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S.Norr - UMD - Fall, 2005 ECE 2006 Lecture for Chapter 4 S.Norr.

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Presentation on theme: "S.Norr - UMD - Fall, 2005 ECE 2006 Lecture for Chapter 4 S.Norr."— Presentation transcript:

1 S.Norr - UMD - Fall, 2005 ECE 2006 Lecture for Chapter 4 S.Norr

2 S.Norr - UMD - Fall, 2005 Circuit Analysis Methods Nodal Analysis: –Applicable to ANY circuit –Uses KCL to determine voltages in cricuit Mesh Analysis: –ONLY applicable to planar circuits –Uses KVL to determine currents in circuit

3 S.Norr - UMD - Fall, 2005 Nodal Analysis Procedure: 1.Select One Node as Reference 2.Assign a Voltage Variable to each remaining Node 3.Apply KCL at each non-reference Node 4.Solve the resulting set of simultaneous equations

4 S.Norr - UMD - Fall, 2005 Grounding Any SINGLE node in a circuit can be grounded without impact on the performance of the circuit. Connecting one node of a circuit to ground provides a Zero Voltage reference at that point Symbols for the Ground Plane:

5 S.Norr - UMD - Fall, 2005 Example of Nodal Analysis Assign a Reference Node

6 S.Norr - UMD - Fall, 2005 Nodal Analysis Example Assign a Voltage to all Other Nodes:

7 S.Norr - UMD - Fall, 2005 Nodal Example (Cont.) Write KCL at One or More Nodes: i 1 + i 2 + i 3 = 0 Re-Write the Currents using Ohm’s Law: i 1 = (Va - 5)/2 ; i 2 = (Va + 3)/4 ; i 3 = (Va – 0)/8 Substitute: (Va - 5)/2 + (Va + 3)/4 + (Va – 0)/8 = 0 Va = 2 Volts

8 S.Norr - UMD - Fall, 2005 Nodal Analysis Example… Use the Resulting Node Voltages to Solve for Currents: Example: i 3 = (Va – 0)/8 = (2 – 0)/8 = 1/4 Amps

9 S.Norr - UMD - Fall, 2005 Example of Nodal Analysis with a Dependent Source Establish a Reference Node:

10 S.Norr - UMD - Fall, 2005 Example of Dependent (Cont.) Assign a voltage at all other nodes:

11 S.Norr - UMD - Fall, 2005 Example of Dependent (Cont.) Applying KCL at Node V 2 : i x + i + 2i = 0

12 S.Norr - UMD - Fall, 2005 Example of Dependent (Cont.) Describe I x using Ohm’s Law: i x = V x /5 ; Vx = V 2 – V 1 = V 2 – 5 Volts i x = (V 2 – 5) /5 Also, Relate i to V 2: V 2 = i * 10 Result: i x = ( i * 10 – 5) /5 = 2 i -1 Substitute back into KCL: i x + i + 2i = 0 2 i -1 + i + 2i = 0 i = 1/5 Amps ; V 2 = 2 Volts

13 S.Norr - UMD - Fall, 2005 MESH Analysis ONLY used with PLANAR circuits –Planar meaning the circuit can be drawn on a two-dimensional plane without any branches crossing over another branch A MESH is a Loop that contains no other Loops within it.

14 S.Norr - UMD - Fall, 2005 MESH Analysis: Assign a current variable to each MESH in a circuit Apply KVL to each Mesh, using Ohm’s law to express each Voltage in terms of the assigned currents Solve the resulting set of simultaneous equations

15 S.Norr - UMD - Fall, 2005 Mesh Example: Assign Mesh Currents:

16 S.Norr - UMD - Fall, 2005 MESH Example… Write KVL in terms of Mesh Currents: Mesh 1: -5 + 2i 1 + 8(i 1 –i 2 ) = 0

17 S.Norr - UMD - Fall, 2005 Mesh Example…. Mesh 2 Equation: 8(i 2 – i 1 ) + 4i 2 – 3 = 0

18 S.Norr - UMD - Fall, 2005 Mesh Example… Solve the Set of Simultaneous Equations: 10i 1 – 8i 2 = 5 -8i 1 + 12i 2 = 3 14i 1 + 0i 2 = 21 i 1 = 3/2 Amps i 2 = 5/4 Amps


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