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Engineering 43
Chp 4
Op Amp Circuits
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Ckts W/ Operational Amplifiers

Why Study OpAmps
At This Point?
1. OpAmps Are Very
Useful Electronic
Components
2. We Have Already Developed The Tools To
Analyze Practical OpAmp Circuits
3. The Linear Models for OpAmps
Include Dependent Sources
– A PRACTICAL Application of Dependent Srcs
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Real Op Amps
LM324 DIP
 Physical Size
Progression of
OpAmps Over
the Years
LMC6294
MAX4240
Maxim (Sunnyvale, CA) Max4241 OpAmp
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Apex PA03
HiPwr OpAmp
 Notice OutPut Rating
• 30A @75 V
 PwrOut → 30A•75V
→ 2.25 kW!
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
 X
 xx
 X
 xx
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
OpAmp Symbol & Model
 The Circuit Symbol
Is a Version of the
Amplifier TRIANGLE
 The Linear Model
• Typical Values
Ri : 105   1012 
OUTPUT RESISTANCE
RO : 1  50
A : 105  107
INPUT RESISTANCE
GAIN
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
OpAmp Power Connections
 BiPolar Power
Supplies
 UniPolar Supply
 For Signal I/O Analysis the Supplies
Need NOT be shown explicitly
• But they MUST be there to actually
Power the Operational Amplifier
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
OpAmp Circuit Model
LOAD
OP-AMP
DRIVING CIRCUIT
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
vi→vo Transfer Characteristics
Linear
Region
vo/vi =
Const
Saturation
 The OUTPUT Voltage Level can NOT exceed the
SUPPLY the Level
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Unity Gain Buffer
Op-Amp BUFFER GAIN
LM324
0.99999
LMC6492
0.9998
MAX4240
0.99995
KVL :  Vs  Ri I  RO I  AOVin  0
 Controlling Variable = Vin  Ri I
KVL : - Vout  RO I  AOVin  0
 Solve For Buffer Gain
Vout
1
 Thus The Amplification

recall AO  
V
Ri
Vs 1 
AO    out  1
RO  AO Ri
VS
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
The Ideal OpAmp
 The IDEAL
Characteristics
• Ro = 0
• Ri = 
• A=
 The Consequences
of Ideality
Ro  0  vo  Av  v 
Ri    i  i  0
A    v  v
Engineering-43: Engineering Circuit Analysis
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i
i
@V3 : V3  VS1  12[V ]
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Voltage Follower
vO  v S
v  v s
 The Voltage Follower
v  v
• Also Called Unity Gain
Buffer (UGB) from Before
vO  v
 Usefulness of UGB
Connection w/o Buffer
vO  vS  iRs
 The SOURCE
Supplies The Power
Engineering-43: Engineering Circuit Analysis
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Buffered Connection
vO  vS
 The Source Supplies
NO Power (the
OpAmp does it)
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Inverting
OpAmp Ckt
 Determine Voltage Gain,
G = Vout/Vin
 Start with Ao
v  0
i  0
v  0
Ao    v  v
 v  0
 Now From Input R
Ri    i  i  0
 Apply KCL at v-
Vs  0 Vout  0

0
R1
R2
Engineering-43: Engineering Circuit Analysis
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 Finally The Gain
Vout
R2
G

Vs
R1
 Next: Examine Ckt w/o
Ideality Assumption
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Replace OpAmp w/ Linear Model
 Consider Again the
Inverting OpAmp
Circuit
1. Identify the Op
Amp Nodes
v
vo
v
 Draw the
Linear Model
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Drawing the OpAmp Linear Model
2. Redraw the circuit
cutting out the
Op Amp
v
3. Draw components
of linear OpAmp
(on the circuit of
step-2)
v
vo
vo
Ri
v
v
RO


Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
A(v  v )
Drawing the OpAmp Linear Model
4. UNTANGLE as
Needed
v
R2
R2
v
 The BEFORE
& AFTER
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
NonIdeal Inverting Amp
 Replace the OpAmp with
the LINEAR Model
b - d
b - a
• Label Nodes for Tracking
 Draw The Linear
Equivalent For Op-amp
 Note the External
Component Branches
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
NonIdeal Inverting Amp cont.
 On The LINEAR Model
Connect The External
Components
 ReDraw Ckt for
Increased Clarity
R2
Engineering-43: Engineering Circuit Analysis
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 Now Must Sweat the
Details
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
NonIdeal Inverting Amp cont.
 Node Analysis
• Note GND Node
 Controlling Variable In
Terms Of Node Voltages
Engineering-43: Engineering Circuit Analysis
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2
 The 2 Eqns in Matrix Form
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Inverting Amp – Invert Matrix
 Use Matrix Inversion to Solve 2 Eqns in 2 Unknowns
• Very Useful for 3 Eqn/Unknwn Systems as well
– e.g., http://www.wikipedia.org/wiki/Matrix_inversion
 The Matrix Determinant 
 Solve for vo
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Inverting Amp – Invert Matrix cont
 Then the System Gain
 Typical Practical Values for the Resistances
• R1 = 1 kΩ
R2 = 5 kΩ
 Then the Real-World Gain
vO
 4.9996994
vS
 Recall The Ideal Case for A→; Then The Eqn at top
vO
 5 1
 5 1
 lim A

 5.0000
vS
1  K  A R2 Ro  1    
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Compare Ideal vs. NonIdeal
v  0
i  0
v  0
 Ideal Assumptions
Ri    i  i  0
A    v  v  0
Engineering-43: Engineering Circuit Analysis
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 Gain for Real Case
• Replace Op-amp By Linear
Model, Solve The Resulting
Circuit With Dep. Sources
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Compare Ideal vs. NonIdeal cont.

i  0

 Ideal Case at Inverting
Terminal
The Ideal Opamp Assumption
Provides an
Excellent
Real-World
Approximation.
Unless Forced to
do Otherwise We
Will Always Use
the IDEAL Model
 Gain for NonIdeal Case
0  vS 0  vO
vO
R2

0

R1
R2
vS
R1
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Example  Differential Amp
 KCL At Inverting Term
 KCL at NONinverting
Terminal
 Assume Ideal OpAmp

 R 
R
R  R 
i  0  vO  1  2 v  2 v1  2  1  1 v  v1 
R1
R1   R2 
 R1 

 By The KCLs
A simple
Voltage
Divider
i  0  v 
Engineering-43: Engineering Circuit Analysis
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R4
R4
v2  v 
v2
R3  R4
R3  R4
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Example  Differential Amp cont
 Then in The Ideal Case
 Now Set External Rs
• R4 = R2
• R3 = R1
 Subbing the Rs
Into the vo Eqn
R2
vO 
(v2  v1 )
R1
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Ex. Precision Diff V-Gain Ckt
 Find vo
 Assume Ideal OpAmp
• Which Voltages are Set?
v1
i  0
v1  v1 , v2  v2
• What Voltages Are Also
Known Due To Infinite
Gain Assumption?
v2
v2
• Now Use The Infinite
Resistance Assumption
 CAUTION: There could be
currents flowing INTO or
OUT of the OpAmps
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Ex. Precision Diff V-Gain Ckt cont
 The Ckt Reduces To Fig. at Right
 KCL at v1
v1
 KCL at v2
va
v2
 Eliminate va Using The above Eqns
and Solve for vo in terms of v1 & v2
• Note the increased Gain over Diff Amp
OpAmp
Current
Engineering-43: Engineering Circuit Analysis
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R2 R1  1  R2 R1  2R2 RG 
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
NONinverting Amp - Ideal
v0
R2
i  0
v  vi
R1
 Ideal Assumptions
• Infinite Gain
v  v
• Infinite Ri with v+ = v1
v  v1  v  v1
Engineering-43: Engineering Circuit Analysis
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 Since i- = 0 Arrive at
“Inverse Voltage Divider”
R1
R1  R2
vi 
v0  v0 
vi
R1  R2
R1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Example  Find Io for Ideal OpAmp
v  12V
v  12V
Ri    i  0
 Ideal Assumptions
A    v  v  12V
Ri    i  i  0
Engineering-43: Engineering Circuit Analysis
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 KCL at v-
12  Vo 12

