Hi,
Two important things to consider:
* input current, including waveform, frequency,
* ADC input voltage range.
The (convertable) ADC input voltage range is limited. Every signal beyond this range can not be converted, you loose infirmation.
Let's say your ADC is powered with 5V and you use this 5V as reference for the ADC and thus your ADC input voltage range is 0V ... 5V.
Now let's go a bit into details (I didn't read the datasheet, the given values are just examples)
Even for an ideal ADC there is a loss of one LSB in range. For a 10 bit ADC the output range is 0...1023.
The upper limit is 1023. The formula to calculate the voltage from the ADC result usually is: V = V_ref × ADC_value / 1024.
Thus the max decodable value is 5V × 1023 / 1024= 4.995V only.
But this missing 5mV most probably is the least problem.
Every ADC has offet error and gain error.
Let's say the offset error is specified with +/-20mV.
If now the input is 100mV, the ADC may decode it as 100mV -20mV = 80mV .... up to 100mV + 20mV = 120mV.
It results in an error of +/-4LSB.
If the offset is positive you will see every change in analog voltage, but you will never see a ADC_value of "0".
If the offset is negative you can't decode input voltages below this. This means if the offset is -15mV and your ADC_input_voltage is somewhere between 0mV and 15mV --> it can't decode this, the result will always be "0".
Similarily with the gain error, but at the upper end of the input range.
Let's say the ADC gain error is +/--0.5%, then the theoretical upper limit is 4.975V ... 5.025V
Thus the decodable limit may be 4.975V or 4076 LSB (ignoring the offset error)..
But worst will be the VCC error. VCC may have initial deviation from ideal 5V and it may be noisy and drift with load curret, time and temperature...resulting in a huge uncertainty of output value.
Then there additionally are the resistor tolerances and drifts....
All in all I recommend no to go to the limits. Leave some headroom to avoid to get into the decoding limits.
This was the ADC side..
***
Now to the input side:
You say 30A.
But what does this 30A mean?
In ideal case it is 30A RMS of an undistorted, noiseless sinewave.
Then the peak values are +/- 30A x sqrt(2). = +/- 42.43A, a total of 84.85V.
But there will be distortion, there will be noise and there will be increased power_ON current.
A filament light bulb may draw easily 5 times nominal current at startup, a motor will dra a lot higher current at startup, a transformer may draw high peak current when gettin into saturation.
Now you need to decide what peak current you are interested to be decoded. No one can decide this for you.
My recommendation is to leave at least 50% headroom. (Including ADC input headroom)
(But for one of my applications even +/- 5 times the RMS value was not sufficient)
Thus for 30A RMS I'd use 45A RMS input range. Means a total range of 127App.
With a CT ratio of 1000 this means 127mApp at the output side.
This 127mA should become 5V ADC input voltage range: R_burden = V/ I = 5V / 127mA = 39 Ohms
Klaus