1. operational amplifier.ppt

7/29/2019 1. Operational Amplifier.ppt

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Operational Amplifier 1

Op-Amp


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Operational Amplifier

An operational amplifier ("op-amp") is

a DC-coupled high-gain electronic

voltage amplifier with a differentialinput and, usually, a single-ended output.

An op-amp produces an output voltage that

is typically hundreds of thousands timeslarger than the voltage difference between

its input terminals

Ref:080114HKN Operational Amplifier 2


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Applications

audio- and video-frequency pre-amplifiers and buffers

differential amplifiers

differentiators and integrators

filters precision rectifiers

precision peak detectors

voltage and current regulators

analog calculators

analog-to-digital converters digital-to-analog converters

voltage clamps

oscillators and waveform generators



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Vd

+

Vo

Rin~inf Rout~0

Input 1

Input 2

Output

+Vcc

-Vcc

Characteristics

Very high differential gain

High input impedance

Low output impedance

Provide voltage changes(amplitude and polarity)

Used in oscillator, filter

and instrumentation

Accumulate a very high

gain by multiple stages

ddo VGV

510saylarge,yver

normallygainaldifferenti:dG


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IC Product

+

1

2

3

4

8

7

6

5

OFFSET

NULL

-IN

+IN

V

N.C.

V+

OUTPUT

OFFSET

NULL

+

1

2

3

4

8

7

6

5

OUTPUT A

-IN A

+IN A

V

V+

OUTPUT B

-IN B

+IN B+

DIP-741 Dual op-amp 1458 device


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Single-Ended Input

+

Vo

~Vi

+

Vo

~Vi

+ terminal : Source

terminal : Ground

0o phase change

+ terminal : Ground

terminal : Source 180o phase change


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Double-Ended Input

~

V1

+

Vo

~V2

+

Vo

~Vd

Differential input

0o phase shift change

between Vo and Vd

VVVd

Qu: What Vo should be if,

V1

V2

(A) (B)

Ans: (A or B) ?


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Distortion

Vd

+

Vo

+Vcc=+5V

Vcc=5V0

+5V

5V

The output voltage never excess the DC

voltage supply of the Op-Amp


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Common-Mode Operation

+

Vo

Vi ~

Same voltage source is applied

at both terminals

Ideally, two input are equally

amplified

Output voltage is ideally zero

due to differential voltage is

zero

Practically, a small output

signal can still be measured

Note for differential circuits:

Opposite inputs : highly amplifiedCommon inputs : slightly amplified

Common-Mode Rejection


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Common-Mode Rejection Ratio (CMRR)

Differential voltage input :

VVVd

Common voltage input :

)(21 VVVc

Output voltage :

ccddo VGVGV Gd : Differential gain

Gc : Common mode gain

)dB(log20CMRR 10c

d

c

d

G

G

G

G

Common-mode rejection ratio:

Note:

When Gd >> Gc or CMRRVo = GdVd

+

NoninvertingInput

InvertingInput

Output


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CMRR ExampleWhat is the CMRR?

Solution :

dBCMRRand

V(2)From

V(1)From

VV

VV

40)10/1000log(20101000

607007060

806006080

702

4010060

2

20100

60401008020100

21

21

cd

cdo

cdo

cc

dd

GG

GGV

GGV

VV

VV

+

100V

20V

80600V

+

100V

40V

60700V

(1) (2)


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Op-Amp Properties(1) Infinite Open Loop gain

- The gain without feedback- Equal to differential gain

- Zero common-mode gain

- Pratically, Gd = 20,000 to 200,000

(2) Infinite Input impedance- Input current ii ~0A

- T- in high-grade op-amp- m-A input current in low-grade op-amp

(3) Zero Output Impedance

- act as perfect internal voltage source- No internal resistance

- Output impedance in series with load

- Reducing output voltage to the load

- Practically,Rout ~ 20-100

+

V1

V2 Vo

+

Vo

i1~0

i2~0

+

Rout

Vo'Rload

outload

loadoload

RR

RVV


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Frequency-Gain Relation

Ideally, signals are amplifiedfrom DC to the highest AC

frequency

Practically, bandwidth is limited

741 family op-amp have a

limited bandwidth of few KHz.

Unity Gain frequencyf1: the

gain at unity

Cutoff frequencyfc: the gain

drop by 3dB from dc gain Gd

(Voltage Gain)

(frequency)

f1

Gd

0.707Gd

fc0

1

GB Product :f1 = Gdfc

20log(0.707)=3dB


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GB ProductExample: Determine the cutoff frequency of an op-amp

having a unit gain frequencyf1 = 10 MHz and voltage

differential gain Gd = 20V/mV

Sol:

Sincef1 = 10 MHzBy using GB production equation

f1 = Gdfc

fc=

f1/

Gd= 10 MHz / 20 V/mV

= 10 106 / 20 103

= 500 Hz

(Voltage Gain)

(frequency)

f1

Gd

0.707Gd

fc0

1

10MHz

? Hz


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Ideal Vs Practical Op-Amp

Ideal Practical

Open Loop gainA 105

BandwidthBW 10-100Hz

Input ImpedanceZin >1M

Output ImpedanceZout 0 10-100

Output Voltage Vout Depends onlyon V

d

= (V+

V

)

Differential

mode signal

Depends slightly

on average input

Vc = (V++V)/2

Common-Mode

signal

CMRR 10-100dB

+

~

AVinVin Vout

Zout=0

I deal op-amp

+

AVinVin Vout

Zout

~Zin

Practical op-amp


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Ideal Op-Amp Applications

Analysis Method :Two ideal Op-Amp Properties:

