6.002
CIRCUITS AND
ELECTRONICS
Introduction and Lumped Circuit Abstraction
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
ADMINISTRIVIA
Lecturer: Prof. Anant Agarwal
Textbook: Agarwal and Lang (A&L)
Readings are important!
Handout no. 3
Web site —
http://web.mit.edu/6.002/www/fall00
Assignments —
Homework exercises
Labs
Quizzes
Final exam
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
Two homework assignments can
be missed (except HW11).
Collaboration policy
Homework
You may collaborate with
others, but do your own
write-up.
Lab
You may work in a team of
two, but do you own write-up.
Info handout
Reading for today —
Chapter 1 of the book
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
What is engineering?
Purposeful use of science
What is 6.002 about?
Gainful employment of
Maxwell’s equations
From electrons to digital gates
and op-amps
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
6.002
Nature as observed in experiments
V
3
6
9
12
…
I
0.1
0.2
0.3
0.4
…
Physics laws or “abstractions”
Maxwell’s
abstraction for
Ohm’s
tables of data
V=RI
Lumped circuit abstraction
+–
R
V
C
L
M
Simple amplifier abstraction
Digital abstraction
Operational
amplifier abstraction
abstraction
Combinational logic
+
-
S
f
Filters
Clocked digital abstraction
Analog system
components:
Modulators,
oscillators,
RF amps,
power supplies 6.061
Instruction set abstraction
Pentium, MIPS 6.004
Programming languages
Java, C++, Matlab 6.001
Software systems 6.033
Operating systems, Browsers
Mice, toasters, sonar, stereos, doom, space shuttle
6.455
6.170
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
Lumped Circuit Abstraction
Consider
The Big Jump
from physics
to EECS
I
V
?
Suppose we wish to answer this question:
What is the current through the bulb?
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
We could do it the Hard Way…
Apply Maxwell’s
Differential form
∂B
Faraday’s
∇× E = −
∂t
∂ρ
Continuity ∇ ⋅ J = −
∂t
Others
ρ
∇⋅E =
ε0
Integral form
∂φ B
∫ E ⋅ dl = − ∂t
∂q
J
⋅
dS
=
−
∫
∂t
q
E
⋅
dS
=
∫
ε0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
Instead, there is an Easy Way…
First, let us build some insight:
Analogy
F
a?
I ask you: What is the acceleration?
You quickly ask me: What is the mass?
I tell you:
m
F
You respond: a =
m
Done !! !
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
Instead, there is an Easy Way…
First, let us build some insight:
Analogy
F
a?
In doing so, you ignored
the object’s shape
its temperature
its color
point of force application
Point-mass discretization
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
The Easy Way…
Consider the filament of the light bulb.
A
B
We do not care about
how current flows inside the filament
its temperature, shape, orientation, etc.
Then, we can replace the bulb with a
discrete resistor
for the purpose of calculating the current.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
The Easy Way…
A
B
Replace the bulb with a
discrete resistor
for the purpose of calculating the current.
A
I
V
+
and I =
V R
R
–
B
In EE, we do things
the easy way…
R represents the only property of interest!
Like with point-mass: replace objects
F
with their mass m to find a =
m
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
The Easy Way…
A
+
V
–
I
R
and
B
I=
V
R
In EE, we do things
the easy way…
R represents the only property of interest!
R relates element v and i
V
I=
R
called element v-i relationship
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
R is a lumped element abstraction
for the bulb.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
R is a lumped element abstraction
for the bulb.
Not so fast, though …
I
A
+ S
A
V
B
–
SB
black box
Although we will take the easy way
using lumped abstractions for the rest
of this course, we must make sure (at
least the first time) that our
abstraction is reasonable. In this case,
ensuring that
V I
are defined
for the element
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
A
+
I
SA
V
V
B
–
I
must be defined
for the element
SB
black box
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
I
must be defined. True when
=
I out of S B
∂q
True only when
= 0 in the filament!
∂t
∫ J ⋅ dS
I into S A
SA
∫ J ⋅ dS
SB
∫ J ⋅ dS − ∫ J ⋅ dS =
SA
from ell
w
x
a
M
SB
IA
∂q
∂t
IB
∂q
=0
I A = I B only if
∂t
So let’s assume this
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
V
Must also be defined.
see
A&L
So let’s assume this too
∂φ B
=0
∂t
outside elements
VAB defined when
So
VAB = ∫AB E ⋅ dl
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
Lumped Matter Discipline (LMD)
Or self imposed constraints:
More in
Chapter 1
of A & L
∂φ B
= 0 outside
∂t
∂q
= 0 inside elements
∂t
bulb, wire, battery
Lumped circuit abstraction applies when
elements adhere to the lumped matter
discipline.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
Demo
only for the
sorts of
questions we
as EEs would
like to ask!
Demo
Lumped element examples
whose behavior is completely
captured by their V–I
relationship.
Exploding resistor demo
can’t predict that!
Pickle demo
can’t predict light, smell
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
So, what does this buy us?
Replace the differential equations
with simple algebra using lumped
circuit abstraction (LCA).
For example —
a
R1
V
+
–
b
R3
R4
d
R2
R5
c
What can we say about voltages in a loop
under the lumped matter discipline?
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
What can we say about voltages in a loop
under LMD?
a
R1
V
b
+
–
R4
R3
d
R2
R5
c
∂φ B
under DMD
∫ E ⋅ dl = − ∂t
0
∫ E ⋅ dl + ∫ E ⋅ dl + ∫ E ⋅ dl = 0
ca
ab
bc
+ Vca + Vab + Vbc
= 0
Kirchhoff’s Voltage Law (KVL):
The sum of the voltages in a loop is 0.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
What can we say about currents?
Consider
I ca
S
a
I da
I ba
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
What can we say about currents?
I ca
S
a
I da
I ba
∂q
∫S J ⋅ dS = − ∂t
under LMD
0
I ca + I da + I ba = 0
Kirchhoff’s Current Law (KCL):
The sum of the currents into a node is 0.
simply conservation of charge
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
KVL and KCL Summary
KVL:
∑ jν j = 0
loop
KCL:
∑jij = 0
node
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 1
6.002
CIRCUITS AND
ELECTRONICS
Basic Circuit Analysis Method
(KVL and KCL method)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Review
Lumped Matter Discipline LMD:
Constraints we impose on ourselves to simplify
our analysis
∂φ B
=0
∂t
∂q
=0
∂t
Outside elements
Inside elements
wires resistors sources
Allows us to create the lumped circuit
abstraction
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Review
LMD allows us to create the
lumped circuit abstraction
i
+
v
Lumped circuit element
power consumed by element = vi
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Review
Review
Maxwell’s equations simplify to
algebraic KVL and KCL under LMD!
KVL:
∑ jν j = 0
loop
KCL:
∑jij = 0
node
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Review
a
R1
+
–
b
R4
R3
R2
d
R5
c
DEMO
vca + vab + vbc = 0
KVL
ica + ida + iba = 0
KCL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Method 1: Basic KVL, KCL method of
Circuit analysis
Goal: Find all element v’s and i’s
1. write element v-i relationships
(from lumped circuit abstraction)
2. write KCL for all nodes
3. write KVL for all loops
lots of unknowns
lots of equations
lots of fun
solve
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Method 1: Basic KVL, KCL method of
Circuit analysis
Element Relationships
For R,
V = IR
For voltage source, V = V0
R
+–
V0
For current source, I = I 0
J
Io
3 lumped circuit elements
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
KVL, KCL Example
a
+
ν1
+
ν 0 = V0
–
R3
b
+
ν2
–
ν4
R1
–
+
–
+
+ν 3 –
R2
–
R4
d
+
ν5
–
R5
c
The Demo Circuit
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Associated variables discipline
i
+
ν
Element e
Current is taken to be positive going
into the positive voltage terminal
Then power consumed
by element e
= νi is positive
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
KVL, KCL Example
a
+
ν 0 = V0
–
i0
L1
+
–
i4
i1 L 2
+
+
ν 4 R4
ν 1 R1
–
–
R3
b i3
d
+ν 3 –
i2
i5
+
+
ν 2 R2
ν 5 R5
–
L3
–
c
The Demo Circuit
L4
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Analyze
ν 0 …ν 5 ,ι0 …ι5
1. Element relationships (v, i )
given v3 = i3 R3
v0 = V0
v4 = i4 R4
v1 = i1 R1
v5 = i5 R5
v2 = i2 R2
12 unknowns
6 equations
2. KCL at the nodes
a: i0 + i1 + i4 = 0
3 independent
b: i2 + i3 − i1 = 0
equations
d: i5 − i3 − i4 = 0
e: − i0 − i2 − i5 = 0 redundant
3. KVL for loops
L1: − v0 + v1 + v2 = 0
3 independent
equations
L2: v1 + v3 − v4 = 0
s
L3: v3 + v5 − v2 = 0
n
o
L4: − v0 + v4 + v5 = 0 redundant ati
s
n
w
u
no
k
eq
n
u
2
2
1
1
/
ugh @#!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Other Analysis Methods
Method 2— Apply element combination rules
A
B
C
R1
R2 R3
G1
G2
V1
V2
+–
+–
…
RN
GN
G1 + G2
+ GN
⇔
V1 + V2
+–
J
J
J
I2
+ RN
1
Gi =
Ri
⇔
D
I1
⇔
⇔
R1 + R2 +
I1 + I 2
Surprisingly, these rules (along with superposition, which
you will learn about later) can solve the circuit on page 8
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Other Analysis Methods
Method 2— Apply element combination rules
I =?
Example
R1
V +
–
R3
R2
I
I
V +
–
R1
V +
–
R2 R3
R2 + R3
R = R1 +
R
R2 R3
R2 + R3
V
I=
R
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Method 3—Node analysis
Particular application of KVL, KCL method
1. Select reference node ( ground)
from which voltages are measured.
2. Label voltages of remaining nodes
with respect to ground.
These are the primary unknowns.
3. Write KCL for all but the ground
node, substituting device laws and
KVL.
4. Solve for node voltages.
5. Back solve for branch voltages and
currents (i.e., the secondary unknowns)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Example: Old Faithful
plus current source
V0
3
+ V e1
– 0
R2
Step 1
R4
e2
R5
J
R1 R
I1
Step 2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Example: Old Faithful
plus current source
V0
3
+ V e1
– 0
R2
R4
e2
R5
J
R1 R
for
I1 convenience,
write
1
Gi =
Ri
KCL at e1
(e1 − V0 )G1 + (e1 − e2 )G3 + (e1 )G2 = 0
KCL at e2
(e2 − e1 )G3 + (e2 − V0 )G4 + (e2 )G5 − I1 = 0
Step 3
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Example: Old Faithful
plus current source
V0
3
+ V e1
– 0
R2
R4
e2
R5
J
R1 R
I1
1
Gi =
Ri
KCL at e1
(e1 − V0 )G1 + (e1 − e2 )G3 + (e1 )G2 = 0
KCL at l2
(e2 − e1 )G3 + (e2 − V0 )G4 + (e2 )G5 − I1 = 0
move constant terms to RHS & collect unknowns
e1 (G1 + G2 + G3 ) + e2 (−G3 ) = V0 (G1 )
e1 (−G3 ) + e2 (G3 + G4 + G5 ) = V0 (G4 ) + I1
2 equations, 2 unknowns
(compare units)
Solve for e’s
Step 4
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
In matrix form:
− G3
⎡ G1V0 ⎤
⎡G1 + G2 + G3
⎤ ⎡ e1 ⎤
=
⎢G V + I ⎥
⎢
G3 + G4 + G5 ⎥⎦ ⎢⎣e2 ⎥⎦
− G3
⎣ 4 0 1⎦
⎣
conductivity
matrix
sources
unknown
node
voltages
Solve
G3
⎡G3 + G4 + G5
⎤ ⎡ G1V0 ⎤
G3
G1 + G2 + G3 ⎥⎦ ⎢⎣G4V0 + I1 ⎥⎦
⎡ e1 ⎤ ⎢⎣
⎢e ⎥ =
(G1 + G2 + G3 )(G3 + G4 + G5 ) − G3 2
⎣ 2⎦
(
)(
) ( )(
)
G +G +G G V + G G V + I
3
4
5 1 0
3 4 0 1
e =
1 G G +G G +G G +G G +G G +G G +G 2 +G G +G G
1 3
1 4
1 5
2 3
2 4
2 5
3
3 4
3 5
e2 =
(G3 )(G1V0 ) + (G1 + G2 + G3 )(G4V0 + I 1 )
G1G3 + G1G4 + G1G5 + G2G3 + G2G4 + G2 G5 + G3 + G3G4 + G3G5
2
(same denominator)
Notice: linear in V0 , I1 , no negatives
in denominator
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
Solve, given
G1 ⎫
1
=
⎬
G5 ⎭ 8.2 K
G2 ⎫
1
⎬=
G4 ⎭ 3.9 K
1
G3 =
1.5 K
I1 = 0
(
)(
)
G G V + G +G +G G V + I
e = 3 10 1 2 3 40 1
2 G + G + G + G + G + G −G 2
1 2 3
3 4 5 3
1
1
1
G +G +G =
+
+
=1
1
2
3 8.2 3.9 1.5
(
G3 + G4 + G5 =
)(
)
1
1
1
+
+
=1
1.5 3.9 8.2
1
1
1
×
+ 1×
3.9 V
e2 = 8.2 1.5
0
1
1− 2
1.5
Check out the
DEMO
e2 = 0.6V0
If V0 = 3V , then e2 = 1.8V0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 2
6.002
CIRCUITS AND
ELECTRONICS
Superposition, Thévenin and Norton
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Review
Circuit Analysis Methods
z KVL:
∑Vi = 0
loop
KCL:
∑ Ii = 0
VI
node
z Circuit composition rules
z Node method – the workhorse of 6.002
KCL at nodes using V ’s referenced
from ground
(KVL implicit in “ (ei − e j ) G ”)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Linearity
R1
V
+
–
R2
e
J
Consider
I
Write node equations –
e −V e
+ −I =0
R1
R2
Notice:
linear in e,V , I
No eV ,VI
terms
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Linearity
R1
Consider
V
+
–
J
R2
Write node equations -e −V e
+ −I =0
R1
R2
Rearrange -⎡1
1⎤
⎢ R + R ⎥e
⎣ 1
2⎦
conductance
matrix
G
=
I
linear in e,V , I
V
+ I
R1
node
linear sum
voltages of sources
e
=
S
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Linearity
Write node equations -e −V e
+ −I =0
R1
R2
Rearrange -⎡1
1⎤
⎢ R + R ⎥e
⎣ 1
2⎦
conductance
matrix
G
or
=
linear in e,V , I
V
+ I
R1
node
linear sum
voltages of sources
e
=
S
R2
R1 R2
e=
V+
I
R1 + R2
R1 + R2
e = a1V1 + a2V2 + … + b1 I1 + b2 I 2 + …
Linear!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Linearity
⇒
Homogeneity
Superposition
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Linearity
⇒
Homogeneity
Superposition
Homogeneity
x1
x2 .
..
y
⇓
αx1
αx2 ..
.
