lecture 6 In The Name of Allah - Ali Karimpourkarimpor.profcms.um.ac.ir/imagesm/354/stories/... · 2017. 5. 21. · lecture 6 Dr. Ali Karimpour, Apr 2017 3 Introduction In manufacturing

lecture 6

Dr. Ali Karimpour, Apr 2017

Instrumentation

In The Name of Allah

Dr. Ali Karimpour

Associate Professor

Ferdowsi University of Mashhad

lecture 6


2

Signal Conditioners and Transmission

Topics to be covered

v Introduction

v Instrumentation Amplifier

v Z V Conversion

v Zero and Span Circuits

v V I and I V Conversion

v V F and F V Conversion

v Filter

v Isolation Circuits

v Cabling

Lecture 6

lecture 6


3

Introduction

In manufacturing process electric signals are interested, but:

Signals may be much too small. It goes down instead of up.

It has undesired dc offset. It may be nonlinear.

Series resistance in the wiring and connectors …

Interference from power sources will introduce 60-Hz and RFI

noises.

Failures, faults, and installation errors may introduce several

hundred volts into the signal.

Difference in earth ground potential can also cause erroneous

reading several times larger than the signal.

This problems combine to make virtually useless the signal

that arrives at the display and controller.

lecture 6


4

Introduction

Fortunately, each of mentioned problems has a solution.

Proper signal conversion and transmission.

Current transmission rather than voltage.

Encoding the information in the frequency of the signal rather than

its amplitude.

Isolation amplifiers and couplers protect circuiting from fault

voltage.

Proper shielding and grounding break ground loop.

Signals may be much too small.

It goes down instead of up.

It has undesired dc offset.

It may be nonlinear.

Series resistance in the wiring

and connectors …

Interference from power sources

will introduce 60-Hz and RFI noises.

Failures, faults, and installation errors

may introduce several hundred volts

into the signal.Difference in earth ground potential

can also cause erroneous reading several

times larger than the signal.

lecture 6


5

Introduction

Different parts of this lecture:.

Proper signal conversion and transmission.

Current transmission rather than voltage.

Encoding the information in the frequency of the signal rather than

its amplitude.

Isolation amplifiers and couplers protect circuiting from fault

voltage.

Proper shielding and grounding break ground loop.

• Instrumentation Amplifier• Zero and Span Circuits

• Current to Voltage Conversion• ……………………………….

• Isolation Circuits

• Cabling

lecture 6


6

Introduction

PVSPe

Drawbacks:

3v ?

I ?

Drawbacks: Small signals buried in large common mode offsets or noise.

e.2u

We can use two voltage follower.

5 v ?

lecture 6


7


v Introduction


v Z V Conversion




v Filter


v Cabling

lecture 6


It is the first

stage of IA.

8

Instrumentation Amplifier

3v ?

I ?

u ?

g

g

R

PVSPI

gI

)2( 10 gg RRIV

)2

1)(( 10

gR

RPVSPV

ground ? 02

2Output VR

R

• A very high

input impedance.

• Setting the gain

by a resistor.

Second stage of IA.

lecture 6


9


gI

0

2

2Output VR

R

0outputReal V

RR

R

wireL

L

wireL

L

RR

R

reductionGain

lecture 6


gI

10


2

2gainNewR

RRwire

?LI

wireL

L

RR

R

reductionGain

By connecting the sense terminal at the load, any nonlinearity

and offsets between the output pin and the load are eliminated.

lecture 6


11


Is it possible to put an offset on the output?

lecture 6


12


Functional Block diagram of AD524 Instrumentation Amplifier

Three laser-trimmed resistors are closely matched to 20-kΩ resistors.

(Value and Temperature matched)

Gain=1

)40000

1(g

RGain

Gain=10

)4440

400001( Gain

Other gains?

lecture 6


13


AD524 Instrumentation Amplifier

dual in-line package (DIP)

Small Outline Integrated Circuit (SOIC)

Leadless chip carriers

lecture 6


14



lecture 6


15



lecture 6


16


Example 6-1: Derive Va, Vb and Vout if variable resistor is 349 ohms.

Answer: Vout=-715 mv

Note that 5 volts was removed from input.

Common-mode rejection ratiocA

GCMRR log20

modecommon

)modecommon (

e

VA

out

c

lecture 6


17


Example 6-2: For previous example derive the common mode error?

According to data sheet CMRR =100 db for gain of 100

100log20 cA

G 510cA

G001.0cA

001.0modecommon

)modecommon (

e

VA

out

c mvVout 5005.0)modecommon (

So common mode error is:

%7.0%100715

5

lecture 6


18


Any difference between IR1 and IR2 is called the offset current.

