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I'm new to analog electronics, and I am stuck on one issue. Let's say I have an input signal, like blue one:

It's an AC input signal

To get a DC form I can rectify it (got a grip on it.) What I need is a smooth (regulated) signal without spikes like the red one:

my precious

Saturation is optional :)

To get this signal I need to use an integrated circuit with op-amp and NF, like this:

s

How do I pick values for each component?

  • R39/R40=gain
  • R40*C37-integral time (not sure)

What excactly are they responsible for? Can you advise where to dig for info?

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  • \$\begingroup\$ Are you sure you need a low-pass filter or integrator? The blue signal seems to be approximately balanced so you won't see much variation at the output. You may want to use a rectifying circuit instead. \$\endgroup\$
    – Sven B
    Apr 12, 2018 at 6:53
  • \$\begingroup\$ Well, if I zoom in the X-scale, it won't be balanced at all, especially for my MC, unfortunately :( \$\endgroup\$
    – VladOv
    Apr 12, 2018 at 8:59
  • \$\begingroup\$ Well, the reason I'm asking is because this circuit will do both: rectifying + low-pass filtering: hyperphysics.phy-astr.gsu.edu/hbase/Electronic/ietron/rect3.gif You can tune how fast it goes down with R*C. \$\endgroup\$
    – Sven B
    Apr 12, 2018 at 9:16

2 Answers 2

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Just so we are clear... you will need a rectifier, otherwise an integrator will yield the average of the signal and you will get zero. Use a precision full wave rectifier.

enter image description here Source: https://circuitdigest.com/electronic-circuits/half-wave-and-full-wave-precision-rectifier-circuit-using-op-amp

The signal will then look like this:

enter image description here

After the signal you can then average it with an filter. The integrating filter you have above will average with a time constant, imagine a square wave, if you run the square wave through the integrator, you'll get something like this. With a more 'rounded' signal you'll get something that looks like you want.

enter image description here

If it's too noisy, you may have to try a higher order filter with two poles, or a different averaging filter. You can try filtering the signal with the time constant in octave or matlab to simulate it.

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    \$\begingroup\$ There is a full-wave version with only one opamp, but I forgot where I saw it (ADI? TI? ...?). It's finicky about matching resistors, though (here a Monte Carlo for 10% R2). Still, for this case it doesn't seem crucial and the other opamp can be saved for a 2nd order Gaussian/Bessel. \$\endgroup\$ Aug 2 at 7:03
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    \$\begingroup\$ Also a rail to rail amplifier with good specs is recommended to avoid signal distortion \$\endgroup\$
    – Voltage Spike
    Aug 2 at 18:01
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In a nutshell, this is an active low-pass filter.

The parallel feedback RC pair (R39, C37) sets the frequency response ( 3dB cutoff freq = 1/(2*PI*R39*C37) in Hertz )

and -(R39/R40) is the low frequency gain

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  • \$\begingroup\$ You mean even if I have rectified AC signal (which is already DC, but jumping), I stil have to think that it's a low-pass filter, but not the integrator? \$\endgroup\$
    – VladOv
    Apr 12, 2018 at 8:58
  • \$\begingroup\$ Well, you don't really have a pure integrator by virtue of the feedback resistor R39. \$\endgroup\$
    – mike65535
    Apr 12, 2018 at 13:10
  • \$\begingroup\$ With an integrator (no discharge resistor R39), the red output line would look like ramps instead of rectangles, and would decay very slowly on the fact the capacitor is only loosing charge through leakage \$\endgroup\$
    – Miron
    Apr 3 at 2:10

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