solving linear equations nodal analysis supernodes (nodal...

Analysis Methods Overview

• Solving Linear Equations

• Nodal Analysis

• Supernodes (Nodal Analysis with Voltage Sources)

• Mesh Analysis

• Supermeshes (Mesh Analysis with Current Sources)

• Introduction to BJT Transistors

This is a very important chapter.

Portland State University ECE 221 Analysis Methods Ver. 1.66 1

Review of Basic Concepts: Current

i4

i5

i3

i2

i1

• What goes in, has to come out

• Kirchhoff’s current law

• Similar to conservation of mass

• Conservation of electrons


Review of Basic Concepts: Voltage

10 V-

+

-

++ - + -

2 kΩ2 kΩ

5 kΩ 7 kΩv1 v2

v3 v4

• The voltage drop from one node to another is the same, nomatter what path is chosen

• Kirchhoff’s voltage law


Resistors in Parallel with Voltage Sources

CircuitRVs vo

-

+

CircuitVs vo

-

+

• What is vo in each case?

• What effect does the resistor have on the current pumped into thecircuit?


Resistors in Series with Current Sources

CircuitIs CircuitIs

Rio io

• What is io in each case?

• What effect does the resistor have on the voltage seen by thecircuit?


Network Terminology

Planar Circuit A circuit that can be drawn on a plane with nocrossing branches

Node Point or portion of a circuit where 2 or more elements arejoined

Essential Node Point or portion of a circuit where 3 or moreelements are joined

Branch Path that connects 2 nodes

Essential Branch Path that connects 2 essential nodes w/o passingthrough an essential node

Loop Path with last node same as starting node that does not crossitself

Mesh Loop that does not enclose any other loops

Note: this isn’t in the text.


Example 1: Terminology

20 V 2 A

R1 R2

R3 R4 R4

R6 R7 R8

35ip

ip

Identify the following informationNodes: Essential Nodes:Branches: Essential Branches:EB’s with Unknown Current: Meshes:


Example 2: Circuit Analysis The Hard Way

10 V

i1 i3

i2 i4

2 mA

1 kΩ 2 kΩ

5 kΩ 10 kΩ

Can solve with KCL & KVL. Four unknowns.


Solving Linear Equations

• Much of our circuit analysis will focus on finding a set of linearequations and solving these equations

• Need as many equations as there are unknowns

• Three possible approaches

– Algebra (elimination, substitution, etc.)

– Cramer’s rule

– Linear algebra

• Last is easiest and least susceptible to errors

• Requires use your scientific calculators


Example 2: Solving Linear Equations

i1 = i2 + i3

i4 = i3 + 2m

10 = 1k i1 + 5k i2

5k i2 = 2k i3 + 10k i4

Rewrite so variables are in consistent order on left side and constantsare on the right side

i1 − i2 − i3 = 0− i3 + i4 = 2m

1k i1 + 5k i2 = 10+ 5k i2 − 2k i3 − 10k i4 = 0


Example 2: Continued (1)

