Signal Conditioning and Transmission

Signal Conditioners and Transmission

2102-487 Industrial Electronics

Overview

Amp V to I I to V Amp

4-20 mA

Motor/Power Arcing Lighting

Noise Source

Interference

Field Control Room

Instrumentation Amplifier (IA)

Instrumentation amplifiers a dedicate differential amplifier with extremely high input impedance. Its gain can be precisely se by a single internal or external resistor. The high common-mode rejection makes IA very useful in recovering small signals buried in large common-mode offsets or noise.•IA consists of Two stages:

•The fist stage: high input impedance and gain control•The second stage: differential amplifier (change differential signal to common to ground

R

R

R+∆R

R

E Vout+

Vdm/2

Vdm/2Vcm +- +-

+-

R-

R+

Equivalent circuit for a Load cell Bridge circuit for sensor application

e1

e2

R

R

Rg

+

-

+

-

Vo

+

-

IA: The First Stage

V1

V2

Rg = gain setting resistor

+−=

go R

ReeV 21)( 21

• Changing Rg will inversely alter the output voltage.

11 eV =

For Ideal op amp, no voltage difference between the inverting and noninverting inputs

22 eV =and

21 eeVgR −=

The voltage across Rg

gR R

eeIg

21 −=

This current must flow through all three resistors because none of the current can flow into the op amp inputs

( )ggRo RRIV += 2

IRg

IA: First + Second stage

+

-+

-Vo

R1

R

R

Rg

+

-

+

-

R1

R1

R1

+−=

go R

ReeV 21)( 211

e2

e1

Vo1

+

-

1oo VV −=

+−=

go R

ReeV 21)( 12-1

+

gRR2121 ee −

V2

V1

1oV

Without loading effect, we can applied the direct multiplication for the cascade system

gRR21Gain +=

+−=

go R

ReeV 21)( 12

The second stage of IA is a unit-gain differential amplifier

+

-

R2

Rg

+

-

+

-

+

-

+V

-V

VrefVref

LoadReference

Output

Sense

R1

R1

R2

R2

R2

IA with Offset (Reference)

+−=

go R

ReeV 21)( 211

e2

e1

refg

o VRReeV +

+−=

21)( 12

refoo VVV +−= 1

-1

+

gRR2121 ee − 1oV

+refV

refg

o VRReeV +

+−=

21)( 12

Vo1

V1

V2

Commercial IA: AD524

AD524 Functional Block Diagram

%20400001 ±

+=

gRG Here Rg = 40, 404, 4.44 kΩ or external

resistor (connect: RG1 – RG2)

Features:Low Noise: 0.3 µVp-p 0.1 Hz to 10 HzLow Nonlinearity: 0.03% (G = 1)High CMRR: 120 dB (G = 1000)Low offset Voltage: 50 µVGain Bandwidth Product: 25 MHzProgrammable Gain of 1, 10, 100, 1000


349Ω

Ex For the circuit, calculate Va, Vb, and Vout as well as the effect on the output of the 5-V common-mode signal.

Vb Va

Vout

V 5supply21 == VVb

V 4.99285V 10349350

349=

Ω+ΩΩ

=aV

( )mV -715V) 5-V 4.99285(100 ==

−= baout VVGV

For a gain of 100, CMRR = 100 dB

cm

dm

GGCMRR log20=

mV 5V 5001.0 =×=cmcmVG

001.0=cmG

%7.0%100mV 715

mV 5=×

The common mode error is


Error Budget Analysis

To illustrate how instrumentation amplifier specification are applied, we will now examine a typical case where an AD524 is required to amplify the output an unbalance transducer. Figure above shows a differential transducer, unbalance by 100 Ω supplying a 0 to 20 mV signal to an AD524C. The output of the IA feeds a 14-bit A/D converter with a 0 to 2 Voltage range. The operating temperature is 25oC to 85oC. Therefore, the largest change in temperature ∆T within the operating range is from ambient to +85oC (85oC - 25oC = 60oC)

Commercial IA: AD5241000ppm = 0.1%

Zero and Span Circuits

The zero and span circuit adjust the output of a transducer to match the levels you want to provide to the controller or display.