Input voltage at the inverting input is e1-IR1RS1

And the voltage at the nonnegative input is e2-IR2RS2

So, assure that the impedance from the inputs to ground(source

impedance) are equal.

lecture 6


19


So, assure that the impedance from the inputs to ground(source

impedance) are equal.

Any way there is difference between IR1 and IR2 so:

Try to keep the source resistance as low as possible.

Assure that the instrumentation amplifier and the sensor have a

common ground.

lecture 6


20


The effects of offset voltage, bias currents, and offset current can be

eliminated by calibration of amplifier.

1. Connect both inputs to

same source.

2. Set the gain to maximum.

3. Adjust the left

potentiometer to give a

zero output.

4. Change the circuit to give

the gain of 1.

5. Adjust the right

potentiometer to give a

zero output.

6. Repeat from 2.

lecture 6


21


v Introduction


v Z V Conversion




v Filter


v Cabling

lecture 6


22

Impedance to Voltage Conversion

• Voltage divider

ref

MF

F

outV

RR

RV

FMRR

ref

M

F

outV

R

RV

Drawbacks:

Source voltage

variation ?

ref

MF

M

outV

RR

RV

ref

MF

outV

RRV

1/

1

Load current ?

Output is not proportional with

the resistor deviation?

lecture 6


23


• Voltage divider

• Wheatstone Bridge

refoutV

R

RRR

RV

1)1(1

1

1,2 RR

Drawbacks: Output is not proportional with the resistor deviation?

refoutVV

4

Output is proportional with the resistor deviation .

lecture 6


24


• Wheatstone Bridge

refoutV

R

RRR

RV

1)1(1

1

1,2 RR

refoutVV

4

1,2 RR

refoutVAV

4

Be sure about source voltage variation ?

Drawbacks: Resistance of wiring

and temperature variation.

lecture 6


25


• Wheatstone Bridge Drawbacks: Resistance of wiring and temperature variation.

refoutVV

4

lecture 6


26


• Wheatstone Bridge Drawbacks: Resistance of wiring and temperature variation.

Dummy sensor

Placement of active and dummy gages for temperature compensation.

refoutVV

4

lecture 6


27


Example 6-3: Determine Vout if

R=240 ohms, Vref=10 V, and

a) Stress causes the upper resistor to

increase by 0.013 ohms.

b) Stress is zero but the upper resistor

increase by 9.4 ohms since of temperature.

c) Stress causes the upper resistor to increase by 0.013 ohms and the

upper resistor increase by 9.4 ohms since of temperature.

Answer of a) Vout= 0.135 mV

Answer of b) Vout= 96 mV ???!!!!

Answer of c) Vout= 96.2 mV ???!!!!

lecture 6


28


Example 6-4:

Determine Vout if

R=240 ohms, E=10 V, and

b) Stress is zero but the right resistors increase by 9.4 ohms since of

temperature.

c) Stress causes the upper resistor to increase by 0.013 ohms and the

right resistors increase by 9.4 ohms since of temperature.

a) Stress causes the upper resistor to increase by 0.013 ohms.

Dummy sensor

Answer of a) Vout= 0.135 mV

Answer of b) Vout= 0 mV it is ok.

Answer of c) Vout= 0.13 mV it is ok.

lecture 6


29


v Introduction


v Z V Conversion




v Filter


v Cabling

lecture 6


30

Zero and Span Circuits

The output of an amplifier rarely matches the levels you want to

provide to the controller, display, or computer.

For example one may need a 10 mV/ib inputs to a digital panel meter,

while a load cell provides a 0.02 mV/ib, and 18 mV output with no

load.

lecture 6


31


Temperature outputs 2.48 to 3.9 V

We need 0 to 5 V for ADC.

bmee inout

VR

Re

R

Re

os

f

in

f

u 1

1 VR

Re

R

Re

os

f

in

f

out 1

Zero and span with inverting amplifier

lecture 6


32


VR

Re

R

Re

os

f

in

i

f

out

Example 6-5: Suppose temperature outputs of a sensor is 2.48 to 3.9 V

derive a circuit that the output be 0 to 5 V.

52.348.29.3

05

i

f

R

Rm

current)(sensor high bemust iR

k330fR

terpotentiomek100k47,k7.93 iR

VR

R

os

f )48.2(52.30 73.8V

R

R

os

f

lecture 6


33


VR

Re

R

Re

os

f

in

i

f

out

Example 6-5(Continue): Suppose temperature outputs of a sensor is

2.48 to 3.9 V derive a circuit that the output be 0 to 5 V.

