Jake - 1 year ago 96

C Question

I'm trying to do the Steinhart-Hart temperature calculation on an Arduino. The equation is

I solved a system of 3 equations to obtain the values of A, B and C, which are:

`A = 0.0164872`

B = -0.00158538

C = 3.3813e-6

When I plug these into WolframAlpha to solve for

`T`

`T=1/(0.0164872-0.00158538*log2(10000)+3.3813E-6*(log2(10000))^3) solve for T`

`T = 298.145 Kelvins = 77 Fahrenheit`

However when I try to use this equation on my Arduino, I get a very wrong answer, I suspect because doubles do not have enough precision. Here's what I'm using:

`double temp = (1 / (A + B*log(R_therm) + C*pow(log(R_therm),3)));`

This returns 222 Kelvin instead, which is way off.

So, how can I do a calculation like this in Arduino?? Any advice is greatly appreciated, thanks.

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Answer Source

Precision is not the main issue. Could even use `float`

and `powf()`

. A thermistor temperature calculation is not *that* accurate. After all the temperature is certainly not better than ±0.1°C accurate. Self heating of the thermistor is a larger factor.

OP's C code assumes log base 2, use log base *e* `log()`

as the constants were derived using log base 2. @Martin R

```
// double temp = (1 / (A + B*log(R_therm) + C*pow(log(R_therm),3)));
double temp = (1 / (A + B*log(R_therm)/log(2) + C*pow(log(R_therm)/log(2),3)));`
```

Sample implementation, that avoids an unnecessary slow `pow()`

call.

```
static const inv_ln2 = 1.4426950408889634073599246810019;
double ln2_R = log(R_therm)*inv_ln2;
double temp = 1.0 / (A + ln2_R*(B + C*ln2_R*ln2_R));
```