Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

BMayer@ChabotCollege.edu • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering 43

Chp 14-1Op Amp Circuits

Ckts W/ Operational Amplifiers OpAmp Utility

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

Real Op Amps Physical Size

Progression of OpAmps Over the Years

Maxim (Sunnyvale, CA) Max4241 OpAmp

LM324 DIP

LMC6294

MAX4240

Apex PA03 HiPwr OpAmp

Notice OutPut Rating• 30A @75 V

PwrOut → 30A•75V→ 2.25 kW!

OpAmp Symbol & Model The Circuit Symbol

Is a Version of the Amplifier TRIANGLE

The Linear Model

MHz 1:1010:

501:1010:

• Typical Values

OUTPUT RESISTANCE

INPUT RESISTANCE

OpAmp InPut Terminolgy The Average of

the two InputVoltages is calledthe Common-ModeSignal

The Difference between the Inputs is called the Differential-Signal, vid

221 vvviCM

OpAmp Power Connections

BiPolar Power Supplies

UniPolar Supply

For Signal I/O Analysis the Supplies Need NOT be shown explicitly• But they MUST physically be there to actually

Power the Operational Amplifier

OpAmp Circuit Model

DRIVING CIRCUIT

LOADOP-AMP

vi→vo Transfer Characteristics

Saturation

The OUTPUT Voltage Level can NOT exceed the SUPPLY the Level

LinearRegionvo/vi = Const

Unity Gain Buffer (FeedBack)

Controlling Variable = IRV iin

Solve For Buffer Gain (I = Vin/R)

V recall1

1 Thus The Amplification1

outO V

Op-Amp BUFFER GAINLM324 0.99999LMC6492 0.9998MAX4240 0.99995

0 inOOis VAIRIRV :KVL0 inOO VAIRoutV- :KVL

FeedBackLoop

The Ideal OpAmp The IDEAL

Characteristics• Ro = 0• Ri = • A = (open loop

gain)• BW =

The Consequences of Ideality ][12:@ 133 VVVV S

0 iiRi

vvAvR oo 0

Summing Point Constraint The MOST important

aspects of Ideality:• Ri = ∞ → i+ = i− = 0

– The OpAmp INput looks like an OPEN Circuit

• A = ∞ → v+ = v−

– The OpAmp Input looks like a SHORT Circuit

This simultaneous OPEN & SHORT Characteristic is called the →

SUMMING POINT CONSTRAINT

Looks Like an OPEN to Current

Looks Like a SHORT to Voltage

Analyzing Ideal OpAmp Ckts1. Verify the presence of NEGATIVE

FeedBack2. Assume the Summing Point Constraint

Applies in this fashion:• i+ = i−= 0 (based on Ri = ∞) • v+ − v− = 0 (based on AO = ∞)

3. Use KVL, KCL, Ohm’s Law, and other linear ckt analysis techniques to determine quantities of interest

Voltage Follower The Voltage Follower

• Also Called Unity Gain Buffer (UGB) from Before

Connection w/o Buffer Buffered Connection

The SOURCE Supplies The Power

The Source Supplies NO Power (the OpAmp does it)

Usefulness of UGB

SO vv sSO iRvv

Inverting OpAmp Ckt

Determine Voltage Gain, G = Vout/Vin

Start with Ao

Now From Input R

Apply KCL at v−

Finally The Gain

0 iiRi

RV outs

Next: Examine Ckt w/o Ideality Assumption

Replace OpAmp w/ Linear Model Consider Again

the Inverting OpAmp Circuit

Draw the Linear Model

1. Identify the Op Amp Nodes

Drawing the OpAmp Linear Model2. Redraw the

circuit cutting out the Op Amp

3. Draw components of linear OpAmp (on the circuit of step-2)

( )A v v

Drawing the OpAmp Linear Model

4. UNTANGLE as Needed

The BEFORE & AFTER

NonIdeal Inverting Amp Replace the OpAmp with

the LINEAR Model• Label Nodes for Tracking

Draw The Linear Equivalent For Op-amp

Note the External Component Branches

NonIdeal Inverting Amp cont. On The LINEAR Model

Connect The External Components

ReDraw Ckt for Increased Clarity

Now Must Sweat the Details

NonIdeal Inverting Amp cont. Node Analysis

• Note GND Node

Controlling Variable In Terms Of Node Voltages

The 2 Eqns in Matrix Form

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

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 9996994.4

Recall The Ideal Case for A→; Then The Eqn at top

0000.5

115lim

RRAKvv

Compare Ideal vs. NonIdeal

Ideal Assumptions Gain for Real Case• Replace Op-amp By Linear

Model, Solve The Resulting Circuit With Dep. Sources

0 iiRi

Compare Ideal vs. NonIdeal cont.

