ADDITIONAL ANALYSIS TECHNIQUES

Transcription

1 ADDTNAL ANALY TECHNQUE LEANNG GAL EEW LNEATY The property hs two equilent definitions. We show nd ppliction of homogeneity. ALY UETN We discuss some implictions of the superposition property in liner circuits. DEEL EENN AND NTN EEM These re two ery powerful nlysis tools tht llow us to focus on prts of circuit nd hide wy unnecessry complexities. MAMUM WE TANFE This is ery useful ppliction of Theenin s nd Norton s theorems. E MED F NDE AND L ANALY DE WEFUL TL T DETEMNE E BEHA F EEY CMNENT N A CCUT The techniques deeloped in chpter ; i.e., comintion series/prllel, oltge diider nd current diider re specil techniques tht re more efficient thn the generl methods, ut he limited pplicility. t is to our dntge to keep them in our repertoire nd use them when they re more efficient. n this section we deelop dditionl techniques tht simplify the nlysis of some circuits. n fct these techniques expnd on concepts tht we he lredy introduced: linerity nd circuit equilence.

2 ME EQUALENT CCUT ALEADY UED 3 LNEATY E MDEL UED AE ALL LNEA. MAEMATCALLY MLE AT EY ATFY E NCLE F UETN. E MDEL T( u u nd ll possile ) Tu for ll possile input y Tu LNEA FF Tu sclrs, pirs u, u AN ALTENATE, AND EQUALENT, DEFNTN F LNEATY LT E UETN NCLE N TW. EMDEL y Tu LNEA FF. T( u u ) Tu Tu, u, u dditiity. T( u) Tu,, u homogeneity NTCE AT, TECHNCALLY, LNEATY CAN NEE BE EFED EMCALLY N A YTEM. BUT T CULD BE DED BY A NGLE CUNTE EAMLE. T CAN BE EFED MAEMATCALLY F E MDEL UED. UNG NDE ANALY F ETE CCUT, NE BTAN MDEL F E FM A f A ECT CNTANNG ALL E NDE LTAGE AND f A ECT DEENDNG NLY N E NDEENDENT UCE. N FACT, E MDEL CAN BE MADE ME DETALED A FLLW. A Bs HEE, A,B, AE MATCE AND s A ECT F ALL NDEENDENT UCE. F CCUT ANALY WE CAN UE E LNEATY AUMTN T DEEL ECAL ANALY TECHNQUE. FT WE EEW E TECHNQUE CUENTLY AALABLE 4

3 A CAE TUDY T EEW AT TECHNQUE DETEMNE edrwing the circuit my help us in recognizing specil cses LUTN TECHNQUE AALABLE?? 5 6

4 UNG HMGENETY Assume tht the nswer is known. Cn we compute the input in ery esy wy?!! f o is gien, then cn e computed using n inerse oltge diider 0 And s using second oltge diider. 4 EQ EQ EQ 4 EQ EQ ole now for the rile o 0 The procedure cn e mde entirely lgorithmic. Gie to o ny ritrry lue (e.g., o = ). Compute the resulting source lue nd cll it s ' ' ' 3. Use linerity. k k ' 0 0, k 4. The gien lue of the source (s) corresponds to k ' Hence the desired output lue is ' ' 0 k0 ' 0 This is nice little tool for specil prolems. Normlly when there is only one source nd in our judgment, soling the prolem ckwrds is ctully esier. 7 LE UNG HMGENETY AUME out [ ] NW UE HMGENETY 6[ ] out [ ] out [ ] [ ] 8

5 LEANNG ETENN CMUTE UNG HMGENETY. UE 6mA.5[ ma] k 6[ ].5[ ma] 3[ ] 0.5[ ma] ma 0.5[ ma] AUME ma UE HMGENETY ma 6mA ma 9 ource uperposition This technique is direct ppliction of linerity. t is normlly useful when the circuit hs only few sources. 0

6 F CLATY WE HW A CCUT W NLY TW UCE - L Due to Linerity L circuit L CNTBUTN BY L CNTBUTN BY L Cn e computed y setting the current L source to zero nd soling the circuit Cn e computed y setting the oltge L source to zero nd soling the circuit UCE UETN Circuit with oltge source set to zero (HT CCUTED) L L L = L Circuit with current source set to zero (EN) Due to the linerity of the models we must he L L L L L The pproch will e useful if soling the two circuits is simpler, or more conenient, thn soling circuit with two sources. L rinciple of ource uperposition We cn he ny comintion of sources. And we cn prtition ny wy we find conenient.

