Tài liệu Circuits & Electronics P8 pptx
Tài liệu Circuits & Electronics P8 pptx
... Fall 2002: Lecture 8
22
One nit …
D
i
S
V
O
v
L
R
VCCS
1
I
v
O
v
For
v
O
>0
,
VCCS
consumes power:
v
O
i
D
For
v
O
<0
,
VCCS
must supply power!
()
2
1
2
−−=
ILSO
vR
K
Vv
()
2
1
2
−=
ID
v
K
i
for ... −−=
IO
vv
()
2
33
110510
2
2
10
−⋅⋅⋅−=
−
I
v
1
I
v
S
V
O
v
6.002 – Fall 2002: Lecture 8
1
6.002
CIRCUITS
AND
ELECTRONICS
Dependent Sources
and Amplifiers
6.002 – Fall 2002: Lecture...
Tài liệu Circuits & Electronics P16 pptx
...
v
P
is particular response to
V
i
cos
ωt
.
6.002 Fall 2000 Lecture
16
11
6.002
CIRCUITS
AND
ELECTRONICS
Sinusoidal Steady State
6.002 Fall 2000 Lecture
16
1
Usual approach…
1 ...
+
–
R
i
C
+
v
I
v
C
–
v
I
(
t
)
= V
i
cos
ω
t
for
t ≥
0 (
V
i
real)
= 0
for
t <
0
v
C
(0)
=
0
for
t =
0
I
v
0
t
6.002 Fall 2000 Lecture
16
4
1
1
1
1
e
c
t
u
r...
Tài liệu Circuits & Electronics P14 pptx
...
1
14
6.002
CIRCUITS
AND
ELECTRONICS
State and Memory
6.002 Fall 2000 Lecture
11
14
Input resistance
R
IN
B
Second attempt buffer
R
IN
store
buffer
d
IN
d
OUT
C
*
5
ln
OH
IN
V
CRT −=
LIN
RR >>
Better, ... + 6 + 5 + 3 + 8
M+
6.002 Fall 2000 Lecture
10
14
A
Stored value leaks away
store
pulse width >>
R
ON
C
Building a memory element …
CR
t
C
L
ev
−
⋅= 5
5
ln
OH
L
V
CRT...
Tài liệu Circuits & Electronics P25 pptx
... 6.002 Fall 2000 Lecture
1
25
6.002
CIRCUITS
AND
ELECTRONICS
Violating the Abstraction Barrier
6.002 Fall 2000 Lecture
5
25
R
i
V
instantaneous ... wires
3. avoid big current swings
V
S
V
S
6.002 Fall 2000 Lecture
6
25
Question: So why did our circuits work?
More in 6.014
t
V5
V5.2
3. Termination
P
a
r
a
l
l
e
l
t
e
r
m
i
n
a
t
i
o
n
D
E
M
O
a
d
d
Tài liệu Circuits & Electronics P24 pptx
... Conversion Circuits (PCC)
Power efficiency of converter important,
so use lots of devices:
MOSFET switches, clock circuits,
inductors, capacitors, op amps, diodes
Reading: Chapter 16 and 4.4 of A & ... segment”:
R
+
–
I
v
+
–
+
–
V6.0
6.0v
I
<
R
O
v
+
–
+
–
I
v
()
R/6.0vi
ID
−=
6.0vv
IO
−=
“Short segment”:
R
+
–
+
–
+
–
V6.0
6.0v
I
≥
I
v
Example
V6.0
+
–
Equivalent
circuit
6.002 Fa...
Tài liệu Circuits & Electronics P23 pptx
... switching
capacitor.
independent of f.
MOSFET ON half
the time.
STATIC
P
DYNAMIC
P
constant time
"RC"
2
T
RR
ONL
>>
>>
Square wave input
f
T
1
=
Demo
Review
In standby mode, half
the gates ... MOSFET (NFET)
D
S
G
on when
v
GS
≥
V
TN
off when
v
GS
<
V
TN
e.g.
V
TN
= 1V
•
P-channel MOSFET (PFET)
on when
v
GS
≤
V
TP
off when
v
GS
>
V
TP
e.g.
V
TP
= -1V
S
D
G
ON...
Tài liệu Circuits & Electronics P15 docx
... conditions to solve for the
remaining constants.
Solving
6.002 Fall 2000 Lecture
19
15
RLC Circuits
See A&L Section 12.2
add
R
no
R
+
–
C
R
L
+
–
)(
tv
)(tv
I
)(
ti
)(
tv
t
Damped sinusoids ... [ZSR]
I
v
0
V
0
t
Let’s solve
I
vv
d
t
vd
LC =+
2
2
For input
6.002 Fall 2000 Lecture
1
15
6.002
CIRCUITS
AND
ELECTRONICS
Second-Order Systems
6.002 Fall 2000 Lecture
11
15
Total soluti...
Tài liệu Circuits & Electronics P13 docx
... Lecture
15
13
Why? Consider
…
1Case
1
R
0
R
pin1
ok
Demo
6.002 Fall 2000 Lecture
1
13
6.002
CIRCUITS
AND
ELECTRONICS
Digital Circuit
6.002 Fall 2000 Lecture
8
13
GSL
CR
t
SSOH
eVVv
−
−=
Or
Find
Tài liệu Circuits & Electronics P12 pdf
...
13
12
t
C
v
V5
V0
VV
I
5
=
VV
O
0
=
5
0
VV
I
0
=
VV
O
5
=
5
0
t
C
v
V5
V0
RC
t
e
−
+ 55
RC
t
e
−
5
RC
=
τ
Remember
demo
B
Examples
6.002 Fall 2000 Lecture
1
12
6.002
CIRCUITS
AND
ELECTRONICS
Capacitors
and First-Order Systems
6.002 Fall 2000 Lecture
8
12
Example…
Method ... −=
()
RC
t
I0IC
eVVVv
−
−+=
6.002 Fall 2000 Lecture
2
12
5V
0V
C
A
B
5V
A
B
C
5
0
5
0
5
0
Reading:
Chapters 9 &...
Tài liệu Circuits & Electronics P11 pdf
... the
same KVL, KCL equations
BOUTA
VVV
++++
"""
so, we can cancel them out
BOUTA
vvv
++++++
""""
1
bouta
vvv
++++
"""
Leaving
2
Since small signal models ... circuit.
Foundations: (Also see section 8.2.1 of A&L)
KVL, KCL applied to some circuit C yields:
III The Small Signal Circuit View
bBoutOUTaA
vVvVvV
+++++++
"""
Replace...
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