Trying to figure out measurement circuit for a PT100 Temp probe - Page 1
Forum section: electronics / projects

author: Yamin
author_rank: Frequent Contributor
timestamp: January 01, 2022, 09:38:02 pm
content: Happy new year all,I was hoping that you guys could help me decipher the workings of the temperature measurement circuit of the unit I'm working on.The circuit is from a soft serve ice cream machine. The quest started when the unit's PT100 temperature probe failed. I was doing some research on whether it was possible to interchange a PT1000 probe in place of a PT100 probe.I didn't come across much about the measurement circuitry, I traced out the one I had.I am having a hard time understanding how its supposed to work. Would greatly appreciate if someone could explain. Also would it be possible to interchange to a PT1000 probe? or does the circuit need some modification?Should I be seeing varying voltage when resistance across the test terminals are changed? (I'm experimenting on a spare board from a discarded unit).The opamp is LM358.Extra note: The power input to the board has 5 pins. 3 pins are tied together to ground and the other two pins are +5V, the same kind of scenario on the signal out as well two pins are tied together. What could be the purpose of tieing the pins together?Thanks in advance.

author: Yamin
author_rank: Frequent Contributor
timestamp: January 01, 2022, 09:56:49 pm
content: Sorry noticed a slight error in the schematic I have drawn. Missed the dot from the terminal connection

author: edpalmer42
author_rank: Super Contributor
timestamp: January 02, 2022, 02:42:18 am
content: PT100 == 100 ohms @ 0C, PT1000 == 1000 ohms @ 0C so no, you can't directly replace one with the other. The other problem is that PT100 probes have differences. You have to match the wiring, typically it's just the number of wires. You also have to match the alpha value of the original probe. If you can't measure the original or get a part number for it, you could be in trouble.Just a thought, are you sure that the original is a PT100 probe? That seems rather fancy, i.e. expensive, for an ice cream maker.Ed

author: Yamin
author_rank: Frequent Contributor
timestamp: January 02, 2022, 08:05:34 am
content: The parts catalogue of the manufacture says its a PT100 probe. Also I did measure the resistance of a working probe at room temperature, I forgot the exact value I shall check when I get back into work.It is a three wire probe, but the third wire is tied to the second wire effectively making it a two wire probe.I thought the PT100 probes were the more common type compared to PT1000 probes.I am very keen on learning how the op amp circuitry works. Further reading on that would be greatly appreciated.

author: nfmax
author_rank: Super Contributor
timestamp: January 02, 2022, 11:52:14 am
content: The probe should measure just under 108 ohms at 20 degrees Celsius. If it is significantly different then it is indeed faulty. What made you suspect it?

author: mawyatt
author_rank: Super Contributor
timestamp: January 02, 2022, 02:29:51 pm
content: The circuit forces ~1ma (3.3V/3.3K) thru PT100, then amplifies the voltage drop across PT100 by ~ 21 (100K/4.7K).Best,

author: Yamin
author_rank: Frequent Contributor
timestamp: January 02, 2022, 07:14:57 pm
content: Quote from: nfmax on January 02, 2022, 11:52:14 amThe unit started to malfunction freezes the mix inside the cylinder. There are two temperature probes I did measure the resistance of both and it did show the same-ish value at room temperature. But when checked with the unit on and it displayed the faulty probe to be way off. I swapped the probes and still the fault followed the probe. I tried with another control board and same the fault follows the probe. I also noticed that the faulty probe 'reacts' very slow to temperature change.Quote from: mawyatt on January 02, 2022, 02:29:51 pmAh thanks very much I'll try to wrap my head around your explanation. Any further reference reading material on a similar circuity?

author: edpalmer42
author_rank: Super Contributor
timestamp: January 02, 2022, 07:49:44 pm
content: Quote from: Yamin on January 02, 2022, 08:05:34 amNo, the third wire is usually significant. PT probes convert resistance to temperature. Lead resistance will corrupt that measurement. Three and four lead probes compensate for lead resistance to give more accurate measurements. Your schematic doesn't show that. Take another look and confirm if the third wire is used. Here's a tutorial on some of the characteristics and differences between probes.https://realpars.com/pt100/If you have the model number of the probe, you should be able to find out its alpha value. With that information, replacement should, hopefully, be possible. Make sure that the replacement probe is rated for use with food.Ed

author: Yamin
author_rank: Frequent Contributor
timestamp: January 02, 2022, 09:10:54 pm
content: Quote from: edpalmer42 on January 02, 2022, 07:49:44 pmThanks for the link, I also have read about the extra wire being used for error compensation.But in this unit the the two wires are crimped together. I have attached a photo of that (note the two white leads are tied together). The control board has 2 channels to measure two temperature probes at different locations .

