Chpter 4: Techniques of Circuit Anlysis Terminology Node-Voltge Method Introduction Dependent Sources Specil Cses Mesh-Current Method Introduction Dependent Sources Specil Cses Comprison of Methods Source Trnsformtions Norton nd Thévenin Equivlent Circuits C4-S1 Chpter 4: Techniques of Circuit Anlysis Mximum Power Trnsfer Superposition Applictions Source Modeling Bipolr Junction Trnsistors Crmer s Rule C4-S 1
Terminology A plnr circuit cn e drwn in plne with no crossing rnches R1 R R1 R V1 R3 R11 R7 R4 R6 V1 R3 R11 R4 R6 R7 C4-S3 Terminology A non-plnr circuit cnnot e redrwn with no overlpping rnches R1 R R16 R0 R1 V1 R18 R19 R7 R3 R11 R4 R6 R17 C4-S4
Terminology R V i i 7 R7 c R3 d R4 e i 8 R8 I1 i 4 i 6 V3 R6 f R5 i 5 g C4-S5 Terminology Nme node essentil node pth rnch essentil rnch loop mesh plnr circuit Definition Point where two or more circuit elements join Node where 3 or more circuit elements join Trce of djoining elements; none included more thn once Pth tht connects two nodes Pth tht connects two essentil nodes without pssing though essentil node Pth with lst node sme s first node Loop tht does not include other loops Circuit tht cn e drwn on plne with no crossing rnches Exmple V -R -R 7 -R 6 R V -R V -R -R 7 -R 6 -R 5 -V 3 V -R -R 7 -R 4 -R 3 C4-S6 3
Simultneous equtions How mny? The numer of unknown currents equls the numer of rnches,, where the current is unknown If n is the numer of nodes, there re n-1 independent equtions from the ppliction of Kirchhoff s current lw Thus, we must pply Kirchhoff s voltge lw to -(n-1) loops or meshes C4-S7 Simultneous equtions How mny? The sme rgument pplies to essentil nodes nd rnches Thus, we cn pply Kirchhoff s current lw to n e -1 nodes nd Kirchhoff s voltge lw to e -(n e -1) loops or meshes. The numer off essentil nodes nd rnches, n e nd e, is lwys less thn or equl to the numer of nodes nd rnches, n nd, respectively C4-S8 4
Disconnected Circuits The previous sttements pertin to connected circuits If circuit hs n nodes, rnches, nd s prts, the current lw cn e pplied n-s times nd the voltge lw -n+s times. Any two seprte prts cn e connected y single conductor. This cuses two nodes to form one node. No current flows in the connection; ny circuit mde up of s prts cn e reduced to connected circuit. C4-S9 Disconnected Circuits R1 V1 R R3 F1 F R4 R5 How mny seprte prts? How mny nodes? How mny independent current equtions? How mny rnches? How mny rnches with unknown current? How mny equtions must e written with the voltge lw Connect the lower node nd repet C4-S10 5
The Node-Voltge Method Strt y lying out the circuit clerly nd mrking the essentil nodes Clculte the numer of voltge equtions needed from n e -1 Select reference node (usully the node with the most rnches) Define the node voltges with respect to the reference voltge A node voltge is voltge rise from the reference node to non-reference node C4-S11 The Node-Voltge Method Generte the node-voltge equtions y writing the current leving ech rnch connected to non-reference node s function of the node voltge nd then sum the current to zero C4-S1 6
The Node-Voltge Method R1 R 1 10Vdc V1 R3 R4 I1 5 10 Adc R1 1 R 1 + + 10Vdc V1 R3 5 v 1 v R4 10 I1 Adc 0 C4-S13 The Node-Voltge Method Consider node 1; the current from the node towrds the source is v1 10 1 The current in the 5Ω resistor is The current in the Ω resistor is v1 5 v v 1 C4-S14 7
