I´m using AD736 to read RMS value from a shunt resistor in a current sensor. In this case, the 220VAC is fired with a triac, so, the sine wave form is not complete (phase angle control). Using an arduino, which allow 5VDC in its analog ports, my question is: What is the input volts limit from current sensor to AD736 and what will be the max output voltage ? In the datasheet, i´m not sure if it is 1V or 2,7V the limit for input voltage.

Hello,

Thanks for the question and sorry for the delay as I was out of office.

There are two responses to this question depending on the polarity of the AD736 input used and the power supply voltage. If you use the inverting input you can apply a little more voltage than if you use the non-inverting input because you're driving the V-to-I resistor directly. If you use the +/-5V specified at the top of the SPECIFICATIONS table, there is more headroom as the internal supply allows for additional headroom, compared to the minimum operating voltage of +2.8/-3.2V.

In either case the max input from your sensor (for a workable conversion) is limited by the peak voltage of the source, in your case the sensor output. For a truncated sine-wave (i.e., current falls to zero prior to termination of half-cycles) you should use the peak value of the waveform and be sure the sensor delivers on that value (for a current-transformer, just set the I-to-V resistor value so as to not exceed +/-3.8V-peak. Then the output will depend on the phase of the triac firing angle. For a maximum (Triac always on) the output will be just the rms input or 3.8/1.414 or 2.7V.

For less current, the output will vary with the sine of t/T, where T is the waveform period and t is the firing time (t is always <T)

I hope this helps.

js