k 00 nf 00nF 0k.2M Problem 22 Problem 23 Problem 24 9 V ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-26

6.9 Soluton to Selected Exercse Problems Problem. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). 9 V Bas: Set = 0 and capactors open. Set = 0 and capactors open. eplace B / B2 oltage dder wth ts Theenn equalent. Assumng BJT n acte, 8k B = 8 k 22 k = 9.9 kω V BB = 22 822 9 = 4.95 V 22k k KVL: V BB = B I B V BE 0 3 I E 4.95 = 9.9 0 3 I E /(β )0.70 3 I E 9 V I E = 4.05 ma I C, I B = I C β = 20.3 µa V BB 22k 8k KVL: 9 = V CE 0 3 I E V CE = 90 3 4 0 3 = 5 V k Snce V CE > V D0 = 0.7, assumpton of BJT n acte s correct. 9 V Bas summary: I E I C = 4.05 ma, I B = 20.3 µa, V CE = 5 V Small-Sgnal: Frst we calculate the small-sgnal parameters: V BB 22k 8k g m = I C = 4 03 = 56 ma/v V T 26 03 r π = β =.28 k r o V A = 50 = 37.0 k g m I C 4 03 k Note that we could hae gnored V CE compared to V A n the aboe expresson for r o. Proceedng wth the small sgnal analyss, we zero bas sources (see crcut). As the nput s at the base and output s at the emtter, ths s a common-collector amplfer (emtter follower). Usng formulas of page 6-2 and notng L, E r o, and E r e : A = g m(r o E L ) g m (r o E L ) = 52 53 B [r π β(r o E L )] = (9.9 k) (.28 k95 k) = 9.42 k ( B ) o = E r o r π B Sg β f l = f p = = 6.4 Ω ( r π β = g m ) 2πC c ( sg ) = 2π 0.47 0 6 (9.9 0 3 0) = 34.2 Hz ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-27

Problem 2. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). Ths s the same crcut as Problem wth excepton of and L. The bas pont s exactly the same. As E L, the amplfer parameters would be the same except f l = f p f p2 = 34.33.39 = 37.6 Hz. Problem 3. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n = 2, β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). Ths crcut s smlar to Problem expect that the transstor s based wth two oltage sources (alues are chosen to ge approxmately the same bas pont). 4 V Bas: Set = 0 and capactors open: k BE-KVL: 0 = V BE 0 3 I E 5 5 V I E = 4.3mA I C, I B = I C β = 2.5 µa 4 V CE-KVL: 4 = V CE 0 3 I E 5 V CE = 90 3 4.3 0 3 = 4.7 V k Bas summary: I E I C = 4.3 ma, I B = 2.5 µa, V CE = 4.7 V 5 V Small-Sgnal: Frst we calculate the small-sgnal parameters: g m = I C = 4.3 03 = 65.4 ma/v V T 26 03 k r π = β g m =.2 k r o V A I C = 50 = 34.9 k 4.3 03 Proceedng wth the sgnal analyss, we zero bas sources (see crcut). As the nput s at the base and output s at the emtter, ths s a common-collector amplfer (emtter follower). The dfference wth Problem s that there s no B ( B = ) whch affects only. A = g m(r o E L ) g m (r o E L ) = 6 62 B [r π (β)(r o E L )] = (.2 k94 k) = 95 k o = E r o r π B Sg β f l = f p2 = = 6.0 Ω ( r π β = g m ) 2π ( L o ) = 2π 0.47 0 6 (00 0 3 6) = 3.39 Hz ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-28

Problem 5. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wthβ = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). Ths crcut s smlar to the crcut of Problem 3 except that the transstor s based wth a current source. 4 V Bas: Set = 0 and capactors open. 4.3mA I E = 4.3 ma I C, I B = I C β BE-KVL: 0 = V BE V E V E = 0.7 V CE-KVL: 4 = V CE V E V CE = 4.7 V = 2.5 µa V EE 4 V V E Bas summary: I E I C = 4.3 ma, I B = 2.5 µa, V CE = 4.7 V Small-Sgnal: Frst we calculate the small-sgnal parameters: 4.3mA V EE g m = I C = 4.3 03 = 65.4 ma/v V T 26 03 r π = β g m =.2 k r o V A I C = 50 = 34.9 k 4.3 03 4.3mA V EE Amplfer esponse: we zero bas sources (the current source becomes an open crcut. As the nput s at the base and output s at the emtter, ths s a common-collector amplfer (emtter follower). The dfference wth problem 3 s that here E. Usng formulas of page 6-2 and notng E L = L r e A = g m(r o E L ) g m (r o E L ) = 4,279 4,280 B [r π β(r o E L )] = (.2 k5.7 M) = 5.7 M o = E r o r π B Sg β f l = f p2 = 2π ( L o ) = 6.0 Ω ( r π β = g m ) = 3.39 Hz Comparng results from Problems through 5 hghlghts the mpact of each element on the amplfer performance as n successe problems, L and C C2 were added, and then B, C C and E were elmnated. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-29

