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OPERATIONAL AMPLIFIER(OP AMP )

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1. Introduction2. Understand the general Op-amp circuit

design Components inside an op.amp.

Characteristics of an op.amp. Circuit symbol and pin configuration

Top view Pin-out configuration Block diagram an op-amp differential amplifier more stages of gain amplifier push-pull amplifier

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3. Understand the differential amplifier4. Understand more stages of gain amplifier5. Understand the push pull amplifier6. Understand the ideal operational amplifier

7. Understand the op-amp configurations Inverting amplifier Non inverting amplifier Summing amplifier Subtractor Differentiator amplifier Integrator amplifier Comparator

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OP - AMP is a solid state device capable of

sensing and amplifying dc and ac input signals.

OP AMP is an amplifier with two inputs

(Differential inputs) and a single output.

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OP AMP consists of 20 transistors 11 resistors

and 1 capacitor.

OP - AMP requires a positive and negative powersupply( Dual power supply ).

This allows the output voltage to swing positiveand negative with respect to ground.

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Op Amp Internal Circuit Diagram

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1.VERY HIGH INPUT RESISTANCE OR EVENINFINITY WHICH PRODUCES NEGLIGIBLE

CURRENT AT THE INPUT.

2.VERY HIGH CURRENT GAIN.

3.VERY LOW OUTPUT IMPEDANCE OR EVENZERO,SO AS NOT TO AFFECT THE OUTPUTOF THE AMPLIFIER BY LOADING.

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Block Diagram (Op Amp)

Internally, the typical Op-Amp has adifferential input, a voltage amplifier and apush pull output.

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An OP

AMP is so named, because it wasoriginally designed to perform mathematical

operations such as addition, subtraction,

multiplication, division, integration,

differentiation etc in analog computer.

Nowadays OP - AMPs are used in analog

computer operations and in timing circuits.

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The Op

Amp is represented by a triangularsymbol .

It has two input and one output terminals.

CIRCUIT SYMBOL AND PIN-OUT CONFIGURATION

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-

+

Inverting Input

Noninverting Input

Output

Positive dc

power supply

Negative dc

power supply

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The terminal with negative sign (-) is called

inverting input.

The terminal with positive sign (+) is callednon inverting input.

The input terminals are at the base and theoutput is at the apex of the triangular

symbol.

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The widely used and most popular type is

Op Amp IC 741.

The top pin on the left side of the notch ispin 1.

The pin number 2 is inverting input and pinnumber 3 is non-inverting input terminal.

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The pin number 6 is the output terminal.

A dc voltage or ac signal connected to pin2 will be 180o out of phase at the output.

A dc voltage or ac signal connected to pin

3 will be in phase at the output.

Pin 4(-) and 7(+) are the power supply

terminals.

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The pin 1 and 5 are Null adjustment pins.

Null adjustment pins are used to null the

output voltage , when equal voltages areinput terminals for perfect balance.

Pin number 8 indicates No connection.

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TOP VIEW OF 8 PIN DIP

Notch

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OP AMP Pin- out configuration

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A single package will often contain severalop-amps

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There are 2 types of application in op-amp

Linear application

Non-linear application

Linear application is where the op-amp operate inlinear region:

Assumptions in linear application:

Input current, Ii = 0

Input voltage: V+=V-

Feedback at the inverting input

Application in op-amp

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Non-linear application is where the op-amp operate innon-linear region

By comparing these two input voltages: positive input

voltages, V

+

and negative input voltage, V

-

where:VO = VCC if V

+ > V-

VO = -VEE if V+ < V-



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Comparator (Pembanding)

Inverter (Penukar)

Audio amplifier (Penguat Audio) Difference Amplifier (Penguat Beza)

Filter (Penapis)

Summing Amplifier (Penguat Jumlahan)


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Inverting Amplifier

Non-Inverting Amplifier

Summing Amplifier

Unity Follower

Difference Amplifier

Integrators

Differentiators

Op-amp Circuit Application

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Vo = A(V+ - V)

Vo/A = V+ - V

Let A --- infinitythen,

V+ - V--- 0

Summary of op-amp behavior

V+ = V

I+ = I = 0

Seems strange, butthe input terminalsto an op-amp act asa short and open atthe same time

