I am interfacing a thermistor with an ADC, and I have found a circuit that, at first glance, appears to suit my needs. A zero and span circuit using a single op amp.

http://www.ti.com/lit/an/slyt173/slyt173.pdf (figure 3, page 3)

The thermistor is in a voltage divider, using a 82k resistor as the fixed resistance. The thermistor's resistance is 3k @ 212F, and 160k at 32F. With 5V across the full divider, this gives 4.8V across the fixed resistor @ 212F, and 1.7V at 32F. I'd like to map 4.8V to 4V, and 1.7V to 0V.

To calculate my required offset and gain, I set up two simultaneous y=mx-b equations:

Vout = m * Vin - b

4.0 = m * 4.8 - b

0 = m * 1.7 - b

Solving for m is as easy as (y2 - y1)/(x2 - x1). plugging m = 1.2903 back into either equation gives you b = 2.1935.

According to the PDF, the equations for the resistors, as they relate to m and b are as follows:

m = (RF + RG)/RG

|b| = (RF/RG) * (R2/(R1+R2))

If m is roughly 4/3, it follows that RF is roughly RG/3, and RF/RG = 1/3.

A little more algebra, and plugging in b = 2.2 makes R2/(R1 + R2) = 2.2 * 3 = 6.6.

There is no combination of R1 and R2 that can make the quotient more than 1.

Does this mean that this circuit can not perform this particular zeroing and spanning?

If so, what are my options to accomplish this?



1 Answer 1


I suggest that you flip over the voltage divider so that you get a range of 0.2V to 3.3V. It's generally better to have one side of the sensor grounded anyhow.

Then you can work out the resistors to get 0~4V for 0.2V to 3.3V.

Another suggestion might be to just connect the op-amp as a unity-gain buffer, and reduce the 82K to 40K so you get 4V at 32°F and 0.348V at 212°F. That gives you an offset zero so you don't suddenly run out of range close to zero C (depending on op-amp offset) and you still get 91% of the resolution.

Speaking of resolution, note that the resolution at high temperature will be much less than the resolution near 32°F, so you should make sure it's good enough for your purposes. It's important to take the op-amp offset into account too, since it greatly affects the high temperature measurement (and the low temperature measurement hardly at all). If one particular temperature is of the most interest, a strategy is to use that value of resistor as the other half of the divider (and use ratiometric conversion) because that's where the resolution will be maximum.

Ratiometric is usually the way to go with resistance measurements because the reference voltage has no effect on the measurement, all that matters is the reference resistor value.

If you do a sensitivity analysis on the method you originally suggested, I think you'll find it adds a lot of error unnecessarily.

To answer your specific title question, yes you can't do this without a voltage higher than 5V to offset the output that much. That's a consequence of the low gain (about 1.3) and the large offset.

  • \$\begingroup\$ Thanks for the excellent answer. Does ratiometric conversion suffer from the same uneven resolution? I don't quite grasp how it works, but I will certainly investigate. This is a temperature probe off of a meat thermometer. I poked around the thermometer's circuits, and see that they are using yet another method to measure. They are measuring the time it takes a known resistance and capacitance to discharge, and comparing that to the thermistors resistance + known capacitance discharge time. I suppose that is ratiometric as well. \$\endgroup\$
    – ToddMuir
    Commented Dec 11, 2014 at 5:28
  • \$\begingroup\$ Yes, the uneven resolution is a general problem with thermistors over a wide range because they are very nonlinear. Measuring oscillator frequency with a thermistor vs. a reference resistor works very well, with the exception that long leads influence the reading because of capacitance. It's the method used in most cheap consumer goods. \$\endgroup\$ Commented Dec 11, 2014 at 5:33

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