73.8VR

R

os

f

12VLet 12/73.8os

f

R

R

terpotentiome500220,454 kandkkRos

kRLetkRRRR composifcomp 56,9.62||||

lecture 6


34


refinout VeRg

e )40000

1(

Span

Zero

Example 6-6: A load cell changes 20 μV/ib

With an 18 mV output at no load.

Design a zero and span circuit which will

Output 0 V dc when there is no load and

will change 10 mV/ib

50002.0/10)40000

1( Rg

100332.80 Rg

refVmVV )18(5000 VVref 9

Zero and span with Instrumentation amplifier

lecture 6


35


v Introduction


v Z V Conversion




v Filter


v Cabling

lecture 6


36

Voltage to Current Conversion

Voltage transmission have some problems:

* Series resistance between the output and load.

* The wire used.

* Temperature dependence.

Current transmission have some properties:

* None will be lost since of wiring resistance.

Types of V2C Conversion

* Floating Load * Grounded Load

lecture 6


37


R

eI in

Remark 1:

satin

loopVe

R

R )1(

Remark 2:

Deriving current?

* Floating Load

Negative current?

lecture 6


38


* Floating Load

Rload inf Vload +V

Rload 0 Vload ein

Remark 3:

lecture 6


39


* Floating Load

Bias

Remark 4: I=0 is natural or circuit failure?

lecture 6


40


Suppose we need

Span

Zero

R

VI R

2

refin

R

eeV

R

eeI

refin

2

refeAeARI )()(2

refeBeBRI )()(2 ))()((2

)()(

AIBI

AeBeR

)()(2 AeARIeref

lecture 6


41


Example 6-7: Suppose we need

VAeARIeref 8.8)5()004.0)(469(2)()(2

100430469001.0)420(2

)5(10

))()((2

)()(

AIBI

AeBeR

lecture 6


42


* Grounded Load

12 eeVV Lout

12 eeVVV LoutRs

sR

eeI 12

Remark 1: satload VeeIR 12

Remark 2: Deriving current?

lecture 6


43


Example 6-8: Design a V2C that span 0 to 1 V to 4 to 20 mA.

For a current 20 mA and ±15 V supplies, what is the maximum load resistance.

s

in

R

eeI 1

sR

e10004.0

sR

e11020.0

Ve 25.01

100225.62 sR satload VeeIR 6.012

5586.012

I

eeVR satload

Span

Zero

lecture 6


44


Buy a converter than build one? XTR110 precision V2C converter.

XTR110 precision V2C converter.

It converts 0-5 V or 0-10 V 4-20 mA or 5-25 mA

It can't give directly required current so it needs an external MOS transistor.

So, it keeps heat outside the XTR110 package to optimise its performance.

lecture 6


45


V2C

C2C

Precision

resistor

divider

network

Precision

resistor

divider

network

Vin1 (10-V full scale)

Vin2 (5-V full scale)

Vref (for offsetting)

2416

21 ininref VVVV

span

ininref

RR

VVV

I 241621

8

Precision

10 V

reference

lecture 6


46


V100

k67.6V5.20

mA6.10

mA20

mA200

mA200

0

V2C

C2C

mA4.00

0 to 20 mA

0

0

lecture 6


47


Adding 4 mA

to output.

Vin 0 to 10 V Iout 4 to 20 mA

lecture 6


48


lecture 6


49


0-10 V

0-10 A

lecture 6


50


Exercise 6-1 : Explain the performance of following system.

Span

Zero

lecture 6


51

Current to Voltage Conversion

Once the current signal gets to the place where it is to be used, it must

be converted back to voltage.

Types of C2V Conversion

* Grounded Load * Floating Load

But it needs suitable spanning.

lecture 6


52

Current to Voltage Conversion* Ground Reference

Effect of load

on output? Zero and Span?Span

Zero

VR

RV

R

RV

os

f

R

i

f

a L V

R

RV

R

RV

os

f

R

i

f

out L

aV

Drawback?

There must be a ground return.

Voltage drop in ground return resistance.

Any change in ground return resistance …..

lecture 6


53


* Floating

zspan

i

f

out VIRR

RV

Effect of load

on output?Span

Zero

z

f

f

b

i

f

a

i

f

out VR

RV

R

RV

R

RV

Common mode producing no difference across the load.

spani RR

lecture 6


54


Example 6-9: Design a floating C2V converter that will convert a 4-20

mA current signal into a 0-10 V ground referenced voltage signal.