i1 − i2 − i3 = 0− i3 + i4 = 2m

1k i1 + 5k i2 = 10+ 5k i2 − 2k i3 − 10k i4 = 0

In Matrix form this becomes⎡⎢⎢⎣

1 −1 −1 00 0 −1 1

1k 5k 0 00 5k −2k −10k

⎤⎥⎥⎦

⎡⎢⎢⎣

i1i2i3i4

⎤⎥⎥⎦ =

⎡⎢⎢⎣

02m100

⎤⎥⎥⎦

or

Ai = b



Ai = b where

A =

⎡⎢⎢⎣

1 −1 −1 00 0 −1 1

1k 5k 0 00 +5k −2k −10k

⎤⎥⎥⎦ i =

⎡⎢⎢⎣

i1i2i3i4

⎤⎥⎥⎦ b =

⎡⎢⎢⎣

02m100

⎤⎥⎥⎦

• Your calculator should be able to solve this directly

• You should only need to enter A and b

• Your calculator will return a vector i

• Simultaneously solves for all the unknown variables

• Much faster than Cramer’s rule or brute-force algrebra

• Read the manuals for your calculators

• This will save you time (homework & exams) and reduce errors



Linear Equations:⎡⎢⎢⎣

1 −1 −1 00 0 −1 11k 5k 0 00 5k −2k −10k

⎤⎥⎥⎦

⎡⎢⎢⎣

i1i2i3i4

⎤⎥⎥⎦ =

⎡⎢⎢⎣

02m100

⎤⎥⎥⎦

Calculator should return:⎡⎢⎢⎣

i1i2i3i4

⎤⎥⎥⎦ =

⎡⎢⎢⎣

+0.909+1.818−0.909+1.091

⎤⎥⎥⎦ mA


Nodal Analysis: Introduction

• There is an another way to solve for currents and voltages

– Easier

– More methodical

– Still based on Ohm’s law, KVL, & KCL

• Nodal analysis is one of two key methods

• Mesh analysis is the other

• We will discuss nodal analysis first

• Based on KCL

• Must understand terminology introduced earlier

• Use to solve for voltages

• All voltages have a common reference point


Nodal Analysis: Step 1 – Identify Essential Nodes

10 V 2 mA

1 kΩ 2 kΩ

5 kΩ 10 kΩ

• Some essential nodes may include portions of the circuit (pieces ofwire)

• Circle the entire node to prevent errors


Nodal Analysis: Step 2 – Pick a Reference

10 V 2 mA

1 kΩ 2 kΩ

5 kΩ 10 kΩ

• Second step is to pick a reference node

• Is often easiest to choose the node that interconnects the mostbranches

• Must be an essential node

• Usually is at bottom of circuit

• Label with the same symbol used for ground


Nodal Analysis: Step 3 – Label Other Essential Nodes

10 V 2 mA

1 kΩ 2 kΩ

5 kΩ 10 kΩ

• Also a bit easier if voltages are labeled

• All voltages are measured relative to the reference node


Nodal Analysis: Step 4 – Apply KCL All Labeled Nodes

10 V 2 mA

1 2

-

+v2

-

+v1

1 kΩ 2 kΩ

5 kΩ 10 kΩ


Nodal Analysis: Step 5 – Solve Linear Equations

Linear Equations:

Solution (from calculator):


Nodal Analysis: Step 6 – Solve for Variables of Interest

10 V 2 mA

1 2

-

+v2

-

+v1

i1 i3

i2 i4

1 kΩ 2 kΩ

5 kΩ 10 kΩ

i1 =i2 =i3 =i4 =


Nodal Analysis: Review of Steps

• Step 1: Identify essential nodes

• Step 2: Pick a reference

– Must be an essential node

– Always label with the ground symbol

– Best to pick essential node with most branches

– Often at the bottom of the circuit diagram

• Step 3: Label other essential nodes

• Step 4: Apply KCL to all labelled nodes except reference node

• Step 5: Solve linear equations

– Generates voltage at each node (relative to reference node)

• Step 6: Solve for variables of interest

– Usually easy after Step 5


Nodal Analysis: Use of Laws

• All three laws are used

• KCL is applied at each labelled node except the reference node

• Ohm’s law is used to determine the current in branches thatcontain resistors

• KVL is used to determine the voltage drop across the resistors


Example 3: Nodal Analysis

144 V

-

+v2

-

+v1 3 A

4 Ω

5 Ω10 Ω

80 Ω

Solve for v1 and v2.


Example 3: Workspace



20 mA

-

+v2

-

+v1

-

+v3 5 V2 kΩ

2.7 kΩ2.7 kΩ

3.3 kΩ

4.7 kΩ

10 kΩ

Solve for v1, v2, and v3.


Example 5: Dependent Voltage Source

50 V-

+

10 Ω

10 Ω

30 Ω 39 Ω 78 Ω

v/5

v

Solve for v.

• What effect does the 10 Ω resistor have on the circuit?

• What is the current flowing through the dependent source?

• How can we apply KCL at the essential nodes without thisinformation?