7 LEANNG EAMLE WE WH T CMUTE E CUENT i = Loop equtions eq [ k] i " eq 6 (3 3) [ k] eq Contriution of nce we know the prtil circuits we need to e le to sole them in n efficient mnner. Contriution of 3 LEANNG EAMLE Compute 0 using source superposition We set to zero the oltge source Current diision hm s lw oltge Diider Now we set to zero the current source 3-6k 3k 0 " [ ] ' " [ ] 4

8 LEANNG EAMLE Compute 0 using source superposition We must e le to sole ech circuit in ery efficient mnner!!! f is known, then o is otined using oltge diider et to zero current source et to zero oltge source cn e otined y series/prllel reduction nd diider - k 4k 8k 8/3 (6) 8/3 The current cn e otined using current diider nd o using hm s lw ' 6k k 6k 6k k ' 0 8 [ ] 7 WHEN N DUBT EDAW! k 4k ma k 6k " 0 k (k 4k) () ma k 6k (k 4k) " 6k ' " 5 mple rolem CMUTE 0 UNG UCE UETN. Consider only the oltge source 0. 5mA 3. Consider only the 4mA source. Consider only the 3mA source Current diider 0. 5mA 03 0 Using source superposition 3mA

9 UETN ALED T -AM CCUT TW UCE. WE ANALYZE NE AT A TME CNTBUTN F. Bsic inerter circuit rinciple of superposition CNTBUTN F Bsic non-inerting mplifier Notice redrwing for dded clrity

10 EENN AND NTN EEM These re some of the most powerful nlysis results to e discussed. They permit hiding informtion tht is not relent nd concentrting on wht is importnt to the nlysis. 9 Low distortion udio power mplifier T MATCH EAKE AND AMLFE, NE HULD ANALYZE CCUT. From reamp (oltge ) To spekers Courtesy of M.J. enrdson T MATCH EAKE AND AMLFE, T MUCH EAE T CNDE EQUALENT CCUT! - ELACE AMLFE BY MLE EQUALENT 0

11 EENN EQUALENCE EEM LNEA CCUT My contin independent nd dependent sources with their controlling riles AT A i LNEA CCUT My contin independent nd dependent sources with their controlling riles AT B i LNEA CCUT AT B AT A Theenin Equilent Circuit for AT A Theenin Equilent ource Theenin Equilent esistnce NTN EQUALENCE EEM LNEA CCUT My contin independent nd dependent sources with their controlling riles AT A i LNEA CCUT My contin independent nd dependent sources with their controlling riles AT B i N N i LNEA CCUT AT B AT A i N N Norton Equilent Circuit for AT A Theenin Norton Equilent ource Theenin Norton Equilent esistnce

12 Exmples of lid nd nlid rtitions 3 UTLNE F F - ersion f Circuit A is unchnged, then the current should e the sme for ANY o. UE UCE UETN DEFNE ECAL i i ic; i CAE: EN CCUT ( i 0) 0 i C i C i i All independent sources set to zero in A C ic i i C HW D WE NTEET EULT? i i i C 4

13 UTLNE F F - ersion LNEA CCUT My contin independent nd dependent sources with their controlling riles AT A i LNEA CCUT My contin independent nd dependent sources with their controlling riles AT B. Becuse of the linerity of the models, for ny rt B the reltionship etween o nd the current, i, hs to e of the form m* i n. esult must hold for eery lid rt B tht we cn imgine n 3. f prt B is n open circuit then i=0 nd f rt B is short circuit then o=0. n this cse 0 m* i C m ic i How do we interpret this? 5 EENN AACH LNEA CCUT My contin independent nd dependent sources with their controlling riles AT A i ANY AT B i For ANY circuit in rt B i AT A MUT BEHAE LKE CCUT This is the Theenin equilent circuit for the circuit in rt A. The oltge source is clled the EENN EQUALENT UCE. The resistnce is clled the EENN EQUALENT ETANCE. 6