VbVmAbI

VaVmAaI

10)(20)(

0)(4)(

zspan

i

f VRaIR

RaV )()(

zspan

i

f VRbIR

RbV )()(

10/ Choose if

RR

5.62)004.002.0(10

010

spanR

kRspan

33ter potentiomemultiturn 50

lecture 6


55


What happen if we reduce Rf/Ri ?

625*20 mA=12.5 V ????

Rf/Ri=1, then Rspan I-V==625 Ω

Rf/Ri=10,

Rf=10Ri=22 kΩ

V

k

kVVz

5.2

)5.62(2.2

220

lecture 6


56


625*20 mA=125 V ????

lecture 6


57


Example 6-10: A V2C converter, has 12 V, Rspan V-I=312 Ω, Imax=20

mA. What is the maximum size Rspan I-V?

IVspanVIspan IRIRV 7.02

I

IRVR IVspan

VIspan

7.2(max)

153

02.0

)312)(02.0(7.212

(max)VIspanR

lecture 6


58


Example 6-11: A modular transducer output 10 to 60 mA of current. The

manufacturer indicates that 100 Ω is the maximum allowable floating

load. Select a gain (Rf/Ri) and Rspan for the C2V converter of figure to

give a -10 to +10 V output.

)()(

)()(

aIbI

aVbVR

R

Rspan

i

f

400

01.006.0

)10(10span

i

fR

R

R

lecture 6


59


v Introduction


v Z V Conversion




v Filter


v Cabling

lecture 6


60

Voltage to frequency conversion

Transmission of current rather than voltage:

It is eliminate the loop resistance error.

Differential nature of floating current transmission allows one to use

an instrumentation amplifier(IA) with high common-mode rejection.

Even best IA may not be able to reduce adequately errors due to

noise picked up in the transmission loop.

Using V2F:

Analog voltage Frequency Current Voltage Analog voltage

lecture 6


61


Analog voltage Frequency Current Analog voltage Frequency Current Voltage Analog voltage Frequency Current Voltage Analog voltage

lecture 6


62


A V2F circuit

1- Input comparator

2- One-shot timer

3- Transistor

4- Switch current source

What is in the chip?

Others are out of chip.

lecture 6


63

Voltage to frequency

conversion

How a V2F works?

lecture 6


64

Voltage to frequency

conversion

In this figure 1-10 V 0-10 kHz

Suggested by manufacturer

Vin1-10 V

fout0-10 kHz

tlow =1.1 RtCt must be less than the

period of the maximum output freq.

10 kΩ resistor at pin3 ?

i≈2/Rs < 200 μA

ttL

sin

out

CRR

RVf

2

47-Ω resistor in series with

CL provides hysteresis for

the comparator, improving

linearity. Why?

lecture 6


65


Example 6-12: Design a V2F converter that the output be 20 kHz when

the input is 5 V.

sf

T 501

max

min

ttlowCRt 1.1 kRandFC tt 8.70047.0

kRandFCLettt

8.60047.0

stlow

35

kCRR

V

fR ttL

in

out

s6.25

1

2terpotentiome1022 kkR

s

min8.0 Tt

low

lecture 6


66


Example 6-13: A pressure transducer measures a pressure from 0 to 100

psig. The output is 1-5 V for a 0-100 psig. It is desired to produce a count

every 20 ms. The count produced at 100 psig must be 100 larger than the

count in 0 psig.

a) What frequency span (in hertz) is needed?

b) What is the output frequency at 100 psig? At 0 psig?

c) What preset could be loaded into an 8-bit binary counter to yield a

count that goes from 0 to 100?

d) Why was a counting period of 20 ms chosen?

Solution: a)

b)

kHzms

countsf 5

20

100

VV 415 VkHzV

kHZ

V

f/25.1

4

5

.25.1,1,0 kHzfsoVVpsigat

.25.6,5,100 kHzfsoVVpsigat

lecture 6


67


Example 6-13(Continue)

c)

d) 1/(20 ms)=50 Hz so:

25)ms20)(s

counts1250(counts

125)ms20)(s

counts6250(counts

23125256preset

1) Accurate timing pulses can be obtained from power line.

2) Any 50-Hz noise on the input voltage are eliminated.

Higher frequency in positive half-cycle and lower frequency in…..

lecture 6


68

Frequency to voltage conversion

The LM131 can also be used as F2V converter.

The output frequency should

be tied to ground to reduce

the noise.

+ input of comparator is tied

to a reference level.

-input of comparator is driven

by the input frequency.

The average value of the

output voltage is proportional with frequency.

lecture 6


69


The average value of the output voltage is proportional with frequency.

