Chapter 2 Properties of resistive circuits
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1 Chaper 2 Properes of resse crcus (oponal) equalen ressance, resse ladder, dualy, dual enes 2.3 Crcus wh conrolled sources conrolled sources (VCVS, CCVS, VCCS, CCVS), equalen ressance 2.4 Lneary and superposon proporonaly prncple, superposon heorem 2.5 Theenn and Noron neworks Theenn s and Noron s heorems, source conerson 2-
2 Oponal. equalen wo-ermnal neworks: wo-ermnal neworks hae he same - characerscs 2. seres-conneced ressors: olage dder seres equalen ressance: R ser R + R R N 3. parallel-conneced ressors: curren dder parallel equalen conducance: G par G + G G N 4. resse ladder: a nework conssng enrely of seres and parallel ressors 5. seres-parallel reducon mehod: o sole branch arables of a resse ladder () sep : fnd he equalen ressance of he ladder nework (2) sep 2: calculae he olage or curren a he nework ermnal (3) sep 3: use KVL, KCL, Ohm s law o calculae branch arables 2-2
3 6. Ex. 2.4 sole ermnal curren, oal dsspaed power p, branch arables x, y () sep R eq 2+ 0//5 8kΩ (2) sep p mW (3) sep 3 x y V V R eq 2-3
4 7. dual neworks: - equaon of nework A - equaon of nework B wh s, nerchanged Seres nework KVL loop eq. equalen ressance olage dder open crcu Parallel nework R + R...) KCL node eq. G + G...) n ( 2 + R ser R R equalen conducance ( 2 + G par Rn curren dder Gn R + R... G G n when R shor crcu when G n n n n G G
5 8. Dual pars 9. Ex. 2.5 dual neworks KVL olage ressance loop seres open ck node olage capacance mpedance KCL curren conducance node parallel shor ck mesh curren nducance admance (2 + 3) (2 + 3)
6 (dual crcu conerson) Sep : place a node a he cener of each loop and he reference node ousde of he gen crcu Sep 2: draw a lne beween he nodes wh each lne crossng an elemen, and replace ha elemen by s dual rule of source: a olage source ha produces a pose (CW) loop curren has as s dual a curren source whose reference drecon s from he ground o nonreference node 2 V pose or CW dual 2A curren source dreced from 0 o 5 A o he ground node 2 ( negae) 5 A or CCW dual 5V olage source 2-6
7 2.3 Crcu wh conrolled sources Bascs. dependen (conrolled) sources: lnear elemen olage gan curren gan (ransformer, olage amplfer) (BJT) ransconducance ransressance (FET) 2-7 (ransformer)
8 (DC ranssor crcu) n ace mode: V 0.7V I I C C BE αi, α: CB curren gan E 0.98 <α< βi, β: CE curren gan B 50 <β< 000 npn ranssor DC equalen model mehod 50 loop : () loop 2: V uA 3 BE loop 3: o 0 IC (2) () 3 ua, node A : + I I 9.5uA 2 I β I.425 ma,(2) V C B o B 2-8 B
9 mehod o IB...() IB 2 9.5uA () o V 2-9
10 2. equalen ressance heorem (ermnal behaor of a nework) For a wo-ermnal load nework conanng ressors and conrolled sources, he equalen ressor s deermned by s ermnal olage and curren. no ndependen source nsde R eq 2-0
11 Dscusson. Ex.2.6 a FET wh g m 5mS, ou? 6 0 g ou ou m g 5 n g m n 25 n Ex.2.7 a VCVS o fnd ( s ) KVL : 2 KVL : s s KCL : s 2 s 9 kω 5kΩ g ou 6kΩ g m ou g 2-
12 3. Ex.2.8 fnd R eq gm R KCL : + g m ( g m + ) R R R R Req g R m f R, c gm Req ( ace) g m 2-2