14 Norton Approch i i i C LNEA CCUT My contin independent nd dependent sources with their controlling riles AT A i ANY AT B i Norton Equilent epresenttion for rt A i C Norton Equilent ource i C Norton 7 ANE EW F EENN AND NTN EEM Theenin i i C i C Norton This equilence cn e iewed s source trnsformtion prolem. t shows how to conert oltge source in series with resistor into n equilent current source in prllel with the resistor. i UCE TANFMATN CAN BE A GD TL T EDUCE E CMLETY F A CCUT 8

15 ource trnsformtion is good tool to reduce complexity in circuit... WHEN T CAN BE ALED!! idel sources re not good models for rel ehior of sources A rel ttery does not produce infinite current when short-circuited. - E MDEL AE EQUALENT WHEN mproed model for oltge source mproed model for current source ource Trnsformtion cn e used to determine the Theenin or Norton Equilent... BUT EE MAY BE ME EFFCENT TECHNQUE 9 EAMLE: LE BY UCE TANFMATN n etween the terminls we connect current source nd resistnce in prllel. The equilent current source will he the lue /3k. The 3k nd the 6k resistors now re in prllel nd cn e comined. n etween the terminls we connect oltge source in series with the resistor. The equilent source hs lue 4mA*k. The k nd the k resistors ecome connected in series nd cn e comined. After the trnsformtion, the sources cn e comined. The equilent current source hs lue 8/4k, nd the comined current source hs lue 4mA. ptions t this point. Do nother source trnsformtion nd get single loop circuit.. Use current diision to compute 0 nd then compute 0 using hm s lw. 30

16 BLEM Compute 0 using source trnsformtion / EQUALENT CCUT 0 eq = 5 r one more source trnsformtion eq 3 eq - eq eq eq eq eq eq eq 4 3 eq 3 ECA F UCE TANFMATN - E MDEL AE EQUALENT WHEN mproed model for oltge source mproed model for current source ource Trnsformtioncn e used to determine the Theenin or Norton Equilent... WE NW EEW EEAL EFFCENT AACHE T DETEMNE EENN NTN EQUALENT CCUT 3

17 A Generl rocedure to Determine the Theenin Equilent i C y short circuit. Determine the Theenin equilent source pen Circuit oltge oltge t - if rt B is remoed hort Circuit Current current through - if rt B is replced i C Theenin Equilent esistnce emoe prt B nd compute the EN CCUT oltge ne circuit prolem LNEA CCUT My contin independent nd dependent sources with their controlling riles AT A i 0. Determine the HT CCUT current, emoe prt B nd compute the HT CCUT current i C econd circuit prolem LNEA CCUT My contin independent nd dependent sources with their controlling riles AT A i C 0 33 AN EAMLE F DETEMNNG E EENN EQUALENT rt B is irrelent. The oltge will e the lue of the Theenin equilent source. To rt B - Wht is n efficient technique to compute the C open circuit oltge? Now for the short circuit current, lets try source superposition. When the current source is open, the current through the short circuit is When the oltge source is set to zero, the current through the short circuit is C C To compute the Theenin resistnce we use C C For this cse, the Theenin resistnce cn e computed s the resistnce from - when ll independent sources he een set to zero. ( ) 0 NDE ANALY s this generl result? 34

18 Determining the Theenin Equilent in Circuits with nly NDEENDENT UCE The Theenin Equilent ource is computed s the open loop oltge The Theenin Equilent esistnce CAN BE CMUTED y setting to zero ll the sources nd then determining the resistnce seen from the terminls where the equilent will e plced - To rt B rt B 3k ince the elution of the Theenin equilent cn e ery simple, we cn dd it to our toolkit for the solution of circuits!! 4k rt B 35 LEANNG BY DNG 5k AT B 6 k (6 ) [ ] k 5k 36

19 LEANNG EAMLE CMUTE o UNG EENN n the region shown, one could use source trnsformtion twice nd reduce tht prt to single source with resistor.... r we cn pply Theenin Equilence to tht prt (iewed s rt A ). The originl circuit ecomes... 4k For the open loop 6 [ ] 8[ ] 3 6 oltge, the prt outside the region is eliminted. And one cn pply Theenin one more time! For open loop oltge use KL 4k 4k *ma nd we he simple oltge diider!! 8 0 6[ ] r we cn use Theenin only once to get oltge diider For the Theenin resistnce 8k rt B For the Theenin oltge we he to nlyze the following circuit MED?? Theenin Equilent of rt A ource superposition, for exmple Contriution of the oltge source Contriution of the current source (k k)*(ma) 8 imple oltge Diider 38

