ECEN 301Discussion #18 – Operational Amplifiers1 DateDayClass No. TitleChaptersHW Due date Lab Due...

ECEN 301Discussion #18 – Operational

Amplifiers1

Date Day ClassNo.

Title Chapters HWDue date

LabDue date

Exam

3 Nov Mon 18 Operational Amplifiers

8.4

LAB 6

4 Nov Tue

5 Nov Wed 19 Binary Numbers 13.1 – 13.2

6 Nov Thu

7 Nov Fri Recitation HW 8

8 Nov Sat

9 Nov Sun

10 Nov Mon 20 Exam Review

LAB 7 EXAM 212 Nov Tue

Schedule…


Amplifiers2

Give to ReceiveAlma 34:28 28 And now behold, my beloved brethren, I say unto you, do not

suppose that this is all; for after ye have done all these things, if ye turn away the needy, and the naked, and visit not the sick and afflicted, and impart of your substance, if ye have, to those who stand in need—I say unto you, if ye do not any of these things, behold, your prayer is vain, and availeth you nothing, and ye are as hypocrites who do deny the faith.


Amplifiers3

Lecture 18 – Operational Amplifiers

Answer questions from last lecture

Continue with Different OpAmp configurations


Amplifiers4

Op-Amps – Open-Loop Model

1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?

–

++

v+

–

+

v–

–

+

vo

–

io

i2

i1–vin

+

–

+

+–

RoutRin

i1

AOLvin

+

vo

–

–

vin

+

v–

v+

NB: op-amps have near-infinite input resistance (Rin) and very small output resistance (Rout)

)(

vvA

vAv

OL

inOLoAOL – open-loop voltage gain

i2


Amplifiers5

Op-Amps – Open-Loop Model


–

+

+–

RoutRin

i1

AOLvin

+

vo

–

–

vin

+

v–

v+

vA

vv

vvA

vAv

OL

o

OL

inOLo

)(i2

Ideally i1 = i2 = 0(since Rin → ∞)

What happens as AOL → ∞ ?→ v– ≈ v+


Amplifiers6

Op-Amps – Closed-Loop Mode

RF

–

+ +

vo

–

i1

RS

vS(t)+–

v+

v–

iF

iS

OLSFOLSFoS

SOL

o

FOL

o

S

S

S

S

F

OLo

F

o

S

OLo

S

S

F

o

S

S

FS

ARRARRvv

RA

v

RA

v

R

v

R

v

R

Av

R

v

R

Av

R

v

R

vv

R

vv

ii

1

/

1

/

1

//



Amplifiers7


OLSFOLSFoS ARRARRvv

1

/

1

/

1RF

–

+ +

vo

–

i1

RS

vS(t)+–

v+

v–

iF

iSNB: if AOL is very large these terms → 0

S

F

S

o

R

R

vv

CLA

:Gain Loop-Closed

NB: if AOL is NOT the same thing as ACL



Amplifiers8


–

+

va RR

–

+

vb RR

–

+

R1

R

R2

vo

1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?

2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?

NB: Current flows through R1 and R2


Amplifiers9


–

+

va RR

–

+

vb RR

–

+

R1

R

R2

vo

1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?


NB: Inverting amplifiers and (RS = RF) → vo = -vi


Amplifiers10

Op-Amps – Closed-Loop Mode1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ?


–

+

R1

R

R2

vo

-va

-vb

21

21

21

21

21

000

R

v

R

vRv

R

v

R

v

R

v

R

v

R

v

R

v

R

vv

R

vv

R

vv

iii

bao

oba

oba

oba

F

iF

i1

i2


Amplifiers11

More OpAmp Configurations


Amplifiers12


The Differential Amplifier: the signal to be amplified is the difference of two signals

–

+ +

vo

–

i1

RF

vS2(t)+–

v+

v–

iFRS

iS

i1vS1(t)

+–

RF

RS


Amplifiers13



021

ii

vv

RF

–

+ +

vo

–

i1

vS2(t)+–

v+

v–

iFRS

iS

i2

vS1(t)+–

RF

RS

NB: an ideal op-amp with negative feedback has the properties

SF ii


Amplifiers14



SF

Fs RR

Rvv

2

:Divider VoltageRF

–

+ +

vo

–

i1

vS2(t)+–

v+

v–

iFRS

iS

i2

vS1(t)+–

RF

RS

S

SS

SSS

R

vvi

Rivvv

1

1

F

oF R

vvi


Amplifiers15



RF

–

+ +

vo

–

i1

vS2(t)+–

v+

v–

iFRS

iS

i2

vS1(t)+–

RF

RS

12

221

SSS

F

FSS

FS

FS

S

S

SFo

FS

FFo

vvR

R

RRR

Rv

RR

v

R

vRv

vRi

vRiv


Amplifiers16

Op-Amps – Level Shifter

Level Shifter: can add or subtract a DC offset from a signal based on the values of RS and/or Vref