AR

is

2002

TR

CR

T

itI

s

ttlow

ave

1.12

in

s

Ltt

Laveavef

R

RCRRIV

2.2

Tf

in

1

This pulse waveform must be filtered to remove the ripple.

lecture 6


70


Example 6-14: A reflective optical sensor is used to encode the velocity

of a shaft. There are six pieces of reflective tape. They are sized and

positioned to produce a 50% duty-cycle wave. The maximum speed is

3000 r/min. Design the frequency-to-voltage converter necessary to

output 10 V at maximum shaft speed. Provide filtering adequate to assure

no more than 10% ripple at 100 r/min(Let RL=100 kΩ)

Hzs

rev

rev

countsf 300

min

60

1

min30006

max

msTtmsHz

Tinpulse

67.15.033.3300

1minmin

msTTThighouthighout

664.28.0)(min)(

msTkRandFCLetCRThighouttttthighout

47.28.6331.1)()(

lecture 6


71


Example 6-14(Continue):

nFCandkRmsCRDDDD

3.31067.15

kRfR

RCRV

sin

s

Ltt

ave48.110

2.2

ok. isit so2001352

AAR

is

Hzs

rev

rev

countsf 10

min

60

1

min1006

sHz

T 1.010

1min

At 100 r/min,

lecture 6


72


Example 6-14(Continue):

fLCR

t

pkouteVv

For 10% ripple,

9.0ln%90

FL

CR

t

pk

out

CR

te

V

vfL

mstTthighoutrpm

53.9747.2100)(100

FR

tC

L

F2.7

)9.0ln( FCPick

F10

lecture 6


73


v Introduction


v Z V Conversion




v Filter


v Cabling

lecture 6


vکاربردهای فیلترuحذف فرکانس های غیر ضروری برای بهبود نسبت سیگنال به نویزuبازیابی سیگنال حاملu به آنالوگو آنالوگکاهش خطای تداخل در مبدل های دیجیتال به

دیجیتالvانواع فیلترها

u نگذرپایین گذر، باال گذر، میان گذر و میانuفیلتر های غیر فعال و فعال

v و می توان از آن ها بهره خیلی کمتر استبارگذاریدر فیلترهای فعال اثرگرفت

74

Filters

lecture 6


R

CVINVOUT

+

-

+

-

75

v 𝑓𝐻 =1

2𝜋𝑅𝐶

v𝑉𝑂𝑈𝑇(𝑠)

𝑉𝐼𝑁(𝑠)=

1

1+𝑅𝐶𝑠

vنکات طراحیu انتخاب خازن در محدودهµF-pFuبدست آوردن مقاومت مورد نیاز برای فرکانس قطع مورد نظرu 1در صورتی که مقاومت بدست آمده در محدودهk-1M نبود خازن را

این محدوده باعث.تغییر می دهیم تا مقاومت در محدوده فوق قرار گیردمی شود اثر بارگذاری در فیلتر کاهش یابد

Passive Low Pass Filter

lecture 6



𝑉𝐼𝑁(𝑠)=

−𝑅2

𝑅1

1

1+𝑅2𝐶𝑠

v 𝑓𝐻 =1

2𝜋𝑅2𝐶

76

Active Low Pass Filter

lecture 6


v 𝑓𝐿 =1

2𝜋𝑅𝐶


𝑉𝐼𝑁(𝑠)=

𝑅𝐶𝑠

1+𝑅𝐶𝑠

vن با سری کرد.برای افزایش مرتبه فیلتر می توان آن ها را با هم سری کرد.فیلترها معادالت مربوط به آن ها به توان طبقات می رسد

v کتر از طبقه اول بسیار کوچامپدانسبه علت اثر بارگذاری باید توجه نمود کهطبقه دوم باشدامپدانس

