Building instrumentation amplifier for PT100 sensor - Page 1
Forum section: electronics / beginners

author: bborisov567
author_rank: Regular Contributor
timestamp: April 15, 2024, 08:44:51 am
content: I am trying to build an instrumentation amplifier for pt100 sensor. My goal is to have 0.01V per deg C at the output. I am using the classical instrumentation amplifier topology and a bridge at the input and LM4040 generating stable 5V. Currently testing with LM324 and R8 is a trim pot, R3 is representing the PT100 sensor, R5 - representing the voltmeter input. The circuit is somewhat working but after sometime the output starts fluctuating probably due to oscillation. Also there is around 5 mV offset on the output with Vdiff+ and Vdiff- shorted. I believe that i can correct that by fine tuning R11. Also i am not sure if the resistor divider is a good idea for generating a dual supply but currently i am looking for the simplest solution. What recommendation could you make on the circuit, mainly regarding stability. I know i can use a dedicated INA ic but the point here is more to learn the design principle, rather than achieving maximum performance.

author: Kleinstein
author_rank: Super Contributor
timestamp: April 15, 2024, 09:34:54 am
content: The 2nd part of the INA should more like use 4 equal resistors. Currently the 2nd part has a gain of 1/10, which is a bit odd.The way the reference is drawn has the reference in sereies to the supply. So the noise of the supply and ref. would add up.A trim pot in the test setup can add instability.The virtual ground is normally using larger resistors and some amplifier to buffer.These days the output signal often goes to a ADC of some kind. To get a ratiometric reading the ADC and bridge drive should use the same voltage reference. With an extrnal voltmeter one would need a relatively low noise reference too. The LM4040 is more like low cost, low power high noise.

author: bborisov567
author_rank: Regular Contributor
timestamp: April 15, 2024, 10:13:37 am
content: For some reason i have switched the places of the 10k and 1k resistor. Thank you for noticing that. The idea was to distribute the gain between the two stages and the second stage to have a gain of 10. Is it required for the second stage to have a unity gain or should I replace the 1k resistor with 100k for gain of 10 like my initial idea. Also could you share some resources or examples for creating a virtual ground?

author: Kleinstein
author_rank: Super Contributor
timestamp: April 15, 2024, 11:14:48 am
content: The 2nd. stage does not have to be gain of 1. Unless one wants high speed, one usually wants relatively low gain for the 2nd stage, as the first part works better with high (e.g. > 10) gain. This starts with the noise of the gain setting resistor. Also gain of the first stage makes resistor mismatch errors in the 2nd stage less relevant.The virtual ground is a pretty standard circuit. No need to be very accurate there.

author: Terry Bites
author_rank: Super Contributor
timestamp: April 16, 2024, 05:46:05 pm
content: A diy INA will kill the PT100's accuracy. Gain and offset woes, CMRR issues and all that drift if you dont use super low tempco precsion resistors.No trimmers allowed!An off the shelf INA will be cheaper than resistors of a suitable quality. eg an INA188I for $5 vs more than handfull of a $1.5 at least 0.1% resisitors.There is never a good resaon to build an INA when you can get much better one ready rolled.Do your errror budget before proceeding.Tip- it can be more effective to drive a very precise (small) current through the PT100 and measure the volts across it.

author: Zero999
author_rank: Super Contributor
timestamp: April 16, 2024, 06:41:43 pm
content: I'm confused the description says the LM324, yet the schematic shows the OP1177.The LM4040 is not connected correctly. As drawn it will simply subtract 5V from the supply voltage, giving 7V. Use it in shunt configuration, as per the data sheet for 5V.As mentioned above, a real, off the shelf instrumentation amplifier IC should be used, rather than a discrete one made with an op-amp. It's difficult and expensive to obtain resistor values to give the required accuracy.How long are the wires to the PT100? Consider using a four wire configuration to eliminate the cable resistance from the equation.A current source would be better than a resistor.