13 2.4 Lneary and superposon Bascs. Lnear resse crcu: crcu conans lnear ressors and ndependen sources. 2. Proporonaly prncple (sngle ndependen source) f yf(x), hen f(kx)kf(x) 3. Superposon heorem (mulple ndependen sources) For a lnear nework, response of a sum of sgnals sum of sgnal responses 4. Conrolled sources are no appled for he superposon heorem. 5. suppressed ndependen olage source s 0 shor crcu suppressed ndependen curren source s 0 open crcu 2-3
14 Dscusson. Ex.2.9 (ladder nework) s 72 use proporonaly prncple o fnd, 2, R eq sep : assume a conenen alue for a branch arable farhes from he source le 2 sep 2: work oward source o oban he source alue x s 3 2 sep 3: calculae he scalng facor K K 9 72 K 8 sep 4: calculae he acual alues by he scalng facor K 3 K 2, 2 K 3 24, K 8 2 s Req 6 2-4
15 Ex.2.0 use superposon o fnd from source of 30V O.C. from source 2 of 3A (6 + 2) + 4 S.C. S.C. O.C. from source 3 of 8A 6 3 ( 8) (2+ 4)
16 Ex.2. use superposon o fnd from source of 30V KCL 9 KVL 30 6(9 ) (4 + 2) from source 2 of 3A KCL 9 KCL Ω 3 4Ω 2 KVL 6(9 ) + 4( 3) A
17 2.5 Theenn and Noron neworks Bascs. source nework: a wo-ermnal nework conanng lnear ressors (conrolled sources) and a leas one ndependen source. 2. Theenn parameers (ermnal behaor of a nework): opencrcu olage oc, shor-crcu curren sc, and Theenn ressance R oc R sc 2-7
18 Dscusson. Theenn s heorem: he equalen crcu of a lnear resse source nework s a olage source oc n seres wh a ressor R. 2. Noron s heorem: he equalen crcu of a lnear resse source nework s a curren source sc n parallel wh a ressor R.,, oc oc 0 sc R 0 sc oc R sc R 2-8 f load SC, 0 R oc sc oc f load OC, 0 sc R R oc sc
19 3. Theenn and Noron equalen crcus are useful o sole he ermnal behaor of a source nework, bu no nernal branches. 4. Ex.2.2 fnd he Theenn parameers OC.. load x x x ( 00 x) KCL x x + 00x oc 80 + ( 00 x ) SC.. load x x x KCL sc R x x 00x oc 6.92Ω sc 2-9
20 5. Ex.2.3 use equalen source crcus o desgn R L n wo cases o ge 24, and 8 Thenn and Noron equalen crcus 20 oc sc 0, R from Teenn equalen crcu RL 24 oc RL 6Ω R + R from Noron equalen crcu R 8 sc RL Ω R + R L L 2-20
21 6. Theenn ressance he equalen ressance of he source nework wh all ndependen sources suppressed. (anoher way o fnd R ) (conans only ressors and dependen sources) R oc sc (ndependen olage source or curren source) oc + R,f oc 0 hen R sc,f sc 0 R Equalen ressance heorem s he specal case of Theenn or Noron heorem. 7. A source nework can be descrbed by any wo of he Theenn parameers: oc, sc, R 2-2
22 8. Ex. 2.4 fnd Theenn parameers sep : 40k easer o fnd sc KVL x 5 x 0 x 0 sc sep 2: fnd R KVL x x x x KCL R 0000 ( conrolled source) oc R sc 30 sep : sep 2: 2-22
23 9. equalen source conerson (ncludng dependen sources) () sources are off same (2) SC.. same s (3) OC.. same s R s s R s s 0. Source conerson can ease crcu analyss. 2-23
24 . Ex. 2.5(ex. 2.9) sole 2 usng equalen source conerson usng 2 2 KCL V 4//6//2 When conrolled sources are presen, source conerson mus no oblerae he deny of any conrolled arables. 2-24
25 2. Ex. 2.6 (ex.2.0) fnd Theenn equalen crcu usng source conerson sep sep 2 sep 3 R L A
(6)(2) (-6)(-4) (-4)(6) + (-2)(-3) + (4)(3) + (2)(-3) = -12-24 + 24 + 6 + 12 6 = 0
Chapter 3 Homework Soluton P3.-, 4, 6, 0, 3, 7, P3.3-, 4, 6, P3.4-, 3, 6, 9, P3.5- P3.6-, 4, 9, 4,, 3, 40 ---------------------------------------------------- P 3.- Determne the alues of, 4,, 3, and 6
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