1. ECE 4991 Electrical and Electronic Circuits Chapter 3

2. Where are we?
Chapter 2 - The basic concepts and practice at analyzing simple electric circuits with sources and resistors
Chapter 3 – More harder networks to analyze and the notion of equivalent circuits
Chapter 4 – Capacitors and inductors added to the mix
Chapter 5 – Analyzing transient situations in complex passive networks
Chapter 8 – New subject – the wonders of operational amplifiers as system elements
Chapter 9 – Introduction to semiconductors – the basics and diodes – more network analysis
Chapter 10 – Bipolar junction transistors and how they work – now you can build your own op amp

3. What’s Important in Chapter 3
Definitions
Nodal Analysis
Mesh Analysis
The Principle of Superposition
Thevenin and Norton Equivalent Circuits
Condition for Maximum Power Transfer

4. 1. Definitions Node voltages Branch currents “Ground” KCL
Nodal Analysis
Mesh currents
KVL
Mesh Analysis
Principle of Superposition
Equivalent circuit
Thevenin theorem
Norton theorem
One-port networks
Source loading

5. 2. Nodal Analysis Used to “analyze” circuits
Solve for currents, voltages, power, etc., throughout circuits
Applies KCL to nodes
Often used in concert with Ohm’s Law

6. Node Method Find nodes – Identify ground node
Label branch currents & node voltages
Node voltages, if not defined by a voltage source, are independent variables
Write KCL for nodes
Solve for unknowns

7. Working with Nodal Analysis
Reference Node Selection (usually Ground)
Define remaining n-1 node voltages.
Apply KCL to obtain Nodal Expressions for node a and b respectively: is – i1 – i2 = 0
i2 – i3 = 0
Node a
Node b iS i1 i2 i3
Node c

8. Working with Nodal Analysis (cont.)
Apply Node Voltage Method
Node a
Node b iS i1 i2 i3
Node c

9. Working with Nodal Analysis (cont.)
Substitution step leads to:
Or equivalently to:

10. Example 3.2
Solving for all unknown currents and voltages in the circuit of Figure 3.5
R1=1kΩ, R2=2kΩ, R3=10kΩ, R4=2kΩ,
I1=10mA, I2=50mA
Node 1
Node 2

11. Example 3.2 (cont.)
Applying KCL at nodes 1 & 2:
After rewriting:

12. Example 3.2 (cont.)
Substituting actual values leads to the system of equations:

13. Working with Nodal Analysis with Voltage Sourcesva vb R1 Vs R2 R3

14. Working with Nodal AnalysisV R1 R3

15. Working with Nodal AnalysisV I R4 R2 R5 R6

16. Working with Nodal Analysis

17. For Next Time Sign onto Blackboard, if still have not
Practice Nodal Analysis
Learn about rest of chapter 3, particularly about mesh analysis

18. 3. Mesh Analysis Also used to “analyze” circuits
Solve for currents, voltages, power, etc., throughout circuits
Applies KVL to meshes
Often used in concert with Ohm’s Law

19. Mesh Method Identify meshes and mesh currents
For n meshes and m current sources, there are n-m independent variables
Write KVL for all meshes with unknown mesh currents
Solve for unknowns I

20. Working with Mesh Analysis
Defining Meshes (use the rule consistently: e.g., clock wise) v1 v3 _ _ + + + + i1 v2 i2 v4 _ _

21. Mesh Analysis Apply KVL Vs-v1-v2 = 0 v2-v3-v4= 0 _ _ + + + + _ _ v1 v3

22. Mesh Analysis
Apply Ohm law: v1 v3 _ _ + + + + i1 v2 i2 v4 _ _

23. Mesh Analysis _ _ + + + + _ _ Rewriting the equations: v1 v3 v4 i1 v2

24. Working with Mesh Analysis

25. Working with Mesh AnalysisV R1 R3

26. Working with Mesh AnalysisV R5 R6

27. Working with Mesh Analysis

28. For Next Time Sign onto Blackboard, if still have not
Keep practicing Nodal Analysis
Practice Mesh Analysis
Learn about rest of chapter 3, particularly about equivalent circuits

