Applied Electronics II · Applied Electronics II Chapter 3: Operational Ampli er Part 1- Op Amp Basics School of Electrical and Computer Engineering Addis Ababa Institute of Technology

Applied Electronics II

Chapter 3: Operational AmplifierPart 1- Op Amp Basics

School of Electrical and Computer EngineeringAddis Ababa Institute of Technology

Addis Ababa University

Daniel D./Abel G.

April 27, 2016

Chapter 3: Operational Amplifier Part 1- Op Amp Basics (AAIT)Chapter Three April 27, 2016 1 / 46

Overview I

1 Introduction

2 The Ideal Op AmpThe Op Amp TerminalsFunction and Characteristics of the Ideal Op Amp

3 The Inverting ConfigurationClosed-Loop GainEffect of Finite Open-Loop GainInput and Output ResistancesAn Important Application - The Weighted Summer

4 The Noninverting ConfigurationThe Closed-Loop GainEffect of Finite Open-Loop GainApplication - The Voltage Follower

5 Difference AmplifiersA Single-Op-Amp Difference Amplifier


Overview II

The Instrumentation Amplifier

6 Integrators and DifferentiatorsThe Inverting IntegratorThe Op-Amp Differentiator

7 DC ImperfectionsOffset VoltageInput Bias and Offset Currents

8 Frequency ResponseFrequency Dependence of the Open-Loop GainFrequency Response of Closed-Loop Amplifiers

9 Large-Signal Operation of Op Amps


Introduction

Introduction

The operational amplifier (Op amps) have been in use for a longtime, their initial applications being primarily in the areas of analogcomputation and sophisticated instrumentation.

Early op amps were constructed from discrete components (vacuumtubes and then transistors, and resistors).

The introduction of integrated circuit (IC) reduced the cost andimproved the performance.

One of the reasons for the popularity of the op amp is its versatility.

IC op amp has characteristics that closely approach the assumed ideal.


The Ideal Op Amp The Op Amp Terminals

The Op Amp Terminals

From a signal point of view the op amp has three terminals: two inputterminals (1 and 2) and one output terminal (3).Most IC op amps require two dc power supplies, as shown in twoterminals, 4 and 5, are brought out of the op-amp package and connectedto a positive voltage VCC and a negative voltage −VEE , respectively.Some times other terminals can include terminals for frequencycompensation and terminals for offset nulling.


The Ideal Op Amp Function and Characteristics of the Ideal Op Amp

Function and Characteristics of the Ideal Op Amp

Op amp is designed to sense the difference between the voltage signalsapplied at its two input terminals and multiply this by a number A.

v3 = A(v2 − v1)

Characteristics of the Ideal Op Amp

Infinite input impedance

Zero output impedance

Zero common-mode gain or, equivalently, infinite common-moderejection

Infinite open-loop gain A

Infinite bandwidth

Question: But, is an amplifier with infinite gain of any use?




An amplifiers input is composed of two components

differential input (vId) - is difference between inputs at inverting andnon-inverting terminals

vId = v2 − v1

common-mode input (vIcm) - is input present at both terminals

vIcm =1

2(v2 + v1)

The input signals v1 and v2

v1 = vIcm − vId/2 and v2 = vIcm + vId/2

Similarly, two components of gain exist

differential gain (A) - gain applied to differential input ONLY

common-mode gain (Acm) - gain applied to common-mode input ONLY




Figure: Equivalent circuit of the ideal opamp.

Figure: Input signals in terms ofdifferential and common-modecomponents.


The Inverting Configuration

The Inverting Configuration

Op amps are not used alone; rather, the op amp is connected to passivecomponents in a feedback circuit.There are two such basic circuit configurations employing an op amp andtwo resistors: the inverting configuration and the noninvertingconfiguration

Figure: The inverting closed-loop configuration.


The Inverting Configuration Closed-Loop Gain

Closed-Loop Gain

Assuming an ideal op amp. How to analyze closed-loop gain for invertingconfiguration of an ideal op-amp?Step 1 Begin at the output terminalStep 2 If vo is finite , then the voltage between the op-amp input terminalsshould be negligibly small and ideally zero.

v2 − v1 =voA

= 0 , because A is ∞

A virtual short circuit between v1 and v2.

