Thévenin’s and Norton’s Theorem

1. Thévenin’s and Norton’s Theorem

2. Objective of Lecture • State Thévenin’s and Norton Theorems. • Chapter 4.5 and 4.6 Fundamentals of Electric Circuits • Demonstrate how Thévenin’s and Norton theorems ca be used to simplify a circuit to one that contains three components: a power source, equivalent resistor, and load.

3. Thévenin’s Theorem • A linear two-terminal circuit can be replaced with an equivalent circuit of an ideal voltage source, VTh, in series with a resistor, RTh. • VTh is equal to the open-circuit voltage at the terminals. • RTh is the equivalent or input resistance when the independent sources are turned off.

4. Circuit Schematic:Thévenin’s Theorem

5. Definitions for Thévenin’s Theorem Linear circuit is a circuit where the voltage is directly proportional to the current (i.e., Ohm’s Law is followed). Two terminals are the 2 nodes/2 wires that can make a connection between the circuit to the load.

6. Definitions for Thévenin’s Theorem + Voc _ Open-circuit voltage Voc is the voltage, V, when the load is an open circuit (i.e., RL = ∞W).

7. Definitions for Thévenin’s Theorem • Input resistance is the resistance seen by the load when VTh = 0V. • It is also the resistance of the linear circuit when the load is a short circuit (RL = 0W).

8. Steps to Determine VTh and RTh • Identify the load, which may be a resistor or a part of the circuit. • Replace the load with an open circuit . • Calculate VOC. This is VTh. • Turn off all independent voltage and currents sources. • Calculate the equivalent resistance of the circuit. This is RTH. • The current through and voltage across the load in series with VTh and RTh is the load’s actual current and voltage in the originial circuit.

9. Norton’s Theorem • A linear two-terminal circuit can be replaced with an equivalent circuit of an ideal current source, IN, in series with a resistor, RN. • IN is equal to the short-circuit current at the terminals. • RN is the equivalent or input resistance when the independent sources are turned off.

10. Definitions for Norton’s Theorem Open-circuit voltage Isc is the current, i, when the load is a short circuit (i.e., RL = 0W).

11. Definitions for Norton’s Theorem • Input resistance is the resistance seen by the load when IN = 0A. • It is also the resistance of the linear circuit when the load is an open circuit (RL = ∞W).

12. Steps to Determine IN and RN • Identify the load, which may be a resistor or a part of the circuit. • Replace the load with a short circuit . • Calculate ISC. This is IN. • Turn off all independent voltage and currents sources. • Calculate the equivalent resistance of the circuit. This is RTH. • The current through and voltage across the load in parallel with IN and RN is the load’s actual current and voltage in the originial circuit.

13. Source Conversion • A Thévenin equivalent circuit can easily be transformed to a Norton equivalent circuit (or visa versa). • If RTh = RN, then VTh = RNIN and IN = VTh/RTh

14. Value of Theorems • Simplification of complex circuits. • Used to predict the current through and voltage across any load attached to the two terminals. • Provides information to users of the circuit.

15. Example #1

16. Example #1 (con’t) Find IN and RN

17. Example #1 (con’t) • Calculation for IN • Look at current divider equation: If RTh = RN= 1kW, then IN = 6mA

18. Why chose RTh = RN? • Suppose VTh = 0V and IN = 0mA • Replace the voltage source with a short circuit. • Replace the current source with an open circuit. • Looking towards the source, both circuits have the identical resistance (1kW).

20. Source Transformation Equations for Thévenin/Norton Transformations VTh = IN RTh IN = VTh/RTh RTh= RN

21. Alternative Approach: Example #1 IN is the current that flows when a short circuit is used as the load with a voltage source IN = VTh/RTh = 6mA

22. Alternative Approach VTh is the voltage across the load when an open short circuit is used as the load with a current source VTh = IN RTh = 6V

23. Example #2 Simplification through Transformation

24. Example #2 (con’t)

25. Example #2 (con’t) Current Source to Voltage Source

26. Example #2 (con’t) Current Source to Voltage Source RTh = 3W VTh = 0.1A (3W) = 0.3V 0.3V

27. Example #2 (con’t) 0.3V

28. Example #2 (con’t) Voltage Source to Current Source RTh = 2W IN = 3V/2W = 1.5A

29. Example #2 - Solution 1 • Simplify to Minimum Number of Current Sources 0.3V

30. Example #2 (con’t) Voltage Source to Current Source RTh = 6W IN = 0.3V/6W = 50.0mA 0.3V

31. Example #2 (con’t)

32. Example #2 (con’t) Current Sources in Parallel Add

33. Example #2 - Solution 2 • Simplify to Minimum Number of Voltage Sources 0.3V

34. Example #2 (con’t) Transform solution for Norton circuit to Thévenin circuit to obtain single voltage source/single equivalent resistor in series with load.

35. PSpice

36. Example #2 - Solution 1

37. Example #2 – Solution 2

38. Summary • Thévenin and Norton transformations are performed to: • Simplify a circuit for analysis • Allow engineers to use a voltage source when a current source is called out in the circuit schematic • Enable an engineer to determine the value of the load resistor for maximum power transfer/impedance matching.