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OP-AMPS in SIGNAL PROCESSING

APPLICATIONS

Filters, Integrators, Differentiators,

and Instrumentation Amplifier

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Sensors :

definition and principles

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Sensors : taxonomies

Measurand physical sensor chemical sensor biological sensor(cf: biosensor)

Invasiveness invasive(contact) sensor noninvasive(noncontact) sensor

Usage type multiple-use(continuous monitoring) sensor disposable sensor

Power requirement passive sensor active sensor

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What is electronics engineering all

about?

Process

Sensors/ transducers

Filters

Amplifiers Feedback

Actuators/transducers

New process

A/D converters Computers

D/A converters

Physicalvariables

ElectricalvariablesV, I, t, f

Physicalvariables

F, P, d, v, t, f

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What is a Signal Processor?

It selects useful partsof a signal

It cleans signal fromimpurities

It compares signals

It converts signals into

forms that can be

easily recognized

The ECG wave in its raw form

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Examples of

biomedical

signals

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Examples of Signal Processors

Basic amplifiers already discussed

Instrumentation amplifier an improveddifferential amplifier

Active filters

Integrators and differentiators

Extra Reading:

Precision rectifiers

Logarithmic amplifiers

Negative capacitance amplifiers

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Signal Conditioning

Ideal operational amplifier

Inverting, non-inverting and instrumentamplifier

Integrator, differentiator

Filters

Examples of signal conditioning module

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Ideal Operational Amplifier

Operational (OP) amplifier is a high-gain dc differentialamplifier. It is made of an integrated circuit chip. It has twoinputs, negative terminal v1 and positive terminal v2 and oneoutput v0. The effective input to the amplifier is the voltagedifference of the two inputs v1-v2.

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Properties of Ideal OP Amplifier

Gain if infinity (A=).

V0 = 0, when v1 = v2

Input impedance is infinity. Output impedance is zero.

Bandwidth is infinitely wide and has no phase

shift.

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Basic Rules for Ideal OP Amplifier

When the OP amplifier output is in its linearrange, the two input terminals are at the same

voltage.

No current flows into either input terminal ofthe OP amp.

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Inverting Amplifier

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Non-inverting Amplifier

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Instrument Amplifier

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Active Filters

Low-pass

High-pass

Band-pass Band-stop (notch)

All-pass (phase-shift)

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Low-pass filter: circuit

i

f

Z

ZG =

V0ViA

-Ri Rf

Cf

GL = Rf/Ri ; f= RfCf; fc = 1/2f

Derive the equation for gain G

What are the important parameters?

)(

)(0

jV

jVG

i

=

i

ff

f

f

R

RCj

CjR

+

=

1

( ) fL

fi

f

iff

f

j

G

jR

R

RCRj

R

+=

+=

+=

11

1

1

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Low-pass filter: characteristics

Gain/GL

f/fc

0.1 1 10 100

0.1

0.01 Slope = -1

1

-3/2

Phase f/fc

-5/4

-

GL = Rf/Ri ;

f= RfCf; fc = 1/2

f

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V0Vi A

-Ri Cf

Integrator ideal: circuit

i

f

i

f

i RCj

ZZ

jVjVG

1

)()(0 ===

Virtual

Ground

i

i

+=

t

iCifi

VdtVCR

V00

1

dt

dVC

R

Vi f

i

i 0==

Express V0 in terms of Vi Write expression for gain G

ffi jCRj 11 ==

Where = RiCf

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Integrator ideal:

characteristics

G/GL

f/fc

0.1 1 10 100

0.1

0.01

0

10

100

Slope = -1

1

-3/2

Phase f/fc

-5/4

-

Draw G/GL versus f/fc

Draw phase versus f/fc

f

fG cc ====

1

== 2702/3

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The integrator with bias currents

Through which component the bias current flows?

