Super Node/Mesh Thevénin/Norton

Super Node/Mesh Thevénin/Norton
Engineering 43 Super Node/Mesh Thevénin/Norton Bruce Mayer, PE Licensed Electrical & Mechanical Engineer

OutLine ReIterate NODE & LOOP/MESH Analysis by way of Examples
SuperNode Method SuperMesh Technique Loops vs Meshes; describe difference Loop & Node Compared Introduction to Thevenin & Norton theorems

OutLine Thevénin theorem: RTH=RN & VTH=VOC
Norton theorem: RN=RTH & IN=ISC Source Transformation Thevenin & Norton theorems for INdependent Source Circuits by DeActivation Dependent Source Circuits by VOC/ISC

Node Analysis (find V by kCl) Example (On Board)
Define a GND Node Label all Non-GND nodes not connected to a Source; i.e., the UNknown Nodes Write the CURRENT Law Eqns at the Unknown Nodes using Ohm’s Law; I = ΔV/R Clear Fractions by Multiplying by the LCD Be sure to Include UNITS Work the Linear Algebra to find Unknown Node Voltages

Loop Analysis (find I by kVl) Define a GND Node
Draw Current Loops or Meshes to NOT be Redundant Pass Thru ALL Circuit Elements Write the VOLTAGE Law Eqns For each Loop/Mesh usingNodes using Ohm’s Law; ΔV=RI Divide by Eqns by the LCM of the I’s Be sure to Include UNITS Work the Linear Algebra to find Unknown Loop/Mesh Currents Example (On Board)

ReCall Node (KCL) Analysis
Need Only ONE KCL Eqn The Remaining Eqns From the Indep Srcs 3 Nodes Plus the Reference. In Principle Need 3 Equations... But two nodes are connected to GND through voltage sources. Hence those node voltages are KNOWN Solving The Eqns

SuperNode Technique Consider This Example
Conventional Node Analysis Requires All Currents at a Node But Have Ckt V-Src Reln More Efficient solution: Enclose The Source, and All Elements In parallel, Inside a Surface. Call That a SuperNode 2 eqns, 3 unknowns... Not Good (IS unknown) Recall: The Current thru the Vsrc is NOT related to the Potential Across it

Supernode cont. Apply KCL to the Surface
The Source Current Is interior to the Surface and is NOT Required Still Need 1 More Equation – Look INSIDE the Surface to Relate V1 & V2 Now Have 2 Equations in 2 Unknowns Then The Ckt Solution Using LCD Technique See Next Slide

Now Apply Gaussian Elim
The Equations Use The V-Source Rln Eqn to Find V2 Mult Eqn-1 by LCD (12 kΩ) SUPERNODE Add Eqns to Elim V2

Find the node voltages And the power supplied By the voltage source
If Vs ABSORBS pwr (as do all the Rs), then from where does the power come? => from the two Current Sources To compute the power supplied by the voltage source We must know the current through node-1 BASED ON PASSIVE SIGN CONVENTION THE POWER IS ABSORBED BY THE SOURCE!!

Illustration using Conductances
Write the Node Equations KCL At v1  At The SuperNode Have V-Constraint v2 − v3 = vA KCL Leaving SuperNode   NOTE: 𝒗 𝟏 ≠𝟎 ‼ Now Have 3 Eqns in 3 Unknowns Solve Using Normal Techniques

Example Find 𝐼 𝑂 Known Node Voltages The SuperNode V-Constraint
Now use KCL at SuperNode to Find V3 Mult by 2 kΩ LCD, collect Terms to Find: 𝐼 𝑂 = − 6 7 V 2 kΩ =643 μA

Numerical Example Find Io Using Nodal Analysis
SUPERNODE Find Io Using Nodal Analysis Known Voltages for Sources Connected to GND Now Notice That V2 is NOT Needed to Find Io 2 Eqns in 2 Unknowns The Constraint Eqn Now KCL at SuperNode By Ohm’s Law

Dependent Sources Circuits With Dependent Sources Present No Significant Additional Complexity The Dependent Sources Are Treated As Regular Sources As With Dependent CURRENT Sources Must Add One Equation For Each Controlling Variable