Problem 6. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wthβ = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). Ths crcut s smlar to the crcut of Problem 5 except that s remoed. The load (00 k resstor) exsts n the bas crcut. Howeer, the bas current n the 00 K resstor s small and does not alter the bas. The output oltage,, howeer, nclude the DC bas oltage. Bas: Set = 0 and capactors open. 4.3mA 4 V V EE 4 V BE-KVL: 0 = V BE V E V E = 0.7 V KCL = V E = 7 µa 00 03 I E = 4.3 0 3 = 4.3 ma V E 4.3mA V EE I C I E = 4.3 ma, I B = I C β CE-KVL: 4 = V CE V E V CE = 4.7 V = 2.5 µa Bas summary: I E I C = 4.3 ma, I B = 2.5 µa, V CE = 4.7 V whch are the exactly the same alues as of those of Problem 5. The amplfer parameters are exactly the same as those of Problem 6 expect f l = 0. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-30

Problem 7. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). 5 V Bas: Set = 0 and capactors open. eplace B / B2 oltagedderwthtstheennequalent: 34k k 00 nf 00 4.7 µ F BE-KVL: B = 5.9 k 34 k = 5.0 k, V BB = 5.9 5 = 2.22 V 5.934 V BB = B I B V BE 50I E 5.9k 50 47 µ F 5 V CE-KVL: 2.22 = 5.0 0 3 I E / β )0.750I E I E = 2.84 ma I C, 5 = 000I C V CE 50I E V CE = 0.5 V > V D0 I B = I C β = 4.2 µa 34k k 5.9k 50 Bas summary: I C I E = 2.84 ma, I B = 4.2 µa, V CE = 0.5 V Small-Sgnal: Frst we calculate the small-sgnal parameters: 5 V k g m = I C V T = 2.84 03 26 0 3 = 09 ma/v r π = β g m =.83 k r o V A I C = 50 = 52.8 k 2.84 03 V BB 34k 5.9k 50 Proceedng wth the small sgnal analyss, we zero bas sources (see crcut). As the nput s at the base and output s at the collector, ths s a common-emtter amplfer wth NO emtter resstor as there s bypass capactor. = g m (r o C L ) = 06 = B r π = 5.0.83 =.34 k A = = sg = 0.93 06 = 99 o = C r o = 0.98 k ( C ) f p = 2πC c ( sg ) = 2π 4.7 0 6 (,3400) = 25.3 Hz ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-3

f p2 = f p3 = f p3 = 2π ( L o ) = 2π00 0 9 (00 0 3 980) 2πC e [ E (/g m ( B sg )/β)] 2πC e [ E 9.67] = 356 Hz f l = f p f p2 f pb = 25.35.9356 = 397 Hz = 5.9 Hz Note that although C b s the largest capactor n the crcut (e.g., 0 tmes larger than C c, f p3 s 0 tmes larger than the other poles. Problem 9. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). Bas: Set = 0 and capactors open. The bas crcut s exactly that of Problem 7 wth B = 5.0 k. Bas summary: I C I E = 2.84 ma, I B = 4.2 µa, V CE = 0.5 V Small-Sgnal: The small-sgnal parameters are also the same as those of Problem 7: g m = 09 ma/v, r π =.83k, and r o = 52.8 k. 00 4.7 µ F 34k 5.9k 5 V k 50 00 nf Proceedng wth the small sgnal analyss, we zero bas sources (see crcut). As the nput s at the base and output s at the collector, ths s a degenerated common-emtter amplfer (.e, wth a emtter resstor): g m( C L ) g m E =.9 = B [r π (β) E ] = 4.8 k ( B ) A = = = 0.98.9 =.87 sg ( )] β E o = C [r o C = k r π E B sg f p = f p2 = 2πC c ( sg ) = 2π 4.7 0 6 (4,80000) = 6.9 Hz 2π ( L o ) 2π ( L c ) 2π00 0 9 (00 0 3 0 3 ) f l = f p f p2 = 6.95.8 = 22.7 Hz = 5.8 Hz ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-32

