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
–
12V
RL
− VS = −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 +
–
R1 + R2
R2
vO
b vIN –
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