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Using the INA330 for temperature measurement rather than PID control

Other Parts Discussed in Thread: INA330


We have to place the gain amplifiers for several sensors in a very tight space. Looking at the INA330, the part count is very low. Now, it is descrbed as an amplifier for temperature control, but looking at the datasheet, it looks like if we adjust the gain resistor to a lower value and lower the control point voltage we could obtain a measurement over a wide temp range (we need -40C to +50C).

Setting it up, though, we are not getting the intended result. In fact, we tried the test circuit and we're still not seeing it work. The swing from 0 (ice tray) to 20 C is only about 0.2V. It doesn't seem to matter what size resistor we use for Rg and the voltage, it is not changing the gain.

We put a precision resistor (10K) for Rset and a 10K thermistor on pin 10, and to ground. Vexcite on pins 2 and 3 is 1 V. We tried external supplies and a voltage divider from Vcc but no difference. One thing unexpected - just putting the part down on the PC board with nothing else but the precision 10K populated, the resistance was measured to be about 4.1K. The thermistor comes up at about 8.4K at room temp.



  • Hello Tom,

    First let me run through some calculations:

    I2=Vexcite/Rset=1V/10k=100uA. This current does not exceed the maximum output current fo the excitation buffers (125uA), so there is no problem here.

    At 0C (32F) a thermistor should be ~32.648k. Therefore I1=Vexcite/Rtherm=1V/32.648k=30.63uA.

    The output equation is VO=(I1-I2)*RG+Vadj=(-69.37u)*200k+2.5=-11.37V.

    So, here's an issue. The output range of the INA330 depends on supply voltage and load. Assuming a single 5V supply, 10kohm load, and using the minimum data sheet value the output swing is 75mV<VO<4.925V.

    If we back-calculate I come up with a maximum Rtherm value of 11.38kohms, which corresponds to ~72F (22.22C).

    Similarly, using the maximum output voltage I calculate the minimum value of Rtherm as 8.92k, which is approximately 81F (27.22C).

    So, to change the range you will need to make a few adjustments. For example, to center the output around the midpoint of your range (41F, or 5C), change Rset to 25kohms and subsequently Vexcite to 2.5V. You then have to calculate a value for RG. Assuming the thermistor impedance is 3.6kohms at 122F (50C), the maximum output swing of the device is 4.925V, and Vadj of 2.5V I calculate an RG value of 4.08k. Given those values we can calculate the output voltage when the thermistor is -40F (~337kohms). VO=((2.5/337K)-100uA)*4.08K+2.5=2.12V. You will notice that this is not linear because the thermistor itself is not linear.

    Hope this helps!