Operational Amplifiers -Introduction - - Dunia ilmiah · · 2014-06-24An operational amplifier...

Operational Amplifiers

-Introduction -

Engr. Muhammad Muizz

By : Engr. Muhammad Muizz Bin Mohd Nawawi

Operational Amplifiers

1. An operational amplifier (op-amp), is a very high gain

differential amplifier with high input impedance and low

output impedance.

2. Provide voltage amplitude changes (amplitude and polarity),

oscillators, filter circuits, and many types of instrumentation

circuits.

3. An op-amp contains a number of differential amplifier stages

to achieve a very high voltage gain.

Basic op-amp. engr. muhammad muizz

Typical Op Amp IC Packages

Dual-in-Line package

(DIP)

Metal Can Package

engr. muhammad muizz

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



Vd

+

Vo

Rin~inf Rout~0

Input 1

Input 2

Output

+Vcc

-Vcc

Operational Amplifier (Op-Amp)

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

510say large,y ver

normally gain aldifferenti : dG

Single-Ended Input

+

Vo

~ Vi

+

Vo

~ Vi

• + terminal : Source

• – terminal : Ground

• 0o phase change

• + terminal : Ground

• – terminal : Source

• 180o phase change


● Single ended

output

Non-

inverting

input

Inverting

input

Single Ended Differential Amplifier

(commonly used in op-amps)


Double-Ended Input

~ V1

+

Vo

~ V2

+

Vo

~Vd

• Differential input

•

• 0o phase shift change

between Vo and Vd

VVVd

What Vo should be if,

V1

V2

(A) (B)

Ans: (A or B) ?

Operational Amplifier engr. muhammad muizz

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 amplified

Common inputs : slightly amplified

Common-Mode Rejection


Common-Mode Rejection Ratio (CMRR) Differential voltage input :

VVVd

Common voltage input :

)(2

1 VVVc

Output voltage :

ccddo VGVGV

Gd : Differential gain

Gc : Common mode gain

)dB(log20CMRR 10

c

d

c

d

G

G

G

G

Common-mode rejection ratio:

Note:

When Gd >> Gc or CMRR

Vo = GdVd

+

Noninverting Input

Inverting Input

Output


CMRR Example What is the CMRR?

Solution :

dB40)10/1000log(20CMRR 10 and 1000

V607007060 (2) From

V806006080 (1) From

V702

40100 V60

2

20100

V6040100 V8020100

21

21

cd

cdo

cdo

cc

dd

GG

GGV

GGV

VV

VV

+

100V

20V

80600V

+

100V

40V

60700V

This method is Not work! Why?

(1) (2)


Op-Amp Properties (1) Infinite Open Loop gain

- The gain without feedback

- Equal to differential gain

- Zero common-mode gain

- Pratically, Ad = 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

load

oloadRR

RVV


Frequency-Gain Relation

• Ideally, signals are amplified from

DC to the highest AC frequency

• Practically, bandwidth is limited

• 741 family op-amp have an limit

bandwidth of few KHz.

• Unity Gain frequency f1: the gain at

unity

• Cutoff frequency fc: the gain drop by

3dB from dc gain Gd

(Voltage Gain)

(frequency)

f1

Gd

0.707Gd

fc0

1

GB Product : f1 = Gd fc

20log(0.707)=3dB


GB Product Example: Determine the cutoff frequency of an op-amp having a unit gain

frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV

Solution:

Since f1 = 10 MHz

By using GB production equation

f1 = Gd fc

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


Ideal Vs Practical Op-Amp

Ideal Practical

Open Loop gain A 105

Bandwidth BW 10-100Hz

Input Impedance Zin >1M

Output Impedance Zout 0 10-100

Output Voltage Vout Depends only

on Vd = (V+V)

Differential

mode signal

Depends slightly

on average input

Vc = (V++V)/2

Common-Mode

signal

CMRR 10-100dB

+

~

AVin

Vin Vout

Zout=0

Ideal op-amp

+

AVin

Vin Vout

Zout

~

Zin

Practical op-amp


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 termainals is zero

For ideal Op-Amp circuit:

(1) Write the kirchhoff node equation at the noninverting terminal V+

(2) Write the kirchhoff node eqaution at the inverting terminal V

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


Noninverting Amplifier

(1) Kirchhoff node equation at V+

yields,

(2) Kirchhoff node equation at V yields,

(3) Setting V+ = V– yields

or

+Vin

Vo

RaRf

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


+

vi

Ra

vov-

v+

Rf

+

vi

Ra

vov-

v+

Rf

R2R1

+

vi

vov-

v+

Rf

+

vi

vov-

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 vR

Rv )1(

Less than unity gain

io vRR

Rv

21

2


Inverting Amplifier


yields,



0V

0_

f

o

a

in

R

VV

R

VV

a

f

in

o

R

R

V

V

Notice: The closed-loop gain Vo/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

Vin

Vo


Summing Amplifier


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

Vo

Rb

Ra

RcVb

Vc


Inverting Integrator Now replace resistors Ra and Rf by complex

components Za and Zf, respectively,

therefore

Supposing

(i) The feedback component is a capacitor C,

i.e.,

(ii) The input component is a resistor R, Za = R

Therefore, the closed-loop gain (Vo/Vin) become:

where

What happens if Za = 1/jC whereas, Zf = R?

Inverting differentiator

in

a

f

o VZ

ZV

CjZ f

1

dttvRC

tv io )(1

)(

tj

ii eVtv )(

+

~

Zf

Za

Vin

Vo

+

~

R

Vin

Vo

C


Integrator Example:

(a) Determine the rate of change

of the output voltage.

(b) Draw the output waveform.

Solution:

+

R

VoVi

0

+5V

10 k

0.01FC

Vo(max)=10 V

100s

(a) Rate of change of the output voltage

smV/

F k

V

50

)01.0)(10(

5

RC

V

t

V io

(b) In 100 s, the voltage decrease

VμssmV/ 5)100)(50( oV

0

+5V

0

-5V

-10V

Vi

Vo


Op-Amp Differentiator

RCdt

dVv i

o

+

R

C

VoVi

0to t1 t2

0

to t1 t2


Operational Amplifiers -Introduction - - Dunia ilmiah · · 2014-06-24An operational amplifier...

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Transcript of Operational Amplifiers -Introduction - - Dunia ilmiah · · 2014-06-24An operational amplifier...