A virtual ground exist at v1.

Step 3 Define current in to inverting input (i1).

i1 =vI − v1

R1=

vI − 0

R1=

vIR1

Step 4 Determine where this current flows?Chapter 3: Operational Amplifier Part 1- Op Amp Basics (AAIT)Chapter Three April 27, 2016 10 / 46

The Inverting Configuration Closed-Loop Gain

Closed-Loop Gain

It cannot go into the op amp, since infinite input impedance draws zerocurrent. i1 will have to flow through R2 to the low-impedance terminal 3.Step 5 Define vo in terms of current flowing across R2.

vo = v1 − i1R2 = 0− vIR1

R2 = −R2

R1vI G = −R2

R1

Figure: The inverting closed-loop configuration.


The Inverting Configuration Effect of Finite Open-Loop Gain

Effect of Finite Open-Loop Gain

How does the gain expression change if open loop gain (A) is not assumedto be infinite?

One must employ analysis similar to the previous.

The voltage at v1 becomes

v2 − v1 =voA

v1 = −voA

The current i1 becomes

i1 =vI − v1

R1=

vI + voA

R1

The output voltage vo becomes

vo = v1 − i1R2 = −voA−

vI + voA

R1R2 vo

(1 +

1 + R2R1

A

)= −v1

R2

R1


The Inverting Configuration Effect of Finite Open-Loop Gain


The Gain will be

GA<∞ =vovi

=−R2/R1

1 +

(1 + R2/R1

A

)

Figure: Analysis of the inverting configuration taking into account the finiteopen-loop gain of the op amp.


The Inverting Configuration Input and Output Resistances

Input and Output ResistancesThe Input Resistance is

Ri =viii

=vi

(vi − v1)/R1=

vivi/R1

= R1

For a Voltage amplification Ri must be large. Then the gain would bereduced, so such configuration suffers from low Ri . Consider the followingcircuit and find the expression of the closed loop gain


The Inverting Configuration Input and Output Resistances

Input and Output Resistances

The closed loop gain

vovi

= −R2

R1

(1 +

R4

R2+

R4

R3

)It can be seen a higher Ri can be achieved without compromising theclosed loop gain.Since the output of the inverting configuration is taken at the terminals ofthe ideal voltage source A(v2 − v1), it follows that the output resistance ofthe closed-loop amplifier is zero.

Ro = 0


The Inverting Configuration An Important Application - The Weighted Summer

An Important Application - The Weighted Summer

Weighted Summer - is a closed-loop amplifier configuration which providesan output voltage which is weighted sum of the inputs.


The Noninverting Configuration

The Noninverting Configuration

The input signal vI is applied directly to the positive input terminal of theop amp while one terminal of R1 is connected to ground.Then the polarity / phase of the output is same as input.

Figure: The noninverting configuration.


The Noninverting Configuration The Closed-Loop Gain

The Closed-Loop Gain

For an ideal case the closed-loop gain by using the previous methods.

vovi

= 1 +R2

R1

Figure: The noninverting configuration.


The Noninverting Configuration Effect of Finite Open-Loop Gain


Assuming the op amp to be ideal except for having a finite open-loop gainA. The closed-loop gain

GA<∞ =vovi

=1 + R2/R1

1 +

(1 + R2/R1

A

)For

A� 1 +R2

R1

the closed-loop gain can be approximated by the ideal value.The percentage error in G resulting from the finite op-amp gain A as

Percentage gain error = − 1 + R2/R1

A + 1 + (R2/R1)

The input impedance Ri of this closed-loop amplifier is ideally infinite,since no current flows into the positive input terminal of the op amp. Theoutput is taken at the terminals of the ideal voltage source thus the outputimpedance Ro of the noninverting configuration is zero.


The Noninverting Configuration Application - The Voltage Follower

The Voltage Follower

The property of high input impedance is a very desirable feature ofthe noninverting configuration.It enables using this circuit as a buffer amplifier to connect a sourcewith a high impedance to a low-impedance load. Buffer amplifier isnot required to provide any voltage gainThis circuit is commonly referred to as a voltage follower, since theoutput “follows” the input.

Figure: a) The unity-gain buffer or follower amplifier. (b) Its equivalent circuitmodel.