Ideal Case at Inverting Terminal

Gain for NonIdeal Case

The Ideal Op-amp Assumption Provides an Excellent Real-World Approximation.

Unless Forced to do Otherwise We Will Always Use the IDEAL Model

Example Differential Amp KCL At Inverting Term

KCL at NONinvertingTerminal

Assume Ideal OpAmp

By The KCLs

2 110 vvRR

RRvi O

40 vRR

A simple Voltage Divider

Example Differential Amp cont Then in The Ideal Case

Now Set External R’s• R4 = R2

• R3 = R1

Subbing the R’s Into the vo Eqn

)( 121

2 vvRRvO

Ex. Precision Diff V-Gain Ckt Find vo

Assume Ideal OpAmps• Which Voltages are

Set?• What Voltages Are Also Known Due To Infinite Gain Assumption?

0i2211 , vvvv

• Now Use The Infinite Resistance Assumption

CAUTION: There could be currents flowing INto or OUTof the OpAmps

Ex. Precision Diff V-Gain Ckt cont The Ckt Reduces To Fig. at Right KCL at v1

KCL at v2

Eliminate va Using The above Eqns and Solve for vo in terms of v1 & v2

• Note the increased Gain over Diff Amp

GRRRRRR 21212 21 OpAmpCurrent

NONinverting Amp - Ideal

Ideal Assumptions• Infinite Gain

Since i− = 0 Arrive at “Inverse Voltage Divider”

• Infinite Ri with v+ = v1

11 vvvv

ii vRRRvv

20 1 :OR

Example Find Io for Ideal OpAmp

Ideal Assumptions KCL at v−

0 iiRi

VvvA 12

o 840212

mAkVI o

O 4.810

Example: TransResistance Ckt The trans-resistance

Amp circuit below performs Current to Voltage Conversion• Find vo/iS

Use the Summing Point Constraint• → v− = 0V• Now by KCL at v−

Node with i− = 0– Notice that iS flows

thru the 1Ω resistor– Thus by Ohm’s Law

ampvolt 11or

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

Example: TransResistance Ckt Notice

that theOpAmpAbsorbsALL of theSourceCurrent

The (Ideal) OpAmp will Source or Sink Current Out-Of or In-To its “nose” to maintain the V-Constraint: v+ = v−

Example: TransConductance Ckt The transonductance

Amp circuit below performs Voltage to Current Conversion• Find io/v1

Use the Summing Point Constraint• → v− = v+

• → i− = i+ = 0

Thus v− = v1 Then

by KCL& Ohm• Notice the Output is

Insensitive to Load Resistance

Example: Summing Amp Circuit

Label Important Quantities

By Virtual Short (v+=v−) Across OpAmp Inputs

Also by Summing Point Contstraint (i+ = i− =0) KCL at vf node

V 0V 0

BAf iii

Use Ohm for iA & iB Also by Ohm Thru Rf

V 0V 0

Recall

Recall iA & iB

Sub for iA & iB

But by Virtual Shortvf=0; Thus after ReArranging

V 0V 0

Affo R

vRvRvv B

Input Resistances: Similarly for RinB

V 0V 0

AAAAAA

RvRvRvvR

RvRvvvvv

BBBBBB

RvRvRvvR

RvRvvvvv

Example: Summing Amp Circuit For the OUTput

Resistance as seen by the LOAD, put the Circuit in a “Black Box”

Thévenizing the Black Box

Recall vo Li

Example: Summing Amp Circuit Thus Have

From the LOAD Perspective Expect

But the Previous analysis vo does NOT depend on RL at ALL

If vL = vo in all cases then have

• Since vo is Not affected by RL, then Ro = 0

vRRvvv

Example: Summing Summary Ai

V 0V 0

v AinA RR

BinB RR

Example: I & V Inputs

Required• Find the expression for vo. • Indicate where and how

Ideal OpAmp assumptions Are Used

Infinite Gain Assumption Fixes v−

Use Infinite Input Resistance Assumption

Apply KCL to Inverting Input

Then Solving

Rvvi o

SSo Rivv

Set Voltages

Example – Find G and Vo

Ideal Assumptions Solving

0 iiRi

SVv SVv _

Yields Inverse Divider

SO VkkkV

VVmVV OS 101.01

Example OpAmp Based I-Mtr Desired Transfer Characteristic = 10V/mA → Find R2

IRRRGVV IIO

NON-INVERTING AMPLIFIER

IRV II

IO 9110110

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

Then the Output Notes on Output Eqn

• Slope = 1+(R2/R1) as Before• Intercept =

−(0.5V)x(4kΩ/1kΩ)

