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I recently bought a board that consists of this circuitry. It came without an instruction whatsoever, so I have to figure out how to measure the temperature based on the output voltage at the To.

enter image description here

The circuit consists of a Wheatstone Bridge of R13-R15 and an NTC thermistor TH1. the voltage between point A and B Vd then go through a differential amplifier U3A and then further amplified by U3B. I don't have a volt meter or oscilloscope with me, what I did is read the To using an Arduino ADC and convert it to Voltage output. In order to calculate the temperature, I need to know the thermistor value based on To output voltage, I sort of know the basic theory on how this circuit works, but not quite able to figure out the relationship between To and the resistance of the thermistor yet.

Here are my questions:

1) What is the gain of this overall circuit? If I read it correct, it has a gain of 12, so Vd = To/12, is this correct?

2) Once I derive the Vd, I assumed that:

Vd = Vref * R14/(R14+R15) - Vref * R15/(R15+TH1)

and further derive from this formula to get the TH1 value. Again, what is the correct formula for getting the resistance value of TH1?

But so far what I get seems to be way off from the correct temperature.

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  • \$\begingroup\$ Gain is more like 20. First stage is 2, second stage is 10. So, total gain of 20. \$\endgroup\$ – JRE Nov 17 at 11:02
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    \$\begingroup\$ @JRE 0030 brain (sleep calls ) says: Gain stage 1 is 1.333 ish & stage 2 = 11 - with the 1.333 only correct when thermistor is 10k. They MAY have thought of R from bridge as 10k / leg so gain =1 but missed (10k//10k) + 10k = 15k Rin. Yes? || Also Vdifferential is referred to Vcc/2 but U3B is ground referenced so I'd expect it to rail due to multip;ing the Vcc/2 component. Yes? (or am I asleep already). \$\endgroup\$ – Russell McMahon Nov 17 at 11:28
  • \$\begingroup\$ U3A is missing a resistor to complete a diff-amp \$\endgroup\$ – JonRB Nov 17 at 11:44
  • \$\begingroup\$ I would've expected a relatively large capacitor in parallel R9, otherwise that's not a very stable reference voltage... \$\endgroup\$ – Marcus Müller Nov 17 at 12:22
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This circuit can be split into four sections

  1. Reference generation
  2. Wheatstone bridge
  3. Differential amplifier
  4. 1st order filter
  5. non-inverting amplifier

Reference Generation

A simple voltage divider from Vcc and buffered to provide a low-impedance source to the bridge:

U2B V+ pin sits at 0.5Vcc and thus Vref ~ 2.5V (accuracy of the rail and resistors dependant)

Wheatstone Bridge

The resistors around the Thermistor (R13,14,15) are 10k so it is safe to assume that the Thermistor is a 10k at some reference temperature.

The thevenin equivalent at this operating point for nodes A & B is therefore 5k with the voltages at these nodes being 1.25V

schematic

simulate this circuit – Schematic created using CircuitLab

Differential Amplifier

Assuming you forgot the 20k from V+ to GND, the differential amplifier stage has a gain of \$ \frac{ 20k}{10k + 5k} \$ = 1.33

schematic

simulate this circuit

1st order filter

This will provide some attenuation at non-DC temperature changes, but assuming the rate of change of thermistor resistors in << the 3dB point of the 1st order filter, this will not change the gain of the circuit (except for a leakage component)

non-inverting amplifier

The final stage is a classic non-inverting amplifier with a gain of 11 ( \$\frac{100k}{10k} + 1\$ )

Overall gain is therefore 14.63 when the temperature is such that the thermistor presents 10k resistance.

As the temperature changes then the bridge impedance will change. This will produce a non-constant impedance and thus the gain of the differential amplifier will also vary with temperature. If this is acceptable this is a cheap implementation. Creating an instrumentation amplifier would produce a more constant impedance and thus gain response. You have an unused OPAMP so this is a possibility

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  • \$\begingroup\$ Thanks for the detail explanation. This gives me a much better understanding of the overall circuit. On Differential Amp, I assumed what you means is 20k/(10k+5k) instead of 0k/(10k+5k). On Non-inverting amplifier, if your explanation is correct, I think the gain is 2(100k/10k)+1, so it is 11, correct? The other half of the OpAmp U2 is used by another part of the circuit that I didn't shown as it is not related to temperature measurement. \$\endgroup\$ – hcheung Nov 17 at 12:27
  • \$\begingroup\$ Thermistor is an NTC thermistor of 10k @25 degree Celsius. \$\endgroup\$ – hcheung Nov 17 at 12:37
  • \$\begingroup\$ Corrected the typo. Definitely not 0k. Also based upon that and missing the final gain being 100k/10k puts the overall gain closer to 13.433 \$\endgroup\$ – JonRB Nov 17 at 12:39
  • \$\begingroup\$ When the bridge is balanced the output (To) should be 0v, this is when TH1=10k. (assuming the extra 20k is included). \$\endgroup\$ – Nedd Nov 17 at 12:47
  • \$\begingroup\$ Last stage gain should be 11, (not 10.1). \$\endgroup\$ – Nedd Nov 17 at 13:09

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