Numerical Example – Dep Isrc
Find Io by Nodal Analysis Notice V-Source Connected to the Reference Node Sub Ix into KCL Eqn KCL At Node-2 Mult By 6 kΩ LCD Controlling Variable In Terms of Node Potential Then Io

Current Controlled V-Source
Find Io SuperNode Constraint Controlling Variable in Terms of Node Voltage Multiply by LCD of 2 kΩ Recall What units of 2000? Hint: by ohm v = Ri Then KCL at SuperNode So Finally

ReCall MESH Analysis Use Mesh Analysis Sub for I1 to Find I2 So Vo

SuperMeshes   Create Mesh Currents
Write Constraint Equation Due To Mesh Currents SHARING Current Sources  Write Equations For Remaining Meshes SUPERMESH  SuperMesh is Loop defined in terms of mesh currents Define A SuperMesh By AVOIDING The Shared Current with a Carefully Chosen Loop

SuperMeshes cont. Write KVL For The SuperMesh as we do NOT know the VOLTAGE Across the 4 mA Current-Source  We Now have 3 Eqns in 3 Unknowns and the Math Model is Complete Solve for I1, I2, I3 using standard techniques SUPERMESH Note that the SuperMesh Eqn is defined in terms of previously established MESH currents KVL

SuperNODE vs SuperMESH
Use superNODE to AVOID a V-source current, IVs, in KCL Eqns Use superMESH to AVOID an I-source, ΔVIs, in KVL Eqns Need CURRENTS in KCL-based NODE-Analysis – do NOT know I thru a V-src Need POTENTIALS in KVL-based LOOP-Analysis – do NOT know V across an I-src

Shared Isrc General Loop Approach
Strategy Define Loop Currents That Do NOT Share Current Sources Even If It Means ABANDONING Meshes For Convenience, Begin by Using Mesh Currents Until Reaching Shared CURRENT Source as V-across an I-source is NOT Known At That Point Define a NEW Loop To Guarantee That The New Loop Gives An Independent Equation, Must Ensure That It Includes Components That Are NOT Part Of Previously Defined Loops Replace MESH-I3 with a LOOP as Mesh-I3 SHARES the 4ma source with Mesh-I2

General Loop Approach cont.
A Possible Approach Create a Loop by Avoiding The Current Source The Eqns for Current Source Loops Note that 𝐼 3 is NOT a Mesh Current The Eqns for 3rd Loop (3 Eqns & 3 Unknowns) The Loop Currents Obtained With This Method Are Different From Those Obtained With A SuperMesh A SuperMesh used Previously Defined Mesh Currents

Game Plan  02Feb16 Skip to Slide 33

Example  Find V Across R’s
For Loop Analysis Note Three Independent Current Sources Four Meshes One Current Source Shared By Two Meshes Careful Choice Of Loop Currents Should Make Only One Loop Equation Necessary Three Loop Currents Can Be Chosen Using Meshes And Not Sharing Any Source Mesh Equations For Loops With I-Sources

Example  Find V Across R’s cont.
KVL for I4 Loop Solve For The Current I4 Using The ISj Now Use Ohm’s Law To Calc Required Voltages Note that Loop-4 does NOT pass thru ANY CURRENT-Sources This AVOIDS the UNknown potentials across the I-sources

Dependent Sources General Approach
Treat The Dep. Source As Though It Were Independent Add One Equation For The Controlling Variable Example at Rt.: Mesh Currents Defined by Sources Mesh-3 by KVL Mesh-4 by KVL

Dependent Sources cont.
The Controlling Variable Eqns Combine Eqns, Then Divide by 1kΩ In Matrix Form Solve by Elimination or Linear-Algebra

Loop & Node Compared Consider the Ckt Find Vo by NODE Analysis
ID Nodes Make a SuperNode Vsrc to GND SuperNode Constraint

Loop & Node Compared (2) KCL at SuperNode Mult. By 1kΩ LCD
The Node 3 KCL Mult. By 1kΩ LCD The SuperNode Eqn

Loop & Node Compared (3) The Controlling Var.
Thus 3 Eqns in Unknowns V1, V2, V3 Recall the GOAL In SuperNode Eqn

