chapters 6 & 7 chapters 6 & 7 networks 1: 0909201-01 networks 1: 0909201-01 15 october 2002...
TRANSCRIPT
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CHAPTERS 6 & 7CHAPTERS 6 & 7
NETWORKS 1: NETWORKS 1: 0909201-010909201-01 15 October 2002 – Lecture 6b
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
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test II
no adjustment mean = 66 14-Fs, 1 A-
w/ 10 points mean = 76 7-Fs, 5 As
0 2 4 6 8
# of student s
A
A-
B+
B
B-
C+
C
C-
D+
D
D-
F
Networ ks I - S econd Test Gr ade Dist r ibut ion
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networks I
Today’s learning objectives – build understanding of the
operational amplifier introduce capacitance introduce inductance
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homework 6
Chapter 6 Pages 244-245 Problems 6.4-3, 6.4-5, 6.4-8
Chapter 7 Pages 296-301 Problems 7.4-3, 7.5-3, 7.6-1, 7.8-3
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new concepts from ch. 6
operational amplifier the ideal operational amplifier nodal analysis of circuits containing ideal op amps design using op amps characteristics of practical op amps
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new concepts from ch. 7
energy storage in a circuit capacitors series and parallel
inductors series and parallel
using op amps in RC circuits
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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.
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OP-AMP symbol and connections
_
+
+–
+–
INVERTING INPUT NODE
NON-INVERTING INPUT NODE
OUTPUTNODEi1
i2
io
vo
v2
v1
NEGATIVE POWER SUPPLY
POSITIVE POWER SUPPLY
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THE OP-AMPFUNDAMENTAL CHARACTERISTICS
_
+
INVERTING INPUT NODE
NON-INVERTING INPUT NODE
OUTPUTNODEi1
i2
io
vo
v2
v1
Ro
Ri
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THE IDEAL OP-AMPFUNDAMENTAL CHARACTERISTICS
_
+
INVERTING INPUT NODE
NON-INVERTING INPUT NODE
OUTPUTNODEi1
i2
io
vo
v2
v1
12
21 00
0
vv
ii
RR oi
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THE INVERTING OP-AMP
_
+
i1
i2
io
vo
v2
v1
Rf
Ri
+–
vs
Node a
1. Write Ideal OpAmp equations.2. Write KCL at Node a.3. Solve for vo/vs
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THE INVERTING OP-AMP
_
+
i1
i2
io
vo
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:
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THE NON-INVERTING OP-AMP
_
+
i1
i2
io
vo
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
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HW example
see HW problems
6.4-1, 6.4-2, 6.4-6
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Op Amp circuit types
for K < 0 for K > 1 for K = 1
special case 0 < K < 1
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What you need to know
Parameters of an Ideal Op Amp Types of Amplification Gain (K) vs. Which nodes and Amps circuits are needed to achieve same How to identify which type of circuit is in use (effect) How to solve simple Op Amp problems
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DEFINITION OF CAPACITANCE
Measure of the ability of a device to store energy in the form of an electric field.
CAPACITOR: IMPORTANT RELATIONSHIPS:
+ –
i
+
+
_
_ d
AC
Cvq
dt
dvCi
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CALCULATING ic FOR A GIVEN v(t)
Let v(t) across a capacitor be a ramp function.
t
v
v
t
tt0t
vCic
ci0tAs
Moral: You can’t change the voltage across a capacitor instantaneously.
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VOLTAGE ACROSS A CAPACITOR
dt
dvCi c
c dtC
idv c
c
dC
idv
tc
c
diC
1dv
t
cc
oc
t
tcc tvdi
C
1v
o
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ENERGY STORED IN A CAPACITOR
divtwt
ccc
dd
dvCvtw
tc
cc
dvvCtw)t(v
)(vcc
2cc Cv
21
tw)t(v
0)(v
2cc Cv
2
1tw
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CAPACITORS IN SERIES
+–
C1 C2 C3
+ v1 - + v2 - + v3 -
i
0tvdiC
1tvdi
C
1tvdi
C
1v o3
t
t3o2
t
t2o1
t
t1 ooo
o3o2o1
t
t321tvtvtvdi
C
1
C
1
C
1v
o
vKVL
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oc
t
teqtvdi
C
1v
o
321eq C
1
C
1
C
1
C
1 o3o2o1oc tvtvtvtv
Capacitors in series combine like resistors in parallel.
CAPACITORS IN SERIES
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CAPACITORS IN PARALLEL
C1 C2 C3
i
i1i2 i3KCL
0dt
dvC
dt
dvC
dt
dvCi 321
dt
dvC
dt
dvCCCi eq321
Capacitors in parallel combine like resistors in series.
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HANDY CHARTELEMENT CURRENT VOLTAGE
R
C
L
R
VI RIV
dt
dvCi c
c dtiC
1v
t
cc
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DEFINITION OF INDUCTANCE Measure of the ability of a device to store
energy in the form of a magnetic field.
INDUCTOR: IMPORTANT RELATIONSHIPS:
i
+ _
LiN dt
diLv
v
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CALCULATING vL FOR A GIVEN i(t)
Let i(t) through an inductor be a ramp function.
t
i
i
t
ttt
iLvL
0
Lvt 0As
Moral: You can’t change the current through an inductor instantaneously.
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CURRENT THROUGH AN INDUCTOR
dt
diLv L
L dtL
vdi L
L
dL
vdi
tL
L
dvL
1di
t
LL
oL
t
tLL tidv
L
1i
o
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ENERGY STORED IN AN INDUCTOR
divtwt
LLL
dd
diLitw
tL
LL
dviLtwti
iLL
)(
)( 2
2
1)(
0)(
LL Litwti
i
2
2
1LL Litw
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INDUCTORS IN SERIES
L1 L2 L3
+ v1 - + v2 - + v3 -
iKVL
0321 dt
diL
dt
diL
dt
diLvi
dt
diL
dt
diLLLv eqi 321
Inductors in series combine like resistors in series.
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INDUCTORS IN PARALLEL
L1 L2 L3v
i1 i2i3
KCL+–
0tidvL
1tidv
L
1tidv
L
1i o3
t
t3o2
2o1
t
t1v
oo
o3o2o1
t
t321v tititidv
L
1
L
1
L
1i
o
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oL
t
teqv tidv
L
1i
o
321eq L
1
L
1
L
1
L
1 o3o2o1oL titititi
INDUCTORS IN PARALLEL
Inductors in parallel combine like resistors in parallel.
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HANDY CHARTELEMENT CURRENT VOLTAGE
R
C
L
R
VI RIV
dt
dvCi c
c dtiC
1v
t
cc
dt
diLv L
L dtvL
1i
t
LL
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OP-AMP CIRCUITS WITH C & L
_
+
i1
i2
io
vo
v2
v1
Cf
Ri
+–
vs
Node a
0
dt
vvdCi
R
vv o1f1
i
1s
12
2
1
vv
0i
0i
dt
vdC
R
v of
i
s
fi
so
CR
v
dt
vd
dtCR
vvd
fi
so dt
CR
vvd
fi
so dtv
CR
1v s
fio
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QUIZ: Find vo= f(vs)
_
+
i1
i2
io
vo
v2
v1
Rf
+–
vs
Node a
Li
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ANSWER TO QUIZ
12
2
1
vv
0i
0i
0R
vvidtvv
L
1
f
o111s
i
0R
vdtv
L
1
f
os
i
dtvL
1
R
vs
if
o dtvL
Rv s
i
fo
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IMPORTANT CONCEPTS FROM CH. 7
I/V Characteristics of C & L.
Energy storage in C & L. Writing KCL & KVL for circuits with C & L. Solving op-amp circuits with C or L in feedback loop. Solving op-amp circuits with C or L at the input.
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1st ORDER CIRCUITS WITH CONSTANT INPUT
+–
t = 0
R1 R2
R3 Cvs
+v(t)-
s321
3 vRRR
R0v
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Thevenin Equivalent at t=0+
Rt
C+–
Voc
+v(t)-
32
32t RR
RRR
s
32
3oc v
RR
RV
KVL 0)t(vR)t(iV toc
i(t)
+ -
0)t(vdt
)t(dvCRV toc CR
V
CR
)t(v
dt
)t(dv
t
oc
t
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SOLUTION OF 1st ORDER EQUATION
CR
V
CR
)t(v
dt
)t(dv
t
oc
t
CR
)t(v
CR
V
dt
)t(dv
tt
oc dtCR
)t(vV)t(dv
t
oc
dtCR
1
)t(vV
)t(dv
toc
dt
CR
1
V)t(v
)t(dv
toc
DdtCR
1
V)t(v
)t(dv
toc
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SOLUTION CONTINUED
DCR
tV)t(vln
toc
DdtCR
1
V)t(v
)t(dv
toc
D
CR
texpV)t(v
toc
CR
texpDexpV)t(v
toc oc
tV
CR
texpDexp)t(v
oct
VCR
0expDexp)0(v
ocV)0(vDexp
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SOLUTION CONTINUED
oct
oc VCR
texpV)0(v)t(v
CR
texpV)0(vV)t(v
tococ
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WITH AN INDUCTOR
+–
t = 0
R1 R2
R3 Lvs
21
s
RR
v0i
i(t)
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Norton equivalent at t=0+
RtIsc
+v(t)-
L i(t)
32
32t RR
RRR
2
ssc R
vI
KCL 0)t(iR
)t(vI
tsc
0)t(idt
)t(diL
R
1I
tsc sc
tt IL
R)t(i
L
R
dt
)t(di
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SOLUTIONsc
tt IL
R)t(i
L
R
dt
)t(di
CR
V
CR
)t(v
dt
)t(dv
t
oc
t
CR
1
L
R
t
t
CR
texpV)0(vV)t(v
tococ
tL
RexpI)0(iI)t(i t
scsc
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HANDY CHARTELEMENT CURRENT VOLTAGE
R
C
L
R
VI RIV
dt
dvCi c
c dtiC
1v
t
cc
dt
diLv L
L dtvL
1i
t
LL
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IMPORTANT CONCEPTS FROM CHAPTER 8
Determining Initial ConditionsSetting up differential equationsSolving for v(t) or i(t)
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