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I have an OPAMP circuit that its output increase steadily without cahnge in the input. I guess the issue is in the transfer function of the circuit and its stability. I want to calculate the transfer function of my circuit but I do not have any idea about that. Could you please help me to calculte the transfer function?

5 volt is generated by a switching regulator called MC34063 and the opamp is a LM358 and I use a UGN3503 Hall Effect sensor enter image description here

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  • \$\begingroup\$ If you have built it and have this problem you need to add component values to your schematic. There is no reason that this circuit should have any significant drift so, add data sheet links to your sensor and the actual op-amp that you used. Then you need to state what the amount of drift is and how regulated your 5 volt supply is. No point calculating the TF until all the stuff I mentioned is clear. \$\endgroup\$
    – Andy aka
    Commented Mar 31 at 14:51
  • \$\begingroup\$ Hello All I add the values of components in my schematic \$\endgroup\$
    – MMB1999
    Commented Mar 31 at 16:01
  • \$\begingroup\$ Ok, it is not overly complicated to extract the transfer function of this arrangement. First, the (+) pin is at a fixed 2.5-V potential or ac-grounded (0 V). You thus have an inverting filter. At dc, when all capacitors are open-circuited, the gain is \$-\frac{R_4}{R_3}\$ and then, you have one pole due to \$R_4C_2\$ and another one with \$R_5C_3\$. \$\endgroup\$ Commented Mar 31 at 17:05
  • \$\begingroup\$ There will be some temperature drift in this circuit, perhaps of the order of +/-100~200uV/°C at the output (a few uV/°C input-referred) but the Hall sensor will likely contribute considerably more. There might also be a problem with your implementation. \$\endgroup\$ Commented Apr 1 at 10:10

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Well, notice that the voltage at \$\displaystyle\text{V}_+\$ is given by:

$$\text{V}_+=\frac{\displaystyle\text{R}_2\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{sC}_1}}{\displaystyle\text{R}_1+\left(\text{R}_2\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{sC}_1}\right)}\cdot\frac{\displaystyle5}{\displaystyle\text{s}}=\frac{\displaystyle\text{R}_2}{\displaystyle\text{R}_2+\text{R}_1\left(\text{sC}_1\text{R}_2+1\right)}\cdot\frac{\displaystyle5}{\displaystyle\text{s}}\tag1$$

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

Now, notice that:

$$\text{I}_{\text{R}_3}=\text{I}_{\text{R}_4\space\text{||}\space\text{C}_2}\space\Longleftrightarrow\space\frac{\displaystyle\text{V}_\text{in}-\text{V}_-}{\displaystyle\text{R}_3}=\frac{\displaystyle\text{V}_--\text{V}_1}{\displaystyle\text{R}_4\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{sC}_2}}\tag2$$

We can also see that:

$$\text{V}_\text{out}=\frac{\displaystyle\frac{\displaystyle1}{\displaystyle\text{sC}_3}}{\displaystyle\text{R}_5+\frac{\displaystyle1}{\displaystyle\text{sC}_3}}\cdot\text{V}_1=\frac{\displaystyle1}{\displaystyle\text{sC}_3\text{R}_5+1}\cdot\text{V}_1\tag3$$

Combining \$\displaystyle\left(2\right)\$, \$\displaystyle\left(3\right)\$ and the fact that for ideal opamps we know that \$\displaystyle\text{V}_+=\text{V}_-\$ gives:

$$\text{V}_\text{out}=\frac{\displaystyle1}{\displaystyle\text{sC}_3\text{R}_5+1}\cdot\left(\text{V}_++\frac{\displaystyle\text{R}_4}{\displaystyle\text{sC}_2\text{R}_4+1}\cdot\frac{\displaystyle\text{V}_+-\text{V}_\text{in}}{\displaystyle\text{R}_3}\right)\tag4$$

Now, you can solve for the transfer function.


Edit, using your values I found:

$$\text{V}_\text{out}=\frac{\displaystyle100000000\left(500\left(48000+47\text{s}\right)-47\text{s}\left(200000+\text{s}\right)\text{V}_\text{i}\right)}{\displaystyle\text{s}\left(100000+\text{s}\right)\left(200000+\text{s}\right)\left(1000+47\text{s}\right)}\tag5$$

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  • \$\begingroup\$ Thanks for your response. but can you explain about 5/S at the first equation? \$\endgroup\$
    – MMB1999
    Commented Mar 31 at 17:09
  • \$\begingroup\$ @MMB1999 Yes, that is the Laplace transform of the voltage source at the left top of your circuit. \$\endgroup\$ Commented Mar 31 at 17:12
  • \$\begingroup\$ ok i am very grateful \$\endgroup\$
    – MMB1999
    Commented Mar 31 at 17:13
  • \$\begingroup\$ @MMB1999 you are more than welcome! \$\endgroup\$ Commented Mar 31 at 17:16
  • \$\begingroup\$ according to your equation the system is stable and does not hava any pole at the right side of the imaginary axis. So what do you think about the increasing the output voltage without increasing in input voltage? \$\endgroup\$
    – MMB1999
    Commented Mar 31 at 20:36
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The ac response of this filter combines a dc gain \$H_0\$ and two poles, located in the left-half plane. The 5-V dc source is there to provide a dc bias as the op-amp is supplied from a positive rail. The (+) is therefore biased at 2.5 V and plays no role in ac. The transfer function is simply obtained by writing the gain of an inverting op-amp, followed by a passive \$RC\$ low-pass filter:

enter image description here

A quick SIMetrix simulation confirms the equation. The dc transfer characteristic - if you apply a dc bias with the input signal - is: \$V_{out}=-V_{in}\frac{R_4}{R_3}+2.5(1+\frac{R_4}{R_3})\$. Any small bias on the input will bring the op-amp in saturation.

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I did not try to find the TF. Run a basic AC analysis to check is there any gain peaking or not. This is just a basic check for stability. For exact picture you need to find the PM of the circuit.

I did not observe any gain peaking.

enter image description here

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