CHAPTERS 5 & 6CHAPTERS 5 & 6

NETWORKS 1: NETWORKS 1: 0909201-010909201-01 8 October 2002 – Lecture 5b

ROWAN UNIVERSITYROWAN UNIVERSITY

College of EngineeringCollege of Engineering

Professor Peter Mark Jansson, PP PEProfessor Peter Mark Jansson, PP PEDEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERINGDEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING

Autumn Semester 2002Autumn Semester 2002

networks I

Today’s learning objectives – apply new methods for reducing

complex circuits to a simpler form equivalent circuits superposition Thévenin’s equivalent Norton’s equivalent

build understanding of maximum power

introduce the operational amplifier

new concepts from ch. 5

electric power for cities - done source transformations - done superposition principle - done Thévenin’s theorem - continue Norton’s theorem maximum power transfer

homework 5

Problems 5.3-1, 5.3-4, 5.3-5, 5.3-6, 5.4-1 review(ed) in lab, 5.4-2, 5.4-6, 5.5-1, 5.5-3, 5.5-9, 5.6-2, 5.6-4, 5.7-1, 5.7-6

Chapter 6 Pages 244-245 Problems 6.4-1, 6.4-2, 6.4-6

next monday’s - test two covers Chapters 3.4-6.4 current division node voltage circuit analysis mesh current circuit analysis when to use n-v vs. m-c source transformations superposition principle Thevenin’s equivalent - Norton’s equivalent maximum power transfer operational amplifiers

Thévenin’s theorem

GOAL: reduce some complex part of a circuit to an equivalent source and a single element (for analysis) THEOREM: for any circuit of resistive elements and energy sources with a terminal pair, the circuit is replaceable by a series combination of vt and Rt

Thévenin equivalent circuit

+

_

vt or voc

Rt

О

О

a

b

Thévenin method If circuit contains resistors and ind. sources

Connect open circuit between a and b. Find voc

Deactivate source(s), calc. Rt by circuit reduction

If circuit has resistors and ind. & dep. sources Connect open circuit between a and b. Find voc

Connect short circuit across a and b. Find isc

Connect 1-A current source from b to a. Find vab

NOTE: Rt = vab / 1 or Rt = voc / isc

If circuit has resistors and only dep. sources Note that voc = 0 Connect 1-A current source from b to a. Find vab

NOTE: Rt = vab / 1

HW example

see HW problem 5.5-1

Norton’s theorem

GOAL: reduce some complex part of a circuit to an equivalent source and a single element (for analysis) THEOREM: for any circuit of resistive elements and energy sources with a terminal pair, the circuit is replaceable by a parallel combination of isc and Rn (this is a source transformation of

the Thevenin)

Norton equivalent circuit

isc

О

О

a

b

Rn = Rt

•

•

Norton method If circuit contains resistors and ind. sources

Connect short circuit between a and b. Find isc

Deactivate ind. source(s), calc. Rn = Rt by circuit reduction

If circuit has resistors and ind. & dep. sources Connect open circuit between a and b. Find voc = vab Connect short circuit across a and b. Find isc

Connect 1-A current source from b to a. Find vab

NOTE: Rn = Rt = vab / 1 or Rn = Rt = voc / isc

If circuit has resistors and only dep. sources Note that isc = 0 Connect 1-A current source from b to a. Find vab

NOTE: Rn = Rt = vab / 1

HW example

see HW problem 5.6-2

maximum power transfer

what is it? often it is desired to gain maximum power transfer for an energy source to a load examples include:

electric utility grid signal transmission (FM radio receiver) source load

maximum power transfer

how do we achieve it?

+

_

vt or vsc

О

О

a

b

Rt

RLOAD

maximum power transfer

how do we calculate it?

tL

L

LtL

s

tL

s

L

RR

dR

dp

RRR

vp

RR

vi

Rip

0

)( 2

2

maximum power transfer theorem

So… maximum power delivered by a source represented by its Thevenin equivalent circuit is attained when the load RL is equal to the Thevenin resistance Rt

efficiency of power transfer

how do we calculate it for a circuit?

max%50/

4)(

2)(

/

22

max

2

inout

L

s

LL

sout

L

s

tL

sssin

inout

pp

R

v

RR

vpp

R

v

RR

vvivp

pp

Norton equivalent circuits

using the calculus on p=i2R in a Norton equivalent circuit we find that it, too, has a maximum when the load RL is equal to the Norton resistance Rn =Rt

HW example

see HW problem 5.7-6

new concepts from ch. 6

electronics operational amplifier the ideal operational amplifier nodal analysis of circuits containing ideal op amps design using op amps characteristics of practical op amps

definition of an OP-AMP

The Op-Amp is an “active” element with a high gain that is designed to be used with other circuit elements to perform a signal processing operation. It requires power supplies, sometimes a single supply, sometimes positive and negative supplies. It has two inputs and a single output.

OP-AMP symbol and connections

_

+

+–

+–

INVERTING INPUT NODE

NON-INVERTING INPUT NODE

OUTPUTNODEi1

i2io

vo

v2

v1

NEGATIVE POWER SUPPLY

POSITIVE POWER SUPPLY

THE OP-AMPFUNDAMENTAL CHARACTERISTICS

_

+

INVERTING INPUT NODE

NON-INVERTING INPUT NODE

OUTPUTNODEi1

i2io

vo

v2

v1

Ro

Ri

THE IDEAL OP-AMPFUNDAMENTAL CHARACTERISTICS

_

+

INVERTING INPUT NODE

NON-INVERTING INPUT NODE

OUTPUTNODEi1

i2io

vo

v2

v1

12

21 00

0

vv

ii

RR oi

THE INVERTING OP-AMP

_

+

i1

i2iovo

v2

v1

Rf

Ri

+–

vs

Node a

1. Write Ideal OpAmp equations.2. Write KCL at Node a.3. Solve for vo/vs

THE INVERTING OP-AMP

_

+

i1

i2iovo

v2

v1

Rf

Ri

+–

vs

Node a

1221 00 vvii

011

1

f

o

i

s

R

vvi

R

vv

i

fso

f

o

i

s

R

Rvv0

R

v

R

v

At node a:

THE NON-INVERTING OP-AMP

_

+

i1

i2iovo

v2

v1

Rf

Ri

+–vs

Node a

1221 00 vvii

00 1

11

f

o

i R

vvi

R

vAt node a:

10

i

fso

f

os

i

s

R

Rvv

R

vv

R

v

HW example

see HW problem 6.4-1

Test Two

next Monday review needed? if so… select 5-6 problems that you would like presented discussed