Slide bab op amp

EE301/2011 Operational Amplifier 1

Chapter 2 :Operational-Amplifier

• Introduction of Operation Amplifier (Op-Amp)

• Understand the Differential Amplifier

• Comparison of ideal and non-ideal Op-Amp

• Non-ideal Op-Amp consideration


Vd

+

Vo

Rin~inf Rout~0

Input 1

Input 2

Output

+Vcc

-Vcc

Operational Amplifier (Op-Amp)

• The circuit symbol for an op-amp is shown to the right, where:

• Input 1: non-inverting input• Input 2: inverting input• Vo: output• +Vcc: positive power

supply• -Vcc: negative power

supply

The power supply pins ( and ) can be labeled in different ways Despite different labeling, the function remains the same

• 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



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

Op- Amp Block Diagram


• The input stage is a differential amplifier. The differential amplifier used as an input stage provides differential inputs and a frequency response down to d.c. Special techniques are used to provide the high input impedance necessary for the operational amplifier.


Differential Amplifier

• The second stage is a high-gain voltage amplifier. This stage may be made from several transistors to provide high gain. A typical operational amplifier could have a voltage gain of 200,000. Most of this gain comes from the voltage amplifier stage.


More Stages of Gain Amplifier

• The final stage of the OP AMP is an output amplifier. The output amplifier provides low output impedance. The actual circuit used could be an emitter follower. The output stage should allow the operational amplifier to deliver several milliamperes to a load. The operational amplifier can be used with three different input conditions (modes). With differential inputs (first mode), both input terminals are used and two input signals which are 180 degrees out of phase with each other are used.


Push-pull amplifier

• Push-pull amplifier An electronic circuit in which two transistors (or vacuum tubes) are used, one as a source of current and one as a sink, to amplify a signal. One device “pushes’’ current out into the load, while the other “pulls” current from it when necessary. A common example is the complementary-symmetry push-pull output stage widely used to drive loudspeakers (see illustration), where an npn transistor can source (push) current from a positive power supply into the load, or a pnp transistor can sink (pull) it into the negative power supply.



Single-Ended Input

+

Vo

~ Vi

+

Vo

~ Vi

• + terminal : Source• – terminal : Ground• 0o phase change

• + terminal : Ground• – terminal : Source• 180o phase change


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) ?


Distortion

Vd

+

Vo

+Vcc=+5V

Vcc= 5V

0

+5V

5V

The output voltage never excess the DC voltage supply of the Op-Amp


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


Common-Mode Rejection Ratio (CMRR)

Differential voltage input :

VVVd

Common voltage input :

)(2

1 VVVc

Output voltage :

ccddo VGVGV

Gd : Differential gainGc : Common mode gain

)dB(log20CMRR 10c

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 ExampleWhat is the CMRR?

Solution :

dBCMRR and

V (2) From

V (1) From

V V

V V

40)10/1000log(20101000

607007060

806006080

702

4010060

2

20100

60401008020100

21

21

cd

cdo

cdo

cc

dd

GG

GGV

GGV

VV

VV

+

100V

20V80600V

+

100V

40V60700V

NB: 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, 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


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 ProductExample: Determine the cutoff frequency of an op-amp having a unit gain frequency f1 = 10 MHz and voltage differential gain Gd = 20V/mV

Sol:

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-AmpIdeal Practical

Open Loop gain A 105

Bandwidth BW 10-100Hz

Input Impedance Zin >1M

Output Impedance Zout 0 10-100

Output Voltage VoutDepends 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 ApplicationsAnalysis 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

ia

fo v

RR

R

R

Rv ))(1(

21

2

Voltage follower io vv

ia

fo v

R

Rv )1(

Less than unity gain

io vRR

Rv

21

2


Inverting Amplifier(1) Kirchhoff node equation at V+

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


Multiple Inputs(1) Kirchhoff node equation at V+

yields,



0V

0_

c

c

b

b

a

a

f

o

R

VV

R

VV

R

VV

R

VV

c

aj j

jf

c

c

b

b

a

afo R

VR

R

V

R

V

R

VRV

+

Rf

Va

VoRb

Ra

RcVb

Vc


Inverting IntegratorNow 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 = RTherefore, the closed-loop gain (Vo/Vin) become:

where

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

ina

fo V

Z

ZV

CjZ f

1

dttvRC

tv io )(1

)(

tjii eVtv )(

+

~

Zf

Za

Vin

Vo

+

~

R

Vin

Vo

C


Op-Amp IntegratorExample:

(a) Determine the rate of changeof 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

+

RC

VoVi

0to t1 t2

0

to t1 t2


Non-ideal case (Inverting Amplifier)

+

~

Rf

Ra

Vin

Vo

3 categories are considering

Close-Loop Voltage Gain Input impedance Output impedance

Equivalent CircuitRf

RaVin

Vo

+

+

R RV

-AV

+

AVin

Vin Vout

Zout

~Zin

Practical op-amp


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

aafafo

a

in

RRAR

RARRRRRRRV

R

V

The Close-Loop Gain, Av

RARRRRRRR

RAR

V

VA

aafaf

f

in

ov


Close-Loop Gain

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

a

fv R

RA

~

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


Input ImpedanceRf

Ra

Vin Vo

+

+

R

-AV

R'

RV

Rf

+

R

-AV

if

V

Input Impedance can be regarded as,RRRR ain //

where R is the equivalent impedance of the red box circuit, that is

fi

VR

However, with the below circuit,

A

RR

i

VR

RRiAVV

of

f

off

1

)()(


Input ImpedanceFinally, we find the input impedance as,

1

11

of

ain RR

A

RRR

RARR

RRRRR

of

ofain )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


Output ImpedanceOnly source-free output impedance would be considered, i.e. Vi is assumed to be 0

Rf

Ra

Vo

+

R

-AV

R

V

io

V

V

Rf

Ra R+

R

-AV

V

i2 i1

(a) (b)

Firstly, with figure (a),

ofafa

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),

RRARRRRRRR

RRRRRRR

i

V

R

afafao

fafao

o

o

out

)1())(1(

)(

is impedance,output The

R and A comparably large,

a

faoout AR

RRRR

)(~

Slide bab op amp

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