(1) The voltage between V+ and V is zero V+ = V

(2) The current into both V+

and V

terminals is zero

For ideal Op-Amp circuit:

(1) Write the kirchhoff node equation at the non-invertingterminal V+

(2) Write the kirchhoff node equation at the invertingterminal V

(3) Set V+ = V and solve for the desired closed-loop gain


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Noninverting Amplifier

(1) Kirchhoff node equation at V+yields,

(2) Kirchhoff node equation at V

yields,

(3) Setting V+ = V yields

or

+VinVo

Ra Rf

iVV

00

f

o

a R

VV

R

V

0

f

oi

a

i

R

VV

R

V

a

f

i

o

R

R

V

V1


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+

vi

Ra

vov-

v+

Rf

+

vi

Ra

vov-

v+

Rf

R2R1

+

vivo

v-

v+

Rf

+

vivo

v-

v+

Rf

R2R1

Noninverting amplifier Noninverting input with voltage divider

i

a

f

o vRR

R

R

Rv ))(1(

21

2

Voltage follower

io vv

i

a

f

o v

R

Rv )1(

Less than unity gain

io vRR

Rv

21

2


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Inverting Amplifier



yields,


0

V

0_

f

o

a

in

R

VV

R

VV

a

f

in

o

R

R

V

V

Notice: The closed-loop gainVo/Vin is

dependent upon the ratio of two resistors,and is independent of the open-loop gain.

This is caused by the use of feedback output

voltage to subtract from the input voltage.

+

~

Rf

Ra

VinVo


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Multiple Inputs



yields,


0V

0_

c

c

b

b

a

a

f

o

R

VV

R

VV

R

VV

R

VV

c

aj j

j

f

c

c

b

b

a

afo

R

VR

R

V

R

V

R

VRV

+

Rf

Va

VoRb

Ra

RcVb

Vc


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Inverting IntegratorNow replace resistorsRa andRfby complex

componentsZa

andZf, respectively, therefore

Supposing

(i) The feedback component is a capacitor C,

i.e.,

(ii) The input component is a resistor R,Za =RTherefore, the closed-loop gain (Vo/Vin) become:

where

What happens ifZa = 1/jCwhereas,Zf=R?Inverting differentiator

in

a

f

o VZ

ZV

CjZf

1

dttvRC

tv io )(1

)(

tjii eVtv )(

+

~

Zf

Za

VinVo

+

~

R

Vin Vo

C


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Op-Amp IntegratorExample:

(a) Determine the rate of change

of the output voltage.

(b) Draw the output waveform.

Solution:

+

RVoVi

0

+5V

10 k

0.01FC

Vo(max)=10 V

100s

(a) Rate of change of the output voltage

smV/

Fk

V

50

)01.0)(10(

5

RC

V

t

V io

(b) In 100 s, the voltage decrease

VssmV/ 5)100)(50( oV

0

+5V

0-5V

-10V

Vi

Vo


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Op-Amp Differentiator

RCdt

dVv io

+

R

C

VoVi0to t1 t2

0

to t1 t2


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Non-ideal case (Inverting Amplifier)

+

~

Rf

Ra

VinVo

3 categories are considering

Close-Loop Voltage Gain

Input impedance Output impedance

Equivalent CircuitRf

Ra

Vin Vo+

+

R RV

-AV

+

AVinVin Vout

Zout

~

Zin

Practical op-amp


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Close-Loop GainRf

RaVin

Vo+

+

R R

-AV

Ra Rf

R

V

V

Vin Vo

Applied KCL at V terminal,

0

f

o

a

in

R

VV

R

V

R

VV

By using the open loop gain,

AVVo

0f

o

f

oo

a

o

a

in

AR

V

R

V

AR

V

AR

V

R

V

fa

aafaf

o

a

in

RRAR

RARRRRRRRV

R

V

The Close-Loop Gain,Av

RARRRRRRR

RAR

V

VA

aafaf

f

in

ov


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Close-Loop Gain

When the open loop gain is very large, the above equation become,

a

f

vR

RA

~

Note : The close-loop gain now reduce to the same formas an ideal case


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Input ImpedanceRf

Ra

VinVo

+

+

R

-AV

R'

RV

Rf

+

R

-AV

if

V

Input Impedance can be regarded as,

RRRR ain //

whereR is the equivalent impedanceof the red box circuit, that is

fi

VR

However, with the below circuit,

A

RR

i

VR

RRiAVV

of

f

off

1

)()(


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Input ImpedanceFinally, we find the input impedance as,

1

11

of

ainRR

A

RRR

RARR

RRRRR

of

of

ain)1(

)(

Since, RARR of )1( , Rin become,

)1(

)(~

A

RRRR

of

ain

Again with )1( ARR of

ain RR ~

Note: The op-amp can provide an impedance isolated from

input to output


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Output ImpedanceOnly source-free output impedance would be considered,

i.e. Vi is assumed to be 0 Rf

Ra

Vo+

R

-AV

RV

io

VV

Rf

Ra R+

R

-AV

V

i2 i1

(a)(b)

Firstly, with figure (a),

o

fafa

ao

af

a VRRRRRR

RRVV

RRR

RRV

//

//

By using KCL,io

= i1

+ i2

o

o

faf

oo

R

AVV

RRR

Vi

)(

//

By substitute the equation from Fig. (a),

RRARRRRRRRRRRRRRR

iV

R

afafao

fafao

o

o

out

)1())(1()(

isimpedance,outputThe

R andA comparably large,

a

fao

outAR

RRRR

)(~

1. operational amplifier.ppt

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