The Node-Voltge Method Thus, t node 1 At node v1 10 v1 v1 v + + = 0 1 5 v v1 v + = 0 10 Solving, we find v 1 = 9.09 V nd v = 10.91 V C4-S15 Dependent Sources If the circuit contins dependent source, the node-voltge equtions must e supplemented with the constrint eqution imposed y the source 5 V1 0 10 H1 + - C4-S16 8
Dependent Sources 5 V1 v 0 1 v 10 H1 + - 0 There re 3 essentil nodes; we need voltge equtions Pick lower node for reference Sum the currents in the two nodes C4-S17 Dependent Sources 5 V1 H1 i φ v 0 10 1 v 8i φ + - 0 v1 0 v1 v1 v + + = 0 0 5 v1 v v 8 v v1 v + + 5 = 0 5 10 v1 = 16 nd v = 10 C4-S18 9
Specil Cse (1) If voltge source is the only element etween etween two essentil nodes, the node-voltge method is simplified Consider 10 V1=100 5 v 1 v 50 I1 5Adc 0 C4-S19 Specil Cse (1) 10 V1=100 5 v 1 v 50 I1 5Adc 0 At node v v v 10 50 1 + 5 But, v 1 = 100 C4-S0 10
Specil Cse (1) 10 V1=100 5 v 1 v 50 I1 5Adc 0 Therefore v 100 v + 5 = 0 10 50 v = 15 C4-S1 Specil Cse () Consider the circuit 5 + - H H V1=50 40 i φ 50 v 1 v v 3 i 100 I1 4Adc 0 There re 4 essentil nodes; we need three equtions C4-S 11
+ - Specil Cse () Two essentil nodes connected y voltge source nd two others connected y current-controlled voltge source. Therefore we need only one eqution nd one constrint 5 H H V1=50 40 50 v 1 v v 3 i 100 I1 4Adc 0 C4-S3 Super nodes In the previous exmple, we cn write the expression tht elimintes the current-controlled voltge source y considering nodes nd 3 to e single node ( supernode) 5 V1=50 v 40 50 1 v v 3 100 I1 4Adc 0 C4-S4 1
Specil Cse () 5 + - H H V1=50 40 50 v 1 v v 3 i 100 I1 4Adc 0 v v1 v + + i = 0 5 50 v3 i 4= 0 100 v = v 3 v 50 + 10 5 C4-S5 Super nodes 5 V1=50 v 40 50 1 v v 3 100 I1 4Adc 0 Applying Kirchhoff s lw to the supernode v v v v 5 50 100 1 3 + + 4= 0 Still hve the sme constrin eqution C4-S6 13
A Trnsistor Amplifier We solved the trnsistor mplifier y the rnch current method in exmple.11 (p 48). There were 6 equtions nd 6 unknowns. We re-exmine it vi the node-voltge method There re 6 known vriles: R 1, R, R C, R E, V CC nd V G Ojective is to compute i B C4-S7 A Trnsistor Amplifier Rc R1 i F1 Vcc F Vg c R Re There re 4 essentil nodes; nodes nd d nd nodes nd c re connected y n independent voltge source d C4-S8 14
A Trnsistor Amplifier Rc R1 i F1 Vcc F Vg c R Re Therefore, we need one eqution for one unknown node voltge ecuse (n e -1)-=1 Use d s reference node d C4-S9 A Trnsistor Amplifier Rc R1 i F1 Vcc F Vg c R Re Comine nodes nd c into supernode d C4-S30 15
A Trnsistor Amplifier Rc R1 i F1 Vcc F Vg c R Re 1 d v v vcc vc + + βi = 0 R R R e C4-S31 But A Trnsistor Amplifier ( ) v = i +βi R c e v = v v c We cn now solve for v to find v 0 ( 1+ β ) e + ( 1 β ) ( ) vccr R vrr 0 1 = RR + + R R + R 1 e 1 We reduced the prolem from one of deling with 6 equtions with 6 unknowns to one of deling with 3 equtions with 3 unknowns C4-S3 16
The Mesh-Current Method The mesh-current method llows us to descrie the circuit in terms of e -(n e -1) equtions The method pplies only to plnr circuits We use meshes ( mesh is loop with no other loops inside it) C4-S33 The Mesh-Current Method Recll the circuit R1 R Vs R6 R8 R3 R7 R5 R4 There re 7 essentil rnches nd 4 essentil nodes We need 7-(4-1) = 4 mesh-current equtions C4-S34 17
