Lecture 260 Shunt-Shunt Feedback (3/5/02) Page 260-1

Transcription

1 Lecture 260 ShuntShunt (3/5/02) Page 2601 LECTURE 260 SHUNTSHUNT FEEDBACK (READING: GHLM ) Objectve The objectve of ths presentaton s: 1.) Show how to dentfy the type of feedback topology 2.) Illustrate the analyss of shuntshunt feedback crcuts Outlne dentfcaton procedure Shuntshunt feedback wth nondeal source and load Examples Summary ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002 Lecture 260 ShuntShunt (3/5/02) Page 2602 IDENTIFICATION OF THE FOUR, SINGLELOOP FEEDBACK TOPOLOGIES TwoTermnal Representaton of a SngleLoop, Negatve System Sgnal Source v Mxng e v e Amplfer of forward gan = a Samplng o Load of reverse gan = f Fg ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002

2 Lecture 260 ShuntShunt (3/5/02) Page 2603 Topology Identfcaton Procedure 1.) Identfy the feedback loop by tracng around the feedforward and feedback path. Also check to see f the feedback s postve or negatve. 2.) Identfy whether or not the mxng network s seres or shunt. If the sgnal source has one termnal on ac ground then: a.) If the nput actve devce has one of ts nput termnals on ac ground, then the mxng network must be shunt. b.) If the sgnal and feedback sources are appled to dfferent nput termnals of the nput actve devce, then the mxng network s seres (ths ncludes dfferental amplfers where two devces form the nput actve devce). c.) If the sgnal source does not have one of ts nput termnals on ac ground or to check the above steps, try to assgn the varables x, x fb, and x e on the schematc n such a manner as to mplement the equaton, x e = x ± x fb If ths equaton can be wrtten usng voltages (currents) then the mxng crcut s seres (shunt). v v e Seres e Shunt ve ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002 v Seres Fg Lecture 260 ShuntShunt (3/5/02) Page 2604 Topology Identfcaton Procedure Contnued 3.) Next dentfy the samplng crcut as seres or shunt. If the load s grounded then, a.) If the out actve devce has one of ts two possble output termnals grounded, then the feedback s shunt. b.) If the output actve devce has nether of ts output termnals on ground and f the output sgnal s taken from one of ts output termnals and the fed back sgnal from the other output termnal, then the feedback s seres. c.) If the load s not grounded or to check the above test, dentfy the load resstor,, and apply the followng test:.) If x fb becomes ero when = 0, then the samplng network s shunt..) If x fb becomes ero when =, then the samplng network s seres. o Seres Shunt Shunt Fg ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002

3 Lecture 260 ShuntShunt (3/5/02) Page 2605 Transstor Examples of Negatve Topology Identfcaton 100K Crcut 1 Crcut 2 2K M1 5K Q2 3K Crcut 3 Crcut 4 Fg ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002 Lecture 260 ShuntShunt (3/5/02) Page 2606 ShuntShunt ncludng Source and Load Resstance Confguraton: s R S v y 22a y 11a y12a y 21a v Basc Amplfer y 11f y 12f y 21f v y 22f s v R S y 11f y 11a y21a v y 22a y 22f RL New Basc Amplfer y 12f New s = a = A = 1af = (y 21a /y 11 y 22 ) 1(y 21a /y 11 y 22 ) y 12f Fg ShuntShunt = y 11 y 12 = y 21 y 22 where for the new basc amplfer, y 11 = v1 =0 = G S y 11a y 11f y 12 = =0 = 0 y 21 = v1 =0 = y 21a y 22 = =0 = G L y 22a y 22f a = y 21a y 11 y 22 and f = y 12f 2 Fg ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002

4 Lecture 260 ShuntShunt (3/5/02) Page 2607 Example 1 Invertng Op Amp Fnd the closedloop transfer functon, A, the closedloop nput resstance, Z f, and the closedloop output resstance, Z of of the shuntshunt confguraton shown. The op amp has a dfferental nput resstance of, voltage gan of a v, and output resstance of o. Soluton Equvalent crcut: Z f Op Amp a v RF Z of o Equvalent ' Z Basc Amplfer o v1 ' R F a v ' ' = v' R o F R F (Use prmed varables to ndcate open loop) ' Z o Fg It s easy to show that f = y 12f = ' ' '=0 = 1 R F The forward gan, a, s a = ' ' = v o ' v 1 ' a v (R F ) R F a R F ' ' = o (R F ) R = F o R F o R F R where F fb ' = 0 R F Fg ut ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002 Lecture 260 ShuntShunt (3/5/02) Page 2608 Example 1 Contnued The loop gan s a T = af = o R F o R F R F The closedloop gan s a R F v o a = 1af = o R F o R F R F (a v R 2 F ) a = ( o R F o R F )( R F ) a v R F 1 o R F o R F R F The closedloop nput mpedance s R F Z f = Z 1T = R F o R F o R F a a v 1 o R F o R F R F The closedloop output mpedance s Z of = Z o 1T = o R F a 1 o R F o R F R F If a v = 200,000, = 2MΩ, and o = 75Ω, R F = 1MΩ, and = 10kΩ, then T = 133,333, Z = 2 1 = 0.667MΩ Z f = 5Ω, Z o 75Ω Z of = 0.563mΩ and A = 999,992Ω ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002

