RTD linearization
in [Design]
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From: Spehro Pefhany on 23 Dec 2009 11:06 On Wed, 23 Dec 2009 00:36:48 -0800, "JosephKK"<quiettechblue(a)yahoo.com> wrote: >On Tue, 22 Dec 2009 14:51:35 -0500, Spehro Pefhany <speffSNIP(a)interlogDOTyou.knowwhat> wrote: > >>On Tue, 22 Dec 2009 12:02:35 -0500, Phil Hobbs >><pcdhSpamMeSenseless(a)electrooptical.net> wrote: >> >>>On 12/22/2009 11:28 AM, John Larkin wrote: >>>> On Mon, 21 Dec 2009 23:38:08 -0800, >>>> "JosephKK"<quiettechblue(a)yahoo.com> wrote: >>>> >>>>> On Sat, 19 Dec 2009 17:39:19 -0800, John Larkin<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>>> >>>>>> On Sat, 19 Dec 2009 17:34:01 -0800, >>>>>> "JosephKK"<quiettechblue(a)yahoo.com> wrote: >>>>>> >>>>>>> On Sun, 13 Dec 2009 09:50:22 -0600, "RogerN"<regor(a)midwest.net> wrote: >>>>>>> >>>>>>>> >>>>>>>> "John Larkin"<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message >>>>>>>> news:p1u7i5thbjmtjvqcj63b291l19rf7ktllp(a)4ax.com... >>>>>>>>> >>>>>>>>> >>>>>>>>> Does anybody remember the value of negative resistance that linearizes >>>>>>>>> a 100 ohm platinum RTD? >>>>>>>>> >>>>>>>>> John >>>>>>>> >>>>>>>> I thought RTD's were supposed to be linear. The 100 ohm resistance being at >>>>>>>> 0 Degrees C and a change of .385 ohms (for a 100 Ohm RTD) per Degree C. I >>>>>>>> read a description of instrumentation for RTD's once. They said they used a >>>>>>>> 1ma current source to the RTD and compared it to the voltage drop with 1ma >>>>>>>> in a 100 Ohm resistor. Low current, 1ma, was used to minimize the RTD >>>>>>>> heating up from power. I know there is a 3rd and sometimes 4th sense wire >>>>>>>> used to compensate for the lead resistance. I don't exactly remember the >>>>>>>> source but I read the information when looking up info for my Allen Bradley >>>>>>>> RTD input card for my PLC 5 rack. >>>>>>>> >>>>>>>> RogerN >>>>>>>> >>>>>>> Platinum RTDs are about total repeatability, a real mantra in the measurement >>>>>>> community. And 393 ppm/K is an exponential, like most all resistance tempcos. >>>>>> >>>>>> Exponential? How so? >>>>>> >>>>>> John >>>>> >>>>> Each degree of temperature change is a multiplier on the degree base before it. >>>>> so the recurrence relation results in an exponential. Thus it can be modeled >>>>> as r' = r(25) * k(1)e^[k(2)*t], an exponential. BTW just look at a R vs T plot. >>>> >>>> What are the coefficients? The curve slopes down. >>>> >>>> I've never seen the platinum RTD curve expressed as an exponential. >>>> The usual formulation is a polynomial. >>>> >>>> www.analogzone.com/acqt_052807.pdf >>>> >>>> http://www.omega.com/temperature/Z/pdf/z251.pdf >>>> >>>> >>>> >>>> John >>>> >>> >>>A curve whose normalized slope is constant is an exponential--for >>>instance, tempco of resistance is generally defined as (1/R)dR/dT, which >>>is d/dT(ln R). If that were really some constant alpha, then ln R would >>>be proportional to T, so R would be something times exp(alpha T). >>> >>>Of course, there's nothing that says the tempco is constant, IOW not all >>>curves are exponential. The RTD curve is a simple rational function of >>>T--it really linearizes beautifully with a bit of negative resistance, >>>with theoretical deviations are of the order of 0.01K over wide >>>temperature ranges. >>> >>>Cheers >>> >>>Phil Hobbs >> >>Maybe 0.1�C over the full temperature range of an RTD (-200 ~ 850�C). >>R = -2311 ohms. >> >>About +/-0.02� for 0 ~ 400�C. >>R = -2515 ohms >> >>And almost nothing for 0 ~ 100�C >>R = -2687 ohms >> >> >So please explain the various correcting "r" and the different performance. The "R" is calculated using a simple equation to exactly zero the linearity error out at mid-scale**. IOW, so Rt(Tmid) - Rt(Tzero)= 0.5* (Rt(Tfs) - Rt(Tzero)), where Rt is the parallel combination of the linearization resistor and the sensor Rt = Rsensor(T)*R/(Rsensor(T) + R) This method of linearization is almost perfect over a relatively narrow span, and pretty good even over the whole usable range of the sensor (of course 100�C -- say -40�C ~ +60�C is a very wide range for human comfort, but only about 10% of the useful range of a general-purpose sensor). ** Since it's an imperfect fit, you have to come up with some kind of criteria for judging what is the 'best' fit. For ranges of a few hundred C or less, it looks like zeroing the error at