Circuit Reduction Techniques


 Lorraine Ferguson
 1 years ago
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1 Crcut Reducton Technques Comnaton of KVLs, KCLs, and characterstcs equatons result n a set of lnear equatons for the crcut arales. Whle the aoe set of equaton s complete and contans all necessary nformaton, een small crcuts requre a large numer of smultaneous equatons to e soled as seen preously. We learned that we can use characterstcs equatons when we are markng the crcut arales and reduce the numer of equatons to e soled to only numer of KCLs and KVLs. We wll learn later two methods, nodeoltage and meshcurrent, whch reduce the numer of equatons to e soled further to ether numer of KCLs or numer of KVLs. Ths s the est we can do n ths drecton. As t s easer to sole smaller sets of equatons (e.g., t s easer to sole two sets of two equatons n two unknown as compared to a set of 4 equatons n 4 unknowns), one can reak up the crcut nto smaller peces and sole each nddually and assemle ack the whole crcut. We can use ths prncple to comne crcut elements and make a much smaller crcut. These technques are descred elow. Recall that we are usng lumped crcut elements,.e., crcut elements communcate to outsde world and other crcut elements only through and. Conersely, the outsde world (the rest of the crcut) communcate wth the crcut element through and. Ths means, for example, that a resstor n a crcut s ewed y the rest of the crcut as a lack ox wth an characterstcs of = R. The rest of the crcut does not know what s nsde the ox. In fact, we can replace the resstor wth any lack ox (contanng whateer) wth the same characterstcs of = R and the rest of the crcut ehaes exactly the same. Alternately, f a lack ox contanng many crcut elements s attached to a crcut and has an characterstcs of =5, we can replace ths lack ox wth a 5 Ω resstor wth no change n the crcut ehaor. Rest of the crcut Rest of the crcut Sucrcut A ox contann some elements Equalent Sucrcut Ths oseraton allows the crcut to e dded nto two or many parts and each soled ndependently. We defne a ox contanng seeral element a sucrcut or a dece. T he aoe fgure shows a twotermnal dece or sucrcut. Note that n many crcut theory text ook (ncludng our textook) crcut and sucrcut are used nterchangealy. MAE140 Notes, Wnter 001 1
2 Sucrcuts play an mportant role n lnear crcut theory. Theenn theorem states that any sucrcut contanng lnear crcut element has an characterstcs of A B = C (where A, B, andc are constants) and t can e reduced to a sucrcut contanng at most two lnear crcut elements (Theenn and Norton Forms). We wll dscuss Theenn Theorem later. Below, we explore sucrcuts n the context of elements that are n seres or n parallel. In each case, we fnd the characterstcs of the sucrcut and use that to fnd the equalent element. Elements n Seres Two elements are called n seres f they and only they share a common node. Alternately, seresconnected elements carry the same current. a A Shared Node KCL: a= B Two Resstors n Seres KCL: = 1 = KVL: 1 =0 : 1 = R 1 1 = R 1 = R = R 1 R1 1 R Susttutng from characterstcs equatons n KVL, we get = R 1 R =(R 1 R ) = R eq R eq = R 1 R So, a sucrcut contanng two resstors n seres has an characterstcs of the form = R eq and s equalent to a resstor, R eq = R 1 R. The aoe can e easly extended: k resstors n seres are equalent to one resstor wth R eq =Σ k j=1 R j. A Resstor n Seres wth a Current Source R KCL: = 1 = s KVL: 1 =0 = R 1 s 1 1 s The characterstcs of ICS states that ts current s s, ndependent of ts oltage ( ). Aoe equatons show that the characterstcs of sucrcut s = s and s ndependent MAE140 Notes, Wnter
