OPERATIONAL AMPLIFIERS Why do we study them at this point???

OPERATIONAL AMPLIFIERS

Why do we study them at this point???

1. OpAmps are very useful electronic components

2. We have already the tools to analyze practical circuits using OpAmps

3. The linear models for OpAmps include dependent sources

TYPICAL DEVICE USING OP-AMPS

OP-AMP ASSEMBLED ON PRINTED CIRCUIT BOARD APEX PA03

PIN OUT FOR LM324

DIMENSIONAL DIAGRAM LM 324 LM324 DIP

LMC6294

MAX4240

CIRCUIT SYMBOL FOR AN OP-AMP SHOWING POWER SUPPLIES

LINEAR MODEL OUTPUT RESISTANCE

INPUT RESISTANCE

GAIN

7 5

12 5

10 10

:

50 1

:

10 10

:

−

Ω

− Ω

Ω

− Ω A

R R

O i

TYPICAL VALUES

CIRCUIT WITH OPERATIONAL AMPLIFIER

DRIVING CIRCUIT

LOAD OP-AMP

TRANSFER PLOTS FOR SOME COMERCIAL OP-AMPS

SATURATION REGION

LINEAR REGION

IDENTIFY SATURATION REGIONS OP-AMP IN SATURATION

CIRCUIT AND MODEL FOR UNITY GAIN BUFFER

= 0 +

+ +

− V

_s

R

_i

I R

_O

I A

_O

V

_in

: KVL

= 0 +

+ R

_O

I A

_O

V

_in

V

out

- : KVL

I R V

_in

=

_i

: VARIABLE G

CONTROLLIN

i O O

s i out

R A R

V R V

+ +

= 1

1 SOLVING

→ 1

⇒

∞

→

S out

O

V

A V

WHY UNIT GAIN BUFFER?

BUFFER GAIN PERFORMANCE OF REAL OP-AMPS

Op-Amp BUFFER GAIN LM324 0.99999 LMC6492 0.9998 MAX4240 0.99995

⇒

∞

= A

) (

0 ⇒ =

₊

−

₋

= v A v v

R

_{O O}

THE IDEAL OP-AMP

⇒

∞

i

= R

∞

=

∞

=

=

⇒ R

_O

0 , R

_i

, A IDEAL

i

+

i

−

THE UNITY GAIN BUFFER – IDEAL OP-AMP ASSUMPTION

v

s

v

₊

=

+

−

= v v

v

OUT

= v

₋

OUT S

v = v

v

OUT

+

−

1

OUT

S

v

⇒ v =

USING LINEAR (NON-IDEAL) OP-AMP MODEL WE OBTAINED

1 1

out s i

O O i

V V R

R A R

=

+ +

PERFORMANCE OF REAL OP-AMPS Op-Amp BUFFER GAIN LM324 0.99999 LMC6492 0.9998 MAX4240 0.99995 IDEAL OP-AMP ASSUMPTION YIELDS EXCELLENT APPROXIMATION!

WHY USE THE VOLTAGE FOLLOWER OR UNITY GAIN BUFFER?

v

s

v

₊

=

+

−

= v v

= v

−

v

_O

S O

v v =

S

O

v

v =

THE VOLTAGE FOLLOWER ACTS AS BUFFER AMPLIFIER

THE SOURCE SUPPLIES POWER

THE SOURCE SUPPLIES NO POWER THE VOLTAGE FOLLOWER ISOLATES ONE CIRCUIT FROM ANOTHER

ESPECIALLY USEFUL IF THE SOURCE HAS VERY LITTLE POWER

CONNECTION WITHOUT BUFFER CONNECTION WITH BUFFER

LEARNING EXAMPLE

s out

V G = V GAIN

THE DETERMINE

= 0 v

+

= 0

∴

=

⇒

∞

= v

₊

v

₋

v

₋

A

_o

= 0 v

−

= 0

=

⇒

∞

= i

₋

i

₊

R

_i

0 0 0

2 1

− =

− +

R V R

V

_{s out}

v

-

@ KCL APPLY

= 0 i

− 1

2

R R V

G V

s

out

= −

=

FOR COMPARISON, NEXT WE EXAMINE THE SAME CIRCUIT WITHOUT THE ASSUMPTION OF IDEAL OP-AMP

REPLACING OP-AMPS BY THEIR LINEAR MODEL

WE USE THIS EXAMPLE TO DEVELOP A PROCEDURE TO DETERMINE OP-AMP CIRCUITS USING THE LINEAR MODELS

1. Identify Op Amp nodes

v

₋

v

₊

v

o

2. Redraw the circuit cutting out the Op Amp

v

₋

v

₊

v

o

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

v

₋

v

₊

v

o

Ri

RO

+

− A v( ₊ −v₋)

