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While reading the MAX6605 datasheet, which is an analog temperature sensor, I ran across the following:

The temperature-to-voltage transfer function has an approximately linear positive slope and can be described by the equation:

VOUT = 744mV + (T ✕ 11.9mV/°C)

where T is the MAX6605’s die temperature in °C. Therefore:

T (°C) = (VOUT - 744mV) / 11.9mV/°C

To account for the small amount of curvature in the transfer function, use the equation below to obtain a more accurate temperature reading:

VOUT = 0.744V + 0.0119V/°C ✕ T(°C) + 1.604 ✕ 10^-6 V/°C^2 ✕ (T(°C))^2

My algebra is terrible. How do I solve for T(°C) from the last equation?

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closed as off topic by The Photon, Brian Carlton, Nick Alexeev, W5VO Nov 15 '12 at 4:47

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Use the quadratic formula. See en.wikipedia.org/wiki/Quadratic_equation – The Photon Nov 14 '12 at 22:24
You can use WolframAlpha to solve for you or to check your answer. – Samuel Nov 14 '12 at 22:48
up vote 2 down vote accepted

The equation is a quadratic in T which can be written more simply as:

 1.604X10^-6T^2 + 0.0119T + (0.744 - VOUT) = 0

where T is in degrees C.

This can be solved using the known formula for quadratic equations but it does become a bit messy. You should probably plot this equation out using Excel or similar software to get a graph of T versus VOUT. Compare it to the linear version to see if the correction is worth the effort over the range of T that you expect in your application. A quick check with a TI-83 calculator shows that the correction is very small. For example, at a temperature of 20C, the linear equation says VOUT would be 0.9820 volts while the quadratic equation says VOUT would be 0.9826 volts. Both equations have VOUT = 0.744 volts at a temperature of 0C. At 50C, VOUT is 1.3350 and 1.3430, respectively. Thus unless you plan on measuring VOUT to better than 1 millivolt, the linear equation should be sufficient.

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The linear equation is my fall back plan. Thanks! – Hair_of_the_Dog Nov 14 '12 at 23:30

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