V. Transistor Amplifiers

Transcription

1 V. Transstor Amplfers 5.1 Introducton In page 1-11, we showed that the response of a twoport networks n a system s completely determned f we sole the smple crcut shown. Furthermore, one can show that f a two-port network contans only lnear elements, the two-port network can be modeled wth a maxmum of four crcut elements. sg R sg 2port Network o For practcal crcut n whch the two-port network does not contan any ndependent sources, the two-port network can be modeled wth 3 elements: the nput resstance, the output resstance, and the oltage transfer functon as s shown below: As the combnaton of the two-port network and the load (dashed box n the crcut) s a two-termnal network, t can be modeled by ts Theenn equalent. Furthermore, as ths box does not contan an ndependent source, V T 0. As such, the box can be modeled as a resstor, called the nput resstance of the two-port network: Input Resstance: R Note that n general R depends on. sg R sg 2port Network o o 2port R L Network To fnd a model for the output port of a two-port network, we assume a oltage s drectly appled to the two-port network (.e., sg and R sg 0). In ths manner, the output port model wll be ndependent of R sg. Agan, as the box contanng and the two-port network s a two-termnal network, t can be modeled wth ts Theenn equalent (see Fgure). We call the Theenn resstance of ths box, the output resstance of the two-port network: Output Resstance: R o o 0 (Theenn Resstance) The Theenn oltage source, V T c the open-loop oltage alue. For an amplfer, the output oltage should be proportonal to. Therefore, f we defne Open-loop Gan: A o c RL ECE65 Lecture Notes (F. Najmabad), Sprng

2 Then, V T c A o,.e., the Theenn oltage can be modeled by a controlled oltage source. Combnng the models for the nput and output ports, we arre at the model for an amplfer whch conssts of three crcut elements as s shown below (left). R Ro A o o sg R sg R Ro A o o RL Voltage Amplfer Model The amplfer crcut model allows us to sole any amplfer confguraton once to compute the three parameters: A o, R and R o. We can then fnd the response of any crcut contanng ths amplfer by utlzng these three parameters (smlar to usng Theenn Theorem to label any two-termnal network wth R T and V T ). For example, for the generc two-port network crcut (crcut rght aboe), we can fnd the response of the amplfer to the presence of R sg and Load: A R o A o sg R R R sg sg sg R R R sg A R R R sg A o R o We see that the open-loop gan A o s the maxmum alue for the amplfer gan A. In addton, to maxmze / sg, we need R and R o 0. A practcal oltage amplfer, thus, s desgned to hae a large R and a small R o (.e., R R sg and R o ). A oltage-controlled oltage source s an deal oltage amplfer as R and R o 0. Smlarly, for a two-stage amplfer: sg R sg Ro1 Ro2 1 R 1 2 L 1 R R 2 A o1 1 A o2 2 o sg R 1 R 1 R sg A o1 R 2 R 2 R o1 A o2 R o2 Note that the nput resstance of the second amplfer, R 2, s the load for the frst amplfer and the output resstance of the frst amplfer, R o1 s R sg for the second one. ECE65 Lecture Notes (F. Najmabad), Sprng

3 We see that n order to maxmze the oltage gan, we need to ensure that the nput resstance of the frst stage s much larger than R sg, the output resstance of the last stage s much smaller than, and the nput resstance of any stage to be much larger than the output resstance of the preous stage: R,N R o,n1 as R of the N th stage appears as the load for N-1 th stage. For sngle-transstor amplfer confguratons (rest of ths secton), we wll see that s smpler to compute A (wth load present) drectly nstead of A o and R o separately. In ths case, the aboe formula for computng total gan of a two-stage amplfer can be smplfed to sg R 1 R 1 R sg A 1 ( R 2 ) A 2 ( ) where A 1 ( R 2 ) s the oltage gan wth R 2 beng the load, etc. An mportant cauton: In general, A o and R o are ndependent of both R sg and (why?). Howeer, R may depend on. Amplfer confguratons n whch R s ndependent of are called unlateral. It s easy, howeer, to ncorporate the dependence of R on by solng any mult-stage amplfer from the load sde toward the sgnal sde: In the aboe fgure, we know the fnal load whch can be used to compute R 2 and R 2 s the load for stage one and ges R 1. Analyss of Transstor Amplfer Crcuts Analyss of a transstor amplfer crcut follows these 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. Analyze the crcut usng a large-sgnal model such as those of page 3-4 or 3-6 for BJT and 3-22/3-23 for MOS. Small Sgnal Response: 1) Compute, r o (and r π for BJT) from bas pont parameters 2) Zero out all bas sources 3) Assume capactors are short crcut. 4) Replace the transstor wth ts small sgnal model. 5) Inspect the crcut. If you dentfy the crcut as a prototype crcut, you can drectly use the formulas for that crcut. Otherwse sole for A, A o, R and R o. Frequency-response: Value of A found n the small sgnal response aboe s called the mdfrequency gan of the amplfer. Couplng and bypass capactors as well as the nternal capactance of transstors ntroduce poles both at low and hgh frequences. We wll ntroduce a method to compute the low-frequency poles. ECE102 nclude a more thorough reew of the amplfer frequency response. ECE65 Lecture Notes (F. Najmabad), Sprng

4 There are four fundamental sngle transstor amplfer confguratons possble and are examned n the followng sectons. Notes: 1) The 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 ports n a MOS s open crcut crcut). As such, we expect that formulas for MOS amplfers would be the same as those of BJT amplfer f we set r π. 3) For MOS crcuts, we use the common approxmaton r o 1 as 2I D V GS V tn, r o V A I D r o 2V A V GS V tn 1 typcally r o s 50 or more. 4) For BJT crcuts, we use the common approxmaton r o 1 as I C nv T, r o V A V CE I C V A I C r o V A V CE nv t 1 typcally r o s seeral thousands. In addton, r π β 1 5) In many text books (e.g., Sedra & Smth), the formulas for BJT amplfers are gen n terms of β & r e (nstead of & r π ) where r e 1 r π β wth r e typcally n 10s or 100 Ω range. Here we keep both form (so we can see the comparson to MOS amplfers) and also dere the formula n r e form. 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 1/r o ECE65 Lecture Notes (F. Najmabad), Sprng