Ex. You may need a 0.01 mV/lb input to a digital panel meter, while the load cell provides a 20 µV/lb reading with 18 mV output with no load.

Ex. A/D converter needs a 0- to 5-V signal, while the temperature transducer output 2.48 to 3.90 V.

Vout

Vin

Vout

Vin

Zero shifted positiv

ely

Zero sh

ifted negati

vely

Original

Original

Span

dou

ble

Span Half

Zero and Span: Inverting summer

+

-Ri

Rf

Rcomp

ROS

+

-R

R

R/2

ein

±V

eu2

eu1

VRR

eRR

eOS

fin

i

fu −−=1 12 uu ee −=

VRR

eRR

eOS

fin

i

fu +=2

Inverting summer Inverting amplifier

Compare this to the eq. of straight line bmxy +=

Here y = eu2, x = ein, m = Rf/Ri and b = Rf/Ros Vein

eout

VRR

bOS

f=

i

f

RR

m =

Zero and Span: Instrument AmplifierEx When the temperature in a process is at its minimum, the sensor outputs 2.48 V. At maximum temperature, it outputs 3.90 V. The A/D converter used to input these data into a computer has the range 0 to 5 V. To provide maximum resolution, you must zero and span the signal from the transducer so that it fills the entire range of converter.

Vin

Vout

VRR

bOS

f=

i

f

RR

m =

Vin = 2.48-3.90 V Vout = 0-5 V

52.3V 48.2V 9.3

V 0V 5(min)(max)(min)(max)

=−−

=−−

=inin

outout

VVVVm

V -8.73V 48.252.30 =×−=−= inout mVVb

The gain, m is set by Rf/Ri , Rf relatively large, so that Riwill not load down the sensors

Let Rf = 330 kΩ Ω=== k 7.9352.3kΩ 330

mR

R fi

Select Rs as a 47 kΩ fixed resistor with a series 100 kΩ potentiometer. (m ~ 2.25-7.02)

b is Rf/ROSV select V = -12 V Ω=−

Ω== k 454

V 73.812V) )(-k 330(

bVR

R fOS

Select Rs as a 220 kΩ fixed resistor with a 500 kΩ potentiometer. (b ~ -18 - -5.5 V)Ω== k 9.62//// 1 OSfcomp RRRR Select Rcomp = 56 kΩ

The resistors in the 2nd stage should be in kΩ range, this will not load the 1st stage, pick R = 2.2 kΩ and so R/2 = 1.1 kΩ

Zero and Span: Instrument Amplifier

+

-

R2

Rg

+

-

+

-

+

-

+V

-V

VrefVref

Reference

Output

Sense

R1

R1

R2

R2

R2

Vout

8

4

5

10

96

7

2

16

13

12

11

3

1 -

+

AD524

span

ein

+

-

Rg

S

R

O

+

-

+V

-V

Vref

+V

-V

-V

Vout

zero

+V

span

zero

( ) refg

out VeeR

V +−

+= 12

400001


Ex The output from a load cell changes 20 µV/lb with an output of 18 mV with no load on the cell. Design a zero and span converter using an instrumentation amplifier which will output 0 Vdc when there is no load, and will change 10 mV/lb.

Vsensor = 20µV/lb x Load + 18 mV

Vout = 10mV/lb x Load = G Vsensor+ Vref8

4

5

10

96

7

2

16

13

12

11

3

1 -

+

AD524

span

Vsensor

+

-

Rg

S

R

O

+

-

+V

-V

Vref

+V

-V

-V

Vout

zero

+V

Vout = 20µV G x Load + 18mV G + Vref

500V/lb 20

mV/lb 10==

µG

( ) refg

out VeeR

V +−

+= 12

400001

500400001 =+=gR

G

V 9−=refV

Select Rg as a 33 Ω fixed resistor with a series 100 Ωpotentiometer. (G ~ 301-1213)

refref VVG +×=+= 500mV 18mV 180

So tie one end of the potentiometer to ground and the other end to –Vsupply. Select resistor and potentiometer values that will put approximately -10 V at one end and -8 V at the other end.

