Chapter 4 Technques of Crcut Analyss 4. Termnology 4.-4.4 The Node-Voltage Method (NVM) 4.5-4.7 The Mesh-Current Method (MCM) 4.8 Choosng NVM or MCM 4.9 Source Transformatons 4.0-4. Théenn and Norton Equalents 4. Maxmum Power Transfer 4.3 Superposton
Oerew Crcut analyss by seres-parallel reducton and -Y transformatons mght be cumbersome or een mpossble when the crcuts are structurally complcated and/or nole wth a lot of elements. Systematc methods that can descrbe crcuts wth mnmum number of smultaneous equatons are of hgh nterest.
Key ponts How to sole a crcut by the Node-Voltage Method and Mesh-Current Method systematcally? What s the meanng of equalent crcut? Why s t useful? How to get the Théenn equalent crcut? What s superposton? Why s t useful? 3
Secton 4. Termnology 4
Defnton 5
Planar crcuts Crcuts wthout crossng branches. Not n contact 6
Example of a nonplanar crcut 7
Identfyng essental nodes n a crcut Number of essental nodes s denoted by n e. 8
Identfyng essental branches n a crcut ab s not!! Number of essental branches s denoted by b e. 9
Identfyng meshes n a crcut 0
Secton 4.-4.4 The Node-Voltage Method (NVM). Standard procedures. Use of supernode
Step : Select one of the n e essental nodes as the reference node Can be ether planar or nonplanar. eference =0 Selecton s arbtrary. Usually, the node connectng to the most branches s selected to smplfy the formulaton.
3 Step : Lst n e - equatons by KCL, Sole them. 0 0, 5 0 Node : Node :, 0 0.6 0.5 0.5.7 V. 0.9 9.09
4 NVM n the presence of dependent sources. 5 0, 8 0 5 0, 5 0 0 Node : Node : Constr ant:, 0 0.6 0. 0.75. A. V, 0 6
Case of falng to dere node equaton When a oltage source (ether ndependent or dependent) s the only element between two essental nodes, the essental branch current s undetermned, fal to apply KCL to ether node! E.g. 3 s undetermned, fal to apply KCL to Nodes, 3. =50 V 3 = +0 3 =? 0 V 5
Soluton : Add an unknown current 50 V 3 3 0 V Node : Node 3: 50 5 50 3 3 4. 00 3 0, 50 5 50 3 00 4() Source constrant: 3 = + 0( )/(5 ) (). 6
Soluton : Use supernode By applyng KCL to a supernode formed by combnaton of two essental nodes, one can get the same equaton wthout the ntermedate step. 50 3 4() 5 50 00 7
Counter example (Example 4.3) G 0-V source s not the only element between Nodes and G, branch current 0V = ( -0)/( ) s stll aalable, KCL can stll be appled to Node, no need to use supernode. 8
Example.: Amplfer crcut () n e =4, 3 unknown oltages. Snce B cannot be dered by node oltages, 4 unknowns. The oltage sources prode constrants: a c V b () V () CC 0 9
Example. () V CC b -V 0 (+) B Apply KCL to Node b: b b V CC 0(3) Apply KCL to Node c: cd ( ) ( ) B B, B c E (4) Use Supernode bc s also OK. 0
Secton 4.5-4.7 The Mesh-Current Method (MCM). Adantage of usng mesh currents as unknowns. Use of supermesh
Branch currents as unknowns Number of essental branches b e =3, number of essental nodes n e =. To sole {,, 3 }, use KCL and KVL to get n e - = and b e -(n e -) =equatons.
Adantage of usng mesh currents as unknowns Can only be planar. 3 elaton between branch currents and mesh currents: = a, = b, 3 = a b. Each mesh current flows nto and out of any node on the way, automatcally satsfy KCL. 3
4 Lst b e -(n e -) equatons by KVL, Sole them Mesh a: Mesh b: () 0 () 3 3 a b b b a a., 3 3 3 3 b a b a
Case of falng to dere mesh equaton When a current source s between two essental nodes (no need to be the only element), the oltage drop across the source s undetermned, fal to apply KVL to ether mesh! 3 E.g. G s undetermned, fal to apply KVL to Meshes a and c. G 5
Soluton : Add an unknown oltage 3 Mesh a: Mesh c: (3 )( a b ) (6 ) a 00, ( )( c b ) 50 (4 ) c. 9 5 6 50() a b c 6
Soluton : Use supermesh By applyng KVL to a supermesh formed by combnaton of two meshes, one can get the same equaton wthout the ntermedate step. ( 3 )( ) ( )( ) 50 (4 ) (6 ) 00, a b c b c a 9 5 6 50() a b c 7
Secton 4.8 The Node-Voltage Method s. the Mesh-Current Method 8
Example 4.6 () Q: P 300 =? 5 meshes, no current source (no supermesh). MCM needs 5 mesh equatons. 4 essental nodes, the dependent oltage source s the only element on that branch ( supernode). NVM needs (4-)-= node equatons. 9
Example 4.6 () Choose the reference node such that P 300 can be calculated by only solng. Apply KCL to Supernode,3: 3 3 56 3 3 8 0() 00 50 00 50 400 500 30
Example 4.6 (3) Node : 3 8 3 0() 300 50 450 500 Source constrant: 50 300 6 3 (3) 3
Example 4.7 () 3 meshes, current sources ( supermeshes). MCM needs mesh equaton. 4 essental nodes, no oltage source s the only element on one branch (no supernode). NVM needs (4-)= 3 node equatons. 3
Example 4.7 () Apply KVL to Supermesh a, b, c: (4 0.8 6 ) a 93; b (.5 7.5 ) b ( 8 ) ( 7.5 ), 0 4 0 93() a b c c 33