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Write node equations at + and - terminals

(Ii=I+ = I-= 0)

Set V+ = V-

Solve for Vo

To analyze an op-amp circuit for linear

operation

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IDEAL OPERATIONAL AMPLIFIERS

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Differential Amplifier circuits In differential amplifier, there are two inputs Vi1 and Vi2 andthere are three outputs (1) at Vo1, (2) at Vo2, and (3) across

between Vo1 and Vo2 Dual supply +VCC and VEE are used so that Vi1 and Vi2 can beconnected to the BJT directly without coupling capacitor

IC1 + IC2 = 2Ic flows

through REthen VRE = 2Icx REVi1

RC

-VEE

+VCC

Vo1

Vo2

RC

RE

Vi2

Q1 Q2AC

AC

VE = -0.7V

VB = 0V dc

+

-

Ce

E

EEC

EEEEREEC

I

mV26r

R2

7.0VI

7.0VV7.0VRI2

CCCC2C1C RIVVV

DC analysis

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Example

In differential amplifier shown,find (1) VB1= VB2 (2) VE1= VE2 (3) IC1 = IC2 and (4) VC1= VC2

VB = 0V dc

VE = -0.7V

Vi1

-9V

+9V

Vo1

Vo2

RC

3.3kVi2

Q1

Q2

AC

AC

3.9k

+-

VC

IC

mA26.1k3.32

7.09

R2

7.0VI

7.0VV7.0VRI2

E

EEC

EEEEEEC

V1.4k9.3mA26.19RIVV CCCCC

V7.0V,V0V EB

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Due to biasing requirements or leakage, asmall amount of current (typically ~10nanoamperes for bipolar op-amps, tens ofpicoamperes for JFET input stages, and only a

few pA for MOSFET input stages) flows into theinputs. When large resistors or sources withhigh output impedances are used in the circuit,these small currents can produce large

unmodeled voltage drops. If the input currentsare matched, and the impedancelooking out of both inputs are matched, thenthe voltages produced at each input will bee ual.

Input Bias Current

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-The output voltage of an op-amp when thedifferential input is zero should be also zero.

- However, due to unavoidable internalimbalances and due to non-zero bias currents, asmall voltage, V

IO, is seen between the

terminals.- ICs provide a means to compensate for this.- This is generally done by connecting anexternal potentiometer to pins designated withOffset Null.- With zero input voltage, the output is set to zeroby adjusting the potentiometer- The pinout for the 741 op-amp (the most

common op-amp IC) is shown next.

Input Offset Voltages

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Input Offset Current

- Input offset current refers to thedifference between the bias currents ofthe amplifier. Again, ideally the two

currents should be equal to obtain azero output voltage. However, there hasto be a difference between the two bias

currents to set the output to zero. Thisdifference is referred to as input offsetcurrent.

- The difference in IBias between the

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Common-mode Gain

Common mode gain is the undesiredgain when the same signal is appliedto both inputs. Ideally it should be

zero.

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Common-mode Rejection Ratio

The ability of a differential amplifier to notpass (reject) the portion of the signalcommon to both the + and - inputs.

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Multistages Amplifier

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Push Pull Amplifier

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Operation (Push Pull)

Most audio power amplifiers use a Class Bconfiguration which employs two commoncollector (emitter-follower) stages where onetransistor provides power to the load during one-

half of the waveform cycle (it pushes) and asecond transistor provides power to the load forthe other half of the cycle (it pulls).

Neither transistor remains on for the entire cycle,giving each transistor time to rest and coolduring the waveform cycle. This makes for amore power-efficient amplifier circuit, but leads

to a distinct type of nonlinearity known as

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Operation (Push Pull)

Distortion occurs because there is a delaybetween the time one transistor turns off and theother transistor turns on.

There will be no output signal until Vin 0.6V .