Problem 0. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). 5 V 34k k 00 nf 00 4.7 µ F 5.9k 270 Bas: Set = 0 and capactors open. Because the 47 µf capactor across the 240 Ω resstor becomes an open crcut, the total E for bas s 270 240 = 50 Ω and the bas crcut s exactly that of Problem 7 (or Problem 9) wth B = 5.0 k. 34k 240 5 V k 47 µ F Bas summary: I C I E = 2.84 ma, I B = 4.2 µa, V CE = 0.5 V Small-Sgnal: The small-sgnal parameters are also the same as those of Problem 7: g m = 09 ma/v, r π =.83k, and r o = 52.8 k. 5.9k 270 240 Proceedng wth the small sgnal analyss, we zero bassources(seecrcut). Asthenputsatthebase and output s at the collector, ths s a degenerated common-emtter amplfer (.e, wth a emtter resstor). For mdband amplfer parameters calacultons, the 47 µf capactor across the 240 Ω resstor becomes a short crcut and the total E for smallsgnal s 270 Ω. 00 4.7 µ F 34k 5.9k 5 V k 270 240 00 nf g m( C L ) = 3.55 g m E = B [r π (β) E ] = 4.6 k ( B ) A = = = 0.98.9 = 3.47 sg ( )] β E o = C [r o C = k r π E B sg f p = 2πC c ( sg ) = 2π 4.7 0 6 (4,60000) = 7.20 Hz ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-33

f p2 = 2π ( L o ) 2π ( L c ) 2π00 0 9 (00 0 3 0 3 ) = 5.8 Hz We need to fnd the pole ntroduced by the 47 µf by-pass capactor, f pb. Although ths confguraton was not ncluded n the formulas for BJT elementary confguraton of page 6-2, we can extend those formulas to coer ths case. The pole ntroduced by the by-pass capactor n the common emtter case s (see fgure below left) f p3 = 2πC e [ E (/g m ( B sg )/β)] 5 V 5 V 34k k 00 nf 34k k 00 nf 00 4.7 µ F 00 4.7 µ F 5.9k e 5.9k E2 e E C e E C e Per our dscusson of Secton 6.7 on how to fnd poles ntroduced by each capactor, E [/g m ( B sg )/β)] s the total resstance seen across the termnal of C e. As can be seen from the crcut (aboe rght), the resstance across C c termnals conssts of two resstors n parallel, E and e. e s the resstance seen between the emtter of the BJT and the ground and s: e /g m ( B sg )/β) fromt he aboe forumla. For the crcut here (defned E = 240 Ω and E2 = 270 Ω), the resstance across C e s made of two resstances n parallel: E and the combnaton of E2 and e, the resstance seen through the emtter of BJT n seres. Thus: f p3 = f pb = 2πC b [ E ( E2 /g m ( B sg )/β)] 2π C e [240 (2709.70.49] = 2π 47 0 6 29 f l = f p f p2 = 7.205.826.2 = 49.2 Hz = 26.2 Hz ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-34

Problem. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). 3V Bas: Set = 0 and capactors open. 2.3k 47 µ F BE-KVL: CE-KVL: 3 = 2.3I E V EB I E = ma I C, I B = I E /(β) = 5 µa 3 = 2.3 0 3 I E V EC 2.3 0 3 I C 3 2.3k 3V 00 nf V EC = 64.6 0 3 0 3 =.4 V 3V Bas summary: I C I E = ma, I B = 5.0 µa, V CE =.4 V 2.3k Small-Sgnal: Frst we calculate the small-sgnal parameters: g m = I C = 03 = 38.5 ma/v V T 26 03 r π = β = 5.26 k r o V A = 50 = 50 k g m I C 03 2.3k 3V Proceedng wth the small sgnal analyss, we zero bas sources. As the nput s at the base andoutputsatthecollector, thssacommon-emtteramplfer. Itdoesnothaeanemtter resstor as 47 µf capactor shorts out E for sgnals. A = g m (r o C L ) = 38.5 0 3 (50 k k 00 k) = 37.9 = B r π = 0.4 k o = C r o = 0.99 k f p = 0 f p2 = f p3 = 2π ( L o ) = 2π00 0 9 (0 5 0 3 ) = 5.8 Hz 2πC e [ E (/g m ( B sg )/β)] = 2πC e [2,300 26] f l = f p f p2 f pb = 05.832 = 48 Hz = 32 Hz ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-35