Ω= 2.80gR

+

- +

-

load

Rwire

Rf

+

-ein

R

Voltage-To-Current Converters

Signal voltage transmission presents many problems. The series resistance between the output of the signal conditioner and the load depends on the distance, the wire used, temperature, conditioner.

inf

o eRR

V

+= 1

owire

VLoadR

Load

+

Rwire ∝ Distance, wire type, temperature

V To I: Floating Load

+

-load

Rwire

+

-ein

R I

I

ein

+

-

+V

-V

The most simple V to I actually is the non-inverting amplifier

ReI in=

Therefore, The resistance in the transmission loop (Rloop = Rwire + Rload) does not affect the transmitted current. However, the output voltage of the op amp is affected by the Rloop

satinloop

out VeR

RV <

+= 1

Rloop must be kept small enough to keep the op amp out of the saturation.

+

-

load

Rwire

R

I

+V

-V

+

-ein

I

Q1

Q2


V to I with current booster

ReI in=

Add transistor forincrease current output

Many transmission standard call for either 20 or 60 mA current. These values are beyond the capabilities of most general-purpose op amps. However, the transistor can be added to increase the transmission current capability.

New op amp with high current output


The signal at the load is inherently differential. So we can use difference or instrumentation amplifier to reject any common-mode noise.

Open and short circuit in the transmission loop can be detected by checking Vout of the non-inverting amplifier (transmission side)

open circuit : Vout = Vsatshort circuit : Vout = ein

However, at the load side, with this circuit , if ein = 0, IL = 0, here it appears that the IL = 0 is the valid signal. If the open or short circuit fault occurs, ILwill fall to zero too. The load would respond as if ein = 0. Therefore a method has been advised to allow the load to differentiate between no signal (circuit failure)

IL = 0 and ein = 0

However, at the load side, with this circuit , if ein = 0, IL = 0, here it appears that the IL = 0 is the valid signal. If the open or short circuit fault occurs, ILwill fall to zero too. The load would respond as if ein = 0. Therefore a method has been advised to allow the load to differentiate between no signal (circuit failure)

IL = 0 and ein = 0

This can be achieved by adding an offset to IL when ein = 0ein = 0: IL > 0

Ex. 4-20 mA standard in current transimission

Rwire

R

Q1

-V+V10 kΩ

1 MΩ

1 MΩ

ein

eref

Iin IL

load

span

zero

-

+

-V

+V


An example of 4-20 mA V to I converter (span + zero adjust)

Re

Re

Ree

I refinrefin

222+=

+= m = 1/2R, b = eref/2R

ein

I

Re

b ref

2=

Rm

21

=


Ex Design an offset voltage-to-current converter that will produce 4 mA with an input of -5 V and 20 mA with an input of 10 V

Vin = -5 - 10 V

V 8.810VmA) 20)( 469(22 =−Ω=−= inref VRIV

Select a 430 Ω fixed resistor with a series 100 Ω potentiometer.

Ω= 469R

I (mA)

Vin

5

10

15

20

4

- 5 5 100

RV

RVI refin

22+=

Iout = 4 - 20 mA

V) 5(V 01mA 4mA 20

(min)(max)(min)(max)

21

−−−

=−−

==inin

outout

VVII

Rm

V To I: Grounded Load

+

-

+

-

e1 e2

R1

R2

R4load

R3

Rs

I

I

VL

R1=R2=R3=R4=R

Vout

The current drive into the load is approximately equal to the current in Rs if RLoad << R2+R4