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Gain/GL

f/fc

0.1 1 10 100

0.1

0.01 Slope = -1

Integrator - practical

V0Vi A

-Ri

Rf

Cf

f

L

fi

f

i j

G

jR

R

jV

jVG

+=

+==

11

1

)(

)(0

Without Rf

With Rf

Rfprovides route for DC bias currents

1Phase f/fc

-3/2

-5/4

-

Without Rf

Low-pass filter Integ.

fcfff fCR 2

1

; ==

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Charge amplifier

vo

Electrode

The piezoelectric sensor generates charge, which is transferred tothe capacitor,C, by the charge amplifier. Feedback resistorR

causes the capacitor voltage to decay to zero

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Charge amplifier: circuit

V0IsC

FET

-

Rf

Cf

IsRdtKdxidtdq ss // ==

Piezoelectricsensor

Is

VirtualGround

IsC = IsR = 0 ===t

ff C

KXdt

dt

Kdx

CVV

00

1

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Step response of a charge amplifierxi

Deflection

vo

vo max

vo max /e

Time

Time

The charge amplifierresponds to a step inputwith an output thatdecays to zero with atime constant=RfCf

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High-pass filter: circuit

V0ViA

-Ri RfCi

i

f

i Z

Z

jV

jVG ==

)(

)(0

GH = -Rf/Ri ; i = RiCi ; fc = 1/2i

Derive the equation for gain G

What are the important parameters?

i

Hi

i

i

i

f

j

Gj

j

j

R

R

+=

+=

11ii

if

ii

f

CRj

CRj

CjR

RG

+

=+

=11

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High-pass filter: characteristics

Gain/GH

f/fc

0.1 1 10 100

0.1

0.01Slope = +1

1

-

Phase f/fc

-3/4

-/2

Draw G/GH

versus f/fc

Draw phase versus f/fc

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Differentiator - ideal

V0ViA

-RfCi

II

dt

dVCRV

dt

dVCi

iif

ii

=

=

0

jCRjG

Cj

R

Z

Z

jV

jVG

if

i

f

i

f

i

==

===1)(

)(0

Slope = +1

fc = 1/2

Phase shift = - 90o

Gain

f/fc

0.1 1 10 100

0.1

0.01

1

10

100

Express V0 in terms of Vi

Write expression for gain G

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Differentiator - practical

V0ViA

-RfCi

Cf

f

f

f

i

ff

if

i

ff

ff

i

f

i j

j

C

C

CRj

CRj

Cj

CjR

CjR

Z

Z

jV

jVG

+

=

+

=

+

===

111

1

1

)(

)(0

In an ideal differentiator, the high

frequency response is limited by theopen-loop gain of the op-amp yieldinga very noisy output voltage.

f= RfCf; fc = 1/2f

Cfis used to limit the hf gain,

hence the hf noise.

The gain at hf: GH = - Ci/Cf

ff

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Differentiator characteristics

Slope = +1

1

-

Phase f/fc

-3/4

-/2

Gain/GH

f/fc

0.1 1 10 100

0.1

0.01

10 Without Cf

With Cf

With Cf

DifferentiatorHigh-pass filter

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Measurement of integrator and

differentiator characteristics

V0V i A-

1k5

15 k

10 nF

Integrator

Form your lab team and assignduties.

Go to the lab bench.

Built the circuit given:

integrator for team 1 anddifferentiator for team 2

Connect the power supply.

Connect the CRO

Do the experiment according tothe procedures in the sheet - 6.

Switch off the PS and signalgenerator

Return your seat

V0ViA

-15 k0.1F

10 nFDifferentiator

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Band-pass filter: circuit

V0ViA-

Ri Rf

Cf

Ci

( )( )

)1)(1()1)(1(

111

1

)(

)(0

fi

iMB

fi

i

i

f

iiff

fi

ii

ff

f

f

i

f

i

jj

jG

jj

j

R

R

G

CRjCRj

RCj

CjR

RCj

CjR

Z

Z

jV

jVG

++=++=

++=

+

+

===

Mid-band gain =GMB = - Rf/Ri

Time-constants: I= RiCi ; f= RfCf

Critical frequencies: fL = 1/2I ; fH = 1/2f

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Band-pass filter: characteristics

Gain/GH

f/fc

0.1 1 10 100

0.1

0.01

Slope = -1Slope = +1

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Band-pass filter (non-inverting)

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Frequency response of bpf

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BPF with bias compensation

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Comparators: Basic Rules

V0=A(V2-V1)

V0 = 0 if V1 = V2 V0 = +VSAT if V1 < V2 V0 = -VSAT if V1 > V2 No current flows into either

input terminals of the op-amp.

