CHAPTERS 5 & 6 CHAPTERS 5 & 6 NETWORKS 1: NETWORKS 1: October 2002 – Lecture 5b ROWAN UNIVERSITY College of Engineering Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn 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, 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, Chapter 6 Pages Problems 6.4-1, 6.4-2, 6.4-6

next monday’s - test two covers Chapters 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 v t and R t

Thévenin equivalent circuit +_+_ v t or v oc RtRt О О a b

Thévenin method If circuit contains resistors and ind. sources Connect open circuit between a and b. Find v oc Deactivate source(s), calc. R t by circuit reduction If circuit has resistors and ind. & dep. sources Connect open circuit between a and b. Find v oc Connect short circuit across a and b. Find i sc Connect 1-A current source from b to a. Find v ab  NOTE: R t = v ab / 1 or R t = v oc / i sc If circuit has resistors and only dep. sources Note that v oc = 0 Connect 1-A current source from b to a. Find v ab  NOTE: R t = v ab / 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 i sc and R n (this is a source transformation of the Thevenin)

Norton equivalent circuit i sc О О a b R n = R t

Norton method If circuit contains resistors and ind. sources Connect short circuit between a and b. Find i sc Deactivate ind. source(s), calc. R n = R t by circuit reduction If circuit has resistors and ind. & dep. sources Connect open circuit between a and b. Find v oc = v ab Connect short circuit across a and b. Find i sc Connect 1-A current source from b to a. Find v ab  NOTE: R n = R t = v ab / 1 or R n = R t = v oc / i sc If circuit has resistors and only dep. sources Note that i sc = 0 Connect 1-A current source from b to a. Find v ab  NOTE: R n = R t = v ab / 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? +_+_ v t or v sc О О a b RtRt R LOAD

maximum power transfer how do we calculate it?

maximum power transfer theorem So… maximum power delivered by a source represented by its Thevenin equivalent circuit is attained when the load R L is equal to the Thevenin resistance R t

efficiency of power transfer how do we calculate it for a circuit?

Norton equivalent circuits using the calculus on p=i 2 R in a Norton equivalent circuit we find that it, too, has a maximum when the load R L is equal to the Norton resistance R n =R t

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 OUTPUT NODE i1i1 i2i2 ioio vovo v2v2 v1v1 NEGATIVE POWER SUPPLY POSITIVE POWER SUPPLY

THE OP-AMP FUNDAMENTAL CHARACTERISTICS _+_+ INVERTING INPUT NODE NON-INVERTING INPUT NODE OUTPUT NODE i1i1 i2i2 ioio vovo v2v2 v1v1 RoRo RiRi

THE IDEAL OP-AMP FUNDAMENTAL CHARACTERISTICS _+_+ INVERTING INPUT NODE NON-INVERTING INPUT NODE OUTPUT NODE i1i1 i2i2 ioio vovo v2v2 v1v1

THE INVERTING OP-AMP _+_+ i1i1 i2i2 ioio vovo v2v2 v1v1 RfRf RiRi +–+– vsvs Node a 1. Write Ideal OpAmp equations. 2. Write KCL at Node a. 3. Solve for v o /v s

THE INVERTING OP-AMP _+_+ i1i1 i2i2 ioio vovo v2v2 v1v1 RfRf RiRi +–+– vsvs Node a At node a:

THE NON-INVERTING OP-AMP _+_+ i1i1 i2i2 ioio vovo v2v2 v1v1 RfRf RiRi +–+– vsvs Node a At node a:

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