0
\$\begingroup\$

I have single supplied sallen-key filter more that have nice 3MHz flat LPF response. I tried to compute the resistance directly by dividing the input voltage by input current which is the voltage and current of source but it seems not ok: enter image description here

I know this "transient analysis" must give me impedance of this frequency. but this blue trace is weird trace!

AC. Analysis: enter image description here I think I have low pass filter then resistance must be constant in pass band, isn't that? why if not?

Filter response: enter image description here What should I do? In transient analysis: There is current lag, use shifted current? or what?

I've also seen somewhere replaced voltage source by current source and simply plot input voltage bode diagram!!! but it's single supplied and we have constraint?!

Buffered RLC filter input impedance in transient analysis (According to "Andy aka"s nice answer): I think AC analysis show input analysis is ultralow, then we need buffer in input and output, isn't it? enter image description here Frequency response that is very good for me: enter image description here

I think this arrangement is correct: enter image description here Why this have high resistance change in pass band(0-3MHz)? then we can't model it as a constant load in our simulation?

\$\endgroup\$
  • 1
    \$\begingroup\$ Make an AC analysis and find the desired voltage-to-current ratio \$\endgroup\$ – LvW Dec 15 '18 at 11:58
  • \$\begingroup\$ @LvW gratefully AC. analysis added by linear scale, but why i didn't get answer in transient analysis? \$\endgroup\$ – mohammadsdtmnd Dec 15 '18 at 12:07
  • 1
    \$\begingroup\$ What do you think will happen when the Current (denominator) goes through zero? \$\endgroup\$ – LvW Dec 15 '18 at 14:59
  • \$\begingroup\$ @LvW Yes, I think exactly what you think :) . But how to extract correct impedance? You mean that is impossible from transient? Do you think the AC analysis extracted the correct impedance? \$\endgroup\$ – mohammadsdtmnd Dec 15 '18 at 15:30
  • 1
    \$\begingroup\$ Yes - I think so. It makes absolutely no sense to search for a ratio of two timely varying quantities. You never can expect a fixed and constant result. \$\endgroup\$ – LvW Dec 15 '18 at 16:15
2
\$\begingroup\$

Do you think the AC analysis extracted the correct impedance?

Yes. it looks likely that the graph starting at about 30 kohm and falling to 3 kohm is precisely the input impedance of the filter design. The only problem is that your filter design is unrealizable because the bare-bones input capacitance of the AD8029 is 2 pF (typically) and your model is attempting to use a capacitor from the non-inverting input to ground of 0.5 pF i.e. your design relies almost entirely on the internal parasitic capacitance of the op-amp being 2 pF (+ 0.5 pF in parallel) and stable. That is a poor way to design a circuit and will lead to disappointment.

You should also rethink having a 120 ohm resistor as the output load because you will get poor results from this op-amp with a load this low in value. If you need to drive a load of 120 ohm, use a suitable line driver circuit added to the output.

But if you are going to do this you might just as well dispense with the AD8029 altogether and use a passive filter approach rather that an op-amp sallen key filter because op-amps used in this configuration never really deliver the goods when operated close to their full bandwidth and it seems like you need that bandwidth.

Use and RLC low pass filter followed by a buffer amp is my opinion.

\$\endgroup\$
  • \$\begingroup\$ TY. Sorry because I was simulated worst case load, even with higher (>1k) load you think passive is better? I have heard that sallen-key is filter with buffer. I have tried to considered that parasitic input capacitor. I mean I decreased input capacitance until filter response fitted on my desired one. But still I can't understand why this is realizable. since it is simulated nicely. I have used 3rd order sallen key calculator in internet.initially it's have very low corner frequency, i decreased resistors and that input capacitor to move corner frequency away,it's anti-alia filter, still bad? \$\endgroup\$ – mohammadsdtmnd Dec 15 '18 at 18:56
  • \$\begingroup\$ Correction on comment: "Still I can't understand why this is unrealizable." I've told wrongly:"realizable" \$\endgroup\$ – mohammadsdtmnd Dec 15 '18 at 19:03
  • \$\begingroup\$ You cannot rely on the 2 pF parasitic capacitance of the opamp. It will vary from device to device and will vary over time and with signal levels and temperature. It is NOT a guaranteed value. The opamp will buffer things nicely and a 1 kohm load will be fine but still the basic problem. I’ve successfully used RLC circuits as anti alias filters at lower frequencies than this so it has to be a strong contender even if preceded by an RC to get the 3rd pole. \$\endgroup\$ – Andy aka Dec 15 '18 at 20:39
  • \$\begingroup\$ I have heard that active filters have higher Q (I think it help sharper edge{but not higher ramp slope}) and maybe better damping factor(I dont know it's uses exactly) and have wider bandwidth, (1) but at least you've contradicted the last one and told it have word BW. and (2) Why less frequency have higher impedance, I think they must not because it is low pass filter and must pass dc to corner frequency in same manner, isn't that? (3) Transient is meaning less in impedance analysis? \$\endgroup\$ – mohammadsdtmnd Dec 16 '18 at 7:40
  • \$\begingroup\$ RLC component Values seems great, except it's resistor, as I can see in: sim.okawa-denshi.jp/en/RLCtool.php . I have edited my comments because I understood my misunderstanding. I've also added RLC analysis. Always thanks for sharing your knowledge and experience. \$\endgroup\$ – mohammadsdtmnd Dec 16 '18 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.