1 /*公式法计算NTC温度值*/ 2 static float FormulaNTCTemperature(float resistance) 3 { 4 float temp; 5 float result=0.0; 6 7 result=resistance/NTC_NOMINAL_RESISTANCE; 8 result=(log(result)/NTC_NOMINAL_CONSTANT)+(1/(NTC_NOMINAL_TEMPERATURE+KELVIN_CONSTANT)); 9 temp=1/result-KELVIN_CONSTANT...
1 /*公式法计算NTC温度值*/ 2 static float FormulaNTCTemperature(float resistance) 3 { 4 float temp; 5 float result=0.0; 6 7 result=resistance/NTC_NOMINAL_RESISTANCE; 8 result=(log(result)/NTC_NOMINAL_CONSTANT)+(1/(NTC_NOMINAL_TEMPERATURE+KELVIN_CONSTANT)); 9 temp=1/result-KELVIN_CONSTANT...
在本篇文章中，我们仅关注一种可以测量温度的传感器。该传感器称为热敏电阻。热敏电阻比其他类型的电阻对...
根据以上公式我们可以实现: 1/*公式法计算NTC温度值*/2staticfloatFormulaNTCTemperature(floatresistance)3{4floattemp;5floatresult=0.0;67result=resistance/NTC_NOMINAL_RESISTANCE;8result=(log(result)/NTC_NOMINAL_CONSTANT)+(1/(NTC_NOMINAL_TEMPERATURE+KELVIN_CONSTANT));9temp=1/result-KELVIN_CONSTANT;1011re...
2 static float FormulaNTCTemperature(float resistance) 3 { 4 float temp; 5 float result=0.0; 6 7 result=resistance/NTC_NOMINAL_RESISTANCE; 8 result=(log(result)/NTC_NOMINAL_CONSTANT)+(1/(NTC_NOMINAL_TEMPERATURE+KELVIN_CONSTANT)); 9 temp=1/result-KELVIN_CONSTANT; ...
temperature = log(((10240000/adc_raw)-10000)); temperature=1 / (0.001129148 + (0.000234125 * temperature) + (0.0000000876741 * temperature * temperature * temperature)); temperature = temperature - 273; This assumes that I have a setup like this: GND|---|10K|---|uC|---|NTC*|---|+...
The formula represents that the relative tolerance on α is equal to the relative tolerance on B-value. 4. Thermal Time constant This is the time period in which the thermistor’s temperature will rapidly change 63.2% of its temperature (T0) difference from ambient temperature (T1). ...
另外对于NTC采样端口的ADC采样精度依然是在手册和资料中没有给出 例如LD PACK+ CELL(X)都已明确给出, NTC 精度我个理解是 +-1度是一个很大的范围ADC采样精度已经远远高于它 内部上啦18K电阻 与外部10K NTC阻止可通过校准手段来达到+-1度 这样理解对吗?
另外对于NTC采样端口的ADC采样精度依然是在手册和资料中没有给出 例如LD PACK+ CELL(X)都已明确给出, NTC 精度我个理解是 +-1度是一个很大的范围ADC采样精度已经远远高于它 内部上啦18K电阻 与外部10K NTC阻止可通过校准手段来达到+-1度 这样理解对吗?
v –1 When electric power P (mW) is spent in ambient e 10 c n B=3450 a temperature T1 and thermistor temperature rises T2 , t B=3900 s i s B=4100 the formula is as follows e R P=C (T2-T1) (3) –2 10 C: Thermal dissipation constant (mW/°C) –20 0 20 40 60 80 ...