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I was trying to compare the differences between passive and active filters and below is for almost same cut off and by using an ideal opamp:

enter image description here enter image description here Above was something I expected.

But then I used a real opamp LM324 and obtain the following outputs:

enter image description here

enter image description here

And for the same circuit if I switch to an opamp called LMC6482 I obtain the following output:

enter image description here

Why with real opamps there is such loss in voltage values or power at very very low freq. unlike in ideal case? Is that something related that Im using single supply? Does the vertical axis show peak-to-peak voltages or amplitudes in LTspice? And why two real opamps give totally different loss?

edit:

The opmap is clipping yes, but why is the amplitude is 0.6V at 0.2Hz in Bode plot which is 1V in transient analysis. And for LMC6482 the amplitude is 0.1V at 0.2Hz in Bode plot which is 1V in transient analysis. There must be a meaning behind these numbers (?) in Bode plots

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  • \$\begingroup\$ They are not ideal opamps, so of course some difference is to be expected. Have a look at the transient response, a common thing is that they are not r2r opamps and clip. \$\endgroup\$ – PlasmaHH Sep 18 '17 at 13:36
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    \$\begingroup\$ Add a negative supply to your opamp say -5 and you will get the desired response. Otherwise you have to bias your design for single supply operation. Alternatively set the DC level of V1 to be say Vcc/2 \$\endgroup\$ – sstobbe Sep 18 '17 at 13:38
  • \$\begingroup\$ At very high frequencies, the passive LPF continues to improve the attenuation. Not so for the active filter, because the OpAmp cannot control Vout at frequencies above UGBW. \$\endgroup\$ – analogsystemsrf Sep 19 '17 at 5:00
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In the first circuit the active Low Pass filter is a 2nd order filter while the passive Low Pass filter is a 1st order filter so I totally expect the curves to be different and they are what they should be.

In the second circuit you're expecting the impossible from that poor LM324 opamp. You give it a single (positive only) supply yet the circuit expects it to be able to output negative voltages as well. That's not going to happen unless you supply the LM324 with a symmetric power supply so for example +5 V and - 5 V.

In your circuit the LM324 cannot amplify properly and that's what we see in the plot as well.

So connect the negative supply rail of the LM324 not to ground but to a -12 V (for example) supply. You have to add another voltage source for that!

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  • \$\begingroup\$ "In your circuit the LM324 cannot amplify properly " Do you mean LMC6482? Why it is too low? \$\endgroup\$ – atmnt Sep 18 '17 at 14:46
  • \$\begingroup\$ What I wrote for the LM324 is also valid for the LMC6482. In fact it is valid for any real opamp IC. Your circuit DC-biases the opamp around ground level (zero Volt) but you only supply it 0 V and +12 V. Due to this, the gain of the opamp will be too low, practically the gain is gone. No gain means that opamp does not work. Either you supply -6 V and +6 V to the opamp so that it can "work" with that 0V DC bias OR you bias the opamp around +6 V, you can do that by adding a 6 Vdc voltage source in series with source V1. \$\endgroup\$ – Bimpelrekkie Sep 18 '17 at 14:53
  • \$\begingroup\$ I strongly recommend that you read Opamps for Everyone: web.mit.edu/6.101/www/reference/op_amps_everyone.pdf to learn more about how to use opamps. \$\endgroup\$ – Bimpelrekkie Sep 18 '17 at 14:56
  • \$\begingroup\$ Ok opmap is clipping yes, but why is the amplitude is 0.6V at 0.2Hz in Bode plot which is 1V in transient analysis. And for LMC6482 the amplitude is 0.1V at 0.2Hz in Bode plot which is 1V in transient analysis. There must be a meaning behind these numbers (?) \$\endgroup\$ – atmnt Sep 18 '17 at 19:02
  • \$\begingroup\$ Actually it's worse than clipping. You're doing an AC analysis which calculates the DC operating point and from that makes a linearized model with frequency dependent behavior. Since the opamp is not properly DC biased it does not behave as it is supposed to so what you get is an unusable and unrealistic linearized model which show almost no signal gain. The numbers mean nothing as it all depends on how the model behaves at a wrong DC biasing. I would never trust a model under such circumstances unless I was absolutely sure it would model the correct behavior. \$\endgroup\$ – Bimpelrekkie Sep 18 '17 at 19:25
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Unrelated to any issues you're seeing, capacitors across an ideal power supply (C1, C2) do absolutely nothing.

Also important to keep in mind, an AC simulation is not just collecting data from a bunch of transient simulations - it is assuming linear operation.

You have powered your op-amp with +12V and 0V, and you are also using a 0V-referenced signal. You need to be careful when using op-amps near their rails. Now the LM324 has an input common mode range that includes the negative rail (0V), and the output can get close to 0V. However, you are asking the amplifier to deliver a voltage lower than 0V, which is a hard feat for any op-amp.

