Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Chp 5.3a Thevenin Norton Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

Thevenin’s & Norton’sTheorems These Are Some Of The Most Powerful Circuit analysis Methods They Permit “Hiding” Information That Is Not Relevant And Allow Concentration On What Is Important To The Analysis

Low Distortion Power Amp to Match Speakers And Amplifier One Should Analyze The Amp Ckt From PreAmp (voltage ) To speakers

Low Dist Pwr Amp cont To Even STAND A CHANCE to Match the Speakers & Amp We Need to Simplify the Ckt Consider a Reduced CIRCUIT EQUIVALENT Replace the OpAmp+BJT Amplifier Ckt with a MUCH Simpler Equivalent The Equivalent Ckt in RED “Looks” The Same to the Speakers As Does the Complicated Circuit

Thevenin’s Equivalence Theorem vTH = Thevenin Equivalent VOLTAGE Source RTH = Thevenin Equivalent Series RESISTANCE Thevenin Equivalent Circuit for PART A

Norton’s Equivalence Theorem iN = Norton Equivalent CURRENT Source RN = Norton Equivalent Parallel RESISTANCE Norton Equivalent Circuit for PART A

Outline of Theorem Proof Consider Linear Circuit → Replace vo with a SOURCE If Circuit-A is Unchanged Then The Current Should Be The Same FOR ANY Vo Use Source Superposition 1st: Inside Ckt-A OPEN all I-Src’s, SHORT All V-Srcs Results in io Due to vo 2nd: Short the External V-Src, vo Results in iSC Due to Sources Inside Ckt-A

Theorem Proof Outline cont. Graphically the Superposition All independent sources set to zero in A Then The Total Current Now DEFINE using V/I Then By Ohm’s Law

Theorem Proof Outline cont.2 Consider Special Case Where Ckt-B is an OPEN (i =0) The Open Ckt Case Suggests Also recall How Do To Interpret These Results? vOC is the EQUIVALENT of a single Voltage Source RTH is the EQUIVALENT of a Single Resistance which generates a Voltage DROP due to the Load Current, i

Theorem Proof –Version 2 Because of the LINEARITY of the models, for any Part B the relationship between vO and the current, i, has to be of the form (Linear Response) Result must hold for “every valid Part B” If part B is an open circuit then i=0 and... If Part B is a short circuit then vO is zero. In this case

Examine Thevenin Approach For ANY Part-B Circuit The Thevenin Equiv Ckt for PART-A → V-Src is Called the THEVENIN EQUIVALENT SOURCE R is called the THEVENIN EQUIVALENT RESISTANCE PART A MUST BEHAVE LIKE THIS CIRCUIT

Examine Norton Approach In The Norton Case Norton The Norton Equiv Ckt for PART-A → The I-Src is Called The NORTON EQUIVALENT SOURCE

Interpret Thevenin & Norton In BOTH Cases This equivalence can be viewed as a source transformation problem. It shows how to convert a voltage source in series with a resistor into an equivalent current source in parallel with the resistor SOURCE TRANSFORMATION CAN BE A GOOD TOOL TO REDUCE THE COMPLEXITY OF A CIRCUIT

Source Transformations Source transformation is a good tool to reduce complexity in a circuit ...WHEN IT CAN APPLIED “IDEAL sources” are NOT good models for the REAL behavior of sources .e.g., A Battery does NOT Supply huge current When Its Terminals are connected across a tiny Resistance as Would an “Ideal” Source These Models are Equivalent When Source X-forms can be used to determine the Thevenin or Norton Equivalent But There May be More Efficient Methods

Example  Solve by Src Xform In between the terminals we connect a current source and a resistance in parallel The equivalent current source will have the value 12V/3kΩ The 3k and the 6k resistors now are in parallel and can be combined In between the terminals we connect a voltage source in series with the resistor The equivalent V-source has value 4mA*2kΩ The new 2k and the 2k resistor become connected in series and can be combined

Solve by Src Xform cont. The Options at This Point After the transformation the sources can be combined The equivalent current source has value 8V/4kΩ = 2mA The Options at This Point Do another source transformation and get a single loop circuit Use current divider to compute IO and then Calc VO using Ohm’s law

PROBLEM Find VO using source transformation EQUIVALENT CIRCUITS Norton Norton 3 current sources in parallel and three resistors in parallel Or one more source transformation

