Regarding an example passive LC ladder filter, when I sweep the source resistance, Rs, the filter characteristics changes as follows as expected:

Enter image description here

On the other hand, if I sweep the load resistance, Rload, it seems the characteristics does not change, but the DC gain changes as follows:

Enter image description here

For ease, I showed the above plots in linear Bode plot instead of dB. So 1 V corresponds to 0 dB.

I'm not trying to build a filter so this is just out of curiosity.

The Rload resistance forms a resistive divider and causes attenuation. If Rload was known to be 1 ohm then I could add a gain stage with a gain of two and compensate for the attenuation. But if Rload is not known and there is no buffer, can there be a feedback between the input and the output which would prevent any DC gain attenuation?

In other words such a feedback which would set the DC gain to zero regardless/varying of Rload so that the frequency response will start from 0 dB at DC. How could that be realized with any behavioral elements (like VCVS) or op-amps in LTspice or any other simulator?

I have written at the beginning of the question that sweeping Rload does not change the filter characteristics (besides DC gain), but am I actually wrong? Because I noticed that the phase and group delay plots vary with Rload, and below is the group delay for different values of Rload:

Enter image description here

I thought the load resistance has no effect on any filter characteristics besides DC gain. Could you also expound on this?

Buffering solved both the DC gain, phase and group delay's dependence to Rload:

Enter image description here

  • \$\begingroup\$ An unity-gain active low-pass filter (eg Sallen-Key)? The configuration already encompasses the referred feedback and in conjunction with the characteristics of the amp. op. (high gain and impedance) is able to produce complex poles (roll-off> 20 dB/decade) without the need to use inductors. For example, Rload could be placed on output or with an adittional voltage follower for buffering. \$\endgroup\$ – Dirceu Rodrigues Jr Mar 23 at 15:12
  • \$\begingroup\$ Not really, I want to implant a feedback which senses the input amplitude and compensates the DC gain attenuation at the output for an unknown Rload. \$\endgroup\$ – user16307 Mar 23 at 15:41
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    \$\begingroup\$ Not only does your filter phase characteristics change with \$R_{load}\$, but if you look at the amplitude plot, the ripple at low \$R_{load}\$ is much less -- I would call that "different characteristics". \$\endgroup\$ – TimWescott Mar 23 at 16:10
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    \$\begingroup\$ L3 does nothing in your circuit with the buffer. \$\endgroup\$ – Spehro Pefhany Mar 23 at 16:31
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    \$\begingroup\$ I know you said you are not building a filter. I just want to point out that these are not practical component values. If you do ever want to implement something like this, you will likely need to use a different technique. If you scale up the frequency and scale down the component values then you may be able to do it with practical inductors and capacitors. \$\endgroup\$ – mkeith Mar 23 at 16:42

Your filter has an output impedance. The load impedance interacts with that output impedance to create a voltage divider.

If you want to eliminate that dependence, then you need a simple voltage follower (buffer). Connect the "nominal" load impedance to the filter, then use a VCVS as an "ideal buffer", controlled by the voltage across that load. The output will be independent of whatever load you put at the VCVS output.

In the real world, use an opamp voltage follower (unity gain).

  • \$\begingroup\$ Thanks could you also address my edit? I thought Rload has no effect on filter characteristics besides DC gain. The phase and group delay looks effected by Rload in simulation. Could you alaso have some comments on this. Thanks! \$\endgroup\$ – user16307 Mar 23 at 16:11
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    \$\begingroup\$ A filter is usually designed for a specific nominal load resistance. That doesn't mean that the output impedance of the filter itself is resistive or even constant -- in fact, it is normally reactive and varies quite a lot with frequency. That's why you should terminate your filter with the resistance it was designed for, and then buffer the output. \$\endgroup\$ – Dave Tweed Mar 23 at 16:14
  • \$\begingroup\$ If I buffer this filter and if now the buffer has a varying load would the characteristics still change depending on the load at the buffer? \$\endgroup\$ – user16307 Mar 23 at 16:15
  • \$\begingroup\$ Not if you have a good buffer. Its whole purpose is to perform that kind of isolation. \$\endgroup\$ – Dave Tweed Mar 23 at 16:16
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    \$\begingroup\$ user16307, there are really two separate questions. A series inductor followed by an infinite impedance ideal buffer has no effect, because the current in the inductor is zero, so the voltage across the inductor is zero, so you might as well replace it with a short circuit. A shunt element WILL still effect the output. But the separate question, is a resistor needed, is more related to how the filter was designed. When you plan to use an output buffer, if your final element is shunt, you can design it so that no resistor is needed (Zout = infinity). \$\endgroup\$ – mkeith Mar 23 at 16:49

