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I want to sample a voltage signal from mains. I found the following circuit but im not sure how it works. The circuit consists of precision amplifier and a buffer which I can't make sense of, the rest is okay. I've did a simulation using a voltage divider and a reference voltage IC but I also would like to understand this scheme and compare between the results. Please help understand this circuit.

Source:

http://www.ti.com/solution/power_quality_meter#

http://www.ti.com/lit/df/tidrfk9/tidrfk9.pdf

enter image description here

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2 Answers 2

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That is not a circuit diagram, but only a block diagram, therefore there is not much we can explain here.

That block diagram simply states the following:

  1. Some fraction of the mains is "sampled" using a transformer, which is also a safety measure to prevent electrical risks.
  2. The output from the transformer is fed to an amplifier/buffer whose output is low-pass filtered (probably to prevent aliasing in the ADC and reduce noise bandwidth).
  3. The filtered signal is converted using an ADC, whose operation depends on a voltage reference Vref. The voltage reference is buffered to avoid "disturbing" the reference source itself.

Nothing more can be said from that block diagram, except for wild guessing.

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  • \$\begingroup\$ Thanks a lot for answering. I'm working on a power quality meter where I want to find reactive power and harmonics. I'm looking for a circuit scheme that sample and deliver these waveforms to a processor. Please provide with some suggestions, thanks. \$\endgroup\$
    – R.M.A
    Jul 4, 2015 at 22:28
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So I've found this document that explains the detailed circuit http://www.ti.com/lit/ug/tidua01/tidua01.pdf

As I mentioned before, I want to perform voltage sampling, in page 9 it mentions the following:

1.4 Potential Transformer and Voltage Transducer With 333-mV Output: The design can also interface with a voltage transformer with rated voltages of 110 V or 230 V extending up to 600 V with output of 333 mV at a rated voltage.

Then on page 20 it mentions the following:

enter image description here

The AC input voltage applied is divided by a resistive potential divider. Multiple resistors are used to increase reliability and also withstand the surge input. The resistor divider is selected to have input impedance is >= 1 MΩ. The input voltage is divided by 1:990 ratio. The inputs are protected for ESD and Surge. The max input is 350 VRMS and the divider output is 350 mV. This is amplified by the non inverting amplifier with X17 gain. There are in total four amplifiers to measure the AC voltage.

The mains voltage is applied to the above voltage divider circuit. The capacitor is there to prevent input surges.

Please take a look and justify my answers. Thanks

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