20 LEANNG EAMLE UE EENN T CMUTE o You he the choice on the wy to prtition the circuit. Mke rt A s simple s possile. ince there re only independent sources, for the Theenin resistnce we set to zero ll sources nd determine the equilent resistnce. rt B For the open circuit oltge, we nlyze the following circuit ( rt A )... ( 4) 0 k 3 Loop Anlysis 6 ma 4k k( ) ma ma 6 3 4k * k * 0 / / 3[ ] The circuit ecomes LEANNG ETENN: UE EENN T CMUTE o AT B 9k 8[ ] ma 3k 6[ ] 3k 6k k EULTNG EQUALENT CCUT 6 k k 4 ( 6 ) 3[ ] 4 4 4k 40

21 LEANNG ETENN: CMUTE o UNG NTN 4k N 3k N ma 3k C C AT B ma CMUTE o UNG EENN AT B N N N k k N 6k 3 4 () [ ] 9 3 k N - k ma 0 3k 3k 4k 4 (6 ) [ ] AMLE BLEM Equilent esistnce: ndependent sources only KL Equilent oltge: Node, loop, superposition 5( ) 0 ( ) Do loops - This is wht we need to get 3 How out source superposition? pening the current source: hort circuiting the oltge source KL

22 AMLE LEM All independent sources All resistors re in prllel!! = k 4k 8k,,, nd to compute Equilent ource... The circuit cn e simplified UCE TANFMATN oltge diider 8k (6 4/ 6)[ ] 8k (8/ 6) k 43 EENN EQUALENT F CCUT W NLY DEENDENT UCE A circuit with only dependent sources cnnot self strt. x ( ) 0 x 0 0 x 0 (ctully tht sttement hs to e qulified it. Wht hppens if? ) F ANY ELY DEGNED CCUT W NLY DEENDENT UCE 0, C 0 This is ig simplifiction!! But we need specil pproch for the computtion of the Theenin equilent resistnce. 0 ince the circuit cnnot self strt, we need to proe it with n externl source. The source cn e either oltge source or current source nd its lue cn e chosen ritrrily! Which one to choose is often determined y the simplicity of the resulting circuit. 44

23 F WE CHE A LTAGE BE... ( ) WE MUT CMUTE CUENT ULED BY BE UCE. ( ) The lue chosen for the proe oltge is irrelent. ftentimes we simply set it to one. 45 F WE CHE A CUENT UCE BE ( ) ( ) We must compute the node oltge KCL 0 The lue of the proe current is irrelent. For simplicity it is often choosen s one. 46

24 LEANNG EAMLE FND E EENN EQUALENT Do we use current proe or oltge proe? f we use oltge proe, there is only one node not connected through source. KCL : k k Controlling rile: LNG, 4 7 E k 3 7 EQUATN 4 k 5 k 0 k Using oltge proe. Must compute current supplied 5 4k 47 LEANNG EAMLE Find the Theenin Equilent circuit t A - B nly dependent sources. Hence = 0 To compute the equilent resistnce we must pply n externl proe. We choose to pply ( ) Controlling rile Conentionl circuit with dependent sources - use node nlysis 3( ) 6 ( ) 0 ( ) 3 6[ ] A 5 0 * / * /5 7 B ( ) ( ma) (0/ 7) k Theenin equilent ma 48

25 AMLE BLEM A B ma MUT FND. AB MED? Loop nlysis = 3 = / = k* 000 ; k * 3 k *( 3) k *( 3 ) 4k *( 3 ) 0 Controlling rile k *( 3 ) oltge cross current proe k *( 3 ) k *( 3 ) Theenin equilent ma The resistnce is numericlly equl to, ut with units of Kohm Theenin Equilent Circuits with oth Dependent nd ndependent ources LNEA CCUT My contin independent nd dependent sources with their controlling riles AT A - We will compute open circuit oltge nd short circuit current i C For ech determintion of Theenin equilent, we will sole two circuits. Any nd ll the techniques discussed should e redily ille; e.g., KCL, KL,comintion series/prllel, node, loop nlysis, source superposition, source trnsformtion, homogeneity The pproch of setting to zero ll sources nd then comining resistnces to determine the Theenin resistnce is in generl not pplicle!! 50