3 of oltage = 1 (as alue of 1 can e anythng). The equalent sucrcut s an ndependent current source wth strength s. A Resstor n Seres wth a Voltage Source R Ths s the Theenn form and cannot e reduced further. s A Voltage Source n Seres wth a Voltage Source s1 s KVL: s1 s =0 = s1 s The characterstcs of IVS states that ther oltages are s1 and s, respectely, ndependent of ther currents. Aoe equaton shows that the characterstcs of sucrcut s = s1 s and s ndependent of current. The equalent sucrcut s an ndependent oltage source wth strength s1 s (algerac sum of two). Note that t s prudent to use KVL to fnd the strngeth of the equalent source. A Voltage Source n Seres wth a Current Source KCL: = s1 = KVL: 1 s =0 = 1 s s1 1 s The characterstcs of ICS states that ts current s s1, ndependent of ts ( 1 ). Aoe equatons show that the characterstcs of sucrcut s = s1 and s ndependent of oltage = 1 s (as alue of 1 can e anythng). The equalent sucrcut s an ndependent current source wth strength s1. Note: A current source n seres wth any element reduced to a current source. Seres element all hae the same current and the current source requres the current through to e equal to ts strength. MAE140 Notes, Wnter
4 A Current Source n Seres wth a Current Source KCL: = s1 = s s1 s A current source n seres wth any element reduces to a current source. Howeer, n the case of two current sources n seres, KCL requres s1 = s. Thus, two current sources can e attached n seres only f s1 = s. If so, the equalent sucrcut s an ndependent current source wth strength s = s1 = s. Ths constrant n allowale crcut confguraton arses ecause we are dealng wth dealzed crcut elements. We wll dscuss real sources later and wll see that two real or practcal current sources can e attached n seres een f they hae dfferent strength (although the result may e a lot of sparks and two urnt out current sources!) Elements n Parallel Two element are called n parallel f they share oth nodes. Alternately, parallelconnected elements hae the same oltage. B a A Share Both Nodes KVL: a = Two Resstors n Parallel KCL: 1 =0 KVL: = 1 = 1 R 1 1 R : 1 = R 1 1 = R Susttutng from characterstcs equatons n KCL, and usng = 1 =,weget = R 1 R = ( 1 1 ) R1 R = R eq 1 R eq = 1 R 1 1 R So, a sucrcut contanng two resstors n parallel has an characterstcs of the form = R eq and s equalent to a resstor, 1/R eq =1/R 1 1/R. The aoe can e easly extended: k resstors n seres are equalent to one resstor wth 1/R eq =Σ k j=11/r j. MAE140 Notes, Wnter
5 A Resstor n Parallel wth a Current Source Ths s the Norton form and cannot e reduced further. We wll show later that Norton and Theenn forms are equalent. 1 R 1 1 s A Resstor n Parallel wth a Voltage Source KCL: 1 =0 KVL: = 1 = s 1 s R 1 1 The characterstcs of IVS states that ts oltage s s, ndependent of ts current,. Aoe equatons show that the characterstcs of sucrcut s = s ndependent of current. The equalent sucrcut s an ndependent oltage source wth strength s. A Current Source n Parallel wth a Current Source KCL: s1 s =0 = s1 s 1 s1 s The characterstcs of ICSs state that ther currents are s1 and s, respectely, ndependent of ther oltage,. Aoe equatons show that the characterstcs of sucrcut s = s1 s ndependent of alue of oltage. The equalent sucrcut s an ndependent oltage source wth strength s = s1 s. A Current Source n Parallel wth a Voltage Source KCL: 1 =0 KVL: = s = 1 s s The characterstcs of IVS states that ts oltage s s, ndependent of ts current, 1. Aoe equatons show that the characterstcs of sucrcut s = s and s ndependent of current = 1 s (as alue of 1 can e anythng). The equalent sucrcut s an ndependent oltage source wth strength s. Note: A oltage source n parallel wth any element reduces to a oltage source. Parallel elements all hae the same oltage and the oltage source requres ts oltage across to e equal to ts strength. MAE140 Notes, Wnter