Simply put, passive filters interact with loads and sources (as you have found). Adding feedback makes it (overall) an active filter.

The suggestion of putting a buffer amplifier on the output is probably a good one. It'll be easy at frequencies (like the 0.1Hz of your filter) where you can use op-amps, far harder at microwave frequencies.

  • \$\begingroup\$ Buffering solved the problem but what do you think about adding termination resistor before the buffer? Would that have any benefit? i.stack.imgur.com/Myeul.png \$\endgroup\$ – user16307 Mar 23 at 16:23
  • \$\begingroup\$ @user16307, if your filter was designed to work (have the desired response) with a certain resistive load, then using that resistor value as a termination in front of the amplifier will make sure your filter has the desired response. If the filter was designed for a high-impedance load, then a resistive load in front of the amplifier will cause the filter response to be something other than what was designed for. \$\endgroup\$ – The Photon Mar 23 at 16:50
  • \$\begingroup\$ An explicit termination, or make sure that your amplifier's input impedance matches the design impedance of the filter. \$\endgroup\$ – TimWescott Mar 23 at 16:59
  • \$\begingroup\$ Tim can you show an example where an LC filter is practically better with a buffer than an active RC filter? considering that there are no CoG/NP0 inductors. Maybe this is just an academic type answer \$\endgroup\$ – Sunnyskyguy EE75 Mar 23 at 17:08
  • \$\begingroup\$ @SunnyskyguyEE75 I would use such an arrangement in a radio. Antenna, filter, amp, diode-ring mixer -- using the amp to isolate the mixer from the filter and visa-versa. Even in a 10MHz circuit you'd be hard pressed to find an op-amp that would be superior to a one-transistor gain stage. \$\endgroup\$ – TimWescott Mar 23 at 17:12

If you wanted a lossless filter at DC then the source impedance must be 0 at DC for a fixed load OR no load, but then not matched impedance.

If you wanted a maximally flat input impedance from DC to almost f -3 dB then it must be a -6 dB lossy Cauer (aka Bessel aka elliptical) filter with the impedance matched at source, filter and load.

If you compute a ladder filter, you can see it is not maximally flat and has ripple without AND with load.

Enter image description here

Top = Bessel Output Response
Middle - Bessel Input response
Bottom = Ladder Filter No Load and with load and ripple
Left= No Load
Right = with load

Active filters have effectively 0 source impedance so they can be made into lossless at DC or with gain.

Bonus Question

The grid has almost zero source impedance, so what does this say about impedance effects on the network?

Is it an accurate model?

Final Question

Do you need a flat input impedance and lossless at DC? If so, then you choose an active Bessel Filter.

  • \$\begingroup\$ Questions should be in the comments to the question, not here in an answer. This is not a forum. If they are rhetorical questions, it is better to put them in a non-question form. \$\endgroup\$ – Peter Mortensen Mar 24 at 10:44
  • \$\begingroup\$ This question is so unrealistic in MHz with excessive tolerance demands on massive coils that cannot possibly be made with high mu cores to anything less than 20% tolerance let alone 5 significant figures. Yes it was rhetorical. \$\endgroup\$ – Sunnyskyguy EE75 Mar 25 at 4:37

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