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Infinite gain for the differential input signal

Zero gain for the common-mode inputsignal

Infinite input impedance

Zero output impedance

Infinite bandwidth

Characteristics of Ideal Op Amps

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(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

load

oloadRR

RVV

Ideal characteristics of Op-Amp

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Input offset voltage

0

when the input terminals are shorted so that , theoutput is a Virtual Ground or vout = 0

Offset current 0

there is assumed to be no leakage or bias currentinto the device

Bandwidth -

the frequency magnitude response is considered tobe flat everywhere with zero phase shift)

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Ideal Practical

Open Loop gainA 105

BandwidthBW 10-100Hz

Input ImpedanceZin >1M

Output ImpedanceZout 0 10-100

Output Voltage Vout Depends onlyon Vd = (V+V)

Differential

mode signal

Depends slightly

on average input

Vc = (V++V)/2Common-Mode

signal

CMRR 10-100dB

+

~

AVin

Vin Vout

Zout=0

Ideal op-amp

+

AVinVin VoutZ

out

~

Zin

Practical op-amp

Ideal Vs Practical Op-Amp

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We are using the two

ideal op ampproperties discussedabove to analyze thiscircuit.

Since the amplifierhas infinite gain, it willdevelop its outputvoltage, Vout, with zeroinput voltage.

INVERTING AMPLIFIER

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INVERTING AMPLIFIER

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Since the differentialinput is zero, the full

input voltage mustappear across Ri, makingthe current in Ri,

Iin = Vin / Ri .

since there is no current

flow into either inputterminal because theinput impedance isinfinite, the current Iin

must also flow in Rf.

INVERTING AMPLIFIER

45

INVERTING AMPLIFIER

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INVERTING AMPLIFIER

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V 0 Ii 0

I1 If Ii

Vs V

R1V Vo

Rf

V V 0

Vo

Vs

Rf

R1

Vo Rf

R1

Vs

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THE NON INVERTING AMPLIFIER

The input signal is appliedto the non inverting input

terminal.

A resistor Ra, is connected

from the inverting inputto the ground.

The feedback resistor Rfis

connected between the

output and the invertinginput.

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+Vin

Vo

RaRf

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Since the potential at theinverting input and that at the

non inverting input are same .

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+Vin

Vo

RaRf

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(R1 + R2 / R1) = VOUT /VA

1 + ( R2 / R1) = VOUT /VIN

A v =1 + ( R2 / R1).

The output and input

voltages are in phase.

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Non Inverting Amplifier

(1) Kirchhoff node equation at V+yields,

(2) Kirchhoff node equation at V

yields,

(3) Setting V+ = V yields

or

+VinVo

RaRf

iVV

00

f

o

a RVV

RV

0

f

oi

a

i

R

VV

R

V

a

f

i

o

R

R

V

V1

The output and input voltages are in phase.

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The Summing Amplifier

The summingamplifier provides

an output equal to

the sum of theinput voltages.

Here we have an

inverting amplifier,

used to sum two

input voltages.

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The input voltagesV1 and V2 are

connected to the

inverting inputthrough the

resistors R1, R2 and

RF

is the feedback

resistance.

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S i A lifi

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Summing Amplifier

This circuit is called

a weighted summer

3

3

2

2

1

1

3

3

2

2

1

1

3

3

2

2

1

1

321

;0

:

;0

:

0

R

V

R

V

R

V

RV

R

V

R

V

R

V

R

VVinsert

R

VV

R

VV

R

VV

R

VVso

Iwhile

IIIII

KCLuse

VV

fo

f

o

f

o

i

RfiRRR

V1

V2

V3

R1

R2

R3

Rf

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Subtrator Amplifier

55

43

21

RR

RR

)(21

2

4VV

R

RVO

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dttvRC

tv

dt

tdvC

R

tv

III

io

i

Ci

)(1

)(

)(0

)( 0

sC

1

Cj

1X

:impedancecetanCapaci

C

Integrator Amplifier

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dt

tdvRCtv

R

tvV

dt

tdv

C

II

i

o

oi

RC

)()(

)()(

Differentiator Amplifier

C t

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Comparator

Vo(V)

10

-5

t

VS(V)

t

Compare V+ and V-

V+=0V-=VS

When:VS>0,V

+>V- so Vo=10VVS

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1VV

O

Unity Follower(Pengikut Voltan)

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Instrumentation Amplifier

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Non-linear application is where the op-ampoperate in non-linear region

By comparing these two input voltages: positive

input voltages, V+

and negative input voltage, V-

where:

VO = VCC if V+ > V-

VO = -VCC if V+ < V-


Non-linear application in op-amp

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chapter3 op amp(1)80911

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