–

+ +

vo

–

RF

Vref+–

v+

v–RSVsensor

AC voltage with DC offset

DC voltage


Amplifiers17


Example1: design a level shifter such that it can remove a 1.8V DC offset from the sensor signal (Find Vref)

RS = 10kΩ, RF = 220kΩ, vs(t) = 1.8+0.1cos(ωt)

–

+ +

vo

–

RF

Vref+–

v+

v–RSVsensor


Amplifiers18




–

+ +

vo

–

RF

Vref+–

v+

v–RSVsensor

Find the Closed-Loop voltage gain by using the principle of superposition on each of the DC voltages


Amplifiers19




–

+ +

vo

–

RF

v+

v–RS

Vsensor+–

DC from sensor:Inverting amplifier

senDCS

F

senDCCLosen

vRR

vAv


Amplifiers20




–

+ +

vo

–

RF

v+

v–RS

Vref+–

DC from reference:Noninverting amplifier

refS

F

refCLoref

vRR

vAv

1


Amplifiers21




refS

FsenDC

S

F

orefosenoDC

vRR

vRR

vvv

1

–

+ +

vo

–

RF

Vref+–

v+

v–RSVsensor


Amplifiers22




V

RR

RRvv

vRR

vRR

SF

SFsDCref

refS

FsDC

S

F

714.110/2201

10/220)8.1(

/1

/

10

–

+ +

vo

–

RF

Vref+–

v+

v–RSVsensor

Since the desire is to remove all DC from the output we require:


Amplifiers23

Op-Amps – Ideal Integrator

The Ideal Integrator: the output signal is the integral of the input signal (over a period of time)

–

+ +

vo(t)

–

i1

CF

RS

vS(t)+–

v+

v–iF(t)

iS(t)

The input signal is AC, but not necessarily sinusoidal

NB: Inverting amplifier setup with RF replaced with a capacitor


Amplifiers24

Op-Amps – Ideal Integrator

The Ideal Integrator: the output signal is the integral of the input signal (over a period of time)

–

+ +

vo(t)

–

i1

CF

RS

vS(t)+–

v+

v–iF(t)

iS(t)

t

SFS

o

FS

So

oF

S

S

FS

dvCR

tv

CR

tv

dt

tdv

dt

tvtvdC

R

tv

titi

)(1

)(

)()(

)()()(

)()(


Amplifiers25

Op-Amps – Ideal IntegratorExample2: find the output voltage if the input is a square wave of amplitude

+/–A with period TT = 10ms, CF = 1uF, RS = 10kΩ

–

+ +

vo(t)

–

i1

CF

RS

vS(t)+–

v+

v–iF(t)

iS(t)

time

vs(t)

A

-A

T/2 T


Amplifiers26

Op-Amps – Ideal IntegratorExample2: find the output voltage if the input is a square wave of

amplitude +/–A with period TT = 10ms, CF = 1uF, RS = 10kΩ

–

+ +

vo(t)

–

i1

CF

RS

vS(t)+–

v+

v–iF(t)

iS(t)

t

SFS

o

t

SSFS

t

SFS

o

dvCR

v

dvdvCR

dvCR

tv

0

0

0

)(1

)0(

)()(1

)(1

)(


Amplifiers27



–

+ +

vo(t)

–

i1

CF

RS

vS(t)+–

v+

v–iF(t)

iS(t)

At

A

Ad

dvCR

vtv

t

t

t

SFS

oo

100

100

1000

)(1

)0()(

0

0

0

NB: since the vs(t) is periodic, we can find vo(t) over a single period – and repeat

2/0 Tt


Amplifiers28



–

+ +

vo(t)

–

i1

CF

RS

vS(t)+–

v+

v–iF(t)

iS(t)

)(100

1002

100

)(1002

100

)(1

2)(

2/

2/

2/

tTA

ATA

dATA

dvCR

Tvtv

t

T

t

T

t

T SFS

oo

NB: since the vs(t) is periodic, we can find vo(t) over a single period – and repeat

TtT 2/


Amplifiers29



–

+ +

vo(t)

–

i1

CF

RS

vS(t)+–

v+

v–iF(t)

iS(t)

timevo(t)