5 Lecture 260 ShuntShunt (3/5/02) Page 2609 Example 2 Transstor Amplfer For the amplfer shown, fnd /, /, and /. Assume that g m = 5mS and r ds = for the MOSFET and r π1 = r π3 =1000Ω and β F1 = β F3 = 100 for the BJTs. Soluton 1.) Fnd the feedback topology and polarty of feedback. The loop conssts of basecollector of, gatedran of, and basecollector of Q3. A postve change at the base of gves a and the gate of, whch gves a at the base of Q3, whch gves a at the base of Q3. feedback s negatve. 2.) The mxng crcut s shunt because only the base termnal of s connected to the nput and feedback. Note that C3 = and B1 = e e =. 3.) The feedback crcut s shunt because the output transstor () has one of ts possble output termnals on ac ground. Also, f = R 4 goes to ero, = 0. 4.) Draw the closedloop crcut and smallsgnal model. = e Q3 R 1 = Ω R 2 = 100Ω R 3 = 20KΩ R4 = Ω R 4 = Ω ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002 = = e = b1 r π1 R 2 Fg Q3 β b1 R3 R 1 = Ω β b3 v gs2 R 2 = 100Ω b3 V CC R 3 = 20KΩ r π3 R 1 R =v 4 o g m2 v gs2 Fg Lecture 260 ShuntShunt (3/5/02) Page Example 2 Contnued 5.) AC openloop model s drawn by, a.) Lookng back nto the feedback network (Q3,R1) to the left wth = 0. b.) Lookng back nto the feedback network (Q3,R1) to the rght wth = 0. The result s, '= ' b1 ' r π1 R 2 β b1 ' R 3 v gs2 ' g m2 v gs2 ' ' β b3 ' b3 ' r π3 R 1 R 4 ' '= ' Fg ) Next, fnd a and f. a = ' ' = v o ' v gs2 ' v gs2 ' ' = g m2 {R 4 [r π3 (1β 3 )R 1 ]}(β 3 R 3 ) = (45.54)(2x10 6 ) = 91.07MΩ f = ' ', ' β 3 B1 = r π3 (1β 3 )R 1 ' = β 3 B3 f = r π3 (1β 3 )R 1 = 0.98mS ( 1/R 1 ) ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002

6 Lecture 260 ShuntShunt (3/5/02) Page Example 2 Contnued 7.) Calculate R and R o. R = ' ' = r π1(1β 1 )R 2 = 11kΩ and R o = ' ' = R 4 [r π3 (1β 3 )R 1 ] = 10kΩ 102kΩ = 9.107kΩ 8.) Fnd /, /, and /. a = 1T = 91.07MΩ x10 3 = Ω (Note: 1 f = 1020Ω) R f = R of = Now, v 2 v = 1 = R 1T = 11kΩ x10 3 = Ω = R o 1T = 9.107kΩ x10 3 = 0.102Ω v = / 1 R = Ω f Ω = V/V ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002 Lecture 260 ShuntShunt (3/5/02) Page SUMMARY We have shown how to dentfy the type of feedback topology Illustrate the analyss of shuntshunt feedback crcuts Procedure 1.) Identfy the type of feedback topology 2.) Draw the schematc for the closedloop amplfer 3.) Break the feedback loop by replacng the feedback network wth the resstance seen lookng nto each end of the feedback network wth the other end shorted. 4.) Draw a schematc of the openloop amplfer wth all varables prmed. 5.) Solve for a, f, R, and R o. 6.) Calculate A and R f and R of by the followng formulas: A = a 1T, R f = R 1T, and R of = R o 1T 7.) If necessary, use the results of 6.) to fnd the desred closedloop transfer functon. ECE 6412 Analog Integrated Crcut Desgn II P.E. Allen 2002