The solutions are to add a DC bias to the input signal, or add a negative power supply, or modify the op-amp circuit to shift the input voltage as well.

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  • \$\begingroup\$ " it is assuming linear operation." What does that mean simply? Can you give an example? \$\endgroup\$ – atmnt Sep 18 '17 at 17:04
  • \$\begingroup\$ @user134429 Sure. Try changing the voltage input to 100V, and notice that the result is basically 100x. It does not check for anything that might distort the output, such as power supply limits. Compare this behavior to a transient simulation. \$\endgroup\$ – W5VO Sep 18 '17 at 17:16
  • \$\begingroup\$ In AC analysis I choose AC Amplitude 1 for the source(meaning 1V). In transient analysis at 0.2Hz for 1V amplitude input the output is also 1V amplitude even though it is clipped at negative voltages the output amplitude is still 1V. But in Bode plot at 0.2Hz there is a huge difference between the input and output voltages. Moreover, the two opamp have totally different outputs at 0.2Hz. So what is the vertical axis in my second Bode plot? I define it as amplitude at signal source(right click). \$\endgroup\$ – atmnt Sep 18 '17 at 17:17
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Other answers have answered most of your questions, but I don't think anyone got this one:

Does the vertical axis show peak-to-peak voltages or amplitudes in LTspice?

Since the AC simulation is linear, the outputs are all proportional to the value you gave for your source (or sources) and it could be whichever you choose.

If the amplitude you specified for your AC source is the RMS voltage, then all of the node voltages and branch currents will be RMS voltages and currents.

If the amplitude you specified for your AC source is the peak-peak voltage, then all of the node voltages and branch currents will be peak-peak voltages and currents.

If the amplitude you specified for your AC source is the actual amplitude (the "\$A\$" in \$A \sin(\omega t + \phi)\$), then all of the node voltages and branch currents will be the amplitudes of those signals.

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  • \$\begingroup\$ That was the heart of the question. But I still dont get it I don't know why:( In AC analysis I choose AC Amplitude 1 for the source(meaning 1V). In transient analysis at 0.2Hz for 1V amplitude input the output is also 1V amplitude even though it is clipped at negative voltages the output amplitude is still 1V. But in Bode plot at 0.2Hz there is a huge difference between the input and output voltages. Moreover, the two opamp have totally different outputs at 0.2Hz. So what is the vertical axis in my second Bode plot? I define it as amplitude at signal source(right click). I really didnt get it \$\endgroup\$ – atmnt Sep 18 '17 at 17:02
  • \$\begingroup\$ The other answers explained why the 2nd active circuit doesn't work: because the bias point is at 0 V, even though LM324 has input common mode range spec down to 0 V, it will probably not perform well at that limit. \$\endgroup\$ – The Photon Sep 18 '17 at 18:09
  • \$\begingroup\$ The AC simulation can only tell you about linear behavior. If the circuit is clipping in the transient sim, the AC sim will give different results. \$\endgroup\$ – The Photon Sep 18 '17 at 18:10
  • \$\begingroup\$ The vertical axis in your 2nd bode plot is whatever you used to define the amplitude of the source in your 2nd circuit. If you meant 1 V to be the RMS, then the plot will show RMS. If you meant 1 V to be peak-peak then the plot will show peak-peak. If the gain of the circuit is 20 (for example), it will be the same whether that's amplitude to amplitude, rms to rms, or peak-peak to peak-peak. \$\endgroup\$ – The Photon Sep 18 '17 at 18:11
  • \$\begingroup\$ 1-) You wrote about defining amplitude. As far as I can see there is only amplitude can be defined in signal source, neither rms nor peak-peak. Where do you have such setting for setting rms instead of the amplitude? 2-) There must be a logic in LTspice which explains why at around 0.2Hz I see 0.5V for LM324 and 0.2V for LMC6482 opamps in AC sim vertical axis of the Bode plot. I still dont know where these numbers come from. Maybe LM324 is clipped thats why it is 0.6V but the LMC6482 is completely unexpected. It is around 0.1V. \$\endgroup\$ – atmnt Sep 18 '17 at 18:39
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In addition to what sstobbe said also note that you are comparing apples with oranges: a 1st order passive filter with a 2nd order active Sallen-Key topology filter.

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  • \$\begingroup\$ In my question Im not comparing active and passive filter Im comparing two active filters; where one of them is made of ideal opamps. If you read it carefully you will see. \$\endgroup\$ – atmnt Sep 18 '17 at 18:58
  • \$\begingroup\$ @user13449: I'm sorry but even after carefully reading this is not clear. Only with your additional comment I can guess what you meant. Next time please put more effort in conveying your thoughts (as you can see I am not the only one who didn't get it). We are not mind readers. BTW: did you take to heart what I've mentioned in the first stentence (comment of sstobbe: your non-ideal OpAmp needs negative supply voltage)? \$\endgroup\$ – Curd Sep 18 '17 at 19:17

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