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The project is design of both LowPass and HighPass sections (2 poles, freq range 10Hz-1.5MHz, Gain=1, moderate Q<1, equiripple filter), variable cutoff freq selected by analog mux (e.g. DG408). Focused on Multiple Feedback and Sallen-Key schemes. Selected first MFB when last cutoff was about 100kHz; now need to push to 1.5 MHz.

Q1: MFB, using OpAmp as integrator, is more demanding in OpAmp performance than SK, correct?

Thought to move to SK in the attempt to exploit better OpAmp bandwidth, that represents a hard constraint (cannot be stretched). Presently, I have selected AD8065: FET input, 145 MHz BW, stable offset. Hard to find better ones for BW, noise, temp stability. Sallen-Key LP (left) and HP (right)

SK has issues of input->output leakage, reduced attenuation in stopband, and some noise problems (minor) synthesis of Sallen-Key issues. What is relevant is the "kick" at high frequency beyond cutoff that is caused by combination of i) increase of OpAmp Zout, ii) mentioned leakage, iii) insufficient OpAmp GBW product.
Texas Instrument paper gives some insight in these limitations (see e.g. Figure3) without clarifying and distinguishing. 3 Zout of AD8065 is 0.01ohm up to 100kHz, then increases 20dB/decade.

Q2: Do you have any ref/info for influence of GBW and Zout on filter response?

  • Zout to be compared to XC2 and R2 in LP and HP, respectively, correct?
  • designed filter has G=1; Zout is given by Analog Devices for OpAmp G=1, that's not the same;
  • found "unofficial" rule for GBW to be > 10 x filter cutoff: too low; imho I would go for a margin of 50-100 times.

Q3: For the said I->O leakage, in EDN article it is suggested to add an input RC stage (going to a 3-pole filter). The added stage is not buffered by additional OpAmp, but seems incorporated in the design of the 2-pole stage. Correct interpretation? Have you ever verified this method?

Thanks!

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    \$\begingroup\$ to answer just your third question. Yes, a third order section has an input RC which gives a real 6dB per octave attenuation of the signal regardless of the opamp high frequency limitations, so you can take those response curves and add another 6dB/octave cutoff to them. That will flatten the 6dB/octave rise as the opamp goes weak, and increase the stopband by a lot. The values of the RC are taken into account in the appropriate design equations, and just come out in the wash. You can only use a 3rd order section when you have a real pole to implement. \$\endgroup\$ – Neil_UK Apr 26 '17 at 11:49
  • \$\begingroup\$ The best is to re-derive the transfer function of the adopted structure while considering an op amp featuring an open-loop gain with low- and high-frequency poles. Not an easy exercise it's true but worth the shot if you want to assess the op amp impact. Check these two papers how2power.com/pdf_view.php?url=/newsletters/1701/articles/… and how2power.com/pdf_view.php?url=/newsletters/1702/articles/… in which the op amp impact on a simple type 2 compensator is studied. Hope this helps. \$\endgroup\$ – Verbal Kint May 16 '17 at 18:21
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About the opamp:

Your opamp has FET-inputs. The parasitic capacitances of FETs depend on drain-source voltage. A non-inverting configuration has the input common mode follow the signal, so input capacitance will vary, this combines with the impedance presented by the circuit to the input to create distortion, which will be dependent on the input resistors values. In an inverting configuration, input common mode is constant, so this does not occur, this might matter in the choice between SK (non-inverting) and MFB (inverting). Some opamps cascode the input FETs, which eliminate this problem. I don't know if yours does. Datasheet only quotes distortion for positive gains... which is already quite low.

And... oops, I came upon the internal schematics in the datasheet and the input FETs are indeed cascoded, so forget what I just wrote ;) (but keep it in mind the next time you select a FET-input opamp for non-inverting application).

Output impedance

In the SK filter, the HF input signal will jump over the opamp through C2 and go straight into the output. Meanwhile the opamp attempts to fight against this, but it has a non-zero output impedance. The amount of feedthrough is thus proportional to C2 and the opamp's closed loop output impedance at this frequency.

Closed-loop output impedance is the open loop output impedance divided by the amount of open loop gain remaining at this frequency, taking into account the feedback network (but you already know that).

Non rail-to-rail output stages are straightforward: the output is an emitter follower, so output impedance can be made quite low, and also constant. Also, this EF stage might be part of a local loop (via the compensation cap) which has bandwidth much higher than the full opamp's GBW. So even when it runs out of GBW the local loop might still keep the output under control...

Rail-to-rail designs are different. The output stage is open-collector (or open-drain) so open loop output impedance will be rather high...

For example this opamp has an open loop Zout of 2k! (fig.9) but strangely it rises at LF. This is because the output stage has local feedback through a cap which stops working at DC. This one has a nice 100 ohms output impedance due to local feedback. But a pair of BJTs at 1mA bias plus emitter resistors would be around 15-20 ohms, much better...

Opamp designers come up with plenty of output stage topologies to try to make RRO opamps as good as the non-RRO ones, and they're being really smart at it, but beware... See figure 4...

Also on this one closed loop Zout rises with frequency but... look at the graph! With G=100 it rises 20dB/decade as expected, but with G=1 it rises 40dB/decade! This is due to having 2 internal feedback loops (nested) and both come into play at G=1, but at G=100 a lot of the OLG in the main loop is spent into actual gain, so there is less remaining to control Zout...

