Operational Amplifiers Supplemental lecture Rick Matthews
Operational Amplifiers
Jan 06, 2016
Operational Amplifiers. Supplemental lecture Rick Matthews. The inverting amplifier. R2 provides negative feedback. The inverting amplifier. R2 provides negative feedback. This means V- is adjusted to V+. The inverting amplifier. R2 provides negative feedback. - PowerPoint PPT Presentation
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Operational Amplifiers
Supplemental lecture
Rick Matthews
The inverting amplifier
• R2 provides negative feedback.
The inverting amplifier
• R2 provides negative feedback.
• This means V- is adjusted to V+.
The inverting amplifier
• R2 provides negative feedback.
• This means V- is adjusted to V+.
• V+ is zero, so V- must be zero, too.
The inverting amplifier
• R2 provides negative feedback.
• This means V- is adjusted to V+.
• V+ is zero, so V- is zero.
1
22
1
/in
out in
I V R
RV IR V
R
I
The inverting amplifier
• R2 provides negative feedback.
• This means V- is adjusted to V+.
• V+ is zero, so V- is zero.
1
22
1
/in
out in
I V R
RV IR V
R
I
The inverting amplifier
• R2 provides negative feedback.
• This means V- is adjusted to V+.
• V+ is zero, so V- is zero.
1
22
1
/in
out in
I V R
RV IR V
R
I
More generally,…
More generally,…
• Whatever sits in the place of R1 serves to create a current I that is a function of Vin.
I=f(Vin)
More generally,…
• Whatever sits in the place of R1 serves to create a current I that is a function of Vin.
• And whatever sits in place of R2 serves to create a voltage Vout that is a second function of I.
I=f(Vin)
Vout= -g(I)
More generally,…
• Whatever sits in the place of R1 serves to create a current I that is a function of Vin.
• And whatever sits in place of R2 serves to create a voltage Vout that is a second function of I.
I=f(Vin)
( )out inV g f V
Vout= -g(I)
Example
2
11
2 2
2
1
/ , so ( ) .
, so ( ) .
( ) .
inin in
R
out in in
VI V R f V
R
V IR g I IR
RV g f V V
R
Example
2
11
2 2
2
1
/ , so ( ) .
, so ( ) .
( ) .
inin in
R
out in in
VI V R f V
R
V IR g I IR
RV g f V V
R
Example
2
11
2 2
2
1
/ , so ( ) .
, so ( ) .
( ) .
inin in
R
out in in
VI V R f V
R
V IR g I IR
RV g f V V
R
Example: Exponentiating amp
exp 1 ,
so ( ) exp 1 .
, so ( ) .
( ) exp 1
exp .
o
in o
R
inout in o
ino
eVI I
kT
eVf V I
kT
V IR g I IR
eVV g f V I R
kT
eVI R
kT
Example: Exponentiating amp
exp 1 ,
so ( ) exp 1 .
, so ( ) .
( ) exp 1
exp .
o
in o
R
inout in o
ino
eVI I
kT
eVf V I
kT
V IR g I IR
eVV g f V I R
kT
eVI R
kT
Example: Exponentiating amp
exp 1 ,
so ( ) exp 1 .
, so ( ) .
( ) exp 1
exp .
o
in o
R
inout in o
ino
eVI I
kT
eVf V I
kT
V IR g I IR
eVV g f V I R
kT
eVI R
kT
Example: Exponentiating amp
2
exp 1 ,
so ( ) exp 1 .
, so ( ) .
( ) exp 1
exp .
o
in o
R
inout in o
ino
eVI I
kT
eVf V I
kT
V IR g I IR
eVV g f V I R
kT
eVI R
kT
Example: Exponentiating amp
exp 1 ,
so ( ) exp 1 .
, so ( ) .
( ) exp 1
exp .
o
in o
R
inout in o
ino
eVI I
kT
eVf V I
kT
V IR g I IR
eVV g f V I R
kT
eVI R
kT
Example: log amp
Box 1 is the resistor.
, so ( ) .
Box 2 is the diode.
exp 1
log 1
log .
( ) log .
Therefore,
( ) log .
in inin
o
o
o
o
inout in
o
V VI f V
R R
eVI I
kT
kT IV
e I
kT I
e I
kT Ig I
e I
VkTV g f V
e I R
Example: log amp
Box 1 is the resistor.
, so ( ) .
Box 2 is the diode.
exp 1
log 1
log .
( ) log .
Therefore,
( ) log .
in inin
o
o
o
o
inout in
o
V VI f V
R R
eVI I
kT
kT IV
e I
kT I
e I
kT Ig I
e I
VkTV g f V
e I R
Example: log amp
Box 1 is the resistor.
, so ( ) .
Box 2 is the diode.
exp 1
log 1
log .
( ) log .
Therefore,
( ) log .
in inin
o
o
o
o
inout in
o
V VI f V
R R
eVI I
kT
kT IV
e I
kT I
e I
kT Ig I
e I
VkTV g f V
e I R
Example: log amp
Box 1 is the resistor.
, so ( ) .
Box 2 is the diode.
exp 1
log 1
log .
( ) log .
Therefore,
( ) log .
in inin
o
o
o
o
inout in
o
V VI f V
R R
eVI I
kT
kT IV
e I
kT I
e I
kT Ig I
e I
VkTV g f V
e I R
Example: log amp
Box 1 is the resistor.
, so ( ) .
Box 2 is the diode.
exp 1
log 1
log .
( ) log .
Therefore,
( ) log .
in inin
o
o
o
o
inout in
o
V VI f V
R R
eVI I
kT
kT IV
e I
kT I
e I
kT Ig I
e I
VkTV g f V
e I R
Example: log amp
Box 1 is the resistor.
, so ( ) .
Box 2 is the diode.
exp 1
log 1
log .
( ) log .
Therefore,
( ) log .
in inin
o
o
o
o
inout in
o
V VI f V
R R
eVI I
kT
kT IV
e I
kT I
e I
kT Ig I
e I
VkTV g f V
e I R
Example: log amp
Box 1 is the resistor.
, so ( ) .
Box 2 is the diode.
exp 1
log 1
log .
( ) log .
Therefore,
( ) log .
in inin
o
o
o
o
inout in
o
V VI f V
R R
eVI I
kT
kT IV
e I
kT I
e I
kT Ig I
e I
VkTV g f V
e I R
A Multiplier
Recall
log( ) log( ) log( )ab a b
Log Amp
Log Amp
SummingAmp
ExponentialAmp
Vin1
Vin2
Vout
log(a)
log(b)
log(a)+log(b) =log(ab)
ab
A Divider
Recall
log( / ) log( ) log( )a b a b
Log Amp
Log Amp
DifferentialAmp
ExponentialAmp
Vin1
Vin2
Vout
log(a)
log(b)
log(a)-log(b) =log(a/b)
a/b
Calculus
R11k
VoutVinC11uF
Differentiator
Calculus
R11k
VoutVinC11uF
VoutVin
C11uF
R11k
Differentiator Integrator
Etc.
• Can you think of a circuit to take cube roots?
• We can fashion sophisticated analog computers this way.
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