5 5.2 Common-Dran and Common-Collector Amplfers Common-Dran or Source Follower Confguraton Crcut shown s the generc small-sgnal crcut of a common-dran 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 source. As the dran s grounded (for small 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 the bas elements) wll reduce to the aboe small-sgnal form of a common-dran amplfer. Some examples are gen below. R G R S V DD R 1 V SS V DD R S o R 2 R S V DD V SS R G R 1 R 2 R G, R G, R S, no Pole from C1 no Pole from C1 We now proceed wth the small-sgnal analyss by replacng the MOS wth ts small-sgnal model. An mportant obseraton s that the resstor R S s parallel to and appears as the load for the transstor. In fact, n many applcatons, R S s replaced by the load (e.g. a speaker, nput of another amplfer crcut). As such, we defne R L R S. The openloop gan of the amplfer s calculated wth R L (both R S and are open crcut) and the output resstance s taken to the left of R S (see page 5-6) R G G gs _ S gs D r o R S R G gs G _ S g m gs r o R S D We compute, A (n the presence of a load) drectly as ths does not complcate the analyss. The open-loop gan s then calculated by settng. Inspectng the crcut, we fnd ECE65 Lecture Notes (F. Najmabad), Sprng

6 that the current gs wll flow n r o R L (from to the ground). Thus, gs Ohm Law: gs (r o R L) (r o R L)( ) A (r o R L ) 1 (r o R L) r o R L r o R L r o R L R L 1 R L where we hae used r o 1 to drop R L compared to r o R L n the denomnator. The open-loop gan can be fnd by settng R L to get r o R L r o and A o r o 1 r o 1 Because A o 1, S G and S follows the nput oltage. Thus, ths confguraton s also called the Source Follower. Fndng R s easy as R G (see crcut) and, therefore, R R G. As R s ndependent of, ths confguraton s unlateral. Note that f R G were not present (see example complete crcut of page 5-5), R. To fnd R 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, we need to attach a x oltage source to the crcut and compute x : R G gs _ G S g m gs r o R L R G gs _ G S g m gs r o x x D R O D KCL: gs x x x gs x r o x r o 1 R o r o 1 1 r o r o R o s typcally small, a few 100 Ω. Note that: A A o 1 R o 1/ R L 1 R L whch s exactly the expresson we had dered before. ECE65 Lecture Notes (F. Najmabad), Sprng

7 In summary, the general propertes of the common-dran amplfer (source follower) nclude an open-loop oltage gan of unty, a large nput resstance (and can be made nfnte n some basng schemes) and a small output resstance. Ths type of crcut s typcally 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 o 1 but R R 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) Note re R o : In some text books (e.g., Sedra & Smth), the output resstance s defned to nclude R D (see crcut). If we call ths resstance to be R o, nspecton of the crcut shows that R o R D R o. R G gs _ G S D gs r o R O R D R O Common-Collector or Emtter Follower Confguraton Crcut shown s the generc small-sgnal crcut of a common-collector 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 emtter. As the collector s grounded (for small 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 small-sgnal form of a common-collector amplfer. Some examples are gen below. R B R E V EE R R 1 E C o 1 V CC V EE R 2 R E V EE V CC R G R 1 R 2 R G, R G, R S, no Pole from C1 no Pole from C1 Smlar to the common-dran confguraton, R E s parallel to and appears as the load for the transstor. We defne R L R E. and the output resstance s taken to the left of R E n the crcut aboe. Proceedng wth the small-sgnal analyss by replacng the BJT wth ts small-sgnal model: ECE65 Lecture Notes (F. Najmabad), Sprng

8 R B B π _ r π E π R E C r o R B B π _ E b r π π r o R E C Inspectng the crcut, we fnd that a total current of b π flows flow n r o R L (from to the ground) where b π /r π : π ( Ohm Law: A π π r π ) (r o R L ) π (r o R L ) (r o R L )( ) r o R L r o R L 1 r o R L r o R L r o R L A R L 1 R L R L R L r e A o r o 1 r o 1 where we hae used r π β 1 n the 2nd equaton to drop π /r π term, and r o 1 to drop R L n the denomnator of the 3rd equaton. Snce r e 1/ n typcally a few tens of Ohms, A 1 unless R L s ery small (tens of Ω). Ths confguraton s also called the Emtter Follower. as e b and e follows the nput oltage. To fnd R / (note r π β): KCL: R B b KVL: b r π ( b π )(r o ) b[r π (1 r π )(r o )] b R B R B r π (1 β)(r o ) R R B r π (1 β)(r o ) R R B [r π (1 β)(r o R L )] Snce R depends on, ths amplfer confguraton s NOT unlateral. Note that when emtter degeneraton basng s used, we need to hae R B (1 β)r E. In ths case, R R B (smlar to the common-dran amplfer n whch R R G ) and the ECE65 Lecture Notes (F. Najmabad), Sprng

9 confguraton becomes unlateral. If R B s not present, the nput resstance s large although t s not nfnte as s the case for the common-dran amplfer. To fnd R o we need to zero out and compute the Theenn Equalent resstance seen at the output termnals: R B B π _ E b r π π r o R L R B B π _ E b r π π r o x x C R O C Because of the presence of the controlled source, we need to attach x oltage source to the crcut and compute x. Notng that π x KCL: x π x x x x x r o r π 1/ r o r π 1 x R o (1/ ) r o r π (1/ ) r e R o x 1/ r o r π snce r π 1 and r o 1. If the output resstance s taken nclude R C, R o R C R o (smlar to the source follower, see fgure n page 5-7). In summary, the general propertes of the common-collector amplfer (emtter follower) nclude an open-loop oltage gan of 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 to amplfy the sgnal current (and power) and dre a load. ECE65 Lecture Notes (F. Najmabad), Sprng