author: MarkT
author_rank: Frequent Contributor
timestamp: April 16, 2024, 06:50:45 pm
content: You need a stiff virtual ground, 50 ohms is not very stiff, it could motor-boat? Normally you use one opamp section to generate virtual ground from a resistor divider more like 10k:10k. Then as the load varies you have the whole of the opamp open loop gain there to keep the ground in place and not providing an unexpected feedback path.

author: David Hess
author_rank: Super Contributor
timestamp: April 17, 2024, 02:05:01 pm
content: Calibration of an RTD is not trivial without a set of reference resistors to substitute for the RTD and represent different temperatures. RTDs are also non-linear adding more than 1C error from -50C to 150C, so if you require more accuracy than this, non-linearity correction should be included.I suggest starting with a simpler circuit that only requires two point calibration like the one below.

author: bborisov567
author_rank: Regular Contributor
timestamp: April 18, 2024, 09:45:35 am
content: To be honest my first schematic was a waste of time. If you are going to do something - do it right! The schematic from David Hess looks really good so i will try building it. LT1006 seems to have better performance than the INA188 and the price is similar. As for the application note schematic - some modifications will be needed. First the excitation current - could 1 mA cause self-heating on a Pt100 sensor? I would like to use the maximum possible current so that noise has less affect on the input. Also i would like to be able to measure negative temperatures. Is that going to be possible with the positive feedback resistor?

author: David Hess
author_rank: Super Contributor
timestamp: April 18, 2024, 02:20:49 pm
content: Quote from: bborisov567 on April 18, 2024, 09:45:35 amThe LT1006 is a single version of the LT1013/LT1014 which are precision versions of the LM324/LM358. Basically what is called for here is a precision single supply operational amplifier, but see below.QuoteRaising the excitation by 10 times for a sensor which is 1/10th the resistance will result in the same self heating in the sensor. If the lower resistance sensor is smaller however, then it will have a higher thermal resistance so greater temperature rise.Any precision current source can be used to replace the LM334 circuit. The 1N457 is used for temperature compensation of the LM334, and is a high conductance low leakage diode. I would use a 2N3904 or similar base-emitter junction in place of the 1N457. Do not use a switching diode like a 1N914 or 1N4148.The bridge resistance going into the operational amplifier is about 500 ohms, so a low noise operational amplifier can be used. This could be a good application for a low noise chopper stabilized operational amplifier.What I would do is work backwards from the noise of the operational amplifier and resistors to figure out what the minimum excitation should be. Roughly, LT1006 noise here is about 0.55uVpp at its input, so with a gain of 92, output noise is about 50 uVpp. Resolution is 10 mV/C, so peak-to-peak noise is about 0.005 degrees C, or just under your 0.01C resolution requirement. So using a 100 ohm RTD will indeed require increasing the current excitation by 10 times to maintain your noise free resolution.A lower noise operational amplifier could replace the LT1006 and considerably relax the excitation requirements. The LT1007 is 0.06uVpp so could use the 100 ohm RTD without raising the excitation current and produce the same output noise. This would have the advantage of lower self heating. There are some even lower noise parts which are suitable.QuoteThis particular design is intended for single supply operation, however this only allows the output to indicate down to 2C as shown. If a dual supply is used, then it will go below 0C and a normal operational amplifier can be used, like a cheap OP-07.I selected this design over many others because it only requires two point calibration. Most linearized designs require three point calibration which requires a good reference thermometer or a triple point cell.

author: bborisov567
author_rank: Regular Contributor
timestamp: April 18, 2024, 04:42:51 pm
content: Since i am going to make a pcb and end up having 5 i might build two devices - one with 1006 and 1mA and one with 1007 and 100uA to see if selfheating with theese currents is an issue. I would to like to start by calculating the total error but the 1 megaohm resistor in the positive feedback is bugging me a bit. How should i include it in the transfer function? My target measuring range would be -40 C to +125 C. For generating dual supply i intend to use the attached circuit, probably with smaller transistors and bigger caps. I have access to a calibrated thermal chamber with the range mention above to test the performace whrn i finally build a device. As far as i understand the two point calibration basically equalizes the bridge (at 5 C) and then sets the gain (at 200 C), right? Also colud you share a design with 3-point calibration and what are its advantages?To sum up the design reqirements:-40 C to 125 C measuring range0.01V per degree C output0.1 C accuracy4 wire Kelvin connection12 V single supply