The Mesh-Current Method R1 R i 4 Vs i 1 i R6 R8 R3 R7 R5 R4 i 3 The mesh current follows the perimeter of the mesh Note tht the currents move clockwise The mesh currents utomticlly stisfy Kirchhoff s current lw ecuse ech flows into nd ech node the mesh psses through The mesh current is NOT NECESSARILY the rnch current! Mesh currents my sometimes not e mesurle C4-S35 The Mesh-Current Method Consider the simple circuit R1 R V1 i 1 i i 3 R3 V For this circuit, e =3 nd n e = There is only one independent current eqution; therefore we need two voltge equtions C4-S36 18
The Mesh-Current Method R1 R V1 i 1 i i 3 R3 V Apply Kirchhoff s current lw to the upper node nd the voltge lw round the two meshes i = i + i 1 3 v = ir + ir 1 1 1 3 3 v = ir ir 3 3 C4-S37 The Mesh-Current Method Solving, we find v = i ( R + R ) ir 1 1 1 3 3 ( ) v = ir + i R + R 1 3 3 Consider sme circuit using mesh-currents R1 R V1 i R3 i V C4-S38 19
The Mesh-Current Method R1 R V1 i R3 i V ( ) ( ) v = ir + i i R 1 1 3 v = i i R + ir 3 Collecting coefficients ( ) ( ) v = i R + R ir 1 1 3 v = ir + i R + R 3 3 C4-S39 The Mesh-Current Method Note tht the results re identicl to the previous method, ut tht we only wrote down two equtions insted of three We lso note tht the rh\nch currents re i i 1 3 = i = i i = i i C4-S40 0
Dependent Sources If the circuit contins dependent source, we must supplement the mesh-current equtions with the constrint eqution for the source Exmple R1 i R R3 V1 i 1 i R4 i 3 H1 + - 15i C4-S41 Dependent Sources R1 i R R3 V1 i 1 i R4 i 3 H1 + - 15i There re six rnches with unknown currents; there re four nodes; -(n-1)=6-(4-1)=3 We need three mesh-current equtions C4-S4 1
Dependent Sources R1 i R R3 V1 i 1 i R4 i 3 H1 + - 15i They re ( 1 ) ( ) ( 1) 1 3( 3) ( ) v = i R + R i i 1 1 4 1 3 0 = R i i + Ri + R i i 0 = R ( i i ) + R i i + 15i i = i i 4 3 1 3 3 1 3 C4-S43 Specil Cses When the rnch contins current source, dditionl mnipultions re required 10 i 3 100 V i 5 A i c 50 V 6 4 e =5 (not 6!); n e =4; we need meshcurrent equtions C4-S44
Specil Cses 10 i 3 100 V i 5 A i c 50 V 6 4 The presence of the known current reduces the numer of equtions y 1 i i = 5 But, we do not know the voltge, v, cross the current source c C4-S45 Specil Cses 10 i 3 100 V i 5 A i c 50 V 6 4 Thus, we write for meshes nd c ( ) 100= 3 i i + v+ 6i ( ) 50= 4i v+ i i c c C4-S46 3
Specil Cses 10 i 3 100 V i 5 A i c 50 V 6 4 Adding 50= 9i 5i + 6i c C4-S47 Supermeshes We cn pproch this prolem in different wy 10 i 3 100 V i i c 50 V 6 Supermesh 4 C4-S48 4
Supermeshes 10 i 3 100 V i i c 50 V 6 4 Supermesh For the supermesh ( ) ( ) 100+ 3 i i + i i + 50+ 4i + 6i = 0 or 50= 9i 5i + 6i c c c C4-S49 A Trnsistor Amplifier Reconsider our simple mplifier circuit Rc R1 i F1 i c Vcc F Vg R i Re C4-S50 5
A Trnsistor Amplifier Rc R1 i F1 i c Vcc F Vg R i Re There re 4 essentil nodes nd 5 essentil rnches; we need 5-(4-1)= mesh-current equtions C4-S51 A Trnsistor Amplifier Rc R1 i F1 i c Vcc F Vg R i Re There is known current tht forces reltionship etween currents i nd i c Thus, only two unknown mesh currents C4-S5 6
A Trnsistor Amplifier Introduce supermesh s follows Rc R1 i ic Vcc Vg R i Re C4-S53 A Trnsistor Amplifier Rc R1 i ic Vcc Vg R i Re R i 1 + v R i cc + R + v 0 e ( ic i ) v0 = 0 R ( i i ) = 0 + E c C4-S54 7
