So a few weeks ago I had to do an experiment and one of the exercises was to measure the bode plot of an op-amp. The instructions told us to measure the voltage and phase difference at the points \$TP_{37}\$ and \$R_{41}\$. During the experiment, one of the instructors told us to use \$TP_{44}\$ instead of \$R_{41}\$.

However, when I look at the schematic of the circuit, I don't understand how these points will give me the correct information. The 2nd op-amp acts like a buffer, and shouldn't change the signal in any way (theoretically), so we were told to ignore it. What's the point then of taking the ratio of \$ TP_{44} \$ and \$TP_{37}\$? To my understanding you're supposed to take the ratio of \$V_{in}\$ (= \$TP_{41}\$) and \$V_{out}\$ (= \$TP_{42} \approx TP_{44} \approx TP_{37})\$. Anyone have an idea on what the correct method is?

Anyway, I found the following transfer function for the op-amp.

$$ H(s) = \frac{-C_1}{C_2} \frac{ \frac{1}{R_1C_1} +s }{ \frac{1}{R_2C_2} + s } \\ C_1 = 220pF, \quad C_2 = 100pF, \quad R_1 = R_2 = 22k\Omega$$


The measured voltages:

Measured voltages

When I take the ratio of the voltages, none of them seems to follow the magnitude of the bode plot. I know that a 100% match is very unlikely, but it doesn't even come close. I've tried different combinations.

Does anyone know what kind of op-amp setup this is? I've looked all over google images for a similar setup, but couldn't find anything. I did however find this same setup in the book "Schaum's Outlines of Electronic Devices and Circuits (2nd ed.), p288". It only confirms my transfer function.

Any advice/help is greatly appreciated.


Some of you pointed out that I should use JP22 as my input, and I agree. I was confused since in the task they asked to look at the signals on TP37 and R41. The input signal should be 2.1V pk-pk.

Also I believe I'm supposed to calculate the bode plot of the whole circuit, not only from the first op-amp. The questions reads as:

"Measure the bode plot for voltage amplification and phase difference between input and output AC from 50Hz to 10MHz."

My apologies for this lack of clarity.

Op-amp circuit


3 Answers 3


Based on the transfer function and the circuit, the contents of the red box is a lead lag filter, depending on the choice of C1 and C2. The difference is dependent on which capacitor is larger. For some frequencies, the gain will be will be unity, and for others, it will be C1/C2. Lead/lag filters also have an effect, of course, on the phase. One causes lead for certain frequencies, the other lag.

You are correct in that the second op amp is a buffer. Most likely the ratio is simply an exercise for you. However, at a certain frequency, an op-amp will begin to act as a low pass filter, although the MCP6022 has a bandwidth of 10MHz. This could mean anything, but I'd expect to see -3dB drop here.

Others may correct me, but I would expect your input to be across JP22. I don't think that the TP41 will tell you much. Your transfer function is certainly from JP22.

I don't think your numbers mean much without the input? But I would suggest making sure that you're converting from Hz to Rad/s for your transfer function.

Good luck and perservere! Please feel free to correct me, I'm a student also.

  • \$\begingroup\$ Thanks for helping! I'll review my work with this new information. \$\endgroup\$
    – space
    Commented Jun 15, 2016 at 2:14
  • \$\begingroup\$ How did you find the -3dB drop? I've been playing around with my measurements and it seems that the second op amp does indeed cause a -3dB drop at 10MHz. \$\endgroup\$
    – space
    Commented Jun 15, 2016 at 22:02
  • 1
    \$\begingroup\$ Neat :D I googled the datasheet. And it calls out that the bandwidth of the op-amps is at 10MHz. Bandwidth is by definition the point at which you start to have a -3dB drop. At some point, all op-amps begin to act as a low pass filter, because they can't keep up with the required slew rate (dV/dt). These particular ones happen to have it at 10MHz. \$\endgroup\$ Commented Jun 16, 2016 at 1:39

You spoke of a "BODE plot of an opamp". I rather think, you need the BODE plot of a frequency-dependent circuit (filter) which contains an opamp. Is this correct? Is this your task? Please note that the shown transfer function is NOT the transfer function of the opamp but for the whole circuit!

With the aim not to disturb the output signal of the first opamp I think you should use the ouput of the unity-gain buffer (second opamp). That is TP37. And do NOT use TP41 as Vin. The correct input is at JP22.

However, what is the meaning of the following: ...none of them seems to follow the magnitude of the bode plot. ?

I think, YOU are creating the plot? I don`t understand. Do you compare your results with another (existing) plot?

EDIT: Now I understand that it is your task to find the theoretical BODE plot of the whole system and to compare it with measured data. For this purpose, it is necessary first to get an impression of the principle magnitude function (in order to know which and how many test points (frequencies) you should take into consideration). In your case, it seems that you need much more measurements in the lower frequency range (up to app. 10kHz), instead of only one point at 50 Hz. Above 10 kHz you will notice a drastic change of magnitude and phase and of, course, here you also need much more measurements. Otherwise you cannot construct the corresponding magnitude and phase plots.

And don`t forget: (1) You need the input-to-output ratio (it makes sense to use 1Vrms as input). (2) It is common to use a log scaled frequency axis.

  • \$\begingroup\$ Yeah I might have been a bit vague about my task. I do believe I'm supposed to calculate the transfer function of the circuit. "Measure the bode plot for voltage amplification and phase difference between input and output AC from 50Hz to 10MHz.", this it what it says in my papers. I'm supposed to measure values and plot them as a bode plot, and then compare them to a theoretical bode plot (H(s)). However when I take Vout/Vin to find the magnitude, none of them really agree with the theoretical bode plot. Do I need to rework my TF then to include the resistors from the buffer? \$\endgroup\$
    – space
    Commented Jun 15, 2016 at 10:32
  • \$\begingroup\$ What is a "theoretical BODE plot"? Different circuits have different BODE plots! Perhaps you want to compare the BODE plot of the whole circuit with the BODE plot of an opamp alone. But why? This makes no sense! \$\endgroup\$
    – LvW
    Commented Jun 15, 2016 at 11:52
  • \$\begingroup\$ No, by theoretical bode plot I meant the bode plot you draw according to the transfer function. I'm supposed to compare that to the effective values I measured. \$\endgroup\$
    – space
    Commented Jun 15, 2016 at 11:55
  • \$\begingroup\$ OK - I see. Yes - this can make sense. And what is the problem now? What is the input voltage - and where did you apply it? \$\endgroup\$
    – LvW
    Commented Jun 15, 2016 at 12:00
  • \$\begingroup\$ @space, read my updated/edited answer. \$\endgroup\$
    – LvW
    Commented Jun 15, 2016 at 12:20

The TF of the red box, from JP22 (input) to TP42 (output) is: \$\small G(s)=-\large \frac{1+2\tau s}{1+\tau s}\$, where \$\small \tau = 2.42\times 10^{-6}\: sec\$. So it's a phase lead network.

  • \$\begingroup\$ Shouldn't it be \$-\frac{1+2\tau s}{1+\tau s}\$ ? \$\endgroup\$
    – space
    Commented Jun 15, 2016 at 19:20
  • \$\begingroup\$ yes, my mistake. \$\endgroup\$
    – Chu
    Commented Jun 15, 2016 at 21:13

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