1

Operational Amplifiers

• Introduction of Operational Amplifier (Op-Amp)

• Characteristics of an Op-Amp

• Comparison of ideal and Non-ideal Op-Amp

• Feedback and Non-Feedback

• Configurations

• Gain; Input Impedance; Output Impedance

• Analysis of ideal Op-Amp Circuits.

2

What is an Operational Amplifier?

• Operational amplifier is an amplifier whose output voltage is

proportional to the negative of its input voltage and that boosts the

amplitude of an input signal, many times, i.e., has a very high

gain.High-gain amplifiers.

• They were developed to be used in synthesizing mathematical

operations in early analog computers, hence their name.

• Typified by the series 741 (The integrated circuit contains 8-pin mini-

DIP, 20 transistors and 11 resistors).

• Used for amplifications, as switches, as filters, as rectifiers, and in

digital circuits.

• Take advantage of large open-loop gain.

• It is usually connected so that part of the output is fed back to the

input.

• Can be used with positive feedback to produce oscillation.

Basics of Differential Amplifier Model

Op Amp is represented by:

A = open-circuit voltage gain

vid = (v+-v-) = differential input signal

voltage

Rid = amplifier input resistance

Ro = amplifier output resistance

The output signal of the amplifier is in

phase with the signal applied at the +

input (non-inverting).

The output signal of the amplifier is 180°

out of phase with the signal applied at the

- input (inverting) terminal.

4

Characteristics of Operational Amplifier

• Very high differential gain

• High input impedance

• Low output impedance

• Wide range of applications:

oscillators, filters and

instrumentation

• Accumulate a very high gain by

cascading multiple stages.

ddo VGV

510 than moresay large,y ver

normally gain aldifferenti : dG

Op-Amp: Power Supply Connections

• Commonly used: dual power supplies

• Common values:

– +15V (+V or V+)

– -15V (-V or V-)

• All output loads connected between output terminal and common ground point

• Usually power supply connections are omitted for ease

Common

ground

point 15 v

15 v

Positive power

supply terminal

Negative power

supply terminal

+

+

-

-

6

The ideal Op-Amp

• Infinite Voltage Gain

– a voltage difference at the two inputs is magnified infinitely

– something like 200,000, means difference between + terminal and terminal is amplified by 200,000!

• Infinite Input Impedance

– no current flows into both inputs

– about 1012 for FET input op-amps

• Zero Output Impedance

– rock-solid independent of load

– roughly to current maximum (usually 5–25 mA)

• Infinitely Fast (Infinite Bandwidth)

– limited to few MHz range

– slew rate limited to 0.5–20 V/s

7

Figure 8.2,

8.3

A Voltage Amplifier Simple Voltage Amplifier Model

inLSinS

Lout

L

inS

inL

Lout

LinLS

inS

inin

AvvvvvRR

R

RR

RAv

RR

RAvvv

RR

Rv

;;

;

8

Op-Amp as an Integrated Circuit

• The integrated circuit operational amplifier evolved soon after

development of the first bipolar integrated circuit.

• The A-709 was introduced by Fairchild Semiconductor in 1965.

• Since then, a vast array of op-amps with improved characteristics,

using both bipolar and MOS technologies, have been designed.

• Most op amps are inexpensive (less than a dollar) and available from a

wide range of suppliers.

• There are usually 20 to 30 transistors that make up an op-amp circuit.

• From a signal point of view, the op-amp has two input terminals and

one output terminal as shown in the following figures.

• The ideal op-amp senses the difference between two input signals and

amplifies the difference to produce an output signal.

• Ideally, the input impedance is infinite, which means that the input

current is zero. The output impedance is zero.