author: edpalmer42
author_rank: Super Contributor
timestamp: January 02, 2022, 09:38:52 pm
content: I'm surprised that they'd use a high quality sensor in such a low quality configuration, but you're right - that's what they've done.By the way, if you can't find the alpha specification for the probe, since you've got two probes you can at least estimate the value by measuring the other sensor. Measure the resistance in an ice/water slurry and it should be 100 ohms. Then use some warm to hot water and measure the resistance and water temperature. Compare the values with known probes or just calculate the value of alpha.Ed

author: nfmax
author_rank: Super Contributor
timestamp: January 02, 2022, 10:34:56 pm
content: Note that the circuit adds a non-linearity to the resistance to voltage transfer function. This compensates for the approximately parabolic variation of resistance with temperature of the sensor. Assuming a PRT to IEC 751, the response comes out at about 1.2mV/K, which seems a bit low. Are you sure about the resistor values?

author: nfmax
author_rank: Super Contributor
timestamp: January 03, 2022, 09:38:58 am
content: The circuit can be analysed 'by hand', rather than using SPICE, to better understand its operation.The 3.3V reference (Vref) is applied to the + input of amplifier A. Assuming a perfect OPAMP, feedback through the PRT (RT) makes the voltage at its - input also equal to Vref. This appears across the 3.3k resistor (Rref), causing a current Vref/Rrefof nominally 1mA (Iref) to flow through it.By Kirchoff's Current Law, Irefmust be equal to the sum of the current through the PRT (IT) and the 'compensation' current through the 47k resistor (which I label R3). Denote the output of amplifier A as VT. The 4.7k/100k divider network multiplies this by a factork(k< 1) and applies it to the + input of amplifier B. Feedback round this amplifier holds its - input at the same voltage, namelykVT.Now the current through the PRT, ITis given by:IT= (VT- Vref)/RTand the compensation current Icompthrough R3is given by:Icomp= (kVT- Vref)/R3Equating the sum of the two currents to Irefgives:Vref/Rref= (VT- Vref)/RT+ (kVT- Vref)/R3Solving for VTgives (after multiplying numerator & denominator by RT, for clarity):VT= Vref(RT(1/R3+ 1/Rref) + 1)/(RTk/R3+ 1)Note that RTappears in both numerator and denominator, but its multiplying factor in the denominator is much larger.This voltage is attenuated by the factorkand then amplified in the output stage, a non-inverting amplifier referenced to Vref, so the final output voltage is:Vout= (kVT- Vref)(R3+ R4)/R3+ Vrefwhere R4is the 100k feedback resistor.Try a few values of RTnear 100 ohms and see how the output voltage varies:100.00 ohms (0˚C) gives 3.13574V98.04 ohms (-5˚C) gives 3.12989V101.95 ohms (+5˚C) gives 3.14156V1.164 mV/˚C for one interval and 1.170 for the other, which is acceptable, if a little insensitive. The OPAMP offset voltages may cause problems, though. What happens to the output voltages in the rest of the system?

author: Yamin
author_rank: Frequent Contributor
timestamp: January 04, 2022, 12:35:21 am
content: Quote from: nfmax on January 03, 2022, 09:38:58 amThank you so much for the detailed explanationtwo quick questions:1.What did you mean by the output voltages in the rest of the system?2.Could you please elaborate on the non-linearityQuote from: nfmax on January 02, 2022, 10:34:56 pmAgain really do appreciate the detailed explanation.Quote from: edpalmer42 on January 02, 2022, 09:38:52 pmThanks for the cool tip.