Difference Amplifiers


A difference amplifier is one that responds to the difference between thetwo signals applied at its input and ideally rejects signals that are commonto the two inputs.

Ideally, the amp will amplify only the differential signal (vId) andreject completely the common-mode input signal (vIcm). However, apractical circuit will behave as below

vo = AdvId + AcmvIcm

The efficacy of a differential amplifier is measured by the degree of itsrejection of common-mode signals in preference to differential signals.

CMRR = 20 log|Ad |Acm

Question: The op amp is itself a difference amplifier; why not just use anop amp?

very high (ideally infinite) gain of the op amp




A difference amplifier is one that responds to the difference between thetwo signals applied at its input and ideally rejects signals that are commonto the two inputs.

Ideally, the amp will amplify only the differential signal (vId) andreject completely the common-mode input signal (vIcm). However, apractical circuit will behave as below

vo = AdvId + AcmvIcm

The efficacy of a differential amplifier is measured by the degree of itsrejection of common-mode signals in preference to differential signals.

CMRR = 20 log|Ad |Acm

Question: The op amp is itself a difference amplifier; why not just use anop amp? very high (ideally infinite) gain of the op amp


Difference Amplifiers A Single-Op-Amp Difference Amplifier

A Single-Op-Amp Difference Amplifier

Analyzing the difference amplifier below using superposition.

vo1 = −R2

R1vI1 vo2 =

R4

R3 + R4

(1 +

R2

R1

)vI2

We have to make the two gain magnitudes equal in order to rejectcommon-mode signals.




R2

R1=

R4

R3 + R4

(1 +

R2

R1

)R2/R1

1 + R2/R1=

R4

R3 + R4=

R4/R3

1 + R4/R3

The condition is obtained when

R2

R1=

R4

R3

Assuming the condition is satisfied, the output voltage

vo =R2

R1(vI2 − vI1)

In addition to rejecting common-mode signals, a difference amplifier isusually required to have a high input resistance. Assuming R4 = R2 andR3 = R1 and applying a differential input.




vId = R1iI + 0 + R1iI

ThusRId =

vIdiI

= 2R1

Note that if the amplifier is required to have a large differential gain (R2/R1),then R1 of necessity will be relatively small and the input resistance will becorrespondingly low, a drawback of this circuit.

Another drawback of the circuit is that it is not easy to vary the differential gain

of the amplifier.Chapter 3: Operational Amplifier Part 1- Op Amp Basics (AAIT)Chapter Three April 27, 2016 24 / 46

Difference Amplifiers The Instrumentation Amplifier


The low-input-resistance problem can be solved by using voltage followersto buffer the two input terminals. But why not get some voltage gain.Solution: using a Noninverting Op Amp.




The output vo

vo =R4

R3

(1 +

R2

R1

)(vI2 − vI1)

The Advantages are

very high input resistance

high differential gain

symmetric gain (assuming that A1 and A2 are matched)

The Disadvantage

Ad and Acm are equal in first stage - meaning that the common-modeand differential inputs are amplified with equal gain

need for matching - if two op amps which comprise stage 1 are notperfectly matched, one will see unintended effects

The Solution is to disconnect the two resistors (R1) connected to node Xfrom ground and connecting them together.




For a differential input applied the gain would remain the same. For a common

mode input voltage vIcm an equal voltage appears at the negative input terminals

of A1 and A2, causing the current through 2R1 to be zero. Thus vo1 and vo2 will

be equal to the input. Thus the first stage no longer amplifies vIcm.Chapter 3: Operational Amplifier Part 1- Op Amp Basics (AAIT)Chapter Three April 27, 2016 27 / 46

Integrators and Differentiators The Inverting Integrator

The Inverting Integrator

By placing a capacitor in the feedback path and a resistor at the input, we obtainthe circuit of below. We shall now show that this circuit realizes the mathematicaloperation of integration. Let the input be a time-varying function vI (t).

The transient description

vO(t) = − 1

CR

t∫0

vI (t)dt − vO(t0)




The steady-state description

Vo(s)

Vi (s)= − 1

sCR

Thus the integrator transfer function has magnitude of 1/ωCR and phaseφ = +90◦

This configuration also known as a Miller integrator has a disadvantage.