Loop & Node Compared (4) Now by Loops The Mesh/Loop Eqns
Note I3 = –2 mA Loop-4: Start with 3 Meshes Add a General Loop to avoid the rt Isrc Note I1 = 2 mA

Loop & Node Compared (5) The Controling Var By Net Current & Ohm’s Law
SubOut Vx in Loop-2 & Loop-4 Eqns: As Before Vx = V2 And V2 is related to the net current From Node-2 to GND After Subbing Find:

Loop & Node Compared (6) Summarize Loops Recall the GOAL
General Comments Nodes (KCL) are generally easier if we have VOLTAGE Sources Loops (kVL) are generally easier if we have CURRENT Sources Using The Loop/Mesh Eqns Find

Game Plan  09Feb16 File: ENGR-43_Lec-02b_Sp16_SuperNM_Thevenin-Norton.pptx Slides 40 → 76 ENGR-43_Lec-02c_Sp16_MaxPwr_SuperPosition.pptx Slides 1→30

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 LEFT → Léon Charles Thévenin (30 March September 1926) • RIGHT → Edward Lawry Norton

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 (Linear) 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 (Source) RESISTANCE Load Driving Circuit Thevenin Equivalent Circuit for PART A

Norton’s Equivalence Theorem
iN = Norton Equivalent CURRENT Source RN = Norton Equivalent Parallel (Source) RESISTANCE Driving Circuit Load POWER Supplied by Part-A is 𝑖∙ 𝑣 𝑂 = POWER Dissipied by Part-B is 𝑖∙ 𝑣 𝑂 Norton Equivalent Circuit for PART A

Examine Thevenin Approach
Driving Circuit Load 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 A LINEAR 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 A LINEAR PART-A MUST BEHAVE LIKE THIS CIRCUIT

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 One time I tested the Thevein/Norton model for AA-Battery. I measured the voltage across the effectively open circuit Voltmeter which produced about 1.55V. I then SHORTED thru an Ameter the same AA-Battery to read about 2A. Thus VS = 1.55 V, IS = 2A. So then RV = RI = 1.55V/2A = 0.78 Ω Source X-forms can be used to determine the Thevenin or Norton Equivalent But There May be More Efficient Methods

White Board → Source Xform
Find 𝑉 𝑂 By Source Transformation Irwin 9e 5-77

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 09Feb16 → SKIP This One In between the terminals we connect a voltage source in series with the resistor The equivalent V-source has value 4mA∙2kΩ = 8 V The new 2k and the 2k resistor become connected in series and can be combined

09Feb16 → SKIP This One Solve by Src Xform cont.
After the transformation the sources can be combined The equivalent current source has value 8V/4kΩ = 2mA 09Feb16 → SKIP This One The Options at This Point Did similar problem on whiteBd Do another source transformation and get a single loop circuit Use current divider to compute IO and then Calc VO using Ohm’s law

09Feb16 → SKIP This One PROBLEM Find VO using source transformation
EQUIVALENT CIRCUITS Norton Norton 09Feb16 → SKIP This One 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

Thevenin w/ Indep. Sources
The Thevenin Equivalent V-Source is computed as the open-loop or open-circuit 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

Source DeActivation Setting a V-source to ZERO to produces a SHORT-Ckt as that is NO Voltage across the Source Setting an I-source to ZERO to produces an OPEN-Ckt as NO Current Flows Thru the Source Note that above applies ONLY to Ckts with ONLY Independent Sources

Src DeActivation Graphically
𝑅 𝑇ℎ = 9/23 kΩ = 391Ω

Src DeActivation by WhtBd
ID Nodes and Resistors ∆ ○ 09Feb16 → SKIP This One □ × □ ∆

09Feb16 → SKIP This One

Thevenin w/ Indep. Sources cont
Since the evaluation of the Thevenin equivalent Resistance for INdependent-Source-Only circuits can be very simple, we can add it to our toolkit for the solution of circuits “Part B” “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 Produces a SHORT

