Give to Receive

ECEN 301 Discussion #18 – Operational Amplifiers 1

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.


Lecture 18 – Operational Amplifiers

Continue with Different OpAmp configurations


Op-Amps – Open-Loop Model1. 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


Op-Amps – Open-Loop Model1. How can v– ≈ v+ when vo is amplifying (v+ - v-) ? 2. How can an opAmp form a closed circuit when (i1 = i2 = 0) ?

–

+

+–

RoutRin

i1

AOLvin

+

vo

–

–

vin

+

v–

v+

vAvv

vvA

vAv

OL

o

OL

inOLo

)(i2

Ideally i1 = i2 = 0(since Rin → ∞)

What happens as AOL → ∞ ?→ v– ≈ v+


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

RAv

RAv

Rv

Rv

RAv

Rv

RAv

Rv

Rvv

Rvv

ii

1/

1/1

//

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



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

RRvv

CLA

:Gain Loop-Closed

NB: if AOL is NOT the same thing as ACL

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



–+

va RR

–+

vb R R

–+

R1R

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



–+

va RR

–+

vb R R

–+

R1R

R2

vo

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


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


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


–+

R1R

R2

vo

-va

-vb

21

21

21

21

21

000

Rv

RvRv

Rv

Rv

Rv

Rv

Rv

Rv

Rvv

Rvv

Rvv

iii

bao

oba

oba

oba

F

iF

i1

i2


More OpAmp Configurations


Op-Amps – Closed-Loop ModeThe 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




021

iivv

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




SF

Fs RR

Rvv

2

:Divider VoltageRF

–

+ +

vo

–

i1

vS2(t) +–

v+

v–

iFRS

iS

i2

vS1(t)+–

RF

RS

S

SS

SSS

Rvvi

Rivvv

1

1

F

oF R

vvi




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

vvRR

RRRRv

RRv

RvRv

vRi

vRiv


Op-Amps – Level ShifterLevel 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


Op-Amps – Level ShifterExample1: 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




–

+ +

vo

–

RF

Vref+–

v+

v–RSVsensor

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




–

+ +

vo

–

RF

v+

v–RS

Vsensor+–

DC from sensor:Inverting amplifier

senDCS

F

senDCCLosen

vRRvAv




–

+ +

vo

–

RF

v+

v–RS

Vref+–

DC from reference:Noninverting amplifier

refS

F

refCLoref

vRR

vAv

1




refS

FsenDC

S

F

orefosenoDC

vRRv

RR

vvv

1

–

+ +

vo

–

RF

Vref+–

v+

v–RSVsensor




V

RRRRvv

vRRv

RR

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:


Op-Amps – Ideal IntegratorThe 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


Op-Amps – Ideal IntegratorThe 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

CRtv

dttdv

dttvtvdC

Rtv

titi

)(1)(

)()(

)()()(

)()(

dttdvCti )()( Recall


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


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)(




–

+ +

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




–

+ +

vo(t)

–

i1

CF

RS

vS(t) +–

v+

v–iF(t)

iS(t)

)(100

1002

100

)(1002

100

)(12

)(

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/




–

+ +

vo(t)

–

i1

CF

RS

vS(t) +–

v+

v–iF(t)

iS(t)

timevo(t)

T/2 T

-50AT


Op-Amps – Ideal DifferentiatorThe 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


Op-Amps – Ideal Differentiator

dttdvCRtv

Rtv

dttvtvdC

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)

dttdvCti )()( Recall


Op-Amps – Closed-Loop ModeCircuit Diagram ACL

Inverting Amplifier

Summing Amplifier

N

nSn

n

F

N

nSnOLno

vRR

vAv

1

1

–+ +

vo

–

+–

vS

RS

RF

SS

F

SCLo

vRRvAv

–+ +

vo

–

+–

+–

+–

RSn

RS2

RS1

vSn

vS2

vS1

RF



Noninverting Amplifier

Voltage Follower

s

sCLo

vvAv

SS

F

SCLo

vRR

vAv

1

–+ +

vo

–

+–

R

RS

RF

vS

–+ +

vo

–+–



Differential Amplifier

12 SSS

Fo vv

RRv

–+ +

vo

–+–

+–

RS

RS

RF

RF

v1

v2



Ideal Integrator

Ideal Differentiator

dttdvCRv S

SFo)(

t

SSF

o dvCR

v )(1

–

+ +vo(t)

–

+–

vS

CS

RF

–+ +

vo(t)–

+–

vS

RS

CF


Op-Amps Example 3: 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)


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

31

621

21

631

111111

0

0:aat KCL

1221

21

21

Soa

SF

oF

a

a

F

aoaS

F

VjVjV

ZV

ZZV

ZZZV

ZVV

ZVV

ZVV

III

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

NB: v+ = v– = vo

and Iin = 0

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


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

021

621

0111

000

:bat KCL

22

2

2

jVV

ZZV

ZV

ZV

ZVV

III

oa

Soa

S

ooa

inS

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



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

656

2

jVV S

o



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)

656

2

jVV

S

o



Op-Amps

+

–

Vo(jω)Iin

ZF=1/jωCF

Z2 v+

v–

IF(jω)

I2(jω)

Z1

I1(jω)ZS

IS(jω)

Vs(jω)


Frequency Response

2nd order Lowpass filter

Instrumentation Op-Amp• Special configuration used for common-mode

voltage rejection)(21 12 VV

RgainRVout

Advantages:- Common-mode voltage rejection- Very high input impedance for V1 and V2

- Set gain with one resistor

Connect sensor with twist pair in differential configuration to minimizeexternal noise pickup. Using groundedshield also helps.

Give to Receive

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