29. 4. The Principle of Superposition
When working with linear circuits, can find the solution for each energy source and combine the results (Principle of Superposition)
Procedure:
Remove all but one energy source
V sources  short wires
I sources  opens
Solve the circuit
Repeat for a different energy source
Add up the solutions

30. SUPERPOSITION: + Superposition Principle:
Section 3.5
The output of a circuit can be found by finding the contribution from each source acting alone and then adding the individual responses to obtain the total response.
iT = i1 + i2 + V1 + V1 - - + iT R i1 R i2 R + + V2 V2 - -

31. SETTING SOURCES EQUAL TO ZERO:
Voltage Source:
In order to set a voltage source to zero, it is replaced by a short circuit. R1 R2 + iS VS R3 - R1 R2
Voltage source set equal to zero iS R3

32. SETTING SOURCES EQUAL TO ZERO:
Current Source:
In order to set a current source to zero, it is replaced by an open circuit. R1 R2 VS + iS R3 - R1 R2
Current source set equal to zero VS + R3 -

33. SUPERPOSITION: Example 10: Calculate VR using superposition: - - 400
200 + + 5V
5mA VR -
250 -

34. SUPERPOSITION: Example 10 cont.:
1. Turn off all independent sources except one and find response due to that source acting alone.
Turning off voltage source:
400
200 +
Voltage source set equal to zero
5mA VR
250 -

35. SUPERPOSITION: Example 10 cont.: VR due to current source only (VR1):
400
i1 = 5mA
Current divider
400  
Vi = i1250 = mA250 = V
VR1 = -Vi
VR1 = V
400
200 + -
5mA VR
250 Vi - i1 +

36. SUPERPOSITION: Example 10 cont.: VR due to voltage source only (VR2):
250
VR2 = 5V
= V
Voltage divider
400  
VR2 = V
400
200 + + + 5V VR
250
VR2 - - -

37. SUPERPOSITION: Example 10 cont.: VR = VR1 + VR2 = -0.588V + 1.471V
400
200 + +
250 5V
5mA VR - -

38. THEVENIN and NORTON CIRCUITS:
Section 3.5
Thevenin and Norton circuits deal with the concept of equivalent circuits.
Even the most complicated circuits can be transformed into an equivalent circuit containing a single source and resistor.
When viewed from the load, any network composed of ideal voltage and current sources, and of linear resistors, may be represented by an equivalent circuit consisting of an ideal voltage source VT in series with an equivalent resistance RT. A A RT
Thevenin Circuit VT + IN
Norton Circuit RN - B B

39. THEVENIN and NORTON CIRCUITS:
Thevenin Equivalent Circuits:
At this point you should be asking yourself two questions; how do we calculate the Thevenin voltage and the Thevenin resistance?
Thevenin Voltage (VT):
The Thevenin voltage is equal to the open-circuit voltage at the load terminals with the load removed. R1 R3 + VS +
VT = VOC R2
VOC RL - -

40. THEVENIN Equivalent CIRCUITS:+ VS + R2
VOC
VT = VOC i - -

41. THEVENIN and NORTON CIRCUITS:
Thevenin Resistance (RT):
The Thevenin resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero). R1 R3 VS + R2 RL - R1 R3
VS set equal to zero
REQ = RT R2

42. THEVENIN and NORTON CIRCUITS:
Example 11:
Find the Thevenin equivalent circuit at terminals ‘A’ and ‘B’:
Thevenin Voltage:
VT = VOC = VAB = V12 = io12
Using mesh analysis: iO = mA
20 5 A + io +
15V
10
12
VOC - - B

43. THEVENIN and NORTON CIRCUITS:
Example 11 cont.:
VT = VOC = mA12
VT = 2.54V
Thevenin Resistance:
Setting all sources equal to zero and looking back into the circuit from terminals “A” and “B”:
RT = [(20 10) + 5] 12
RT = 
20 5 A + io +
15V
10
12
VOC - - B

