OP-AMPS in SIGNAL PROCESSING APPLICATIONS

Filters, Integrators, Differentiators, and Instrumentation Amplifier

Sensors :definition and principles

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

What is electronics engineering all about?Process

Sensors/ transducers

Filters

Amplifiers Feedback

Actuators/transducers

New process

A/D converters Computers

D/A converters

Physical variables

Electrical variablesV, I, t, f

Physical variables

F, P, d, v, t, f

What is a Signal Processor?

• It selects useful parts of a signal

• It cleans signal from impurities

• It compares signals

• It converts signals into forms that can be easily recognized

The ECG wave in it’s raw form

Examples of biomedical

signals

Examples of Signal Processors

• Basic amplifiers – already discussed

• Instrumentation amplifier – an improved differential amplifier

• Active filters

• Integrators and differentiators

• Extra Reading:– Precision rectifiers

– Logarithmic amplifiers

– Negative capacitance amplifiers

Signal Conditioning

• Ideal operational amplifier

• Inverting, non-inverting and instrument amplifier

• Integrator, differentiator

• Filters

• Examples of signal conditioning module

Ideal Operational Amplifier

Operational (OP) amplifier is a high-gain dc differential amplifier. It is made of an integrated circuit chip. It has two inputs, negative terminal v1 and positive terminal v2 and one output v0. The effective input to the amplifier is the voltage difference of the two inputs v1-v2.

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.

Basic Rules for Ideal OP Amplifier

• When the OP amplifier output is in its linear range, the two input terminals are at the same voltage.

• No current flows into either input terminal of the OP amp.

Inverting Amplifier

Non-inverting Amplifier

Instrument Amplifier

Active Filters

• Low-pass

• High-pass

• Band-pass

• Band-stop (notch)

• All-pass (phase-shift)

Low-pass filter: circuit

i

f

ZZ

G −=

V0 Vi A +

- Ri Rf

Cf

GL = Rf/Ri ; τf = RfCf ; fc = 1/2πτf

Derive the equation for gain G

What are the important parameters?

)()(0

ωω

jVjVG

i

=

i

ff

f

f

R

RCj

CjR

+

−=ω

ω

1

( ) f

L

fi

f

iff

f

jG

jRR

RCRjR

ωτωτω +=

+−=

+−=

111

1

Low-pass filter: characteristics

Gain/GL

f/fc

0.1 1 10 100

0.1

0.01 Slope = -1

1

-3π/2

Phase f/fc

-5π/4

-π

GL = Rf/Ri ; τf = RfCf ; fc = 1/2πτf

V0 Vi A +

- Ri Cf

Integrator – ideal: circuit

i

f

i

f

i RCj

ZZ

jVjVG

ωωω

1

)()(0 −=−==

VirtualGround

i

i

∫ +−=t

iCifi

VdtVCR

V00

1

dtdVC

RVi f

i

i 0==

Express V0 in terms of Vi Write expression for gain G

ffi jCRj ωτω11

−=−=

Where τ = RiCf

Integrator – ideal: characteristics

G/GL

f/fc

0.1 1 10 100

0.1

0.01

0

10

100

Slope = -1

1

-3π/2

Phase f/fc

-5π/4

-π

Draw G/GL versus f/fc

Draw phase versus f/fc

ffG cc ====

ωω

ωτ1

°−=−= 2702/3πθ

The integrator with bias currents

Through which component the bias current flows?

Gain/GL

f/fc

0.1 1 10 100

0.1

0.01 Slope = -1

Integrator - practical

V0 Vi A +

- Ri

Rf

Cf

f

L

fi

f

i jG

jRR

jVjVG

ωτωτωω

+=

+−==

111

)()(0

Without Rf

With Rf

Rf provides route for DC bias currents 1Phase f/fc

-3π/2

-5π/4

-π

Without Rf

Low-pass filter Integ.

fcfff fCR

πττ

21; ==

Charge amplifier

vo

Electrode

The piezoelectric sensor generates charge, which is transferred to the capacitor, C, by the charge amplifier. Feedback resistor Rcauses the capacitor voltage to decay to zero

Charge amplifier: circuit

V0 IsC

FET +

-

Rf

Cf

IsR

dtKdxidtdq ss // ==

Piezoelectric sensor

Is

VirtualGround

IsC = IsR = 0 ∫ −=−=−=t

ff CKXdt

dtKdx

CVV

001

Step response of a charge amplifierxi

Deflection

vo

vo max

vo max /e

Time

Time

The charge amplifier responds to a step input with an output that decays to zero with a time constant τ = RfCf

High-pass filter: circuit

V0 Vi A +

- Ri Rf Ci

i

f

i ZZ

jVjVG −==

)()(0

ωωGH = -Rf/Ri ; τi = RiCi ; fc = 1/2πτi

Derive the equation for gain G

What are the important parameters?