26 EAMLE Use Theenin to determine o pen circuit oltge hort circuit current olution to the prolem A ( k) ptions??? 0 " rt B Guidelines to prtition: rt A should e s simple s possile. After rt A is replced y the Theenin Equilent, we should he ery simple circuit. The dependent sources nd their controlling riles must remin together. Constrint t super node ( ) ( ) KCL t super node 0 k Eqution for controlling rile ole A k ( /k ) Negtie resistnces for some s k k k k k k C 8mA k k C [ k] 4 ( /k ) etting ll sources to zero nd comining resistnces will yield n incorrect lue!!!! But setting ll independent source to zero nd dependent source lie yield correct lue. 5 Find o using Theenin pen circuit oltge Method??? uper node hort Circuit Current 000 KCL KL C ( 3 ) ma 0 k 6k k C ma ( 3 ) /(6k) 0. 5mA (3/ 4) C k ( 3/8)[ ] The equilent circuit k KL (3/ 4)[ Controlling rile k 0 (3/8)[ ] (3/ 4) The equilent resistnce cnnot e otined y short circuiting the sources nd determining the resistnce of the resulting interconnection of resistors. - k ] 5

27 EAMLE - - Liner Model for Trnsistor x g m x 3 C The lterntie for mixed sources, pen circuit oltge gm3 x hort circuit current C g m x Equilent esistnce C x g 3 m C g m 3 etting ll independent source to zero nd dependent source lie yield sme result. 53 AMLE BLEM supernode Mixed sources. Must compute oc nd sc pen circuit oltge hort circuit current KL KCL t super node 0 The two 4k resistors re in prllel 0 [ ] C 4 4k *( / 4) [ ] 6k * 0 C KCL t supernode C C FNAL ANWE C ma 7 (/7) ma 7 C k 54

28 AMLE BLEM hort circuit current ingle node C Mixed sources! Must compute open loop oltge nd short circuit current pen circuit oltge For x use oltge diider For use KL ( ( ) ( 4 /3) ( ) ( )( /3) 4 3 KCL@x x We need to compute x s 0 KCL gin cn gie the short circuit current C FNAL ANWE C ) C 4( ) 4( ) 3 C 55 LEANNG EAMLE FND AND LT,, WHEN 0 0k DATA T BE LTTED 4k k Using ECEL to generte nd plot dt EENN EQUALENT EAMLE x[khm] oc[] th[khm] oc[] UNG ECEL oc[] 4 th[khm] x[khm] 56

29 57 LEANNG EAMLE FND AND LT,, WHEN 0 0k DATA T BE LTTED 4 4k 6 4 4k Using MATLAB to generte nd plot dt» x=[0:0.:0]'; %define the rnge of resistors to use» oc=-6*x./(x4); %the formul for oc. Notice "./"» th=4*x./(4x); %formul for Theenin resistnce.» plot(x,oc,'o', x,th,'md')» title('ung MATLAB'), %proper grphing tools» grid, xlel('x(khm)'), ylel('olts/khms')» legend('oc[]','th[khm]') 58

30 A ME GENEAL EW F EENN EEM LNEA CCUT My contin independent nd dependent sources with their controlling riles AT A i UUAL NTEETATN LNEA CCUT with ALL independent sources set to zero AT A i NTEETATN ALE EEN WHEN E AE ELEMENT NCLUDE NDUCT AND CAACT MAMUM WE TANFE Courtesy of M.J. enrdson From reamp (oltge ) To spekers The simplest model for speker is resistnce EAKE MDEL BAC MDEL F E ANALY F WE TANFE 60