A Trnsistor Amplifier Rc R1 i ic Vcc i B Vg R i Re βi i B B = i = i i c i C4-S55 A Trnsistor Amplifier Solving, we find i i v0r = R R + 1 1 v0r = R R + vccr vcc( 1+ β ) RE ( 1+ β ) RE ( R1 + R ) 1 ( 1+ β ) REvcc ( 1+ β ) R ( R + R ) E 1 C4-S56 8
Comprison of Methods The dvntge of oth the node-voltge method nd the mesh-current method is the reduction of the numer of simultneous equtions to e solved You must e very systemtic to orgnize nd write the equtions Which method is preferred? There is no cler-cut nswer. C4-S57 Comprison of Methods To decide which technique to use, sk the following questions: Does one of the methods led to fewer simultneous equtions to solve? Does the circuit contin supernodes? If so, using the node-voltge method reduces the numer of simultneous equtions. Does the circuit contin supermeshes? If so, using the mesh-current method reduces the numer of simultneous equtions. Will solving some portion of the circuit give the requested solution? IF so, which method is most efficient for solving just the pertinent portion of the circuit. C4-S58 9
Comprison of Methods Alterntively, one my look t this from two points of view: The nture of the network The informtion required C4-S59 Nture of the Network Networks tht contin mny seriesconnected elements, voltge sources, or supermeshes re more suitle for mesh nlysis Networks with prllel-connected elements, current sources, or supernodes re more suitle for node nlysis A circuit with fewer nodes thn meshes is etter nlyzed with node nlysis C4-S60 30
Nture of the Network A circuit with fewer meshes thn nodes is etter nlyzed with mesh nlysis Mesh nlysis is the only method to use in nlyzing trnsistor circuits Mesh nlysis is often not esy to use in nlyzing op mp circuits (there is no direct wy to otin the voltge cross n op mp) For non-plnr circuits, node nlysis is the only wy-mesh nlysis pplies only to plnr circuits C4-S61 Informtion Required If node voltges re required, it my e expedient to pply node nlysis If mesh or rnch currents re required, it my e expedient to pply mesh nlysis C4-S6 31
Exmple Find the power dissipted in the 300 Ω resistor in the following circuit 300 i 1 150 100 50 500 56 V 00 50 i1 + - 400 18 V There re five meshes, nd thus five mesh-current equtions plus the constrint eqution [8-(4-1)=5] C4-S63 Exmple 300 i 1 150 100 50 500 c 56 V 00 50 i1 + - 400 18 V There re 4 essentil nodes nd thus only 3 node-voltge equtions We cn crete supernode out of the dependent voltge source, thus reducing the prolem to two node-voltge equtions nd constrint eqution C4-S64 3
Exmple 300 i 1 150 100 50 500 c 56 V 00 50 i1 + - 400 18 V There re two logicl choices for the reference: either t or on the seline (where there re most components connected) C4-S65 Exmple 300 i 1 150 100 50 500 c 56 V 00 50 i1 + - 400 18 V Using the seline s the reference v v 56 v v v vc + + + = 0 00 150 100 300 vc vc + 18 vc v vc v + + + = 0 400 500 50 300 v v v v v = i = = c c 50 1 50 300 6 C4-S66 33
Source Trnsformtions A source trnsformtion llows voltge source in series with resistor to e replced y current source in prllel with the sme resistor R Vs Is R C4-S67 Source Trnsformtions Wht is the reltionship etween v s nd i s so tht the trnsformtion is trnsprent to lod resistor R L connected etween the terminls nd? Clerly, so long s the current tht psses through R L is the sme, the two circuits re equivlent. C4-S68 34