i

Hi

i

i

i

f

jGj

jj

RR

ωτωτ

ωτωτ

+=

+−=

11ii

if

ii

f

CRjCRj

CjRR

Gω

ω

ω +−=

+−=

11

High-pass filter: characteristicsGain/GH

f/fc

0.1 1 10 100

0.1

0.01Slope = +1

1

-π

Phase f/fc

-3π/4

-π/2

Draw G/GH versus f/fc

Draw phase versus f/fc

Differentiator - ideal

V0 Vi A +

- Rf Ci

I

I

dtdVCRV

dtdVCi

iif

ii

−=

=

0

ωτωωω

ω

jCRjGCj

RZZ

jVjVG

if

i

f

i

f

i

−=−=

−=−== 1)()(0

Slope = +1

fc = 1/2πτ

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

Differentiator - practical

V0 Vi A +

- Rf Ci

Cf

f

f

f

i

ff

if

i

ff

ff

i

f

i jj

CC

CRjCRj

Cj

CjRCjR

ZZ

jVjVG

ωτωτ

ωω

ω

ω

ω

ωω

+−=

+−=

+−=−==

111

1

1

)()(0

In an ideal differentiator, the highfrequency response is limited by theopen-loop gain of the op-amp yieldinga very noisy output voltage.

τf = RfCf ; fc = 1/2πτf

Cf is used to limit the hf gain, hence the hf noise.

The gain at hf: GH = - Ci/Cf

Differentiator characteristics

Slope = +1

1

-π

Phase f/fc

-3π/4

-π/2

Gain/GH

f/fc

0.1 1 10 100

0.1

0.01

10 Without Cf

With Cf

With Cf

Differentiator High-pass filter

Measurement of integrator and differentiator characteristics

V0 Vi A +

- 1k5

15 k

10 nF

Integrator

• Form your lab team and assign duties.

• Go to the lab bench.

• Built the circuit given: integrator for team 1 and differentiator for team 2

• Connect the power supply.

• Connect the CRO

• Do the experiment according to the procedures in the sheet - 6.

• Switch off the PS and signal generator

• Return your seat

V0 Vi A +

- 15 k 0.1µF

10 nF Differentiator

Band-pass filter: circuit

V0 Vi A +

- 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

jjjG

jjj

RR

G

CRjCRjRCj

CjR

RCj

CjR

ZZ

jVjVG

ωτωτωτ

ωτωτωτ

ωωω

ω

ω

ω

ωω

++=

++−=

++−=

+

+

−=−==

Mid-band gain =GMB = - Rf/Ri

Time-constants: τI= RiCi ; τf = RfCf

Critical frequencies: fL = 1/2πτI ; fH = 1/2πτf

Band-pass filter: characteristics

Gain/GH

f/fc

0.1 1 10 100

0.1

0.01

Slope = -1Slope = +1

Band-pass filter (non-inverting)

Frequency response of bpf

BPF with bias compensation

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

V2 A

+

-

Simple comparator (inverting)

V0 Vref A

+

- R1

R2

Vi

V-

Output stays at +VSAT if V-<0

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

Output goes to -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

VSA

T

Comparator with hysteresis

V0 Vref A

+

- R1

R2

Vi

V-

R3 R4

V+

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

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

Output stays at +VSAT if V-<V+

Output goes to -VSAT if V->V+

Characteristics

V0 Vref A

+

- R1

R2

Vi

V-

R3 R4

V+

V0

Vi

10V

10V -10V

-10V VL

+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

VL

Vhys

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

Instrumentation Amplifier

Overall Circuit Structure

• Overall differential gain is given by:

R4

R3

vo

R3

R4

R2

va

R2

R1

vb

VS+

VS+

VS+

1

221RRG +=α

3

4

RRG =β

+==

3

4

1

221RR

RRGGGD βα

DifferenceAmplifier

Input Conditioner

VS–

VS–

VS–

In-Amp: Input Conditioner

• Note that iR1 = iR2a = iR2b. Then the diff. output voltage 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''RR

vvvvG

ab

abD +=

−−

=

11Rivv Rab =−VS–

VS–

In-Amp: Input Conditioner

• This input conditioner does not amplify common-mode signals!

• Brief proof of principle:– Since vb – va = iR1R1:

– The output voltages are thus equal to:

01 =⇒== Rcmba ivvv

cmba vvv == ''

R2

va

R2

R1

vb

VS+

VS+

va'

vb'

iR1

iR2a

iR2b

VS–

VS–

In-Amp: Difference Amplifier

1) From voltage divider principles:

2) Note that iR1a = iR2a. Thus:

R2

R1

vbvo

va

R1

R2

VS+

+

=+21

2

RRRvv b iR1a

iR2a

12 Rvv

Rvv ao −

=− −−

1122

11Rvv

RRRv ao −

+=⇒ −

VS–

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

1

2

1

2 )(RR

vvvGvv

RRv

ab

odabo =

−=⇒−=⇒

+

=−21

2

RRRvv b

112

2

12

12

2 Rvv

RRR

RRRR

Rv a

bo −

+

+=

VS–