Problem 2. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). Ths s the same crcut as that of Problem expect that the transstor s based wth a current source Bas: Set = 0 and capactors open. From the crcut I E = ma ma 3V 47 µ F BE-KVL: I E = ma I C, V E = V EB = 0.7 V I B = I C β = 5 µa 2.3k 3V 00 nf CE-KVL: V E = V CE 2.3 0 3 I C 3 = V CE 0.7 3V V CE =.4 V ma Bas summary: I C I E = ma, I B = 50 µa, V CE =.4 V. Small-Sgnal: As the bas pont s exactly the same as that of problem, we hae: g m = 38.5 ma/v, r π = 5.26k, and r o = 50 k. 2.3k 3V Amplfer response: The only dfference wth problem s that E n ths crcut. E only appears n f p3 but E = does not change results: A = 37.9, = 0.4 k, o = 0.99 k, and f l = 05.832 = 48 Hz. Problem 3. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth β = 200 and V A = 50 V. Ignore the Early effect n basng calculatons). Bas: Set = 0 and capactors open. 2.5 V BE-KVL: B = 2 k 3 k = 6.24 k, V BB = 2 2.5 =.2 V 23 V BB = B I B V BE 50I E 0nF 000nF k 400 3k 2k 00nF.2 = 6.24 0 3 I E /(β)0.7400i E 2.5 V CE-KVL: I E =.6 ma I C, 2.5 = 000I C V CE 400I E I B = I C β = 5.8 µa k 3k V CE = 2.5,400.6 0 3 = 0.88 V 400 2k Bas summary: I C I E =.6 ma, I B = 5.8 µa, V CE = 0.88 V ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-36

Small-Sgnal: Frst we calculate the small-sgnal parameters: g m = I C V T =.6 03 26 0 3 = 44.6 ma/v r π = β g m = 4.48 k r o V A I C = 50 = 29 k.6 03 Proceedng wth the small sgnal analyss, we zero bas sources. As the nput s at the emtter and output s at the collector, ths s a common-base amplfer (note L = ). = g m (r o C L ) = 44.6 0 3 (29 k k ) = 44.26 = E r π ( C L )/r o g m = k 4.48 k 22.4 = 2.8 Ω ( /g m ) o = C [r o (g m ( E r π sg ))] C = k f p = f p2 = 2πC c ( sg ) = 2π000 0 9 (2.8) 2π ( L o ) = 0 = 7.30 khz CB B [r π (β)( sg E )] = 6.24 k 84.9 k 5.8 k f p3 = 2πC b CB == = 274 Hz 2π 00 0 9 5.8 03 f l = f p f p2 f pb = 7,3000274 = 7.57 khz Note the small nput resstance of ths amplfer and correspondng large f p. ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-37

Problem 4. Fnd the bas pont and amplfer parameters of ths crcut (V tn = 4 V, V tp = 4 V, µ p C ox (W/L) = µ n C ox (W/L) = 0.4 ma/v 2, and λ = 0.0 V. Ignore the channel-wdth modulaton effect n basng calculatons.) Bas: Snce I G = 0:.3M 8 V 0k 0.5 M V G =.3 M0.5 M 8 = 5 V G =.3 M 500 k = 36 k 00 500k Assume PMOS s n the acte state, 8 V I D = 0.5µ p C ox (W/L)V 2 OV.3M 0k SG-KVL: 8 = 0 4 I D V SG V G = 0 4 I D V OV V tp V G 0 4 0.5 0.4 0 3 V 2 OV V OV 845 = 0 2V 2 OV V OV 9 = 0 V G 500k Negate root s unphyscal, V OV =.89 V and V SG = V OV V tp = 5.89 V. I D = 0.5 0.4 0 3 V 2 OV = 0.7 ma SD-KVL: 8 = V SD 0 4 I D V SD = 80 4 0.7 0 3 = 0.9 V Snce V SD = 0.9 V OV =.89 V, our assumpton of PMOS n acte s justfed. Bas summary: I D = 0.7 ma, V OV =.89 V, V SG = 5.89 V, V SD = 0.9 V. Small-Sgnal: Frst we calculate the small-sgnal parameters: g m = 2I D = 2 0.7 03 V OV.89 r o = λi D = = 0.75 ma/v = 4 k 0.0 0.7 03 Proceedng wth the small sgnal analyss, we zero bas sources. As the nput s at the gate and output s at the source, ths s a common-dran amplfer (source follower). = g m(r o S L ) g m (r o S L ) = 6.4 7.4 = 0.866 = G = 36 k ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-38