midscale is pretty reasonable-- an S-shaped error curve that is centered on zero error and crosses the zero line at zero, full-scale and mid scale. OTOH, over the full range, it yields asymmetrical peak errors so it may not be optimal. Sometimes cost functions can be pretty strange- mathematicians like to use sum (or integral) of error squared, because it's easy to differentiate and find a closed form solution for the minimum error-- in some cases. Instrument designers, OTOH, are aiming for specs like worst-case error PLUS maybe there should be no systemic error at calibration points and/or some easy-to-check point like room temperature (or body temperature) for a temperature instrument or ~21% for an oxygen meter. Nobody in the real world is likely to request a sensor or instrument with a maximum integral of error squared over some range. ;-) Anyway, these days we can often use a processor or FPGA to apply simple brute-force numerical techniques and reduce linearization errors to negligible levels in relation to the error budget-- or even below the noise floor. Thermocouples have ugly (but fairly small) nonlinearity... nothing a set of double-precision polynomials curve-fitted to the values can't handle. The same method can be used to linearize RTDs when you have a universal-input instrument-- no need to be clever for that part of the design... in fact it could be counterproductive.
From: JosephKK on 23 Dec 2009 03:34 On Tue, 22 Dec 2009 12:02:35 -0500, Phil Hobbs <pcdhSpamMeSenseless(a)electrooptical.net> wrote: >On 12/22/2009 11:28 AM, John Larkin wrote: >> On Mon, 21 Dec 2009 23:38:08 -0800, >> "JosephKK"<quiettechblue(a)yahoo.com> wrote: >> >>> On Sat, 19 Dec 2009 17:39:19 -0800, John Larkin<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>> >>>> On Sat, 19 Dec 2009 17:34:01 -0800, >>>> "JosephKK"<quiettechblue(a)yahoo.com> wrote: >>>> >>>>> On Sun, 13 Dec 2009 09:50:22 -0600, "RogerN"<regor(a)midwest.net> wrote: >>>>> >>>>>> >>>>>> "John Larkin"<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message >>>>>> news:p1u7i5thbjmtjvqcj63b291l19rf7ktllp(a)4ax.com... >>>>>>> >>>>>>> >>>>>>> Does anybody remember the value of negative resistance that linearizes >>>>>>> a 100 ohm platinum RTD? >>>>>>> >>>>>>> John >>>>>> >>>>>> I thought RTD's were supposed to be linear. The 100 ohm resistance being at >>>>>> 0 Degrees C and a change of .385 ohms (for a 100 Ohm RTD) per Degree C. I >>>>>> read a description of instrumentation for RTD's once. They said they used a >>>>>> 1ma current source to the RTD and compared it to the voltage drop with 1ma >>>>>> in a 100 Ohm resistor. Low current, 1ma, was used to minimize the RTD >>>>>> heating up from power. I know there is a 3rd and sometimes 4th sense wire >>>>>> used to compensate for the lead resistance. I don't exactly remember the >>>>>> source but I read the information when looking up info for my Allen Bradley >>>>>> RTD input card for my PLC 5 rack. >>>>>> >>>>>> RogerN >>>>>> >>>>> Platinum RTDs are about total repeatability, a real mantra in the measurement >>>>> community. And 393 ppm/K is an exponential, like most all resistance tempcos. >>>> >>>> Exponential? How so? >>>> >>>> John >>> >>> Each degree of temperature change is a multiplier on the degree base before it. >>> so the recurrence relation results in an exponential. Thus it can be modeled >>> as r' = r(25) * k(1)e^[k(2)*t], an exponential. BTW just look at a R vs T plot. >> >> What are the coefficients? The curve slopes down. >> >> I've never seen the platinum RTD curve expressed as an exponential. >> The usual formulation is a polynomial. >> >> www.analogzone.com/acqt_052807.pdf >> >> http://www.omega.com/temperature/Z/pdf/z251.pdf >> >> >> >> John >> > >A curve whose normalized slope is constant is an exponential--for >instance, tempco of resistance is generally defined as (1/R)dR/dT, which >is d/dT(ln R). If that were really some constant alpha, then ln R would >be proportional to T, so R would be something times exp(alpha T). > >Of course, there's nothing that says the tempco is constant, IOW not all >curves are exponential. The RTD curve is a simple rational function of >T--it really linearizes beautifully with a bit of negative resistance, >with theoretical deviations are of the order of 0.01K over wide >temperature ranges. > >Cheers > >Phil Hobbs Interesting, i listen to you on this but not to JL.
