ESP-12 Project (is the esp enough)

Now that this is actually coming together, I have a followup question:

The tap above is 1v, then going through a 100k resistor then in line with the CT and the ADS. This is to offset the incoming negative voltage to minimum of 0V in the case that it will never drop below 0V (GND).

  1. Do i understand that correctly?
  2. If that is the case, let's say I put a PZCT-02 to J2 (diagram above) it has 1000/1 ratio and 10 Ohm burden resistor built in. According to "Irms = (((voltage / 2.0) / resistor) * Turns) * 0.707" Irms = 34.5amps. I am assuming voltage is 1V because that is the most I can offset. Am i understanding that correctly? or do I still go off of the ADS voltage of 3.3v? which will be ok for my application? I am measuring MAX of 65 Amps

Yes. The ~1volt offset is there so the ADS never sees any negative voltages.
Together with a PGA gain for a resolution of ~2volt, each input can swing between 0 and 2volt without clipping.
That's 4volt peak/peak in differential mode.
(1volt AC is 2.8volt peak/peak).

Thank you.

I ordered 2 SCT-013-000V (100A/1V) to be my CTs. So hooking them up in differential mode, Ill realistically get something between 1V and 1.65V

I dont think i understand the differential part just yet. Let me try to explain it in my own words.

the 1V max coming off the CT is either coming off the + or the - depending on if the sine wave is above 0 or below. so doing differential, let's say pin 0 is - and pin 1 is + and there is a load of 50 amps coming through. that would indicate that at any given time, there is +500mV either coming to pin 0 or 1. and 0V going to the other pin? if so, the difference would be 500mV. I assume that is peak to peak? a sample elsewhere in the wave, lets say 1/2 way, would show +250mV on pin 0, and -250mV on pin 1? which would give the same differential value?

I think you get most of it.
One of the inputs gets more positive, and the other one equally but more negative.
Average voltage on both inputs (AC zero crossing) stays at that 1volt bias.
The A/D works out the difference between the two inputs.

Haven't seen this done this way on this forum, so make sure you keep us posted.