10 5.3 Common-Source and Common-Emtter Amplfers Common-Source Confguraton Crcut shown s the generc small-sgnal crcut of a common-dran 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 small sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the common-source amplfer. R G R D If source degeneraton basng s used for the common-source confguraton, a resstor R S should be present. A by-pass capactor s typcally used so that small sgnals by-pass R S, effectely makng the source grounded for small sgnal as s shown. R D R G C R b S We replace the MOS wth ts small-sgnal model. In ths confguraton, R D s parallel to and appears as the load for the transstor. As such, we defne R L R D and the output resstance s taken to the left of R D (see below). Inspecton of the crcut shows that gs. Also, a current of gs flows n r o R L (from the ground to ). Ohm Law: gs (r o R L ) A (r o R L ) A o r o R G G gs _ gs S D r o R D The negate sgn n the gan s ndcate of a 180 phase shft n the output sgnal. Inspectng the crcut, we fnd R / R G (unlateral amplfer). To fnd R o, we set 0. As gs 0, the controlled current source becomes an open crcut and R o r o. If the output resstance s taken to nclude R D (see dscusson n page 5-7), R o R D R o. R G G gs _ gs S D r o R O R L In summary, the general propertes of the common-source amplfer nclude a large open-loop oltage, a large nput resstance (and can be made nfnte wth some basng schemes) but a medum output resstance. ECE65 Lecture Notes (F. Najmabad), Sprng

11 Common-Emtter Confguraton Crcut shown s the generc small-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 small sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the common-emtter amplfer. R B R C Smlar to the common-dran confguraton, f emtter degeneraton basng s used, a resstor R E should be present wth a by-pass capactor. Ths capactor effectely make the emtter grounded for small sgnal as s shown. R C R B C R b E We now replace the BJT wth ts small sgnal model. In ths confguraton, R C s parallel to and appears as the load for the transstor (R L R C ) and the output resstance s taken to the left of R C n the crcut aboe. C 1 C B C 2 g m π R B r π r R π o C _ E Inspecton of the crcut shows that π. Also, a current of π flows n r o R L (from the ground to ). Ohm Law: π (r o R L) A (r o R L ) r o R L r e A o r o r o r e The negate sgn n the gan s ndcate of a 180 phase shft n the output sgnal. From the crcut, we fnd R / R B r π (a unlateral amplfer) To fnd R o, we set 0. As π 0, the controlled current source becomes an open crcut and R o r o. Smlarly, R o R C R o R C r o. C 1 B C g π m R B r π π r o _ E R L In summary, the general propertes of the common-emtter amplfer nclude a large openloop oltage, a medum nput resstance and a medum to large output resstance. R O ECE65 Lecture Notes (F. Najmabad), Sprng

12 5.4 Common-Source and Common-Emtter Amplfers wth Degeneraton Common-Source Confguraton wth a Source Resstor Crcut shown s the generc small-sgnal crcut of a common-source amplfer wth degeneraton. Note that the nput s appled at the gate and the output s taken at the dran smlar to a common-dran amplfer but a source resstor s now present. R D R G R S We replace the MOS wth ts small-sgnal model. Smlar to the common-dran amplfer, R D s parallel to and appears as the load for the transstor (R L R D and the output resstance s taken to the left of R D n the crcut aboe). Usng nodeoltage method: C 1 C G D 2 gs R G gs r o R D _ S R S Node s : Node : s s gs 0 R S r o R L s r o gs 0 s R S R 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: A r o ( ) r o R S (1 r o ) r o ( ) r o R S r o A R L 1 R S R L/r o A o r o If r o s large compared to R L and/or f R S and R L of the same order (.e., R L /R S r o ), we can drop the last term to fnd: A 1 R S The amplfer gan s substantally reduced wth the presence of R S but t has become much less senste to change n. Inspectng the crcut we fnd R / R G, smlar to a common-source amplfer. ECE65 Lecture Notes (F. Najmabad), Sprng

13 To fnd R 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. R G G gs _ gs S D r o R L R G G gs _ gs S D r o x x R S R O R S By KCL, a current of x gs should flow n r o and x should flow n R S. Snce gs R S x : KVL : x r o ( x gs ) R S x r o ( x R S x ) R s x x (r o r o R S R S ) R o x x r o R S r o R S r o (1 R S ) In summary, source degeneraton has led to an amplfer wth a lower gan whch s less senste to transstor parameters and seeral other benefts (e.g., larger band-wdth) whch are beyond the scope of ths course. Common-Emtter Confguraton wth an Emtter Resstor Crcut shown s the generc small-sgnal crcut of a common-emtter amplfer wth degeneraton (.e., wth 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. R C R B R E We now replace the BJT wth ts small-sgnal model. Smlar to the common-dran amplfer, R C s parallel to and appears as the load for the transstor (R L R C ) and the output resstance s taken to the left of R C n the crcut aboe. Usng node-oltage method and notng π e : C 1 C B 2 b C g m π RB r π π r o R C _ E R E Node e : Node : 0 e e e π R E r π r o 0 e g m π 0 r o e e 0 R E r π ECE65 Lecture Notes (F. Najmabad), Sprng

14 The thrd equaton s the sum of the frst two. Fndng e from the thrd equaton and substtutng n the 2nd equaton, we get: A R L R E (1 R L /r o)(1 R E /r π ) R L A R E r e ( /r o)(r E r π )/β A o r o r o 1 R E /r π r e R E /β R L R E (1 R L /r o)(r E r π )/β If (R L/r o )/β 1 (a ery good approxmaton), we can drop the last term to fnd: A R L 1 R E R L R E r e whch s the expresson often used. Note that the amplfer gan s reduced wth the presence of R E but t has become substantally less senste to any change n β (only through r e ). From the crcut we fnd R / R B ( / b ). The exact formulaton for / b s cumbersome. A good approxmaton whch leads to a smple expresson s r o R E. In ths case, r o can be remoed from the crcut and b r π ( b π )R E b r π ( b β b )R E b [r π (1 β)r E ] R R B [r π (1 β)r E ] To fnd R 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. By KCL, current x π wll flow through r o and current x wll flow through R E r π : C 1 B b C π R B r π π _ E R E r o R O R L r π _ π C π E R E r o x x π x (R E r π ) KVL: x ( x π )r o x (R E r π ) x [r o r o (R E r π )] ( R o r o [1 (R E r π )] r o 1 βr ) E r π R E ECE65 Lecture Notes (F. Najmabad), Sprng