A = = sg = 0.866 o = g m S =.33 k 0 k =.7 k f p = /[2πC c ( sg )] = 0.94 Hz f p2 = /[2π ( L o )] = 3.39 Hz f l = f p f p2 = 4.33 Hz Problem 5. Fnd the bas pont and amplfer parameters of ths crcut (V tn = 4 V, V tp = 4 V, µ p C ox (W/L) = µ n C ox (W/L) = 0.4 ma/v 2, and λ = 0.0 V. Ignore the channel-wdth modulaton effect n basng calculatons. 3V Answer: 0k I D = 0.7 ma, V OV =.89 V, V SG = 5.89 V, V SD = 0.9 V. / = 0.866, =, o =.7 k, f p = 0, f p2 = 3.35 Hz, f l = 4.3 Hz. 00 5 V Problem 6. Fnd the bas pont and amplfer parameters of ths crcut (V tn = 4 V, V tp = 4 V, µ p C ox (W/L) = µ n C ox (W/L) = 0.4 ma/v 2, and λ = 0.0 V. Ignore the channel-wdth modulaton effect n basng calculatons.) Ths crcut s also smlar to that of problem 4 expect that t s based wth a current source: V SS 0.7 ma I D = 0.7 ma I D = 0.5µ p C ox (W/L)V 2 OV 0.7 0 3 = 0.5 0.4 0 3 V 2 OV V OV =.88 V 5 V V SG = V OV V tp = 5.88 V V SS V SG = V S V G = 5.88 V S = 5.88 V 0.7 ma V SD = V S V D = 5.88(5) = 0.9 V (> V OV =.89) Bas summary: I D = 0.7 ma, V OV =.88 V, V SG = 5.88 V, V SD = 0.9 V. 5 V ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-39

V SS g m = 2I D = 2 0.7 03 V OV.89 r o = λi D = = 0.75 ma/v = 4 k 0.0 0.7 03 Ths s a common-dran amplfer (source follower). Note G =, s =. 5 V = g m(r o S L ) g m (r o S L ) = 43.9 44.9 = 0.98 = G = A = = sg = 0.98 o = g m S =.33 k =.33 k f p = /[2πC c ( sg )] = 0 f p2 = /[2π ( L o )] = 3.39 Hz f l = f p f p2 = 3.39 Hz Problem 7. Fnd the bas pont and amplfer parameters of ths crcut (V tn = 4 V, V tp = 4 V, µ p C ox (W/L) = µ n C ox (W/L) = 0.4 ma/v 2, and λ = 0.0 V. Ignore the channel-wdth modulaton effect n basng calculatons. 5 V Answer: I D = 0.7 ma, V OV =.88 V, V SG = 5.88 V, V SD = 0.9 V. / = 0.866, =, o =.7 k, f p = 0, f p2 = 3.35 Hz, f l = 4.3 Hz. 0.7 ma Amp esponse: A = 0.98, =, o =.33 k, and f l = 3.39 Hz. V SS ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-40

