热敏电阻（103AT-11）样品项目

这一次，实际上，不久前，我偶然发现了一个热敏电阻的样本项目。
虽然电路似乎很容易，但计算并不像我预期的那么容易。
根据热敏电阻的网页，我们可以计算出热敏电阻的电阻如下方程。
RT：在温度T（K）的阻力
R0：T0电阻
B：热敏电阻相关的常数。对于103AT-11，它是3435±1%。
转换从TK（Kelvin）Tc（摄氏度）是TC = TK - 273.15
除双方通过R0，我们得到
替换E到日志
用B分隔两边
从数据表该热敏电阻的电阻值10.0k在25摄氏度，
所以如果我们更换R0 R25和T0 T25（克列温为25C）方程是现在
移动1 / T25从右至左
左侧缩小
然后根据改革方程
最后我们得到
在下面的图中
RT＝R1 x VT/（VDDA -VT）
如果我们正常的ADC值（ADin）输入电压0V ~ VDDA V至0 ~ 1
RT＝R1 x AdIn／（1 -埃丁）
现在我们可以编写一个代码
RT＝R1*ADIN／（1 -埃丁）；
K = 1 /（log（RT / R25）/ B＋1／T25）；
t＝K－273.15；
长话短说，这就是全部！
（AKA，附上的项目）
摩托罗拉
tempsensor180422.cyprj.archive01.zip
542.6 K



     以下为原文
  This time, actually a while ago, I came across a chance to make a sample project of thermistor.

Although the circuit seemed to be easy, the calculation was not as easy as I was expecting.
According to the web page of the thermistor, we can calculate the resistance of  a thermistor with following equation.


Rt : Resistance at temperature t (kelvin)
R0: Resistance at T0
B: Thermistor related constant. For 103AT-11 it is 3435 +/- 1%
Conversion from Tk (Kelvin) to Tc (Celsius) is Tc = Tk - 273.15

Dividing both sides by R0, we get


Replacing e to log

Dividing both sides by B

From the Datasheet the resistance value of this thermistor is 10.0K at 25 degree Celsius,
so if we replace R0 to R25 and T0 to T25 (Klevin for 25c) the equation  is now

Moving 1/T25 from the right side to the left side

Reducing the left side

Then reforming the equation according to T

Finally we get

In the figure below

Rt = R1 x Vt / (VDDA -Vt)


If we normalize the ADC value (ADin)  of input voltage 0V ~ VDDA V to 0.0 ~ 1.0

Rt = R1 x ADin / (1 - ADin)

Now we can write a code

Rt = R1 * ADin / (1 - ADin) ;
K = 1.0 / (log(Rt / R25) / B + 1.0/T25) ;
T = K - 273.15 ;

To make the long story short, here is the whole enchilada!
(aka, project attached...)

moto

•   TempSensor180422.cyprj.Archive01.zip 542.6 K