Lecture 2 Review

• Methods of Analysis

—Nodal analysis

• Without Voltage Sources

• With Voltage Sources

—Mesh analysis

• Without Current Sources

• With Coltage Sources

—Analysis by Inspection

Quiz

An open-loop gain: 2105

Input resistance: 2 M Output resistance: 50 Find the closed-loop gain:

s

o

v

v

Lecture 3 DC Circuits Circuit Theorems: to simplify circuit analysis

Lecture Objectives

• Linear circuit

• Superposition

• Equivalent Circuits

—Simple Circuits

—Y- Transforms

—Source transformation

—Thevenin's theorem

—Norton's theorem

• Maximum power transfer

Linear circuit

Linear Circuit Input(excitation),i Output(response), v

• Linearity Property

—Homogeneity:

—Additivity:

iRv

000101 )( , kvRikRkivkii

21212121 )( , vvRiRiRiiviii

Example: Linearity

Assume I0 = 1 A,… the source current Is =? 5 A

Since Is = 15 A = 3 × 5A, according to linearity (homogeneity), I0 = ? 3 A

3V

5V

8V

1A 2A

3A

6V

14V

2A

5A

Superposition: several sources

• Linear circuit: Additivity

• Voltage or Current = algebraic sum of v or i due to each independent source acting alone, respectively.

• Can we use superposition for power? No! Why?

If there are N sources, you’ll need to do similar calculation N times.

Example: Superposition

Use the superposition theorem to find v in the circuit:

v = 10 V

Voltage source 0 V Short circuit

Current source 0 A Open circuit

Turn off

How about dependent sources?

Equivalence of circuits

• Series-parallel combination • Y- transformation • Source Transformation • Thevenin’s Theorem • Norton’s Theorem

a

b

What elements are in the box?

What is the behaviour of the circuit? Voltage-Current relation

Voltage-current relation at terminals a and b is identical to that of its equivalent circuit.

• Series circuits

— Current: elements in series carry the same current

— Resistance in series: the sum of the individual resistances

• Parallel circuits

— Voltage: elements in parallel have the same voltage

— Conductance in parallel: the sum of the individual conductances

Simple resistive circuits

A

LR

One path I L2 L1

211121 RR

A

L

A

L

A

LLRT

two paths

A=A1+A2

A1

A2

21

21

2121

21

111111

,

GGG

RRL

A

L

A

L

AA

R

AA

LR

T

T

T

V

L=L1+L2

Equivalent

- Y Transformation

cba

ba

cba

ac

cba

cb

RRR

RRR

RRR

RRR

RRR

RRR

3

2

1

Y T

Y- Transformation

Example

Obtain the equivalent resistance Rab for the circuit and use it to find current i

Source transformation

SouceCurrent A Source VoltageA

Check: Voltage Source Current Source Turn-off source R

Open Circuit

Short Circuit

sv Ris

R

vs

si

Replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa.

R

R

That’s what the external sources ‘feel’

this part of the circuit. That’s how this part of circuit acts on

the load as a ’source’.

Example

Use source transformation to find vo in the circuit

Thevenin's theorem

Other elements - fixed

Load – variable

Vth : Open circuit voltage

RTh : the input resistance at the terminals when the independent sources are turned off

Example

Find the Thevenin equivalent circuit of the circuit, to the left of the terminals a-b. Then find the current through RL = 6, 16, and 36

Norton's theorem

Maximum power transfer

• Minimizing power dissipated in the process of transmission and distribution • Maximize the power delivered to a load

LRip 2LTh

Th

RR

Vi

L

LTh

ThL R

RR

VRip

2

2

Apendix

Op Amp - typical packages

Looking from the top

Pin 1 is always to the left of the notch or dot.

741 General-purpose: Fairchild Semiconductor … Intel

For package information: http://www.intersil.com/design/packages/

1. An active circuit element 2. Perform mathematical

operations

Feedback path - example

An open-loop gain: 2105

Input resistance: 2 M Output resistance: 50 Find the closed-loop gain:

s

o

v

v

Nodal analysis

Feedback: negative feedback

9999699.1s

o

v

v

Insensitive to A

This is tedious.

Ideal Op Amp

• Infinite open-loop gain, • Infinite input resistance, • Zero output resistance,

A

iR

0oR

Virtual open circuit i1 = i2 =0

Virtual close circuit vd = 0; v1 = v2

Idea Op Amp - Example

Rework it using the ideal op amp model

v1

v2

3

1

3

1

10201010

os vvvv

021 vv

2

1020

0

1010

033

s

o

os

v

v

vv

9999699.1s

o

v

v

Working with non-ideal: Negligibly small error results from assuming ideal op amp characteristics Virtual open circuit: i1 = i2 =0 Virtual close circuit: vd = 0; v1 = v2

i2 =0

i1 =0

Node 1:

Idea Op Amp: 021 ii

Idea Op Amp:

Lecture 4 DC Circuits Capacitors and Inductors