9

Figure 8.4

Operational Amplifier Model Symbols and Circuit Diagram

Op-Amp Stages with Pin-outs of IC741

2

4

7

6

3

11

Ideal Versus Real Op-Amps

Characteristics Ideal Op-Amp Typical Op-Amp

Input Resistance Infinity 106 (bipolar)

109 - 1012 (FET)

Input Current 0 10-12 – 10-8 A

Output Resistance 0 100 – 1000

Operational Gain Infinity 105 - 109

Common Mode Gain 0 10-5

Bandwidth Infinity Attenuates and phases at high

frequencies (depends on slew rate)

Temperature Independent Bandwidth and gain

http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/opampcon.html#c1

12

Single Inputs Op-Amp

• + Terminal : Source

• – Terminal : Ground

• 0o Phase change

• + Terminal : Ground

• – Terminal : Source

• 180o Phase change

13

Double Inputs Op Amp

• Differential input

•

• 0o phase shift change

between Vo and Vd

VVVd

Queation: What Vo should be if,

V1

V2

(A) (B)

Ans: (A or B) ?

14

Distortion / Saturation

• The output voltage never exceeds the DC voltage supply of the Op-Amp.

• Practical Op Amps have limited output voltage and current ranges.

• Voltage: Usually limited to a few volts less than power supply span.

• Current: Limited by additional circuits (to limit power dissipation or protect

against accidental short circuits).

15

A Practical Application: Why Feedback

• Self-balancing mechanism, which allows the amplifier to

preserve zero potential difference between its input

terminals.

• A practical example that illustrates a common application of

negative feedback is the thermostat. This simple

temperature control system operates by comparing the

desired ambient temperature and the temperature measured

by the thermometer and turning a heat source on and off to

maintain the difference between actual and desired

temperature as close to zero as possible.

Impedances

RIN

ROUT

VIN AVIN VOUT

• The amplifier measures voltage across RIN, then generates a voltage

which is larger by a factor A

• This voltage generator, in series with the output resistance ROUT, is

connected to the output port.

• A should be a constant (gain is linear)

Add an input - a source voltage VS plus source impedance RS

Impedances

RIN

ROUT

VIN AVIN VOUT

Note the voltage divider RS + RIN.

VIN=VS(RIN/(RIN+RS)

We want VIN = VS regardless of source impedance

So want RIN to be large.

The ideal amplifier has an infinite input impedance!

VS

RS

Add a load - an output circuit with a resistance RL

Impedances

Note the voltage divider ROUT + RL.

VOUT=AVIN(RL/(RL+ROUT))

Want VOUT=AVIN regardless of load

We want ROUT to be small.

The ideal amplifier has zero output impedance!

RIN

ROUT

VIN AVIN VOUT VS

RS

RL

Feedback Feedback refers to connecting the output of the amplifier to

its input

19

20

Op-Amps for Math

• Inverting

• Non-Inverting

• Summing

• Differencing

• Integrating

• Differentiating

21

Figur

e 8.5

Common Op-Amp Configurations

Inverting Amplifier

SvSR

FRoutv

FRvA

outv

FR

outv

SRvA

outv

SR

Sv

vviniFiSi

iniFR

voutvFi

SR

vSvSi

iniFiSi

;0;

0;;

Voltage Gain of Inverting Amplifier

• The negative voltage gain

implies that there is a 1800 phase

shift between both dc and

sinusoidal input and output

signals.

• The gain magnitude can be

greater than 1 if R2 > R1

• The gain magnitude can be less

than 1 if R1 > R2

•The inverting input of the Op

Amp is at ground potential

(although it is physically

connected to ground) and is said

to be at virtual ground.

0ov22

i1

isv RRs

1

svsi

R

But is= i2 and v- = 0 (since vid= v+ - v-= 0)

and

1

2

svov

R

R

vA

Input and Output Resistances

Rin

vsis

R1 since v0

Rout is found by applying a test current

(or voltage) source to the amplifier

output and determining the voltage (or

current) after turning off all

independent sources. Hence, vs = 0

11i

22ixv RR

But i1=i2

)12

(1ixv RR

Since v- = 0, i1=0. Therefore vx = 0

irrespective of the value of ix .

0out

R

Inverting Amplifier: An Example

• Problem: Design an inverting amplifier for

• Given Data: Av= 20 dB, Rin = 20k,

• Assumptions: Ideal op amp

• Analysis: Input resistance is controlled by R1 and voltage gain is set

by R2 / R1.

and Av = -100

A minus sign is added since the amplifier is inverting.

