[email protected] ENGR-43_Lec-04_Op-Amps.ppt 1 Bruce Mayer, PE Engineering-43: Engineering...

[email protected] • ENGR-43_Lec-04_Op-Amps.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Chp 4Chp 4Op Amp Op Amp CircuitsCircuits



Ckts W/ Operational AmplifiersCkts 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



Real Op AmpsReal Op Amps Physical Size

Progression of OpAmps Over the Years

Maxim (Sunnyvale, CA) Max4241 OpAmp

LM324 DIP

LMC6294

MAX4240



Apex PA03 Apex PA03 HiPwr OpAmpHiPwr OpAmp

Notice OutPut Rating• 30A @75 V

PwrOut → 30A•75V→ 2.25 kW!



X xx

X xx



OpAmp Symbol & ModelOpAmp Symbol & Model The Circuit Symbol

Is a Version of the Amplifier TRIANGLE

The Linear Model

75

125

1010:

501:

1010:

A

R

R

O

i

• Typical Values

OUTPUT RESISTANCE

INPUT RESISTANCE

GAIN



OpAmp Power ConnectionsOpAmp 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



OpAmp Circuit ModelOpAmp Circuit Model

DRIVING CIRCUIT

LOAD

OP-AMP



vvii→v→voo Transfer Characteristics Transfer Characteristics

Saturation

The OUTPUT Voltage Level can NOT exceed the SUPPLY the Level

LinearRegionvo/vi = Const



Unity Gain BufferUnity Gain Buffer

Controlling Variable = IRV iin

Solve For Buffer Gain

O

iOO

is

out A

RARRV

Vrecall

1

1 Thus The Amplification

1S

outO V

VA

Op-Amp BUFFER GAINLM324 0.99999LMC6492 0.9998MAX4240 0.99995

0 inOOis VAIRIRV :KVL

0 inOO VAIRoutV- :KVL



The Ideal OpAmpThe Ideal OpAmp The IDEAL

Characteristics• Ro = 0

• Ri =

• A = The Consequences

of Ideality

][12:@ 133 VVVV S

i

i

vvA

0 iiRi

vvAvR oo 0



Voltage FollowerVoltage 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 Inverting OpAmp CktOpAmp 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

v

vvAo

0 iiRi

000

21

R

V

R

V outs

1

2

R

R

V

VG

s

out

Next: Examine Ckt w/o Ideality Assumption



Replace OpAmp w/ Linear ModelReplace 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 ModelDrawing the OpAmp Linear Model

2. 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 ModelDrawing the OpAmp Linear Model

4. UNTANGLE as Needed

2R

v

v

The BEFORE & AFTER

2R



NonIdeal Inverting AmpNonIdeal 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.NonIdeal Inverting Amp cont. On The LINEAR Model

Connect The External Components

ReDraw Ckt for Increased Clarity

Now Must Sweat the Details

2R



NonIdeal Inverting Amp cont.NonIdeal Inverting Amp cont. Node Analysis

• Note GND Node

Controlling Variable In Terms Of Node Voltages

The 2 Eqns in Matrix Form

2



Inverting Amp – Invert MatrixInverting 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 contInverting 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

9996994.4S

O

v

v

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

0000.5

1

15

1

15lim

2

oA

S

O

RRAKv

v



Compare Ideal vs. NonIdealCompare 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.Compare Ideal vs. NonIdeal cont.

Ideal Case at Inverting Terminal Gain for NonIdeal Case

0i

1

2

21

000

R

R

v

v

R

v

R

v

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 Example Differential Amp 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 vvR

R

R

Rv

R

Rv

R

Rvi O

243

42

43

40 vRR

Rvv

RR

Rvi

A simple Voltage Divider



Example Example Differential Amp Differential Amp contcont

Then in The Ideal Case

Now Set External Rs• R4 = R2

• R3 = R1

Subbing the Rs Into the vo Eqn

)( 121

2 vvR

RvO



Ex. Precision Diff V-Gain CktEx. Precision Diff V-Gain Ckt Find vo

Assume Ideal OpAmp• 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 OUT of the OpAmps



Ex. Precision Diff V-Gain Ckt contEx. 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 - IdealNONinverting 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 vR