4. Redraw as needed

R

2

v

₋

v

₊

INVERTING AMPLIFIER: ANALYSIS OF NON IDEAL CASE

NODE ANALYSIS

CONTROLLING VARIABLE IN TERMS OF NODE VOLTAGES

USE LINEAR ALGEBRA

Ω

= Ω

=

= 10

, 10

, 10

8

5

O

i

R

R

A : AMP -

OP

TYPICAL

₁

= 1 Ω ,

₂

= 5 Ω ⇒ = − 4 . 9996994

S O

v k v

R k

R = ∞ ⇒ = − 5 . 000

S O

v

A v

= 0 i

−

= 0

=

⇒

∞

= i

₋

i

₊

R

_i

− +

=

⇒

∞

= v v

A

= 0 v

+

= 0 v

−

KCL @ INVERTING TERMINAL

0 0 0

2 1

− =

− +

R v R

v

_{S O}

1 2

R R v

v

s

O

= −

⇒

THE IDEAL OP-AMP ASSUMPTION PROVIDES EXCELLENT APPROXIMATION.

(UNLESS FORCED OTHERWISE WE WILL ALWAYS USE IT!) GAIN FOR NON-IDEAL CASE

SUMMARY COMPARISON: IDEAL OP-AMP AND NON-IDEAL CASE

NON-IDEAL CASE

REPLACE OP-AMP BY LINEAR MODEL SOLVE THE RESULTING CIRCUIT WITH DEPENDENT SOURCES

LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER

THE OP-AMP IS DEFINED BY ITS 3 NODES. HENCE IT NEEDS 3 EQUATIONS

THINK NODES!

KCL AT V_ AND V+ YIELD TWO EQUATIONS

(INFINITE INPUT RESISTANCE IMPLIES THAT i-, i+ ARE KNOWN)

OUTPUT CURRENT IS NOT KNOWN

DON’T USE KCL AT OUTPUT NODE. GET THIRD EQUATION FROM INFINITE GAIN ASSUMPTION (v+ = v-)

LEARNING EXAMPLE: DIFFERENTIAL AMPLIFIER

NODES @ INVERTING TERMINAL

NODES @ NON INVERTING TERMINAL

IDEAL OP-AMP CONDITIONS

2 4 3

4 2

4 3

0

4

v

R R

v R R v

R v R

i ⇒ = +

= +

⇒

=

_{+ −}

+

 

 



   −

 



 +

=

 −



 



 +

=

_{− − 1}

2 1 1

2 1

1 2 1

2

1

1 v v

R R R

v R R v R

R v

_O

R

) (

,

_{2 1}

1 2 1

3 2

4

v v

R v R

R R

R

R = = ⇒

_O

= −

LEARNING EXAMPLE: USE IDEAL OP-AMP

v

O

+

FIND

v

1

1−

v

2+

v

2−

v

1

v

o

2

v

o 1

v

m

2

v

m

6 NODE EQUATIONS + 2 IDEAL OP-AMP

2 2

1 1

v v

v v

=

=

+ +

2 2

1 1

m m

v v

v v

=

=

−

−

FINISH WITH INPUT NODE EQUATIONS…

USE INFINTE GAIN ASSUMPTION

− +

−

+

=

_{1 2}

=

₂

1

v v v

v

2 2

1

1

v v v

v

_m

=

_m

=

∴

USE REMAINING NODE EQUATIONS

0 0 :

@

1 2 1

2 1

2 01 1

1

− + =

− +

− +

R v v

R v v

R v

v v

^o

G m

0 0

:

@

2 2 1

2

1 2 2

2

− + − + + =

R v R

v v

R v v v

G o

m

ONLY UNKWONS ARE OUTPUT NODE VOLTAGES VARIABLE 1

REQUIRED FOR

SOLVE v_o = v_o

LEARNING EXTENSION

FIND I

_O

. ASSUME IDEAL OP - AMP V

v

₊

= 12 V

v

A

_O

= ∞ ⇒

₋

= 12

= 0

⇒

∞

= i

₋

R

_i

V v

₋

= 12

2 0 12 12

: 12 − + =

−

k k

v V

^o

KCL@ ⇒ V

_o

= 84 V

k mA I

_O

V

^o

8 . 4

10 =

=

∴

“inverse voltage divider”

i

i

v

R R v R

R v R

v R

1 2 1

0 0

2 1

1

+

= + ⇒

=

= 0 i

−

INFINITE INPUT RESISTANCE NONINVERTING AMPLIFIER - IDEAL OP-AMP LEARNING EXTENSION

v

+

v

_ ^o

v

SET VOLTAGE

v

₊

= v

₁

INFINITE GAIN ASSUMPTION

1

1 v v

v

v ₊ = ⇒ ₋ =

R

₂

R

₁

v

i

v

₋

=

v 0

FIND GAIN AND INPUT RESISTANCE - NON IDEAL OP-AMP

NOW RE-DRAW CIRCUIT TO ENHANCE CLARITY. THERE ARE ONLY TWO LOOPS DETERMINE EQUIVALENT CIRCUIT USING LINEAR MODEL FOR OP-AMP

v

₊

v

₋

v

O

Ri

− +

R

O

( )

A v

₊

− v

₋

COMPLETE EQUIVALENT FOR MESH ANALYSIS

− + v

O

v

i

= v

₊

− v

₋

MESH 1

MESH 2

CONTROLLNG VARIABLE IN TERMS OF LOOP CURRENTS

) (

_{1 2}

1 2

2

i R i i

R

v

_O

= − + −

INPUT RESISTANCE

1 1

i

R

_in

= v

^GAIN

i O

v G = v

MESH 1

MESH 2

CONTROLLNG VARIABLE IN TERMS OF LOOP CURRENTS

MATHEMATICAL MODEL

) (

_{1 2}

1 2

2

i R i i

R

v

_O

= − + −

REPLACE AND PUT IN MATRIX FORM

 

 



= 

 

 



 



 





+ +

−

− +

) 0 (

)

(

₁

2 1 2

1 1

1 2

1

v

i i R

R R

R AR

R R

R

O i

 

 



 



 





+ +

−

−

= +

 

 





⁻

0 )

( )

(

¹ ₁

2 1

1

1 2

1 2

1

v

R R

R R

AR

R R

R i

i

O i

THE FORMAL SOLUTION

) (

) )(

( R

₁

+ R

₂

+ R

_O

R

₁

+ R

₂

+ R

₁

AR

_i

− R

₁

=

∆

 

 





+

−

−

+

= +

) (

) (

) (

2 1

1

1 2

1

R R

R AR

R R

R Adj R

i

O

 

 



 



 





+

−

−

+ +

= ∆

 

 





0 ) (

) (

)

1 (

₁

2 1

1

1 2

1 2

1

v

R R

R AR

R R

R R

i i

i

O

THE SOLUTIONS

1 2

1

1

R R R v

i

^O

∆ +

= +

∆

−

= − (

₁

)

2

R i AR

ⁱ

∞ ???

→ A

1 1 2

1 1

2 1

1

2 2 1

1 1

) )(

( )

(

) (

R v AR

R v R

R R

R R

i R R

i R v

i O

O

∆

− + +

∆ +

= +

+

−

=

R

i

AR

A → ∞ ⇒ ∆ →

₁

R

_in

→ ∞

1 2 1

1

R

R R

v

G = v

^O

→ +

A SEMI-IDEAL OP-AMP MODEL

This is an intermediate model, more accurate than the ideal op-amp model but simpler than the linear model used so far

, 0,

i O O

R = ∞ R = A = A

Non-inverting amplifier and semi-ideal model

0 i

₊

i

₋

⇒ = = v

₊

≠ v

₋

!!

Replacement Equation

( )

O O e O

v = A v = A v

₊

− v

₋

v

₊

v

₋

e in

v = v

2 1

1

(as before)

( ) (replaces )

S

O

O S O

v v

R R

v v

R

A v v v v v

+

−

− + −

=

= +

− = =

1

! 2

1 ;

O O

S O

v A R

v A β R R

⇒ = β =

+ +

actual gain-ideal gain ideal gain

GE = ₁ ¹

A

O

β

= +

i S

− + v

O

R

+

−

+ -

v S

Sample Problem

Find the expression for Vo. Indicate where and how you are using the Ideal OpAmp assumptions

Set voltages?

