Power Meter IEEE 754 problem

Projects Networking, Protocols, and Devices
1

Dear all,

I am using Arduino Uno+ MAX485 module to communicate with an digital power meter.

The communication is successfully established, but I have a problem with the register value.

For example, when I want to read Voltage, the Voltage data is actually store in 2 registers(4 bytes). After some research I think should be IEEE 754 format. What I need to do is just to translate it into float by using the following code:
float y = (float)&x;

However, my problem is, since my acquired data from meter is in unsigned int and from two separated registers. How do I combine the two value from two registers and convert them into HEX and employ the above function?

Or is there any other simpler way to operate?

Thanks for ur help!

Regards,
Wei

2

The gcc compiler uses the IEEE 754 floating point format. That is what Arduino uses.

You have to set the bytes in the right order, and use it as a float.
It will be easy when using a 'union', but it can also be done with pointers.
Here are a few solutions: http://www.avrfreaks.net/forum/convert-float-ieee754-and-backwards

3

Hi Koepel,

Thanks for your advice, I have tried pointer method but the result always shows 0.
Should I doubt the two register is not in IEEE 754 format?
Below is the data from Serial Monitor:
"
Phase 1 Current:

regs[9]: 0
regs[10]: 16120 <----The actual current shows on power meter is 161.20Amp
d=16120
y=0.00
"

Someone suggested me to times a 0,01 to the regs[10] and ignore the regs[9] data. This method is acceptable if I eventually cannot solve this problem.

The code has attached below:

//regs[9] and regs1[10] is the Phase 1 Current data from two of the registers of the Power Meter.
   Serial.println("Phase 1 Current:");
   Serial.print("regs[9]: ");
   Serial.println(regs[9]); //L1 Current
   Serial.print("regs[10]: ");
   Serial.println(regs[10]); //L1 Current
   int c = regs[9]<<16;
   uint32_t d = c + regs[10];   //bitshift and combine regs[9] and regs[10]
   Serial.print("d=");
   Serial.println(d);
   float y = *(float*)&d;
   Serial.print("y=");
   Serial.println (y, 2);
   delay(1000);
4

as d is a uint32_t you can cast it to (float) and divide by 100.0

   Serial.print("d=");
   Serial.println(d);
   float y = (float)d/100.0f;
   Serial.print("y=");
   Serial.println (y, 2);

in fact you don't need the cast as the 100.0f would cause d to be converted to float
gives

d=16120
y=161.20
5

Hi horace,

Thanks and it works!

Btw, do you think this is necessary?

int c = regs[9]<<16;
uint32_t d = c + regs[10];   //bitshift and combine regs[9] and regs[10]

Should I ignore the regs[9] and take only regs[10] value?
Does this mean that this is actually doesn't matter with IEEE 754 standard?

Attach the IEEE 754 format:

[color=#222222]An IEEE 754 [i]format[/i] is a "set of representations of numerical values and symbols". A format may also include how the set is encoded.[/color]
[color=#222222]A format comprises:[/color]
Finite numbers, which may be either base 2 (binary) or base 10 (decimal). Each finite number is described by three integers: s = a sign (zero or one), c = a significand (or 'coefficient'), q = an exponent. The numerical value of a finite number is
  (−1)s × c × bq
where b is the base (2 or 10), also called radix. For example, if the base is 10, the sign is 1 (indicating negative), the significand is 12345, and the exponent is −3, then the value of the number is (−1)1 × 12345 × 10−3 = −1 × 12345 × .001 = −12.345.Two infinities: +∞ and −∞.Two kinds of NaN: a quiet NaN (qNaN) and a signaling NaN (sNaN). A NaN may carry a payload that is intended for diagnostic information indicating the source of the NaN. The sign of a NaN has no meaning, but it may be predictable in some circumstances.

Thank you!