Example 4.7 (3) By the two current source constrants: b c a b 0.4 0.5(3) 0.4 c 0.8 c () 34
Secton 4.9 Source Transformatons 35
Source transformatons A oltage source s n seres wth a resstor can be replaced by a current source s n parallel wth the same resstor or ce ersa, where s s = 36
37 Proof of source transformaton For any load resstor L, the current and oltage between termnals a, b should be consstent n both confguratons.., s L L L s. ) // (, L L s L s s L s L L L + +
edundant resstors Why? Why? 38
Example 4.9 () Q: () o =? () P 50V =? 39
Example 4.9 () To fnd o, transform the 50-V oltage source nto a 0-A current source. 40
Example 4.9 (3) Now o s smply the oltage of the total load. o ( A) (0 ) 0 V. 4
Example 4.9 (4) P 50V has to be calculated by the orgnal crcut: 0 V P 50V 50 V (50 0) V. A, 5 5 (. A) (50 V).8 kw. (Power extracton) 4
Secton 4.0, 4. Théenn and Norton Equalent. Defnton of equalent crcut. Methods to get the two Théenn equalent crcut parameters Th, Th 3. Methods to get Théenn resstance Th alone 4. Example of applcatons 43
Théenn equalent crcut A general crcut For any load resstor L, the current and oltage between termnals a, b should be consstent n both confguratons. 44
Method to get V Th and Th Fnd open crcut oltage oc =V Th Fnd short crcut current sc Th oc sc Note: Ths method s nald when the crcuts contan only dependent sources. 45
Example: Calculatng oc Let termnals a, b be open, no current flows through the 4- resstor, oc =. c Open By NVM, the only unknown s the node oltage. Apply KCL to Node c: 5 3, 5 0 3 V V Th. 46
Example: Calculatng sc Let termnals a, b be short, sc = /(4 ). c Short By NVM, the only unknown s node oltage. Apply KCL to Node c: 5 5 sc 3, 0 4 (4 ) 4 A. 6 V, 47
Example: Calculatng Th and Norton crcut The Théenn resstance s: Th = V Th / sc = (3 V)/(4 A) = 8. = The Norton equalent crcut can be dered by source transformaton: 48
Method to get V Th and Th Use a seres of source transformatons when the crcut contans only ndependent sources. Same as by oc, sc. 49
Method to get Th alone For crcuts wth only ndependent sources, Step : Deactate all sources: () Voltage source Short, () Current source Open. Step : Th s the resstance seen by an obserer lookng nto the network at the desgnated termnal par. E.g. Th = ab = [(5 )//(0 )]+(4 ) = 8. 50
Method to get Th alone For crcuts wth or wthout dependent sources: Step : Deactate all ndependent sources; Step : Apply a test oltage T or test current T source to the desgnated termnals; Step 3: Calculate the termnal current T f a test oltage T source s used and ce ersa; Step 4: Get the Théenn resstance by Th T T. 5
Example 4.: Fnd Th of a crcut () Step : Deactate all ndependent sources: Short 5
Example 4. () Step : Assume a test oltage source T : Step 3: Calculate termnal current T ( T ): T T 3T T 0. 5 k 00 Step 4: Th = T / T = 00. 53
Amplfer crcut soled by equalent crcut () B =? To fnd the equalent crcut that dres termnals b and d (the nput of a BJT transstor), we can redraw the crcut as f t were composed of two stages. 54
Amplfer crcut soled by equalent crcut () For termnals b, d, the left part has an open-crcut oltage:, Th CC The resstance wth V CC gets shorted s: Th // 55
Amplfer crcut soled by equalent crcut (3) The left part can be replaced by a Théenn crcut. Wth ths, we can apply KVL to loop bcdb: Th Th Th B V ( ) BE, VTh V0 B ( ) 0 E. 56
Secton 4. Maxmum Power Transfer 57
Formulaton () Consder a crcut (represented by a Théenn equalent) loaded wth a resstance L. The power dsspaton at L s: p L V To fnd the alue of L that leads to maxmum power transfer, perform derate: dp d L V Th ( Th Th L ) ( Th Th L L L ( 4 ) L Th, L ). 58
Formulaton () When the derate equals zero, p s maxmzed: dp d L ( V Th L Th p Th max ( L Th L ) Th L ) ( L Th, V L Th Th ( Th Th L L Th ( 4 ) The maxmum transfer power s: L ) Th 0, V 4 Th Th L. ) 0, 59
Secton 4.3 Superposton 60
What s superposton? In a crcut consstng of lnear elements only, superposton allows us to actate one ndependent source at a tme and sum the nddual oltages and currents to determne the actual oltages and currents when all ndependent sources are acte. Superposton s useful n desgnng a large system, where the mpact of each ndependent source s crtcal for system optmzaton. 6
Example () Q:,,3,4 =? Deactated 5 A, 0 A, 3 4 5 A. 6
Example () Deactated A, 4 A, 6 A, 6 A. 3 4 63
64 Example (3). A, 6 5 A, 6 5 6 A, 4 0 7 A, 5 4 4 4 3 3 3 The -A (0-V) source s more mportant n determnng,, ( 3, 4 ).
Key ponts How to sole a crcut by the Node-Voltage Method and Mesh-Current Method systematcally? What s the meanng of equalent crcut? Why s t useful? How to get the Théenn equalent crcut? What s superposton? Why s t useful? 65