-3
\$\begingroup\$

I need to measure an magnitude of an AC signal with a constant frequency. The problem is offset of the AC signal varies. What I am thinking is filtering the signal and than using Delta Sigma ADC to measure it. However, I haven't experienced any Those ADC's before. Is that a good approach or should I think different solutions.

\$\endgroup\$
  • 1
    \$\begingroup\$ [Olin's edited questions from the answer] To get answers more than just generalities as above, you have to answer: What is the AC frequency? What is the bandwidth of the DC offset? What is the voltage range of the signals? What resolution and accuracy do you need the magnitude result to be? Is the AC waveshape known, like is it always a sine? This is equivalent to asking what harmonic content it has. How often do you need a updated magnitude value? \$\endgroup\$ – Oli Glaser Jul 12 '12 at 22:01
  • \$\begingroup\$ Magnitude needs to be defined as RMS, weighted average, Peak or Peak to Peak. and also dynamic range. To reject drifting DC is not hard with a highpass filter as long as the range of DC "noise" and constant f "signal" is defined. I presume you want best "signal to noise" ratio or SNR and since it is sine. Peak is adequate then convert to RMS. Eliminating the noise is key to accurate results. in this case any ADC works \$\endgroup\$ – Sunnyskyguy EE75 Jul 12 '12 at 23:38
1
\$\begingroup\$

You are getting ahead of yourself. You can't pick a technology before knowing what you want to do.

Delta-sigma is one of several easily available A/D technologies. They can be very high resolution at the expense of speed. However, there is no way to tell whether that is a good tradeoff in your case from the information given.

As for the DC offset, you basically need a high pass filter. Those also can be realized in various ways, including a simple C,R up front or software later. The advantage of putting the high pass filter at the input before the A/D is that it reduces the range the A/D has to have to only the peak to peak AC level.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.