learning objectives notes/lesson4et438a.pdf · learning objectives lesson4et438a.pptx 2 after this...

8/27/2015

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ET 438a

Automatic Control Systems Technology

1lesson4et438a.pptx

Learning Objectives

lesson4et438a.pptx 2

After this presentation you will be able to:

Develop mathematical relationships for a sensor that has a linear output.

Convert linear mathematical equations into block diagrams that represent sensor scaling circuits.

Adjust the output range of a sensor using operational amplifier circuits.

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Scaling Sensor Outputs


x= transducer input(measured value)

VT = transducer outputvoltage

KT = transducer gain(slope)

Ks = scalar gainVs = scalar output

x1 x2

VT2

VT1

KT

x

VT

KS

ST

o

i

= =output span

input span

Span = max. value - min. value

Sensor Scaling Case 1: No Offset in Scalar or

Sensor


Transducer gain formula: VT = KTx

Required scalar gain:

KS

Ss

d

T

= =desired span

transducer span

V K K xs s T= ⋅ ⋅

Scalar output formula

V K Vs s T= ⋅

x

VT

Ks

KT

SensorKT

Block Diagram

ScalarKs

VT Vs

Input OutputVT = KTx Vs = KsVT

x

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Example 4-1: Case 1 Sensor


A pressure transducer has a usable range P (x) range of 0-50 psig (lb/in2

gauge). It has a voltage output range of 0-1.25 Vdc over the pressure range. Scale the output to a range of 0-10 Vdc

Find KTi

oT

S

S

spaninput

spanoutput K == V/psig 025.0

psig 050

V 025.1KT =

−−

=

Find scalar gainT

ds

S

S

span transducer

span desiredK == V/V 8

V 025.1

V 010K s =

−−

=

KT=0.025 V/psig

Ks=8 V/VVT VsP

Sensor Scaling Case 2: Offset in Sensor No

Offset in Output


KT

Ks

Sensor gain formula: VT = KTx + b

Where b = transducer offset

Scalar gain formula must subtract offset

bKVKV

xKKV

bKbKxKKV

bK)bxK(KV

sTss

Tss

ssTss

sTss

−=

=

−+=

−+=Correctform of scalar output

Scalar equation

x

VT

b

Block diagram

SensorKT∙x+b

ScalarKs∙VT- Ks∙ b

VT Vsx

VT=KT∙x+b Vs=Ks∙VT- Ks∙ b

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Example 4-2: Offset Sensor Output

Range


A temperature transducer has a range of 0-100 C (input x) It has a voltage output range of 1-5 Vdc (VT) over this temperature range. Scale the output to a range of 0-10 Vdc. Find the transducer gain, KT, and the offset,b. Find the scalar relationship required to get the desired output range. Draw a block diagram of this sensor/scalar system that includes the mathematical relationship derived above.

Find KT

i

oT

S

S

spaninput

spanoutput K == V/C 04.0

C 0 - 100

V 15KT =

−=

Find b from point slope form of line )xx(KVV 1T1TT −=−

1x04.0V

)0x(04.01V

T

T

+=

−=−

b=1.0

Example 4-2: Solution (2)


Find scalar gain KS

Ss

d

T

= =desired span

transducer span

S

Sd

T

=−−

= =10 0

5 1

10

42 5

V

V V / V = K s.

V K V K b

V V

s s T s

s T

= −

= −2 5 2 5 1. . ( )

Scalar formula V Vs T= −2 5 2 5. .

Transducer 0.04x+1

VT= 0.04x+1

x

VTVs

Input

Output

Scalar2.5VT-2.5

Vs = 2.5VT-2.5

Block Diagram

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Sensor Scaling Case 3: Transducer with

no offset, Output offset


KT

Ks

VT

x

c = scalar offset can be +- value

Transducer gain formula:VT = KTx

Scalar formula must add a constant

cVKV Tss +=

T

ds

S

S

span transducer

span desiredK ==

To find c, use point slope form using scalar points

)VV(KVV 1TTs1ss −=−

c

Example 4-3: Case 3 Sensor Scaling


A pressure transducer has an input range of 0 - 25 psig (x) and an output range of 0 - 1 V (VT) Find the scaling equation to convert this range into the desired range of -5 V to +5 Vdc) Find transducer gain and scalar gain formulas. Draw the block diagram of the complete system

Find KT KS

ST

o

i

= =output span

input spanKT =

−=

1 0004

V

25 - 0 psig V / psig.