15 In summary, emtter degeneraton has led to an amplfer wth a lower gan whch s much less senste to transstor parameters, a substantally larger nput resstance and a somewhat larger output resstance. 5.5 Common-Gate and Common-Base Amplfers Common-Gate Confguraton Crcut shown s the generc small-sgnal crcut of a commongate amplfer. Note that the nput s appled at the source and the output s taken at the dran. As the gate s grounded (for small sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the common gate amplfer. If the base has to based to a DC alue (for example usng oltage dder crcut shown), a by-pass capactor s added to short the base for small sgnals as s shown wth R G R 2 R 1. R D R S We replacng MOS wth ts small sgnal model. In ths confguraton, R D s parallel to and appears as the load for the transstor. As such, we defne R L R D and the output resstance s taken to the left of R D n the crcut aboe. Notng that gs and wrtng the node equaton at : C b R 1 R2 R D R S R L r o ( ) 0 ( 1 1 ) 1 r o r o r o A (R L r o) r o r o S R S G g m gs _ gs D ro R D A o r o D To fnd R, t s easer to wrte R R S 1 (see crcut). By KCL at node S, current 1 gs wll flow n r o and current 1 wll flow n R L R D. Thus: S 1 R S g m gs _ gs G ro R D ( 1 gs )r o 1 R L 1 r o r o 1 R L ECE65 Lecture Notes (F. Najmabad), Sprng

16 1 r o R L 1 r o 1 R L /r o R R S 1 R L /r o As can be seen, the common-gate amplfer s NOT unlateral.e., R depends on the load. To fnd R o, we set 0. As gs 0, the controlled current source becomes an open crcut and R o r o (Note that R S s shorted out). S RS G g m gs In summary, the general propertes of the common-gate amplfer nclude 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-source confguraton but a much lower nput resstance). _ gs D ro R O R L Common-Base Confguraton Crcut shown s the generc small-sgnal crcut of a commonbase amplfer. Note that the nput s appled at the source and the output s taken at the dran. As the gate s grounded (for small sgnal), t s the common termnal of nput and output. Thus, ths crcut s called the common-gate amplfer. If the base has to based to a DC alue (for example usng oltage dder crcut shown), a by-pass capactor s added to short the base for small sgnals as s shown wth R B R 2 R 1. R C R E We replace the BJT wth ts small sgnal model. In ths confguraton, R C s parallel to and appears as the load for the transstor. As such, we defne R L R C and the output resstance s taken to the left of R C n the crcut aboe. Notng that π and wrtng the node equaton at : C b R 1 R2 R C R E R L r o π 0 ( 1 1 ) 1 r o r o r o A (R L r o ) A o r o r o r o _ r π π B E π R E C r o R C ECE65 Lecture Notes (F. Najmabad), Sprng

17 To fnd R, t s easer to wrte R (R E r π ) 1 (see crcut). By KCL at node E, current 1 π wll flow n r o and current 1 wll flow n R L R C. Thus (settng π ) by KVL: ( 1 π )r o 1 R L 1 r o r o 1 R L 1 r o R L 1 r o 1 R L /r o R R E r π 1 R L /r o As can be seen, the common-base amplfer s NOT unlateral.e., R depends on the load. To fnd R o, we set 0. As π 0, the controlled current source becomes an open crcut and R o r o (Note that R E r π s shorted out). 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). _ π B E 1 R E E _ π B π r π π R E C r o r π C r o R C R O R L 5.6 Summary of Amplfer Confguratons The common-source (CS) and common-emtter (CE) amplfers hae a hgh gan and are the man confguraton n a practcal amplfer. Ignorng bas resstors R G or R 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 102). 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), Sprng

18 5.7 Low Frequency Response 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 c1 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 R o A o R R R sg 1/(sC c1 ) R R R sg s s ω p1, R R R sg A s s ω p1 2πf p1 ω p1 R sg V sg Cc1 V R 1 C c1 (R R sg ) Ro A V o V o RL 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 c1 has ntroduced a low-frequency pole and the amplfer gan falls at low frequences. One may be tempted to compute the mpact of the couplng capactor at the output n a smlar manner. Fgure below s for a common-dran amplfer (wth R D appearng as the load, see fgure n page 5-5). It s straghtforward to show (left as an exercse): V sg R sg V R Ro A V o Cc2 RD V o RL V o V sg R R R sg A s s ω p2 2πf p2 ω p2 1 C c2 ( R D R o ) For unlateral amplfers, f p2 calculated aboe s the pole ntroduced by C c2. Howeer, f the amplfer s NOT unlateral, R depends on the load and the expresson of R wll nclude C c2 as ths capactor s part of the load ( R L R D ( 1/sC c2 )). As such, we need to compute the R /(R R sg ) term to fnd the pole ntroduced by C c2. Ths ssue s specally mportant for amplfers wth small R,.e., common-gate and common-base amplfers. We can stll use the aboe formula, howeer, f we replace R o (output resstance wth 0) wth R out (output resstance wth sg 0). ECE65 Lecture Notes (F. Najmabad), Sprng

19 Smlarly, f a by-pass capactor s present (see page 5-8), t wll ntroduce yet another pole, f p3. The follownethod allows one to compute the poles by nspecton. 1. Zero out V sg. 2. Consder each capactor separately (.e., assume all other capactors are short) 3. Compute, R, the total resstance between the termnals of the capactor. The pole ntroduced by that capactor s gen by f p 1 2πC R 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 s ω p1 s s ω p2 s s ω 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 p1 f p2 f p3 Exercse: Show that the method aboe ges the poles correspondng to C c1 and C C2 (for unlateral amplfer) as was calculated preously. The next two pages nclude a summary of formulas for elementary transstor confguratons. These formulas are correct wthn approxmaton of r o 1 and β 1 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 a factor of ten) than the others. For example, n a common emtter amplfer, we often fnd that R C and R C r o. Then, r o R C R C and the gan formula can be smplfed to A R C /r e. ECE65 Lecture Notes (F. Najmabad), Sprng

