Active filter with Op Amp has higher gain than expected

Juan Martinez

Joined Nov 18, 2022
6
Hello everyone, I'm new here. I'm a physics teacher and in a lab we built an active filter based around an Op Amp. I attach a picture of the circuit, I think it is one of the canonical configurations.

The students obtained the bode plots using an osciloscope, I attach two examples from different groups. In all cases, the measured gain values were higher than those obtained by the theoretical transfer function. Does anyone have any idea what could cause this offset? I attach the phase part of the plots as well,which show better agreement.

Thank you all for your time.

Best,
Juan

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ericgibbs

Joined Jan 29, 2010
17,131
hi Juan,
Welcome to AAC.
Please post a diagram of the circuit, we can then use LTSpice to compare our results.
E

crutschow

Joined Mar 14, 2008
31,491
Need the actual circuit with all part values including the op amp.
Otherwise it's just a guess.

Juan Martinez

Joined Nov 18, 2022
6
Hello! Thank you. I poster a Diagram in one of the attachments, I include it here in the text.
They used 1kohm and 2.2kOhm resistors as R1 and R2 respectively. 150nF and 68nF for C1 and C2 respectively. The simulated curves were with the real values measured with a multimeter: 2140 and 999 ohms. Capacitances were not measured. The OpAmp model is LM741
Thanks!

Part of the idea was for them to see that the OpAmp is not ideal and thus there is a current even though there is no load. In resonance it was estimated to be 0.2mA. Maybe the non ideal component has something to do?

crutschow

Joined Mar 14, 2008
31,491
So what is the non-ideal component?
Need the op amp type and it's operating power connections.

ericgibbs

Joined Jan 29, 2010
17,131
hi Juan,
This what a basic LM741 shows.
E

Juan Martinez

Joined Nov 18, 2022
6
So what is the non-ideal component?
Need the op amp type and it's operating power connections.
I know the OpAmp is LM741, if there is more information required, I don't have it, as the OpAmps were in the lab already and I don't know brand or have a specific data sheet. They were operated with +- 15Vcc.

Maybe I was wrong in calling them non ideal (I think it got lost in translation, in Spanish it works better). I meant to say "real" as in non perfect virtual ground, draws power (non infinite input impedance) and or non 0 output impedance.

Thank you!

ericgibbs

Joined Jan 29, 2010
17,131
hi Juan,
This is a plot with +/-15V supply and the frequency sweep from 10Hz through 1MHz.

It is close to your result.
E

crutschow

Joined Mar 14, 2008
31,491
Below is my simulation which is essentially the same as Eric's, and seems close to your actual results.
Since the op amp non-idealities should have little affect on the response at that frequency, it would seem your calculation of the theoretical transfer function is somehow not correct.

Audioguru again

Joined Oct 21, 2019
5,549
The old 741 opamp is 54 years old and has a fairly low maximum output current. C2 and R2 are the load on the original circuit.
Try building it with 10k and 22k resistors and 6.8mF and 15nF capacitors.
Do not use less than 10k for its output load resistor.

WBahn

Joined Mar 31, 2012
28,149
I know the OpAmp is LM741, if there is more information required, I don't have it, as the OpAmps were in the lab already and I don't know brand or have a specific data sheet. They were operated with +- 15Vcc.

Maybe I was wrong in calling them non ideal (I think it got lost in translation, in Spanish it works better). I meant to say "real" as in non perfect virtual ground, draws power (non infinite input impedance) and or non 0 output impedance.

Thank you!
You are confusing a couple of concepts. An ideal opamp may have infinite input impedance and zero output impedance, but this does NOT mean that it draws no power. It is, after all, an ideal voltage-controlled voltage source. It will produce a voltage at the output by either delivering or absorbing however much power is required to make that happen.

Juan Martinez

Joined Nov 18, 2022
6
Below is my simulation which is essentially the same as Eric's, and seems close to your actual results.
Since the op amp non-idealities should have little affect on the response at that frequency, it would seem your calculation of the theoretical transfer function is somehow not correct.

View attachment 281017
Thank you an Eric for the simulations, seems I need to learn LTSpice, looks awesome. Here I attach a graph I made from the students results, which with the transfer function calculated yields the same as your plots.

I think this shows it more clearly, the "overshoot" they had near resonance. As yours, the model implemented from the transfer function yields a peak around 3.35dB whereas the students measured a gain over 5dB. But maybe it is an error in the measurments, and at higher frequency their results are really inaccurate. Another problem I speculate they could have is with the Slew Rate of the OpAmp, maybe the amplitud of the wavesource was too high to work in that frequency. I unfortunately don't have much experience with this circuits, so I guess I'll never know if it was an error in measurements or something that could actually happen.

On something not related to my original problem, but got me curious, does anyone have any idea what could cause the second order zero at around 100MHz in Eric's plot? Because I got as a transfer function:

With resonant frequency wo=1/(R1*C1)=1/(R2*C2) which lacks that "bump" that causes the shift from slope -20dB/dec to 20dB/dec.

Sorry this shifted into general doubts, and I don't know if it is appropiate, but it got me really curious.

Thank you all for your time!

Juan Martinez

Joined Nov 18, 2022
6
You are confusing a couple of concepts. An ideal opamp may have infinite input impedance and zero output impedance, but this does NOT mean that it draws no power. It is, after all, an ideal voltage-controlled voltage source. It will produce a voltage at the output by either delivering or absorbing however much power is required to make that happen.
Hey! Maybe I expresed myself poorly (and I am also definitely confused about electronics xD). I meant to say that the V+ and V-, at least as I understand, should in theory not draw any current, because of the infinite impedance. So the Op amp would not draw power from the source at Vi, but of course it would from the DC supply connected to +-Vcc. Is that correct?

