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I've been lurking through these forums for years but just now had the urge to ask for help.

I'm creating a thermometer (using an NTC thermistor) for my college project. I began to analyze my sensor in terms of its linearity, temperature range, etc (specifications in general) ... For this project, I decided to use a temperature range from 55ºC to 150ºC (celsius).

Afterwards, I implemented my sensor in a wheatstone bridge. In order to do so, I defined my lowest temperature (55ºC) resistance value (2989 Ohms) and my highest temperature (150ºC) resistance value (182.6 Ohms). I proceeded to find the value of the other 3 resistances so I could end up with a [0 ---- 5] voltage at the end of my bridge (beeing 0V to 182.6 Ohms and 5V to 2989).

My biggest problem now is implementing an I.A.:

My current wheatstone bridge voltage ranges from -13mV to 5.05V which is fine (tho I'm currently designing a gain = 2x I.A. and I'll then throw it back to my voltage range by using a summing amplifier in order to reduce circuit's noise). Thing is, I'm not able to have 2x my bridge's voltage at the end of my I.A. (even tough I'm following the I.A. equation in order to find it's gain).

As you can see, I simply defined the same value for every resistance besides the middle one (this should mean that (1+(2*Any-other-resistance / Rmiddle)) * (V+ - V-) = 2(V+ - V-), which should define a voltage range of [-0,0026 ---- 10,1]V).

Note: I'm using TL084's to build my I.A. (tho I'm testing some other components like TL082 or LM324)

There's a picture attached as I know I messed up explaining this. Any help would be appreciated. I'm really stuck and I have no idea what's the problem (also thought it might be balancing the bridge or having high voltage on my op amps but atm I'm just throwing possibilities into the air without any thinking).

The circuit

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  • \$\begingroup\$ I suggest that you use a 1N4148 diode as a temperature sensor. When forward biased at 1mA constant current the forward voltage drop of the diode will vary nicely linear with temperature. \$\endgroup\$ – Michael Karas Apr 5 at 2:03
  • \$\begingroup\$ Welcome to SE EE! Do you really need to build the instrumentation amplifier by yourself? It can be quite challenging. There are tons of it available in the online store such as AD620 for 0.8 USD here. It will give you better accuracy, precision, time productivity, and also less hassle. \$\endgroup\$ – Unknown123 Apr 5 at 2:10
  • \$\begingroup\$ Answering to Michael, I don't "need" to use an ntc but I already did so much work with it (it may seem useless or little to no work but I am actually studying this for the first time, so small things matter a lot atm :p). Thanks nontheless! \$\endgroup\$ – Jonaas18 Apr 5 at 2:17
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    \$\begingroup\$ Unknown123, thanks a lot! Yes I have to do it all by myself as Its required that I linearize the sensor by myself. I can't use sensors which are already treated voltage wise, and the same applies to the I. A.. Seems Hard but fun imo. \$\endgroup\$ – Jonaas18 Apr 5 at 2:19
  • \$\begingroup\$ Well good luck then! Also, please use mention feature to notify the person, else they won't notice if you're replying or not. \$\endgroup\$ – Unknown123 Apr 5 at 2:32
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In your above example, you can clearly see that U2A is saturated. You should stay well away from the power supply rails (several volts is usually enough).

With instrumentation amplifiers it is not enough that the inputs are within common mode range and the output is not saturated, you also have to ensure that the internal nodes do not saturate. The cause of the saturation is your large common mode voltage (almost 12V) relative to supply voltage and the presence of gain in the input stage. You are not the first to be bitten by this.

Maybe the easiest fix is to remove R8 and increase R4 & R6 to get your desired output voltage.

By the way, your feedback resistors are a bit on the low side, you can increase them by 10:1 and avoid loading the outputs so much with little effect on accuracy.

And on a more advanced topic, which you may have already considered, but I'll put it out here anyway, you have a LOT of current going through your NTC thermistor. The self-heating may be more significant than you would like. You can alter the bridge values or reduce the excitation voltage and add more gain back later. In the situation shown the power dissipation is about 10.5mW, which could represent a self-heating error of several degrees C or more (assuming a small leaded thermistor in air). Sensor makers are by times prone to quoting figures under favorable circumstances such as in briskly moving liquid.

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  • \$\begingroup\$ Thanks for the reply Spehro. A question: how can I increase my R4/R8 relation if I remove R8? Also, what do you mean by having feedback resistors on the low side? Does that mean that I should use resistances of higher value? Wouldn't that change the gain equation? \$\endgroup\$ – Jonaas18 Apr 5 at 2:20
  • \$\begingroup\$ It should read increase R4 and R6, I mis-read the small print, sorry about that. R4=R6 for a differential amplifier with gain. \$\endgroup\$ – Spehro Pefhany Apr 5 at 2:30
  • \$\begingroup\$ To answer your second question, if you take the circuit you have, and increase all the 1K to 10K and the 2K to 20K it will not change the gain equation (and it will saturate still) because the gain is dependent on resistor ratios, not absolute values. That is a degree of freedom. On the low side you are limited by how much you want to load the op-amp output (and maybe by power consumption) on the high side, by errors due to offset currents, and maybe by noise considerations. \$\endgroup\$ – Spehro Pefhany Apr 5 at 2:32
  • \$\begingroup\$ Thanks. I will try it tomorrow, aswell as using the tl084. As for the voltage / current on my ntc, self heating isn't much of an issue as we won't use it for a continuous a ammount of time. \$\endgroup\$ – Jonaas18 Apr 5 at 2:34
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    \$\begingroup\$ Got it. Well mate, I will compile everything you said on a checklist and work from there. I will update if anything goes wrong but nontheless, thanks a lot for your help. You are insane, to say the least! \$\endgroup\$ – Jonaas18 Apr 5 at 2:48

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