6 A Voltage Source n Parallel wth a Voltage Source KVL: = s1 = s 1 s1 s A oltage source n parallel wth any element reduces to a oltage source. Howeer, n the case of two oltage sources n parallel, KVL requres s1 = s. Thus, two oltage sources can e attached n parallel only f s1 = s. If so, the equalent sucrcut s an ndependent oltage source wth strength s = s1 = s. Ths constrant n allowale crcut confguraton arses ecause we are dealng wt dealzed crcut element. We wll dscuss real sources later. Summary of Twotermnal Equalent Sucrcuts Seres Parallel R 1 and R Resstor (R eq = R 1 R ) Resstor (1/R eq =1/R 1 1/R ) R and IVS ( s ) Theenn Form IVS ( s ) R and ICS ( s ) ICS ( s ) Norton Form IVS ( s1 )andivs( s ) IVS ( s = s1 s ) IVS ( s = s1 = s ) IVS ( s1 )andics( s ) ICS ( s ) IVS ( s ) ICS ( s1 )andics( s ) ICS ( s = s1 = s ) ICS ( s = s1 s ) Theenn and Norton forms are equalent. Connecton s allowed only f s1 = s or s1 = s. Theenn and Norton Forms and Source Transformaton Theenn and Norton forms are equalent. One can replace one wth the other. Ths s called source transformaton and s helpful n rearrangng other elements n the crcut and sometme arrng at more element eng n seres and parallel. Watch out for polartes of the IVS and the ICS and follow the dagrams on the left! RT N RN T Theenn Form Norton Form Equalent f R =R T N and = T N R T MAE140 Notes, Wnter
7 Three or More Element n Seres Poston of elements n seres can e nterchanged wthout any effect on the crcut. The reason s that KCL ensures that all elements carry the same current, the oltage across each element s unquely set y ts characterstcs and alue of, and the oltage oer all of the elements, = a c = c a s the same f we nterchange the poston of the elements. A B C a c C A B c a To smplfy crcuts wth 3 or more elements n seres: 1. Check f there s a current source. If so, all seres elements can e replaced wth a current source of the same strength. Note that f there are more than one current source, you should check for llegal connectons of two current sources n seres.. Rearrange elements and group resstors and oltage sources together. Replace resstors wth a resstor, R eq =ΣR j and oltage sources wth a oltage source wth strength s =Σ sj. It s prudent to use KVL to ensure that you get the correct algerac sum of Σ sj. Example: Four elements are n seres. One current source exsts. The equalent element s a current source. 10 Ω 5 V 0 Ω A A Example: Four elements are n seres. No current source exsts. Group resstors and IVSs together. 10 Ω 10 Ω 5 V 0 Ω V 0 Ω 5 V V s R eq =100=30Ω KVL: 5 s =0 s =3V 30 Ω 3 V MAE140 Notes, Wnter
8 Three or More Element n Parallel A a c B C B C A c a Poston of elements n parallel can e nterchanged wthout any effect on the crcut ecause the deal wres can e stretched wthout any mpact on the crcut. (Imagne lftng element A out of the paper, mong t ehnd element C and puttng t ack on the paper.) To smplfy crcuts wth 3 or more elements n parallel: 1. Check f there s a oltage source. If so, all parallel elements can e replaced wth a oltage source of the same strength. Note that f there are more than one oltage source, you should check for llegal connectons of two oltages sources n parallel.. Rearrange elements and group resstors and current sources together. Replace resstors wth a resstor, 1/R eq =Σ 1/R j and current sources wth a current source wth strength s =Σ sj. It s prudent to use KCL to ensure that you get the correct algerac sum of Σ sj. Applcaton of Crcut Reducton Technques Crcut reducton technques are powerful methods to smplfy the crcut as they reduce the numer of elements (and therefore, the numer of equatons to e soled smultaneously). Howeer, when seeral crcut element are comned, the crcut arales assocated wth those elements are lost n the process of transformaton. In prncple, one should sole the smplfes crcut and fnd the remanng crcut arales. Then, one should go ack to the orgnal crcut to fnd the lost crcut arales. Two examples elow show ths process. Because of ths extra step, crcut reducton technques do not always lead to smpler crcuts and they should e used judcously. MAE140 Notes, Wnter
