EVAL-ADE7880EBZ, different issues


I have recently purchased an Eval-ADE7880EBZ which I would like to test as an Energy Meter. However, I am a novice in the field (especially in 3-phase AC systems, current transormers, etc). So I have a few questions that I would like to have answered before damaging anything.

I have current transformers with a transformation ratio of 100/5A. If I have understood correctly if I want the maximum range to be 5Arms (for the 500mVrms), I have to add the following resistors to the input network of the current measurement :

R1 = R2 = 0.5 * 0.5 / sqrt(2) * 20 / 5 = 0.707 Ohm
=> This will give me 0.5Vrms if my current measured on the phase is 5Arms.

In order to take a small margin, I can use a 0.68 Ohm resistor.
=> This will give me 0.5Vrms if my current measured on the phase is 5.2Arms.

1. Is it correct ?

2. Do I need to pay attention to the power dissipation by this resistor? (For example if I have a voltage of 230V/AC with a load consuming 0.1A/AC, do I need to have an input resistor capable of dissipating 230V/AC*0.1A/AC /20(transform ratio) = 1.15W? Or don't I need a resistor that can dissipate a large load?

3. I have tried to connect the development kit to a three-phase source (4 wires, U1,U2,U3,Un). I haven't put a current transformer so I don't measure the current at the moment, I just wanted to measure the RMS voltages with the "ADE7880_Eval_front_panel.vi" software and RMS voltage function. I impose as input a standard three phase 50Hz, 230VAC signal. I have no load at the moment (open circuit). The results I get on the graphical application are however superfluous. Is there anything I am doing wrong?

4. If I want to add a load on the phases, do I have to use power resistors? For example, if I have a three-phase signal (4 wires) of 230VAC, with a load of 1KOhm, it must be able to dissipate P = 230^2 / 1k = 52.9W? Or is it that in an AC system, there are not the same problems as in DC for power dissipation?

Thank you in advance for your help,

Best regards,


  • +1
    •  Analog Employees 
    on Sep 29, 2020 8:13 PM 1 month ago

    1. yes your calculations are correct. IF you uses a ct with a turns ration of 2000:1 you can reduce your burden power dissipation. do you need a 5A secondary for your application? do you plan on using different turns ratios like 1000:5 or 500:5 for different max currents? 

    2. yes power dissipation is important as well as how much power the ct can put out. can it drive 5 amps into .68 ohms? yes it will get hot at full scale

    3. I don't know how you have the connections and register values. Is the run bit set? if you have 1Meg and 1K resister divider and you apply 230v you should have 230mv at the adc input.   At fullscale the RMS is 3,766,572d at 353.5mvrms at the adc. 

    230  / 353.5*  3,766,572 = approx 2450669d in the Vrms register. You can read the Vrms register in the read all registers window it is faster.

    4. I am not sure of your last question. if you add a load to your ct  or shunt and it is 230vRMS to neutral If you add resistors your power factor is 1 Vrms * IRMS  * pf(1)= power.  230 ohms / 5 amps  = 

    I use a 3KW(per phase) resistor bank for some of my testing and it gets hot. Resistive light bulbs work and if you don't mind the power factor capacitors loads don't get hot

    Rms AC and DC are equivalent  230dc and 230v RMS  with a 1K load dissipate the same power. 52W

    You can also use phantom loads that don't get hot like below



    I use MTE source meter  and Rotek accurate sources.


  • Thank you, you have answered most of my questions.

    I would need a few details:

    For question 4, I was talking about adding loads on the primary side, for each of the 3 phases (currently I don't have loads as can be seen on the diagram). But you have answered my question, I must obviously use a load adapted in power. For example, if I1, I2, I3 = 5A, and I have a load of 2 Ohms, the load must be able to dissipate 50W (R*i^2). I think that's correct?

    I currently have 100/5A current transformers (MBS EASK 31.4 - 100/5A 2.5VA - 0.2 KI), can I also use them for small currents? For example to convert 0.5A to 25mA? Or is this only effective for large currents?

    I am going to see the phantom loads in detail.

    I use now the "Omicron CMC 256" plus as energy source.