αy
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Linearity
⇒
Homogeneity
Superposition
Superposition
x1a
x2 a .
.
.
ya
x1b
x2 b .
.
yb
.
⇓
x1a + x1b
x2 a + x2 b .
..
y a + yb
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Linearity
⇒
Homogeneity
Superposition
Specific superposition example:
V1
0
0
V2
y1
y2
⇓
V1 + 0
0 + V2
y1 + y2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Method 4: Superposition method
The output of a circuit is
determined by summing the
responses to each source
acting alone.
s
e
c
r
u
so
t
n
e
nd
e
p
e
i nd
only
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
i
V =0 +
–
i
+
v
+
v
-
short
I =0
J
i
i
+
v
+
v
-
-
open
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Back to the example
Use superposition method
V
+
–
e
R2
J
R1
I
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Back to the example
Use superposition method
V acting alone
e
R1
V
I = 0 eV =
R2
+
–
I acting alone
R2
V
R1 + R2
e
R2
V =0
sum
J
R1
I
R1 R2
eI =
I
R1 + R2
superposition
R2
R1 R2
e = eV + eI =
V+
I
R1 + R2
R1 + R2
Voilà !
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Demo
salt
water
constant
+
–
?
+
–
output shows
superposition
sinusoid
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Yet another method…
Consider
By superposition
v = ∑ α mVm + ∑ β n I n + Ri
m
n
no
resistance
units
units
By setting
∀n I n = 0, ∀mVm = 0,
i = 0
i = 0
i
+
v
-
J
y network
r
a
r
t
i
N
Arb
resistors
Vm
In
+
–
J
i
also
independent
of external
excitement &
behaves like
a resistor
All
∀n I n = 0,
∀mVm = 0
independent of external
excitation and behaves like a
voltage “ vTH ”
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Or
v = vTH + RTH i
As far as the external world is concerned
(for the purpose of I-V relation),
“Arbitrary network N” is indistinguishable
from:
RTH
Thévenin
equivalent
network
vTH
RTH
+ vTH
–
+
v
J
N
i
-
open circuit voltage
at terminal pair (a.k.a. port)
resistance of network seen
from port
( Vm ’s, I n ’s set to 0)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Method 4:
The Thévenin Method
J
i
N
+
–
+
–
+
v
-
E
+
v
E
Thévenin equivalent
RTH
+ vTH
–
i
-
Replace network N with its Thévenin
equivalent, then solve external network E.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Example:
+
V
–
R2
J
i1 R1
I
i1 R1
RTH
+
V
–
VTH
i1 =
+ I
–
V − VTH
R1 + RTH
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
VTH :
VTH = IR2
RTH :
RTH = R2
+
VTH
-
R2
+
RTH
-
R2
J
Example:
I
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Graphically,
v = vTH + RTH i
i
1
RTH
v
vTH
“V ”
OC
− I SC
Open circuit
(i ≡ 0)
v = vTH
VOC
Short circuit
(v ≡ 0)
− vTH
i =
RTH
− I SC
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
in recitation,
see text
Method 5:
The Norton Method
J
+
–
+
–
+
v
-
IN
J
i
RTH = RN
Norton
equivalent
IN =
VTH
RTH
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
Summary
„Discretize matter
LMD
Physics
LCA
EE
„
R, I, V
Linear networks
„
Analysis methods (linear)
KVL, KCL, I — V
Combination rules
Node method
Superposition
Thévenin
Norton
„
Next
Nonlinear analysis
Discretize voltage
…
101100 …
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 3
6.002
CIRCUITS AND
ELECTRONICS
The Digital Abstraction
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Review
z Discretize matter by agreeing to
observe the lumped matter discipline
Lumped Circuit Abstraction
zAnalysis tool kit: KVL/KCL, node method,
superposition, Thévenin, Norton
(remember superposition, Thévenin,
Norton apply only for linear circuits)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Today
Discretize value
Digital abstraction
Interestingly, we will see shortly that the
tools learned in the previous three
lectures are sufficient to analyze simple
digital circuits
Reading: Chapter 5 of Agarwal & Lang
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
But first, why digital?
In the past …
Analog signal processing
R1
V0
R2
V1 +
–
V1 and V2
+
–
V2
might represent the
outputs of two
sensors, for example.
By superposition,
V0 =
R2
R1
V1 +
V2
R1 + R2
R1 + R2
If R1 = R 2 ,
V1 + V2
V0 =
2
The above is an “adder” circuit.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Noise Problem
t
add noise on
this wire
Receiver:
huh?
… noise hampers our ability to distinguish
between small differences in value —
e.g. between 3.1V and 3.2V.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Value Discretization
Restrict values to be one of two
HIGH
LOW
5V
0V
TRUE
FALSE
1
0
…like two digits 0 and 1
Why is this discretization useful?
(Remember, numbers larger than 1 can be
represented using multiple binary digits and
coding, much like using multiple decimal digits to
represent numbers greater than 9. E.g., the
binary number 101 has decimal value 5.)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Digital System
sender
noise
VN
VS
VR
VN = 0V
receiver
VS
VR
5V “0” “1” “0” HIGH
“0” “1” “0”
5V
t
2.5V
0V
LOW
0V
t
2.5V
With noise
VS
VN = 0.2V
“0” “1” “0”
5V
“0” “1” “0”
0.2V
t
t
2.5V
VS
2.5V
t
0V
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Digital System
Better noise immunity
Lots of “noise margin”
For “1”: noise margin 5V to 2.5V = 2.5V
For “0”: noise margin 0V to 2.5V = 2.5V
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Voltage Thresholds
and Logic Values
5V
1
1
sender
0
1
2.5V receiver
0
0
0V
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
But, but, but …
What about 2.5V?
Hmmm… create “no man’s land”
or forbidden region
For example,
5V
1
sender
3V
2V
0
1
VH
forbidden
region
receiver
VL
0
0V
“1”
V
“0”
0V
H
5V
V
L
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
But, but, but …
Where’s the noise margin?
What if the sender sent 1: V H ?
Hold the sender to tougher standards!
5V
1
V
0H
1
V
IH
sender
V
IL
0
receiver
0
V
0L
0V
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
But, but, but …
Where’s the noise margin?
What if the sender sent 1: V H ?
Hold the sender to tougher standards!
5V
1
V
0H
1
sender
Noise margins
V
IH
receiver
V
IL
0
0
V
0L
0V
“1” noise margin: V
- V
IH
0H
“0” noise margin: VIL - V
0L
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
5V
V
0H
V
IH
V
IL
V
0L
0V
5V
V
0H
V
IH
V
IL
V
0L
0V
0
1
0
1
sender
t
0
1
0
1
receiver
t
Digital systems follow static discipline: if
inputs to the digital system meet valid input
thresholds, then the system guarantees its
outputs will meet valid output thresholds.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Processing digital signals
Recall, we have only two values —
1,0
Map naturally to logic: T, F
Can also represent numbers
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Processing digital signals
Boolean Logic
If X is true and Y is true
Then Z is true else Z is false.
Z = X AND Y
X, Y, Z
are digital signals
“0” , “1”
Z = X • Y
Boolean equation
X
Y
AND gate
Z
Truth table representation:
X Y Z
0
0
1
1
0
1
0
1
0
0
0
1
Enumerate all input combinations
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Combinational gate
abstraction
„ Adheres to static discipline
„ Outputs are a function of
inputs alone.
Digital logic designers do not
have to care about what is
inside a gate.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Demo
X
Y
Z
Noise
X
Y
Z
Z = X • Y
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Examples for recitation
X
t
Y
t
Z
t
Z = X • Y
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
In recitation…
Another example of a gate
If (A is true) OR (B is true)
then C is true
else C is false
C = A + B
A
B
Boolean equation
OR
C
OR gate
More gates
B
B
Inverter
X
Y
Z
NAND
Z = X • Y
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
Boolean Identities
X • 1 = X
X • 0 = X
X + 1 = 1
X +0 = X
1 = 0
0 = 1
AB + AC = A • (B + C)
Digital Circuits
Implement:
B
C
output = A + B • C
B•C
output
A
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 4
6.002
CIRCUITS AND
ELECTRONICS
Inside the Digital Gate
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
Review
The Digital Abstraction
z Discretize value 0, 1
z Static discipline
meet voltage thresholds
sender
VOH
VOL
receiver
VIH
VIL
forbidden
region
Specifies how gates must be designed
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
Review
Combinational gate abstraction
outputs function of input alone
satisfies static discipline
A
B
C
NAND
A B
0 0
0 1
1 0
1 1
C
1
1
1
0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
For example:
a digital circuit
Demo
A⋅ B
A
B
D
C
D = (C ⋅ (A ⋅ B ))
3 gates here
„ A Pentium III class microprocessor
is a circuit with over 4 million gates !!
„ The RAW chip (http://www.cag.lcs.mit.edu/raw)
being built at the
Lab for Computer Science at MIT
has about 3 million gates.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
How to build a digital gate
Analogy
l ike
power
supply
A
(li
taps
s)
e
h
c
t
i
ke sw
B
C
if A=ON AND B=ON
C has H20
else
C has no H20
Use this insight to build an AND gate.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
How to build a digital gate
OR gate
A
C
B
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
Electrical Analogy
C
B
A
V +
–
Bulb C is ON if A AND B are ON,
else C is off
Key: “switch” device
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
Electrical Analogy
equivalent ckt
Key: “switch” device
control
in
C =0
in
out
C
in
out
C=1
3-Terminal device
if C = 0
else
out
short circuit between in and out
open circuit between in and out
For mechanical switch,
control
mechanical pressure
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
Consider
VS
RL
RL
VOUT
+ VS
–
VOUT
IN
C
C
VS = “1”
OUT
VS
VOUT
C =0
Truth table for
C VOUT
0 1
1 0
VS
VOUT
C =1
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
What about?
VS
Truth table for
c1 c2 VO
0 0 1
0 1 1
1 0 1
1 1 0
VOUT
c1
c2
Truth table for
VS
VOUT
c1
c2
c1 c2 VO
0 0 1
0 1 0
1 0 0
1 1 0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
What about?
can also build compound gates
VS
D
A
C
D = (A ⋅ B) + C
B
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
The MOSFET Device
Metal-Oxide
Semiconductor
Field-Effect
Transistor
drain
D
G
gate
≡
S
source
3 terminal lumped element
behaves like a switch
G : control terminal
D, S : behave in a symmetric
manner (for our needs)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
The MOSFET Device
Understand its operation by viewing it
as a two-port element —
out k
k
c
e
Ch extboo l
the t s interna
for it ture.
iG
c
u
r
t
s
D
iDS
G
+
vGS
–
vDS
S
–
G
D
iDS
on
vGS ≥ VT
S
D
off
G
vGS < VT
S
+
VT ≈ 1V typically
“Switch” model (S model) of the MOSFET
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
Demo
Check the MOS device
on a scope.
i
DS
+
vDS
+
vGS
–
–
iDS
vGS ≥ VT
vGS < VT
vDS
iDS vs vDS
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
A MOSFET Inverter
VS = 5V
RL
vOUT
A
B
IN
A
B
Note the power of abstraction.
The abstract inverter gate representation
hides the internal details such as power
supply connections, RL, GND, etc.
(When we build digital circuits, the
and
are common across all gates!)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
Example
vOUT
5V
vOUT
vIN
0V V
T
v IN
=1V
5V
The T1000 model laptop desires gates that satisfy
the static discipline with voltage thresholds. Does
out inverter qualify?
1:
VOL = 0.5V
VIL = 0.9V
VOH = 4.5V
VIH = 4.1V
sender
5
4.5 V
OH
receiver
5
4.1
0.9
0.5 VOL
0:
0
0
Our inverter satisfies this.
1
VIH
VIL
0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
E.g.:
Does our inverter satisfy the static
discipline for these thresholds:
VOL = 0.2V
VIL = 0.5V
VOH = 4.8V
VIH = 4.5V
yes
x
VOL = 0.5V
VIL = 1.5V
VOH = 4.5V
VIH = 3.5V
no
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
Switch resistor (SR) model
of MOSFET
…more accurate MOS model
D
D
G
G
G
S
D
vGS < VT
S
RON
vGS ≥ VT S
e.g. RON = 5 KΩ
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
SR Model of MOSFET
D
D
G
G
G
vGS < VT
S
S
MOSFET
S model
RON
vGS ≥ VT S
MOSFET
SR model
vGS ≥ VT
vGS ≥ VT
iDS
D
iDS
1
RON
vGS < VT
vGS < VT
vDS
vDS
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
Using the SR model
VS
RL
RL
vOUT
+ VS
–
vOUT
IN
C
C
VS = “1”
OUT
Truth table for
VS
RL
vOUT
C VOUT
0 1
1 0
RON
C =0
VS
RL
C =1
vGS ≥ VT
vOUT
RON
Choose RL, RON, VS such that:
V R
v
= S ON ≤ V
OL
OUT R
+R
L
ON
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 5
6.002
CIRCUITS AND
ELECTRONICS
Nonlinear Analysis
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
Review
Discretize matter t LCA
m1 X KVL, KCL, i-v
m2 X Composition rules
m3 X Node method
m4 X Superposition
m5 X Thévenin, Norton
any
circuit
linear
circuits
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
Review
Discretize value t Digital abstraction
X Subcircuits for given “switch”
setting are linear! So, all 5
methods (m1 – m5) can be
applied
VS
VS
A =1
B =1
RL
RL
C
A
C
RON
B
RON
SR MOSFET Model
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
Today
Nonlinear Analysis
X Analytical method
based on m1, m2, m3
X Graphical method
X Introduction to incremental analysis
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
How do we analyze nonlinear
circuits, for example:
Hypothetical
nonlinear
D
device
(Expo Dweeb ☺)
iD
V
+
vD
-
+
–
+ vD -
D
iD
iD
iD = aebvD
a
vD
0,0
(Curiously, the device supplies power when vD is negative)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
Method 1: Analytical Method
Using the node method,
(remember the node method applies for linear or
nonlinear circuits)
vD − V
+ iD = 0
R
iD = aebvD
2 unknowns
1
2
2 equations
Solve the equation by
trial and error
numerical methods
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
Method 2: Graphical Method
Notice: the solution satisfies equations
1 and 2
iD
2
iD = aebvD
a
vD
iD
V vD
1 iD = −
R R
V
R
1
slope = −
R
vD
V
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
Combine the two constraints
iD
V
1
R
~ 0 .4
a
¼
called “loadline”
for reasons you
will see later
vD
~ 0.5
e.g.
V
1
V =1
vD = 0.5V
R =1
1
a=
4
b =1
iD = 0.4 A
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
Method 3: Incremental Analysis
Motivation: music over a light beam
Can we pull this off?
iD
+
vD LED
light
intensity
I D ∝ iD
vI music signal
iR
vI (t ) +
–
t
vI (t )
iD (t )
light
AMP
iR ∝ I R
light intensity IR
in photoreceiver
LED: Light
Emitting
expoDweep ☺
iR (t )
sound
nonlinear
linear
problem! will result in distortion
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
Problem:
The LED is nonlinear
distortion
iD
iD
vD
vD = vI
t
vD
t
iD
vD
t
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
If only it were linear …
iD
iD
vD
vD
t
it would’ve been ok.
