Sallen-Key input Z

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...

Frank beat me to it. He almost got the wording the same as I would have, too.

To work right your Sallen-Key wants the node between the two R3's to be driven by a source whose Thevinen equivalent resistance is equal to the right-side R3, or 8K. You've got it set up to be 16K (R1 in parallel to R2 works out to 8K, then put another 8K in series).

You could probably get your intended result by replacing the left-side R3 with a wire, and making sure that C1 is big enough to work well into 8K at your desired low-frequency roll-off.

It was Leon Charles Thevenin, not Thevinen

:-)

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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.

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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.

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

Did the original filter design use two 8K resistors? If so, the impedance of the R1-R2 divider is in series with that 8K and is messing up the filter response.

Get rid of R3. The output impedance of the R1-R2 divider is already

The only waveform that matters is at the filter output. Expect intermediate stuff to be ugly.

Thanks. My spell checker barfs on both of them. Well, it did until just now, when I added your spelling to its dictionary.

The capacitors appear as ac-short circuits to the step edge. So the junctio n of R1 and R2 sees a short at the other end of R3, making the edge step at the junction input step x (R3||R2)/(R1+(R3||R2))=input step x 0.1 . As t he capacitors begin to charge the junction voltage exponentially charges to input 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.

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.

You would in mine. The advantage of an S-K filter is that it can be made to have a DC gain of precisely 1, and that gain can have a zero tempco. Most other filters have gain that depends on resistor ratios, usually lots of resistor ratios. That's nasty when an instrument has to have ppm DC accuracy.

but say

Whether it clips also depends on the signal range and the power supplies. A little peaking isn't necessarily lethal, and one can shuffle the stages if that helps.

R1 and R2 sees a short at the other end of R3, making the edge step at the 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 to input 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.

in more charge, the circuit draws less current from the source.

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

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.

ion of R1 and R2 sees a short at the other end of R3, making the edge step at the 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 to input x R2/(R1 + R2)= input x 0.2 . This is about what your scope imag e is showing. Dominant time constant is [sqrt(R1||R2) + R3/sqrt(R1||R2)]^2 x C2, which approximately agrees with image.

s take in more charge, the circuit draws less current from the source.- Hid e quoted text -

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

George H.

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

George H.

Or do it right, using a quad op-amp package... Sallen-Key sucks >:-}

What transfer function are you trying to realize? ...Jim Thompson

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

(Hence the voltage divider on the input.)

George H.

tion of R1 and R2 sees a short at the other end of R3, making the edge step at the junction input step x (R3||R2)/(R1+(R3||R2))=input step x 0.1 . A s the capacitors begin to charge the junction voltage exponentially charges to input x R2/(R1 + R2)= input x 0.2 . This is about what your scope ima ge is showing. Dominant time constant is [sqrt(R1||R2) + R3/sqrt(R1||R2)]^2 x C2, which approximately agrees with image.

rs take in more charge, the circuit draws less current from the source.

"The Journey is the reward"

eff.com

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.

ide quoted text -

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

George H.