20 Summary of Elementary MOS Confguratons Common Dran (Source Follower): A (R S ) 1 (R S ) R R G R o 1 r o 1 R o R S R o f p2 1/[2πC c2 ( R S R out )] R out 1 R G R S Common Source: A (r o R D ) R R G R o r o R o R D R o R D f p2 1/[2πC c2 ( R D R out )] f pb 1 2πC b [R S (1/ )] R out r o R G R S C b Common Source wth Source Resstance: (R D ) A 1 R S (R D )/r o R R G R D R o r o (1 R S ) R o R D R o R G f p2 1/[2πC c2 ( R D R out )] R out r o (1 R S ) R S Common Gate: A (r o R D ) R R S [1/ (R D )/ r o ] R 1 C b R D R o r o R o R D R o f p2 1/[2πC c2 ( R D R out )] R out r o (1 R S ) R S f pb 1/[2πC b R G ] f l Σ j f pj and f p1 1/[2πC c1 (R R sg )]. ECE65 Lecture Notes (F. Najmabad), Sprng

21 Summary of Elementary BJT Confguratons Common Collector (Emtter Follower): A (R E ) 1 (R E ) R E R E r e R R B [r π (1 β)(r o R E )] R o r π 1 1 r e f p2 1/[2πC c2 ( R E R out )] Common Emtter: R o R E R o A (r o R C ) r o R C r e R R B r π R o r o R o R C R o f p2 1/[2πC c2 ( R C R out )] f pb 1 2πC b [R E (r e (R B R sg )/β)] R B ( ) rπ R B R sg R out r o r e 1 β R out r o R E R B R C R E C b Common Emtter wth Emtter Resstor: A (R C ) R C 1 g m R E R E r e R R R B [r π (1 β)r E ] B ( R o r o [1 (R E r π )] r o 1 βr ) E R o r π R R C R o E ) βr E f p2 1/[2πC c2 ( R C R out )] R out r o (1 r π R E R B R sg Common Base: R C R E A (r o R C ) r o R C r e R R E r π [1/ (R C )/( r o )] R B C b R C R o r o R o R C R o ) βr E f p2 1/[2πC c2 ( R C R out )] R out r o (1 r π R E R B R sg RE f pb 1/[2πC b R CB ] R CB R B [r π (1 β)(r sg R E )] ECE65 Lecture Notes (F. Najmabad), Sprng

22 5.8 Exercse Problems Problem 1 to 3: Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Problem 4: If V cos(ωt) n the crcut of Problem 3, what s the maxmum alue of V for ths crcut to work properly? Problem 5 to 14: Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. 9 V 9 V 9 V 18k 0.47 µ F 18k 0.47 µ F 0.47 F 4 V µ 0.47 µ F o 22k 0.47 µ F 0.47 µ F 22k 22k 18k 5 V Problem 1 Problem 2 Problem 3 Problem 5 15 V 15 V 4 V 4 V 34k 100 nf 5.9k µ F 0.47 µ F o o 4.7 µ F 4.7 µ F 100 nf 4.3mA 4.3mA 5.9k µ F 34k V EE V EE Problem 6 Problem 7 Problem 8 Problem 9 15 V 15 V 34k 100 nf 3V V EE 34k 100 nf 4.7 µ F 2.3k 47 µ F 1mA 47 µ F 4.7 µ F 5.9k k µ F 2.3k 3V 100 nf 2.3k 3V 100 nf Problem 10 Problem 11 Problem 12 Problem 13 ECE65 Lecture Notes (F. Najmabad), Sprng

23 Problem Fnd the bas pont and amplfer parameters of ths crcut (V tn 4 V, V tp 4 V, k (W/L) 0.4 ma/v 2, Ignore the channel-wdth modulaton effect n basng calculatons. Fnd the saturaton oltages for ths amplfer. Problem Fnd the bas pont and amplfer parameters of ths crcut (V tn 1 V, V tp 1 V, k p (W/L) k n (W/L) 0.8 ma/v2, λ 0.01 V 1 ). Ignore the channel-wdth modulaton effect n basng calculatons. Fnd the saturaton oltages for ths amplfer. 2.5 V 18 V 10nF 1 µ F 13k 1.3M 0.47 µ F 0.47 µ F 13V 0.47 µ F o V SS 0.71 ma 0.47 µ F k 500k 5V 5V Problem 14 Problem 15 Problem 16 Problem V 15 V 15V 5 V 0.71 ma 0.47 µ F 1.8M 1.2M 1 µ F 1.8M 1.2M 1 µ F 1.2M 1.8M 1 µ F V SS Problem 18 Problem 19 Problem 20 Problem 21 9V 6V 1 µ F 15V 1.2M 1.8M o Problem 22 Problem 23 Problem 24 Problem M 1.2M 15 V 1 µ F 6V 9V ECE65 Lecture Notes (F. Najmabad), Sprng

24 Problems 27 to 29: Fnd the bas pont of each BJT, the oerall gan ( / ), and the lower cut-off frequency of ths amplfer parameters (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). 2.5 V 15 V 15 V 1M 10nF 33k 4.7 µ F 2k Q1 18k 0.47 µ F Q2 33k 4.7 µ F 2k Q1 Q2 o 1M 6.2k k 6.2k 500 Problem 26 Problem 27 Problem V 15k 3.6k 1.5k 4.7 µ F 2.7k Q1 510 Q2 510 Problem 29 ECE65 Lecture Notes (F. Najmabad), Sprng