v S

v ₊ =

Use infinite gain assumption v

₋

= v

_S

v

−

Use infinite input resistance assumption and apply KCL to inverting input

= 0 + −

⁻

R v i

_S

v

^o

S S

o

v Ri

v = −

= 0 i

−

v

+

i

S

− + v

O

R

+

−

Sample Problem

1. Locate nodes 2. Erase Op-Amp

3. Place linear model

4. Redraw if necessary

+ -

v

o

v

+

v

−

_R

R

_O

R

_i

i

_S

A(v

₊

- v

_-

)

TWO LOOPS. ONE CURRENT SOURCE. USE MESHES DRAW THE LINEAR EQUIVALENT CIRCUIT AND WRITE THE LOOP EQUATIONS

i

1 ²

i

MESH 1

i

₁

= i

_s

MESH 2

R

_i

( i

₂

− i

_S

) + ( R + R

_O

) i

₂

+ A ( v

₊

− v

_{_}

)

CONTROLLING VARIABLE

v

₊

− v

_{_}

= R

_i

( i

₂

− i

₁

) R

i

− +

R

O

( )

A v

₊

− v

₋

S

= V

V

O

AND GAIN

FIND

V

S

v

₊

=

V

S

v

_{_}

=

= 0 i

−

“INVERSE VOLTAGE DIVIDER”

V

O

V

S ²

R

R

1 S

O

V

k k V k

1 1 100 +

=

= 101

=

S O

V G V

V V

mV

V

_S

= 1 ⇒

_O

= 0 . 101

LEARNING EXTENSION

LEARNING EXAMPLE UNDER IDEAL CONDITIONS BOTH CIRCUITS SATISFY

1 2

8 4

V

O

= V − V

1 2

If 1 2 , 2 3

dc supplis are 10

V V V V V V

V

≤ ≤ ≤ ≤

±

DETERMINE IF BOTH IMPLEMENTATIONS PRODUCE THE FULL RANGE FOR THE OUTPU

1 2

X

2

V = V − V

1 2

1V ≤V ≤ 2 , 2V V ≤V ≤ 3V ⇒ −1V ≤V_X ≤ 2V

V

_X

OK!

O

4

X

V = V

1 V V

_X

2 V 4 V V

_O

8 V

− ≤ ≤ ⇒ − ≤ ≤ V OK

_O

!

8

1

V

Y

= − V

1 V ≤ V

1

≤ 2 V ⇒ − 16 V ≤ V

_Y

≤ − 8 V

EXCEEDS SUPPLY VALUE.

THIS OP-AMP SATURATES!

POOR IMPLEMENTATION

COMPARATOR CIRCUITS

Some REAL OpAmps require a “pull up resistor.”

ZERO-CROSSING DETECTOR

LEARNING BY APPLICATION

OP-AMP BASED AMMETER

NON-INVERTING AMPLIFIER

1

1

2

R G = + R

I R V

_I

=

_I

I R R

GV R

V

_{O I}

 

_I

 



 +

=

=

1

1

2

1

^B

4 (design eq.)

A

R + R =

LEARNING EXAMPLE

DC MOTOR CONTROL - REVISITED

CHOOSE NON-INVERTING AMPLIFIER (WITH POWER OP-AMP PA03)

Constraints:

V

_M

≤ 20 V

Power dissipation

in amplifier

≤ 100mW

Simplifying assumptions:

R

_i

≈ ∞ , R

_O

≈ ⇒ 0

Significant power losses Occur only in Ra, Rb

Worst case occurs when Vm=20

(20 )

2 MX

100

A B

P V mW

R R

= ≤

+ ⇒ R

_A

+ R

_B

≥ 4000 Ω 3

B

A

R R =

One solution:

R

_B

= 3 k Ω , R

_A

= 1 k Ω

Standard values at 5%!

DESIGN EXAMPLE: INSTRUMENTATION AMPLIFIER

DESIGN SPECIFICATIONS

1 2

10 V

O

G = V V =

−

• “HIGH INPUT RESISTENCE”

• “LOW POWER DISSIPATION”

• OPERATE FROM 2 AA BATTERIES

ANALISIS OF PROPOSED CONFIGURATION

O X Y

V = V − V

1 2 1

1

@ :V V V V^X 0

A R R

− + − =

2 1 2

2

@ :V V V V^Y 0

B R R

− + − =

1 2 2 1

1

1

2

1

1 2

O

R R R R

V V V V V

R R R R

   

=   +   −   +   + −

SIMPLIFY DESIGN BY MAKING R1 = R2 ¹

1 2

1 2 ( )

O

V R V V

R

 

⇒ =   +   −

DESIGN EQUATION:

