linearization arrangements for rtds

SUGATA MUNSHI

Deptt. Of Electrical Engg.

Jadavpur University

of 25

LINEARIZATION ARRANGEMENTS FOR RTDs

FOUR-WIRE OHMMETER METHOD

[ ]

0

2

210

,

....1)(

ttt

where

tttRIRItv n

ntoreftref

−=∆

∆++∆+∆+== ααα

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Jadavpur University

of 25

� Thus, while the relation between vo and Rt is

linear, that between v0 and t is nonlinear.

� r is of manganin or constantan� low tempco.

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Jadavpur University

of 25

0

With r connected,

( ) ( ) tref eq ref

t

rRv t I R t I

r R= =

+

� Vo vs. t is linear → Req vs. t is linear.

� Aim: To linearize vo vs. t characteristic over tl to

tu °°°°C. → linearize Req vs. t.

� Condition to be fulfilled : ∆∆∆∆R 1 = ∆∆∆∆R2 .

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Jadavpur University

of 25

That is,

( ) ( ) ( ) ( )

,

2 ( ) ( ) ( ) (1)

eq m eq l eq u eq m

eq m eq u eq l

R t R t R t R t

or

R t R t R t

− = −

= +

( ) ( )

2 (2)

22 1 1

,

2 2

2 1

,

(

1

= 2

m l u l m m l u

m l u

m l u

m l u

m l u

m l u

l m

l u m u

t t t

t t t

t t t

t t t t t t t t t t

t

t t t

t t t

t t t

R R R

rR rR rR

r R r R r R

rR rR rR

r R r R r R

or

r R r R r R

Finally

R R R R R R Rr

R R R R R

= ++ + +

− = − + −+ + +

= ++ + +

+ − + −=

+ − − (3)

) ( )m m lt t tR R− −

� If Rt vs. t characteristic is linear,

SUGATA MUNSHI


Jadavpur University

of 25

as expecte

( ) ( )

d, .

u m m lt t t tR R R R

r

− = −

∴ = ∞

LINEARIZING ARRANGEMENT FOR RTDs LINEARIZING ARRANGEMENT FOR RTDs LINEARIZING ARRANGEMENT FOR RTDs LINEARIZING ARRANGEMENT FOR RTDs

HAVING RESISTANCEHAVING RESISTANCEHAVING RESISTANCEHAVING RESISTANCE----TEMPERATURE CURVE TEMPERATURE CURVE TEMPERATURE CURVE TEMPERATURE CURVE

THAT IS CONVEX UPWARD:THAT IS CONVEX UPWARD:THAT IS CONVEX UPWARD:THAT IS CONVEX UPWARD:

� If Rt vs. t characteristic has progressively

decreasing slope ( αααα2 is -ve)

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Jadavpur University

of 25

( ) ( )

, since numerator of the

expression for

Value of r obtai

r is +ve

ned is -v

.

e

u m m lt t t tR R R R− < −

∴

Example: Platinum RTD .

� The problem is solved by using an op-amp based

const. current source with –ve internal resistance.

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( )

0 1 0

0

0

3 1

1 3 1 3 11

3 3

0

Applying KCL at o/p terminal,

(1)

or,

(2

)

,

)

(3

t t

t

r

t

V R R R

V V V V V

R R R

V

R R R

V

o

V RV

R R R

V V

r

+

+ −

++ +

=

− −+ =

−

−

=

( )1 3

1 2

1 2 3 2 3

( ) (4)ef

ref

V RR R V

R R R R R+ + =

+ + +

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Jadavpur University

of 25

3

1 3

3 20

1

0

1

1

3

3 2

2

3

3

1

2

; where,

Effective linearizing resistan

, r has a negative valu

,

( )

ce

(

e

)

=

.

reftref r

tr

ef

ef

t

t

VR rV I I

R r R

and

R Rr

R R

Finally

R RR

V R R

If R

VRR

RR

R

R

R

= =+

− =

=−

+ −

<

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Jadavpur University

of 25

LINEARIZATION ARRANGEMENTS FOR NTC

THERMISTORS:

PRINCIPLE PRINCIPLE PRINCIPLE PRINCIPLE ::::

⌦ Desired: Linear relation between S &

T over certain temp. range.

