# Sallen-Key input Z

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

1. ### George HeroldGuest

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 MilesGuest

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. ### JoergGuest

<nitpick_mode>

It was Leon Charles Thevenin, not Thevinen

</nitpick_mode>

4. ### George HeroldGuest

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. ### George HeroldGuest

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. ### Spehro PefhanyGuest

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. ### misoGuest

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. ### Spehro PefhanyGuest

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. ### George HeroldGuest

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. ### George HeroldGuest

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

George H.

12. ### George HeroldGuest

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

George H.

13. ### George HeroldGuest

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

(Hence the voltage divider on the input.)

George H.

14. ### George HeroldGuest

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. ### George HeroldGuest

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. ### misoGuest

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. ### George HeroldGuest

Wow! Dat's a lot of opamps!

George H.

19. ### George HeroldGuest

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. ### George HeroldGuest

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