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I am using a sensor that outputs a 0-10 VAC (alternating current) and I need to convert this signal into a signal within 0-3.3VDC (direct current) to feed into an ADC.

What is the best way to go about doing this?

Is there an integrated circuit/breakout board that will take care of this conversion?

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    \$\begingroup\$ What is "0-3.3VDC" ?? If it is direct, it has to some specific value. \$\endgroup\$
    – Eugene Sh.
    Commented Jul 2, 2015 at 15:47
  • \$\begingroup\$ Mention the frequency of AC signal. You would rather use voltage dividers and feed directly to ADC \$\endgroup\$
    – User323693
    Commented Jul 2, 2015 at 16:02
  • \$\begingroup\$ Simple resistive dividers in parallel with capacitor dividers may serve the job \$\endgroup\$
    – User323693
    Commented Jul 2, 2015 at 16:03
  • \$\begingroup\$ electronics.stackexchange.com/questions/53515/… similar \$\endgroup\$
    – User323693
    Commented Jul 2, 2015 at 16:05
  • \$\begingroup\$ Do you want the peak of the signal converting to dc or maybe the rms of the ac signal. Maybe something else but certainly a frequency range is needed. \$\endgroup\$
    – Andy aka
    Commented Jul 2, 2015 at 17:19

2 Answers 2

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Why not an active peak detector? Something like this would do it, though you'll need to adjust the time constant on the output filter. This is assuming you're actually coming from the same ground, i.e. your AC wave is biased around 5V.

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If the input voltage is mains frequency and you wish a DC voltage reading, you can divide the input voltage down by about 5:1 and use a circuit such as this one from Burr-Brown:

enter image description here

Preferably use a CMOS type RRIO op-amp. R1 = R2 = (say) 100K and R3 = 49.9K.

Add an RC low pass filter afterward and buffer it to get smooth DC (and you could amplify it a bit with the buffer after filtering to get more of the range of the ADC). So only three op-amps and a few discrete parts.

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