What do we do?
Zen is the answer
… next lecture!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 6
6.002
CIRCUITS AND
ELECTRONICS
Incremental Analysis
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
Review
Nonlinear Analysis
X Analytical method
X Graphical method
Today
X Incremental analysis
Reading: Section 4.5
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
Method 3: Incremental Analysis
Motivation: music over a light beam
Can we pull this off?
iD
+
vD LED
light
intensity
I D ∝ iD
vI music signal
iR
vI (t ) +
–
t
vI (t )
iD (t )
light
AMP
iR ∝ I R
light intensity IR
in photoreceiver
LED: Light
Emitting
expoDweep ☺
iR (t )
sound
nonlinear
linear
problem! will result in distortion
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
Problem:
The LED is nonlinear
distortion
iD
iD
vD
vD = vI
t
vD
t
iD
vD
t
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
Insight:
iD
small region
looks linear
(about VD , ID)
ID
VD
vD
DC offset
or DC bias
Trick:
vi (t ) +
–
vI
VI
+
–
iD = I D + id
+
vD LED
vD = VD + vd
VI
vi
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
Result
iD
id
ID
vD
VD
vd
very small
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
Result
vD = vI
vd
vD
VD
t
iD
id
iD
~linear!
ID
t
Demo
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
The incremental method:
(or small signal method)
1. Operate at some DC offset
or bias point VD, ID .
2. Superimpose small signal vd
(music) on top of VD .
3. Response id to small signal vd
is approximately linear.
Notation:
iD = I D + id
total
DC
small
variable offset superimposed
signal
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
What does this mean
mathematically?
Or, why is the small signal response
linear?
nonlinear
iD = f (vD )
We replaced
large DC
vd
vD = VD + ΔvD
increment
about VD
using Taylor’s Expansion to expand
f(vD) near vD=VD :
iD = f (VD ) +
+
df (vD )
⋅ ΔvD
dvD vD =VD
1 d 2 f (v D )
2
⋅
Δ
v
D +"
2
2! dvD v =V
D
D
neglect higher order terms
because ΔvD is small
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
iD ≈ f (VD ) +
d f (v D )
⋅ ΔvD
d vD vD =VD
constant
w.r.t. ΔvD
constant w.r.t. ΔvD
slope at VD, ID
We can write
X : I D + ΔiD ≈ f (VD ) +
d f (v D )
⋅ Δ vD
d vD vD =VD
equating DC and time-varying parts,
I D = f (VD )
operating point
d f (v D )
ΔiD =
⋅ ΔvD
d vD vD =VD
constant w.r.t. ΔvD
so, Δ iD ∝ ΔvD
By notation,
Δ iD = id
Δ v D = vd
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
In our example,
iD = a e
bv D
From X : I D + id ≈ a e bVD + a e bVD ⋅ b ⋅ vd
Equate DC and incremental terms,
I D = a ebVD
operating point
aka bias pt.
aka DC offset
id = a ebVD ⋅ b ⋅ vd
id = I D ⋅ b ⋅ vd
constant
small signal
behavior
linear!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
Graphical interpretation
operating point
I D = a ebVD
id = I D ⋅ b ⋅ vd
A
slope at
VD, ID
iD
ID
id
B
VD
operating
point
vd
vD
we are
approximating
A with B
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
graphically
mathematically
now, circuit
We saw the small signal
Large signal circuit:
ID
VI
+
LED VD
-
+
–
I D = a ebVD
Small signal reponse: id = I D b vd
+ vd -
behaves like:
id
R=
small signal circuit:
1
ID b
id
vi
+
–
+
vd
-
1
I Db
Linear!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 7
6.002
CIRCUITS AND
ELECTRONICS
Dependent Sources
and Amplifiers
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Review
„ Nonlinear circuits — can use the
node method
„ Small signal trick resulted in linear
response
Today
„ Dependent sources
„ Amplifiers
Reading: Chapter 7.1, 7.2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Dependent sources
Seen previously
+
i
Resistor
Independent
Current source
v
–
+
R
v –
i
I
v
i=
R
i=I
2-terminal 1-port devices
New type of device: Dependent source
iI
i
O
+
control
port
vI
f ( vI )
–
+
vO
output
port
–
2-port device
E.g., Voltage Controlled Current Source
Current at output port is a function of voltage
at the input port
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Dependent Sources: Examples
Example 1: Find V
+
R V
–
independent
current
source
I = I0
V = I0R
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Dependent Sources: Examples
Example 2: Find V
voltage
controled
current
source
+
R V
–
K
I = f (V ) =
V
iI
+
+
R V
–
f (vI ) =
K
vI
iO
+
vI
vO
–
–
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Dependent Sources: Examples
Example 2: Find V
voltage
controled
current
source
+
R V
–
K
I = f (V ) =
V
e.g. K = 10-3 Amp·Volt
R = 1kΩ
K
V = IR = R
V
or V 2 = KR
or V = KR
= 10 −3 ⋅ 10 3
= 1 Volt
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Another dependent source example
RL
iIN
vI +
–
iD
+
+
vIN
vO
–
–
e.g.
VS +
–
iD = f (vIN )
iD = f (vIN )
K
2
= (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
Find vO as a function of vI .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Another dependent source example
VS
RL
iIN
vI +
–
iD
+
+
vIN
vO
–
–
iD = f (vIN )
e.g.
iD = f (vIN )
K
2
= (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
Find vO as a function of vI .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Another dependent source example
VS
RL
vI
vI
+
–
vO
K
2
iD = (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
Find vO as a function of vI .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Another dependent source example
VS
RL
vI
vI
+
–
vO
K
2
iD = (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
KVL
− VS + iD RL + vO = 0
vO = VS − iD RL
K
2
vO = VS − (vI − 1) RL
2
vO = VS
for vI ≥ 1
for vI < 1
Hold that thought
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Next, Amplifiers
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Why amplify?
Signal amplification key to both analog
and digital processing.
Analog:
AMP
IN
Input
Port
OUT
Output
Port
Besides the obvious advantages of being
heard farther away, amplification is key
to noise tolerance during communcation
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Why amplify?
Amplification is key to noise tolerance
during communcation
No amplification
useful
signal
1 mV
nois
e
10 mV
huh?
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Try amplification
e
nois
AMP
not bad!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Why amplify?
Digital:
Valid region
5V
5V
VIH IN
VIL
0V
5V
OUT
Digital System
IN
5V
VOL
OUT
V OH
VIH
VIL
0V
0V
VOH
t
V OL
0V
t
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Why amplify?
Digital:
Static discipline requires amplification!
Minimum amplification needed:
VIH
VIL
VOH
VOL
VOH − VOL
VIH − VIL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
An amplifier is a 3-ported device, actually
Power port
Input
port
iO
iI
+v
– I
Amplifier
+ v Output
– O port
We often don’t show the power port.
Also, for convenience we commonly observe
“the common ground discipline.”
In other words, all ports often share a
common reference point called “ground.”
POWER
IN
OUT
How do we build one?
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Remember?
VS
RL
vI
vI
+
–
vO
K
2
iD = (vIN − 1) for vIN ≥ 1
2
iD = 0
otherwise
KVL
− VS + iD RL + vO = 0
vO = VS − iD RL
K
2
vO = VS − (vI − 1) RL
2
vO = VS
for vI ≥ 1
for vI < 1
Claim: This is an amplifier
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
So, where’s the amplification?
Let’s look at the vO versus vI curve.
mA
e.g. VS = 10V , K = 2 2 , RL = 5 kΩ
V
K
2
vO = VS − RL (vI − 1)
2
2 −3
2
3
= 10 − ⋅10 ⋅ 5 ⋅ 10 (vI − 1)
2
vO = 10 − 5 (vI − 1)
vO
VS
2
ΔvO
1
ΔvO
>1
Δv I
ΔvI
vI
amplification
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
Plot vO versus vI
vO = 10 − 5 (vI − 1)
2
0.1 change
in vI
Demo
vI
vO
0.0
1.0
1.5
2.0
2.1
2.2
2.3
2.4
10.00
10.00
8.75
5.00
4.00
2.80
1.50
~ 0.00
1V change
in vO
Gain!
Measure vO .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
One nit …
vO
What
happens
here?
1
vI
Mathematically,
K
2
vO = VS − RL (vI − 1)
2
So
is mathematically predicted behavior
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
One nit …
vO
K
2
vO = VS − RL (vI − 1)
2
What
happens
here?
vI
1
However, from
K
2
iD =
(vI − 1)
2
VS
for vI ≥ 1
RL
vO
VCCS
iD
For vO>0, VCCS consumes power: vO iD
For vO<0, VCCS must supply power!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
If VCCS is a device that can source
power, then the mathematically
predicted behavior will be observed —
vO
K
2
i.e. vO = VS − RL (vI − 1)
2
vI
where vO goes -ve
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
If VCCS is a passive device,
then it cannot source power,
so vO cannot go -ve.
So, something must give!
Turns out, our model breaks down.
K
2
iD = (vI − 1)
2
will no longer be valid when vO ≤ 0 .
e.g. iD saturates (stops increasing)
and we observe:
Commonly
vO
1
vI
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 – Fall 2002: Lecture 8
6.002
CIRCUITS AND
ELECTRONICS
MOSFET Amplifier
Large Signal Analysis
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Review
„ Amp constructed using dependent source
control a
a′
port
„
DS
b output
b ′ port
Dependent source in a circuit
+
–
a +
b
v
i = f (v )
a′ –
b′
„ Superposition with dependent sources:
one way tleave all dependent sources in;
solve for one independent source at a
time [section 3.5.1 of the text]
„ Next, quick review of amp …
Reading: Chapter 7.3–7.7
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Amp review
VS
RL
vO
VCCS
vI
K
2
iD = (vI − 1)
2
+
–
for vI ≥ 1V
= 0 otherwise
vO = VS − iD RL
K
(vI − 1)2
2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Key device Needed:
v
A
B
i = f (v )
voltage controlled
current source
C
Let’s look at our old friend, the MOSFET …
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Key device Needed:
Our old friend, the MOSFET …
First, we sort of lied. The on-state behavior of the
MOSFET is quite a bit more complex than either the
ideal switch or the resistor model would have you believe.
D
G
vGS < VT
D
S
S
?
V
G
vGS ≥ T
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Graphically
Demo
iDS
+
vGS
–
iDS
egio
n
iDS
vGS ≥ VT
vGS < VT
vGS < VT
vDS
S MODEL
vDS = vGS − VT
vGS 1
Saturation
region
vGS 2
vGS3
...
vGS ≥ VT
Trio
de r
iDS
v+DS
–
vDS
SR MODEL
vGS < VT Cutoff
vDS
region
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Graphically
iDS
+
vGS
–
iDS
iDS
egio
n
S MODEL
vGS 2
vGS3
...
vGS < VT
Saturation
region
Trio
de r
vGS ≥ VT
vDS
vDS = vGS − VT
vGS 1
iDS
vGS ≥ VT
vGS < VT
v+DS
–
vDS
SR MODEL
vGS < VT
vDS
when
vDS ≥ vGS − VT
Notice that
MOSFET
behaves like a
current source
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
MOSFET SCS Model
When
vDS ≥ vGS − VT
the MOSFET is in its saturation region, and the
switch current source (SCS) model of the MOSFET is
more accurate than the S or SR model
D
G
vGS < VT
S
D
D
G
S
G
vGS ≥ VT
iDS = f (vGS )
K
2
= (vGS − VT )
2
S
when
vDS ≥ vGS − VT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Reconciling the models…
iDS
iDS
vGS ≥ VT
vGS < VT
vDS
S MODEL
for fun!
vGS < VT
Saturation
region
vGS 2
vGS3
...
vGS ≥ VT
Trio
de r
egio
n
iDS
vDS = vGS − VT
vGS 1
vDS
SR MODEL
for digital
designs
vGS < VT
vDS
SCS MODEL
for analog
designs
When to use each model in 6.002?
Note: alternatively (in more advanced courses)
vDS ≥ vGS − VT
vDS < vGS − VT
use SCS model
use SR model
or, use SU Model (Section 7.8 of A&L)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Back to Amplifier
VS
vI
AMP
vO
VS
RL
vI
G
D
S
vO
K
2
iDS = (vI − VT )
2
in saturation
region
To ensure the MOSFET operates as a VCCS,
we must operate it in its saturation region
only. To do so, we promise to adhere to the
“saturation discipline”
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
MOSFET Amplifier
VS
RL
vI
G
D
S
vO
K
2
iDS = (vI − VT )
2
in saturation
region
To ensure the MOSFET operates as a VCCS,
we must operate it in its saturation region
only. We promise to adhere to the
“saturation discipline.”
In other words, we will operate the amp
circuit such that
vGS ≥ VT and vDS ≥ vGS – VT
vO ≥ vI – vT
at all times.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Let’s analyze the circuit
First, replace the MOSFET with its
SCS model.
VS
RL
vO
G
vGS = vI
+
–
+
vI
–
D
K
2
iDS = (vI − VT )
2
S
A
for vO ≥ vI − VT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Let’s analyze the circuit
VS
RL
vO
G
vGS = vI
+
–
+
vI
–
D
iDS =
K
(vI − VT )2
2
A
for vO ≥ vI − VT
S
(vO = vDS in our example)
1
Analytical method: vO vs vI
vO = VS − iDS RL
B
K
2
or vO = VS − (vI − VT ) RL for vI ≥ VT
2
v ≥ v −V
O
vO = VS
I
T
vI < VT
(MOSFET turns off)
for
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Graphical method vO vs vI
K
2
From A : iDS = (vI − VT ) ,
2
vO ≥ vI − VT
2
for
⇓
2iDS
vO ≥
K
⇓
K 2
iDS ≤ vO
2
VS v0
−
B : iDS =
RL RL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
2
Graphical method vO vs vI
K 2
K
2
A : iDS = (vI − VT ) , for iDS ≤ vO
2
2
VS vO
=
−
i
B : DS
RL RL
iDS
K 2
iDS ≤ vO
2
VS
RL
B
Lo
ad
A
li n
e
vI
= vGS
VS
Constraints
A
and
B
vO
must be met
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
2
Graphical method vO vs vI
iDS
VS
RL
iDS ≤
K 2
vO
2
A
B
vI
VI
I DS
VO
VS
vO
Constraints A and B must be met.
Then, given VI, we can find VO, IDS .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Large Signal Analysis
of Amplifier
(under “saturation discipline”)
1
vO versus vI
2
Valid input operating range and
valid output operating range
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Large Signal Analysis
vO versus vI
1
vO
K
2
VS − (vI − VT ) RL
2
VS
VT
vO = vI − VT
gets into
triode region
vI
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Large Signal Analysis
What are valid operating ranges
under the saturation discipline?
vI ≥ VT
Our
K 2
iDS ≤ vO
Constraints vO ≥ v I − VT
2
2
iDS
VS
RL
iDS ≤
K 2
vO
2
K
(vI − VT )2
2
vI
VS vO
iDS =
−
RL RL
iDS =
VS
?
vO
vI = VT
vO = VS and iDS = 0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
2
Large Signal Analysis
What are valid operating ranges
under the saturation discipline?