2 R

₁

= 9 R

VO

V1

V2

G

USE LARGE RESISTORS FOR LOW POWER

e g R . ., = 100 k Ω , R

1

= R

2

= 450 k Ω

1

;

2

A B

V = V V = V

Infinite gain

MAX4240

DESIGN EXAMPLE

DESIGN SPECIFICATION ^O 10

in

V V =

Power loss in resistors should not exceed

100mW

_{2 2}

V Vin V

− ≤ ≤

when

Design equationS: ²

1

1 10

O

in

V R

V = + R =

1 2

2

1 2

(20 )

R R

100

P V mW

R R

+

≤ =

+

Max Vo is 20V

Solve design equations (by trial and error if necessary)

2

9

1

R R

⇒ =

1 2

4

R R k

⇒ + ≥ Ω

1

400

R ≥ Ω

DESIGN EXAMPLE

IMPLEMENT THE OPERATION

V

_O

= − 0.9 V

₁

− 0.1 V

₂

DESIGN CONSTRAINTS

• AS FEW COMPONENTS AS POSSIBLE

• MINIMIZE POWER DISSIPATED

• USE RESISTORS NO LARGER THAN 10K Given the function (weighted sum with sign change) a basic weighted adder may work

ANALYSIS OF POSSIBLE SOLUTION

1 2

1 1 2

0

@ :

^O

0

V

V V V

V R R R

+

−

=

− − − =

^{1 2}

1 2

O

R R

V V V

R R

⇒ = − −

2 1

2

9 0.1

R R

R R

=

=

DESIGN EQUATIONS

2 1

R > R > R

SOLVE DESIGN EQUATIONS USING TRIAL AND ERROR IF NECESSARY

2

10 ,5.6 ,...

R = k k

ANALYZE EACH SOLUTION FOR OTHER CONSTRAINTS AND FACTORS; e.g.

DO WE USE ONLY STANDARD COMPONENTS?

DESIGN EXAMPLE

DESIGN 4-20mA TO 0 – 5V CONVERTER 1. CONVERT CURRENT TO VOLTAGE USING A RESISTOR

I

V

^I

= RI

⇒

CANNOT GIVE DESIRED RANGE!

2. CHOOSE RESISTOR TO PROVIDE THE 5V CHANGE

… AND SHIFT LEVELS DOWN!

5 0

312.5 0.020 0.004

MAX MIN

MAX MIN

V V

R I I

− −

= = = Ω

− −

1.25 ≤ V

_I

≤ 6.25

MUST SHIFT DOWN BY 1.25V (SUBSTRACT 1.25 V)

2

1 2

I

V R V

R R

+

=

+

1

1 2

(

_{O SHIFT}

)

_SHIFT

V R V V V

R R

−

= − +

+

( )

2

1

O I SHIFT

V R V V

= R −

V

₊

= V

₋

⇒

DOES NOT LOAD PHONOGRAPH

1000 OF

TION AMPLIFICA

AN PROVIDES IT

THAT SO

DETERMINE R

₂

, R

₁

LEARNING BY DESIGN

) 1

)(

1 (

1 2

1

R

R V

V

_O

= +

LEARNING EXAMPLE

UNITY GAIN BUFFER

COMPARATOR CIRCUITS

T

T

e

R = 57 . 45

^−0.0227

ONLY ONE LED IS ON AT ANY GIVEN TIME

MATLAB SIMULATION OF TEMPERATURE SENSOR

WE SHOW THE SEQUENCE OF MATLAB INSTRUCTIONS USED TO OBTAIN THE PLOT OF THE VOLTAGE AS FUNCTION OF THE TEMPERATURE

»T=[60:0.1:90]'; %define a column array of temperature values

» RT=57.45*exp(-0.0227*T); %model of thermistor

» RX=9.32; %computed resistance needed for voltage divider

» VT=3*RX./(RX+RT); %voltage divider equation. Notice “./” to create output array

» plot(T,VT, ‘mo’); %basic plotting instruction

» title('OUTPUT OF TEMPERATURE SENSOR'); %proper graph labeling tools

» xlabel('TEMPERATURE(DEG. FARENHEIT)')

» ylabel('VOLTS')

» legend('VOLTAGE V_T')

v

1

v

₊

=

= 0 i

−

v

1

v

₋

=

KCL @ v_

EXAMPLE OF TRANSFER CURVE SHOWING SATURATION

THE TRANSFER CURVE

IN LINEAR RANGE THIS SOURCE CREATES THE OFFSET

OFFSET

OUTPUT CANNOT EXCEED SUPPLY (10V)