# 一、原理

/* NTC热敏电阻公式 Rt = R * exp(B*(1/T1-1/T2))
Rt：在T1下的电阻值
T1/T2: 指的是K度，即开尔文温度, T=273.15 + 摄氏度
R: 在T2下的标称阻值，比如在25摄氏度10000欧, T2=273.15 + 25
B:热敏电阻一个参数， 比如3950
exp：e的n次方

Rt = 10000 * exp(3950*(1/(273.15+t1) - 1/298.15))
t1=1/(ln(Rt/10000)/3950+1/298.15)-273.15
*/

我的外部电路如下：

J2是电池插座， 三根线， 中间的就是NTC， BAT_DET连接处理器的ADC采样引脚， 分压电阻阻值47k， 参考电压1.8v， 所以：

Rt = Vadc * Rf / (Vin-Vadc)
Rf: 分压电阻  对应电路47k
Vin： 参考电压 1.8v

# 二、示例代码

下面就演示温度从零下-10度到60度 热敏电阻阻值以及根据采样电压反推热敏电阻的温度， 需要注意的是，代码需要包含math.h函数库的支持， 同时， 数学ln()的叫法在C库是log()

/* gcc test.c -lm */#include<stdio.h>
#include <math.h>

int main()
{　　int t1;

/* 温度从零下-10度到60度 热敏电阻阻值 */
for(t1=-10; t1<60; t1++) {
Rt = 10000 * exp(3950*(1/(273.15+t1) - 1/298.15));
}

/* 根据采样电压反推热敏电阻的温度 */
t1=1/(ln(Rt/10000)/3950+1/298.15)-273.15;
}

return 0;
}

-10 = 996.1667819082mv
-9 = 970.8171351943mv
-8 = 945.5463090268mv
-7 = 920.3938859522mv
-6 = 895.3981112286mv
-5 = 870.5956912898mv
-4 = 846.0216150692mv
-3 = 821.7089994324mv
-2 = 797.6889594676mv
-1 = 773.9905039042mv
0 = 750.6404554807mv
1 = 727.6633956682mv
2 = 705.0816327966mv
3 = 682.9151923094mv
4 = 661.1818276246mv
5 = 639.8970498688mv
6 = 619.0741746119mv
7 = 598.7243836333mv
8 = 578.8567997063mv
9 = 559.4785723909mv
10 = 540.5949728601mv
11 = 522.2094958620mv
12 = 504.3239670152mv
13 = 486.9386537600mv
14 = 470.0523784230mv
15 = 453.6626319968mv
16 = 437.7656873959mv
17 = 422.3567110965mv
18 = 407.4298722257mv
19 = 392.9784483112mv
20 = 378.9949270420mv
21 = 365.4711035231mv
22 = 352.3981726282mv
23 = 339.7668161643mv
24 = 327.5672846610mv
25 = 315.7894736842mv
26 = 304.4229946520mv
27 = 293.4572401925mv
28 = 282.8814441428mv
29 = 272.6847363289mv
30 = 262.8561923090mv
31 = 253.3848782846mv
32 = 244.2598914134mv
33 = 235.4703957662mv
34 = 227.0056541849mv
35 = 218.8550563020mv
36 = 211.0081429835mv
37 = 203.4546274551mv
38 = 196.1844133667mv
39 = 189.1876100427mv
40 = 182.4545451581mv
41 = 175.9757750675mv
42 = 169.7420930061mv
43 = 163.7445353679mv
44 = 157.9743862542mv
45 = 152.4231804741mv
46 = 147.0827051655mv
47 = 141.9450001930mv
48 = 137.0023574679mv
49 = 132.2473193230mv
50 = 127.6726760645mv
51 = 123.2714628124mv
52 = 119.0369557321mv
53 = 114.9626677478mv
54 = 111.0423438230mv
55 = 107.2699558814mv
56 = 103.6396974382mv
57 = 100.1459779998mv
58 = 96.7834172883mv
59 = 93.5468393367mv
100.0000000000mv = 57
200.0000000000mv = 37
300.0000000000mv = 26
400.0000000000mv = 18
500.0000000000mv = 12
600.0000000000mv = 6
700.0000000000mv = 2
800.0000000000mv = -2
900.0000000000mv = -6

# 三、其他

a. 数学中对数用log表示，ln表示以e为底数， C库使用log()却表示数学的ln， 如果要表示数学的logab, 由于等效数学的lnb/lna， 即等效C代码log(b)/log(a)

c. 由于我是要在驱动实现这个功能， 内核没有包含这个math.h和libgcc.a库， 所以参考网上实现了个差不多精度函数：

double ln(double a)
{
int N = 15;
int k,nk;
double x,xx,y;

x = (a-1)/(a+1);
xx = x*x;
nk = 2*N+1;
y = 1.0/nk;
for(k=N;k>0;k--) {
nk = nk - 2;
y = 1.0/nk+xx*y;
}

return 2.0*x*y;
}/* https://blog.csdn.net/mike190267481/article/details/7404702 */

    printf("%.10f, %.10f\n", ln(0.1), log(0.1));
printf("%.10f, %.10f\n", ln(1), log(1));
printf("%.10f, %.10f\n", ln(5), log(5));
printf("%.10f, %.10f\n", ln(10), log(10));
printf("%.10f, %.10f\n", ln(15), log(15));
printf("%.10f, %.10f\n", ln(20), log(20));

-2.3023645999, -2.3025850930
0.0000000000, 0.0000000000
1.6094377510, 1.6094379124
2.3023645999, 2.3025850930
2.7053425934, 2.7080502011
2.9856609824, 2.9957322736

Linux内核使用浮点运算问题

posted @ 2019-03-28 16:23  Vedic  阅读(1584)  评论(0编辑  收藏