25 5.9 Soluton to Selected Exercse Problems Problem 1. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Bas: 9 V Set 0 and capactors open. Replace R 1 and R 2 wth ther Theenn equalent: 18k R B 18 k 22 k 9.9 kω V BB V 0.47 µ F 22k KVL: V BB R B I B V BE 10 3 I E I B I E 1 β I E 201 ( ) I E V BB R B 9 V I E 4 ma I C, I B I C β 20 µa KVL: 9 V CE 10 3 I E V CE V Bas summary: I E I C 4 ma, I B 20 µa, V CE 5 V Small-Sgnal: Frst we calculate the small-sgnal parameters: I C nv T r π β 2.60 k r e 1 13 Ω r o V A V CE k I C µ F o 9.9k 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 5-21 and notng, R E r o, and R E r e : A R E R E r e R E R E r e 1 R R B [r π (1 β)r E ] (9.9 k) (203.6 k) 9.9 k R B R o r e 13 Ω 1 f l f p1 2πC c1 (R R sg ) 1 34 Hz 2π ECE65 Lecture Notes (F. Najmabad), Sprng

26 Problem 2. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Ths s the same crcut as Problem 1 wth excepton of C c2 and. The bas pont s exactly the same. As R E, the amplfer parameters are the same except f l f p1 f p2 37 Hz. Problem 3. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Ths crcut s smlar to Problem 1 expect that the transstor s based wth two oltage sources (alues are chosen to ge the same bas pont). Bas: Set 0 and capactors open: 4 V 5 V 0.47 µ F o BE-KVL: 0 V BE 10 3 I E 5 4 V I E 4.3mA I C, I B I C β 21.5 µa CE-KVL: 4 V CE 10 3 I E 5 V CE V Bas summary: I E I C 4.3 ma, I B 21.5 µa, V CE 4.7 V 5 V Small-Sgnal: Frst we calculate the small-sgnal parameters: I C nv T r π β 2.42 k r e Ω r o V A V CE k I C µ F 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 5-21 and notng R E, R E r o, and R E r e : A R E R E r e R E R E r e 1 R R B [r π (1 β)r E ] r π (1 β)r E k R o r e 12 Ω 1 f l f p2 2πC c2 ( R E r e ) 1 2πC c2 ( r e ) Hz 2πC c2 ECE65 Lecture Notes (F. Najmabad), Sprng

27 Problem 4: If V cos(ωt) n the crcut of Problem 3, what s the maxmum alue of V for ths crcut to work properly? For the amplfer to work properly, the BJT has to reman n the acte state,.e, E I E e > 0 and CE V CE ce > V D0 0.7 V. The frst equaton ges a mnmum alue for e (or C > 0 ges a mnmum alue for c ). Combnaton of the 2nd equaton and CE-KVL ges a maxmum alue for e (or c ). Lmts on can be found from the two lmts for e (e.g., n ths problem 10 3 e as R E ). /A then ges the range for : E I E e > 0 e > I E 4.3 ma 10 3 e > 4.3 V CE-KVL 4 CE 10 3 E E < 9 CE CE > V D E 10 3 (I E e ) < 8.3 V 10 3 e < 4.0 V 4.3 < A < 4.0 V 4.3 < < 4.0 V V < 4.0 V Problem 5. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly). Ths s the PNP analog of crcut of Problem 2. Bas summary: I E I C 4 ma, I B 20 µa, V EC 5 V Small-Sgnal: /Ω, r e 13 Ω, r π 2.60k, and r o 38.8 k. Amp response: A 1, R 9.9 k, R o 13 Ω, and f l f p1 f p2 37 Hz. Max sgnal: 4.3 < A < < < 4.0 V < 4.0 V Problem 6. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. Ths crcut s smlar to the crcut of Problem 3 except that the transstor s based wth a current source. Bas: Set 0 and capactors open. 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 21.5 µa 4 V 4.3mA V EE 4 V 0.47 µ F o Bas summary: I E I C 4.3 ma, I B 21.5 µa, V CE 4.7 V V E 4.3mA V EE ECE65 Lecture Notes (F. Najmabad), Sprng

28 Small-Sgnal: Frst we calculate the small-sgnal parameters: I C nv T r π β 2.42 k r e Ω r o V A V CE k I C µ F o Amplfer Response: 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) wth R E. Usng formulas of page 5-21 and notng R E r e A R E R E r e r e 1 R R B [r π (1 β)r E ] f l f p2 R o r e 12 Ω 1 2πC c2 ( R E r e ) 1 2πC c2 ( r e ) Hz 2πC c2 BJT s n acte f C E > 0 and CE > V D0. The current source makes the problem slghtly complcated ( e flows through the 100 k resstor whle I E s fed by the current source). As such, CE-KVL degenerates nto two equatons, one for small sgnal and one for basng: CE-KVL (SS) 0 ce 10 5 e and CE V CE ce > V D0 0.7 ce 10 5 e > 0.7 V CE 4.0 V 10 5 e < 4.0 V E I E e > 0 e > I E 4.3 ma 10 5 e > 430 V 430 < A < 4.0 V 430 < < 4 V V < 4.0 V Note that the aboe lmt s correct ONLY for an deal current source (that s why the lower lmt for s so large). In practcal applcatons, the current source wll mpose a condton on E (e.g., mnmum oltage on the collector of a current mrror). Problem 7. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. Ths crcut s smlar to the crcut of Problem 6 except that C c2 s remoed. Bas: Set 0 and capactors open. BE-KVL: 0 V BE V E V E 0.7 V 4 V 4.3mA V EE o 4 V 4.3mA V EE V E 1 ECE65 Lecture Notes (F. Najmabad), Sprng