RRvv

RR

Rv

1

2100

21

1



Example Example Find I Find Ioo for Ideal OpAmp for Ideal OpAmp

Ideal Assumptions KCL at v-

0 iiRi

VvvA 12

Vv 12

0 iRi

Vv 12

VVkk

Vo

o 8402

12

12

12

mAk

VI oO 4.8

10



Find G & RFind G & Rinin for NonIdeal Case 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



Find G & RFind G & Rinin for NonIdeal Case cont for NonIdeal Case cont The Equivalent Circuit for

Mesh Analysis Add The External

Components

iR

v

v

OvOR

)( vvA1v

1R

2R

Now Re-draw Circuit To Enhance Clarity • There Are Only Two

Loops

Ov



Find G & RFind G & Rinin for NonIdeal Case cont for NonIdeal Case cont 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 & RG & Rinin for NonIdeal Case cont 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

i

vRin

1v

vG O

0)(

)( 1

2

1

211

121 v

i

i

RRRRAR

RRR

Oi

Ov



G & RG & Rinin for NonIdeal Case cont for NonIdeal Case cont The Matrix-Inversion Soln

Invert Matrix as Before• Find Determinant, • Adjoint Matrix

0)(

)( 1

1

211

121

2

1 v

RRRRAR

RRR

i

i

Oi

)())(( 112121 RARRRRRRR iO

)()(

)(

211

121

RRRAR

RRRRAdj

i

O

Then the Solution

0)()(

)(1 1

211

121

2

1 v

RRRAR

RRRR

i

i

i

O

Solving for Mesh Currents

121

1 vRRR

i O

)( 1

2

RARi i



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

By Mesh Currents

Ov

1121

1211

22111

))(()(

)(

vRARRR

vRRRR

iRRiRv

iO

O

This Looks Ugly• How Can we Simplify?

– Recall A → Ri →



G & RG & Rinin for NonIdeal Case cont 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

))(()(

vR

RRv

RAR

ARRRv

vARRR

vRARRR

v

vRARRR

vRRRR

v

i

iO

iiO

iOO

Finally then G:

1112121 )())(( RARRARRRRRRRLim iiOA

1

21

1 R

RR

v

vG O

And The Expression for Rin: inR Infinite Input

Resistance is GOOD



ExampleExample

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

R

vvi oS

SSo Rivv

Set Voltages



ExampleExample 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 contExample 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



Example – Find G and VExample – Find G and Voo

Ideal Assumptions Solving

0 iiRi

SVvvA

SV

SVv SVv _

0i

Yields Inverse Divider

OV

SV2R

1R

SO Vk

kkV

1

1100

101S

O

V

VG

VVmVV OS 101.01



Key to OpAmp Ckt Analysis Key to OpAmp Ckt Analysis IIOAOA

Remember that the “Nose” of the OpAmp “Triangle” can SOURCE or SINK “Infinite” amounts of Current

IOA = ± ∞

|IOA,max| = Isat



ComparatorComparator Ideal Comparator and Transfer Characteristic

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



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

IRR

RGVV IIO

1

21

NON-INVERTING AMPLIFIER

1

21R

RG

IRV II

kR

R

RkR

R

R

mA

V

I

VI

O 911011

102

1

2

1

2



Offset & SaturationOffset & 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 & SaturationExample – 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 WorkWhiteBoard 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



Key to OpAmp Ckt Analysis Key to OpAmp Ckt Analysis IIOAOA

Remember that the “Nose” of the OpAmp “Triangle” can SOURCE or SINK “Infinite” amounts of Current

IOA = ± ∞



Unity Gain BufferUnity Gain Buffer

Controlling Variable = IRV iin

Solve For Buffer Gain by KVL

O

iOO

is

out A

RARRV

Vrecall

1

1 Thus The Amplification

1S

outO V

VA

0 inOOis VAIRIRV :KVL

Op-Amp BUFFER GAINLM324 0.99999LMC6492 0.9998MAX4240 0.99995

0 inOO VAIRoutV- :KVL

[email protected] ENGR-43_Lec-04_Op-Amps.ppt 1 Bruce Mayer, PE Engineering-43: Engineering...

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Transcript of [email protected] ENGR-43_Lec-04_Op-Amps.ppt 1 Bruce Mayer, PE Engineering-43: Engineering...