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So here's the circuit: enter image description here

And here's my attempt:

enter image description here

V1, R6 and R7 form a voltage divider so half of V1 voltage goes to the output of U3. C4 ac-couples the filter, blocking any dc level in the signal source.

In Ltspice when i try to run this it tells me that it failed to find DC operating point for AC analysis. Not sure why i get this message.

Also, i'm trying to study the impact of the input offset voltages of the second opamp. As shown in the image below, left part is the first stage of the filter and right part is my attempt.

enter image description here

As i am working with an ideal op amp i said that the current that goes to the non inverting input is 0, same as the inverting input. I ignored R3 and C3 and got the equation on the right.

Is it right? If not is it because i ignored R3 and C3?


ADDED:

So if I put C4 in front of U3 everything works ok. I get the same response as if I used both sources instead of only one, I don't know why it doesn't work when C4 is left of U3.

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  • \$\begingroup\$ Put a fairly large capacitor between R5 and ground. This will not only put the dc level output of U1 at mid-supply but it will also reduce the offset at the output of U1 (due to input offset voltage) to be equal to the input offset voltage rather than being equal to the input offset voltage multiplied by the non-inverting gain. \$\endgroup\$
    – user173271
    Nov 13, 2021 at 21:12
  • \$\begingroup\$ So if i put C4 in front of U3 everything works ok. I get the same response as if i used both sources instead of only one, i dont know why it doesn't work when C4 is left of U3. \$\endgroup\$
    – Scipio
    Nov 16, 2021 at 2:34

3 Answers 3

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In your single supply circuit, you have R5 connected to "zero volt" ground. It should be connected to your virtual ground (2.5v).

Ideally, you would create a low-impedance virtual ground with a U4 wired as a follower (a 2.5v voltage divider connected to "+" input and connect output to "-" input). That U4 output becomes your virtual ground. Besides connecting R5 to thst virtual ground, you could connect C3 and C2 to the virtual ground. This would prevent some small pops at power-up.

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  • \$\begingroup\$ Not ! using the Inverting input requires a negative bias for R5. So he has to use two inverting amplifiers with Vss/2 for the Vin+ \$\endgroup\$ Nov 13, 2021 at 21:19
  • \$\begingroup\$ So you're saying the original circuit should connect R5 to a mid-point voltage either? I don't understand how a 2.5v bias on a 5v single supply is any different than a zero-volt "bias" on the original circuit's bipolar supply (for R5). \$\endgroup\$ Nov 13, 2021 at 21:39
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If you replace R5 with a voltage divider of 30.8K + 30.8K from +5 to ground you will maintain the nominal 2.5V output to the output of U1 and thence to the output, since the filter has a DC gain of +1.

However There is DC gain of 5.45 so mismatches in resistor values due to tolerances could result in about +/-275mV of output offset if you use 1% resistors for R6, R7, and the above-mentioned two resistors (which replace R5). That could be avoided by adding another op-amp to act as a rail splitter but there are potential issues with that.

Always run a DC operating point analysis and a transient analysis before trying to run an AC analysis. If your amplifiers clip or are railed, the results will be useless from the latter.

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Vos output offset = Vio * Av + Iio * R.thevenin

You need a Vcc/2 bias on Vin+ to use a 0V based input on the inverting amplifier to use a single supply. Then both virtual ground inputs stay at Vcc/2. Since you have 2 stages, two similar inverting amp's results in a non-inverting output biased at the output by Vcc/2 plus the Voutput offset from the above equation, using the datasheet for;

Vio = Input offset voltage @ Temp. and Vcc given for worse case and nom.
Iio = Input offset current ....ditto

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