29 KCL 1 V E 7 µa I E ma I E 3 ma I C, I B I C β CE-KVL: 4 V CE V E V CE 4.7 V 15 µa Bas summary: I E I C 4.3 ma, I B 21.5 µa, V CE 4.7 V whch s the exactly the same as of that of Problem 6. The small-sgnal, amplfer parameters, and maxmum and are exactly the same as those of Problem 6 expect f l 0 (.e., ths amplfer can amplfy DC sgnals). Problem 8. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. Bas: 15 V Set 0 and capactors open. Replace R 1 and R 2 wth ther Theenn equalent: 34k 100 nf R B 5.9 k 34 k 5.0 k, V BB V 4.7 µ F 5.9k µ F BE-KVL: V BB R B I B V BE 510I E I B I E 1 β I E 201 ( ) I E k 15 V CE-KVL: I E 3 ma I C, I B I C β V CC 1000I C V CE 510I E 15 µa V CE 15 1, V Bas summary: I E I C 3 ma, I B 15 µa, V CE 10.5 V Small-Sgnal: Frst we calculate the small-sgnal parameters: I C nv T r π β 3.47 k r e Ω r o V A V CE I C k 5.9k V 4.95V 5.0k 510 ECE65 Lecture Notes (F. Najmabad), Sprng

30 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). Usng formulas of page 5-21 and notng R C, R C r o, R E r e, and R sg 0: 4.7 µ F 5.0k 510 A r o R C r e R C r e 58 R R B r π k R out R o r o 53.5 k 1 f p1 2πC c1 (R R sg ) Hz 2π f p2 f pb 1 2πC c2 ( R C R out ) 1 2πC c2 ( R c ) 1 2πC c Hz 1 2πC b [R E (r e (R B R sg )/β)] Hz 2πC b [R E 17.3] f l f p1 f p2 f pb Hz To fnd the maxmum ampltude for, we note that c R C 10 3 c (as R C ): CE-KVL (I C C ) CE 510I E As e does not flow n the R E 510 Ω resstor because of the 47 µf by-pass capactor, CE-KVL CE 15 1, 510I C 10 3 c c CE > V D0 0.7 c < 9.8 ma 10 3 c < 9.8 V C I C c > 0 c > I C 3 ma 10 3 c > 3 V 3 < 58 < 9.8 V 0.17 < < V Problem 9. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. Ths s the PNP analog of crcut of Problem 8. Bas summary: I E I C 3 ma, I B 15 µa, V EC 10.5 V Small-Sgnal: /Ω, r e 17.3 Ω. r π 3.47k, and r o 53.5 k. Amp response: A 58, R 2.05 k, R o 53.5 k, and f l f p1 f p2 f pb 113 Hz. ECE65 Lecture Notes (F. Najmabad), Sprng

31 Problem 10. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. Bas: 15 V Set 0 and capactors open. The bas crcut s exactly that of Problem 8 wth R B 5.0 k. Bas summary: I E I C 3 ma, I B 15 µa, V CE 10.5 V. Small-Sgnal: The small-sgnal parameters are also the same as those of Problem 8: /Ω, r e 17.3 Ω. r π 3.47k, and r o 53.5 k. 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). Usng formulas of page 5-21 and notng R C, R C r o, and R E r e : 34k 4.7 µ F 5.9k k 100 nf 15 V A R C R C 1.90 R E r e R E r e R R B [r π (1 β)r E ] R B [(1 β)r E ] R B 5.0 k R o r o [1 (R E r π )] r o [ (510 3, 470)] M ) βr E R out r o (1 26.6r o 1.4 M r π R E R B R sg f p1 f p2 1 2πC c1 (R R sg ) Hz 2π µ F 1 2πC c2 ( R C R out ) 1 2πC c2 ( R c ) 1 2πC c Hz f l f p1 f p Hz 5.0k 5.9k To fnd the maxmum ampltude for, we need to fnd c as c R C 10 3 c (as R C ). Then, (note compared to Problem 8, e now flows n the R E 510 Ω resstor:) CE-KVL (I C C ) CE 510(I E e ) CE 15 1, 510I C 1, 510 c , 510 c CE > V D0 0.7 c < 6.5 ma 10 3 c < 6.5 V C I C c > 0 c > I C 3 ma 10 3 c > 3 V 3 < 1.9 < 6.5 V 3.42 < < 1.58 V ECE65 Lecture Notes (F. Najmabad), Sprng

32 Problem 11. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. Bas: Set 0 and capactors open. Because the 47 µf capactor across the 240 Ω resstor becomes an open crcut, the total R E for bas s Ω and the bas crcut s exactly that of Problem 8 wth R B 5.0 k. Bas summary: I E I C 3 ma, I B 15 µa, V CE 10.5 V. Small-Sgnal: The small-sgnal parameters are also the same as those of Problem 8: /Ω, r e 17.3 Ω. r π 3.47k, and r o 53.5 k. 34k 4.7 µ F 5.9k 15 V nf 47 µ F 15 V Proceedng wth the small sgnal analyss, we zero bas sources (see crcut). In ths case, the 47 µf capactor across the 240 Ω resstor becomes a short crcut and the total R E for small-sgnal s 270 Ω. As the nput s at the base and output s at the collector, ths s a degenerated commonemtter amplfer (.e, wth a emtter resstor). Usng formulas of page 5-21 and notng R C, R C r o, and R E r e : 34k 5.9k A R C R C 3.48 R E r e R E r e R R B [r π (1 β)r E ] R B [(1 β)r E ] R B 5.0 k R o r o [1 (R E r π )] r o [ (270 3, 470)] k ) βr E R out r o (1 15.4r o 826 k r π R E R B R sg f p1 f p2 1 2πC c1 (R R sg ) Hz 2π k 4.7 µ F 1 2πC c2 ( R C R out ) 1 2πC c2 ( R c ) Hz 2πC c2 5.9k 15 V 100 nf 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 5-21, we can extend those formulas to coer ths case. ECE65 Lecture Notes (F. Najmabad), Sprng