Source Trnsformtions For the voltge source i L vs = R+ R For the current source (using the current divider) i L L R = i R+ R L s Thus i s vs = R C4-S69 Source Trnsformtions Wht is the effect of resistor in prllel with the voltge source or resistor in series with the current source? Nothing! Neither ffect the current through the lod resistor nd thus, my e ignored! C4-S70 35
Source Trnsformtions R R Vs Rp Vs Rs Is R Is R C4-S71 Exmple 4 6 5 6 V 30 0 40 V 10 Find the power generted/sored y the 6 V source. The 40 V source is uninteresting replce y 5Ω resistor in prllel with n 8 A current source. Replce the resulting prllel 0Ω nd 5Ω resistors y 4Ω resistor C4-S7 36
Exmple 4 6 6 V 30 4 8 A 10 Replce the 4Ω resistor nd the current source with 4Ω resistor in series with 3 V source. Recognize tht it does not mtter where the 3 V source is in the chin of resistors Replce the 3 V source nd 0Ω resistor with 1.6 A current source nd prllel 0Ω resistor C4-S73 Exmple 4 6 V 30 0 1.6A Replce the prllel resistors with 1Ω resistor Replce the 1.6 A current source nd 1Ω resistor with series resistor nd 1 x 1.6 = 19. V source C4-S74 37
Exmple 4 1 6 V i 19. V There is voltge rise from the 6 V source to the 19. V source; the current thus flows from the 19. V source to the 6 V source nd is 19. 6 i = = 0.85 A 4+ 1 C4-S75 Exmple The 6 V source is thus sink nd dissiptes 6 x 0.85 = 4.95 W C4-S76 38
Thévenin nd Norton Equivlents Sometimes we re interested only in wht hppens t pir of terminls (e.g, n pplince plugged into the wll). We wnt to know only how the voltge nd current supplied t the terminls s we vry the lod. Thévenin nd Norton equivlents re circuit techniques tht focus on the terminl ehvior. C4-S77 Thévenin nd Norton Equivlents A Thevenin equivlent circuit is n independent voltge source V Th in series with resistor R Th RTh VTh If the lod resistnce is infinitely lrge, then the open circuit voltge is V Th Thus, we clculte V Th y clculting the open circuit voltge in the originl circuit C4-S78 39
Thévenin nd Norton Equivlents RTh VTh If the ld resistnce is zero, we hve short circuit. This must e the sme s the short circuit current in the ctul circuit. Thus, R Th v = i Th sc C4-S79 Exmple 5 4 5 0 + 3 v1 To compute the Thévenin equivlent voltge, we must compute v 1 Using the lower node s reference, we solve single node-voltge eqution v1 5 v1 + 3= 0 5 0 v = 3 1 - C4-S80 40
Exmple 5 4 + 5 0 3 v i sc To compute the Thévenin equivlent resistnce, we must compute i sc This is esy if we cn find v Write the node-voltge eqution v 5 v v + 3+ = 0 5 0 4 v = 16 - C4-S81 Exmple Thus nd R Th 16 i sc = = 4 4 vth 3 = = = 8 i 4 sc 8 3 C4-S8 41
Norton Equivlent Circuit The Norton equivlent circuit is found y performing source trnsformtion on the Thévenin equivlent The Norton current is just i sc in prllel with the the Thévenin equivlent resistnce Note tht you cn often mke effective use of source trnsformtions to derive either the Norton or Thévenin equivlent C4-S83 Exmple Find the Thévenin equivlent for 000 i + 5 + - + - 3v 0i v - 5 Note tht the current etween the two meshes is zero! The Thévenin voltge is the potentil drop cross the 5Ω resistor C4-S84 4