 0  Vo  84V
12k
2k
Vo
 IO 
 8.4mA
10k
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Find G & Rin for NonIdeal Case
 The OpAmp Model
v

Ri

v
RO
vO
A( v   v  )
 Add Input Source-V
 Determine Equivalent
Circuit Using Linear
Model For Op-Amp
Engineering-43: Engineering Circuit Analysis
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v1


v

Ri

v
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
RO
vO
A( v   v  )
Find G & Rin for NonIdeal Case cont
 Add The External
Components
 The Equivalent Circuit for
Mesh Analysis
R2

v1

v

Ri
v
RO

R1
vO
A( v   v  )
 Now Re-draw Circuit To
Enhance Clarity

vO

• There Are Only Two Loops
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Find G & Rin for NonIdeal Case cont
 Now The Mesh Eqns
R2
• Mesh-1
2

• Mesh-2
v1
 The Controlling Variable
in Terms of Loop
Currents

v
1


Ri
R1
v
RO
A( v   v  )
 Eliminating vi
vO   R2i2  R1 (i1  i2 )
 Continue Analysis on
Next Slide
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
vO
G & Rin for NonIdeal Case cont
 The Math Model From
Mesh Analysis

vO
vO   R2i2  R1 (i2  i1 )

 The Input-R and Gain
v1
Rin 
i1
vO
G
v1
 Then the Model in
Matrix Form →
Engineering-43: Engineering Circuit Analysis
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 R1
( R1  R2 )
  i1  v1 
 AR  R ( R  R  R ) i    0 
1
1
2
O  2 
 
 i
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
G & Rin for NonIdeal Case cont
 The Matrix-Inversion Soln
1
 R1
 v1 
 i1  ( R1  R2 )
i    AR  R ( R  R  R )  0 
1
1
2
O   
 2  i
 Invert Matrix as Before
• Find Determinant, 
• Adjoint Matrix
  ( R1  R2  RO )( R1  R2 )  R1( ARi  R1)
R1 
( R  R2  RO )
Adj   1

  ( ARi  R1 ) ( R1  R2 )
 Then the Solution
R1  v1 
 i1  1 ( R1  R2  RO )
i      ( AR  R ) ( R  R )  0 
 2

i
1
1
2  
Engineering-43: Engineering Circuit Analysis
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 Solving for Mesh Currents
R1  R2  RO
i1 
v1

i2 
 ( ARi  R1 )
 Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
G & Rin for NonIdeal Case cont
 The (Long) Expression for The input Resistance
 By Mesh Currents
vO  R1i1  ( R1  R2 )i2

R1 ( R1  R2  RO )
( R  R2 )( ARi  R1 )
v1  1
v1


 This Looks Ugly
• How Can we Simplify?

vO
– Recall
 A→
 Ri → 
Engineering-43: Engineering Circuit Analysis
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
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
G & Rin for NonIdeal Case cont
 Use A→ in Expression for 
  Lim A ( R1  R2  RO )( R1  R2 )  R1 ( ARi  R1 )  ARi R1
 Now Since For Op-Amps Ri→ Also, then vo
R1 ( R1  R2  RO )
( R1  R2 )( ARi  R1 )
vO 
v1 
v1


( R1  R2 )( ARi  R1 )
( R1  R2 )( ARi )
vO  0 
v1 
v1


( R  R2 )( ARi )
( R  R2 )
vO  1
v1  1
v1
ARi R1
R1
vO
R1  R2
 Finally then G:
G

v1
R1

 And The Expression for Rin: Rin  
Engineering-43: Engineering Circuit Analysis
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Infinite Input
Resistance
is GOOD
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Example
 Set Voltages
R
i  0
v



v
vO
iS
+
-
vS
 Required

• Find the expression for vo.
• Indicate where and how
Ideal OpAmp assumptions
Are Used
Engineering-43: Engineering Circuit Analysis
37
v  vS
 Infinite Gain Assumption
Fixes vv  vS
 Use Infinite Input
Resistance Assumption
 Apply KCL to Inverting
Input
 Then Solving
vo  v
iS 
0
R
vo  vS  Ri S
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Example
R
 Required