This V to I actually is a derivative of difference amplifier

Using superposition: Vout due to e1 = -e1Vout due to e2 = e2Vout due to VL = VL

LL

Vouteouteoutout

VeeV ee

VVVVL

+=++=

++=

1221 --

|||21

ss

LoutRL R

eeR

VVIIs

12 −=

−=≈

As in the case of the floating load: 12 eeIRV loadSat −+>

V To I: Grounded Load

s

ref

s

in

s

refinL R

eRe

Ree

I −=−

≈

An example of 4-20 mA V to I converter (span + zero adjust)

+

-

-V

+V

+

-ein load

100 kΩ

I

VL

RbRa

100 kΩ

100 kΩ

100 kΩ

Rs

eref

span

zero

m = 1/R, b = -eref/R

ein

I

s

ref

Re

b −=

sRm 1

=

V To I: XTR110

•4 to 20 mA Transmitter•Selectable input/output ranges: 0-5V or 0-10V Inputs

4-20mA, 0-20mA, 5-25mA Outputs• Required an external MOS transistor to transmit current to load

Precision resistive network

V to I converter

I to I converter(current mirror)

Reference voltage

V To I: XTR110

R115kΩ

R320kΩ

R410kΩ

R25kΩ

16.25kΩ

R5

VIN1(10V)

VREF IN

R76250Ω4mAspan

R61562.5Ω

16mAspan

+

-

+

-

R8500Ω

R950Ω

VIN2(5V)

VCC 13.5-40V

RL

Io4-20mA

16

1

3

14

89 10

5

3

4

IoIo/10

Va

Io/10

Vb

XTR110External MOS

666 R

VRVI ab

R ==

The current IR6 is approximately equal to IR8

68 RR II ≈

Since there is no voltage difference between the op amp inputs

98 RR VV =

89

889 10 R

RRo I

RRIII ===

6

10RVI a

o =

V To I: XTR110

( )span

refinino R

VVVI

16/2/4/10 21 ++=

The voltage at Va is defined by precision divider network and the voltage at Vin1, Vin2 and Vref . Using superposition

Va due to Vin1 = Vin1/4Va due to Vin2 = Vin2/2Va due to Vref = Vref/16

Therefore, the relation of the output current can be shown as

16/2/4/ 21 refinina VVVV ++=

This IC allows us to change the value of Rspan by using R6, R7 and external resistors so therefore the Io span can be set to 16mA, 4mA or arbitrary value.Ex if Vin2 = 0, Vref = 10 V, Vin1 varies from 0-10V and Use Rspan = R6 = 1562.5Ω

At Vin1 =0 V; Io = 4 mA at zero Vin1

At Vin1 =0 V; Io = 20 mA at Vin1 = 10 V Zero+span = 4mA + 16 mA

Zero = 4 mA

V To I: XTR110

Va

R115kΩ

R25kΩ

16.25kΩ

R5

VIN1

VREF IN

VIN2

5

3

4

R320kΩ

R410kΩ

Va

R115kΩ R2

5kΩ

16.25kΩ

R5

VIN1

R320kΩ

R410kΩ

R115kΩ R2

5kΩ

16.25kΩ

R5

VIN2

R410kΩ

R320kΩ

Va

R115kΩ

R25kΩ

16.25kΩ

R5

VREF IN

R320kΩ

R410kΩ

Va

IN141

IN121543

2154

)////()////( VVRRRRR

RRRRVa =++

+= IN22

1IN2

21534

2153

)////()////( VVRRRRR

RRRRVa =++

+= REF16

1 REF

21

2

43215

43

//)//(// VV

RRR

RRRRRRRVa =

+++=

1 2 3

1 2 3+ +

REF161

IN221

IN141 VVVVa ++=

kΩ 27=inR kΩ 22=inR kΩ 20=inR

V To I: XTR110

XTR110

15

12

3

4

59 2

16

1

13

14

0 to 10 V

4 to 20 mA Out

VLRL

1 µF

15 V

A basic 4 to 20 mA transmitter

Pin connections for standard XTR110 input voltage/output current ranges

V To I: XTR110

V 6400 += inout VV

mV 10±=sensorV V 102 −=sensorV mA 204 −=outI

Ω=

1250)4/(10 1in

outVI

Ω+

=1250

4/)6400(10 sensorout

VI

Current-To-Voltage ConvertersGrounded Load I to V converter

+

-R2

Rf

Rcomp

Ros

+

-R

R

R/2

+

-

4 to20 mA

RL Vout

+

-

±V

Current is converted into a voltage by RL

Span and Zero circuit

Voltage follower is inserted to avoid the loading effect

The grounded load converter has many problems with the common ground (use by many device) especially the noise source. To reduce this problem, we use a floating load.

