Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

[email protected] • ENGR-43_Lec-14a_IDeal_Op_Amps.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

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:

75

125

BWA

RR

O

i

• Typical Values

OUTPUT RESISTANCE

INPUT RESISTANCE

GAIN



OpAmp InPut Terminolgy The Average of

the two InputVoltages is calledthe Common-ModeSignal

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

ov

221 vvviCM

idOo

id

vAv

vvv

and

21



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)

O

iOO

is

out A

RARRV

V recall1

1 Thus The Amplification1

S

outO V

VA

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

i

i

vvA

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

svv

vv

vvO

SO vv

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

0v

0v

0i

0

vvvAo

0 iiRi

000

21

RV

RV outs

1

2

RR

VVGs

out

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

v

v

ov



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)

v

v

ovv

v

ov

OR

( )A v v

iR



Drawing the OpAmp Linear Model

4. UNTANGLE as Needed

2R

v

v

The BEFORE & AFTER

2R

vvve



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

b - a

b - d



NonIdeal Inverting Amp cont. On The LINEAR Model

Connect The External Components

ReDraw Ckt for Increased Clarity

Now Must Sweat the Details

2R

vvve



NonIdeal Inverting Amp cont. Node Analysis

• Note GND Node

Controlling Variable In Terms Of Node Voltages

The 2 Eqns in Matrix Form

2

vvve

1vve



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

S

O

vv

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

0000.5

115

115lim

2

oA

S

O

RRAKvv



Compare Ideal vs. NonIdeal

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

Model, Solve The Resulting Circuit With Dep. Sources

0i

0v

0v

0 iiRi

0 vvA



Compare Ideal vs. NonIdeal cont.

Ideal Case at Inverting Terminal

Gain for NonIdeal Case

0i

1

2

21

000RR

vv

Rv

Rv

S

OOS

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

1

2

1

1

21

1

2

1

2 110 vvRR

RRv

RRv

RRvi O

243

42

43

40 vRR

RvvRR

Rvi

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?

2v

1v

2v

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

1v

2v

av

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”

vv

0i

R 2

R 1

ivv

0v

• Infinite Ri with v+ = v1

11 vvvv

ii vRRRvv

RRRv

1

2100

21

1

ivRRv

1

20 1 :OR



Example Find Io for Ideal OpAmp

Ideal Assumptions KCL at v−

0 iiRi

VvvA 12

Vv 12

0 iRi

Vv 12

VVkk

Vo

o 840212

1212

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

V 0

V 0

ampvolt 11or

1 V 0

SO

SO

iv

ivSi



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

IIO

IIO

GRvi

Rv

Rvi

1or

0

1

11

1v

0V

LR

0

0

Oi



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

Ai

Bi

fi

Lifv

V 0V 0

BAf iii




Use Ohm for iA & iB Also by Ohm Thru Rf

Ai

Bi

fi

Lifv

V 0V 0

B

B

B

B

B

fBB

A

A

A

A

A

fAA

Rv

Rv

Rvv

i

Rv

Rv

Rvv

i

0

0

BAffo

BAf

fffo

iiRvv

iii

iRvv

So

Recall




Recall iA & iB

Sub for iA & iB

But by Virtual Shortvf=0; Thus after ReArranging

Ai

Bi

fi

Lifv

V 0V 0

B

BB

A

AA R

viRvi

B

B

A

Affo R

vRvRvv B

B

fA

A

fo v

RR

vRR

v




Input Resistances: Similarly for RinB

Ai

Bi

fi

Lifv

V 0V 0

AAAA

AinA

AAAAAA

AinA

AAAAA

AAA

A

AinA

RRRv

vR

RvRvRvvR

RvRvvvvv

ivR

11

0

BBBB

BinB

BBBBBB

BinB

BBBBB

BBB

B

BinB

RRRv

vR

RvRvRvvR

RvRvvvvv

ivR

11

0



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

Li

Lv

BB

fA

A

fo v

RR

vRR

v



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

Li

LvBB

fA

A

f vRR

vRR

ooL

LL v

RRRv

00

ooL

LL

oL

LLoL

RvRRv

vRRvvv



Example: Summing Summary Ai

Bi

fi

Lifv

V 0V 0

BB

fA

A

fo v

RR

vRR

v AinA RR

BinB RR

0oR



Example: I & V Inputs

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

Ideal OpAmp assumptions Are Used

Infinite Gain Assumption Fixes v−

Si

Ov

R

+-

Sv

v 0i

v

Svv

Svv

Use Infinite Input Resistance Assumption

Apply KCL to Inverting Input

Then Solving

0

Rvvi o

S

SSo Rivv

Set Voltages



Example – Find G and Vo

Ideal Assumptions Solving

0 iiRi

SVvvA

SV

SVv SVv _

0i

Yields Inverse Divider

OV

SV2R

1R

SO VkkkV

11100

101S

O

VVG

VVmVV OS 101.01



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

IRRRGVV IIO

1

21

NON-INVERTING AMPLIFIER

1

21RRG

IRV II

kR

RRR

RRk

mAV

IV

IO 9110110

110

21

2

1

2



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)

1vv

0i

1vv

1vvvA



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

[email protected]

Engineering 43

Appendix





29.4 k

29.4



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

O

iOO

is

out A

RARRV

V recall1

1 Thus The Amplification1

S

outO V

VA

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

iR

v

v

OvOR

)( vvA

Add Input Source-V

iR

v

v

OvOR

)( vvA1v

The OpAmp Model



Example Required

• Draw The Linear Equivalent Circuit

• Write The Loop Equations

1. Locate Nodes

Si

Ov

R

+-

ov

v

v

Ri

RO

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

+-

ov

v

v R

R O

R iiS

A (v + - v - )

1i2i

sii 1

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

iR

v

v

OvOR

)( vvA1v

1R

2R

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

Ov



Find G & Rin for NonIdeal Case Now The Mesh Eqns

• Mesh-1

• Mesh-2

iR

v

v

OvOR

)( vvA1v

1R

2R

)( 21122 iiRiRvO

1

2

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

1

1

ivRin

1vvG O

0)(

)( 1

2

1

211

121 vii

RRRRARRRR

Oi

Ov



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

Invert Matrix as Before• Find Determinant, • Adjoint Matrix

0)()( 1

1

211

121

2

1 vRRRRAR

RRRii

Oi

)())(( 112121 RARRRRRRR iO

)()()(

211

121

RRRARRRRR

Adji

O

Then the Solution

0)()(

)(1 1

211

121

2

1 vRRRAR

RRRRii

i

O

Solving for Mesh Currents

121

1 vRRRi O

)( 1

2RARi i



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

By Mesh Currents

Ov

1121

1211

22111

))(()()(

vRARRRvRRRRiRRiRv

iO

O

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

11

211

1

21

121

1121

1121

1211

)())((

))(())((0

))(()(

vRRRv

RARARRRv

vARRRvRARRRv

vRARRRvRRRRv

i

iO

iiO

iOO

Finally then G:

1112121 )())(( RARRARRRRRRRLim iiOA

1

21

1 RRR

vvG O

And The Expression for Rin: inR Infinite Input

Resistance is GOOD

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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