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Lesson 27 – Capacitors and Inductors II (Sections 55 and 57) (CLOs 62 and 63)
This is the second of two lessons on Capacitors and Inductors. This lesson discusses combining multiple devices
and introduces two new operational modules, the integrator and differentiator.
A computer exercise on an Opamp
integrator and differentiator can be useful.
This is an easy lesson.
You might even find time to give a quiz – or catch up if you’re behind.
There are three
topics to discuss.
The first is how to combine Inductors and Capacitors, the second involves two additional Op
Amp modules – integrators and differentiators, and the third is how to analyze
L
and
C
circuits at dc.
You can derive the relationship for series and parallel
L
’s and
C
’s or you can simply tell your students that –
L
’s
behave like
R
’s and
C
’s behave like
G
’s. (OK – just say the reverse of
R
’s).
Do a few examples, but students
really do not have difficulty with this.
In developing the integrator and differentiator, it is a GREAT opportunity to review node voltage analysis and the
iv
relationships for inductors and capacitors.
Some students want to write
i
= (
v
1

v
2
)
/C
.
It may have not sunk in
that the
iv
characteristics of capacitors and inductors are differential/integral relationships.
Use this effort to
bring out this fact.
After you derive the relationship, it would be useful to show the block diagram of both to
include the inverting and weighting part of the derivation.
Figure 617 in the text summaries all of the OpAmp
modules studied thus far.
They need either to memorize these relationships or know how to derive them.
Suggest doing an example of how an integrator or differentiator works.
Example 610 is somewhat similar to the
computer exercise (below).
Oldtimers (whoever they are) might tell students – especially Aerospace majors who
define aircraft or satellite performance by a set of coupled differential equations (e.g. yaw, pitch and roll) – that
differential equations were once solved exclusively using analog computers.
Example 611 shows how a circuit
could actually solve a firstorder integral equation.
Finally using the derivative equations show the behavior of capacitors and inductors at dc.
Do an example with
inductors and capacitors driven by a dc source.
Actually, redraw the circuit with capacitors replaced with opens
and inductors with shorts.
This will be useful when we develop transfer functions for filters.
Finally, you might upgrade the summary table to include the new stuff from this lesson.
Device/Model
⇒
Resistance – R
Inductance – L
Capacitance – C
Units
⇒
ohms,
Ω
henrys,
H
farads,
F
Circuit Symbol
+
_
v
R
i
R
R
+
_
v
L
i
L
L
+
_
v
C
i
C
C
Voltage Equation
v
R
= i
R
R
v
L
= L di
L
/dt
v
C
=
(1
/C
)
0
∫
t
i
C
dt + v
C
(0
+
)
Current Equation
i
R
= v
R
G = v
R
/R
i
L
=
(1
/L
)
0
∫
t
v
L
dt + i
L
(0
+
)
i
C
= C dv
C
/dt
Power Equation
p
R
= i
R
×
v
R
p
L
= i
L
×
v
L
p
C
= i
C
×
v
C
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 Spring '08
 MAURICIODEOLIVERIA

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