Problem 8. Fnd the bas pont and amplfer parameters of ths crcut (V tn = V, V tp = V, µ p C ox (W/L) = µ n C ox (W/L) = 0.8 ma/v 2, and λ = 0.0 V. Ignore the channel-wdth modulaton effect n basng calculatons. Bas: Snce I G = 0: 5 V.8M 0k.2 M V G =.2 M.8 M 5 = 6 V G =.2 M.8 M = 720 k 00 00nF.2M 0k 00 nf µ F Assume NMOS s n the acte state, 5 V I D = 0.5µ p C ox (W/L)V 2 OV.8M 0k GS-KVL: V G = V GS 0 4 I D = V OV V t 0 4 I D 0 4 0.5 0.8 0 3 V 2 OV V OV 6 = 0 V G 4V 2 OV V OV 5 = 0.2M 0k Negate root s unphyscal, V OV =.0 V and V GS = V OV V t = 2.0 V. I D = 0.5 0.8 0 3 V 2 OV = 0.40 ma DS-KVL: 5 = 0 4 I D V DS 0 4 I D V DS = 7 V (> V OV =.0) Bas summary: I D = 0.40 ma, V OV =.0 V, V GS = 2.0 V, V DS = 7.0 V. Small-Sgnal: g m = 2I D V OV = 2 0.4 0 3 = 0.8 ma/v r o = λi D = = 250 k 0.4 03 As the nput s at the gate and output s at the collector, ths s a common-source amplfer. There s no S because of the by-pass capactor. = g m (r o D L ) = 0.8 0 3 (250 k 0 k 00 k) = 7.02 = G = 720 k A = = sg = 7.02 o = D r o = 0 k 00 k = 9.09 k ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-4

f p = /[2πC c ( sg )] = 2.2 Hz f p2 = /[2π ( L o )] = 4.6 Hz f p3 = 2πC s [ S (/g m )] = 2πC s [0 k.25 k] = = 43 Hz 2π 0 6. 03 f l = f p f p2 f pb = 2.24.643 = 60 Hz Problem 2. Fnd the bas pont and amplfer parameters of ths crcut (V tn = V, V tp = V, µ p C ox (W/L) = µ n C ox (W/L) = 0.8 ma/v 2, and λ = 0.0 V. Ignore the channel-wdth modulaton effect n basng calculatons. Bas: Snce I G = 0: 5 V.2M 0k.8 M V G =.2 M.8 M 5 = 9 V G =.2 M.8 M = 720 k 00 00nF.8M 0k 00 nf Assume PMOS s n the acte state, 5 V I D = 0.5µ p C ox (W/L)V 2 OV.2M 0k SG-KVL: 5 = 0 4 I D V SG V G = 0 4 I D V OV V tp V G 0 4 0.5 0.8 0 3 V 2 OV V OV 59 = 0 V G 4V 2 OV V OV 5 = 0.8M 0k Negate root s unphyscal, V OV =.0 V and V SG = V OV V tp = 2.0 V. I D = 0.5 0.8 0 3 V 2 OV = 0.40 ma SD-KVL: 5 = 0 4 I D V SD 0 4 I D V SD = 7 V (> V OV =.0) Bas summary: I D = 0.40 ma, V OV =.0 V, V SG = 2.0 V, V SD = 7.0 V. Small-Sgnal: g m = 2I D V OV = 2 0.4 0 3 = 0.8 ma/v r o = λi D = = 250 k 0.4 03 ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-42

As the nput s at the gate and output s at the collector, ths s a common-source amplfer wth S. = g m ( D L ) g m S ( D L )/r o = = 7.27 9.04 = 0.805 = G = 720 k A = = sg = 0.805 o = D [r o (g m S )] = 0 k ( D ) f p = /[2πC c ( sg )] = 2.2 Hz f p2 = /[2π ( L o )] = 4.5 Hz f l = f p f p2 = 6.7 Hz 0.8 0 3 (0 k 00 k) 0.8 0 3 0 4 (0 k 00 k)/(250 k) Note one needs to choose D to be seeral tmes S for ths amplfer to hae a gan larger than unty. Problem 24. Fnd the bas pont and amplfer parameters of ths crcut (V tn = V, V tp = V, µ p C ox (W/L) = µ n C ox (W/L) = 0.8 ma/v 2, and λ = 0.0 V. Ignore the channel-wdth modulaton effect n basng calculatons. 5 V Answer: 0k.8M I D = 0.40 ma, V OV =.0 V, V GS = 2.0 V, V SD = 7.0 V. 00nF / = 0.866, =, o =.7 k, f p = 0, f p2 = 3.35 Hz, f l = 4.3 Hz. 00nF 0k.2M Amp esponse (common-gate amp): A = 7.7, =. k, o = 0 k, and f l =.45 0 3 22 =.47 khz, ECE65 Lecture Notes (F. Najmabad), Wnter 202 6-43