Transcript Operational Amplifiers

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.
I  Vin / R1
Vout
R2
  IR2   Vin
R1
I
The inverting amplifier
• R2 provides negative
feedback.
• This means V- is
adjusted to V+.
• V+ is zero, so V- is
zero.
I  Vin / R1
Vout
R2
  IR2   Vin
R1
I
The inverting amplifier
• R2 provides negative
feedback.
• This means V- is
adjusted to V+.
• V+ is zero, so V- is
zero.
I  Vin / R1
Vout
R2
  IR2   Vin
R1
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.
Vout= -g(I)
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.
Vout   g  f (Vin ) 
Vout= -g(I)
I=f(Vin)
Example
Vin
I  Vin / R1 , so f (Vin )  .
R1
VR2  IR2 , so g ( I )  IR2 .
Vout
R2
  g  f (Vin )    Vin .
R1
Example
Vin
I  Vin / R1 , so f (Vin )  .
R1
VR2  IR2 , so g ( I )  IR2 .
Vout
R2
  g  f (Vin )    Vin .
R1
Example
Vin
I  Vin / R1 , so f (Vin )  .
R1
VR2  IR2 , so g ( I )  IR2 .
Vout
R2
  g  f (Vin )    Vin .
R1
Example: Exponentiating amp

 eV
I  I o exp 
 kT

 
  1 ,
 

 eV
so f (Vin )  I o exp 
 kT

VR  IR, so g ( I )  IR.
Vout
 
  1 .
 

 eVin  
  g  f (Vin )    I o R exp 
  1
 kT  

 eVin 
  I o R exp 
.
 kT 
Example: Exponentiating amp

 eV
I  I o exp 
 kT

 
  1 ,
 

 eV
so f (Vin )  I o exp 
 kT

VR  IR, so g ( I )  IR.
Vout
 
  1 .
 

 eVin  
  g  f (Vin )    I o R exp 
  1
 kT  

 eVin 
  I o R exp 
.
 kT 
Example: Exponentiating amp

 eV
I  I o exp 
 kT

 
  1 ,
 

 eV
so f (Vin )  I o exp 
 kT

VR  IR, so g ( I )  IR.
Vout
 
  1 .
 

 eVin  
  g  f (Vin )    I o R exp 
  1
 kT  

 eVin 
  I o R exp 
.
 kT 
Example: Exponentiating amp

 eV
I  I o exp 
 kT

 
  1 ,
 

 eV
so f (Vin )  I o exp 
 kT

VR2  IR, so g ( I )  IR.
Vout
 
  1 .
 

 eVin  
  g  f (Vin )    I o R exp 
  1
 kT  

 eVin 
  I o R exp 
.
 kT 
Example: Exponentiating amp

 eV
I  I o exp 
 kT

 
  1 ,
 

 eV
so f (Vin )  I o exp 
 kT

VR  IR, so g ( I )  IR.
Vout
 
  1 .
 

 eVin  
  g  f (Vin )    I o R exp 
  1
 kT  

 eVin 
  I o R exp 
.
 kT 
Example:
log
amp
Box 1 is the resistor.
Vin
Vin
I
, so f (Vin )  .
R
R
Box 2 is the diode.

 eV
I  I o exp 
 kT

 
  1
 
 I

kT
V 
log   1
e
 Io 
 I 
kT

log   .
e
 Io 
 I 
kT
g (I ) 
log   .
e
 Io 
Therefore,
Vout   g  f (Vin )   
V 
kT
log  in  .
e
 Io R 
Example:
log
amp
Box 1 is the resistor.
Vin
Vin
I
, so f (Vin )  .
R
R
Box 2 is the diode.

 eV
I  I o exp 
 kT

 
  1
 
 I

kT
V 
log   1
e
 Io 
 I 
kT

log   .
e
 Io 
 I 
kT
g (I ) 
log   .
e
 Io 
Therefore,
Vout   g  f (Vin )   
V 
kT
log  in  .
e
 Io R 
Example:
log
amp
Box 1 is the resistor.
Vin
Vin
I
, so f (Vin )  .
R
R
Box 2 is the diode.

 eV
I  I o exp 
 kT

 
  1
 
 I

kT
V 
log   1
e
 Io 
 I 
kT

log   .
e
 Io 
 I 
kT
g (I ) 
log   .
e
 Io 
Therefore,
Vout   g  f (Vin )   
V 
kT
log  in  .
e
 Io R 
Example:
log
amp
Box 1 is the resistor.
Vin
Vin
I
, so f (Vin )  .
R
R
Box 2 is the diode.

 eV
I  I o exp 
 kT

 
  1
 
 I

kT
V 
log   1
e
 Io 
 I 
kT

log   .
e
 Io 
 I 
kT
g (I ) 
log   .
e
 Io 
Therefore,
Vout   g  f (Vin )   
V 
kT
log  in  .
e
 Io R 
Example:
log
amp
Box 1 is the resistor.
Vin
Vin
I
, so f (Vin )  .
R
R
Box 2 is the diode.

 eV
I  I o exp 
 kT

 
  1
 
 I

kT
V 
log   1
e
 Io 
 I 
kT

log   .
e
 Io 
 I 
kT
g (I ) 
log   .
e
 Io 
Therefore,
Vout   g  f (Vin )   
V 
kT
log  in  .
e
 Io R 
Example:
log
amp
Box 1 is the resistor.
Vin
Vin
I
, so f (Vin )  .
R
R
Box 2 is the diode.

 eV
I  I o exp 
 kT

 
  1
 
 I

kT
V 
log   1
e
 Io 
 I 
kT

log   .
e
 Io 
 I 
kT
g (I ) 
log   .
e
 Io 
Therefore,
Vout   g  f (Vin )   
V 
kT
log  in  .
e
 Io R 
Example:
log
amp
Box 1 is the resistor.
Vin
Vin
I
, so f (Vin )  .
R
R
Box 2 is the diode.

 eV
I  I o exp 
 kT

 
  1
 
 I

kT
V 
log   1
e
 Io 
 I 
kT

log   .
e
 Io 
 I 
kT
g (I ) 
log   .
e
 Io 
Therefore,
Vout   g  f (Vin )   
V 
kT
log  in  .
e
 Io R 
A Multiplier
Recall
log(ab)  log(a )  log(b)
log(a)
Vin1
Log Amp
ab
Summing
Amp
Vin2
log(a)+log(b)
=log(ab)
Log Amp
log(b)
Exponential
Amp
Vout
A Divider
Recall
log( a / b)  log( a)  log(b)
log(a)
Vin1
Log Amp
a/b
Differential
Amp
Vin2
log(a)-log(b)
=log(a/b)
Log Amp
log(b)
Exponential
Amp
Vout
Calculus
Differentiator
R1
1k
Vin
C1
1uF
Vout
Calculus
Vin
Differentiator
Integrator
R1
1k
C1
1uF
C1
1uF
Vout
Vin
R1
1k
Vout
Etc.
• Can you think of a circuit to take cube
roots?
• We can fashion sophisticated analog
computers this way.