K 2
iDS ≤ vO
2
iDS
K
2
iDS = (vI − VT )
2
vI
V
v
iDS = S − O
RL RL
vO
− 1 + 1 + 2 KRLVS
vI = VT +
KRL
− 1 + 1 + 2 KRLVS
vO =
KRL
VS vO
iDS =
−
RL RL
vI = VT
vO = VS and iDS = 0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
Large Signal Analysis
Summary
1
vO versus vI
vO = VS −
2
K
(vI − VT )2 RL
2
Valid operating ranges under the
saturation discipline?
Valid input range:
vI : VT
to
− 1 + 1 + 2 KRLVS
VT +
KRL
corresponding output range:
vO : VS to
− 1 + 1 + 2 KRLVS
KRL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 9
6.002
CIRCUITS AND
ELECTRONICS
Amplifiers -Small Signal Model
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Review
„ MOSFET amp
VS
RL
vO
vI
iDS
„ Saturation discipline — operate
MOSFET only in saturation region
„Large signal analysis
1. Find vO vs vI under saturation discipline.
2. Valid vI , vO ranges under saturation discipline.
Reading: Small signal model -- Chapter 8
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Large Signal Review
1
vO vs vI
K
(vI − 1)2 RL
2
valid for vI ≥ VT
and
vO ≥ vI – VT
K 2
(same as iDS ≤ vO )
2
vO = VS −
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Large Signal Review
2
Valid operating ranges
VS
vO
5V
corresponding
interesting
region for vO
vO > vI − VT
vO = vI − VT
vO < vI − VT
1V
vI
VT
1V
2V
“interesting” region
for vI . Saturation
discipline satisfied.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
But…
5V
VS
vO
vO = vI − VT
vO 1V
vI
VT
1V
Demo
vI
2V
Amplifies alright,
but distorts
vI
vO
t
Amp is nonlinear … /
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Small Signal Model
vO
~ 5V VS
Focus on this line segment
(VI , VO )
~ 1V
vI
VT
1V
~ 2V
2
K (vI − VT )
vO = VS −
RL
2
Amp all right, but nonlinear!
Hmmm … So what about our linear amplifier ???
Insight:
But, observe vI vs vO about some
point (VI , VO) … looks quite linear !
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Trick
ΔvO
vo
VO
vi
(VI ,VO )
looks
linear
VI
ΔvI
™
Operate amp at VI , VO
Æ DC “bias” (good choice: midpoint
of input operating range)
™ Superimpose small signal on top of VI
™ Response to small signal seems to be
approximately linear
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Trick
ΔvO
vo
VO
vi
(VI ,VO )
looks
linear
VI
ΔvI
™
Operate amp at VI , VO
Æ DC “bias” (good choice: midpoint
of input operating range)
™ Superimpose small signal on top of VI
™ Response to small signal seems to be
approximately linear
Let’s look at this in more detail —
I graphically
II mathematically
III from a circuit viewpoint
next
week
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
I Graphically
We use a DC bias VI to “boost” interesting input
signal above VT, and in fact, well above VT .
VS
RL
interesting
input signal
ΔvI +
–
VI +
–
vO
Offset voltage or bias
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Graphically
VS
RL
interesting
input signal
vO
ΔvI +
–
VI +
–
VS
vO
operating
point
VO
0
VI , VO
vO = vI − VT
vI
VT
Good choice for operating point:
midpoint of input operating range
VI
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Small Signal Model
aka incremental model
aka linearized model
Notation —
Input:
vI = VI + vi
total
DC small
variable bias signal (like ΔvI)
bias voltage aka operating point voltage
Output: vO = VO + vo
Graphically,
vI
vO
vi
vo
VI
VO
0
t
0
t
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
II Mathematically
(… watch my fingers)
RL K
2
vO = VS −
(vI − VT ) VO = VS − RL K (VI − VT )2
2
2
“
substituting vI = VI + vi
vi << VI
RL K
vO = VS −
( [VI + vi ] − vT )2
2
= VS −
RL K
( [VI − VT ] + vi )2
2
(
RL K
[VI − VT ]2 + 2 [VI − vT ]vi + vi 2
= VS −
2
RL K
VO + vo = VS −
(VI − VT )2 − RL K (VI − VT ) vi
2
From “,
)
vo = − RL K (VI − VT ) vi
gm
related to
VI
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Mathematically
vo = − RL K (VI − VT ) vi
gm
related to
VI
vo = − g m RL vi
For a given DC operating point voltage VI,
VI – VT is constant. So,
vo = − A vi
constant w.r.t. vi
In other words, our circuit behaves like a linear amplifier
for small signals
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
Another way
RL K
vO = VS −
(vI − VT )2
2
(
)
R K
⎡
⎤
2
L
v −V
⎢VS −
⎥
I
T
2
d ⎢⎣
⎥⎦
vo =
dv
I
⋅ vi
v =V
I
I
slope at VI
vo = − RL K (VI − VT ) ⋅ vi
g m = K (VI − VT )
A = − g m RL
amp gain
Also, see Figure 8.9 in the course notes
for a graphical interpretation of this result
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
More next lecture …
Demo
iDS
load line
input signal response
operating point
VI
VO
vO
How to choose the bias point:
1. Gain component g m ∝ VI
2. vi gets big Æ distortion.
So bias carefully
3. Input valid operating range.
Bias at midpoint of input operating
range for maximum swing.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 10
6.002
CIRCUITS AND
ELECTRONICS
Small Signal Circuits
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
Review:
„ Small signal notation
vA = VA + va
total operating
point
„
small
signal
vOUT = f (vI )
d
vout =
f (vI )
⋅ vi
dv I
v I =VI
VS
„
vI = VI + vi
vi
VI
RL
vO = VO + vo
+
–
+
–
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
Review:
I Graphical view
(using transfer function)
vO
behaves linear
for small
perturbations
vI
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
Review:
II Mathematical view
K (vI − VT )
vO = VS −
RL
2
2
⎡V − K v − V 2 R ⎤
( I T ) L⎥
S
⎢
d ⎣
2
⎦
vo =
dvI
⋅ vi
v I =VI
vo = − K (VI − VT ) RL ⋅ vi
gm
related to VI
constant for fixed
DC bias
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
How to choose the bias point,
using yet another graphical view
based on the load line
i DS
i DS <
Demo
K 2
vO
2
V S vO
i
=
load line DS R − R
L
L
input signal
response
VI
VO
− 1 + 1 + 2 KR LV S
v I = VT +
KR L
vO
v I = VT
Choosing a bias point:
1. Gain
g m RL ∝ VI
2. Input valid operating range for amp.
3. Bias to select gain and input swing.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
III The Small Signal Circuit View
We can derive small circuit equivalent
models for our devices, and thereby conduct
small signal analysis directly on circuits
e.g. large signal
circuit model
for amp
vI +
–
R
VS
vOUT
K
2
iD = (vI − VT )
2
+
–
1
We can replace large signal models with
small signal circuit models.
Foundations: Section 8.2.1 and also in the
last slide in this lecture.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
Small Signal Circuit Analysis
1
Find operating point using DC bias
inputs using large signal model.
2
Develop small signal (linearized)
models for elements.
3
Replace original elements with small
signal models.
Analyze resulting linearized circuit…
Key: Can use superposition and other
linear circuit tools with linearized
circuit!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
Small Signal Models
A
MOSFET
large
signal
D
vGS
Small signal?
iDS =
K
(vGS − VT )2
2
S
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
Small Signal Models
A
MOSFET
large
signal
D
vGS
K
2
iDS = (vGS − VT )
2
Small signal:
K
2
iDS = (vGS − VT )
2
S
∂ ⎡K
2⎤
ids =
(
v
−
V
)
⋅ v gs
GS
T
⎢
⎥
∂vGS ⎣ 2
⎦ vGS =VGS
ids = K (VGS − VT ) ⋅ v gs
ids is linear in vgs !
gm
D
small
signal
v gs
ids = K (VGS − VT ) v gs
S
ids = g m v gs
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
B
DC Supply VS
large
signal
vS = VS
iS
+ vS = VS
–
Small signal
∂VS
vs =
⋅ is
∂iS iS = I S
is +
vs
–
vs = 0
DC source behaves
as short to small
signals.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
C
Similarly, R
large
signal
iR +
vR
R
–
v R = R iR
vr =
∂ ( RiR )
⋅ ir
∂iR iR = I R
vr = R ⋅ ir
small
signal
ir +
vr
R
–
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
Amplifier example:
Large signal
RL
vO
+ v
– I
Small signal
RL
vo
+ V
– S
+ vi
–
iDS
K
2
iDS = (vI − VT )
2
ids
ids = K (VI − VT ) ⋅ vi
K
2
vO = VS − (vI − VT ) RL
2
ids RL + vo = 0
vo = −ids RL
vo = − K (VI − VT )RL ⋅ vi
= − g m RL ⋅ vi
Notice, first we need to find operating
point voltages/currents.
Get these from a large signal analysis.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
III The Small Signal Circuit View
To find the relationship between the small signal parameters of
a circuit, we can replace large signal device models with
corresponding small signal device models, and then analyze the
resulting small signal circuit.
Foundations: (Also see section 8.2.1 of A&L)
KVL, KCL applied to some circuit C yields:
" + v A + " + vOUT + " + vB + "
1
Replace total variables with
operating point variables plus small signal variables
" + VA + v a " + VOUT + vout + VB + vb + "
Operating point variables themselves satisfy the
same KVL, KCL equations
" + VA
" + VOUT
+ VB
+"
so, we can cancel them out
Leaving
"
+ va "
+ vout
+ vb + "
2
But 2 is the same equation as 1 with small signal
variables replacing total variables, so 2 must reflect same
topology as in C, except that small signal models are used.
Since small signal models are linear, our linear tools will now
apply…
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 11
6.002
CIRCUITS AND
ELECTRONICS
Capacitors
and First-Order Systems
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
Motivation
Demo
5V
5V
B
C
A
5V
0V
5
A
0
5
Expect this, right?
But observe this!
B
0
5
Expected
Observed
C
0
Reading:
Chapters 9 & 10
Delay!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
The Capacitor
D
n-channel MOSFET
symbol
G
S
drain
gate
m+
e+
t +
a+
l +
+
n
o
x
i
d
e
source
s
i
l n-channel
p
i MOSFET
n-channel c
o
n
n
D
G
CGS
S
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
Ideal Linear Capacitor
+
+ A
++++
-----
E
d
EA
d
obeys DMD!
total charge on
capacitor
= +q − q = 0
C=
i
C
q
+
v
–
q = C v
coulombs farads volts
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
Ideal Linear Capacitor
i
q
C
q = C v
+
v
–
dq
i=
dt
d (Cv )
=
dt
dv
=C
dt
⎡ E = 1 Cv 2 ⎤
⎢⎣
⎥⎦
2
A capacitor is an energy storage device
Æ memory device Æ history matters!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
Analyzing an RC circuit
Thévenin Equivalent:
vI (t ) +
–
R
C
+
vC (t )
–
Apply node method:
vC − vI
dvC
+C
=0
R
dt
dvC
+ vC = vI
RC
dt
t ≥ t0
vC (t0 ) given
units
of time
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
Let’s do an example:
+
v I (t )
R
+
–
C
vC (t )
–
vI (t ) = VI
vC (0 ) = V0 given
dvC
RC
+ vC = VI
dt
X
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
Example…
vI (t ) = VI
vC (0 ) = V0 given
RC
dvC
+ vC = VI
dt
X
vC (t ) = vCH (t ) + vCP (t )
total homogeneous
particular
Method of homogeneous and particular
solutions:
1
Find the particular solution.
2 Find the homogeneous solution.
3 The total solution is the sum of
the particular and homogeneous
solutions.
Use the initial conditions to solve
for the remaining constants.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
1 Particular solution
dvCP
+ vCP = VI
dt
RC
vCP = VI
RC
works
dVI
+ VI = VI
dt
0
In general, use trial and error.
vCP : any solution that satisfies the
original equation X
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
2 Homogeneous solution
dvCH
RC
+ vCH = 0
dt
Y
vCH : solution to the homogeneous
equation Y
(set drive to zero)
vCH = A e st
assume solution
of this form. A, s ?
dA e st
+ A e st = 0
RC
dt
R CA s e st + A e st = 0
Discard trivial A = 0 solution,
Characteristic equation
R C s +1 = 0
s= −
or
1
RC
vCH = Ae
−t
RC
RC
called time
constant
τ
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
3 Total solution
vC = vCP + vCH
vC = VI + A e
−t
RC
Find remaining unknown from initial
conditions:
at t = 0
Given,
vC = V0
so,
V0 = VI + A
or
A = V0 − VI
thus
vC = VI + (V0 − VI ) e
also
iC = C
−t
RC
dvC
(V − VI )
e
=− 0
dt
R
−t
RC
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
vC = VI + (V0 − VI ) e
−t
RC
vC
VI
V0
0
t
RC
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
Examples
vC
vC
5V
5V
5 + 5e
−t
RC
5e
t
0V
VO = 0V
VI = 5V
5
0
−t
RC
t
0V
VO = 5V
VI = 0V
5
0
τ = RC
Remember
B
demo
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 12
6.002
CIRCUITS AND
ELECTRONICS
Digital Circuit
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Review
vI
vI +
–
VI
+
vC
–
R
C
t
0
vC (0 ) = VO
vC = VI + (VO − VI ) e
−t
RC
1
vC
VI
time constant RC
t
VO
RC
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Let’s apply the result to
an inverter.
B
A
X
First, rising delay tr at B
VS
VS
A
vA
5V
0
1 Æ 0 at A
B
CGS
X
t
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
First, rising delay tr at B
VS
VS
A
B
vA
5V
CGS
X
0
1 Æ 0 at A
t
vB
5V
ideal
observed
t
0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
First, rising delay tr at B
VS
VS
A
B
vA
5V
CGS
X
0
1 Æ 0 at A
t
5V
VOH
rising delay of X
vB
0
tr
t
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Equivalent circuit for 0Æ1 at B
vI = VS
RL
+
–
CGS
vI = VS
vB (0 ) = 0
From
+
vB
–
for t ≥ 0
1
vB = VS + (0 − VS ) e
−t
RL CGS
Now, we need to find t for which
vB = VOH .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Or
vOH = VS − VS e
Find tr :
VS e
−t r
RL CGS
−t
RL CGS
= VS − VOH
VS − VOH
− tr
= ln
RL CGS
VS
VS − VOH
t r = − RL CGS ln
VS
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Or
vOH = VS − VS e
Find tr :
VS e
−t r
RL CGS
−t
RL CGS
= VS − VOH
VS − VOH
− tr
= ln
RLCGS
VS
VS − VOH
t r = − RL CGS ln
VS
e.g.
RL = 1K
VS = 5V
CGS = 0.1 pF
VOH = 4V
t r = −1 × 10 3 × 0.1 × 10 −12 ln
= 0.16 ns
5−4
5
RC = 0.1 ns !