According to Taylor’s theorem, S(T) can

be expanded into an infinite series about a

reference temperature Tr as,

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Jadavpur University

of 25

2 3

( ) ( ) ( ) ( ) ( ) .......2! 3!

,

( )

r

r r r r

r

nn

r n

T T

h hS T S T hS T S T S T

where h T T

d SS T

dT=

′ ′′ ′′′= + + + +

=

=

∼

⌦ For above series to converge, magnitude of terms in

increasing powers of h should decrease rapidly with ascent

of power of h.

⌦ Practical systems�� sufficient to consider up to term in

h3 .

⌦ Major contribution to nonlinearity comes from term in

h2.

⌦ If S'' (Tr ) is set to zero, then, if contribution from h3

term is negligibly small,

( ) ( ) ( )r r

S T S T hS T′+�

⌦ Reasonable linearity of S(T) vs. T curve. Usually Tr =

Tm ��midpoint of temp. range of interest.

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⌦ S'' (Tm ) is set to zero by selecting linearizing ckt.

parameter(s).Then S(T) has an inflection at T=Tm . Thus

quasi-linearization is achieved.

SHUNT LINEARIZATION SCHEME:SHUNT LINEARIZATION SCHEME:SHUNT LINEARIZATION SCHEME:SHUNT LINEARIZATION SCHEME:

Here,

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Jadavpur University

of 25

0

0 eq

0 m eq m

V (T) vs. T linear

( ) ( ) ( )

R (T) vs. T is linear

i.e.

V (

(1)

T )=0 R (T )=0

Tref eq ref

T

rRS T V T I R T I

r R

∴ →

′′ →

=

′′

= =+

( ) ( )

( ) ( )

( )

( )

2

2 2

22 2 2

4

3 2 2 2

3

( )( ) (2)

" 2( )

r 2 =

T T T T Teq

T T

T T T T

eq

T

T T T T

T

rR r R rR R r RR T

r R r R

r R r R r R r RR T

r R

R r R R r R

r R

′ ′ ′+ −′ = =

+ +

′′ ′+ − +′′ =

+

′′ ′′ ′+ −

+

( )

2 2

3

r 2 = (3)

T T T T

T

rR R R R

r R

′′ ′′ ′+ −

+

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Jadavpur University

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0

0

2

1 1

2

Hence to have ( ) = 0, we should make

2

(4)

Considering

2

,

,

m

m

m

m

m m m m

T

eq m

T T T T

T T

T T

T

T

T T

R T

rR R R R

R R e

Rr R

R

R R

β

−

′′

′′ ′ ′′= −

∴

=

′= −

′′

= 0

0

1 1

, (5)

mT Te

β

−

2

2

4 3 3

2

3

.

(6)

22

,

2

m m

m m

T T

T T

m

T T

m

T T T T

m

R RT

or

R R R RT T T T

or

R RT

R RT T

β

β β

β

β

β β

β

′ = −

′′ = + = +

′ = −

′′ = +

(

7)

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Jadavpur University

of 25

m

m

m

m

22

4Tm

T

T

3Tm m

From equations (4), (5), (6) and (7),

β2R

Tr = R

ββR 2

T T

2r =

Finally,

R (8) 2

m

m

T

T

β

β

−

+

−

+

∴

SERIES SERIES SERIES SERIES LINEARIZATION SCHEME:LINEARIZATION SCHEME:LINEARIZATION SCHEME:LINEARIZATION SCHEME:

r = Linearizing resistance

� As T increases→ V0 increases.

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Jadavpur University

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0

0 2

( ) ( ) (1)

V (T) = (2)( )

T

T

T

rS T V T E

r R

REr

r R

= =+

′′ −

+

2 2

0 4

0 m

2 2

2

( ) 2 ( )V (T) = (3)

( )

To have V (T )=0, we should have,

( ) 2 ( )

,

2 =

2 r =

2

m m m m

m

m

m

m

T T T T

T

T T T T

T

T

mT

T

m

R r R R r REr

r R

R r R R r R

or

Rr R

R

TR

T

β

β

′′ ′ + − + ′′ −+

′′

′′ ′+ = +

′

+

−′′

∴−

(4)

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Jadavpur University

of 25

LINEARIZATION USING LOG-NETWORK

0 0

0

,

Output of network is,

( ) ( ) (1)

.

i Teq

TTeq

T

i

ref

V IR

where

rRR

r R

VS T V T K n

V

K Scale factor

=

=+

= = −

=

�

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Jadavpur University

of 25

0 0 0

0 0

,

( ) ( )( ) ( )

=( )

,

is a constant.