V0

V1

V2A

+

-

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Simple comparator (inverting)

V0Vref A

+

-

R1

R2

Vi

V-

Output stays at +VSAT if V-0

+10V

-10V

V0

Time

-Vref*R1/(R1+R2)

Vi*R2/(R1+R2)

V0

Vi

10V

10V-10V

-10V-Vref*R1/R2

-

VSAT

VSAT

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Comparator with hysteresis

V0Vref A

+

-

R1

R2

Vi

V-

R3R4

V

V- = (Vi*R2+Vref*R1)/(R1+R2)

V+= V0*R3 /(R3+R4)

Output stays at +VSAT if V-V+

V

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CharacteristicsV0

Vref A

+

-

R1

R2

Vi

V-

R3R4

V+

V0

Vi

10V

10V-10V

-10VVL

+VSAT

-VSAT

VH

Vhys

VH = -Vref*R1/R2 VSAT*R3/(R3+R4)

VL = -Vref*R1/R2 + VSAT*R3/(R3+R4)

+10V

-10V

V0

Time

-Vref*R1/(R1+R2)

Vi*R2/(R1+R2)

VH

VLVhys

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Op-Amps in Reality

Fabricated using integrated-circuit technologies

Inside an op-amp:

1) Transistors

2) Parasitic capacitors

3) Internal resistors

Op-amp chip configuration that we will use:Quad-channel (i.e. 4-in-1)

14 pins in the op-amp chip

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Instrumentation Amplifier

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Overall Circuit Structure

Overall differential gain is given by:

R4

R3

vo

R3

R4

R2

va

R2

R1

vb

VS+

VS+

VS+

1

221R

RG +=

3

4

R

RG =

+==

3

4

1

221R

R

R

RGGGD

DifferenceAmplifier

InputConditioner

VS

VS

VS

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In-Amp: Input Conditioner

Note that iR1 = iR2a = iR2b. Then the diff. outputvoltage is given by:

The diff. input voltage is equal to:

The differential gain is thus equal to:

R2

va

R2

R1

vb

VS+

VS+

va'

vb'

iR1

iR2a( )211 2'' RRivv Rab +=

iR2b

1

221''

R

R

vv

vvG

ab

abD +=

=

11Rivv Rab =VS

VS

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In-Amp: Input Conditioner

This input conditioner does not amplifycommon-mode signals!

Brief proof of principle:

Since vb v

a= iR1R1:

The output voltages are thusequal to:

01 === Rcmba ivvv

cmba vvv == ''

R2

va

R2

R1

vb

VS+

VS+

va'

vb'

iR1

iR2a

iR2b

VS

VS

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In-Amp: Difference Amplifier

1) From voltage divider principles:

2) Note that iR1a = iR2a. Thus:

R2

R1

vbvo

va

R1

R2

VS+

+=+

21

2

RR

Rvv b iR1a

iR2a

12 R

vv

R

vv ao =

1122

11

R

vv

RRR

v ao

+=

VS

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In-Amp: Difference Amplifier

3) Note that v

= v+. So from the voltage divider

expression:

4) Substituting the above into the

output voltage expression, we get:

R2

R1

vbvo

va

R1

R2

VS+iR1a

iR2a

22 )(Rv

GvvR

v o ===

+= 212

RR

R

vv b

112

2

12

12

2 R

vvRR

R

RR

RR

R

v ab

o

+

+=

VS