From: JosephKK on 23 Dec 2009 03:36 On Tue, 22 Dec 2009 14:51:35 -0500, Spehro Pefhany <speffSNIP(a)interlogDOTyou.knowwhat> wrote: >On Tue, 22 Dec 2009 12:02:35 -0500, Phil Hobbs ><pcdhSpamMeSenseless(a)electrooptical.net> wrote: > >>On 12/22/2009 11:28 AM, John Larkin wrote: >>> On Mon, 21 Dec 2009 23:38:08 -0800, >>> "JosephKK"<quiettechblue(a)yahoo.com> wrote: >>> >>>> On Sat, 19 Dec 2009 17:39:19 -0800, John Larkin<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>> >>>>> On Sat, 19 Dec 2009 17:34:01 -0800, >>>>> "JosephKK"<quiettechblue(a)yahoo.com> wrote: >>>>> >>>>>> On Sun, 13 Dec 2009 09:50:22 -0600, "RogerN"<regor(a)midwest.net> wrote: >>>>>> >>>>>>> >>>>>>> "John Larkin"<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message >>>>>>> news:p1u7i5thbjmtjvqcj63b291l19rf7ktllp(a)4ax.com... >>>>>>>> >>>>>>>> >>>>>>>> Does anybody remember the value of negative resistance that linearizes >>>>>>>> a 100 ohm platinum RTD? >>>>>>>> >>>>>>>> John >>>>>>> >>>>>>> I thought RTD's were supposed to be linear. The 100 ohm resistance being at >>>>>>> 0 Degrees C and a change of .385 ohms (for a 100 Ohm RTD) per Degree C. I >>>>>>> read a description of instrumentation for RTD's once. They said they used a >>>>>>> 1ma current source to the RTD and compared it to the voltage drop with 1ma >>>>>>> in a 100 Ohm resistor. Low current, 1ma, was used to minimize the RTD >>>>>>> heating up from power. I know there is a 3rd and sometimes 4th sense wire >>>>>>> used to compensate for the lead resistance. I don't exactly remember the >>>>>>> source but I read the information when looking up info for my Allen Bradley >>>>>>> RTD input card for my PLC 5 rack. >>>>>>> >>>>>>> RogerN >>>>>>> >>>>>> Platinum RTDs are about total repeatability, a real mantra in the measurement >>>>>> community. And 393 ppm/K is an exponential, like most all resistance tempcos. >>>>> >>>>> Exponential? How so? >>>>> >>>>> John >>>> >>>> Each degree of temperature change is a multiplier on the degree base before it. >>>> so the recurrence relation results in an exponential. Thus it can be modeled >>>> as r' = r(25) * k(1)e^[k(2)*t], an exponential. BTW just look at a R vs T plot. >>> >>> What are the coefficients? The curve slopes down. >>> >>> I've never seen the platinum RTD curve expressed as an exponential. >>> The usual formulation is a polynomial. >>> >>> www.analogzone.com/acqt_052807.pdf >>> >>> http://www.omega.com/temperature/Z/pdf/z251.pdf >>> >>> >>> >>> John >>> >> >>A curve whose normalized slope is constant is an exponential--for >>instance, tempco of resistance is generally defined as (1/R)dR/dT, which >>is d/dT(ln R). If that were really some constant alpha, then ln R would >>be proportional to T, so R would be something times exp(alpha T). >> >>Of course, there's nothing that says the tempco is constant, IOW not all >>curves are exponential. The RTD curve is a simple rational function of >>T--it really linearizes beautifully with a bit of negative resistance, >>with theoretical deviations are of the order of 0.01K over wide >>temperature ranges. >> >>Cheers >> >>Phil Hobbs > >Maybe 0.1°C over the full temperature range of an RTD (-200 ~ 850°C). >R = -2311 ohms. > >About +/-0.02° for 0 ~ 400°C. >R = -2515 ohms > >And almost nothing for 0 ~ 100°C >R = -2687 ohms > > So please explain the various correcting "r" and the different performance.
