2012/9/17 Methods of Analysis •Introduction •Nodal Analysis •Nodal Analysis with Voltage Sources •Mesh Analysis •Mesh Analysis with Current Sources •Nodal Analysis vs. Mesh Analysis •Applications Introduction •Nodal Analysis –Based on KCL. •Mesh Analysis –Based on KVL. •Linear algebra is applied to solve the resulting simultaneous equations. – Ax = B or x = A-1B. 1 2012/9/17 Nodal Analysis •Circuit variables = node voltages –KVL is automatically satisfied. •Steps to analyze an n-node network –Select a reference node (as ground), assign voltages v1, v2, …, vn-1 for the remaining n-1 nodes. –Use Ohm’ sl awto express currents of resistors. –Apply KCL to each of the n-1 nodes. –Solve the resulting equations. Earth ground Chassis ground Case Study * At node 1, applying KCL gives I1 I 2 i1 i2 (1) * At node 2, applying KCL gives I 2 i2 i3 (2) * Applying Ohm' s law gives v 0 i1 1 or i1 G1v1 R1 v v i2 1 2 R2 or i2 G2 v1 v2 v 0 i3 2 R3 or i3 G3v2 Assign vn (1) I1 I 2 G1v1 G2 v1 v2 (3) (2) I 2 G2 v1 v2 G3v2 G G2 1 G2 (4) G2 v1 I I 1 2 G2 G3 v2 I 2 2 2012/9/17 Nodal Analysis with Voltage Sources •If a voltage source is connected between a nonreference node and the reference node (or ground). –The node voltage is defined by the voltage source. –Number of variables is reduced. –Simplified analysis. Cont ’ d •If a voltage source is connected between two nonreference nodes. –IS is difficult to define. –I t ’ sdi f f i c ul tt os ol v et hepr obl e m byus i ngKCL. I IS VS V V VS - I S I-V curve of a voltage source 3 2012/9/17 Cont ’ d •Analysis Strategy –The two nodes form a supernode (a closed boundary). –Eq. 1: Apply KCL to the supernode. –Eq. 2: Apply KVL to derive the relationship between the two nodes. Supernode Case Study with Supernode v1 10 V (1) Applying KCL to the supernode, i1 i4 i2 i3 v1 v2 v1 v3 2 4 v2 0 v3 0 (2) 8 6 Applying KVL to the supernode, v2 v3 5 (3) 3 variables solved by 3 equations. 4 2012/9/17 Example 1 Supernode 2 i1 i2 7 v 2 v1 2 Example 2 Supernode Supernode i 3 i1 i2 i1 i3 i4 i5 v1 v 2 20 v 3 v4 3v x 5 2012/9/17 What is a mesh? •A mesh is a loop that does not contain any other loop within it. Mesh Analysis •Circuit variables = mesh currents –KCL must be satisfied. ( How ??? ) •Steps to analyze an n-mesh network –Assign mesh currents i1, i2, …, in . –Use Ohm’ sl awto express voltages of resistors. –Apply KVL to each of the n meshes. –Solve the resulting equations. 6 2012/9/17 Mesh Analysis •Applicable only for planar circuits. •An example for nonplanar circuits is shown below. Case Study For mesh 1, applying KVL gives V1 R1i1 R3 i1 i2 0 R1 R3 i1 R3i2 V1 For mesh 2, applying KVL gives R2i2 V2 R3 i2 i1 0 R R3 1 R3 R3 i1 V1 R2 R3 i2 V2 R3i1 R2 R3 i2 V2 7 2012/9/17 Mesh Analysis with Current Sources •If a current source exists only in one mesh. –The mesh current is defined by the current source. –Number of variables is reduced. –Simplified analysis. Cont ’ d •If a current source exists between two meshes. –VS is difficult to define. –I t ’ sdi f f i c ul tt os ol vet hepr obl e m byus i ngKVLf ore a c h mesh. I + VS _ IS V I I S - VS I-V curve of a current source 8 2012/9/17 Cont ’ d •Analysis Strategy –A supermesh is resulted. –Eq. 1: Apply KVL to the supermesh. –Eq. 2: Apply KCL to derive the relationship between the two mesh currents. Supermesh KCL : i2 i1 I S Excluded Example 1 i1 i2 i2 5 A Applying KVL for mesh 1, 10 4i 6 i1 i2 0 i1 2 A 9 2012/9/17 Example 2 Supermesh Applying KVL to the supermesh, 20 6i1 10i2 4i2 0 Applying KCL to node 0, i2 i1 6 6i1 14i2 20 i1 3.2 A, i2 2.8 A Example 3 Supermesh =i2-5 =i2-3Io • Applying KVL to the supermesh • Applying KCL to node P • Applying KCL to node Q • Applying KVL to mesh 4 4 variables solved by 4 equations 10 2012/9/17 How to choose? •Nodal Analysis –More parallel-connected elements, voltage sources, or supernodes. –Nnode < Nmesh –If node voltages are required. •Mesh Analysis –More series-connected elements, current sources, or supermeshes. –Nmesh < Nnode –If branch currents are required. Applications: Transistors •Bipolar Junction Transistors (BJTs) •Field-Effect Transistors (FETs) 11 2012/9/17 Bipolar Junction Transistors (BJTs) •Current-controlled devices I E I B I C (KCL) VCE VEB VBC 0 (KVL) VBE 0.7 V I C I B ( ~ 100) I C I E (0 1) I E 1 I B 1 DC Equivalent Model of BJT 12 2012/9/17 Example of Amplifier Circuit I C I B 13