VRR

IRRR

VOS

fL

i

fout += m = RfRL/Ri, b = Rf/ROSV

I To V Converters: Floating Load

refspani

fout VIR

RR

V += m = RfRL/Ri, b = Vref

+

-

Rf

R4

+

-

Ri

RiRspan

Rf

Rpot

+V

-V

Vout

+

-

4-20 mA

Vref

To prevent the loading effect be sure that Rspan << Rior Instrumentation amplifier can be used in stead

Ex Design a floating current-to-voltage converter that will convert a 4- to 20-mA current signal into 0- to 10-V ground-referenced voltage signal. If voltage-to-current converter, has +V = 12 V, Rspan V-I =312 Ω, Imax = 20 mA What is the maximum allowable Rspan I-V

Iin = 4 - 20 mA Vout = 0-10 V

Ω=−−

=−−

=− 625mA 4mA 02V 0V 01

(min)(max)(min)(max)

inin

outoutVspanI

i

f

IIVVR

RR

Choose Rf/Ri =10 This seem arbitrary now, but we’ll come to check it again

Find Vref : ( )( ) V 5.25.62mA 4kΩ 2.2kΩ 220 −=Ω−=−= IR

RR

VV spani

foutref

refspani

fout VIR

RR

V +=

Ω=− 5.62VspanIR

Since R i >> Rspan I-V, pick Ri = 2.2 kΩ Rf = 22 kΩ


IspanVVspanI IRIRV −− +++=+ 7.02

Therefore, 62.5 Ω resistor would work. However, if we had chosen R f/Ri =1 then Rspan I-V = 625 Ω, which is too large. The op Amp in the voltage-to-curren converter would saturate before 20 mA would be reached.


Rspan I-V

I-

+

-V

+V

Rspan V-I

+-

+- +

-

IRspan I-V

IRspan V-I

+

-

I

0.7 V

2 V

Ω=−−−

=− 153mA 20

)312)(mA 20(7.0212(max)VspanIR

Voltage-To-Frequency Converters

To provide the high noise immunity, digital transmission must be used.

V to F will convert the analog voltage from the sensor or signal conditioner to a pulse train. The pulse width is constant but the frequency varies linearly with the applied voltage

inVf ∝

V to FVin fout

fout

Vin

Ideal V to F characteristics

V-To-F: LM131

Suitable for various applications:precision V to F convertersimple low-cost circuits for A/D conversionprecision F to V converterlong-term integrationlinear frequency modulation and demodulation

LM131


RS

3

Comparator

+

- One shottimer

Switchedcurrentsource

RL

CL

Rt Ct

Frequencyoutput

Vlogic

8VCC

V1Input

voltage7

1 6

25

4

ab

Vin

Vx

Vout

fout High Vin → high fout Adjustment Low Vin → low foutAdjust High Vin → high fout

t

t

t

t

Basic V to F block diagram

ttL

Sino CRR

RVf 1V 09.2

=ttL

Sino CRR

RVf 1V 09.2

=

Vx


RS

3

Comparator

+

- One shottimer

Switchedcurrentsource

RL

CL

Rt Ct

Frequencyoutput

Vlogic

8VCC

VinInput

voltage7

1 6

25

4

Determine the timer of one shot

Phase I: Charge CL with Iref within the specific time TOS from the one shot timer. At the end of this phase, VC reaches the value assigned as Vx

Determine Iref

ttOS CRT 1.1=

Phase II: CL discharge through Iref until VC is equal to Vin. The time in this phase is defined as TII

