tak boleh harap
TRANSCRIPT
-
7/29/2019 tak boleh harap
1/6
Lecture 11, Slide 1EECS40, Fall 2003 Prof. King
Lecture #11
ANNOUNCEMENTS
Homework Assignment #4 will be posted today
Midterm #1: Monday Sept. 29th (11:10AM-12:00PM)
closed book; one page (8.5x11) of notes & calculator allowed
covers Chapters 1-5 in textbook (HW#1-4)
Midterm Review Session: Friday 9/26 7-9PM, 277 Cory
Extra office hours:
Steve: 9/26 from 12-2PM
Farhana: 9/27 from 1-3PM, 9/28 from 9-11AM
Practice problems and old exam are posted online
OUTLINE Review: op amp circuit analysis
The capacitor (Chapter 6.2 in text)
Lecture 11, Slide 2EECS40, Fall 2003 Prof. King
Review: Op Amp Circuit Analysis
Procedure:
1. Assume that the op amp is ideala) Apply KCL at (+) and () terminals, noting ip = 0 & in = 0
b) Note that vn = vpc) Write an expression forvo
2. Calculate vo
3. Check: Is the op-amp operating in its linear region?IfV vo V
+,, then the assumption is valid.
If calculated vo
> V+, then vo is saturated at V+
If calculated vo
< V, then vo is saturated at V
+
+
vn
+
vp
ip
inio
+
vo
+
+
vn
+
vp
ip
inio
+
vo
-
7/29/2019 tak boleh harap
2/6
Lecture 11, Slide 3EECS40, Fall 2003 Prof. King
Op Amp Circuit Analysis Example
Consider the following circuit:
Assume the op amp is ideal.
a) Calculate vo ifvs = 100 mV
b) What is the voltage gain vo/vs of this amplifier?c) Specify the range of values ofvs for which the
op amp operates in a linear mode
+
+vo
+
vs
in
+vp
+
vn
10 k
1 k10 V
10 V
Lecture 11, Slide 4EECS40, Fall 2003 Prof. King
Op Amp Circuit Analysis Example contd.
What if the op amp is not ideal?
Ri= 10 k
Ro = 1 k
A = 103
+
vs
10 k
1 k
+
vo
+
Ro
A(vpvn)
Ri
+vp
+
vn
-
7/29/2019 tak boleh harap
3/6
Lecture 11, Slide 5EECS40, Fall 2003 Prof. King
Re-draw the circuit
& analyze:
KCL @ node a:
KCL @ node b:
+
vs
1 k+
vn
+103(vn)
1 k10 k
10 k
+
vo
1087.9
-
7/29/2019 tak boleh harap
4/6
Lecture 11, Slide 7EECS40, Fall 2003 Prof. King
The Capacitor
Two conductors (a,b) separated by an insulator:
difference in potential = Vab=> equal & opposite charge Qon conductors
Q = CVab
where Cis the capacitance of the structure,
positive (+) charge is on the conductor at higher potential
Parallel-plate capacitor: area of the plates =A
separation between plates = d
dielectric permittivityof insulator =
=> capacitanced
AC
=
(stored charge in terms of voltage)
Lecture 11, Slide 8EECS40, Fall 2003 Prof. King
Symbol:
Units: Farads (Coulombs/Volt)
Current-Voltage relationship:
or
Note: vcmust be a continuous function of time
+
vc
ic
dt
dCv
dt
dvC
dt
dQi c
cc +==
C C
(typical range of values: 1 pF to 1 F)
-
7/29/2019 tak boleh harap
5/6
Lecture 11, Slide 9EECS40, Fall 2003 Prof. King
Voltage in Terms of Current
)0()(1)0(
)(1)(
)0()()(
00
0
c
t
c
t
cc
t
c
vdttiCC
Qdtti
Ctv
QdttitQ
+=+=
+=
Lecture 11, Slide 10EECS40, Fall 2003 Prof. King
You might think the energy stored on a capacitor is QV,
which has the dimension of Joules. But during charging,
the average voltage across the capacitor was only half the
final value ofV.
Thus, energy is .2
2
1
2
1CVQV =
Example: A 1 pF capacitance charged to 5 Volts
has (5V)2 (1pF) = 12.5 pJ
Stored Energy
-
7/29/2019 tak boleh harap
6/6
Lecture 11, Slide 11EECS40, Fall 2003 Prof. King
=
=
==
=
=
=
==Final
Initial
c
Final
Initial
Final
Initial
ccc
Vv
Vv
dQvdttt
tt
dt
dQVv
Vv
vdtivw
2CV212CV
21
Vv
Vv
dvCvw InitialFinalFinal
Initial
cc
=
=
==
+
vc
ic
A more rigorous derivation
Lecture 11, Slide 12EECS40, Fall 2003 Prof. King
Integrating Amplifier
)0()(1
)(0
C
t
INo vdttvRC
tv +=
+
vo
R in
+
vp
+
vn
ic
C
vC +
vin