# Op-amp with feedback

Hope someone might be able to help, if you look at the circuit attached. Why is the feedback of resistor and cap not being fed to ground? What purpose does this serve. I was thinking its probably to filter out unwanted high frequencies but unsure?? :surprised

Thanks

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Hello,

It seems to me that the system is doing some sort of filtering.

To know what type of filtering, you can write the transfer function from input (sensor input) to output (voltage after capacitor C3).

The output is AC coupled, this removes the DC offset of the sensor signal at the output.

Assume ideal opamp.

In the very high frequency case the caps are shorts, so IC1 acts like a voltage follower. The whole thing acts like a voltage source whose output is V(3) and has a 10k impedance.

In the DC case the caps are opens. This is kind of a weird case and the final settling voltage is probably going to be dependent on the intial conditions of node 2 and 6 but I think in general IC1 will be able to make nodes 2 and 3 equal (V(C2,6)=V(3) V(C1,R1,R3)=0 thus I(R1,R3)=0 and everything is stable and IC1 is a follower).

So IC1 is a notch filter and the whole thing is a high pass filter?

At a first glance it looks like a integrator ;) But, actually C in parallel with R is used to control the bandwidth of the amplifier so without c2 it will be a non-inverting low-pass shelving amplifier, but together with a C2 & R it makes it a high-pass shelving amplifier, so i say it is a bandpass!

Other than that capacitors in parallel with a feedback resistor are often used to improve the speed of a comparator, for instance, by increasing the amount of feedback at high frequencies.

Initially I thought bandpass too but I think it will pass a DC signal, so then it wouldn't be bandpass. Also, I think one wants IC1 to pass DC so the AC coupling stage has a bias point.

Anyway, I'll just wait for someone else to do the math. I should really be working on my own circuits. ;)

Actually, now I think Antoker is right.

I was thinking that it was trying to remove middle frequencies from the signal.

But now that I really look at it I think the IC1 stage amplifies signals in a band by 1+R3/R1 and just passes everything else as a copy (i.e. unity gain at very high and low frequencies). And the R2 and C3 stage serves as AC coupling.

Averagesupernova
Gold Member
It looks to me like it is made to amplify an AC signal that has a DC offset on it but at the same time NOT amplify the DC offset.

it's a filter. C3 is large and basically a DC blocking cap. as long as the output is conneted to a decently high-impedance input, i would not worry about R2 and C3. if not (and the input impedance of whatever it is that your output is connected to is in the ballpark of R2) you will have to know that input impedance to get a complete analysis of the circuit.

now combine C1 and R3 together into a common impedance (call it ZF, "F" for "feedback") using impedances in parallel and combine C2 and R1 (in series) into a common impedance called Z1. then it is just a linear op-amp circuit in non-inverting configuration and the (frequency dependent) gain is 1 + ZF/Z1 (with the gain at DC being 0 due to C3).

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It looks to me like it is made to amplify an AC signal that has a DC offset on it but at the same time NOT amplify the DC offset.

That is quite obvious, especially if you look at the output stage, since it includes a capacitor, hence the dc is being blocked, so it doesn't matter if there is a dc offset.

Actually, now I think Antoker is right.

I was thinking that it was trying to remove middle frequencies from the signal.

But now that I really look at it I think the IC1 stage amplifies signals in a band by 1+R3/R1 and just passes everything else as a copy (i.e. unity gain at very high and low frequencies). And the R2 and C3 stage serves as AC coupling.

Basically I remembered how each of shelving filters look like, and since they are together that makes a bandpass filter, as long as a HP stage comes first. I have not done any calculations either, I'm just too tired :zzz:

And another thing I've noticed, what's the point of a rc-series at the output? Is it to match opamp to the load at the specified frequency? It just can't be a simple dc-block, since input of the opamp is not biased at one half of the psu, by a divider and there is no virtual ground.. brrr, nevermind just thinking out loud

Ok, I wasn't too tired anyway, so I've written a transfer function for this amplifier s-domain, excluding output rc-series circuit, here is what a came up with:

I started with a normal transfer function for typical non-inverting amp, replaced resistive parts with a complex impedances, re-arranged and then I came up with the following transfer function:

$$H(s) = \frac{1+10.2\cdot s + 0.1\cdot 10^{-1} \cdot s^{2}}{(1+0.1\cdot s)\cdot(1+\frac{1}{10}\cdot s)}$$

Then, since it is hard to analyze this function by inspection I ported it to matlab and got a bode-plot for teh function, here is a result.

Still it looks, like a bandpass, since I didn't include the output stage so it wont be a correct result, since rc-series will introduce an additional zero/pole to the system.

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Averagesupernova