We can found that 1/(TOS + TII) = f ∝ Vin


Phase I: Charge CL with Iref within the specific time TOS determined from the one shot timer. At the end of this phase, VC reaches the value assigned as Vx

Phase II: CL discharge through Iref until VC is equal to Vin. The time in this phase is defined as TII

RL CLIref

Initial condition: VC(0) = Vin

( ) Lreft

LrefinC RIeRIVtV +−= − τ/)(

Assume RLCL >> TOS ττ /1/OS

T Te OS −≈−

LLCR=τ

( )τOS

inLrefinOSCxTVRIVTVV −+== )(

RLCL

I

Initial condition: VC(0) = Vxτ/)( t

xC eVtV −= LLCR=τ

Since RLCL >> TOS, Therefore Vx /Vin ~ 1

τ/)( IITxIICin eVTVV −==

in

xII V

VT lnτ=

1ln −≈in

x

in

x

VV

VV

−= 1

in

xII V

VT τ

1

2


Combine eq.(1) and (2)

Substitute Iref = 1.9 V/ Rs ,TOS = 1.1RtCt

in

OSLrefIIOS V

TRITTT =+=

OSLref

in

TRIV

Tf ==

1

( ) ttL

sin

ttL

sin

CRRRV

CRRRVf

1

V .0921.1 V 9.1==

Ex Design a voltage-to-frequency converter that will output 20 kHz when the input is 5 V.

s 50kHz 2011

maxmin µ===

fT

At maximum frequency, the minimum period is

( ) ( )( )( )( )( )Ω=

ΩΩ== k 6.25

V 5µF .00470k .86k 100kHz 20V 2V 2 0

in

ttLs V

CRRfR

1.1 ttlow CRt =

pick Ct = 0.0047 µF (somewhat arbitrary, but suggested by manufacturer)

V To F Converters

We have to set the pulse width no wider than about 80% of the minimum period, otherwise, at the higher frequencies the pulse width may approach or exceed the period, which will not work.

pick Rt = 6.8 kΩ, since this is not a critical parameter, as long as it is not too big. This give tlow = 35 µs. Pick RL = 100 KΩ.

Select a 22 kΩ fixed resistor with a series 10 kΩ potentiometer.

( )( )( )( ) Ω=== k 7.7

µF .004701.1µs 058.0

1.1 t

lowt C

tR


ino VfVoltkHz 1

=

Simple V to F converter

ttL

Sino CRR

RVf 1V 09.2

=

Manufacturer’s recommended, RL = 100 kΩ, Rt = 6.8 kΩ, Ct = 0.01 µF and RS = 14.2 kΩ

From µA 200V 9.1<=

SRI

Low-pass filter

Ex Design a voltage-to-frequency converter that will output 20 kHz when the input is 5 V.

s 50kHz 2011

maxmin µ===

fT

At maximum frequency, the minimum period is

( ) ( )( )( )( )( )Ω=

ΩΩ== k 6.25

V 5µF .00470k .86k 100kHz 20V 2V 2 0

in

ttLs V

CRRfR

1.1 ttlow CRt =

pick Ct = 0.0047 µF (somewhat arbitrary, but suggested by manufacturer)

V To F Converters

We have to set the pulse width no wider than about 80% of the minimum period, otherwise, at the higher frequencies the pulse width may approach or exceed the period, which will not work.

pick Rt = 6.8 kΩ, since this is not a critical parameter, as long as it is not too big. This give tlow = 35 µs. Pick RL = 100 KΩ.

Select a 22 kΩ fixed resistor with a series 10 kΩ potentiometer.