author: nfmax
author_rank: Super Contributor
timestamp: January 05, 2022, 10:31:45 pm
content: Quote from: Yamin on January 04, 2022, 12:35:21 amAnswers:I mean to ask, does the output voltage from this circuit get read by an ADC, with or without additional amplification, or is it processed by purely analogue circuitry, such as differential amplifiers, comparators, integrators or the like.A very simple theory of electrical conduction in metals predicts that the resistivity of a metal should be directly proportional to (absolute) temperature, falling to zero at 0K. The great advantage of platinum as a thermometer is that, over a very wide range of temperatures, it follows quite closely this simple behaviour, much more so than other metals. Deviations from linearity arise from several different causes:Thermal expansion of the metal causes the number of conduction electrons per unit volume to vary with temperatureThe presence of impurity atoms (typically trace amounts of other platinum group transition elements) causes additional, temperature-independent scattering of the conduction electronsCrystal defects such as vacancies, dislocations & grain boundaries, also cause temperature-independent scatteringThe effect of these is toreducethe linear temperature coefficient of resistivity below what it would be in a perfectly pure, perfect crystal of platinum, giving an approximately quadratic variation of resistance with temperature. This was originally investigated and described by Callendar around the end of the 19th Century.The quadratic is specified as a function of R(0), the resistance of the thermometer (PRT) at 0˚C:R(t) = R(0)(1 + At+ Bt2)wheretis the thermometer temperature in ˚C, and A & B are constants depending on the purity and state of anneal of the platinum wire in the thermometer. Platinum today can be produced in much higher purity than was available to Callendar, so the constants in use today are not the same as those he determined.The 'alpha' value is used to quantify the purity and state of anneal of the platinum. It is simply the average fractional temperature coefficient of resistance over the temperature range 0˚C to 100˚C:alpha = (R(100˚C) - R(0))/R(0) * 1/100Note that the alpha value is not used when calculating temperature from resistance! PRTs entered industrial use in the UK in the 1930's, at which time typical alpha values were around 0.00385˚C-1. They only became widespread in the USA during the 1940's & 50's, when alpha values around 0.00391˚C-1could be attained. Subsequent standardisation in both countries 'fixed' these values, with wire made from platinum of higher purity being doped during manufacture to reduce alpha to the required value. The IEC 751 standard calls for the lower alpha value, with A = 3.90802 x 10-3and B = -5.802 x 10-7. Different standards bodies use different A & B values for the higher alpha value, e.g. OIML has A = 3.96868 x 10-3and B = -5.8677 x 10-7.If your ice cream machine is American, it may use probes with the higher alpha value: if it is Italian (or indeed almost any other nationality) it will use the IEC standard.The non-linearity introduced in the PRT conditioning circuit is an approximation to theinverseof the Callendar quadratic, giving a linear variation of output voltage with temperature. The limits of error (especially since the circuit does not attempt to correct for lead resistance) are such that it hardly matters which alpha value is used, though the scale factor and output voltage at 0˚C will differ somewhat. Generally, individual circuit calibration must be used when accuracy is important - possibly the 3.3V supply may be adjustable.The whole approach of linearising the PRT response in analogue circuitry is now outdated. It is far preferable to accurately measure the PRT resistance and use a processor to compute temperature. The computation can then include the van Dusen C coefficient (which I ignored above) to give better accuracy below 0˚C, and allow for calibration using factors specific to the individual PRT being used. With individually calibrated laboratory platinum thermometers (notthe fragile SPRTs used in metrology work) calibration uncertainties of around ±20mK are achievable over a limited temperature range.IEC specifies class A and class B tolerances for the PRT probes themselves. A class A PRT has a tolerance of ±0.15˚C at 0˚C, increasing linearly with temperature deviation from 0˚C, to ±0.35˚C at 100˚C and -100˚C. A class B probe has corresponding tolerances of ±0.3˚C and ±0.8˚C. Other tolerance classes are available but not specified by the standard. Note these tolerancesexcludeany errors introduced in measuring the resistance of the probe and deriving temperature from it.PRT probes are generally very stable and drift free, so that the calibration error of an individual probe will not change significantly over time. Consequently, in industrial applications, class B probes are almost invariably used, as any errors can be adjusted out when commisioning the system.Edited to add a (hopefully) helpful document about industrial PRTs