At ω = 0, the magnitude of the integrator transfer function is infinite.This indicates that at dc the op amp is operating with an open loop.

Solution: By placing a very large resistor in parallel with the capacitor,negative feedback is employed to make dc gain “finite”.




The integrator transfer function becomes

Vo(s)

Vi (s)= − RF/R

1 + sCRF

The lower the value we select for RF , the higher the corner frequency will be and

the more nonideal the integrator becomes. Thus selecting a value for RF presents

the designer with a trade-off between dc performance and signal performance.Chapter 3: Operational Amplifier Part 1- Op Amp Basics (AAIT)Chapter Three April 27, 2016 30 / 46

Integrators and Differentiators The Op-Amp Differentiator

The Op-Amp DifferentiatorInterchanging the location of the capacitor and the resistor of the integratorcircuit results in the circuit which performs the mathematical function ofdifferentiation.

The transient description

vO(t) = −CRdvI (t)

dt


Integrators and Differentiators The Op-Amp Differentiator

The Op-Amp Differentiator

The steady-state description

Vo(s)

Vi (s)= −sCR

Thus the integrator transfer function has magnitude of ωCR and phaseφ = −90◦

This configuration as a differentiator has a disadvantage.

Differentiator acts as noise amplifier, exhibiting large changes inoutput from small (but fast) changes in input. As such, it is rarelyused in practice.

When the circuit is used, it is usually necessary to connect a small-valuedresistor in series with the capacitor. This modification, unfortunately, turnsthe circuit into a nonideal differentiator.


DC Imperfections Offset Voltage

Offset VoltageNow we consider some of the important nonideal properties of the op amp.What happens If the two input terminals of the op amp are tied together andconnected to ground.

Ideally since vid = 0, we expect vO = 0

In practice a finite dc voltage exists at the output.



Offset Voltage

The causes of VOS is unavoidable mismatches in the differential stage ofthe op amp. It is impossible to perfectly match all transistors.General-purpose op amps exhibit VOS in the range of 1 mV to 5 mV. Also,the value of VOS depends on temperature.

Analysis to determine the effect of the op-amp VOS on their performance isthe same for both inverting and the noninverting amplifier configurations.

VO = VOS

[1 +

R2

R1

]Chapter 3: Operational Amplifier Part 1- Op Amp Basics (AAIT)Chapter Three April 27, 2016 34 / 46


Offset VoltageHow to reduced Offset Voltageoffset nulling terminals A variable resistor (if properly set) may be used to reduce

the asymmetry present and, in turn, reduce offset.

capacitive coupling A series capacitor placed between the source and op amp maybe used to reduce offset, although it will also filter out dc signals.


DC Imperfections Input Bias and Offset Currents

Input Bias and Offset Currents

In order for the op amp to operate, its two input terminals have to besupplied with dc currents, termed the input bias currents, IB .

IB =IB1 + IB2

2

IOS = |IB1 − IB2|

input offset currents, IOS - the difference between bias current at bothterminals.The resulting output voltage

VO = IB1R2 u IBR2



Input Bias and Offset CurrentsTo reduce the value of the output dc voltage due to the input bias currents,logically it is ↓ R2 but higher R2 needed for gain.The solution is introducing a resistance R3 in series with the noninverting input.

The output voltage when calculated

VO = −IB2R3 + R2(IB1 − IB2R3/R1)

Assuming IB1 = IB2 = IB

VO = IB [R2 − R3(1 + R2/R1)]



Input Bias and Offset Currents

Thus we can reduce VO to zero by selecting R3 such that

R3 =R2

1 + R2/R1=

R1R2

R1 + R2

We conclude that to minimize the effect of the input bias currents, oneshould place in the positive lead a resistance equal to the equivalent dcresistance seen by the inverting terminal.

This is the case for op amps constructed using bipolar junction transistors(BJTs). Those using MOSFETs in the first (input) stage do not draw anappreciable input bias current; nevertheless, the input terminals shouldhave continuous dc paths to ground.


Frequency Response Frequency Dependence of the Open-Loop Gain

Frequency Dependence of the Open-Loop GainThe differential open-loop gain A of an op amp is not infinite; rather, it is finiteand decreases with frequency.