( )( )( )( ) Ω=== k 7.7

µF .004701.1µs 058.0

1.1 t

lowt C

tR

Frequency-To-Voltage Converters

Comparator

RL

8

+

-One shot

timer

Rt

Ct

7

Iout

VCC

VCC

1

Vin

RD

CD

Connect to low pass filter

The amplitude of pulse current output is

Vin

Iout

i

T

The average voltage output is

ins

LttLeavave f

RRCRRIV 1.1 V 9.1 ×==The average current output is

sRi V .91

=

TRCR

TitI

s

ttave

1.1V .91 ×==

Frequency-To-Voltage Converters

inave fVkHz

V 1=

Simple F to V converter

inS

Lttave f

RRCRV 1.1V 9.1 ×=

RL = 100 kΩ, Rt = 6.8 kΩ, Ct = 0.01 µF and RS = 14.2 kΩ

From

Low pass filter for output current

High pass filter for input frequency

Ex 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 shaft 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

The maximum frequency is Hz 300s 60

min 1min

rev 0003rev

counts 6max =××=f

Select

F To V Converters

ms 33.3Hz 300

1min ==T

( )( ) ms 664.2ms 33.38.08.0 min)( ==≤ TT highout

Select Ct = 0.33 µF ( )( )( )( ) Ω=== k 7.7

µF .004701.1µs 058.0

1.1 t

lowt C

tR

ms 67.15.0 min == TTpulse

pick Rt = 6.8 kΩ, This give tout(high) = 2.47 ms. We must set 5RDCD << 1.67 ms, so pick RD = 10 kΩ.

)()1.0(5 highoutDD tCR ≤ nF 3.3≤DC

At 100 r/min

F To V Converters

µA 200µA 135k 4.81V .91V .91

<=Ω

==sR

i

Filtering is accomplished by RL and CF

For 10% ripple,

FLCR=τ

( ) µF 25.9)9.0ln(k 100

ms 53.97=

Ω−≥FC

ins

Lttave f

RRCRV 1.1 V 2 ×=

( )( )( )( )( )( )Ω=

ΩΩ=

×= k 8.14

V 10Hz 003k 001µF .330k .861.1V 21.1 V 2

ave

inLtts V

fRCRR

Select Rs as a 10 kΩ fixed resistor with a series 10 kΩ potentiometer.

Check

Hz 10s 60

min 1min

rev 001rev

counts 6=××=f

s 1.0Hz 011

min ==T ms 05.0=pulseT

τ/tpkout eVV −=

FLCRt

pk

out eVV /9.0 −≥=

)9.0ln(LF R

tC −≥

Select CF = 10 µF

ms 97.53ms 2.47-ms 100)(o10Hz ==−= highuttTtand, we have

Cabling• Cables are important because they are the longest parts of the system

and therefore act as efficient antennas that pick up and/or radiate noise.

• There are three coupling mechanisms that can occur between fields and cables, and between cables (crosstalk).

• Capacitive or electric coupling (interaction of electric field and circuit)

• Inductive or magnetic coupling (interaction between of the magnetic field of two circuits.

• Electromagnetic coupling or radiation (RF interference)

Magnetic Coupling• When a current flows in a closed circuit, it produce a magnetic flux, φ

which is proportional to the current.

• When current flow in one circuit produces a flux in a second circuit, there is a mutual inductance M12 between circuits 1 and 2

• The voltage VN induced in circuit to 2 due to the current I1

LI=φ

1

1212 I

M φ=

112IMjdtdiMVN ω== Assume i varies sinusoidal with time

φ12 ∝ the current, geometry, and orientation

Magnetic coupling between two circuits

Magnetic Coupling

• Separate the sensitive, input signal condition from the other portions of the electronics: Digital signal (high f), high power circuit, ac power (50 or 60 Hz line)

• place low level signal on the separate card• board layout• separate low level signal conduit or race way, from ac power line, communication or digital cable conduits.

• Reduce loop area (twisted pair)

• Use magnetic shield (effective at high frequency)

Magnetic Coupling

Capacitive Coupling• The potential different between two conductors generates a proportional

electric field. This will result in the redistribution or movement (current) of charge by external voltage source

Capacitive coupling between two conductors Capacitive coupling with shield

1212

12

)(1V

CCRjRCjV

GN ++

=ω

ω

12212

12

)(1V

CCCRjRCjV

SGN +++

=ω

ω

*The shield must be grounded.

