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

2. 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

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

4. 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

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

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

7. 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

8. 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

9. 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

10. 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

11. 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

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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. 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

20. 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

21. 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

22. 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?

23. 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

24. 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”

25. 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

26. 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

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

28. 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

29. 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

30. 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

31. 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

32. 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

33. 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

34. 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

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

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

37. Example
Xform
KVL
Deactivate Srcs for RTH
Use Loops for VTH

38. 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

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