Example – Offset & Saturation Note how “Offset” Source Generates a

NON-Zero Output When v1 = 0 The Transfer Characteristic for This Circuit

IN LINEAR RANGE

−2V Offset

“Saturates” at “Rail” Potential

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

All Done for Today

What’san

OpAmp?

Bruce Mayer, PERegistered Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering 43

Appendix

29.4 k

Unity Gain Buffer Source-Driven Circuit • Vs = 60mV• Rs = 29.4 kΩ• RL = 600 Ω

Key to OpAmp Ckt Analysis IOA Remember that the “Nose” of the

OpAmp “Triangle” can SOURCE or SINK “Infinite” amounts of Current

IOA = ± ∞

Unity Gain Buffer

Controlling Variable = IRV iin

Solve For Buffer Gain by KVL

V recall1

1 Thus The Amplification1

outO V

0 inOOis VAIRIRV :KVL

Op-Amp BUFFER GAINLM324 0.99999LMC6492 0.9998MAX4240 0.99995

0 inOO VAIRoutV- :KVL

Comparator Ideal Comparator and Transfer Characteristic

“Zero-Cross” Detector → Heart of Solid State Relay Cnrtl

Find G & Rin for NonIdeal Case

Determine Equivalent Circuit Using Linear Model For Op-Amp

)( vvA

Add Input Source-V

)( vvA1v

The OpAmp Model

Example Required

• Draw The Linear Equivalent Circuit

• Write The Loop Equations

1. Locate Nodes

2. Place the nodes in linear circuit model

Example cont3. Add Remaining

Components to Complete Linear Model

Examine Circuit• Two Loops• One Current Source

Use Meshes• Mesh-1

• Mesh-2

A (v + - v - )

0)()()( _22 vvAiRRiiR OSi

• Controlling Variable

)( 2_ Si iiRvv

DONE• But Could Sub for (v+-v-)

and solve for i2

Find G & Rin for NonIdeal Case The Equivalent Circuit for

Mesh Analysis Add The External

Components

)( vvA1v

Now Re-draw Circuit To Enhance Clarity • Be Faithful to the NODES• There Are Only Two Loops

Find G & Rin for NonIdeal Case Now The Mesh Eqns

• Mesh-1

• Mesh-2

)( vvA1v

)( 21122 iiRiRvO

The Controlling Variable in Terms of Loop Currents

Eliminating vi

Continue Analysis on Next Slide

G & Rin for NonIdeal Case cont The Math Model From

Mesh Analysis

The Input-R and Gain

Then the Model in Matrix Form →

)( 12122 iiRiRvO

1vvG O

121 vii

RRRRARRRR

G & Rin for NonIdeal Case cont The Matrix-Inversion Soln

Invert Matrix as Before• Find Determinant, • Adjoint Matrix

0)()( 1

1 vRRRRAR

)())(( 112121 RARRRRRRR iO

)()()(

RRRARRRRR

Then the Solution

1 vRRRAR

RRRRii

Solving for Mesh Currents

1 vRRRi O

2RARi i

G & Rin for NonIdeal Case cont The (Long) Expression for The input Resistance

By Mesh Currents

))(()()(

vRARRRvRRRRiRRiRv

This Looks Ugly• How Can we Simplify?

– Recall A → Ri →

G & Rin for NonIdeal Case cont Use A→ in Expression for

Now Since For Op-Amps Ri→ Also, then vo

)())((

))(())((0

))(()(

RARARRRv

vARRRvRARRRv

vRARRRvRRRRv

Finally then G:

1112121 )())(( RARRARRRRRRRLim iiOA

And The Expression for Rin: inR Infinite Input

Resistance is GOOD

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

pe engineering

short circuit

vthe opamp input

n s j o h n s o n c

buffer gain

open circuita

open loop gainbw

ca max4241 opamp lm324

Documents

Bruce Mayer, PE Registered Electrical & Mechanical Engineer....

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer...