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Below is the filter with ideal op amps and its frequency response:

enter image description here enter image description here

Below is the filter with real op amps and its frequency response:

enter image description here enter image description here

Very different responses from both circuits... The frequency response needs to be the same as the one with ideal op amps, i dont why this is happening in the case with the real op amps.

Here is the first stage with an ideal op amp:

enter image description here

enter image description here

And here is the first stage with a real op amp:

enter image description here

enter image description here

Why isn't the frequency response the same?

Edit -

This appears to be the answer to my problem:

enter image description here

Edit2:

Ac analysis : it's much better but not perfect

enter image description here

Added a voltage buffer, it got a bit better but still not the same as in the ideal case, why?

enter image description here

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    \$\begingroup\$ What happens if you add a -5V supply for VEE? The input is centered at the negative rail. Do a transient analysis and you'll see the issues. \$\endgroup\$
    – qrk
    Nov 30, 2022 at 17:23
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    \$\begingroup\$ Your op-amps are acting as half wave rectifiers plus you are driving the op-amps into saturation and that ruins there frequency response. \$\endgroup\$
    – Andy aka
    Nov 30, 2022 at 17:30
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    \$\begingroup\$ The input offset ideally needs to be halfway between the rails. \$\endgroup\$
    – Andy aka
    Nov 30, 2022 at 17:31
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    \$\begingroup\$ @G0tBlackOps Alternatively, you can ground VEE, as you originally did, and add a 2.5 V voltage source in series with the signal source. \$\endgroup\$
    – qrk
    Nov 30, 2022 at 17:38
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    \$\begingroup\$ A real opamp cannot produce an output below its negative rail. Most opamps are also unhappy with inputs below their negative rail. You're feeding an AC source (so a voltage which swings both positive and negative) into an opamp with its negative rail tied to ground. \$\endgroup\$
    – brhans
    Nov 30, 2022 at 18:01

1 Answer 1

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The simulator first does a DC analysis to determine the small signal gain of the op amps. Your op amps have no negative supply and the input bias is 0 V, so the small signal gain is very low because they are operating outside their linear range (specifically, the output voltage cannot quite get to ground).

The simulator then ignores any nonlinearities and does a small signal analysis swept over the requested frequency range. You can set the AC input amplitude to a ridiculously large voltage and it will have no effect on the simulation.

Ideal op amps can handle any input voltage and produce any output voltage, so their operation is not affected by supply and bias voltages.

In reality you will need to keep the signal voltages inside the common mode input range and output supply rails. The TLV6001 has 'rail to rail' inputs and outputs, so the best bias point (for maximum signal amplitude) is half the supply voltage.

If using a single supply you can create a 'virtual ground' at 2.5 V with a voltage divider and capacitor, like this:-

schematic

simulate this circuit – Schematic created using CircuitLab

The capacitor needs to be large enough to keep the 'virtual ground' impedance low at the operating frequencies. For the high pass stages you can just connect the lower ends of R9 and R11 to this 2.5 V bias point.

The low pass stage is bit trickier because it is getting bias voltage from the signal source. You should isolate it with a coupling capacitor and then feed bias voltage through a resistor like this:-

schematic

simulate this circuit

Alternatively you could switch the order so the low pass stage is after one of the high pass stages. Since the high pass stage outputs 2.5 V DC you can couple the signal directly to the low pass stage for bias.

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  • \$\begingroup\$ Instead of using a resistor why cant we connect virtual ground directly to the non inverting input? Why doesn't the low pass also outputs 2.5V? Perhaps the 100nF capacitor? Perhaps i could just put the two first high pass filters then the low pass filter so i only need to make one connection to virtual ground in the first high pass filter \$\endgroup\$
    – Scipio
    Dec 1, 2022 at 1:19
  • \$\begingroup\$ I saw some other circuit where one virtual ground equal to the one you made had a voltage buffer next to it, what's the purpose of that here? I placed that circuit in the new edit in this post \$\endgroup\$
    – Scipio
    Dec 1, 2022 at 1:25
  • \$\begingroup\$ The buffer makes the virtual ground 'stiffer' without needing low value resistors (that draw high current) and/or a large capacitor. It's especially useful in circuits that might have a large DC 'ground' current. Downside is op amp response at high frequency can cause noise and instability, and of course you need an extra op amp. \$\endgroup\$ Dec 1, 2022 at 4:44
  • \$\begingroup\$ "nstead of using a resistor why cant we connect virtual ground directly to the non inverting input?" - because it would short out the signal. "Why doesn't the low pass also outputs 2.5V?" - it does, but only when the + input is biased at 2.5V. This on no good for a following high pass stage because it is AC coupled at the input. "Perhaps i could just put the two first high pass filters then the low pass filter so i only need to make one connection to virtual ground in the first high pass filter" - yes, except each high pass filter will need it's own connection to virtual ground. \$\endgroup\$ Dec 1, 2022 at 4:51

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