QI have a doubt about the voltage to be applied to the input pins VP,VN ÷
IAP,IAN . The ADE7953 can be powered at 3.3Vdc and can accept ±250mV at IAP,
IAN and ±500mV at VP, VN pins in single ended configuration.
In my design I want to connect VN to GND (through 1Kohm resistor) and apply a
±500mV sine wave to VP pin, the same for IAN (connected directly to ground) and
IAP (±250mV sine wave).
The question is related to the fact that the device is powered at 3.3Vdc and
can accept a voltage of -250mV to IAP and a voltage of -500mV to VP pins
respect to GND (DGND and AGND are shorted togheter). How is this possible? I
think that the answer is in the "-VREF" voltage showed in figure 36 of the
data-sheet but I don't find any information about the "-VREF". Can you give me
AThe schematic should be OK. You can get accustomed to the IC if you take an
ADE7953 evaluation board and configure the board.
The ADE7953 accepts positive and negative signals at its inputs, even if there
is only a positive voltage coming from the voltage reference circuit. ADI has a
patent on this approach (attached). The abstract of the patent offers some
explanations. Details on this could be only provided by the designers but
ADE7953 should be viewed as a black box that has some knobs (registers) that
you can manipulate.