T T

ref T T

T

T

ref

or

IrR GrRS T V T K n K n

V r R r R

rRK n K nG

r R

where

IG

V

= = − = −

+ +

− −

+

=

� �

� �

( ) ( )2

2

00 0

200 2

( )

...................... (2)

( ) ( ) ( 2 ) (3)( )

TT T

T T TT

T T T T T

T T

K rRr R r RV T K

rR R r Rr R

K rV T R R r R R r R

R r R

′′ −+′∴ = − =

+ +

−′′ ′′ ′ = + − + ′′ +

Thus for making 0 ( ) 0mV T′′ = the following condition

has to be fulfilled.

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Jadavpur University

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( )

2

2

2

( ) ( 2 )

,

2 (4)

m m m m m

m m m m

m m m

T T T T T

T T T T

T T T

R R r R R r R

or

R R R Rr

R R R

′′ ′+ = +

′ ′′−=

′′ ′−

0

0

1 1

2

3

Recalling that,

,

2

m

m

m m

m m

T T

T T

T T

m

T T

m m

R R e

R RT

and

R RT T

β

β

β β

−

=

′ = −

′′ = +

We get,

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Jadavpur University

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22 2

4 3

22 2

3 4

2 2

2

22

=

2

2

( 2 )

m m m

m m

m

mT

T T T

M M M

T T

m m m

T

m m

m

m

m

m

R R RT T T

r

R RT T T

RT T

T

R Tr

T

T

β β β

β β β

β β

β

β

β

− +

=

+ −

− +

=

+

−

−

∴

SUGATA MUNSHI


Jadavpur University

of 25

IMPLEMENTATION OF LOG-AMPLIFIER

CIRCUIT :

When a transistor is placed in a circuit such that

the collector voltage VC = 0, the collector current

is given by the equation

19

exp 1

,

Current transfer ratio between emitter & collector 1

Reverse saturation current of base-emitter diode.

q= Magnitude of electronic charge=1.6 10 .

Boltzma

EC ES

a

ES

qVI I

KT

where

I

C

K

α

α

−

−−

≈

=

×

=

�

23nn's Constant =1.38 10 / .

Ambient temperature in Kelvin.a

J Kelvin

T

−×

=

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Jadavpur University

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-13

ESTypically I 10 & 1,

&

exp

C ES

EC ES

a

A

I I

qVI I

KT

α≈ ≈

∴

−

∵

�

�

0 C

0 0

In the above ckt.

and I

a CE

ES

iE i

a i i

ES ref

KT IV n

q I

ee V I

R

KT e ee n K n

q I R V

∴−

= = =

∴ = − = −

� �

� �

Note: Diode D protects T against excessive reverse

base-emitter voltage.

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Jadavpur University

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SUGATA MUNSHI


Jadavpur University

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2

2

0

3

2

=Constant ( if

( )( )

R )

a

Z

T

i

ES

Z

T

KT V TV

V VI r

rR RR

R

Tq R

r

nI

−

=

+

=

+

∴ �

� �

For the log ckt. considered:

⌦ 0 .a

K T∝ Changes by 0.3% /ºC in vicinity of

27ºC.

⌦ Main temp. error is due IES which

approximately doubles for every 10ºC change

in temp.

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Jadavpur University

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Temperatre Compensated Log-Amplifier:

Q1 and Q2 are matched transistors.

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( )

10 0

2

1

1 11 0 0

1 1

2 22 0 0

2 1

110 0

1 2

0 0

2 V ( ) &

,

V ,

,

( )

Ca E

ES ES

Cb E

ES ES

ESb a

ES

i r

i

ref

ef

I VV V K n K n

I R I

I VV V K n K n

I R I

R IVV A V V AK n

R I V

VV AK n

V

V TV AK

If V T V

th

n

o

V

r

en

α α

α α

α

α

=

=

= = − = −

= = − = −

∴ = − = ×

= =

� �

� �

�

�

�

linearization arrangements for rtds

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