Arduino 4

Arduino 4-20ma精度(Arduino 4-20ma Accuracy)

我使用2线电流回路将压力变送器Dwyer MS-121（范围从-100 Pa到+ 100Pa）连接到Arduino。 对于分流电阻，我使用220欧姆。 所以要从电压到压力进行计算，这就是我所做的：阅读是我从分流电阻读取的

(((((float)reading * 0.0049) - 0.88) / 3.52) * 200) - 100)

0.0049用于知道实际电压，然后从4-20mA（4mA * 220Ohm）的最低范围中获得0.88，从4-20ma（20mA和220Ohm）的最高范围中获得3.52，并且减去0.88以得到0到3.52（所以我可以通过百分比来计算）。 那么200因为我的压力变送器的范围（-100到+ 100Pa）。 最后一个是减去100，所以我可以得到一个很好的-100Pa到+ 100Pa的范围。

现在，我的压力变送器也有显示。 我从显示屏得到的值是-19.4Pa，我从Arduino得到的值是-21.1Pa。 它相差1,7Pa或0.85％。 我很好奇为什么arduino的价值不准确，是我的计算还是我的电路？

I am connecting a pressure transmitter Dwyer MS-121(Have a Range of -100 Pa to +100Pa) to Arduino using 2 Wires Current Loop. For the shunt resistor, I am using 220Ohm. So to calculate from voltage to pressure, this is what I do: reading is what I read from the shunt resistor

(((((float)reading * 0.0049) - 0.88) / 3.52) * 200) - 100)

0.0049 is used to know the real voltage, then 0.88 from the bottom range of 4-20mA(4mA * 220Ohm), 3.52 from the top range of 4-20ma(20mA & 220Ohm) and substracted by 0.88 to get a range from 0 to 3.52(So I can calculate it by percentage). Then 200 because the range of my pressure transmitter ( -100 to +100Pa). And the last is substracting it by 100 so I can get a nice range of -100Pa to +100Pa.

Now, My pressure transmitter has display too. The value I got from display is -19.4Pa and I got the value from Arduino is -21.1Pa. It has a difference by 1,7Pa or 0,85%. I am curious why the value from arduino is inaccurate, is it my calculation, or my circuit?

最满意答案

您的220欧姆电阻可能仅指定为精确到5％。 所有的模拟读数都是相对于Arduino的5V电源而言的，它本身可能只有5％的准确度。 所以你的8.8％结果（不知道你是如何得到0.85％）在预期的范围内，只考虑这两个可能的误差来源。

您至少正在计算一个计算误差：0.0049仅仅是一个近似值，实际电压转换系数为5/1024。 那里3.5％的折扣。

Your 220 ohm resistor is probably only specified as accurate to 5%. And all of your analog readings are relative to the Arduino's 5V power supply, which itself is probably only accurate to 5%. So your 8.8% result (not sure how you got 0.85%) is within the expected range, just considering these two possible sources of error.

You are making at least one calculation error: that 0.0049 is just an approximation, the actual voltage conversion factor is 5/1024. That's 3.5% off right there.