Sampling DC offset of a composite signal

The signals are differential, each centered at 1.75 V. The AC signal superimposed on them can go as high as 0.8 Vpp, so the overall voltage is always positive. The bandwidth of the signal is from about 3 kHz to

I'm not so used to working down at audio frequencies. Taking your idea 4 (since it seems an active filter is needed), I can make a 1st order filter with an op amp in the inverting configuration (say 1 Mohm in parallel with 0.1 uF in the feedback path and a 1 Mohm resistor going into the inverting terminal). That would give me about 0.8 mV of ripple, not even taking into account part tolerances. I am feeding a dc offset into the non-inverting terminal, so the fact that it's an inverting amp is ok.

I'll probably need a second order filter -- my max suggested error in common mode voltage is 1 mV.

Thanks for the ideas.

Stephen

Yes. A resister and a capacitor LP filter!!!

Kevin Aylward snipped-for-privacy@anasoft.co.uk

I read in sci.electronics.design that Stephen Boulet wrote (in ) about 'Sampling DC offset of a composite signal', on Mon, 20 Dec 2004:

How about a low-pass filter?

"Stephen Boulet" wrote

If you sample fast enough all the filtering/integrating can be done in software.

However, I am not quite sure what you mean by 'only positive voltages'. After you subtract the DC offset from the signal the remaining AC will be negative half the time. If the system is strictly analog you will need to have a virtual ground somewhere.

"Stephen Boulet" wrote

Ah, that's different. I thought you had a DC offset voltage to be gotten rid of.

If you want the AC w/o the DC why not just use a differential amplifier? A _real_ diff. amp works better that the usual 1-trick pony. You will need to reference to a virtual ground, though ...

If you want the DC without the AC why not just add the two signals.

A small 1:1 audio transformer will remove any common mode DC, and is possibly the most sensible solution.

To eliminate common mode errors due to the DC I would suggest using capacitive coupling.

View the following ascii-art schematic with courier or a similar fixed-width font.

diff in + ----------+ | / R Low offset voltage op-amp. / | | | +---------+------------|+ _______ Vout | | +--|- / | --- C | | / | / R --- | |/ | | +----------+ / | | --- GND diff in - ----------+

Vout will be your DC offset, if I understand you right.

This is a simple low-pass filter with a time constant of RC/2. So you should choose R and C such that the frequency is well below the lowest frequency of interest. Also, you have to be careful that R is not too big because op-amps do have input bias current, and you want the maximum input bias current (see the datasheet) to be much less than the DC current flowing through R. So you want an op-amp with a low offset voltage and a low bias current.

What the heck. We might as well choose R and C.

We probably don't want to use a capacitor any bigger than 10 uF. So we'll set C to 10 uF and choose R based on that. We'll set f to 300 Hz, since that is one tenth of your cutoff frequency. This will assure minimal attenuation at frequencies of interest.

f = 2 * pi * R/2 * C R = f / (pi * C) R = 300 / (pi * 10uF) R = 30 / (pi * 1uF)

This is around 10 * 1,000,000 which is 10 Megohms. You will need to find an op-amp with a very low input bias current, or choose an even larger capacitor.

So, tentatively, we have: R = 10 Megohm C = 10 uF

You could also use less uF or megohms, and accept more attenuation near the low end of your band. Or you could cascade multiple op-amp low-pass stages to get good performance with more reasonable values for R and C. For example, you could put your cutoff frequency at 1000 Hz, and go through three stages. You might want to put together a spreadsheet to try different scenarios.

And you should double-check all my calculations. I've been known to make mistakes!

HTH

--Mac

Use a DC-servo made from a precision opamp and a first order LP(to maintain stability) feeding back into the reference input of the existing amp. You will need 2circuits for both sides of the bridge. You do not need a second order filter. If there is some ripple because of the finite attenuation, you will not change the DC-level, but attenuate the lower frequencies, just like a high-pass. actually if you choose the input capacitor a little bigger, you can compensate for this.

That is very true. the time constant is R*C fg= 1/(2pi*R*C) with C= 0.1uF and 100Hz I get here 15k915, just take 15k.

I read in sci.electronics.design that Stephen Boulet wrote (in ) about 'Sampling DC offset of a composite signal', on Mon, 20 Dec 2004:

With one op-amp you can make a third-order filter. But be careful which op-amp you choose. And audio op-amp won't make a good filter at 75 kHz. You need a faster one.

And the problem with a low pass filter is...?

You have two inputs V1= VDC + Vac/2 and V2= VDC-Vac/2 and Vac is the "signal"? -the only possible interpretation of "signals are differential." So what the hell do you want to pick off, VDC or Vac? If it's VDC then you've been given your answer, a summer. Or is it Vac, in which case you use an ac-coupled differential amplifier. What is your problem? Is this a single supply application you can't figure out?- or homework? Take it to sci.electronics.basics.

Uhh, you analog guys is hard to please ;-)

Al

Boy, I really screwed that one up. Thanks for correcting me. The only thing I got right was using R/2 instead of the usual R, because, if you look at my schematic, the resistors are in parallel with respect to the common mode input.

So you should use: fq= 1/(pi*R*C)

--Mac