9 Example: Fnd f R 1 = R =1kΩand s =5V. Resstors R 1 and R are n seres, so we can replace them wth an equalent resstor R eq = R 1 R =kω. The resultant crcut s shown. From the pont of ew of the rest of the crcut (.e., IVS) nothng has changed and so the current remans the same. Howeer, the crcut arales n the reduced part, 1 and do not appear n the reduced crcut. To fnd, we frst sole the reduced crcut to fnd : KVL: s R eq =0 = s = 5 =.5 ma R eq, 000 s R1 1 R s R1 R We then use the alue of n the orgnal crcut to fnd : = R =1, =.5 V Example: Fnd 0 and 1. In ths crcut, we hae two sources n parallel and two sources n seres. The prolem unknowns are oltage and current of the sources that are n parallel. So, t s not prudent n the frst step to comne them. Rather, we comne the two sources n seres. A current source n seres wth a oltage source reduces to a current source wth the same strength as s shown. 1 can now e found y KCL and 0 y KVL: KCL: = 0 1 = 5 A A 100 V 10 A 100 V V 15 A 15 A KVL: =0 0 = 100 V Note that f we had reduced the two sources n parallel n the orgnal crcut, we would hae reached a crcut whch was tral and not helpful n fndng 0 and 1 (try t!). So, crcut reducton should e used judcously. Example elow shows a crcut that can e soled wthout any crcut reducton and, n fact, crcut reducton makes the soluton more dffcult. Example: In the crcut aoe, fnd 0 and. Both 0 and an e found y KVL: KVL: =0 0 = 100 V KVL: =0 = 140 V MAE140 Notes, Wnter 001 0
10 Some Practcal Resste Crcuts Voltage Dder: The two resstors can e replaced y an equalent resstor, R eq = R 1 R.Thus: s = R eq = s R eq 1 = R 1 = R 1 R eq s = R = R R eq s s R1 R 1 Also, 1 = R 1 R Ths crcut s called a oltage dder as the two resstors dde the oltage of the IVS etween them proportonal to ther alues. Ths crcut can e extended y addng more resstors to the crcut and get more reference oltages. Ths crcut s used extensely n electronc crcuts. The asc reason s that power supples are ulky and/or expense. Typcally, one power supply wth one oltage s proded. On the other hand, more than one oltage may e needed for the crcut to operate properly. Example: A attery operated rado has a 9 V attery. Part of rado crcuts requre a 6 V supply. Desgn a oltage dder crcut to supply 6 V oltage to these crcuts. The desred crcut s the oltage dder crcut aoe wth s =9Vand = 6 V. Then, = R R eq s 6= R R 1 R 9 R R 1 R = 6 9 Ths s one equaton n two unknowns and one s free to choose one parameter. For example, choosng R 1 =1kΩwegetR =kω. Voltage dders are affected y the load current drawn from them (see fgure). The oltage dder formula can only e used for the crcut shown on the rght f l (proe t!). We wll dscuss the mpact of the load on oltage dders when we dscuss real sources. s R1 1 L R Load MAE140 Notes, Wnter 001 1
11 Current Dder: The two resstors can e replaced y an equalent resstor, 1/R eq =1/R 1 1/R.Thus: = R eq s R 1 1 R = 1 R 1 1 = R 1 = R eq R 1 s = R = R = R eq R s Also, 1 = R R 1 Ths crcut s called a current dder as the two resstors dde the current of the ICS etween them (nersely proportonal to ther alues). Ths crcut can e extended y addng more resstors to the crcut and get more reference currents. Wheatstone Brdge A typcal Ohmmeter measures the resstance of a resstor y usng the Ohm s Law. It apples a known oltage of s across the resstor, measures the current flowng through the resstor, and ts dal are set to conert