ECE 4991 Electrical and Electronic Circuits Chapter 3

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

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

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

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

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

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

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

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

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

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

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

Working with Nodal Analysis with Voltage Sources va vb R1 Vs R2 R3

Working with Nodal Analysis V R1 R3

Working with Nodal Analysis V I R4 R2 R5 R6

Working with Nodal Analysis

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

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

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

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

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

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

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

Working with Mesh Analysis

Working with Mesh Analysis V R1 R3

Working with Mesh Analysis V R5 R6

Working with Mesh Analysis

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

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

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 - -

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

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 -

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

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 -

SUPERPOSITION: Example 10 cont.: VR due to current source only (VR1): 400 i1 = 5mA Current divider 400 + 200 + 250 Vi = i1250 = 2.353mA250 = 0.588V VR1 = -Vi VR1 = -0.588V 400 200 + - 5mA VR 250 Vi - i1 +

SUPERPOSITION: Example 10 cont.: VR due to voltage source only (VR2): 250 VR2 = 5V = 1.471V Voltage divider 400 + 200 + 250 VR2 = 1.471V 400 200 + + + 5V VR 250 VR2 - - -

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

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

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 - -

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

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

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 = 211.27mA 20 5 A + io + 15V 10 12 VOC - - B

THEVENIN and NORTON CIRCUITS: Example 11 cont.: VT = VOC = 211.3mA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 = 5.92  20 5 A + io + 15V 10 12 VOC - - B

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 - -

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

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

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 -

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

Norton Equivalent Circuit

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

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

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

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

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

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

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

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

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