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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.
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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, R Th. V Th is equal to the open-circuit voltage at the terminals. R Th is the equivalent or input resistance when the independent sources are turned off.
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Circuit Schematic: Thévenin’s Theorem
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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.
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Definitions for Thévenin’s Theorem Open-circuit voltage V oc is the voltage, V, when the load is an open circuit (i.e., R L = ∞ ). + V oc _
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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 (R L = 0 ).
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Steps to Determine V Th and R Th 1. Identify the load, which may be a resistor or a part of the circuit. 2. Replace the load with an open circuit. 3. Calculate V OC. This is V Th. 4. Turn off all independent voltage and currents sources. 5. Calculate the equivalent resistance of the circuit. This is R TH. The current through and voltage across the load in series with V Th and R Th is the load’s actual current and voltage in the originial circuit.
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Norton’s Theorem A linear two-terminal circuit can be replaced with an equivalent circuit of an ideal current source, I N, in series with a resistor, R N. I N is equal to the short-circuit current at the terminals. R N is the equivalent or input resistance when the independent sources are turned off.
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Definitions for Norton’s Theorem Open-circuit voltage I sc is the current, i, when the load is a short circuit (i.e., R L = 0 ).
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Definitions for Norton’s Theorem Input resistance is the resistance seen by the load when I N = 0A. It is also the resistance of the linear circuit when the load is an open circuit (R L = ∞ ).
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Steps to Determine I N and R N 1. Identify the load, which may be a resistor or a part of the circuit. 2. Replace the load with a short circuit. 3. Calculate I SC. This is I N. 4. Turn off all independent voltage and currents sources. 5. Calculate the equivalent resistance of the circuit. This is R TH. The current through and voltage across the load in parallel with I N and R N is the load’s actual current and voltage in the originial circuit.
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Source Conversion A Thévenin equivalent circuit can easily be transformed to a Norton equivalent circuit (or visa versa). If R Th = R N, then V Th = R N I N and I N = V Th /R Th
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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.
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Example #1
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Example #1 (con’t) Find I N and R N
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Example #1 (con’t) Calculation for I N Look at current divider equation: If R Th = R N = 1k , then I N = 6mA
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Why chose R Th = R N ? Suppose V Th = 0V and I N = 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 (1k ).
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Source Transformation Equations for Thévenin/Norton Transformations V Th = I N R Th I N = V Th /R Th R Th = R N
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Alternative Approach: Example #1 I N is the current that flows when a short circuit is used as the load with a voltage source I N = V Th /R Th = 6mA
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Alternative Approach V Th is the voltage across the load when an open short circuit is used as the load with a current source V Th = I N R Th = 6V
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Example #2 Simplification through Transformation
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Example #2 (con’t)
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Current Source to Voltage Source
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Example #2 (con’t) R Th = 3 V Th = 0.1A (3 ) = 0.3V 0.3V Current Source to Voltage Source
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Example #2 (con’t) 0.3V
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Example #2 (con’t) R Th = 2 I N = 3V/2 = 1.5A Voltage Source to Current Source
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0.3V Example #2 - Solution 1 Simplify to Minimum Number of Current Sources
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Example #2 (con’t) R Th = 6 I N = 0.3V/6 = 50.0mA 0.3V Voltage Source to Current Source
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Example #2 (con’t)
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Current Sources in Parallel Add
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Example #2 - Solution 2 Simplify to Minimum Number of Voltage Sources 0.3V
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Example #2 (con’t) Transform solution for Norton circuit to Thévenin circuit to obtain single voltage source/single equivalent resistor in series with load.
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PSpice
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Example #2 - Solution 1
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Example #2 – Solution 2
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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.
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