Loading in 5 sec....

Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollegePowerPoint Presentation

Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

Download Presentation

Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

Loading in 2 Seconds...

- 67 Views
- Uploaded on
- Presentation posted in: General

Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Engineering 43

Chp 3.1bNodal Analysis

Bruce Mayer, PE

Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

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

- Find Vo
- To Start
- 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 V2to Find Vo

- Now KCL at Node 1

- Known Node Potential

- 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

The LCDs

*2k

*2k

*2k

(1)

(2)

(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

Consider ThisExample

Conventional Node Analysis Requires All Currents At A Node

SUPERNODE

- But Have Ckt V-Src Reln

- More Efficient solution:
- Enclose The Source, And All Elements In Parallel, Inside A Surface.
- Call That a SuperNode

- Enclose The Source, And All Elements In Parallel, Inside A Surface.

- 2 eqns, 3 unknowns...Not Good
- Recall: The Current thru the Vsrc is NOT related to the Potential Across it

Apply KCL to the Surface

SUPERNODE

- The Source Current Is interior To The Surface And Is NOT Required

- Now Have 2 Equations in 2 Unknowns
- Then The Ckt Solution Using LCD Technique
- See Next Slide

The Equations

- Use The V-Source Rln Eqn to Find V2

SUPERNODE

- Mult Eqn-1 by LCD (12 kΩ)

- Add Eqns to Elim V2

Find the node voltages

And the power supplied

By the voltage source

To compute the power supplied by the voltage source We must know the current through it: @ node-1

BASED ON PASSIVE SIGN CONVENTION THE

POWER IS ABSORBED BY THE SOURCE!!

Write the Node Equations

KCL At v1

- At The SuperNode Have V-Constraint
- v2−v3= vA

- KCL Leaving Supernode

- Now Have 3 Eqnsin 3 Unknowns
- Solve Using Normal Techniques

Find Io

Known Node Voltages

SUPERNODE

- The SuperNode V-Constraint

- Now KCL at SuperNode

- Or

- Lets Turn on the Lights for 5-7 min
- Students are invited to Analyze the following Ckt
- Hint: Use SuperNode

- Determine the OutPut Current, IO

Find Io Using Nodal Analysis

Known Voltages for Sources Connected to GND

SUPERNODE

- 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

Write the Node Eqns

Set UP

Identify all nodes

Select a reference

Label All nodes

supernode

- Nodes Connected To Reference Through A Voltage Source

- Eqn Bookkeeping:
- KCL@ V3
- KCL@ SuperNode,
- 2 Constraint Equations
- One Known Node

- Voltage Sources In Between Nodes And Possible Supernodes

- Choose to Connect V2 & V4

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

- 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

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

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

Simplify the LCD Eqn

- By Ohm’s Law

Find Io

Supernode Constraint

- Controlling Variable in Terms of Node Voltage

- Multiply by LCD of 2 kΩ

- Recall

- Then

- KCL at SuperNode

- So Finally

IX

- Let’s Work This Problem

- Find the OutPut Voltage, VO