Exmple 000 i + 5 + - + - 3v 0i v - 5 ( )( ) v = v = 0i 5 = 500i Th 5 3v 5 3v i = = 000 000 v = 5 Th Th C4-S85 Exmple 000 i 5 + - + - 3v 0i 5 The control voltge goes to zero when the short is pplied All the current now ppers in the short isc = 0i C4-S86 43
Exmple 000 i 5 + - + - 3v 0i 5 The current controlling the dependent current source is 5 i = =.5 ma 000 i = 0.5 = 50 ma sc ( ) C4-S87 Exmple The Thévenin resistnce is R Th vth 5 = = = 100 i 0.050 sc The Thévenin equivlent is 100 5 C4-S88 44
Thévenin Equivlent There re two other techniques for deriving Thévenin equivlent If the circuit contins only independent sources, dectivte the sources nd clculte the resistnce s seen y looking into the network t the designted terminls To dectivte source, short circuit voltge sources nd open circuit current sources C4-S89 Exmple Consider the previous exmple 5 4 5 0 + 3 v1-5 4 0 C4-S90 45
Exmple The Thévenin resistnce is R Th 1 = R = 4+ = 4+ 4= 8 1 1 + 0 5 The Thévenin voltge is computed s efore C4-S91 Thévenin Equivlent If the circuit contins dependent sources, dectivte ll the independent sources nd then pply test voltge source or test current source to the Thévenin terminls The Thévenin resistnce is the rtio of the voltge cross the test source to the current delivered y the test source C4-S9 46
Exmple Consider the circuit 000 i + 5 + - + - 3v 0i v - 5 000 i 3vth + - + - 0i 5 i T VT C4-S93 Exmple To find the Thévenin resistnce, we find From the revised figure R Th v = i T T or R Th vt it = + 0i 5 3vT i = 000 vt 1 = = = 100 i 1 ( 3) T + 0 5 000 C4-S94 47
Mximum Power Trnsfer We re often interested in the power trnsferred from circuit to lod. We re thus interested in The efficiency of trnsfer The mount of power trnsferred Mximum power trnsferred my e descried in terms of lod resistor ttched to circuit where the circuit is written in terms of the Thévenin equivlent C4-S95 Mximum Power Trnsfer RTh VTh RL The power trnsferred to the lod resistor is p = i R L v Th = RTh R + L R L C4-S96 48
Mximum Power Trnsfer To find the mximum power trnsferred to the lod resistor, we set ( R + R ) = R ( R + R ) ( RTh + RL) RL ( RTh + RL) 4 ( + ) dp = vth = 0 drl RTh RL Th L L Th L R L = R Th The mximum power is trnsferred if the lod resistnce is equl to the Thévenin equivlent resistnce C4-S97 Superposition Liner systems oey the principle of superposition which sttes tht whenever liner system is excited (driven) y more thn one independent source of energy, the totl response is the sum of the individul responses This principle cn e used to solve circuit prolems if it leds to fewer, simpler equtions C4-S98 49
Superposition Exmple: 6 10 V 3 4 1 A We wish to find the rnch currents. Look first t the currents from the 10 V source y dectivting the current source C4-S99 Superposition We dectivte the current source y open-circuiting it 6 v 1 10 V 3 4 v1 10 v1 v1 + + = 0 6 3 + 4 v = 3 1 C4-S100 50
Superposition 6 v 1 10 V 3 4 The rnch currents re 10 30 i 1 = = 15 6 30 i = = 10 3 30 i 3 = = 5 6 C4-S101 Superposition Now, we look t the rnch currents from the current source Here, we short-circuit the voltge source 6 v 3 v 4 3 4 1 A C4-S10 51
Superposition 6 v 3 v 4 3 4 1 A v3 v3 v3 v4 + + = 0 3 6 v4 v3 v4 + + 1= 0 4 v = 1 3 v = 4 4 C4-S103 Superposition The rnch currents re: v 1 6 6 v 1 4 3 3 v v 1+ 4 v 4 6 4 4 3 i 1 = = = 3 i = = = 3 4 i 3 = = = 4 i 4 = = = The rnch currents with oth sources ctive re i = 15= = 17 1 i = 10 4= 6 i = 5+ 6= 11 3 i = 5 6= 1 4 6 C4-S104 5
