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Value of xWATT & XVAR registers

Question asked by larsulrikhune on Dec 17, 2014
Latest reply on Dec 19, 2014 by hmani

Hi there,


I am testing the ADE7878 evaluation board, has build the evaluation board in a PC cabinett and has hooked the ADE7878 processor to an MSP430 development board via SPI, and the measurement values are send to a MODBUS Master simulator and updated once pr. second.


I have attached the pictures of the setup in lab. 20141217_133635.jpg


As input to the voltage, I have connected a tone-generator, with a 50 Hz. pure sine 0,5 V. P. (FULL_SCALE ADC input)

As input to the current, i have made a bypass, so the same voltage is present at the current inputs


As seen on the scope picture, the amplitude and phase of the current and voltage channels are the same.(There are some anti-aliasing shadow-effect seen on the picture - in real life, there is only two waves, and not four as seen in the picture, sorry)


The read-out of the current and voltage RMS values are identical, and very close to specified full-scale value of 4,191,910

(I measure about 3.923.000 on all 6 channels..)


I have calculated VLEVEL with the following parameters:


VLEVEL = VFS/VN x 491520 = 400/230 x 491520 = 282624

I have written the VLEVEL register with 282624 (0x00045000)

VN - Phase-neutral voltage: 230 VoltVFS - Phase-phase RMS voltage: 400 Volt


Now comes my question,


How comes that I at AWATT measure -1837290??


If I set my current input reference as 400 Amps is full-scale ADC-inpu and eauqals 4,191,910..


How do i then relate the power calculation, considered a cos-phi=1, current and voltage in phase?


Can someone please help me with how to relate the WATT register to real-world signals?


I see that the formula for AWATT, assuming only the fundemental part should be: 


AWATT=Vrms x Irms x COS(V-I) x PMAX x (1/2^4)


If I enter the values from V & I rms with cos-phi=1 and Pmax = 33,516,139 i get 3,233E+19 which is a bit bigger than can fit in a 24-bit signed value...


The dynemaic range of a 24-bit signed register should be 8.388.606 (2^23)


How do Pmax fit into a 24-bit signed register value?


How come i dont get 33.516.139 at full scale input?

What am i missing here?


I than You in advance,

Kind regards, Lars Ulrik Hune