Solving Op-Amp with Feedback - Get Help Here!

• birdy_uk
In summary, it seems that the circuit is filtering the sensor input to remove high frequencies. It is not clear what purpose this serves. Additionally, the circuit includes a capacitor and resistor to create a capacitor-resistor-opamp (CRO) circuit, which is used to amplify an AC signal with a DC offset.

birdy_uk

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??

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.

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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Averagesupernova said:
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.

es1 said:
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 won't be a correct result, since rc-series will introduce an additional zero/pole to the system.

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• bode.jpg
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Antoker, I was not referring to C3 on the output. That is most obviously a DC blocking capacitor. I was referring to C2 making the feedback 100% for DC which puts whatever DC voltage that is on the + input on the output. C3 doesn't do much for you if the DC gain of the amplifier causes the output to go to the power supply rail. This is why C2 is there. Technically yes it does form a bandpass filter because C1 offers more negative feedback as the frequency goes up. I would guess the gain approaches 1 at a frequency less than 5 to 10 hertz, maybe less. Not sure what the high end is, by just looking at it I'm sure the gain drops below 1 before you get to 300 hertz. One reason for R2 being there might be because opamps don't work well with capacitive loads. We have no idea what this thing drives so one assumption would be to have a series resistor there to prevent a capacitive load to ground from causing the opamp to go into oscillation.

Concerning the output stage, I would guess that the large resistor will swamp any Xc the capacitor contributes within the passband. If you included the output stage I doubt it would look much different.

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Averagesupernova, sorry, my bad. :) Yeah, I think also that output rc, won't do much about the gain/frequency response of the opamp.

1. How do I determine the feedback resistance value in an op-amp circuit?

The feedback resistance value in an op-amp circuit can be determined by using the formula Rf = (Vout/Vin - 1) * R1, where Vout is the desired output voltage, Vin is the input voltage, and R1 is the input resistance value.

2. What is the purpose of using feedback in an op-amp circuit?

The purpose of using feedback in an op-amp circuit is to stabilize the output voltage and reduce the effects of changes in input voltage on the output. It also helps to control the gain and bandwidth of the circuit.

3. How do I calculate the gain of an op-amp circuit with feedback?

The gain of an op-amp circuit with feedback can be calculated by using the formula A = -Rf/R1, where A is the gain and Rf and R1 are the feedback and input resistance values respectively.

4. What is the difference between positive and negative feedback in an op-amp circuit?

Positive feedback in an op-amp circuit amplifies the input signal, while negative feedback reduces the output signal. Positive feedback is used in oscillators and comparators, while negative feedback is used in amplifiers and filters.

5. How do I troubleshoot an op-amp circuit with feedback that is not working correctly?

If an op-amp circuit with feedback is not working correctly, you can troubleshoot by checking the input and output voltages, ensuring the correct values for resistors and capacitors are used, and checking for any loose connections or faulty components. You can also simulate the circuit using software to identify any potential issues.

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