33 The pole ntroduced by the by-pass capactor n the common emtter case s (see fgure) f pb 1 2πC b [R E (r e (R B R sg )/β)] R C Per our dscusson of page 5-19 on how to fnd poles ntroduced by each capactor, R E (r e (R B R sg )/β)] s the total resstance seen across the termnal of C b. As can be seen from the crcut, the resstance across C b termnals conssts of two resstors n parallel, R E and R e, the resstance seen through the emtter of the BJT: R e r e (R B R sg )/β) from the formula aboe. For the crcut here (defned R E1 240 Ω and R E2 270 Ω), the resstance across C b s made of two resstances n parallel: R E1 and the combnaton of R E2 and R e, the resstance seen through the emtter of BJT n seres. Thus: f pb f pb 1 2πC b [R E1 (R E2 r e (R B R sg )/β)] 1 2πC b [240 (270 17] Hz 2π f l f p1 f p Hz R B 34k 4.7 µ F 5.9k R e R E R E2 R E1 15 V R e C b 100 nf 47 µ F To fnd the maxmum ampltude for, we need to fnd c as c R C 10 3 c (as R C ). Then, (note e now flows n 270 Ω resstor whle I E flows n a 510 Ω resstor): CE-KVL (I C C ) CE 510I E 270 e CE 15 1, 510I C 1, 270 c , 270 c CE > V D0 0.7 c < 7.7 ma 10 3 c < 7.7 V C I C c > 0 c > I C 3 ma 10 3 c > 3 V 3 < 3.5 < 7.7 V 2.2 < < 0.86 V ECE65 Lecture Notes (F. Najmabad), Sprng

34 Problem 12. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. Bas: Set 0 and capactors open. BE-KVL: 3 2.3I E V EB I E 1 ma I C, I B I E 1 β 5 µa CE-KVL: I E V EC I C 3 V EC V Bas summary: I E I C 1 ma, I B 50 µa, V CE 1.4 V 3V 2.3k 2.3k 3V 3V 2.3k 100 nf 47 µ F Small-Sgnal: Frst we calculate the small-sgnal parameters: I C nv T r π β 10.4 k r e 1 52 Ω r o V A V CE I C k k 3V Proceedng wth the small sgnal analyss, we zero bas sources. As the nput s at the base and output s at the collector, ths s a common-emtter amplfer. It does not hae an emtter resstor as 47 µf capactor shorts out R E for small sgnals. Usng formulas of page 5-21 and notng R C, R C r o, and R E r e : A r o R C R C 44.2 r e r e R R B r π 10.4 k R out R o r o 151 k f p1 0 f p2 f pb 1 2πC c2 ( R C R out ) 1 2πC c2 ( R c ) 15.9 Hz 1 2πC b [R E (r e (R B R sg )/β)] 1 2πC b [R E 52] 66.6Hz f l f p1 f p2 f pb Hz ECE65 Lecture Notes (F. Najmabad), Sprng

35 To fnd the maxmum ampltude for, we need to fnd c as c R C c (as R C ). Then, (note only I E flows n R E 2.3 k resstor): CE-KVL (I C C ) CE I E 3 CE I C c c CE > V D0 0.7 c < 0.3 ma c < 0.7 V C I C c > 0 c > I C 1 ma c > 2.3 V 2.3 < 44.2 < 0.7 V < < V Problem 13. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. Ths s the same crcut as that of Problem 13 expect that the transstor s based wth a current source Bas: Set 0 and capactors open. From the crcut I E 1 ma BE-KVL: I E 1 ma I C, V E V EB 0.7 V I B I C β 5 µa CE-KVL: V E V CE I C 3 V CE 0.7 V CE 1.4 V V EE 1mA 2.3k 3V V EE 1mA V E 100 nf 47 µ F Bas summary: I E I C 1 ma, I B 50 µa, V CE 1.4 V. Small-Sgnal: As the bas pont s exactly the same as that of problem 12, we hae: /Ω, r e 52 Ω, r π 10.4k, and r o 151 k. Amplfer response: The only dfference wth problem 12 s that R E n ths crcut: A 44.2, R 10.4 k, R o 151 k, and f l Hz. Maxmum ampltude for : note only I E flows n the current source whle e flows through the by-pass capactor. As such, the results are smlar to those of problem 12: 2.3 < 44.2 < 0.7 V or < < V. 2.3k 3V ECE65 Lecture Notes (F. Najmabad), Sprng

36 Problem 14. Fnd the bas pont and amplfer parameters of ths crcut (S BJT wth n 2, β 200 and V A 150 V. Ignore the Early effect n basng calculatons). Fnd the maxmum ampltude of for the crcut to work properly. Bas: Set 0 and capactors open. 2.5 V R B 12 k 13 k 6.24 k, V BB V BE-KVL: V BB R B I B V BE 510I E I B I E 1 β I E 201 ( ) I E nF 1 µ F V 13k 13k 12k CE-KVL: I E 1.16 ma I C, I C V CE 400I E I B I C β 5.8 µa k V CE 2.5 1, V Bas summary: I E I C 1.16 ma, I B 5.8 µa, V CE 0.88 V Small-Sgnal: Frst we calculate the small-sgnal parameters: I C nv T r π β 8.97 k r e Ω r o V A V CE I C k 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. Usng formulas of page 5-21 and notng, and R C r o : A (r o R C ) R C 22.3 R R E r π [1/ (R C )/( r o )] R E r π (1/ ) R 400 8, Ω ) βr E R out r o (1 r π R E R B R sg R out r o [ /(8, )] 9.54r o 1.2 M f p1 1/[2πC c1 (R R sg )] 3.95 khz R o r o 129 k ECE65 Lecture Notes (F. Najmabad), Sprng

37 f p2 1 2πC c2 ( R C R out ) 0 R CB R B [r π (1 β)(r sg R E )] 6.24 k 8.97 k 3.68 k 1 f pb 432 Hz 2πC b R CB f l f p1 f p2 f pb 3, khz Note the small nput resstance of ths amplfer and correspondng large f p1. To fnd the maxmum ampltude for, we need to fnd c as c R C 10 3 c (as R C ): CE-KVL (I C C ) CE 400(I E e ) CE 2.5 1, 400I C 1, 400 c , 400 c CE > V D0 0.7 c < 0.13 ma 10 3 c < 0.18 V C I C c > 0 c > I C 1.16 ma 10 3 c > 1.16 V 1.16 < 22.3 < 0.18 V 52 < < 8 mv ECE65 Lecture Notes (F. Najmabad), Sprng