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Since i wont be able to provide negative voltage to the negative power rail of the OPAMP, i wanted to give my signal a slight positive DC offset and execute the amplification process in the positive region. Since capacitors in the filter circuit might as well serve as coupling capacitors, i am really confused about where to put the dc offset voltage divider. The setup below is supposed to filter signals below 130 Hertz and amplify the signal ten times but does neither. Any idea on how to make the design work is appreciated.

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Op-amp datasheet

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    \$\begingroup\$ R12 does not connect to ground - only to R11 and the - input. \$\endgroup\$ – WhatRoughBeast Sep 13 '16 at 20:05
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    \$\begingroup\$ First, the connection between V- and R11, R12 looks like a mistake in the schematic, is that so? \$\endgroup\$ – Timo Sep 13 '16 at 20:06
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    \$\begingroup\$ A sallen key filter with this much gain will straight out oscillate so it's pointless figuring out where to put the resistors. \$\endgroup\$ – Andy aka Sep 13 '16 at 20:30
  • \$\begingroup\$ Can you please elaborate andy ? \$\endgroup\$ – Dogus Ural Sep 13 '16 at 20:44
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try

schematic

simulate this circuit – Schematic created using CircuitLab

You've rather got in a mess with which resistors are biasing and which doing signal.

Now I've finished my tea, and thought a bit more, and read the comments ...

R1-R2 provides mid rail bias of the op amp, and the right impedance to ground for the filter network.

It looks like the intention of your R9,10 was to provide the bias, but then you've connected them in parallel to your inverting input. What you need is a network of two resistors that simulataneously provides the right voltage, and the right impedance to ground. My R1,2 provide mid rail, but it would be equally as easy to use for instance 1.6k and 4.7k to bias to 0.25x rail with 1.2k impedance to ground.

C3 is a better more generic solution for getting the DC offset into the feedback path, which in this case of a high pass filter, is obviously the case. It will still work if you ask for gain of 1000, which separate biassing probably won't. If correct gain and offset is important down to DC, then R5 should be replaced by a 2k resistive divider like R1R2.

Your R7 looks like it's part of a Sallen Key filter, but unfortunately the gain is wrong. With an equal component SK, a gain of unity gives you quite low Q. As you increase the gain, the Q increases, becoming infinite at a gain of 3. In fact such an equal component gain 3 SK filter is called a Wein Bridge Oscillator! Gains above 3 just give you instability and crashing into the rails.

R3,7 may be the clever bit. It undoes the x10 gain of the opamp, and presents the signal to the caps' mid point at the impedance of R3//R7, the 1.2k in your original. This will give you back the response of a unity gain equal component sallen key, followed by x10 gain, with one opamp.

However, while such a filter is stable, high pass, and has ultimate rejection at 40dB/decade, the passband is not one of the recognised responses. This may or may not matter. You could increase the gain at the filter by reducing R3 and increasing R7 a little, while keeping their parallel impedance 1.2k, to choose responses in the Butterworth or other families.

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  • \$\begingroup\$ Neil, put a capacitor in series with R5 or it won't work at all. Ha ha I see - you have R4 to do this but with this much gain it'll sing! \$\endgroup\$ – Andy aka Sep 13 '16 at 20:30
  • \$\begingroup\$ I didn't want to redesign his filter, just set his biassing straight, which is why the comment. I think if the feedback to the caps' mid point is taken from a pot down from the output with the right output impedance, that will straighten out the filter and still provide the x10 gain. With only x10 DC gain, a difference in mid point R1R2 and R4R5 and some input offset will still be contained within the rails. Could be more elegant, but it was tea-time! A C to ground rather than R4R5 would have been better. \$\endgroup\$ – Neil_UK Sep 13 '16 at 21:30
  • \$\begingroup\$ i am just curious if its a good practice to wire the voltage divider at that point of the circuit, how would it effect the gain etc \$\endgroup\$ – Dogus Ural Sep 14 '16 at 10:29
  • \$\begingroup\$ A schematic doesn't identify 'points' of the circuit, but only nets. The voltage divider goes to the positive supply, which as I pointed out earlier, is an AC ground. If it isn't, it should be, and if it still isn't, then you can decouple the bias chain suitably so it doesn't put noise from the positive rail onto your signal node, or suffer uncertain impedance to ground. It goes to the +ve rail to avoid the need for another supply. And BTW, it affects gain, not effects gain \$\endgroup\$ – Neil_UK Sep 14 '16 at 11:24

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