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Falling Delay tf
Falling delay tf is
the t for which vB falls to VOL
Equivalent circuit for 1 Æ 0 at B
vB (0 ) = VS
(5V )
RL
VS +
–
+
CGS vB
–
RON
X
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Falling Delay tf
Equivalent circuit for 1 Æ 0 at B
vB (0 ) = VS
(5V )
RL
VS +
–
+
CGS vB
–
RON
X
Thévenin replacement …
RTH
VTH +
–
+
CGS vB
–
RTH = RL || RON
RON
VTH = VS
RON + RL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
From
1
vB = VTH + (VS − VTH ) e
−t
RTH CGS
Falling decay tf is
the t for which vB falls to VOL
−t f
VOL = VTH + (VS − VTH ) e RTH CGS
or
VOL − VTH
t f = − RTH CGS ln
VS − VTH
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
V −V
t f = − RTH CGS ln OL TH
VS − VTH
e.g.
RL = 1K
VS = 5V
CGS = 0.1 pF
VOL = 1V
RTH ≈ 10Ω,
RON = 10Ω
VTH ≈ 0V
t f = −10 ⋅ 0.1 ⋅10
= 1.6 ps
−12
1
ln
5
RC = 1 ps !
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
For recitation: Slow may be better
Problem
chip
pin 2
pin 1
v
CL
v:
ideal
observed
slow!
So the engineers decided to speed it up…
RL
RON
made RL small
made RON small
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
For recitation: Slow may be better
Problem
chip
pin 2
pin 1
v
CL
v:
ideal
…
observed
slow!
but, disaster!
v:
observed
expected
VIL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Why? Consider
Case
1
…
Demo
R1
pin1
R0
ok
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Why? Consider
Case 2
…
Demo
CP
R1
pin1
pin2
R0
R2
crosstalk!
CP
R
model for crosstalk:
+
v
+
–
–
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
Case 3
…
6.002 expert saw the solution
R1
CP
R0
R2
+
–
slower transitions!
Detailed analysis in recitation.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
13
6.002
CIRCUITS AND
ELECTRONICS
State and Memory
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
Review
Recall
vI +
–
v I = VI
for
R
C
+
vC
–
vC (0)
t ≥0
−t
vC = VI + (vC (0)− VI ) e RC
1
Reading: Sections 10.3, 10.5, and 10.7
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
This lecture will dwell on the
memory property of capacitors.
For the RC circuit in the previous slide
vI
vI
VI
t ≥0
t
0
vC
VI
−t
vC = VI + (vC (0)− VI ) e RC
vC (0 )
0
t
Notice that the capacitor voltage for t ≥ 0 is
independent of the form of the input voltage
before t = 0 . Instead, it depends only on the
capacitor voltage at t = 0 , and the input voltage
for t ≥ 0 .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
State
State : summary of past inputs relevant
to predicting the future
q=CV
for linear capacitors,
capacitor voltage V
is also state variable
state variable, actually
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
State
Back to our simple RC circuit 1
vC = f (vC (0 ), vI (t ))
vC = VI + (vC (0 ) − VI ) e
−t
RC
Summarizes the past input relevant
to predicting future behavior
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
State
We are often interested in circuit
response for
zero state
vC (0) = 0
zero input
vI (t) = 0
Correspondingly,
zero state response or ZSR
vC = VI − VI e
−t
RC
2
zero input response or ZIR
vC = vC (0 ) e
−t
RC
3
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
One application of STATE
DIGITAL MEMORY
Why memory?
Or, why is combinational logic insufficient?
Examples
Consider adding 6 numbers on your
calculator
2+9+6+5+3+8
M+
“Remembering” transient inputs
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
Memory Abstraction
A 1-bit memory element
d IN
store M
d OUT
The
6.004
view
$
The
NEC
View
¥
☺
Remembers input when store goes high.
Like a camera that records input (dIN) when the
user presses the shutter release button.
The recorded value is visible at dOUT .
d IN
store
remembers the 1
d OUT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
Building a memory element …
A First attempt
dIN
*
dOUT
C
storage
node
store
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
Building a memory element …
A
vC d
dIN
*
store = 1
OUT
C
vC d
dIN
*
store = 0
OUT
C
vC
Stored value leaks away
vC = 5 ⋅ e
RL
5V
VOH
t
−t
RL C
V
T = − RL C ln OH
5
from 2
T
store pulse width >> RON C
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
Building a memory element …
B Second attempt
dIN
buffer
dOUT
*
C
RIN
buffer
store
Input resistance RIN
VOH
T = − RIN C ln
5
RIN >> RL
Better, but still not perfect.
Demo
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
Building a memory element …
C Third attempt
buffer + refresh
store
dIN
dOUT
*
store
C
Does this work?
No. External value can
influence storage node.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
Building a memory element …
D Fourth attempt
buffer + decoupled
refresh
store
dIN
dOUT
*
C
store
Works!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
A Memory Array
4-bit memory
IN
store
Address
OUT
Decoder
00
01
A
d IN
S M
d OUT
B
d IN
S M
d OUT
A
C
d IN
S M
d OUT
B
D
d IN
S M
d OUT
C
a0 a1
2
Address
10
11
IN store
D
OUT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
Truth table for decoder
a0
0
0
1
1
a1
0
1
0
1
A
1
0
0
0
B
0
1
0
0
C
0
0
1
0
D
0
0
0
1
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
Agarwal’s top 10 list on memory
10
9
8
7
6
5
I have no recollection, Senator.
I forgot the homework was due today.
Adlibbing ≡ ZSR
I think, therefore I am.
I think that was right.
I forgot the rest …
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 14
6.002
CIRCUITS AND
ELECTRONICS
Second-Order Systems
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Second-Order Systems
5V
5V
Demo
2KΩ
50Ω
2KΩ
S
A
+
–
C
B
large
loop
CGS
Our old friend, the inverter, driving another.
The parasitic inductance of the wire and
the gate-to-source capacitance of the
MOSFET are shown
[Review complex algebra appendix for next class]
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Second-Order Systems
5V
5V
Demo
50Ω
2KΩ
2KΩ
S
C
A
+
–
Relevant circuit:
B
large
loop
2KΩ
CGS
L
5V +
–
B
CGS
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Observed Output
2kΩ
5
vA
t
0
vB
2kΩ
t
0
vC
t
0
Now, let’s try to speed up our inverter by
closing the switch S to lower the effective
resistance
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Observed Output ~50Ω
5
vA
t
0
vB
0
50Ω
t
vC
t
0
Huh!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
First, let’s analyze the LC network
i (t )
L
vI (t )
+
–
C
+
v(t )
–
Node method:
dv
i (t ) = C
dt
Recall
di
vI − v = L
dt
t
dv
1
v
v
dt
C
(
−
)
=
∫ I
L −∞
dt
1
(v I − v )
L
1 t
(vI − v) dt = i
∫
L −∞
d 2v
=C 2
dt
d 2v
LC 2 + v = vI
dt
time2
v, i state variables
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Solving
Recall, the method of homogeneous and
particular solutions:
1
Find the particular solution.
2
Find the homogeneous solution.
L
4 steps
3
The total solution is the sum of the
particular and homogeneous.
Use initial conditions to solve for the
remaining constants.
v = vP (t ) + vH (t )
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Let’s solve
d 2v
LC 2 + v = vI
dt
For input
V0
vI
t
0
And for initial conditions
v(0) = 0 i(0) = 0 [ZSR]
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
1
Particular solution
d 2 vP
LC 2 + vP = V0
dt
is a solution.
vP = V0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
2
Homogeneous solution
Solution to
d 2 vH
LC 2 + vH = 0
dt
Recall, vH :
solution to homogeneous
equation (drive set to zero)
Four-step method:
A Assume solution of the form*
vH = Ae st
, A, s = ?
so,
B
LCAs 2 e st + Ae st = 0
s2 = −
characteristic
equation
1
LC
1
s=±j
LC
C Roots
j = −1
1
ωo =
LC
s = ± jω o
General solution,
D
vH = A1e jωot + A2 e − jωot
Differential equations are commonly
solved by guessing solutions
*
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
3
Total solution
v(t ) = vP (t ) + vH (t )
v( t ) = V0 + A1e jωot + A2 e − jωot
Find unknowns from initial conditions.
v(0) = 0
0 = V0 + A1 + A2
i ( 0) = 0
dv
i (t ) = C
dt
i( t ) = CA1 jωo e jωot − CA2 jωo e − jωot
so,
0 = CA1 jωo − CA2 jωo
or,
A1 = A2
− V0 = 2 A
V0
A1 = −
2
so,
V0 jωot
v( t ) = V0 − (e + e − jωot )
2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
3
Total solution
Remember Euler relation
e jx = cos x + j sin x
(verify using Taylor’s
expansion)
e jx + e − jx
= cos x
2
so,
v( t ) = V0 − V0 cos ωot
where
i( t ) = CV0ωo sin ωot
1
ωo =
LC
The output looks sinusoidal
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
v(t )
Plotting the Total Solution
2V0
V0
0
π
π
3π
2
2
CV0ωo
0
2π
ωo t
i (t )
π
π
2
3π
2
2π
ωo t
− CV0ωo
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Summary of Method
1
Write DE for circuit by applying
node method.
2
Find particular solution vP by guessing
and trial & error.
3
Find homogeneous solution vH
A Assume solution of the form Aest .
B Obtain characteristic equation.
C Solve characteristic equation
for roots si .
D Form vH by summing Ai esit
terms.
4
Total solution is vP + vH ,
solve for remaining constants using
initial conditions.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Example
What if we have:
L
iC
+
C vC
–
vC (0) = V
iC (0) = 0
We can obtain the answer directly from
the homogeneous solution (V0 = 0).
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Example
L
iC
+
C vC
–
vC (0) = V
iC (0) = 0
We can obtain the answer directly from
the homogeneous solution (V0 = 0).
vC ( t ) = A1e jωot + A2 e − jωot
vC (0) = V
V = A1 + A2
iC (0) = 0
0 = CA1 jωo − CA2 jωo
or
V
or A1 = A2 =
2
V jω o t
vC = (e + e − jωot )
2
vC = V cos ωot
iC = −CV ωo sin ωot
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Example
vC
V
2π
ωo t
CVωo iC
2π
ωo t
− CVωo
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
Energy
EC
C:
1
CV 2
2
1
2
CvC
2
2π
ωo t
EL
1
2
L : LiC
2
1
CV 2
2
2π
Notice
ωo t
1
1
1
2
2
CvC + LiC = CV 2
2
2
2
Total energy in the system is a constant,
but it sloshes back and forth between the
Capacitor and the inductor
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
RLC Circuits
R
L
vI (t ) +
–
i (t )
C
+
v(t )
–
v(t )
no R
add R
t
Damped sinusoids with R – remember demo!
See A&L Section 12.2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
15
6.002
CIRCUITS AND
ELECTRONICS
Damped Second-Order
Systems
6.002
Fall 03
1
Damped Second-Order Systems
5V
5V
2KΩ
50Ω
2KΩ
S
C
A
B
+
–
large
loop
CGS
Remember this Demo
Our old friend, the inverter, driving another.
The parasitic inductance of the wire and
the gate-to-source capacitance of the
MOSFET are shown
[Review complex algebra appendix in Agarwal & Lang
for next class]
6.002
Fall 03
2
Damped Second-Order Systems
5V
5V
50Ω
2KΩ
2KΩ
S
C
A
B
+
–
large
loop
Relevant circuit:
5V +
–
6.002
Fall 03
2KΩ
CGS
L
B
CGS
3
Observed Output
2kΩ
5
vA
0
t
vB
2kΩ
t
0
vC
0
t
Now, let’s try to speed up our inverter by
closing the switch S to lower the effective
resistance
6.002
Fall 03
4
Observed Output ~50Ω
5
vA
0
t
vB
50Ω
0
t
vC
0
t
Huh!
6.002
Fall 03
5
In the last lecture, we started by
analyzing the simpler LC circuit to
build intuition
i (t )
L
vI (t )
6.002
+
–
Fall 03
C
+
v(t )
–
6
In the last lecture…
We solved
d 2v 1
1
+
v
=
vI
2
dt
LC
LC
For input
VI
vI
0
t
And for initial conditions
v(0) = 0 i(0) = 0 [ZSR]
6.002
Fall 03
7
In the last lecture…
Total solution
v(t ) = VI − VI cosω t
o
where
1
LC
ωo =
v(t )
2VI
LC
VI
vI
0
t
i (t )
L
v I (t )
6.002
+
–
Fall 03
C
+
v (t )
–
8
Today, we will close the loop on our
observations in the demo by
analyzing the RLC circuit
R
L
vI (t ) +
–
i (t )
C
+
v(t )
–
v(t )
2VI
LC
VI
vI
0
add R
t
Damped sinusoids with R – remember demo!
See A&L Section 13.6
6.002
Fall 03
9
Let’s analyze the RLC network
vA
vI (t )
L
i (t )
+
v(t )
–
R
+
–
C
Node method:
vA :
1 t
vA − v
∫ (vI − v A ) dt =
L −∞
R
vA − v
dv
=C
R
dt
v:
Recall
element rules
L:
vL = L
t
di
dt
1
vL dt = i
∫
L −∞
d 2 v R dv 1
1
+
+
v=
vI
2
dt
L dt LC
LC
6.002
Fall 03
C:
dvC
iC = C
dt
v, i state variables
10
Let’s analyze the RLC network
vA
vI (t ) +
–
L
i (t )
R
C
+
v(t )
–
Node method:
1 t
vA − v
∫ (v I − v A ) dt =
L −∞
R
vA :
vA − v
dv
=C
R
dt
v:
1
d 2v
( vI − v A ) = C 2
L
dt
1
d 2v
( vI − v A ) = 2
dt
LC
dv
v A = RC + v
dt
dv
1
d 2v
( vI − RC − v ) = 2
LC
dt
dt
d 2 v R dv 1
1
+
+
v
=
vI
2
dt
L dt LC
LC
6.002
Fall 03
11
Solving
Recall, the method of homogeneous and
particular solutions:
1
Find the particular solution.
2
Find the homogeneous solution.
L
4 steps
3
The total solution is the sum of the
particular and homogeneous.
Use initial conditions to solve for the
remaining constants.
v = vP (t ) + vH (t )
6.002
Fall 03
12
Let’s solve
d 2 v R dv 1
1
+
+
v
=
vI
2
dt
L dt LC
LC
For input
VI
vI
0
t
And for initial conditions
v(0) = 0 i(0) = 0 [ZSR]
6.002
Fall 03
13
1
Particular solution
d 2 vP R dvP
1
1
+
+
vP =
VI
2
dt
L dt LC
LC
vP = VI
6.002
Fall 03
is a solution.
14
2
Homogeneous solution
Solution to
1
d 2 vH R dvH
+
+
vH = 0
2
dt
LC
dt
L
Recall, vH : solution to homogeneous
equation (drive set to zero)
Four-step method:
A Assume solution of the form
vH = Ae st
, A, s = ?