author: Kleinstein
author_rank: Super Contributor
timestamp: April 18, 2024, 05:09:02 pm
content: For PT100 at current of 1 mA can be ok if the sensor is not a very small one. One still has some self heating, but it can be acceptable. 100 µA and this onle 10 mV at the sensor is rather low.For the current source is very much depends on how the voltage is read. An ADC in the circuit would use the same reference voltage and thus work ratiometric.With an external voltmeter there would ideally be a way to also measure current, or the voltage over a reference resistor.A correction for the linearity could often be done on the digital side instead of some analog, approximate correction and more cal points.The LT1006 is made for relatively high impedance signal sources and is not ideal for a PT100.For the noise, the speed of the readings and the time window can also be relevant. The relevant bandwidth could be different from the 0.1-10 Hz found in datasheets. Often one would be slower and this would favor AZ amplifiers. Here many types are 5 V max and one may need a voltage regulator for the amplifier.

author: bborisov567
author_rank: Regular Contributor
timestamp: April 18, 2024, 05:20:52 pm
content: I intend to use it with external voltemeter, otherwise i would use something like ADS1220.

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: April 18, 2024, 05:29:51 pm
content: You can get pretty good performance from a PT100 with no amplifier - use an ADC with a differential reference input, and connect that across a precision resistor in series with the PT100, and a low-noise current source.

author: David Hess
author_rank: Super Contributor
timestamp: April 18, 2024, 06:49:04 pm
content: Quote from: bborisov567 on April 18, 2024, 04:42:51 pmThe positive feedback is a common way to add a small square function to correct for curvature in the response. It works by adjusting the excitation to the resistive sensor, so as the output increases, the excitation slightly increases also. For this kind of design it is the difference between 1.5C error and 0.1C or better error.QuoteThat might work but the class-b output stage can be a problem. If the current requirements are only for this circuit, then just drive the ground with the output from the 741 directly, and add like a 10 solid tantalum or 100 microfarad aluminum electrolytic directly across the output of the operational amplifier. This should not oscillate because the large value of capacitance will "swamp" the output of the operational amplifier. The ESR of a solid tantalum or aluminum electrolytic capacitor helps this work.Another way to make a virtual ground at low currents is with a couple of diodes in series or a low value zener diode. The virtual ground does not need to conduct much current in this circuit, and if it varies a little bit that does not matter. A dual supply operational amplifier will be happy with just a couple volts of negative supply.QuoteYou can really use any two points. For this circuit if modified to go below zero, I would use an ice bath and a boiling water bath, with correction for atmospheric pressure which could create as much as 0.5C error or more.QuoteA three point calibration adjusts for the zero, gain, and for the non-linearity, but requires 3 reference temperatures. This can get the error to down below 0.05C, or even to 0.005C. Linear Technology published different circuits for this including some 4-wire ones. I will post an ultra-linear 4-wire example below, but it requires split supplies. Also shown is an example where precision resistors are used for calibration.I was thinking of using a pair of precision thermisters to do a three point calibration of an RTD circuit, but have not because I want to get some precision resistors to represent the three points first. Precision thermistors can get within 0.05C or maybe could be calibrated to be even better over a limited range.

author: bborisov567
author_rank: Regular Contributor
timestamp: April 19, 2024, 10:49:01 am
content: @mikeselectricstuff - I started doing the calculation - scaled the LM334 resistors for 1 mA and got to the bridge. Then i found out that with 1 mA excitation current and 100 ohm resistors i already get 0.1V/C from the bridge. Now i start asking myself - is there a point for an op amp since I will be feeding the output of the bridge to a multimeter which is going to have 1 Megaohm input impedance at worst. Except for linearization is there another reason to use op amp?