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Methods of Analysis Instructor: Chia-Ming Tsai Electronics Engineering National Chiao Tung University Hsinchu, Taiwan, R.O.C. Contents • • • • • • • 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, x=A-1B Nodal Analysis • Circuit variables = node voltages • Steps to determine node voltages – Select a reference node, assign voltages v1, v2,…, vn-1 for the remaining n-1 nodes – Use Ohm’s law to express currents of resistors – Apply KCL to each of the n-1 nodes – Solve the resulting equations Symbols for Reference Node (Ground) Used in this course Case Study Applying Ohm' s law gives v1 0 i1 or i1 G1v1 R1 i2 v1 v2 R2 or i2 G2 v1 v2 i3 v2 0 R3 or i3 G3v2 At node 1, applying KCL gives I1 I 2 i1 i2 I1 I 2 G1v1 G2 v1 v2 At node 2, applying KCL gives I 2 i2 i3 I 2 G2 v1 v2 G3v2 G1 G2 G2 G2 v1 I1 I 2 G3 G3 v2 I 2 Assign vn 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 Continued • If a voltage source is connected between two nonreference nodes – The two nodes form a supernode – Apply KCL to the supernode (similar to a closed boundary) – 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 2 v2 0 v3 0 (2) 2 2 Applying KVL to the supernode, v2 v3 5 (3) 3 variables solved by 3 equations Example 1 Example 2 What is a mesh? • A mesh is a loop that does not contain any other loop within it. Mesh Analysis • Circuit variables = mesh currents • Steps to determine mesh currents – – – – Assign mesh currents i1, i2,…, in Use Ohm’s law to express voltages of resistors Apply KVL to each of the n meshes Solve the resulting equations Continued • 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 R3i1 R2 R3 i2 V2 R1 R3 R3 R3 i1 V1 R2 R3 i2 V2 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 Continued • If a current source exists between two meshes – A supermesh is resulted Excluded – Apply KVL to the Supermesh supermesh – Apply KCL to derive the relationship between the two mesh i2 i1 I S currents Example 1 i1 i2 i2 5 A Applying KVL for mesh 1, 10 4i 6i1 i2 0 i1 2 A 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 • • • • 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 How to choose? • Nodal Analysis – More parallel-connected elements, current sources, or supernodes – Nnode < Nmesh – If node voltages are required • Mesh Analysis – More series-connected elements, voltage sources, or supermeshes – Nmesh < Nnode – If branch currents are required Applications: Transistors • Bipolar Junction Transistors (BJTs) • Field-Effect Transistors (FETs) Bipolar Junction Transistors (BJTs) • Current-controlled devices I E I B IC (KCL) VCE VEB VBC 0 (KVL) VBE 0.7 V IC I B ( ~ 100) IC I E (0 1) I E 1 I B 1 DC Equivalent Model of BJT Example of Amplifier Circuit