CHAPTERS 5 & 6 CHAPTERS 5 & 6 NETWORKS 1: NETWORKS 1: 0909201-01 0909201-01 8 October 2002 – Lecture 5b ROWAN UNIVERSITY ROWAN UNIVERSITY College of Engineering College of Engineering Professor Peter Mark Jansson, PP PE Professor Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING Autumn Semester 2002 Autumn Semester 2002
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. networks I. Today’s learning objectives – - PowerPoint PPT Presentation
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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
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