77

VIN VOUT

+

-

+

-

RC

Passive High Pass Filter

lecture 6



𝑉𝐼𝑁(𝑠)=

−𝑅2𝐶𝑠

1+𝑅1𝐶𝑠

v 𝑓𝐿 =1

2𝜋𝑅1𝐶

78

Active High Pass Filter

C

R2

R1

+

-VIN

VOUT

lecture 6


v 𝑓𝑐1 =1

2𝜋𝑅𝐻𝐶𝐻

v 𝑓𝑐2 =1

2𝜋𝑅𝐿𝐶𝐿

v𝑅𝐻

𝑅𝐿

lecture 6



𝑉𝐼𝑁(𝑠)=

−1

1+𝑅𝐿𝐶𝐿𝑠∙−𝑅𝐻𝐶𝐻𝑠

1+𝑅𝐻𝐶𝐻𝑠

v 𝑓𝐿 =1

2𝜋𝑅𝐻𝐶𝐻

v 𝑓𝐻 =1

2𝜋𝑅𝐿𝐶𝐿

v 𝑓𝐿 ≫ 𝑓𝐻80

Active Band Pass Filter

CH

RH

RH

+

-

VIN

VOUT

RF

RF

+

-

RL

RL

+

-

CL

lecture 6


v 𝑅1 =𝜋𝑅

10

v 𝐶1 =10𝐶

𝜋

v 𝑓𝐶 =1

2𝜋𝑅𝐶

v 𝑓𝑛 = 0.785𝑓𝑐81

R

C

R

C

R1C1

VINVOUT

v 𝑓𝐻 = 0.187𝑓𝑐v 𝑓𝐿 = 4.57𝑓𝑐

Passive Band Stop Filter

lecture 6


82


𝑉𝐼𝑁(𝑠)=

−1

1+𝑅𝐿𝐶𝐿𝑠+

−𝑅𝐻𝐶𝐻𝑠

1+𝑅𝐻𝐶𝐻𝑠

v 𝑓1 =1

2𝜋𝑅L𝐶L

v 𝑓2 =1

2𝜋𝑅H𝐶H

v 𝑓1 ≪ 𝑓2

Active Band Stop Filter

VOUT

RF

+

-

RL

RL

+

-

CL

CH

RH

RH

+

-

VIN

RF

RF

lecture 6


Active Filters

Sallen-Key Topology

83

vاختار پیاده سازی فیلترهای پایین گذر، میان گذر و باال گذر با سSallen-Key

v𝑉𝑂𝑈𝑇

𝑉𝐼𝑁=

𝑍3𝑍4

𝑍1𝑍2+𝑍3 𝑍1+𝑍2 +𝑍3𝑍4

lecture 6


84

Low Pass High Pass

Band Pass

Active Filters

Sallen-Key Topology

lecture 6


85


v Introduction


v Z V Conversion




v Filter


v Cabling

lecture 6


86

Isolation Circuits

lecture 6


87

Isolation Circuits

Even with high-quality IA, with proper grounding and cabling and use

I or F transmission,

• Ground loops or common ground connections,

• Extremely high common-mode voltages,

• Very low failure current requirements

All of these can be solved by isolation circuits.

Common of IA and sensor common must be connected, but

ground connection are made primarily for personnel safety.

Common-mode voltages that exceed the power supply voltages

not only cause an IA to fail, but would pass it to ……………

In cardiac monitoring, the conditioner must be able to withstand a

defibrillator pulse of 5-kV and continue to process the patient’s

heartbeat properly.

lecture 6


88

Isolation Circuits

More tolerate More accurate

lecture 6


89

Isolation Circuits

Type of isolation Amplifiers

• Transformer coupled amplifiers

• Optically coupled amplifiers

• Optical coupling for On/Off applications

lecture 6


90

Isolation CircuitsTransformer coupled amplifiers

Two or three port inputs. (Vin hi, Vin low and common)

Input

Output

Power

Three isolated ground.

Isolated output voltage

for sensor.

lecture 6


91

Isolation CircuitsTransformer coupled amplifiers

lecture 6


92

Isolation Circuits

Dc Ac

Ac DcAc Dc

Input modulation

Output

Transformer coupled amplifiers

EMI

Reduction

lecture 6


93

Isolation Circuits

lecture 6


94

Isolation Circuits

Transformer coupled amplifiers are expensive and bulky.

The carrier generates EMI if set at a high frequency.

The carrier limits the amplifier BW if set at a low frequency.

Optically coupled isolation amplifiers replace the transformer

with a LED and a pair of photodiodes.

The EMI is removed and frequency response is increased.

lecture 6


95

Isolation Circuits

Comparing Transformer coupled versus optically coupled amplifiers.

lecture 6


96

Isolation CircuitsOptically coupled amplifiers

You must provide separate power for input and output part.

i.e. use an IC with separate dc/dc converter.

lecture 6


97


Drawback?

Nonlinearity.

Two isolated ground.Vin Light in LED I2

lecture 6


98


D1 and D2 are

carefully matched.

1IRv

Gin

2IRv

kout

in

G

k

outv

R

Rv

This technique (an LED and a pair of photodiodes) reduce nonlinearity.

lecture 6


99

Isolation CircuitsOptical coupling for On/Off applications

TIL112 Optical coupler is a much simpler and less expensive alternate

to isolation amplifier.