Find Ks KS

Ss

d

T

= =desired span

transducer span

S

Sd

T

=− −−

= =5 5

1 0

10

110

( ) V

V V/ V=Ks

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Use point-slope form of line to find the value of c

)VV(KVV 1TTs1ss −=− 5) ,1()V,V( 1s1T =

Can use either pointin pair defining range

5V10V

)1V(105V

Ts

Ts

−=

−=−

Scalar gain formulac=-5

Transducer 0.04x

VT= 0.04x

x

VTVs

Input

Output

Scalar10VT-5

Vs = 10VT-5

Block Diagram

Example 4-4 Scalar Equations


Range of linear temperature transducer is 32- 212 F with a transducer gain of 10 mV/F. The desired output of the transducer for the range of temperature is 0 - 10 Vdc. Find the gain formula.

Find span of transducer

0.32

2.12

Move origin to (32, 0.32)

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Example 4-4 Solution (2)


Compute sensor equation Shift origin to (32 F, 0.32 V)So b=0.32

Check equation at data points



lesson4et438a.pptx

Transducer 0.01x

VT= 0.01T

T

VTVs

InputTemp

Output

Scalar5.56VT-1.779

Vs = 5.56VT-1.779

Block Diagram

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Example 4-5


A linear temperature transducer has an input range of -20 C to 50 C and a gain of KT = 20 mV/C. The desired output range is 0 - 5 Vdc. The transducer output voltage is bipolar (+-). Find the scaling equation.

Compute the transducer and scalar span

Scalar formula

Find b graphically. Must shift origin to(-20 C, -0.4 V)



Find scalar gain Check scalar equation at data points

Transducer 0.02x

VT= 0.02T

T

VTVs

InputTemp

Output

Scalar3.571VT-1.429

Vs = 3.571VT-1.429

Block Diagram

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Practical Realization of Scalar Equations

Using OP AMPs


Scaling without offset: use inverting or non-inverting amps to implement Ks

in

fs

R

RK

−=

For inverting amps

V0

Rf

Rin

Vcc

Sensor

+=

1

fs

R

R1K

For non-inverting amps


Using OP AMPs


For transducers with offset use inverting and summation amps

Stage 2 OP AMP changes

sign if Rf1 = Rin AV = -1

Vs

Stage 1Stage 2

Rf

Rb

RT

Rf1

Vb

Rin

VT

b

b

fT

T

fo V

R

RV

R

RV

−+

−=

b

b

fT

T

fb

b

fT

T

fs V

R

RV

R

R)1(V

R

RV

R

RV +=−

−+

−=

Stage 1 Overall gain

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Using OP AMPs


Equate the sensor with offset formula to the OP AMP gain derived on the last slide

( ) b

b

fT

T

fsTss V

R

RV

R

RbKVKV +=−= so b

b

fs

T

fs V

R

RbK

R

RK ==

Vs

Stage 1Stage 2

Rf

Rb

RT

Rf1

Vb

Rin

VT

Example 4-6: Implementing Scalar Equation

with OP AMPs


Design an OP AMP circuit that will implement the scalar equation from Example 4-5 Assume Rf1 = Rin = 100 kΩ Rf = 470 kΩ

V Vs T= +3 571 1429. .

Scalar relationship for Example 4-5

Vs

Stage 1Stage 2

Rf

Rb

RT

Rf1

Vb

Rin

VT

From scalar equationKs= 3.571

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Value of RT is not a standard value. Use a standard value near the computed value in series with a potentiometer and calibrate circuit. Use -0.4 and 1.0 V and adjust the potentiometer

127 k 10 k



Compute the value of Rb

Simpler design method: Let Rb=Rf

So….

bs

b

f

VbK

1R

R

=

= Set Vb to the numerical value of Ksb

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Completed design

Vs

Stage 1Stage 2

Rf=470 k

Rb=657.8 k

RT=131.6 k

Rf1= 100 k

Vb

Rin=100 k

VT

ET 438a

Automatic Control Systems Technology


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