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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. Thévenin’s Theorem A linear two-terminal circuit can be replaced with an equivalent circuit of an ideal voltage source, V Th, in series with a resistor, RTh. V Th is equal to the open-circuit voltage at the terminals. RTh is the equivalent or input resistance when the independent sources are turned off. Circuit Schematic: Thévenin’s Theorem 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. 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). VOC VTh Definitions for Thévenin’s Theorem Input resistance is the resistance seen by the load when V Th = 0V. It is also the resistance of the linear circuit when the load is a short circuit (RL = 0W). Rin RTh VTh iSC 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 V Th. Turn off all independent voltage and currents sources. Calculate the equivalent resistance of the circuit. This is RTH. 1. 2. 3. 4. 5. The current through and voltage across the load in series with V Th and RTh is the load’s actual current and voltage in the originial circuit. 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. Definitions for Norton’s Theorem Open-circuit voltage Isc is the current, i, when the load is a short circuit (i.e., RL = 0W). I SC I N 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). Rin RN VOC I N 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. 1. 2. 3. 4. 5. 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. Source Conversion A Thévenin equivalent circuit can easily be transformed to a Norton equivalent circuit (or visa versa). If RTh = RN, then V Th = RNIN and IN = V Th/RTh 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. Example #1 Example #1 (con’t) Find IN and RN Example #1 (con’t) Calculation for IN Look at current divider equation: I load Req Rload Rload RN 1 IN IN Rload RN Rload RN 2mA IN 2kW RN If RTh = RN= 1kW, then IN = 6mA Why chose RTh = RN? Suppose V Th = 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). Source Transformation Equations for Thévenin/Norton Transformations V Th = IN RTh IN = V Th/RTh RTh= RN 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 Alternative Approach V Th is the voltage across the load when an open short circuit is used as the load with a current source VTh = IN RTh = 6V Example #2 Simplification through Transformation Example #2 (con’t) Example #2 (con’t) Current Source to Voltage Source Example #2 (con’t) Current Source to Voltage Source RTh = 3W VTh = 0.1A (3W) = 0.3V 0.3V Example #2 (con’t) 0.3V Example #2 (con’t) Voltage Source to Current Source RTh = 2W IN = 3V/2W = 1.5A Example #2 - Solution 1 Simplify to Minimum Number of Current Sources 0.3V Example #2 (con’t) Voltage Source to Current Source RTh = 6W IN = 0.3V/6W = 50.0mA 0.3V Example #2 (con’t) Example #2 (con’t) Current Sources in Parallel Add Example #2 - Solution 2 Simplify to Minimum Number of Voltage Sources 0.3V Example #2 (con’t) Transform solution for Norton circuit to Thévenin circuit to obtain single voltage source/single equivalent resistor in series with load. PSpice Example #2 - Solution 1 Example #2 – Solution 2 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.