Zout will also depend on load, current going through the output devices, and voltage headroom. Output being closer to the rails means the output device which is conducting has a tiny Vce (or Vds) so it loses speed, gm, etc, it performs worse.

Thus, Zout of a RRO opamp will increase as the output nears the rails.

This means HF feedthrough in your SK filter is also dependent of how close the output is to the rails.

In the curve you posted, "opamp B" has much higher GBW than "opamp A" but it also has much worse HF leakage! Thus "B" has much higher open loop Zout. Thus I would suspect "B" to be a not very smart, perhaps a bit old RRO design, whereas "A" would either be a smarter modern RRO with local feedback on the output, or a simple non-RRO topology.

Q1: MFB, using OpAmp as integrator, is more demanding in OpAmp performance than SK, correct?

Not sure what "demanding" means, but the MFB has a passive RC at the input which should mean lower HF leakage, see here for comparisons.

Q2: Do you have any ref/info for influence of GBW and Zout on filter response?

See above.

Zout to be compared to XC2 and R2 in LP and HP, respectively, correct?

Yes.

designed filter has G=1; Zout is given by Analog Devices for OpAmp G=1, that's not the same;

Well, at HF we can simplify: C1 shorts IN+ to GND so the opamp is a voltage follower (G=1) with its input grounded. So the Zout curve from the datasheet is usable. Output voltage is entirely dependent on how Zout handles the current injected into it via C2.

But remember the Zout curve is for the output at mid-supply, if the output is close enough to the rails that the output devices get squeezed and near quasi-saturation (for bipolars) then the curve no longer holds, and only a measurement will tell the truth.

Q3: For the said I->O leakage, in EDN article it is suggested to add an input RC stage (going to a 3-pole filter). The added stage is not buffered by additional OpAmp, but seems incorporated in the design of the 2-pole stage. Correct interpretation? Have you ever verified this method?

Hmmm... you'd have to make sure it doesn't change the input impedance too much, and doesn't modify the filter response. That depends on the output impedance of the source, so can't say without knowing that. You can also put a RC (or LRC) passive filter on the output.

You could hack C2 a bit, put a L+R in series to increase its impedance at HF, it should happen at a frequency much higher than the cutoff in order not to change the response though, but it could work.

Next episode!

I loaded up the AD8065 model in spice (AD gives it on their website). It's a subcircuit, not a default opamp model, so let's hope it is accurate. The open loop gain matches the datasheet. Closed loop output impedance is higher in the model (5R at 10MHz vs 1R datasheet). So it'll be pessimistic.

I don't know your component values but... The opamp has input capacitance of 4.5pF so C1 has to be larger than that, I'll select 22pF so chip to chip variation in input capacitance doesn't change the cutoff too much... in any case the average input capcitance of the opamp should be substracted, according to spice C1=19pF does the trick.

I use low values for R3/R4 to prevent peaking, and add a small cap across R4.

Now the issue is that C2 is now 2.2nF which injects quite a bit of current into the ouput, so HF leakage is quite important. Sallen-key is black. Ideal opamp model with zero Zout is red.

enter image description here

Since this software has a filter designer, I also went for MFB. C14 has to be larger than the opamp input cap, so again I went for 33p, but this resulted in R29/R30 being too low, so tweaked the values and eyeballed it. It is not exactly the same transfer function, but for purposes of illustration it'll do! And HF leakage is much lower (blue trace). So MFB looks better in this case, I guess.

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  • \$\begingroup\$ Why JEFT OpAmp. The reason is that filter resistor values are quite large and in the HP their value changes when selecting a different cutoff frequency. This would cause dramatic change of overall offset if using a BJT amplifier, caused by the input bias and offset current. And, yes, the input stage design and distortion looked very good. \$\endgroup\$ – andrea Sep 5 '17 at 14:06
  • \$\begingroup\$ I have preferred "traditionally" low output impedance and avoided rail-to-rail OAs: I don't need rail-to-rail because the nominal is +/2 V with +/- 5V of supply; over-range is +/- 4V; I'm a bit hard head and do not "trust" rail-to-rail OAs when dealing with things like Zout and distortion. To this aim I have limited output voltage swing; the filter is always loaded by about 1k that should be an optimal value for distortion without impacting with offset and current noise of the downstream stage. \$\endgroup\$ – andrea Sep 5 '17 at 14:10
  • \$\begingroup\$ OK! You're right, in these conditions you should avoid all the pitfalls. This opamp looks good btw, nice part. Did you build it and measure the circuit? \$\endgroup\$ – peufeu Sep 5 '17 at 14:30
  • \$\begingroup\$ Thumbs up, thanks. > I made a mistake: the output stage has a gain of 2 (to limit overshooting with steep edges), so the swing for the filter is +/- 1V and over-ranges to +/2 V, that's even better for distortion... probably I will never see such a problem .... \$\endgroup\$ – andrea Sep 5 '17 at 14:47
  • \$\begingroup\$ The circuit is going to be done when I have finished other stuff that we need for site measurements in September. So, hopefully pcb comes from the manufacturer mid of October and I test it within November. I will post goods and bads :/ \$\endgroup\$ – andrea Sep 5 '17 at 14:49

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