Phototransistor current is proportional to

LED current, but it is nonlinear.

So it is not suitable for analog transmission.

No current in LED phototransistor is off.

More than 20 mA phototransistor is on.

in LED

So digital data can transmit through this coupler.

lecture 6


100


How to derive R1 ?

lecture 6


101


1- ADC with optical isolation

There are two common ways of using optical couplers to transmit digital data.

± V1, and Vlogic 1must all be provided,

referenced to the

input common.

Vlogic 2 must be

provided,by a

different common.

It has a high

gathering rate. But it

is expensive since:

1- A/D converter.

2- A/D is in input

side.

lecture 6


102


2- V2F converter with optical isolation

There are two common ways of using optical couplers to transmit digital data.

If time is not an

important issue.

As the A/D

isolation scheme,

sensor, signal

conditioner and

appropriate power

Supplies are all

On the input side.

However, the expensive A/D converter has been replaced by much smaller,

much cheaper V2F converter.

lecture 6


v The input is duty-cycle modulated and transmitted digitally

across the barrier

v The output section receives the modulated signal, converts

it back to an analog voltage and removes the ripple

component inherent in the demodulation

103

Capacitive Isolation Amplifier

Modulator DemodulatorInput

signal

Output

signal

Capacitiveisolation

lecture 6


v ISO124

104

Ix

Iin Ic

D

B

C

Capacitive Isolation Amplifier

lecture 6


v Ripple noises are removed

v It avoids device noise, radiation noise and conducted noise

v High immunity to magnetic noise

v Useful for analog systems

v high gain stability and linearity

v Supports faster data transition compared to optical isolation

105

Capacitive Isolation Advantages

lecture 6


106


v Introduction


v Z V Conversion




v Filter


v Cabling

lecture 6


107

Cabling

Even with the best transducers, amplifiers, filters, transmission, and

isolation it is entirely possible that the signal received by controller

is junk.

Proper cabling can shield against interference from the large

magnetic and electric fields produced in all manufacturing

environment.

Proper grounding of the shields, circuits, and power supplies will

minimize the impact of differing ground potentials.

Without proper shielding and grounding even techniques, even the

most expensive …..

Proper cabling will allow even a cheap system to realize its full potential.

lecture 6


108

Cabling

Current flow leads to magnetic field.

Magnetic field is 50 Hz or Dc but we have problem with 50 Hz and

switching in the Dc one.

Radio frequency of digital signal of the processor also produce

significant magnetic field.

So place small signal analog circuitry on a separate card from

computer and power electronics.

Separate this card from others or put it an a small magnetic shield box.

Absorbing magnetic field is in dB.

Never, never run 220-V ac power and low level analog cables in the

same …

lecture 6


109

Cabling

Absorbing magnetic field is in dB. A 20 dB one change the ratio to 10.

In audio frequency steel is the choice but in radio frequency…..

In fact, there are several coating that can sprayed onto plastic cases

which effectively work (in high frequency).

lecture 6


110

Cabling

For effective shielding against electrostatic field try to use a conductive

shield.

Shield must be connected to infinite source.

Do not use the shield as the signal common.

Use a shield for signal common as well.

Use a shielded pair or twin-axial, never simple coaxial cable.

lecture 6


111

Grounding

Effective electrostatic shield must be grounded. But the signal common

must not be connected to the shield.

Connection to ground at some points, is it good or bad?

1- Shield is designed

for this current?

2- Magnetic field of this current?

lecture 6


112

Grounding

Disconnect from some point.

Use a 100 pF parasitic capacitance.

MpFHz

Xc

32)100)(50(2

1

lecture 6


113

Grounding

If disconnection is not possible?

Sometimes the computer input low signal, from the signal conditioner

is tied to earth by manufacturer for safety reasons.

Disconnect this

lecture 6


114

Grounding

If Disconnection is not possible? Try it by following:

In fact the isolation amplifier (or digital optical coupler) breaks the

ground loop that had exited through electronics.

In fact, above configuration is probably the best general approach to

shielding and grounding.

The only remaining element in the data acquisition channel to be

connected is the power supply.

lecture 6


115

Grounding

What about power supply?

In every sensitive systems,

this error may be a

significant part of the

signal from the transducer.

The solution is a power

supply transformer with

shielded secondary.

lecture 6


116

Grounding

The solution is a

power supply

transformer with

shielded secondary.

In this configuration

any vnoise causes

current to flow

along the shield, not

along the signal

common.

lecture 6


117

Grounding

An interesting problem.

Suppose it is

100 cm.

lecture 6


118

Grounding

What do you do with ground connection of the other electronics?