31 MAMUM WE TANFE L UCE L (LAD) Techniclly we need to erify tht it is indeed mximum The lue of the mximum power tht cn e trnsferred is - L (mx) 4 L L ; L L L L L For eery choice of L we he different power. How do we find the mximum lue? Consider L s function of L nd find the mximum of such function d d L L L L 3 L L 4 L L et the deritie to zero to find extreme points. For this cse we need to set to zero the numertor. 0 L L * L The mximum power trnsfer theorem The lod tht mximizes the power trnsfer for circuit is equl to the Theenin equilent resistnce of the circuit. NLY N CAE WE NEED T CMUTE E EENN LTAGE 6 LEANNG EAMLE DETEMNE L F MAMUM WE TANFE We need to find the Theenin resistnce t -. The circuit contins only independent sources... 4k 3k 6k k,6k 6k esistnce for mximum power trnsfer M f we MUT find the lue of the power tht cn e trnsferred EN we need the Theenin oltge!!! loop: ma loop:3 k( ) 6k 3 0 3[ ] [ ma] 9k 3 3 KL : 4k * 6k * 0[ ] 00[ ] 5 [ mw ] 4 M 4*6k 6 6

32 LEANNG EAMLE DETEMNE c d This is mixed sources prolem Now the short circuit current L AND MAMUM. Find the Theenin equilent t -. ememer tht for mximum power trnsfer L M 4...And it is simpler if we do Theenin t c - d nd ccount for the 4k t the end loop: 4mA ' loop : k k 4k( ) 0 ' Controlling rile: " 0 WE 4mA 8[ ] C 4mA TANFEED k ememer now where the prtition ws mde M L 6k 8 8 [ mw ] [ ] 4*6 3 mw 63 LEANNG EAMLE EAMNE WE, UTUT LTAGE AND CUENT A FUNCTN F ETANCE N N N UT N UT N 64

33 LEANNG BY ALCATN FND E EENN EQUALENT F E UNKNWN ELEMENT UNG A ET AND A LTMETE EQUALENT F ELEMENTUCE.4 (mesured in open circuit).6 0.8mA.4 k MEAUED AC TET ET 65 DEGN EAMLE ossile Circuit mplement the fine/corse djustment 0 TUNE CAE FNE um of terms suggests superposition gins less thn one suggest oltge diider Circuits for superposition TUNE C TUNE F ( ) ( ) 0 CAE NFNTE BLE LUTN. UE E CTEA LU ENGNEENG JUDGMENT, e. g., k (resonle choice) 900, 9k FNE DEGN EQUATN EQ AND EE UNKNWN! 66

34 DEGN EAMLE DEGN AN ATTENUAT AD DEGN EQUATN N UT 50 UT 0 N L ( ) 50 ( ) 50 UT L Dependent Eqs! L UT L UT UT 0 LNG E EQUATN YELD 0.83, Anlysis of olution requires specil, high ccurcy resistors smll resistnce my imply lrge power dissipted my require lrge power rting to oid heting L 67 DEGN EAMLE roposed solution DEGN A CCUT T EALZE E EQUATN [ in olts, in ma] ANALY F E EQUEMENT sum of oltge nd current gins lrger thn one inerting ANALY F ED CCUT 0 (infinite gin) A: 0 ANALY F LUTN k is stndrd resistor 667 is k k uses stndrd components! 000 E MED superposition Norton (see ook)

35 DEGN EAMLE F 00 F 00 F CFM Design of sensor F UE A EE ET W EACH FAN T ENE CUENT DE AN NDCATN F TTAL AFLW CNTANT LTAGE D N ENNG ET CANNT ECEED % F NMNAL 4 FAN LTAGE = 50CFM F E NDCAT Design of ndictor Adder inerter nerter ENE 35.mA F 0.00F 5 6 F ENE ENE F CFM CFM ENE F 0.00 F F ENE CFM CFM ENE 4.96 DEGN EQUATN! 0.W ENE F ENE 3 69 LEANNG BY DEGN CUENT ELAD EN tt 50 5 KCL@ : k A 50k 0 tt NT MUT G HGH WHEN CUENT ECEED 9A A 50( ) 50 sense DEGN EQUEMENT tt 9A A 6 50 sense sense 9( A)

36 LEANNG BY DEGN DEGN EQUATN 0 (0.5) 5 (.0) ref ref 9 5 ref 9 CHE 0k DEGN EQUEMENT NUT UTUT DETEMNE ref,, - ref 0 ref GENEATE ref AND LATE LTAGE DDE Anlyzing circuit using superposition 7