From: John Larkin on 23 Dec 2009 17:31 On Wed, 23 Dec 2009 00:34:35 -0800, "JosephKK"<quiettechblue(a)yahoo.com> wrote: >On Tue, 22 Dec 2009 12:02:35 -0500, Phil Hobbs <pcdhSpamMeSenseless(a)electrooptical.net> wrote: > >>On 12/22/2009 11:28 AM, John Larkin wrote: >>> On Mon, 21 Dec 2009 23:38:08 -0800, >>> "JosephKK"<quiettechblue(a)yahoo.com> wrote: >>> >>>> On Sat, 19 Dec 2009 17:39:19 -0800, John Larkin<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >>>> >>>>> On Sat, 19 Dec 2009 17:34:01 -0800, >>>>> "JosephKK"<quiettechblue(a)yahoo.com> wrote: >>>>> >>>>>> On Sun, 13 Dec 2009 09:50:22 -0600, "RogerN"<regor(a)midwest.net> wrote: >>>>>> >>>>>>> >>>>>>> "John Larkin"<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote in message >>>>>>> news:p1u7i5thbjmtjvqcj63b291l19rf7ktllp(a)4ax.com... >>>>>>>> >>>>>>>> >>>>>>>> Does anybody remember the value of negative resistance that linearizes >>>>>>>> a 100 ohm platinum RTD? >>>>>>>> >>>>>>>> John >>>>>>> >>>>>>> I thought RTD's were supposed to be linear. The 100 ohm resistance being at >>>>>>> 0 Degrees C and a change of .385 ohms (for a 100 Ohm RTD) per Degree C. I >>>>>>> read a description of instrumentation for RTD's once. They said they used a >>>>>>> 1ma current source to the RTD and compared it to the voltage drop with 1ma >>>>>>> in a 100 Ohm resistor. Low current, 1ma, was used to minimize the RTD >>>>>>> heating up from power. I know there is a 3rd and sometimes 4th sense wire >>>>>>> used to compensate for the lead resistance. I don't exactly remember the >>>>>>> source but I read the information when looking up info for my Allen Bradley >>>>>>> RTD input card for my PLC 5 rack. >>>>>>> >>>>>>> RogerN >>>>>>> >>>>>> Platinum RTDs are about total repeatability, a real mantra in the measurement >>>>>> community. And 393 ppm/K is an exponential, like most all resistance tempcos. >>>>> >>>>> Exponential? How so? >>>>> >>>>> John >>>> >>>> Each degree of temperature change is a multiplier on the degree base before it. >>>> so the recurrence relation results in an exponential. Thus it can be modeled >>>> as r' = r(25) * k(1)e^[k(2)*t], an exponential. BTW just look at a R vs T plot. >>> >>> What are the coefficients? The curve slopes down. >>> >>> I've never seen the platinum RTD curve expressed as an exponential. >>> The usual formulation is a polynomial. >>> >>> www.analogzone.com/acqt_052807.pdf >>> >>> http://www.omega.com/temperature/Z/pdf/z251.pdf >>> >>> >>> >>> John >>> >> >>A curve whose normalized slope is constant is an exponential--for >>instance, tempco of resistance is generally defined as (1/R)dR/dT, which >>is d/dT(ln R). If that were really some constant alpha, then ln R would >>be proportional to T, so R would be something times exp(alpha T). >> >>Of course, there's nothing that says the tempco is constant, IOW not all >>curves are exponential. The RTD curve is a simple rational function of >>T--it really linearizes beautifully with a bit of negative resistance, >>with theoretical deviations are of the order of 0.01K over wide >>temperature ranges. >> >>Cheers >> >>Phil Hobbs > >Interesting, i listen to you on this but not to JL. You still haven't put numbers on your exponential RTD thing. John
From: JosephKK on 24 Dec 2009 00:34