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

Let’s do this one on the WhtBoard
Find: Vt & Rt This is MQ-02e

09Feb16 → SKIP This One 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 09Feb16 → SKIP This One Deactivate the 6V and 2mA Source for RTH

09Feb16 → SKIP This One Thevenin Example cont
For the open circuit voltage we analyze the circuit at Right (“Part A”) Use Loop/Mesh Analysis 09Feb16 → SKIP This One Finally The Equivalent Circuit Then VOC

7kΩ 6V CALCULATE Vo USING NORTON COMPUTE Vo USING THEVENIN PART B

DEpendent & INdependent Srcs
MUST Find The Open Circuit Voltage And Short Circuit Current Solve Two Circuits (Voc & Isc) For Each Thevenin Equivalent Any and all the techniques may be used; e.g., KCL, KVL, combination series/parallel, node & loop analysis, source superposition, source transformation, homogeneity Setting To Zero All Sources And Then Combining Resistances To Determine The Thevenin Resistance is, in General, NOT Applicable!!

Example Recognize Mixed sources The Open Ckt Voltage Solve for VTH
Must Compute Open Circuit Voltage, VOC, and Short Circuit Current, ISC The Open Ckt Voltage Solve for VTH Use V-Divider to Find VX The Short Ckt Current Note that Shorting a-to-b Results in a Single Large Node For Vb Use KVL (Sub VX) Now VTH = VX − Vb

Example cont Need to Find Vx KCL at Single Node Then RTH
Va = Vb = VX Need to Find Vx KCL at Single Node Then RTH Solving For Vx The Equivalent Circuit KCL at Node-b for ISC What are the units of “a” => conductance = A/V

09Feb16 STOP HERE

Illustration Use Thevenin to Determine Vo Partition Guidelines
“Part B” Use Thevenin to Determine Vo Partition Guidelines “Part-B” Should be as Simple As Possible After “Part A” is replaced by the Thevenin equivalent should result in a very simple circuit The DEpendent srcs and their controlling variables MUST remain together Constraint at SuperNode KCL at SuperNode Use SuperNode to Find Open Ckt Voltage

Illustration cont The Controlling Variable
Solving 3 Eqns for 3 Unknowns Yields At Node-A find VA=0 → The Dependent Source is a SHORT Yields Reduced Ckt Now Tackle Short Circuit Current

Illustration cont Using the Reduced Ckt Now Find RTH
Finally the Solution Note: Some ckts can produce NEGATIVE RTH Setting All Sources To Zero And Combining Resistances Will Yield An INCORRECT Value

Numerical Example Find Vo using Thevenin Define Part-A
Find VOC using SuperNode Super node Apply KVL KVL With The Controlling Variable Find

Numerical Example cont
Next The Short Circuit Current MINUS 3V Using VOC & ISC The ONLY Value That Satisfies the Above eqns The Equiv. Ckt KCL at Top Node Recall Dep Src is a SHORT

Note on Example The Equivalent Resistance CanNOT Be Obtained By DeActivating The Sources And Determining The Resistance Of The Resulting InterConnection Of Resistors Suggest Trying it → Rth,wrong = 2.5 kΩ Rth,actual = 0.75 kΩ Req

EXAMPLE: Find Vo By Thevenin
“Part B” Select Partition Use Meshes to Find VOC KVL for V_oc By Dep. Src Constraint Now KVL on Entrance Loop In The Mesh Eqns Solve for VOC The Controlling Variable

Find Vo By Thevenin cont
Now Find ISC The Mesh Equations Then Thevenin Resistance The Controlling Variable Use Thevenin To Find Vo Solving for I1 Find Again Find ISC by Mesh KVL

Illustration The Merthod for Mixed Sources The Open Ckt Voltage
For the Short Ckt Current

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

All Done for Today More on Source Xforms

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

Example Find Vo To Start Notice Need Only V1 and V2 to Find Vo
Identify & Label All Nodes Write Node Equations Examine Ckt to Determine Best Solution Strategy R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2k Is1 =2mA, Is2 = 4mA, Is3 = 4mA, Vs1 = 12 V Notice Need Only V1 and V2 to Find Vo Now KCL at Node 1 Known Node Potential