44. THEVENIN and NORTON CIRCUITS:
Check
THEVENIN and NORTON CIRCUITS:
Example 12:
Find the Thevenin equivalent circuit seen by the load RL:
VT: VT = VOC = V700 = 4mA700
VT = 2.8V
4mA
400
500 + +
10mA
VOC
200
700
V700 RL - -

45. THEVENIN and NORTON CIRCUITS:
Example 12 cont.:
Thevenin Resistance: Set all sources equal to zero:
RT: RT = 500  RT = 1.2k
400
500 + +
VOC
200
700
V700 RL - -

46. THEVENIN and NORTON CIRCUITS:
Example 12 cont.: VT
Thevenin Resistance:
RT =
iSC
700
iSC = 4mA
= 2.33mA
Current Divider
700 
RT = 2.8V/2.33mA
RT = 1.2k
500
400
4mA +
iSC
10mA
200
700
V700 -

47. THEVENIN and NORTON CIRCUITS:
Norton Equivalent Circuits:
Norton Current (IN):
The Norton current is equal to the short-circuit current at the load terminals with the load removed. R1 R3
iSC VS +
IN = iSC R2 -

48. Norton Equivalent Circuit
iSC VS + i1 R2 - i2
Mesh Analysis:
Node Analysis:

49. Norton Equivalent Circuit

50. THEVENIN and NORTON CIRCUITS:
Norton Resistance (RN):
The Norton resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero). R1 R3
iSC VS + R2 - R1 R3
VS set equal to zero
REQ = RN R2

51. THEVENIN and NORTON CIRCUITS:
Summary:
Thevenin Equivalent Circuit:
The Thevenin voltage is equal to the open-circuit voltage at the load terminals with the load removed.
The Thevenin resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero) or VT/iSC.
VT = VOC = VAB (with load removed)
RT = VT/iSC = REQ as seen by RL

52. THEVENIN and NORTON CIRCUITS:
Summary:
Norton Equivalent Circuit:
The Norton current is equal to the short-circuit current at the load terminals with the load removed.
The Norton resistance is the equivalent resistance seen by the load with all independent sources removed (set equal to zero) or VT/IN.
IN = ISC (with load removed)
RN = RT VOC/IN = REQ as seen by RL

53. Source Transformation
Source transformation allows for the conversion of an ideal voltage source in series with a resistor to an ideal current source in parallel with a resistor and vice versa.
As previously seen, any circuit can be transformed to its Thevenin or Norton equivalent circuit at the load resistance RL. Therefore, a voltage source in series with a resistor (Thevenin) can be transformed to a current source in parallel with a resistor (Norton) and the V-I characteristics at the terminals “A” “B” will be the same. A A R1 IS
Rest
of Circuit
Rest
of Circuit + VS
Source Transformation R1 - B B

54. Source Transformation
Source transformation allows for the conversion of an ideal voltage source in series with a resistor to an ideal current source in parallel with a resistor and vice versa. A R1
Rest
of Circuit +
VS = ISR1 - B
Source Transformation A
Rest
of Circuit
IS = VS/R1 R1 B

55. SOURCE TRANSFORMATION:
Source transformation allows for the conversion of an ideal voltage source in series with a resistor to an ideal current source in parallel with a resistor and vice versa. R1 R3 + VS R2 R4 -
Voltages across and currents through R2, R3, and R4 are the same for both circuits!
IS = VS/R1 R3 R1 R2 R4

56. SOURCE TRANSFORMATION:
Example 13:
Use source transformation and current divider rule to calculate io:
1.5k
1k iO + 8V
2k
300 -

57. SOURCE TRANSFORMATION:
Example 13 cont.:
Converting the voltage source in series with the 1.5k resistor to a current source in parallel with a resistor we have the following circuit:
Same V-I characteristics
1k iO
8V/1.5k = 5.33mA
1.5k
2k
300

58. SOURCE TRANSFORMATION:
Example 13 cont.:
1/1.3k
iO =
5.33mA
1/1.5k + 1/2k + 1/1.3k
iO = 2.12mA
1k iO
5.33mA
1.5k
2k
300