Source Xform Summary These Models are Equivalent Source X-forms can be used to determine the Thevenin or Norton Equivalent Next Review Several Additional Approaches To Determine Thevenin Or Norton Equivalent Circuits

Determine the Thevenin Equiv. vTH = OPEN CIRCUIT Voltage at A-B if Part-B is Removed and Left UNconnected iSC = SHORT CIRCUIT Current at A-B if Voltage at A-B is Removed and Replaced with a Wire (a short) Then by R = V/I

Graphically... Then One circuit problem 1. Determine the Thevenin equivalent source Remove part B and compute the OPEN CIRCUIT voltage Second circuit problem 2. Determine the SHORT CIRCUIT current Remove part B and compute the SHORT CIRCUIT current Then

Example  Find Thevenin Equiv. Part B is irrelevant. The voltage Vab will be the value of the Thevenin equivalent source. Find VTH by Nodal Analysis: Iout = 0 For Short Circuit Current Use Superposition When IS is Open the Current Thru the Short When VS is Shorted the Current Thru the Short

Example – Find Thevenin cont Find the Total Short Ckt Current Find Thevenin Resistance To Find RTH Recall Then RTH In this case the Thevenin resistance can be computed as the resistance from a-b when all independent sources have been set to zero Is this a GENERAL Result?

Thevenin w/ Indep. Sources The Thevenin Equivalent V-Source is computed as the open loop voltage The Thevenin Equivalent Resistance CAN BE COMPUTED by setting to zero all the Independent sources and then determining the resistance seen from the terminals where the equivalent will be placed

Thevenin w/ Indep. Sources cont “Part B” Since the evaluation of the Thevenin equivalent can be very simple, we can add it to our toolkit for the solution of circuits “Part B”

Thevenin Example Find Vo Using Thevenin’s Theorem “PART B” Find Vo Using Thevenin’s Theorem Identify Part-B (the Load) Break The Circuit At the Part-B Terminals Deactivate 12V Source to Find Thevenin Resistance

Thevenin Example cont. Note That RTH Could be Found using ISC Then by I-Divider By Series-Parallel R’s Finally RTH Then Itot Same As Before

Thevenin Example cont.2 Finally the Thevenin Equivalent Circuit And Vo By V-Divider

Thevenin Example Find Vo Using Thevenin’s Theorem Alternative: apply Thevenin Equivalence to that part (viewed as “Part A”) Deactivating (Shorting) The 12V Source Yields in the region shown, could use source transformation twice and reduce that part to a single source with a resistor. Opening the Loop at the Points Shown Yields

Thevenin Example cont. Then the Original Circuit Becomes After “Theveninizing” For Open Circuit Voltage Use KVL Result is V-Divider for Vo Apply Thevenin Again Deactivating The 8V & 2mA Sources Gives

Thevenin Example  Alternative Can Apply Thevenin only once to get a voltage divider For the Thevenin Resistance Deactivate Sources “Part B” For the Thevenin voltage Need to analyze this circuit Find VOC by SuperPosition

Thevenin Alternative cont. Open 2mA Source To find Vsrc Contribution to VOC Short 12V Source To find Isrc Contribution to VOC Thevenin Equivalent of “Part A” A Simple Voltage-Divider as Before

Thevenin Example Use Thevenin To Find Vo “Part B” Use Thevenin To Find Vo Have a CHOICE on How to Partition the Ckt Make “Part-B” As Simple as Possible Deactivate the 6V and 2mA Source for RTH

Thevenin Example cont For the open circuit voltage we analyze the circuit at Right (“Part A”) Use Loop/Mesh Analysis Finally The Equivalent Circuit Then VOC

Numerical Example Find Vo Using Thevenin’s Theorem “PART B” Find Vo Using Thevenin’s Theorem First, Identify Part-B Deactivate (i.e. Short Ckt) 6V & 12V Sources to Find RTH

Numerical Example cont. Use Loop Analysis to Find the Open Circuit Voltage The Resulting Equivalent Circuit Finally the Output

CALCULATE Vo USING NORTON PART B COMPUTE Vo USING THEVENIN PART B

Example Xform KVL Deactivate Srcs for RTH Use Loops for VTH

Example cont. OR, Use Superposition to Find Thevenin Voltage First Open The Current Source Next Short-Circuit the Voltage Source Using I-Divider Find Isrc Contribution by KVL Add to Find Total VTH KVL

WhiteBoard Work Let’s Work This Problem Find Vo by Source Transformation 7e prob 4.20