CHAPTERS 5 & 6

1. CHAPTERS 5 & 6 NETWORKS 1: 0909201-01 8 October 2002 – Lecture 5b ROWAN UNIVERSITY College of Engineering Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn Semester 2002

2. 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

3. 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

4. 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 244-245Problems 6.4-1, 6.4-2, 6.4-6

5. 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

6. 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 vtand Rt

7. Thévenin equivalent circuit Rt a О + _ vt orvoc b О

8. 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

9. HW example • see HW problem 5.5-1

10. 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 iscand Rn (this is a source transformation of the Thevenin)

11. Norton equivalent circuit • a О isc Rn = Rt • b О

12. 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

13. HW example • see HW problem 5.6-2

14. 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

15. maximum power transfer • how do we achieve it? a О Rt + _ vt orvsc RLOAD b О

16. maximum power transfer • how do we calculate it?

17. 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

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

19. 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

20. HW example • see HW problem 5.7-6

21. 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

22. 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.

23. _ + + – + – OP-AMP symbol and connections INVERTING INPUT NODE v1 OUTPUT NODE i1 vo io i2 v2 POSITIVE POWER SUPPLY NON-INVERTING INPUT NODE NEGATIVE POWER SUPPLY

24. THE OP-AMPFUNDAMENTAL CHARACTERISTICS INVERTING INPUT NODE _ + Ri v1 OUTPUT NODE i1 vo io Ro i2 v2 NON-INVERTING INPUT NODE

25. INVERTING INPUT NODE v1 OUTPUT NODE i1 _ + vo io i2 v2 NON-INVERTING INPUT NODE THE IDEAL OP-AMPFUNDAMENTAL CHARACTERISTICS

26. _ + + – THE INVERTING OP-AMP Rf Ri Node a v1 i1 vo io vs i2 v2 1. Write Ideal OpAmp equations. 2. Write KCL at Node a. 3. Solve for vo/vs

27. _ + + – THE INVERTING OP-AMP Rf Ri Node a v1 i1 vo io vs i2 v2 At node a:

28. _ + THE NON-INVERTING OP-AMP Rf Ri Node a v1 i1 vo io i2 v2 + – vs At node a:

29. HW example • see HW problem 6.4-1

30. Test Two • next Monday • review needed? • if so… select 5-6 problems that you would like presented discussed