Applictions-Source Modeling An idel voltge source provides constnt output voltge independent of the lod resistnce; n idel current source provides constnt current independent of the lod Prcticl voltge nd current sources hve internl resistnces or source resistnces C4-S105 Source Modeling Rs Vs Is Rp Prcticl sources ecome idel s R s 0 nd R p C4-S106 53
Source Modeling For voltge source with lod resistnce R l cross its terminls RL vl = vs R + R L v s is the unloded source voltge s C4-S107 Source Modeling We see tht s the lod resistor ecome smll compred to the internl resistnce of the source, the voltge cross the lod decreses rpidly C4-S108 54
In the cse of current source Source Modeling i L = R p R p + R L i S where i s is the unloded source current C4-S109 Source Modeling In this cse, we see tht unless the lod resistnce is smll compred to the internl resistnce of the source, there is significnt decrese in the current delivered to the lod resistor C4-S110 55
Source Modeling To find the vlue of v s nd R s Mesure the open circuit voltge v oc =v s Connect lod nd djust R L until you mesure lod voltge tht is exctly ½ the open circuit voltge vs v L = Under thee conditions, R L =R Th =R s C4-S111 Exmple The terminl voltge of source is 1 V when connected to W lod. When the lod is disconnected, the terminl voltge rises to 1.4 V. Wht is the source voltge nd the internl resistnce? Wht is the voltge when n 8 Ω lod is connected? Solution V s =V oc =1.4 V C4-S11 56
With the lod pplied from which i L p R L L Exmple v = R L L 1 7 = = Ω vl 1 1 = = = A R 7 6 L 1.4 1.0 Rs = =.4 Ω 1 6 C4-S113 Exmple With n 8 Ω lod pplied 8 v L = 1.4= 9.54 V 8+.4 This exmple clerly indictes tht the lrger the lod resistor is compred to the source resistnce, the closer the voltge cross the lod pproches the open circuit source voltge C4-S114 57
Applictions-The BJT C4-S115 Crmer s Rule Given set of liner equtions x + x + x + L x = 11 1 1 13 3 1n n 1 x + x + x + L x = M 1 1 3 3 n n x + x + x + L x = n1 1 n n3 3 nn n n present them in mtrix form 11 1 13 L 1 n x1 1 1 3 n x 1 31 3 33 3n x1 = 3 M O M M L x n1 n1 n1 nn 1 n C4-S116 58
Crmer s Rule Crmer s Rule sttes tht 1 x1 = x x M x 3 n = 3 = n = C4-S117 Crmer s Rule Where L 11 1 13 1n 1 3 n = 31 3 33 3n M O L n1 n1 n1 nn L 1 1 13 1n 3 n = 1 3 3 33 3n M O L n n1 n1 nn C4-S118 59
Crmer s Rule The determinnts my e evluted s follows 1 = 11M11 1M1 + 13M13 L+ ( 1) + n M1n where M ij is the minor otined y eliminting the i th row nd the j th column Note tht you cn lso expnd long the first column = M M + M L+ ( 1) n Mn + 1 11 11 1 1 31 31 1 C4-S119 Exmple Consider the circuit R3 4 1 R1 R 3 8 F1 I1 3 Gin= R4 4 C4-S10 60
Exmple Using the node voltge method v1 v3 v1 v 3+ = 0 4 v v1 v v v3 + + = 0 4 8 v3 v1 v3 v v1 v + + = 0 4 8 These simplify to 3v v v = 1 1 3 4v + 7v v = 0 1 3 v 3v + v = 0 1 3 C4-S11 In mtrix nottion Exmple 3 1 v1 1 4 7 1 v = 0 3 1 v3 0 Using Crmer s rule 7 1 1 1 = 3 ( 4) + 3 1 3 1 7 1 = 3(7 3) + 4( 5) + (+ 7) = 1 0+ 18= 10 C4-S1 61
Exmple 1 1 7 1 1 = 0 7 1 = 1 = 1(7 3) = 48 3 1 0 3 1 3 1 1 1 3 1 4 1 = 4 0 1 = 0 4 1 = 1 = 1( 4+ ) = 4 1 0 1 0 1 3 1 1 3 4 7 3 = 4 7 0 = 0 4 7 = 1 = 1(1 14) = 4 3 3 0 0 3 v = 48 10= 4.8 V, v = 4 10=.4 V, nd v = 4 10=.4 V 1 3 C4-S13 6