B
Form the characteristic equation
f(s)
C Find the roots of the characteristic equation
s1 , s2
D General solution
vH = A1e s1t + A2 e s2t
6.002
Fall 03
15
2
Homogeneous solution
1
d 2 vH R dvH
+
+
vH = 0
2
dt
LC
dt
L
Solution to
A Assume solution of the form
vH = Ae st
, A, s = ?
so,
As2est +
B
R
1
Asest +
Aest = 0
L
LC
R
1
s + s+
=0
L
LC
characteristic
equation
s + 2αs + ω o = 0
ωo =
1
LC
α=
R
2L
2
2
C Roots
2
s1 = −α + α 2 − ω 2 o
s2 = −α − α 2 − ω 2 o
D
General solution
vH = A1e
6.002
Fall 03
⎛⎜ −α + α 2 −ω 2 o ⎞⎟ t
⎝
⎠
+ A2 e
⎛⎜ −α − α 2 −ω 2 o ⎞⎟ t
⎝
⎠
16
3
Total solution
v(t ) = vP (t ) + vH (t )
v(t ) = VI + A1e
⎛⎜ −α + α 2 −ω 2 o ⎞⎟ t
⎝
⎠
+ A2 e
⎛⎜ −α − α 2 −ω 2 o ⎞⎟ t
⎝
⎠
Find unknowns from initial conditions.
v(0) = 0 : 0 = VI + A1 + A2
i (0) = 0 :
dv
i (t ) = C
dt
(
)
CA (− α − α − ω )e
= CA1 − α + α 2 − ω 2 o e
2
so,
2
⎛⎜ −α + α 2 −ω 2 o ⎞⎟ t
⎝
⎠
+
⎛⎜ −α − α 2 −ω 2 o ⎞⎟ t
⎝
⎠
2
o
(
) (
0 = A1 − α + α 2 − ω 2 o + A2 − α − α 2 − ω 2 o
)
Mathematically: solve for unknowns,
done.
6.002
Fall 03
17
Let’s stare at this a while longer…
⎛ α 2 −ω 2 o ⎞⎟ t
−αt ⎜⎝
⎠
v(t ) = VI + A1e e
⎛ − α 2 −ω 2 o ⎞⎟ t
−αt ⎜⎝
⎠
+ A2 e e
3 cases:
α > ωo
Overdamped
v(t ) = VI + A1e
α < ωo
−α1t
v(t ) = VI + A1e e
−αt
= VI + A1e e
α = ωo
6.002
+ A2 e
−α 2 t
v
t
Underdamped
⎛ j ω 2 o −α 2 ⎞⎟ t
−αt ⎜⎝
⎠
= VI + K1e
VI vI
−αt
jω d t
⎛⎜ − j ω 2 −α 2 ⎞⎟ t
o
−αt ⎝
⎠
+ A2 e e
−αt − jωd t
+ A2e e
cosωd t + K 2e
−αt
sin ωd t
ωd = ω 2 o − α 2
e jωd t = cosωd t + j sin ωd t
Critically damped
Later…
Fall 03
18
Let’s stare at underdamped a while longer…
α < ωo
Underdamped contd…
v(t ) = VI + K1e−αt cosωd t + K 2e−αt sin ωd t
v(0) = 0 : K1 = −VI
dv
i (0) = 0 : i (t ) = C
dt
= −CK1αe−αt cosωd t − CK 2ωd e−αt sin ωd t
− CK1αe−αt sin ωd t + CK 2ωd e−αt cosωd t
0 = − K1α + K 2ωd
Vα
K2 = − 1
ωd
v(t ) = VI − VI e
−αt
α −αt
cosωd t − VI
e sin ωd t
ωd
Note: For R = 0
⇒α = 0
v(t ) = VI − VI cosωot
Same as LC as expected
6.002
Fall 03
19
Let’s stare at underdamped a while longer…
α < ωo
Underdamped contd…
v(t ) = VI − VI e
−αt
α −αt
cosωd t − VI
e sin ωd t
ωd
Remember, scaled sum of sines (of the same
frequency) are also sines! -- Appendix B.7
ωo −αt ⎛
−1 α ⎞
⎟⎟
v(t ) = VI − VI
e cos⎜⎜ ωd t − tan
ωd
ωd ⎠
⎝
v(t )
2VI
LC
VI
vI
0
6.002
add R
t
Fall 03
20
α < ωo
Underdamped contd…
v(t ) = VI − VI e
−αt
α −αt
cosωd t − VI
e sin ωd t
ωd
Remember, scaled sum of sines (of the same
frequency) are also sines! -- Appendix B.7
ωo −αt ⎛
−1 α ⎞
⎟⎟
v(t ) = VI − VI
e cos⎜⎜ ωd t − tan
ωd
ωd ⎠
⎝
v(t )
2VI
LC
VI
vI
0
add R
t
v
α = ωo
Critically damped
underdamped
criticallydamped
overdamped
t
Section 13.2.3
6.002
Fall 03
21
Remember this? Closed the
loop…
5
vA
0
t
vB
50Ω
0
t
vC
0
t
See example 12.9 on page 664 of the A&L textbook
for inverter-pair analysis
6.002
Fall 03
22
Intuitive Analysis
See Sec. 12.7 of A&L textbook
ωo −αt ⎛
−1 α ⎞
⎟⎟
⎜
v
(
t
)
V
V
e
cos
t
tan
ω
=
−
−
Underdamped
I
I
d
⎜
ωd
ωd ⎠
⎝
v(t )
e −αt
“ringing”
VI
0
2π
t
ωd
Characteristic
equation
s2 +
R
1
s+
=0
L
LC
s 2 + 2αs + ω 2 o = 0
ωd : Oscillation frequency
α : Governs rate of decay
ωd = ω 2 o − α 2
VI : Final value
v(0) : Initial value
Q=
6.002
ωo
: Quality factor (approximately
2α
the number of cycles of ringing)
Fall 03
23
Intuitive Analysis
See Sec. 12.7
of A&L textbook
Ringing stops
after Q cycles V I
v(t )
VI
v(0)
i (0)
is –ve
so v(t)
must
drop
?
0
period
2π
t
ωd
Characteristic
equation
s2 +
R
1
s+
=0
L
LC
s 2 + 2αs + ω 2 o = 0
ωd = ω 2 o − α 2
i(t )
L
vI +–
6.002
Q=
R
C
Fall 03
+
v(t )
–
ωo
2α
given
i (0) -ve
v(0) +ve
24
6.002
CIRCUITS AND
ELECTRONICS
Sinusoidal Steady State
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Review
„
We now understand the why of:
v
5V
R
L
C
„Today, look at response of networks
to sinusoidal drive.
Sinusoids important because signals can be
represented as a sum of sinusoids. Response to
sinusoids of various frequencies -- aka frequency
response -- tells us a lot about the system
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Motivation
For motivation, consider our old friend,
the amplifier:
V
S
+
–
vi
VBIAS
+
–
vO
vC
R
Demo
CGS
Observe vo amplitude as the frequency of the
input vi changes. Notice it decreases with
frequency.
Also observe vo shift as frequency changes
(phase).
Need to study behavior of networks for
sinusoidal drive.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Sinusoidal Response of RC Network
Example:
vI +
–
iC
R
+
vC
–
vI (t ) = Vi cos ω t
=0
for t ≥ 0 (Vi real)
for t < 0
vC (0) = 0
for t = 0
vI
t
0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Our Approach
Example:
vI +
–
iC
R
+
vC
–
Effort
Determine vC(t)
agony
Usual
approach
sneaky approach
very
sneaky
t
Th
is
le
ct
ur
11 e
:0
0
11
:2
0
12
N
ex
:0
0
t
le
ct
ur
e
easy
!
e
m
e
g
l
u
d
In
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Let’s use the usual approach…
1
Set up DE.
2
Find vp.
3
Find vH.
4
vC = vP + vH,
solve for unknowns
using initial conditions
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Usual approach…
1
Set up DE
RC
dvC
+ vC = vI
dt
= Vi cos ω t
That was easy!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
2
Find vp
dvP
RC
+ vP = Vi cos ωt
dt
First try:
vP = A
Æ nope
Second try:
vP = A cos ωt
Æ nope
Third try:
vP = A cos(ωt + φ ) frequency
amplitude phase
− RCAω sin(ωt + φ ) + A cos(ωt + φ ) = Vi cos ωt
− RCAω sin ωt cos φ − RCAω cos ωt sin φ +
A cos ωt cos φ − A sin ωt sin φ
= Vi cos ωt
..
.
gasp !
works, but trig nightmare!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Let’s get sneaky!
Find particular solution to another input…
dvPS
+ vPS = vIS
RC
dt
= Vi e st
st
Try solution vPS = V p e
RC
dV p e st
(S: sneaky :-))
+ V p e st = Vi e st
dt
sRCV p e st + V p e st = Vi e st
Nice
property
of
exponentials
( sRC + 1 )V p = Vi
Vp =
Vi
1 + sRC
Vi
⋅ e st
Thus, vPS =
1 + sRC
is particular solution to Vi e st
ly
Vi
⋅ e jω t
1 + jωRC
easy!
jω t
solution for Vi e
where we replace s = jω
complex amplitude
Vp
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
2
Fourth try to find vP…
using the sneaky approach
Fact 1: Finding the response to
Vi e jω t
was easy.
Fact 2: vI = Vi cos ωt
= real[Vi e jω t ] = real[vIS ]
from Euler relation,
e jω t = cos ωt + j sin ωt
real
part
vI response vP
vIS response vPS
real
part
an inverse superposition argument,
assuming system is real, linear.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
2 Fourth try to find vP…
so,
complex
vP = Re[vPS ] = Re[V p e jωt ]
⎡ Vi
⎤
= Re ⎢
⋅ e jω t ⎥
⎣1 + jωRC
⎦
Vi (1 − jωRC ) jω t ⎤
⎡
= Re
⋅e
⎢⎣ 1 + ω 2 R 2C 2
⎥⎦
Vi
⎡
j φ jω t ⎤
= Re ⎢
⋅ e e ⎥ , tan φ = −ωRC
2 2 2
⎣ 1+ω R C
⎦
Vi
⎡
j ( ωt +φ ) ⎤
= Re ⎢
⋅
e
⎥⎦
⎣ 1 + ω 2 R 2C 2
vP =
Vi
1+ω R C
2
2
2
⋅ cos( ωt + φ )
Recall, vP is particular response to Vi cos ωt .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
3
Find vH
Recall,
vH = Ae
−t
RC
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
4
Find total solution
vC = vP + vH
vC =
Vi
1+ω R C
2
2
2
cos( ωt + φ ) + Ae
−
t
RC
where φ = tan −1 ( −ωRC )
Given vC(0) = 0 for t = 0
so,
Vi
A=−
cos(φ )
2 2 2
1+ ω R C
Done!
Phew !
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Sinusoidal Steady State
We are usually interested only in the
particular solution for sinusoids,
i.e. after transients have died.
Notice when t → ∞, vC → vP as e
vC =
Vi
1+ω R C
2
2
2
−
t
RC
cos( ωt + φ ) + Ae
−
→0
t
RC
0
where φ = tan −1 ( −ωRC )
Vi
A=−
cos(φ )
2 2 2
1+ ω R C
Vp
∠Vp
Described as
SSS: Sinusoidal Steady State
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Sinusoidal Steady State
All information about SSS is contained
in Vp , the complex amplitude!
Steps 3 , 4
were a waste of
time!
Vi
Vp =
1 + jωRC
Recall
Vp
1
=
Vi 1 + jωRC
Vp
Vi
magnitude
=
Vp
phase φ : ∠
Vi
Vp
Vi
1
1 + ω 2 R 2C
=
jφ
e
where
2
φ = tan −1 − ωRC
1
1 + ω 2 R 2C 2
= − tan −1 ωRC
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Sinusoidal Steady State
Visualizing the process of finding the
particular solution vP
Vi cos ωt
drive
D.E.
+
nightmare
trig.
V p cos[ωt + ∠V p ]
particular
solution
algebraic
take
equation
real
+
part
complex
algebra V p e jω t
sneak
in
Vi e jωt
drive
the sneaky path!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Magnitude Plot
transfer function
Vp
H ( jω ) =
Vi
Vp
Vi
Vp
Vi
=
1
1 + ω 2 R 2C 2
1
log
scale
log
scale
1
ω=
RC
ω
From demo: explains vo fall off
for high frequencies!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
Phase Plot
φ = tan −1 − ωRC
φ =∠
Vp
Vi
ω=
0
−
−
1
RC
ω
log scale
π
4
π
2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
16
6.002
CIRCUITS AND
ELECTRONICS
The Impedance Model
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Review
„
Sinusoidal Steady State (SSS)
Reading 13.1, 13.2
vI = Vi cos ωt +
–
C
+
vO
–
„
Focus on steady state, only care
about vP as vH dies away.
„
Focus on sinusoids.
SSS
„
R
Sinusoidal Steady State (SSS)
Reading 13.1, 13.2
Reading: Section 13.3 from course notes.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
vP
Review
Vi cos ωt
1
usual
circuit
model
sneak
in
Vi e jωt
drive
set
up
DE
complex
algebra
V p cos[ωt + ∠V p ]
nightmare
trig.
Vp
2
The Sneaky Path
3
vH
take 4
real total
part
V p e jω t
Vi
1 + jωRC
Vp contains all the information we need:
Vp
∠V p
Amplitude of output cosine
phase
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Review
vO = V p cos(ωt + ∠V p )
Vp
Vi
Vp
=
1
= H ( jω ) transfer
function
1 + jωRC
remember
demo
1
Vi
1
2
1
1
ωRC
1 + ω 2 R 2C 2
Bode plot
∠
Vp
ω=
Vi
ω
1
ω=
RC
break frequency
0
1
RC
ω
⎛ − ωRC ⎞ π
tan −1 ⎜
⎟ −
⎝ 1 ⎠ 4
−
π
2
The Frequency View
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Is there an even simpler way
to get Vp ?
Vi
Vp =
1 + jωRC
Divide numerator and denominator by jωC.
1
V p = Vi
jω C
1
+R
jω C
Hmmm… looks like a voltage divider
relationship.
ZC
V p = Vi
ZC + R
Let’s explore further…
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Impedance Model
Is there an even simpler way to get Vp ?
Consider:
+
vR
–
+
vC
–
iR
i R = I r e jω t
vR = RiR
R
vR = Vr e jω t
Vr e jω t = RI r e jω t
Resistor
jω t
iC
iC = I C e
C
vC = VC e jω t
Vr = RI r
dvC
iC = C
dt
I C e jω t = CVC jωe jω t
1
IC
j ωC
ZC
di
vL = L L
dt
Capacitor
+
vL
–
iL
i L = I l e jω t
L
vL = Vl e jω t
VC =
Vl e jω t = LI l jωe jω t
Inductor
Vl = jωL I l
ZL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Impedance Model
In other words,
capacitor
Ic
+
Vc
–
ZC
Vc = Z C I c
1
ZC =
j ωC
impedance
inductor
resistor
Il
+
Vl
–
+
Vr
–
ZL
Ir
ZR
Vl = Z l I l
Z l = j ωL
Vr = Z r I r
Zr = R
For a drive of the form Vc e jωt ,
complex amplitude Vc is related to the
complex amplitude Ic algebraically,
by a generalization of Ohm’s Law.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Back to RC example…
R
+
C vC
–
vI +
–
Impedance model:
ZR = R
Ic
+
Vc
–
Vi +
–
1
ZC =
jωC
1
ZC
jωC
Vc =
Vi =
Vi
1
ZC + Z R
+R
jωC
Vc =
1
Vi
1 + jωRC
Done!
