1. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 1 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Chp 5.3b Thevénin Norton

2. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 2 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Thevénin’s Equivalence Theorem  Thevenin Equivalent Circuit for PART A  v TH = Thévenin Equivalent VOLTAGE Source  R TH = Thévenin Equivalent SERIES RESISTANCE

3. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 3 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Norton’s Equivalence Theorem  Norton Equivalent Circuit for PART A  i N = Norton Equivalent CURRENT Source  R N = Norton Equivalent PARALLEL RESISTANCE

4. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 4 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example (reminder)  DeActivate The Sources  Note That All Three Resistors Are in Parallel  Next Find The Open Ckt Voltage as V TH = V oc  Find the Thévenin Equivalent at Terminals a-b Note That the Ckt Drawing Can be Simplified Parallel

5. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 5 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example cont.  ReDrawing The Circuit  Using Voltage Divider SOURCE Xform 8/6 kΩ 4/3 kΩ = = 3mA x 4kΩ/3 a b  

6. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 6 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Thevenin w/ Only Dependent Srcs  A circuit with only dependent sources cannot SELF-START Actually that statement has to be qualified a bit. What happens if a=R 1 +R 2 ? –I x can take ANY Value  For ANY PROPERLY Designed Circuit With ONLY Dependent Sources KVL

7. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 7 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Thevenin w/ Dependent Srcs cont  V TH = 0 is a BIG Simplification  But Need A Special Approach To Find the Thevenin Equivalent Resistance  Since The Circuit Cannot Self Start, PROBE It With An EXTERNAL Source The PROBE Can Be Either A VOLTAGE Source Or A CURRENT Source Whose Value Can Be Chosen ARBITRARILY –Which One To Choose Is Often Determined By The Simplicity Of The Resulting Circuit

8. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 8 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Voltage Probe  If a VOLTAGE Probe is Chosen, Then Must Find the CURRENT Supplied by The Probe V-source  Since V P is Arbitrary, Usually Set it to 1.00000

9. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 9 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Current Probe  If a CURRENT Probe is Chosen, Then Must Find the VOLTAGE Generated by The Probe I-source  The Value for I P is Arbitrary, Usually Set it to 1.00000

10. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 10 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example  Find the Thevenin Equivalent Circuit at A-B  Use A CURRENT or VOLTAGE Probe? Using a Voltage Probe Results In Only One Node Not Connected to GND Through a Source  Apply the V-Probe, and Analyze by KCL at V 1

11. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 11 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example cont.  The Controlling Variable  Solving the Eqns  To Determine R TH need to Calc Probe Current  Calc R TH using V P & I P 933Ω

12. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 12 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Dependent Source Example  Find the Thevenin Equivalent circuit at A-B  Only Dependent Sources, Thus V TH = 0  Apply a CURRENT Probe to Determine The equivalent resistance  Have a“Conventional” circuit with dependent sources use node analysis Let I P = 1 mA = 1x10 -3  By KCL at V 1 and V 2

13. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 13 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Dependent Source Example cont  Now the Controlling Variable  Sub for I x in V 1 KCL  Multiply LCD Against Both Resulting KCL Eqns  Eliminate V 1  Then  And From the Ckt Observe V P = V 2  With I P = 1 mA 1.43 kΩ

14. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 14 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Illustration  Find Thevenin Equivalent At Terminals A-B Using a 1 mA Current Probe  Use LOOP Analysis to Find V BA = V P  Carefully Choose Loops  Now Loop Eqns

15. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 15 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Illustration cont  The Controlling Variable  The Potential Across the Current Probe by KVL on I 2 loop Recall V P = V B − V A  Now the R TH  The Thevenin Equivalent  Solving for the Values I 1 = I P /2 = 0.5 mA I 2 = I P = 1 mA I 3 = 0 V P = 2 V R TH = 2 kΩ

16. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 16 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis DEpendent & INdependent Srcs  Find The Open Circuit Voltage And Short Circuit Current  Solve Two Circuits (V oc & I sc ) 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!!

17. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 17 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Illustration  Use Thevenin to Determine V o  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 sources and their controlling variables MUST remain together  Use SuperNode to Find Open Ckt Voltage “Part B”  Constraint at SuperNode  KCL at SuperNode

18. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 18 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Illustration cont  The Controlling Variable  Solving 3 Eqns for 3 Unknowns Yields  Now Tackle Short Circuit Current  At Node-A find  V A =0 → The Dependent Source is a SHORT Yields Reduced Ckt

19. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 19 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Illustration cont  Using the Reduced Ckt  Now Find R TH  Setting All Sources To Zero And Combining Resistances Will Yield An INCORRECT Value  Finally the Solution  Note Some Values of “a” Result in NEGATIVE R TH

20. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 20 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example  Find V o using Thevenin  Define Part-A  Find V OC using SuperNode Super node KVL  Apply KVL  With The Controlling Variable Find

21. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 21 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Numerical Example cont  Next The Short Circuit Current  The ONLY Value That Satisfies the Above eqns  KCL at Top Node Recall Dep Src is a SHORT MINUS MINUS 3V  Using V OC & I SC  The Equiv. Ckt

22. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 22 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 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 → R th,wrong = 2.5 kΩ –R th,actual = 0.75 kΩ R eq

23. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 23 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis EXAMPLE: Find V o By Thevenin  Select Partition  Use Meshes to Find V OC “Part B” KVL for V_oc  In The Loop Eqns  The Controlling Variable  By Dep. Src Constraint  Solve for V OC  Now KVL on Entrance Loop

24. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 24 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Find V o By Thevenin cont  Now Find I SC  The Mesh Equations  The Controlling Variable  Solving for I 1 Find Again  Find I SC by Mesh KVL  Then Thevenin Resistance  Use Thevenin To Find Vo

25. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 25 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis Illustration  The Merthod for Mixed Sources  For the Short Ckt Current  The Open Ckt Voltage

26. BMayer@ChabotCollege.edu ENGR-43_Lec-05-3b_Thevein-Norton_Part-b.ppt 26 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis WhiteBoard Work LLet’s Work an Example on the Board FFind the NORTON Equivalent at A-B