Av dB

20log

10Av

, Av

1040dB/20dB100

k201 in

RR

AvR

2R1

R2100R

12M

25

Design Example: Design an inverting amplifier with a closed loop

voltage of Av = -5. Assume the op-amp is driven by a sinusoidal soorce, vs

= 0.1 sin t volts, which has a source resistance of Rsr = 1 k and which

supply a maximum current of 5 A. Assume that the frequency is low.

k 100205)1(52y Accordingl

.51

2 and k 19 be should 1

kΩ 206105

1.0

(max)

(max)(min)1 write then weA,5 max)( If

1Therefore .resistance source therepresents

.1 :sresistance twomeans example in this (

RsrRR

RsrR

RvAR

si

svsrRRsi

RsrR

svsisrR

srRRsRsRsR

svsi

26

To Solve Ideal Op-Amp Circuit

• If the noninverting terminal of the op-amp is at ground potential, then

the inverting terminal is at virtual ground. Sum currents at this point,

assuming zero current enters the op-amp itself.

• If the noninverting terminal of the op-amp is not at ground potential,

then the inverting terminal voltage is equal to that at the noninverting

terminal. Sum currents at the inverting terminal node, assuming zero

current enters the op-amp itself.

• For an ideal op-amp circuit, the output voltage is determined from

either step 1 or step 2 above and is independent of any load connected

to the output terminal.

27

Inverting Amplifier with a T-Network

R1

R2 R3

vx

R4

0

0 -

+ vo

i1

i2 i3

i4

vs

)2

3

4

31(1

2

get weequations above theCombing

342;342

)1

2(220

R

R

R

R

R

R

sv

ovvA

R

ovxv

R

xv

R

xviii

R

RsvRixv

28

Design Example: An op-amp with a T-network is to be used as a preamplifier for a

microphone. The maximum microphone output voltage is 12 mV (rms) and the

microphone has an output resistance of 1 k. The op-amp circuit is to be designed

such that the maximum output voltage is 1.2 V (rms). The input amplifier resistance

should be fairly large but all resistance values should be less than 500 k.

k 1.384R and k 40032

k 50 be will)effective 1( resistance total then thek 1 sr and k 49 1set weIf

ncalculatio in the resistance source theof value theinclude should We

5.104

3;8)4

31(8100

81

3

1

2 choose weIf

1

3)4

31(1

2

100012.0

2.1

RR

RRR

R

R

R

R

R

R

R

R

R

R

R

R

R

RvA

vA

29

Figure

8.7

Summing Amplifier

N

n

S

S

Fout

F

outF

S

S

n

FN

n

n

n

n

vR

Rv

R

vi

NnR

vi

iiii

1

21

,...2,1......

...

30

Figu

re

8.8,

8.9

Non-inverting Amplifier Voltage Follower

S

F

S

out

R

R

v

v1

outSvv

Non-inverting Amplifier

The input signal is applied to the non-inverting input terminal.

A portion of the output signal is fed back to the negative input terminal.

Analysis is conducted by relating the voltage at v1 to input voltage vs and

output voltage vo.

Non-inverting Amplifier Voltage Gain, Input Resistance and Output Resistance

Since i-= 0 and

But vid = 0

Since i+=0

21

1ov

1v

RR

R

1v

idvsv

1vsv

1

211

21

svov

1

21svov

R

R

R

RRvA

R

RR

i

sv

inR

Rout is found by applying a test current source to the amplifier output

after setting vs = 0. It is identical to the output resistance of the inverting

amplifier i.e. Rout = 0.

Non-inverting Amplifier: An Example

• Problem: Determine the output voltage and current for the given non-

inverting amplifier.

• Given Data: R1= 3k, R2 = 43k, vs= +0.1 V

• Assumptions: Ideal op amp

• Analysis:

Since i-=0,

Av1R

2R1

143k

3k15.3

voAvvs(15.3)(0.1V)1.53V

A3.33k3k43

V53.1

12

ovoi

RR

34

Design Example: Design a noninverting amplifier with a closed loop gain

of Av = 5. The output voltage is limited to -10 V vo +10 V and the

maximum current in any resistor is limited to 50 A

Answer: R1 = 40 k, R2 = 160 k

The Unity Gain Amplifier or “The Buffer”

Buffer is a special case of the non-inverting amplifier with infinite R1 and zero R2. Hence Av = 1.