All our old friends apply!
KVL, KCL, superposition…
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Another example, recall series RLC:
Remember, we want only the steady-state
response to sinusoid
Ir
L
Vi e jω t
Vi +
–
Vi cos ωt
C
R
Vi Z R
Vr =
Z L + ZC + Z R
+
Vr
–
Vr e jω t
Vr cos(ωt + ∠Vr )
Vi R
Vr =
1
j ωL +
+R
jωC
Vr =
Vi jωCR
− ω 2 LC + 1 + jωCR
We will study this and other functions
in more detail in the next lecture.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Big Picture…
V p cos[ωt + ∠V p ]
Vi cos ωt
usual
circuit
model
set
up
DE
nightmare
trig.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Big Picture…
V p cos[ωt + ∠V p ]
Vi cos ωt
usual
circuit
model
Vi e jωt
drive
set
up
DE
nightmare
trig.
complex
algebra
take
real
part
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
The Big Picture…
V p cos[ωt + ∠V p ]
Vi cos ωt
usual
circuit
model
Vi e jωt
drive
set
up
DE
nightmare
trig.
complex
algebra
impedance-based
circuit model
take
real
part
complex
algebra
No D.E.s, no trig!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Back to
Ir
Vr
jωRC
=
Vi 1 + jωRC − ω 2 LC
Vi
L
+
–
C
R
+
Vr
–
Let’s study this transfer function
Vr
jωRC
=
Vi 1 + jωRC − ω 2 LC
(
jωRC
1 − ω 2 LC ) − jωRC
=
⋅
2
(1 − ω LC ) + jωRC (1 − ω 2 LC ) − jωRC
Vr
=
Vi
ωRC
(1 − ω LC ) + (ωRC )
2
2
2
Observe
Low ω : ≈ ωRC
R
High ω : ≈
ωL
ω LC = 1 : ≈ 1
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
Graphically
Vr
=
Vi
ωRC
(1 − ω LC ) + (ωRC )
2
2
2
Low ω : ≈ ωRC
R
High ω : ≈
ωL
ω LC = 1 : ≈ 1
Vr
Vi
“Band Pass”
1
R
ωL
ωRC
1
LC
ω
Remember this trick to sketch the form of
transfer functions quickly.
More next week…
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 17
6.002
CIRCUITS AND
ELECTRONICS
Filters
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Review
R
vI +
–
C
+
vC
–
ZC
+
Vc
–
ZR
Vi +
–
ZC
Vc =
⋅ Vi
ZC + Z R
1
Vc
1
jωC
=
=
1
Vi
+ R 1 + jωRC
j ωC
Reading: Section 14.5, 14.6, 15.3 from A & L.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
A Filter
ZR
Vi +
–
+
Vc
–
ZC
ZC
1
⋅ Vi =
Vc =
ZC + Z R
1 + jωRC
Vc
H (ω ) =
Vi
1
“Low Pass Filter”
ω
Demo
with audio
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Quick Review of ImpedancesJust as
I ab
A
R1
+
Vab
R2
B
I ab
A
R1
–
+
Vab
j ωL
B
Vab
RAB =
= R1 + R2
I ab
Vab
Z AB =
= R1 + jωL
I ab
–
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Quick Review of Impedances
Similarly
A
Z AB = R1 + Z C || R2 + Z L
R1
R2
C
L
= R1 +
Z C R2
+ ZL
Z C + R2
= R1 +
R2
+ jωL
1 + jωCR2
B
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
We can build other filters by
combining impedances
Z (ω )
L
Z
R
C
ω
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
We can build other filters by
combining impedances
Z (ω)
L
Z
R
ω
C
H (ω )
HPF
High Pass Filter
+
–
ω
H (ω )
LPF
Low Pass Filter
ω
+
–
H (ω )
HPF
+
–
ω
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Check out:
C
L
+
R Vr
–
Vi +
–
Intuitively:
Vr
1
Vi
C
k
bloc
eq
r
f
w
s lo
L bloc
ωo =
1
ω RC
(1 − ω LC ) + (ω RC )
2
freq
ω
LC
R
Vr
=
1
Vi
jω L +
+R
jω C
j ω RC
=
1 − ω 2 LC + j ω RC
Vr
=
Vi
ks hig
h
2
2
At resonance,
ω = ωo
and
ZL + ZC = 0,
so Vi sees
only R!
More later…
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
What about:
Vlc
+
Vi +
–
Vlc
Vi
1
L
–
C
R
Band Stop Filter
C open
L open
ω
Check out Vl and Vc in the lab.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Another example:
R
+
L
Vi +
–
C
Vo
–
Vo
Vi
ort
h
s
L
BPF
Cs
ωo
ho
rt
ω
Application: see AM radio coming up shortly
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
AM Radio Receiver
antenna
R
Vi +
–
L
C
demodulator
amplifier
Thévenin
antenna
model
crystal radio demo
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
AM Receiver
R
Vi +
–
L
C
demodulator
amplifier
filter
signal
strength
10 KHz
WBZ
News
Radio
f
540 …1000 1010 1020 1030 … 1600 KHz
“Selectivity” important —
relates to a parameter “Q” for the filter. Next…
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Selectivity:
Look at series RLC in more detail
C
L
Vi +
–
Recall,
Vr
Vi
+
Vr
–
R
Vr
R
=
Vi R + jω L + 1
jω C
1
higher Q
1
2
Δω
bandwidth
ω
ωo
ωo
Define Q =
Δω
quality factor
high Q ⇒ more selective
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Quality Factor Q
Q=
ωo
Δω
ωο:
R
Vr
=
Vi R + jω L +
1
=
1
L
1 ⎞
⎛
1 + j⎜ ω −
⎟
jω C
ω
R
CR
⎠
⎝
at ωο =0
1
ωo =
LC
Δω ?
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Quality Factor Q
ωo
Q=
Δω
Δω :
Note that abs magnitude is
when
Vr
=
Vi
i.e. when
1
2
1
1
=
⎛ L
1 ⎞ 1 ± j1
1 + j⎜ ω −
⎟
⎝ R ω CR ⎠
1
ωL
−
= ±1
R ω CR
ω2 ∓
ωR
L
−
1
=0
LC
Looking at the roots of both equations,
R 1 R2
4
ω1 =
+
+
2 L 2 L2 LC
R 1
ω2 = −
+
2L 2
R2
4
+
L2 LC
R
Δω = ω1 − ω2 =
L
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Quality Factor Q
ωo
Q=
Δω
Q=
ωo
R
L
=
ωo L
1
ωo =
LC
R
The lower the R (for series R),
the sharper the peak
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
Quality Factor Q
Another way of looking at Q :
energy stored
Q = 2π
energy lost per cycle
1
2
L Ir
= 2π 2
1 2 2π
Ir R
ω0
2
ωo L
Q=
R
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
18
6.002
CIRCUITS AND
ELECTRONICS
The Operational Amplifier
Abstraction
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Review
„ MOSFET amplifier — 3 ports
+
+
vO output
port
–
+
input
vI
port
–
VS
power
port
–
„ Amplifier abstraction
VS
+
+
vI
–
–
+
v
– O
vI
vO
Function of vI
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Review
vI
vO
Function of vI
„ Can use as an abstract building block for
more complex circuits (of course, need
to be careful about input and output).
„ Today
Introduce a more powerful amplifier
abstraction and use it to build more
complex circuits.
Reading: Chapter 15 from A & L.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Operational Amplifier
Op Amp
VS
input
port
power
port
+
+
–
output
port
–
+
–
−VS
More abstract representation:
+
vIN
–
+
–
vOUT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Circuit model (ideal):
vO
+
i=0
v+
+
v
–
v–
–
i=0
+
–
Av
A→∞
i.e.  ∞ input resistance
 0 output resistance
 “A” virtually ∞
 No saturation
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Using it…
12V
+
–
VS = 12V
vO
+
vIN
–
RL
− VS = −12V
–
12V +
Demo
12V
− 10 μV
vO active region
saturation
10μV
− 12V
vIN
A ~ 106
but unreliable,
temp. dependent
(Note: possible confusion with MOSFET saturation!)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Let us build a circuit…
Circuit: noninverting amplifier
v+
v−
vIN +
–
+
vO
–
R1
R2
Equivalent circuit model
+
i=0
vIN +
–
op amp
v+
v
−
vO
+ A(v + − v − )
–
R1
–
i=0
R2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Let us analyze the circuit:
Find vO in terms of vIN, etc.
vO = A(v + − v − )
R2 ⎞
⎛
= A⎜ vIN − vO
⎟
R1 + R2 ⎠
⎝
⎛
AR2 ⎞
vO ⎜ 1 +
⎟ = AvIN
⎝ R1 + R2 ⎠
AvIN
vO =
AR2
1+
R1 + R2
What happens when “A” is very large?
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Let’s see… When A is large
AvIN
AvIN
≈
vO =
AR2
AR2
1+
R1 + R2 R1 + R2
≈ vIN
Suppose
(R1 + R2 )
A = 10 6
R1 = 9 R
R2 = R
R2
gain
10 6 ⋅ vIN
vO =
10 6 R
1+
9R + R
10 6 ⋅ vIN
=
1
6
1 + 10 ⋅
10
vO ≈ vIN ⋅ 10
Demo
Gain:
„ determined by resistor ratio
„ insensitive to A, temperature, fab variations
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Why did this happen?
Insight:
5V
v+
vIN +
–
5V
v−
10V
+
A
–
6V
6V
negative
feedback
–
i =0
12V
vO = 2vIN
R
vO
2
R
e.g. vIN = 5V
Suppose I perturb the circuit…
(e.g., force vO momentarily to 12V somehow).
Stable point is when v+ ≈ v- .
Key: negative feedback Æ portion of
output fed to –ve input.
e.g. Car antilock brakes
Æ small corrections.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Question: How to control a
high-strung device?
Antilock brakes
is it
turning?
no
di
s
yes
release apply
Michelin
it’s
all about
control
c
yes/no
k
c
a
db
e
e
f
v. v. powerful brakes
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
More op amp insights:
Observe, under negative feedback,
⎛ R1 + R2 ⎞
⎟vIN
⎜
R1 ⎠
v
→0
v+ − v− = O = ⎝
A
A
v+ ≈ v−
We also know
i+ ≈ 0
i -≈ 0
Æyields an easier analysis method
(under negative feedback).
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Insightful analysis method
under negative feedback
v+ ≈ v−
i+ ≈ 0
i− ≈ 0
g vO = vIN
a vIN
+
vIN +
–
b vIN
R1 + R2
R2
vO
–
c vIN
R1 f
e i=0
vIN
d
R2
vIN
R2
R2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Question:
a vIN v +
+
vIN
+
–
b vIN
v
−
–
c vIN
vO ?
vO ≈ vIN
or
R1 + R2
vO = vIN
R2
with R1 = 0
R2 = ∞
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
Why is this circuit useful?
+
vIN +
–
vO
–
vO ≈ vIN
Buffer
voltage gain = 1
input impedance = ∞
output impedance = 0
current gain = ∞
power gain = ∞
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 19
6.002
CIRCUITS AND
ELECTRONICS
Operational Amplifier Circuits
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 20
Review
„ Operational amplifier abstraction
+
 ∞ input resistance
–
 Gain “A” very large
 0 output resistance
„ Building block for analog systems
„ We will see these examples:
Digital-to-analog converters
Filters
Clock generators
Amplifiers
Adders
Integrators & Differentiators
Reading: Chapter 15.5 & 15.6 of A & L.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 20
Consider this circuit:
R2
i
i
R1
v2 +
–
v1 +
–
R2
+
v = v1
R1 + R2
≈ v−
v2 − v −
i=
R1
R1
v− –
v+ +
R2
+
vOUT
–
vOUT = v − − iR2
v2 − v −
=v −
⋅ R2
R1
−
R2
⎡ R2 ⎤
= v ⎢1 + ⎥ − v2
R1
⎣ R1 ⎦
−
R2
R1 + R2
R2
= v1
⋅
− v2
R1 + R2
R1
R1
=
R2
(v1 − v2 )
R1
subtracts!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 20
Another way of solving —
use superposition
v1 → 0
v2 → 0
R1
R2
R1
v2 +
–
R2
v1 +
–
–
v+ +
–
vOUT2
+
R2
R1
R1 || R2
R2
vOUT2 = − v2
R1
vOUT1
R1 + R2
vOUT1 = v ⋅
R1
+
v1 ⋅ R2 R1 + R2
=
⋅
R1 + R2
R1
= v1
vOUT = vOUT1 + vOUT2
R2
= (v1 − v2 )
R1
R2
R1
Still subtracts!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 20
Let’s build an intergrator…
vI +
–
+
vO
–
∫ dt
Let’s start with the following insight:
i
+
i +
–
C
vO
–
t
1
vO = ∫ i dt
C −∞
vO is related to ∫ i dt
But we need to somehow convert
voltage vI to current.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 20
First try… use resistor
+ vR –
vI +
–
i
+
R
C
vO
vI
→i
R
–
But, vO must be very small compared
to vR, or else
vI
i≠
R
When is vO small compared to vR ?
dv
larger the RC,
RC O + vO = vI
dt
smaller the vO
vR
dvO
when RC
>> vO
for good
dt
integrator
dvO
≈ vI
RC
ωRC >> 1
dt
t
1
or
vO ≈
vI dt
∫
Demo
RC −∞
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 20
There’s a better way…
i
Notice
i
–
+
v − ≈ 0V under negative feedback
vI
i=
so,
R
–
R
vI +
–
+
vC
vI
+ –
R
–
vO = −vC
R
+
vI –
+
t
+
vO
1 vI
– vO = − ∫ dt
C −∞ R
We have our integrator.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 20
Now, let’s build a differentiator…
+
vO
–
d
dt
vI +
–
Let’s start with the following insights:
i
vI
+
–
C
dvI
i=C
dt
dvI
i is related to
dt
But we need to somehow convert current
to voltage.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 20
Differentiator…
Recall
i
i
–
+
R
i
–
+
vO
–
+
0V
i
C
vI +
–
+ vC –
Demo
R
–
+
vO = −iR
current
to
voltage
vO
vI = vC
dvI
i=C
dt
vO = − RC
dvI
dt
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 20
6.002
CIRCUITS AND
ELECTRONICS
Op Amps Positive Feedback
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Negative vs Positive Feedback
Consider this circuit — negative feedback
vIN
R1
R2
–
R1
vIN +
+
R2
–
+
vOUT = − vIN
–
R1
is
s
ly age
a
an t p
e x
se ne
on
+
R2
vOUT = − vIN ”
“
–
R1
and this — positive feedback
R2
vIN +
–
R1
+
–
What’s the difference?