the measured alue of current nto the alue of resstance y usng R = s / measured. (Ths s why one cannot measure the alue of a resstor whle t s attached n a crcut, Ohmmeter works only f the resstor s not attached to anythng ut the meter.) A typcal dgtal multmeter measure resstance wthn an accuracy of aout 1%. In some cases, hgher accuracy s needed. Resstor rdges are used for ths purpose and they are made of two oltage dder crcuts put n parallel wth each other. The rdges operate ased on the fact that whle t s dffcult to measure the dfference etween 1 and 1.01 V or 1 and 1.01 A dstnctly (they are only 1% apart and wthn the accuracy of meter), t s easy to measure 0.01 V. A most wdely used rdge s the Wheatstone rdge shown. It conssts of two oltage dder crcuts wth a oltmeter measurng the oltage etween ponts A and R1 RA B (denoted y m ). We note from oltage dder formulas: A s m B = R s R 1 R B = R B s R A R B R B RB MAE140 Notes, Wnter 001
12 ( ) R R B KVL: m B =0 m = B = s R 1 R R A R B A Wheatstone rdge s used n two modes. To measure the alue of a resstor accurately, and to montor to change n a resstance accurately. Measurng resstance accurately: Suppose the unknown resstance s R B. T wo known and accurate resstors R 1 and R A are chosen. An accurate ut arale resstor s used for R. Typcally, a resstor ox or a decade ox s used. The dals on the ox swtch some accurate resstor n parallel or n seres to get the desred resstor alue accurately. The rdge s setup and s powered up. The arale resstor R s ared untl the meter read zero oltage for m. Then: ( ) R R B m = s =0 R 1 R R A R B R R B = R 1 R R 1 R R A R B R R 1 = R A R R B R B = R A R R 1 = R A R B R B Note that we do not need to know alue of s to fnd R B.Thsway,R B s measure wthn the accuracy of resstors,r 1, R,andR A. Measurng resstance changes accurately: In certan sensors, the resstance of the sensor changes proportonal to external forces or condtons. For example, a stran gauge measures the elongaton (stran) of a sold materal caused y appled forces (stress). A typcal stran gauge conssts of a thn flm of conductng materal deposted on an nsulatng sustrate and onded to a test memer. When the test artcle s under stress, ts dmenson changes (e.g., ts length L changes to L L. The resstance of the stran gauge (the conductng flm) change accordng to R =R G L L where R G s the resstance of gauge when no stress s appled, R s the change n the gauge resstance, and factor of comes from the fact that as the materal s elongated ts cross secton s reduced. Usually, we lke to measure stran alues, L/L, that can e as small as 10 4 (whch means the changes n gauge resstance s of the same order and cannot e measured y a smple ohmmeter). MAE140 Notes, Wnter 001 3
13 Wheatstone rdge s used to measure R n the followng confguraton. The stran gauge, R G s put n place of R. R 1 s replaced y a smlar stran gauge whch s under no stress and s used as a reference. The resstances R A and R B are chosen such that rdge s alanced ( m = 0) when no stress s appled. Followng the equatons for a alanced rdge, we get: R 1 R 1 R G = R B R A R B Typcally, R 1 = R G whch mples R A = R B. When stress s appled to the system, the stran gauge resstance changes to R = R G R and a oltage m appears on the rdge. Then, ( ) R G R m = s R 1 R G R R B R A R B Usng R 1 = R G and R A = R B,weget: ( RG R m = s R G R 1 ) m = s R G R R G R (R G R) R m = s (R G R) R m = s 4R G where n the last equaton, we hae gnored R n the denomnator snce R <<R G.Usng the relatonshp etween R and stran ( L), we get L L = R R G = m s Note that n ths applcaton an accurate oltage source, s, s needed. MAE140 Notes, Wnter 001 4
+ + +   This circuit than can be reduced to a planar circuit
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