It provides an excellent electrical isolation while maintaining the signal

voltage level. The ideal buffer requires no input current and can drive any desired load resistance without loss of signal voltage.

Buffer is used in many data acquisition system applications.

36

Figure

8.10

Differential Amplifier

)(12

1

2

2

21

2

21

1

vvR

Rv

vvRR

Rv

vv

R

vv

R

vv

out

out

Differential Amplifier Using Op-Amp

v v

11

1

v vi

R

01

2

v vi

R

+

-1R

2R

1vov

v

v1i

1i

2v

1R

2R

22

1 2

Rv v

R R

01

1 2

v vv v

R R

2 21 2 2 0

1 2 1 2

1 2

R Rv v v v

R R R R

R R

Differential Amplifier Using Op-Amp

+

-1R

2R

1vov

v

v1i

1i

2v

1R

2R

2 21 2 2 0

1 2 1 2

1 2

R Rv v v v

R R R R

R R

2

2 2 20 1 2 2

1 1 2 1 1 2

R R Rv v v v

R R R R R R

2 2 20 1 2

1 1 2 1

1R R R

v v vR R R R

20 2 1

1

Rv v v

R

Differential Amplifier Applications

• Very useful if you have two inputs corrupted with the same noise

• Subtract one from the other to remove noise, remainder is signal

• Applications: Electrocardiagram to measure the potential difference

between two points on the body

The AD624AD is an instrumentation amplifier - this is a high gain, dc coupled

differential amplifier with a high input impedance and high CMRR (the chip

actually contains a few Op Amps)

http://www.picotech.com/applications/ecg.html

Difference Amplifier: An Example

• Problem: Determine vo

• Given Data: R1= 10k, R2 =100k, v1=5 V, v2=3 V

• Assumptions: Ideal op amp. Hence, v-= v+ and i-= i+= 0.

• Analysis: Using dc values,

Adm

R

2R1

100k

10k10

VoAdmV1V

2

10(53)

Vo20.0 V

Here Adm is called the “differential mode voltage gain” of the difference amplifier.

Finite Common-Mode Rejection Ratio

(CMRR)

A(or Adm) = differential-mode gain

Acm = common-mode gain

vid = differential-mode input voltage

vic = common-mode input voltage

A real amplifier responds to signal

common to both inputs, called the

common-mode input voltage (vic).

In general,

voAdmvidAcmvic

Adm

Admvid

vic

CMRR

CMRR AdmAcm

and CMRR(dB) 20log10

(CMRR)

An ideal amplifier has Acm = 0, but for a

real amplifier,

21id

v

icvv

22id

v

icvv

voAdm(v

1v

2)Acm

v1v

22

voAdm(vid

)Acm(vic

)

Finite CMRR: An Example

• Problem: Find output voltage error introduced by finite CMRR.

• Given Data: Adm= 2500, CMRR = 80 dB, v1 = 5.001 V, v2 = 4.999 V

• Assumptions: Op amp is ideal, except for CMRR. A CMRR in dB of 80 dB corresponds to a CMRR of 104.

• Analysis:

The output error introduced by finite CMRR is 25% of the expected ideal output.

vid5.001V4.999V

vic 5.001V4.999V

25.000V

voAdmvid

vic

CMRR

2500 0.0025.000

104

V6.25V

In the " ideal" case, voAdmvid5.00 V

% output error 6.255.00

5.00100%25%

uA741 CMRR Test Differential Gain

Differential Gain Adm = 5 V/5 mV = 1000

uA741 CMRR Test Common Mode Gain

Common Mode Gain Acm = 160 mV/5 V = 0.032

CMRRAdm

Acm

1000

.032 3.125x104

CMRR(dB) 20log10 CMRR 89.9 dB

Op-Amp circuit Analysis

47

48