Instead of -4 V

optional

?bv

lecture 6


119

Grounding

Ideally, every stage throughout the data acquisition and control system

would have its own, separate ground return.

This is impractical, so group your circuits according to:

• The type and size of signal

• The magnitude of ground return currents

• The circuits sensitivity

lecture 6


120

Grounding

lecture 6


121

Appendix: Instrumentation Amplifier

AD620The AD620 is a monolithic

instrumentation amplifier based on

a modification of the classic three op

amp approach. Absolute

value trimming allows the user to

program gain accurately

(to 0.15% at G = 100) with only one

resistor. Monolithic

construction and laser wafer

trimming allow the tight matching

and tracking of circuit components,

thus ensuring the high level

of performance inherent in this

circuit.

lecture 6


122


The input transistors Q1 and Q2 provide a single differential pair bipolar input for high precision (Figure 36), yet offer 10× lower input bias current thanks to Super ϐeta processing.Feedback through the Q1-A1-R1 loop and the Q2-A2-R2 loop maintains constant collector current of the input devices Q1 and Q2, thereby impressing the input voltage across the external gain setting resistor RG. This creates a differential gain from the inputs to the A1/A2 outputs given by G = (R1 + R2)/RG + 1. Theunity-gain subtractor, A3, removes any common-mode signal, yielding a single-ended output referred to the REF pin potential.

lecture 6


123


The internal gain resistors,

R1 and R2, are trimmed to

anabsolute value of 24.7 kΩ,

allowing the gain to be

programmed accurately with

a single external resistor.

The gain equation is then

𝐺 =49.4𝑘Ω

𝑅𝑔+ 1

𝑅𝑔 =49.4𝑘Ω

𝐺 − 1

lecture 6


124

Pressure MeasurementAlthough useful in many bridge applications, such as weigh scales, the AD620 is especially

suitable for higher resistance pressure sensors powered at lower voltages where small size

and low power become more significant. Figure 38 shows a 3 kΩ pressure transducer bridge

powered from 5 V. In such a circuit, the bridge consumes only 1.7 mA. Adding the AD620

and a buffered voltage divider allows the signal to be conditioned for only 3.8 mA of total

supply current.


lecture 6


125


Furthermore, the low bias currents and low current noise, coupled with the low

voltage noise of the AD620, improve the dynamic range for better performance.

The value of capacitor C1 is chosen to maintain stability of the right leg drive loop.

Proper safeguards, such as isolation, must be added to this circuit to protect the

patient from possible harm.

Medical ECG

The low current noise

of the AD620 allows

its use in ECG

monitors (Figure 39)

where high source

resistances of 1 MΩ

or higher are not

uncommon.

lecture 6


126

Precision V-I Converter

The AD620, along with another op

amp and two resistors,

makes a precision current source

(Figure 40). The op amp

buffers the reference terminal to

maintain good CMR. The

output voltage, VX, of the AD620

appears across R1, which

converts it to a current. This current,

less only the input bias

current of the op amp, then flows

out to the load.


I L =Vx

R1=[(𝑽 𝑰𝑵+) – (𝑽 𝑰𝑵 – )] 𝑮

𝑹𝟏

lecture 6


127

GROUNDING

Since the AD620 output voltage is

developed with respect to the

potential on the reference terminal, it

can solve many grounding problems

by simply tying the REF pin to the

appropriate “local ground.” To

isolate low level analog signals from

a noisy digital environment, many

data-acquisition components have

separate analog and digital ground

pins (Figure 45).


It would be convenient to use a single ground line; however, current through ground

wires and PC runs of the circuit card can cause hundreds of millivolts of error.

Therefore, separate ground returns should be provided to minimize the current flow

from the sensitive points to the system ground. These ground returns must be tied

together at some point, usually best at the ADC package shown in Figure 45.

lecture 6


References

v Industrial Control Electronics J.M. Jacob, Prentice-Hall,

1989

v AD620 and ISO124 datasheet

v Fundamentals of measurements in instrumentations by

prof hamid.taghirad and s.ali salamati

128

Some Useful websites for the course

http://saba.kntu.ac.ir/eecd/ecourses/instrumentation.htm

v http://profsite.um.ac.ir/~shoraka/Instrumentation.htm

v http://karimpor.profcms.um.ac.ir/index.php/courses/10328

Documents

lecture 6 In The Name of Allah - Ali Karimpourkarimpor.profcms.um.ac.ir/imagesm/354/stories/... · 2017. 5. 21. · lecture 6 Dr. Ali Karimpour, Apr 2017 3 Introduction In manufacturing