LTC1966 RMS-DC Converter issue

Hi,
I have an AC source where I want to connect it to LTC1966 to measure RMS voltage.

1. I am not getting the expected output. Can I use a battery (5V) as the supply voltage and connect my AC source to IN1 and IN2 of LTC1966 (no common ground)?
IF NOT,
2. I have a 6-0-6 transformer which I can use for this. But I do not understand how to use the same transformer for both supply and AC input.

Can anyone be a help for me, please?

FYI:
I am using this breakout board.

Regards,
Lakmal

lakmalp said:

Can I use a battery (5V) as the supply voltage and connect my AC source to IN1 and IN2 of LTC1966 (no common ground)?

Click to expand...

No.

lakmalp said:

I have a 6-0-6 transformer which I can use for this. But I do not understand how to use the same transformer for both supply and AC input.

Click to expand...

You use the transformer with a full-wave rectifier circuit with center-tap grounded (not a bridge) to generate Vdd power for the circuit.
Then use a resistive divider from one of the transformer outputs to drop the peak voltage to 1V, (0.7Vrms).

This voltage is then capacitively coupled to the input, with a bias voltage to the other input as shown below from the data sheet (please read it):

Thank you for your reply and I built the circuit as you instructed. But I get 0V at the output. I adjusted the potentiometer so that it gives 0.3V AC. Can you please check whether what I have built is what you meant. LTC1966 breakout board has built-in caps and resistors. I used LM2596 buck converter to get 5V. I have attached the circuit diagram I built.

Schematic:


LM2596:


LTC1966:

Below is an LTspice simulation.
Check that all connections to and voltages at the pins are correct.

crutschow Thank you very much for your assistance. Excuse me for my childish questions. You have V(in) and V(in1) in the above graph. Can you explain what are these voltage patterns? I thought when the voltage (AC) at the in1 is 1Vp-p and the output should be 0V -> 5V (Vdd). Maybe I haven't understood the datasheet.

P.S:
I'll simulate this myself first. Then I can understand. Thanks.

lakmalp said:

You have V(in) and V(in1) in the above graph. Can you explain what are these voltage patterns?

Click to expand...

V(in) is the sinewave input from your transformer pot.
V(in1) is the input at the output of the input capacitor, which is apparently offset 2.5V by the 2.5v bias voltage at In2.

lakmalp said:

I thought when the voltage (AC) at the in1 is 1Vp-p and the output should be 0V -> 5V (Vdd).

Click to expand...

No.
The filtered output DC voltage is the RMS value of the input voltage.
Curiously I could not find the gain value explicitly stated in the data sheet but it is shown in the figure below where the output is 200mVdc for a 200mVac RMS input for crest factors below 4 (a sinewave has a crest factor of 1.4).

crutschow Thanks a lot. I am really impressed by your commitment to helping someone to broaden his knowledge. I think now I am comfortable enough to continue. I got a lot to read just after seeing your answer. Thanks again...