• Grounding is one of the primary ways of minimizing unwanted noise and pick up. (also bad grounding can cause the serious problem of interference)

• A well-designed system can provide the protection against unwanted interference and emission, without any addition per-unit cost to the product

• Grounds category: safety grounds (earth ground)signal grounds

A signal ground (better definition) is a low-impedance path for current to return to the source.

Grounding Problems

A signal ground is normally defined as an equipotential point or plane that serves as a reference potential for a circuit or system (can not be realized in practical systems)

Ground SystemGround System can be divided into three categories.

single-point groundsmultipoint groundshybrid grounds (frequency dependence)

1 2 3 1 2 3 1 2 3

Two types of single-point grounding connection

Series connection Parallel connection Multipoint connection

Single-Point Ground System

1 2 3R1 R2 R3

I3I1+ I2+ I3 I2+ I3

I3I2I1

A series ground system is undesirable from a noise standpoint but has the advantage of simple wiring

A B C

1 2 3R1

R2

I2

I1

I3

R3

A parallel ground system provides good noise performance at low frequency but is mechanical cumbersome.

A

B

C

33

232

1321

)()(

RIVVVRIIVV

RIIIV

BAC

AB

A

++=++=

++=

33

22

11

RIVRIVRIV

C

B

A

===


Ex Calculate Va and Vb resulting from the input signal and from the effects of the ground returns current under the following circumstances(a) Actuator off, Ic = 5 mA, Ip = 20 mA, Ia = 0;(b) Actuator on, Ic = 8 mA, Ip = 35 mA, Ia = 1 A;

The voltage, Vb is determined by the 40-mV input signal and any voltage developed across the 50 mΩ resistance by the ground return current.

Where Va = (Ic+Ip+Ia)50mΩ

Single Point Grounding System

ab VV

Ω++

Ω−=

k 01MΩ 11)mV 40(

k 01MΩ 1

(a) Actuator off

V 87.3)mV 25.1)(101()mV 40)(100( −=+−=bV

mV 1.25)m 0)(50mA 20mA (5 =Ω++=aV

The error is about 3% compare to the correct value (-100)(40 mV) = -4 V

(a) Actuator off

V 25.1)mV 52)(101()mV 40)(100( =+−=bV

mV 25)m A)(50 1mA 20mA (5 =Ω++=aV(b) Actuator on

The large return current has raised the non-inverting input to 52 mV, and this causes the unacceptable error.


Multiple-Point Ground SystemGround loops can occur when the multiple ground points are separated by a large distance and connected to the ac power ground, or when low-level analog circuits are used.

Plant ControlBuilding

0-2000VGround difference

signal wiring

0-100 m

Multiple-Point Ground System

circuit1

circuit2

Ground loop

VN

VG

circuit1

circuit2

VN

VG

A ground loop between two circuits

A ground loop between two circuits can be broken by inserting a common-mode choke

Multiple-Point Ground System

A ground loop between two circuits can be broken by inserting an optical coupler

A ground loop between two circuits can be broken by inserting a transformer

circuit1

circuit2

VN

VG

circuit1

circuit2

VN

VG

Isolation Amplifiers

General concepts: Isolation Amplifier provides three important advantages over normal amplifiers.

Safety issues for some industrial and medical applications: signal common isolation and common ground can be achieved

Extremely high common-mode voltage tolerance (in general amp this is normally less than power supply)

Very low failure currents

Transformer-coupled Amplifiers

Block diagram of AD289 isolation amp

Symbol of transformer-coupled amp.

Optically Coupled Amplifiers

Transformer-coupled isolation amplifiers are expensive, bulky, bandwidth limited and slow response. Optically-coupled isolation amplifiers have less non-linear and isolation.

Optically Coupled Amplifiers

Symbol of optically coupled amp. Simplified schematic diagram

G

in

RVII == 21

inG

KKout V

RRRIV == 2

Documents

Signal Conditioning and Transmission