• Draw The Linear

Equivalent Circuit
• Write The Loop Equations
vO
iS
vo
v


RO
Ri
+
-
1. Locate Nodes
2. Place the nodes in
linear circuit model
v
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
v
Example cont
3. Add Remaining
Components to Complete
Linear Model
R
i2
i1
iS
RO
Ri
+
-
 Examine Circuit
• Two Loops
• One Current Source
 Use Meshes
• Mesh-1 i1  is
vo
v
 DONE
• Mesh-2
Ri (i2  iS )  ( R  RO )i2  A(v  v_ )  0
• But Could Sub for (v+-v-)
and solve for i2
• Controlling Variable
v  v_  Ri (i2  iS )
Engineering-43: Engineering Circuit Analysis
39
A(v + - v -)
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Example – Find G and Vo
v  VS
VS 
VO
v _  VS
i  0
VS
R2
R1
 Ideal Assumptions
A    v  v  VS
Ri    i  i  0
 Yields Inverse Divider
Engineering-43: Engineering Circuit Analysis
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 Solving
VO 
VO
G
 101
VS
100k  1k
VS
1k
VS  1mV  VO  0.101V
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Key to OpAmp Ckt Analysis  IOA
 Remember that the “Nose” of the
OpAmp “Triangle” can SOURCE or
SINK “Infinite” amounts of Current
IOA = ± ∞
|IOA,max| = Isat
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Comparator
 Ideal Comparator and Transfer Characteristic
 “Zero-Cross” Detector → Heart of Solid State Relay Cnrtl
Engineering-43: Engineering Circuit Analysis
42
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Example  OpAmp Based I-Mtr
 Desired Transfer Characteristic = 10V/mA → Find R2
NON-INVERTING AMPLIFIER
R
G 1 2
R1
VI  RI I
 R 
VO  GVI  1  2  RI I
 R1 
 R 
VO 10V  R2 

 1   RI  10k  1  2   R2  9k
I 1mA  R1 
 R1 
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Offset & Saturation
 A NonInverting Amp
• But How to Handle This???
 Start w/ KCL at v• Assume Ideality
• Then at Node Between the
1k & 4k Resistors
v  v1
i  0
v  v1
A    v  v  v1
 Then the Output
Engineering-43: Engineering Circuit Analysis
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 Notes on Output Eqn
• Slope = 1+(R2/R1) as Before
• Intercept =
−(0.5V)x(4kΩ/1kΩ)
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Example – Offset & Saturation
 Note how “Offset” Source Generates a
Non-Zero Output When v1 = 0
 The Transfer Characteristic for This Circuit
“Saturates” at
“Rail” Potential
IN LINEAR RANGE
−2V Offset
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
WhiteBoard Work
 Let’s Work a Unity
Gain Buffer Problem
• Vs = 60mV
• Rs = 29.4 kΩ
• RL = 600 Ω
• Find Load Power
WITH and withOUT
OpAmp UGB
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Key to OpAmp Ckt Analysis  IOA
 Remember that the “Nose” of the
OpAmp “Triangle” can SOURCE or
SINK “Infinite” amounts of Current
IOA = ± ∞
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt
Unity Gain Buffer
Op-Amp BUFFER GAIN
LM324
0.99999
LMC6492
0.9998
MAX4240
0.99995
 Controlling Variable = Vin  Ri I
KVL :  Vs  Ri I  RO I  AOVin  0
KVL : - Vout  RO I  AOVin  0
 Solve For Buffer Gain by KVL
Vout
1
 Thus The Amplification

recall AO  
V
Ri
Vs 1 
AO    out  1
RO  AO Ri
VS
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04_Op-Amps.ppt