On Wed, 23 Dec 2009 11:06:16 -0500, Spehro Pefhany <speffSNIP(a)interlogDOTyou.knowwhat> wrote: >On Wed, 23 Dec 2009 00:36:48 -0800, >"JosephKK"<quiettechblue(a)yahoo.com> wrote: > >>On Tue, 22 Dec 2009 14:51:35 -0500, Spehro Pefhany <speffSNIP(a)interlogDOTyou.knowwhat> wrote: >> >>>On Tue, 22 Dec 2009 12:02:35 -0500, Phil Hobbs >>><pcdhSpamMeSenseless(a)electrooptical.net> wrote: >>> >>>>On 12/22/2009 11:28 AM, John Larkin wrote: >>>>> On Mon, 21 Dec 2009 23:38:08 -0800, >>>>> "JosephKK"<quiettechblue(a)yahoo.com> wrote: >>>>> <snip> >>>> >>>>A curve whose normalized slope is constant is an exponential--for >>>>instance, tempco of resistance is generally defined as (1/R)dR/dT, which >>>>is d/dT(ln R). If that were really some constant alpha, then ln R would >>>>be proportional to T, so R would be something times exp(alpha T). >>>> >>>>Of course, there's nothing that says the tempco is constant, IOW not all >>>>curves are exponential. The RTD curve is a simple rational function of >>>>T--it really linearizes beautifully with a bit of negative resistance, >>>>with theoretical deviations are of the order of 0.01K over wide >>>>temperature ranges. >>>> >>>>Cheers >>>> >>>>Phil Hobbs >>> >>>Maybe 0.1°C over the full temperature range of an RTD (-200 ~ 850°C). >>>R = -2311 ohms. >>> >>>About +/-0.02° for 0 ~ 400°C. >>>R = -2515 ohms >>> >>>And almost nothing for 0 ~ 100°C >>>R = -2687 ohms >>> >>> >>So please explain the various correcting "r" and the different performance. > > >The "R" is calculated using a simple equation to exactly zero the >linearity error out at mid-scale**. IOW, so >Rt(Tmid) - Rt(Tzero)= 0.5* (Rt(Tfs) - Rt(Tzero)), where Rt is the >parallel combination of the linearization resistor and the sensor >Rt = Rsensor(T)*R/(Rsensor(T) + R) > >This method of linearization is almost perfect over a relatively >narrow span, and pretty good even over the whole usable range of the >sensor (of course 100°C -- say -40°C ~ +60°C is a very wide range for >human comfort, but only about 10% of the useful range of a >general-purpose sensor). > >** Since it's an imperfect fit, you have to come up with some kind of >criteria for judging what is the 'best' fit. For ranges of a few >hundred C or less, it looks like zeroing the error at midscale is >pretty reasonable-- an S-shaped error curve that is centered on zero >error and crosses the zero line at zero, full-scale and mid scale. > >OTOH, over the full range, it yields asymmetrical peak errors so it >may not be optimal. Sometimes cost functions can be pretty strange- >mathematicians like to use sum (or integral) of error squared, because >it's easy to differentiate and find a closed form solution for the >minimum error-- in some cases. Instrument designers, OTOH, are aiming >for specs like worst-case error PLUS maybe there should be no systemic >error at calibration points and/or some easy-to-check point like room >temperature (or body temperature) for a temperature instrument or ~21% >for an oxygen meter. Nobody in the real world is likely to request a >sensor or instrument with a maximum integral of error squared over >some range. ;-) > >Anyway, these days we can often use a processor or FPGA to apply >simple brute-force numerical techniques and reduce linearization >errors to negligible levels in relation to the error budget-- or even >below the noise floor. Thermocouples have ugly (but fairly small) >nonlinearity... nothing a set of double-precision polynomials >curve-fitted to the values can't handle. The same method can be used >to linearize RTDs when you have a universal-input instrument-- no need >to be clever for that part of the design... in fact it could be >counterproductive. > OK. I see now, it works much like the changes in quartz crystal cuts in the 1960s through 1980s, with doubly rotated AT (DRAT) and later SC cuts. Set things up to create a "flat spot/range" in the tempco curve shape. |