Consider what happens when there is a pertubation…
Positive feedback drives op amp into saturation:
vOUT → ±VS
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Static Analysis of Positive Feedback Ckt
R2
v IN
v IN +
–
+
–
R1
+
–
vOUT
v + R2
R1
v
−
vOUT
+
–
A(v + − v − )
vOUT = A(v + − v − )
= Av +
⎡ v − vIN
⎤
= A⎢ OUT
⋅ R1 + vIN ⎥
⎣ R1 + R2
⎦
=
AR1vIN
AR1
vOUT −
+ AvIN
R1 + R2
R1 + R2
⎡
⎡
AR1 ⎤
R1 ⎤
=
−
vOUT ⎢1 −
v
A
1
⎥ IN ⎢ R + R ⎥
⎣
⎣ R1 + R2 ⎦
1
2⎦
⎡1 − R1 ⎤
⎢ R +R ⎥
R
1
2
vOUT = ⎢
⎥ ⋅ AvIN = − 2 vIN
R1
⎢ − AR1 ⎥
⎢⎣ R1 + R2 ⎥⎦
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Representing dynamics of op amp…
v+
+
–
v−
+
v*
R
C
(v + − v − )
vo
+
–
Av*
–
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Representing dynamics of op amp…
Consider this circuit and let’s analyze its
dynamics to build insight.
R2
R1
+
vo
–
R4
R3
Circuit model
R1
v+
v−
R3
vo
A
R2
+
–
R
C
(v + − v − )
+
v*
+
–
–
+
vo
–
R4
Let’s develop equation representing time
behavior of vo .
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Dynamics of op amp…
vo = Av
*
vo
or v =
A
*
dv* *
RC
+ v = v+ − v_
dt
vo R1
+
= γ vo
R1 + R2
vo R3
−
γ vo
v =
=−
R3 + R4
v+ =
RC dvo vo
+ = v+ − v_
A dt A
+
= ( γ − −γ ) vo
neglect
or
dvo ⎡ 1
A − + ⎤
+
+
( γ − γ ) vo = 0
⎢
⎥⎦
dt ⎣ RC RC
dvo
A − +
+
( γ − γ ) vo = 0
dt RC
time −1
or
dvo vo
RC
+ = 0 where T = − +
dt T
A( γ − γ )
vo ( 0 ) = 0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Consider a small disturbance to vo
(noise).
+
−
if γ > γ
if
if
+
γ > γ−
+ −
γ = γ
vo
K
T is positive
vo = K e
−
t
T
stable
T is negative
vo = K e
t
T
unstable
T is very large
vo = K
neutral
unstable
neutral
stable
t
disturbance
Now, let’s build some useful circuits with
positive feedback.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
One use for instability: Build on the
basic op amp as a comparator
+ VS
v+
v
−
+
vo
–
− VS
+ VS
vo
v+ − v−
0
− VS
vo
−
v →0
v
+
t
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Now, use positive feedback
vi
–
vo
+
R2
vo R1
v =
R1 + R2
R1
+
+
v = 7.5
vo = 15
vi
( vi = v − ) > 7.5
e.g. R1 = R2
VS = 15
v− < v+
v − < −7.5
v − > 7.5
vo = −15
v − = −7.5
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Now, use positive feedback
vi
–
vo
+
R2
vo R1
v =
R1 + R2
R1
+
VS R1
v =
R1 + R2
+
vo = +VS
15
vi
( vi = v − ) > v +
e.g. R1 = R2
VS = 15
v− < v+
v − < −7.5
v − > 7.5
vo = −VS − 15 v − =
− VS R1
R1 + R2
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
vo
15
VS
hysteresis
− 7 .5
Demo
0
vi
7 .5
− 15
− VS
Why is hysteresis useful?
vi
v
o
e.g., analog
to digital
7.5
t
− 7.5
Demo
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Without hysteresis
vi
7.5
analog
to digital
vo
vi
t
− 7 .5
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Oscillator — can create a clock
R
vC
–
C
vo
+
R1
vo
2
R1
vo
VS
VS
2
v+
v−
vC
VS
2
− VS
−
Demo
v
t
−
v+
Assume
vo = VS
vC = 0
at t = 0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
Clocks in Digital Systems
„ We built an oscillator using an op amp.
t
can use as a clock
„ Why do we use a clock in a digital system?
(See page 735 of A & L)
1
1
0
sender
receiver
clock
a
1,1,0?
b
When is the signal valid?
common timebase -- when to “look” at a signal
(e.g. whenever the clock is high)
Æ Discretization of time
one bit of information associated with
an interval of time (cycle)
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
21
6.002
CIRCUITS AND
ELECTRONICS
Energy and Power
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Why worry about energy?
-
small batteries
Æ good
Today:
„ How long will the battery last?
in standby mode
in active use
„ Will the chip overheat and self-destruct?
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Look at energy dissipation in
MOSFET gates
VS
R
+
+
vI
–
C
vO
–
C: wiring capacitance and
CGS of following gate
Let us determine
standby power
active use power
Let’s work out a few related examples first.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Example 1:
I
V +
–
Power
R
+
V
–
V2
P = VI =
R
Energy dissipated in time T
E = VIT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Example 1:
for our gate
VS
VS
RL
RL
vO
vI high
vO
vI low
RON
RON
2
VS
P=
RL + RON
P=0
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Example 2:
Consider
R1
S1
S2
VS +
–
R2
C
T
T1
T2
S1 closed S1 open
S 2 open
S 2 closed
t
Find energy dissipated in each cycle.
Find average power P.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
T1 : S1 closed, S2 open
i
VS +
–
assume
vC = 0 at t = 0
R1
+
vC
–
C
vC
i
VS
R1
VS
t
VS
e
R1
−t
R1C
t
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Total energy provided by source during T1
T1
E = ∫ VS i dt
0
T1
−t
2
VS R1C
=∫
e dt
R1
0
2
VS
=−
R1C e
R1
−t T1
R1C
0
−T1
⎛
⎞
2
R
C
1
= C VS ⎜ 1 − e ⎟
⎜
⎟
⎝
⎠
≈ C VS if T1 >> R1C
2
I.e., if we wait long enough
1
2
C VS stored on C ,
2
1
2
E1 = C VS dissipated in R1
2
Independent
of R!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
T2 : S2 closed, S1 open
+
vC
–
C
R2
Initially, vC = VS
(recall T1 >> R1C)
So, initially,
1
2
energy stored in capacitor = CVS
2
Assume T2 >> R2C
So, capacitor discharges ~fully in T2
So, energy dissipated in R2 during T2
1
2
E2 = CVS
2
E1, E2 independent of R2 !
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Putting the two together:
Energy dissipated in each cycle
E = E1 + E2
1
1
2
2
= CVS + CVS
2
2
E = CVS
2
energy dissipated in
charging & discharging C
Assumes C charges and discharges fully.
Average power
E
P=
T
CVS
=
T
2
= CVS f
2
1
frequency f =
T
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Back to our inverter —
VS
RL
vO
vIN
RON
C
What is P for the following input?
vIN
T
2
T
2
T
t
1
T=
f
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Equivalent Circuit
RL
VS +
–
C
RON
What is P for the following input?
vIN
T
2
T
2
T
t
1
T=
f
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
What is P for gate?
We can show (see section 12.2 of A & L)
2
P=
2
VS
RL
2
+ CVS f
2( RL + RON )
(RL + RON )2
when RL >> RON
2
VS
2
P=
+ CVS f
2 RL
r
e
b
m
e
m
re
P STATIC
independent of f.
MOSFET ON half
the time.
e
b
m
e
rem
r
P DYNAMIC
related to switching
capacitor
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
What is P for gate?
when RL >> RON
2
VS
2
P=
+ CVS f
2 RL
In standby mode,
half the gates in a
chip can be
assumed to be on.
So P STATIC per
gate is still VS2 .
In standby mode,
fÆ0,
so dynamic power
is 0
2RL
Relates to standby
power.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
Some numbers…
a chip with 106 gates clocking
C =1f F
at 100 MHZ
RL = 10 kΩ
f = 100 × 10 6
VS = 5 V
25
6⎡
−15
6⎤
10
25
100
10
+
×
×
×
P = 10 ⎢
4
⎥⎦
⎣ 2 × 10
= 10 6 [1.25 milliwatts + 2.5 microwatts ]
problem !
1.25KW!
must get rid of this
2.5W
not bad
α VS 2
α f
reduce VS
next
lecture
5 V → 1V
2.5 W → 150 mW
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
22
6.002
CIRCUITS AND
ELECTRONICS
Energy, CMOS
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
Review
„
VS
RL
2
VS
P=
RL + RON
vO
vI
RON
„
T1: closed
T2: open R
open
closed
S1
S2
1
VS +
–
C
R2
1
T = T1 + T2 =
f
2
P = CVS f
Reading: Section 11.5 of A & L.
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
Review
VS
Inverter —
RL
vO
vI
RON
C
1
Square wave input
T=
f
2
V
2
P = S + CVS f
2 RL
Demo
P STATIC
independent of f.
MOSFET ON half
the time.
P DYNAMIC
RL >> RON
T
>>" RC"
2
time constant
related to switching
capacitor.
In standby mode,
fÆ0,
so dynamic power is 0
In standby mode, half
the gates in a chip can
be assumed to be on.
So P STATIC per gate is
still VS2 .
2RL
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
Review
2
P=
VS
2
+ CVS f
2 RL
Chip with 106 gates clocking at 100 MHz
C = 1 f F, RL = 10 KΩ , f = 100 × 10 6 , VS = 5 V
2
⎡
5
6
−15
2
6⎤
P = 10 ⎢
+ 10 × 5 × 100 × 10 ⎥
3
gates ⎣ 2 × 10 × 10
⎦
= 10 6 [1.25 milliwatts + 2.5 μ watts ]
1.25KWatts
problem !
• independent of f
• also standby power
(assume ½ MOSFETs
ON if f Æ 0)
• must get rid of this!
+
2.5Watts
not bad
• αf
• αVS2
reduce VS
5VÆ1V
2.5VÆ150mW
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
How to get rid of static power
Intuition:
VS
i
VS
RL
RL
vO high
vI high
vO low
vI low
RON
MOSFET
off
idea !
VS
vI high
vO low
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
New Device PFET
• N-channel MOSFET (NFET)
D
on when vGS ≥ VTN
off when vGS < VTN
e.g. VTN = 1V
G
S
• P-channel MOSFET (PFET)
S
G
on when vGS ≤ VTP
off when vGS > VTP
e.g. VTP = -1V
D
5V
ON when
less than 4V
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
Consider this circuit:
VS
vI
G
+
–
G
S
D
D
S
PU = pull up
vO
PD = pull down
works like an inverter!
IN
OUT
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
Consider this circuit:
works like an inverter!
OUT
IN
vI = 5V (input high)
vI = 0V (input low)
VS = 5V
VS = 5V
RON p
+
vI = 5V
–
vO
RON n
= 0V
+
vI = 0V
–
vO
= 5V
Complementary
MOS
(our previous logic was called “NMOS”)
Called “CMOS logic”
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
Key: no path from VS to GND!
no static power!
Let’s compute P DYNAMIC
VS
vI
T
vI
vO
closed for
vI low
closed for
vI high
RON p
VS +
–
From
t
1
f =
T
C
RON n
C
2
P = CVS f
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
For our previous example —
C = 1 f F, VS = 5 V , f = 100 MHz , 1
2
P = CV S f
= 10 − 15 × 5 2 × 100 × 10 6
= 2 . 5 μwatts per gate
P = 2 . 5 μwatts for 10 6 gate chip
Gates
f
P
106
100 ~2.5
MHz watts
Pentium?
2x106
300
~15
MHz watts
PII?
2x106
600
~30
MHz watts
PII?
8x106
~240
1.2 GHz watts
PIII?
25x106
~1875
3 GHz watts
PIV?
“keep
all
else
same”
!
p
s
ga
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
How to reduce power
A VS
5V Æ 3V Æ 1.8V Æ 1.5V
~PIV Æ 170 watts Æ better, but high
and use big heatsink
B Turn off clock when not in use.
C Change VS depending on need.
Æ Æ next time:
power supply
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
CMOS Logic
NAND:
VS
A
A B
0 0
0 1
1 0
1 1
B
Z
1
1
1
0
Z
A
B
5V
0V
G
S
on
D
5V
5V
G
S
off
D
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
In general, if we want to implement F
e.g. F = A ⋅ B = A + B
VS
short when
A = 0 or B = 0,
open otherwise
short
when F
is true,
else open
A
B
Z
short
when F
is true,
else open
short when
A · B is true,
else open
r
e
b
m
reme gan’s law
r
o
M
e
D
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture
23
6.002
CIRCUITS AND
ELECTRONICS
Violating the Abstraction Barrier
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
Case 1: The Double Take
Problem
R
VO
“0” Æ “1” Vi
expected
observed
VO “1”
VO “1”
huh?
“0”
t
t
“0”
in forbidden region!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
(a) DC case
R
VO
V1
Vi
Vi = 5V DC
VO = 5V DC V1 = 5V DC
very high
impedance,
like open
circuit
OK
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
(b)
Step
R
VO
V1
Vi
very high
impedance,
like open
circuit
5V Vi
b.1
0V
b.3
5V
t
t=0
VO
not ok!
VO = 2.5V
t=0
b.2
2T
t
5V V1
looks ok!
t=0 T
t
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
2.5
....
R
5
R→
Vi
characteristic
impedance
instantaneous R divider
finite propagation speed
of signals
5V
5V
5V
0
0
2T
0 T
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
Question: So why did our circuits work?
5V V1
rce
u
o
S
“
on”
i
t
a
n
i
Term
1. Look only at V1
0
O
M
E
D
2. Keep wires short
O
e
DEM mall wir
s
us e
3. Termination
O
DEM at the
R
add
e nd
0
t
T
5V VO
0
0
le l
l
a
r
a
P
tion
a
n
i
term
5V VO
2.5V
t
0
t
More in 6.014
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
Case 2: The Double Dip
Problem Æ strange spikes on supply
0
V
1
1
0
OK
driving a 50 Ω
resistor!
0
V
driving a 50 Ω
resistor!
input
Why?
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
Drop across inductor
Ldi
dt
VS
V
Inverter current
v inductor
VS
solution
1. short wires
2. low inductance wires
3. avoid big current swings
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
Case 3: The Double Team, or,
Slower may be faster!
Problem
a given chip
worked,
but was slow.
ideal
C
actual
Let’s try speeding it up by using stronger
drivers
ideal
ω
L
actual
Disaster!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
Why?
DEMO
Consider
ok
C
DEMO
R1
R0
R2
dV
α
dt
dV
C
dt
crosstalk!
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
How does this relate to chip?
Solution
DEMO
small dV
dt
Load output!
— put cap on outputs of chip
— jitter edges
— slew edges
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
Case 4: The Double Jump
Careful abstraction violation for the
better…
Recall
Vo
expect
Vi
Vo
Vi
but, observe
Vo
Vi
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25
Case 4: The Double Jump
Careful abstraction violation for the
better…
5V
Vi
5V + 3V
5V
0V
3V
So, pullup has
stronger drive
as output rises
Cite as: Anant Agarwal and Jeffrey Lang, course materials for 6.002 Circuits and Electronics, Spring 2007. MIT
OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6.002 Fall 2000
Lecture 25