Example cont. At Node 2 At Node 4
R1 = 1k; R2 = 2k, R3 = 1k, R4 = 2k Is1 =2mA, Is2 = 4mA, Is3 = 4mA, Vs = 12 V To Solve the System of Equations Use LCD-multiplication and Gaussian Elimination

Example cont. The LCDs Now Add Eqns (2) & (3) To Eliminate V4
*2kΩ (1) *2kΩ (2) *2kΩ (3) Now Add Eqns (2) & (3) To Eliminate V4 (4) Now Add Eqns (4) & (1) To Eliminate V2 BackSub into (4) To Find V2 Find Vo by Difference Eqn

Complex SuperNode Write the Node Eqns Set UP
Identify all nodes Select a reference Label All nodes Nodes Connected To Reference Through A Voltage Source Eqn Bookkeeping: V3 SuperNode 2 Constraint Equations One Known Node Voltage Sources In Between Nodes And Possible Supernodes Choose to Connect V2 & V4

Complex SuperNode cont.
Now KCL at Node-3 supernode Vs2 Vs3 Now KCL at Supernode Take Care Not to Omit Any Currents Vs1 Constraints Due to Voltage Sources 5 Equations 5 Unknowns → Have to Sweat Details

Dep V-Source Example Find Io by Nodal Analysis
Notice V-Source Connected to the Reference Node SuperNode Constraint KCL at SuperNode Controlling Variable in Terms of Node Voltage Mult By 12 kΩ LCD

Dep V-Source Example cont
Simplify the LCD Eqn By Ohm’s Law

Numerical Example Select Soln Method Select Mesh Currents
Loop Analysis 3 meshes One current source Nodal Analysis 3 non-reference nodes One super node Both Approaches Seem Comparable → Select LOOP Analysis Specifically Choose MESHES Select Mesh Currents Write Loop Eqns for Meshes 1, 2, 3 by KVL

Numerical Example cont
We Seek Vo, Thus Using Ohm’s Law Need only Find I3 Simplify: Divide Loop Eqns by 1kΩ I Coeffs Become NUMBERS Voltages Converted to mA Note That I1 = IS and Sub into Loop Eqns Substitute into The Remaining Loop Eqns See Next Slide

Numerical Example cont.2
The Loop-2 and Loop-3 Eqns Then by Ohm’s Law

Find Vo- Compare Mesh vs. Loop
Using MESH Currents Using LOOP Currents Treat The Dependent Source As One More Voltage Source Mesh-1 & Mesh-2 Loop-1 & Loop-2

Compare Loop vs. Mesh cont.
Using MESH Currents Using LOOP Currents Now Express The Controlling Variable In Terms Of MESH or LOOP Currents Solving

Compare Loop vs. Node cont.
Using MESH Currents Using LOOP Currents Solutions Finally Notice The Difference Between MESH Current I1 and LOOP Current I1 even Though They Are Associated With The Same Path The Selection Of LOOP Currents Simplifies Expression for Vx and Computation of Vo

Dep Isrc Not Shared by Mesh
Treat The Dependent Source As A Conventional Source Equations For Meshes With Current Sources Express The Controlling Variable, Vx, In Terms Of Loop Currents Then KVL on The Remaining Loop (I3)

Dep Isrc Not Shared by Mesh
Asked to Find Only Vo Need Only Determine I3 The Dep Src Eqns From KVL Eqn for I3 Thus

Find Vo Using Mesh Analysis
Draw the Mesh Currents Controlling Variable In Terms Of Loop Currents Write KVL Mesh Eqns For Mesh-1 & Mesh-2

Find Vo Using Mesh Analysis cont
Substitute & Collect Terms Solve for I2 Finally Vo

WhiteBoard Work Let’s Work This Problem Find the OutPut Voltage, VO IX
Prob e => Vo = (4/3)V = 1.333V

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 (Source or Rat’sNest Generated) 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 for Ckt-A Then By Ohm’s Law

Theorem Proof Outline cont.2
Consider Special Case Where Ckt-B is an OPEN (i =0) The Open Ckt Eqn Suggests Also recall Think y = mx + b 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

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?

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

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

Super Node/Mesh Thevénin/Norton

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