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Sallen-Key input Z

Discussion in 'Electronic Design' started by George Herold, Nov 28, 2012.

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  1. Sallen-Key input Z?

    Grumble, It looks like I (sorta) screwed up again.
    I’ve got a Sallen-Key low pass (fc ~200kHz).
    Here’s circuit fragment.


    |\
    | \ 1uF
    | >--C1--+ 100pF
    | / | +---C2---------+
    40k R1 | |\ |
    +---R3--+--R3--+--|+\ |
    10k R2 8k 8k | | >-‘
    | C2
    GND |
    GND

    (I left out the gain setting resistors)

    I’ve got a nice step coming out of the first opamp.
    but this weirdo shaped step when I look between R1 and R2.
    Here’s a ‘scope shot.

    http://bayimg.com/AAGKOaaei
    The upper trace is the output of the whole lowpass. (six poles)
    and the lower trace the voltage step between R1 and R2.

    If I decrease R1 and R2 by a factor of ten the few microsecond tail
    becomes ten times smaller in amplitude.

    Can anyone help me understand this?
    Is the input impedance of the Sallen-Key time dependent?

    Thanks,

    George H.
     
  2. Frank Miles

    Frank Miles Guest

    Time dependent - no. Frequency dependent - of course!

    It's probably pretty common to think that the Sallen-Key filter has a
    high input impedance - after all, the signal is (indirectly) connected to
    the noninverting input, right?

    As you've just demonstrated, what works at DC is grossly different for
    the LPF nearing the cutoff frequency. If you do the analysis you'll find
    that circuit Zin drops to about R3 (in your drawing) at cutoff, and
    (assuming normal Q range) doesn't get much better above that.

    If you need to drive a Sallen-Key filter with a nonzero source impedance
    you would ideally redevelop the design equations to include that
    impedance. Then you can obtain the performance specs that you need.
    It's not that hard!

    HTH...
     
  3. Joerg

    Joerg Guest

    <nitpick_mode>

    It was Leon Charles Thevenin, not Thevinen

    </nitpick_mode>

    :)
     
  4. OK thanks guys. I've got no problem seeing the frequency dependent
    input Z.

    There's 16k ohm / 100pF LP even without the opamp and feedback C. But
    the shape of the response didn't 'jive' with what I expected. So I
    went and LTspiced the circuit... well I left out the opamp and
    grounded the opamp end of the feedback C. So just a pair of LP
    RC's.

    I got a step response just like I posted. (As you all would have
    guessed.)

    Well I've learned my new thing for the day, time to go home :^)

    George H.
     
  5. Dang... I'm just getting my head around the step response.
    It's just like

    Vin->--R---R---C->Gnd and probing between the two R's.

    I clearly need to go and play more with square waves and RC's.

    Oh and about the nice idea of recomputing my first R3 to make it all
    come out right.
    That's a great idea, unfortunately I was trying to do a bit much with
    this section of circuit and the 10k resistor is actually a pot that
    adjusts the amplitude.. so I'm a bit f'ed.

    George H.
     
  6. Probably easiest to just buffer with an op-amp follower after the
    divider. Just one part to hack in.
     
  7. Guest

    The capacitors appear as ac-short circuits to the step edge. So the junction of R1 and R2 sees a short at the other end of R3, making the edge step atthe junction input step x (R3||R2)/(R1+(R3||R2))=input step x 0.1 . As the capacitors begin to charge the junction voltage exponentially charges toinput x R2/(R1 + R2)= input x 0.2 . This is about what your scope image is showing. Dominant time constant is [sqrt(R1||R2) + R3/sqrt(R1||R2)]^2 x C2, which approximately agrees with image.
    Don't really want to get into definitions, but obviously as the capacitors take in more charge, the circuit draws less current from the source.
     
  8. miso

    miso Guest

    You do realize that all nodes of an active filter need to be examined
    for gain, even if you are not using any of the intermediate nodes.

    You would never find a Sallen Key in any volume production item, but say
    for a leapfrog ladder design filter, it is important that the gain at
    each node be adjusted so that all nodes peak at 0db, assuming the output
    is designed for 0db.

    Generally what is done is you sweep the input over a sufficient range
    and note the gain at each op amp. For a ladder filter, you can set the
    gain at each node independently. For a Salen Key, you lack that freedom,
    but you do need to insure that no node is above 0dB. That is, you can't
    expect the output not to clip if you have gain inside the network.

    For high order filters, the Sallen Key is too sensitive to component
    tolerance. You should be using some flavor of multiple feedback unless
    the tolerances are very sloppy.
     
  9. On Wed, 28 Nov 2012 15:19:52 -0800 (PST), the renowned

    The damping ratio for the first stage as shown (16K effective
    resistance for the input resistor) is actually slightly more than 1,
    (rather than exactly one if it was 10K), but he's seeing oscillatory
    step response.



    Best regards,
    Spehro Pefhany
     
  10. Yeah, driving home I figure I'll turn the first opamp into a buffer,
    and then change a few resistors in the next (2) stages.. a four pole
    butterworth.

    No one will care but me....

    George H.
     
  11. Thanks Fred, I was confused by the waveform... but I've got it now.

    George H.
     
  12. Yeah sloppy and low volume. ~20 per year, but hopefully for many
    years.

    George H.
     
  13. Well, this is not 'really' an SK... It's got some resistor dependent
    gain.

    (Hence the voltage divider on the input.)

    George H.
     
  14. It was 'designed' to be a 6-pole butterworth.... I'm not sure what my
    input Z 'screw-up' does to the step response.... (The first stage was
    the lowest gain stage of the three.)

    George H.
     
  15. Grin... too late for that.
    And I hate opamp quad packs...
    I can never get things nice and tight. :^)

    George H.
     
  16. Guest

    The design I'm working on uses sixteen quads per board. Thirty-two
    duals wouldn't cut it. Getting the design tight is simply a matter of
    a little work (and two sides). In fact, thirty-two duals would be
    *far* worse. You can pack a lot of 0402s in a small space.
     
  17. miso

    miso Guest

    I don't know what you call volume. I'm talking at least a million units.

    Power supply and signal range limitation are obvious. What you need to
    do is engineer the filter so that there is no gain at any op amp output
    at any frequency. I don't know how I can make this any clearer. That is
    how it is done if you do active filter design properly.

    In addition, you insure no node is drooping either. That increases
    noise. Dynamic range adjusting is filters 101. But in analog, often
    close is good enough as long as nobody is out there doing a better job.

    If you want a true 0dB loss, use the Fluke scheme. LTC and Maxim both
    made chips violating the patent. ;-) Incidentally by design, this filter is not dynamic range adjusted at
    every node. The FB pin must peak.
     
  18. Wow! Dat's a lot of opamps!

    George H.
     
  19. Hi Jim, I'll make a small confession. I don’t have a good feel for
    how zero’s effect the filter response.
    (Off the top of my head, I’m not even sure how to add a zero to a
    filter. Does it have to be an active filter to add a zero... more
    than just an RL or RC? )

    So sure if you write down some differential equation I can point to
    the zeros. (Well as long as it’s not some ugly monstrosity.) And
    yeah, the ‘gain’ goes to zero at the zero frequency. But more than
    that I don’t know. I understand “all pole” filters in terms of simple
    harmonic motion, with resonant frequencies and damping factors. Are
    there the equivalent of damping factors in zeros?
    Maybe someone can suggest a good book or webpage?

    George H.
     
  20. Oh, I found this,
    http://www.jhu.edu/~signals/explore/index.html
     
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