• Ans: One extra variable

• Implies we need an extra equation


Example 5: Continued

50 V-

+

10 Ω

10 Ω

30 Ω 39 Ω 78 Ω

v/5

v

Solve for v.


Nodal Analysis and Supernodes

• Supernodes eliminate the need to introduce an extra variable(unknown current)

• Necessary when a voltage source is between two labeled nodes(excluding reference node)

• Still need to use voltage source to generate one of the equations


Example 6: Dependent Source Continued

50 V-

+

10 Ω

10 Ω

30 Ω 39 Ω 78 Ω

v/5

v

Solve for v. Use a supernode.


Example 7: Dependent Voltage Source

20 V

+ -

1 Ω2 Ω 4 Ω

20 Ω 40 Ω 80 Ω 3.125v

v

35iφ

iφ

Find the power developed by the 20 V source.



11 mA

i1

20 Vi2

10 Vi3

250 Ω

500 Ω

1 kΩ

25 kΩ

Solve for i1, i2, and i3.



1 A

3i

i

-

+v

1 Ω

1 Ω

2 Ω

2 Ω4 Ω

Solve for v.


Mesh Analysis: Introduction

• Recall: There is an easier way to solve for currents and voltagesthan applying KVL and KCL directly

• Nodal analysis is one of two key methods

• Mesh analysis is the other

– Applies KVL to solve for currents

– More abstract

– Work with imaginary currents

– Only applies to planar circuits


Mesh Analysis: Step 1 – Label Meshes

40 V 64 V

ia icib

1.5 Ω2 Ω

3 Ω 4 Ω

45 Ω

Find the branch currents ia, ib, and ic.

• Recall: A mesh is a loop that does not enclose any other loops


Mesh Analysis: Step 2 – Apply KVL to Each Mesh

40 V 64 V

ia icib

1.5 Ω2 Ω

3 Ω 4 Ω

45 Ω


Mesh Analysis: Step 3 – Solve Linear Equations

[50 −45−45 50.5

] [i1i2

]=

[4064

]

i1 = 9.8 A

i2 = 10 A


Mesh Analysis: Step 4 – Solve for Variables of Interest

40 V 64 V

ia icib

1.5 Ω2 Ω

3 Ω 4 Ω

45 Ω

ia =ib =ic =


Mesh Analysis: Review of Steps

• Step 1 – Label Meshes

• Step 2 – Apply KVL to Each Mesh

• Step 3 – Solve Linear Equations

• Step 4 – Solve for Variables of Interest

– Usually easy after Step 3

• Limitation: Only works with planar circuits


Example 10: Mesh Analysis

12 V

110 V 70V

2 Ω

3 Ω

4 Ω

6 Ω

10 Ω 12 Ω

Find the total power developed in the circuit.



18 V 15 V3 A

2 Ω

3 Ω

6 Ω

9 Ω

Find the total power dissipated.

• Problem: What is the voltage across the 3 A source?

• Solutions

1 Add it as a variable

2 Use a supermesh

• Second option requires less work



18 V 15 V3 A

2 Ω

3 Ω

6 Ω

9 Ω

Find the total power dissipated. Add a variable.



18 V 15 V3 A

2 Ω

3 Ω

6 Ω

9 Ω

Find the total power dissipated. Use a supermesh.



200 V

4.3 id

ie

ib

id

ia

ic

10 Ω

10 Ω

25 Ω

50 Ω

100 Ω

Find the branch currents ia – ie.



1.5 mA

8 V

2 kΩ

3 kΩ

4 kΩ

4 kΩ 4 kΩ

5 kΩ

7 kΩ

3iα

iα

Solve for iα


Nodal versus Mesh Analysis

• You should know how to do both

• Which is more efficient depends on the problem

• Will learn which to use with experience

• Nodal analysis used more often

• On exams, I will specify which method to use

Concise Summary:

Nodal Analysis Mesh AnalysisMethod KCL KVLSolve For Node Voltages Mesh Currents“Super” Conditions Voltage Sources Current Sources


solving linear equations nodal analysis supernodes (nodal...

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