T/2 T

-50AT


Amplifiers30

Op-Amps – Ideal Differentiator

The Ideal Differentiator: the output signal is the derivative of the input signal (over a period of time)

–

+ +

vo(t)

–

i1

CS

RF

vS(t)+–

v+

v–

iF(t)

iS(t)

The input signal is AC, but not necessarily sinusoidal

NB: Inverting amplifier setup with RS replaced with a capacitor


Amplifiers31

Op-Amps – Ideal Differentiator

dt

tdvCRtv

R

tv

dt

tvtvdC

titi

SSFo

F

oSS

FS

)()(

)()()(

)()(

NB: this type of differentiator is rarely used in practice since it amplifies noise

The Ideal Differentiator: the output signal is the derivative of the input signal (over a period of time)

–

+ +

vo(t)

–

i1

CS

RF

vS(t)+–

v+

v–

iF(t)

iS(t)


Amplifiers32


Circuit Diagram ACL

Inverting Amplifier

Summing Amplifier

N

nSn

n

F

N

nSnOLno

vR

R

vAv

1

1

–

+ +vo

–

+–

vS

RS

RF

SS

F

SCLo

vRR

vAv

–

+ +vo

–

+–

+–

+–

RSn

RS2

RS1

vSn

vS2

vS1

RF


Amplifiers33


Circuit Diagram ACL

Noninverting Amplifier

Voltage Follower

s

sCLo

v

vAv

SS

F

SCLo

vR

R

vAv

1

–

+ +vo

–

+–

R

RS

RF

vS

–

+ +vo

–

+–


Amplifiers34


Circuit Diagram ACL

Differential Amplifier

12 SSS

Fo vv

R

Rv

–+ +

vo

–

+–

+–

RS

RS

RF

RF

v1

v2


Amplifiers35


Circuit Diagram ACL

Ideal Integrator

Ideal Differentiator

dt

tdvCRv SSFo

)(

t

SSF

o dvCR

v )(1

–

+ +vo(t)

–

+–

vS

CS

RF

–

+ +vo(t)

–

+–

vS

RS

CF


Amplifiers36

Op-Amps Example3: find an expression for the gain if vs(t) is sinusoidal

CF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F

+

–

vo(t)i1

CF

R2 v+

v–

iF(t)

i2(t)

R1

i1(t) CS

iS(t)

vs(t)


Amplifiers37

Op-Amps

+

–

Vo(jω)Iin

ZF=1/jωCF

Z2 v+

v–

IF(jω)

I2(jω)

Z1

I1(jω)ZS

IS(jω)

Vs(jω)

Node a

Node b

3

1

62

1

2

1

63

1

111111

0

0

:aat KCL

1221

21

21

Soa

SF

oF

a

a

F

aoaS

F

Vj

Vj

V

ZV

ZZV

ZZZV

Z

VV

Z

VV

Z

VV

III

1. Transfer to frequency domain2. Apply KCL at nodes a and b

NB: v+ = v– = vo

and Iin = 0

Example3: find an expression for the gainCF = 1/6 F, R1 = 3Ω, R2 = 2Ω, CS = 1/6 F


Amplifiers38

Op-Amps

+

–

Vo(jω)Iin

ZF=1/jωCF

Z2 v+

v–

IF(jω)

I2(jω)

Z1

I1(jω)ZS

IS(jω)

Vs(jω)

Node a

Node b

02

1

62

1

0111

00

0

:bat KCL

22

2

2

jVV

ZZV

ZV

Z

V

Z

VV

III

oa

Soa

S

ooa

inS

1. Transfer to frequency domain2. Apply KCL at nodes a and b



Amplifiers39

Op-Amps

033 jVV oa

+

–

Vo(jω)Iin

ZF=1/jωCF

Z2 v+

v–

IF(jω)

I2(jω)

Z1

I1(jω)ZS

IS(jω)

Vs(jω)

1. Transfer to frequency domain2. Apply KCL at nodes a and b3. Express Vo in terms of Vs

Soa VjVjV 235

65

62

j

VV So



Amplifiers40

Op-Amps

+

–

Vo(jω)Iin

ZF=1/jωCF

Z2 v+

v–

IF(jω)

I2(jω)

Z1

I1(jω)ZS

IS(jω)

Vs(jω)

1. Transfer to frequency domain2. Apply KCL at nodes a and b3. Express Vo in terms of VS

4. Find the gain (Vo/VS)

65

62

jV

V

S

o


ECEN 301Discussion #18 – Operational Amplifiers1 DateDayClass No. TitleChaptersHW Due date Lab Due...

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