It is high at dc, but falls off at a rather low frequency.

Internal compensation - is the presence of internal passive components(caps) which cause op-amp to demonstrate STC low-pass response.

Frequency compensation - is the process of modifying the open-loop gain toincrease stability.

Figure: Open-loop gain of a general-purpose internally compensated op amp.


Frequency Response Frequency Dependence of the Open-Loop Gain

Frequency Dependence of the Open-Loop Gain

The gain of an internally compensated op-amp may be expressed as shownbelow

The transfer function in Laplace domain: A(s) =A0

1 + s/ωb

The transfer function in Frequency domain: A(ω) =A0

1 + ω/ωb

The transfer function for high frequnecy: A(ω) ≈ A0ωb

ω

Magnitude gain for high frequnecy: |A(ω)| ≈ A0ωb

ω=ωt

ω

Unity gain occurs at ωt ωt = A0ωb


Frequency Response Frequency Response of Closed-Loop Amplifiers

Frequency Response of Closed-Loop Amplifiers

The effect of limited op-amp gain and bandwidth on the closed-looptransfer functions of the inverting configurations.Step 1 Define closed-loop gain of an inverting amplifier with finiteopen-loop gain (A)

Vo

Vi=

−R2/R1

1 + (1 + R2/R1)/A

Step 2 Insert frequency-dependent description of A

Vo

Vi=

−R2/R1

1 +1 + R2/R1(

A0

1 + s/ωb

) =−R2/R1

1 +(

1+R2/R1

A0

)+ s

ωb

(1+R2/R1

A0

)

Step 3 Assume A0 � 1 + R2/R1

Vo

Vi=

−R2/R1

1 + s(1+R2/R1)ωt


Frequency Response Frequency Response of Closed-Loop Amplifiers

Frequency Response of Closed-Loop Amplifiers

By using the same methods the effect of limited op-amp gain andbandwidth on the closed-loop transfer functions of the noninvertingconfigurations.

Vo

Vi=

1 + R2/R1

1 + s(1+R2/R1)ωt

3dB frequency - is the frequency at which the amplifier gain is attenuated3dB from maximum (aka. dc value).

ω3dB =ωt

1 + R2/R1


Large-Signal Operation of Op Amps


The following are limitations on the performance of op-amp circuits whenlarge output signals are present.

1 Output Voltage SaturationOp amps operate linearly over a limited range of output voltages. Ifsupply voltage +/- 15V is vO will saturate around +/- 13V.

2 Output Current LimitsAnother limitation on the operation of op amps is that their outputcurrent is limited to a specified maximum. If the circuit requires a largercurrent, the op-amp output voltage will saturate at the levelcorresponding to the maximum allowed output current.

3 Slew RateSlew Rate is the maximum rate of change possible at the output of areal op amp.

SR =dvodt

max




If slew rate is less than rate of change of input it becomes problematic.Slewing occurs because the bandwidth of the op-amp is limited, so theoutput at very high frequencies is attenuated.



Large-Signal Operation of Op Amps4 Full-Power Bandwidth

Op-amp slew-rate limiting can cause nonlinear distortion in sinusoidalwaveforms.Assume a unity-gain follower with a sine-wave input

vI = Vi sinωt

The rate of changedvIdt

= ωVi cosωt

Now if ωVi exceeds the slew rate of the op amp, the output waveformwill be distorted in the manner shown.




Full-power bandwidth (fM) is the frequency at which an output sinusoidwith amplitude equal to the rated output voltage of the op amp begins toshow distortion due to slew-rate limiting.

SR = ωMVoMax fM =SR

2πVoMax

Maximum output voltage (VoMax) - is equal to (AvI ).

Output sinusoids of amplitudes smaller than VoMax will show slewratedistortion at frequencies higher than fM

At a frequency ω higher than fM , the maximum amplitude of theundistorted output sinusoid is

Vo = VoMax

(ωM

ω

)Chapter 3: Operational Amplifier Part 1- Op Amp Basics (AAIT)Chapter Three April 27, 2016 46 / 46

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Applied Electronics II · Applied Electronics II Chapter 3: Operational Ampli er Part 1- Op Amp Basics School of Electrical and Computer Engineering Addis Ababa Institute of Technology