This leads me to a second question, that may be related to the answer by Audioguru Gain. In my circuit, as there is no load in the output, if the Op Amp was ideal, there should not be any current in the circuit, right? Because V+ and V- draw no current and the output is in open circuit. Then the current that is present, which we could measure (around 0.2mA) is because the OpAmp is a real component and it absorbs some current?

Thank you!

ericgibbs

Joined Jan 29, 2010
17,131
hi Juan,
Some Transistor models do not include all the datasheet parameters, so there are some limitations in simulations.
I have re-run the sim using a frequency sweep of 10Hz through 100k [ limit in earlier plot was 1MHz], so the plot is similar to Carl's.

BTW: LTSpice is a free download, steep learning curve, but there are many users at AAC who can help.
https://www.analog.com/en/design-center/design-tools-and-calculators/ltspice-simulator.html
E

ericgibbs

Joined Jan 29, 2010
17,131
This leads me to a second question, that may be related to the answer by Audioguru Gain. In my circuit, as there is no load in the output, if the Op Amp was ideal, there should not be any current in the circuit, right? Because V+ and V- draw no current and the output is in open circuit. Then the
#

Hi Juan,
Consider the power ie: current used in just powering the internal circuitry of the 741.

E

crutschow

Joined Mar 14, 2008
31,491
In my circuit, as there is no load in the output, if the Op Amp was ideal, there should not be any current in the circuit, right?
There is an AC load from the feedback elements.

MrAl

Joined Jun 17, 2014
9,757
Hello everyone, I'm new here. I'm a physics teacher and in a lab we built an active filter based around an Op Amp. I attach a picture of the circuit, I think it is one of the canonical configurations.

The students obtained the bode plots using an osciloscope, I attach two examples from different groups. In all cases, the measured gain values were higher than those obtained by the theoretical transfer function. Does anyone have any idea what could cause this offset? I attach the phase part of the plots as well,which show better agreement.

Thank you all for your time.

Best,
Juan

Hello,

Something looks strange with the time axis in your first frequency plot.

The theoretical peak frequency occurs at:
Fpk=1/sqrt(R1*R2*C1*C2)

and at that peak frequency the peak amplitude gain is:
AmpPeak=(C1*R2)/(C2*R1+C1*R1)

and in db it is the log base 10 of that then multiplied by 20.

With your chosen values for the two resistors and capacitors that means the center frequency is:
fc=1062.450500 Hz

and the peak is:
amppeak=3.6011489db

Now you specified that the two resistor values are a little bit off, but the way they are different means the peak would be slightly lower, which is not what you are seeing. If the two capacitors were significantly off, like say by about 50 percent, then you would see the higher amplitude you are seeing. This isnt too likely but it could be true.
The other thing is we dont see any input specs from the actual experiment setup. We need to know the input voltage measured as accurately as possible. This is especially around 1kHz.

A quick guess, and this is just a guess, is that the students might be measuring the input AC voltage with a meter that does not respond well at 1kHz and is made mostly for measuring line voltages at 50 or 60Hz, maybe up to 400Hz (aircraft).

So to be sure of what is happening with measurements like this, the test setup has to be specified exactly too and that means the make and model number of the test equipment has to be listed along with component values and measurements, which would include the cap value measurements as well using the best equipment you can get or perhaps even using a frequency generator and oscilloscope.

So a couple more things have to be specified here.

Another word, when comparing graphs in log amplitude we have to be very careful to check the vertical scale very carefully. The first graph is not labeled very well so it would be easy to think that two measurements are nearly the same when they are very much different, by as much as two fold.

If you take all this into consideration we should be able to figure this out. Most important i think is the input voltage measurements and the make and model of the equipment being used or the specs on that piece of equipment.

Last edited:

Juan Martinez

Joined Nov 18, 2022
6
Hello,

Something looks strange with the time axis in your first frequency plot.

The theoretical peak frequency occurs at:
Fpk=1/sqrt(R1*R2*C1*C2)

and at that peak frequency the peak amplitude gain is:
AmpPeak=(C1*R2)/(C2*R1+C1*R1)

and in db it is the log base 10 of that then multiplied by 20.

With your chosen values for the two resistors and capacitors that means the center frequency is:
fc=1062.450500 Hz

and the peak is:
amppeak=3.6011489db

Now you specified that the two resistor values are a little bit off, but the way they are different means the peak would be slightly lower, which is not what you are seeing. If the two capacitors were significantly off, like say by about 50 percent, then you would see the higher amplitude you are seeing. This isnt too likely but it could be true.
The other thing is we dont see any input specs from the actual experiment setup. We need to know the input voltage measured as accurately as possible. This is especially around 1kHz.

A quick guess, and this is just a guess, is that the students might be measuring the input AC voltage with a meter that does not respond well at 1kHz and is made mostly for measuring line voltages at 50 or 60Hz, maybe up to 400Hz (aircraft).

So to be sure of what is happening with measurements like this, the test setup has to be specified exactly too and that means the make and model number of the test equipment has to be listed along with component values and measurements, which would include the cap value measurements as well using the best equipment you can get or perhaps even using a frequency generator and oscilloscope.

So a couple more things have to be specified here.

Another word, when comparing graphs in log amplitude we have to be very careful to check the vertical scale very carefully. The first graph is not labeled very well so it would be easy to think that two measurements are nearly the same when they are very much different, by as much as two fold.

If you take all this into consideration we should be able to figure this out. Most important i think is the input voltage measurements and the make and model of the equipment being used or the specs on that piece of equipment.
Hello, thank you for your answer. Yes, I think the students made a mistake in the scale of the first graph (can't check cos they are in recess now), but limiting the analisis to the following, which is a plot of their experimental data with a theoretical curve I made that matches the LTSpice simulations:

I get a similar thing, a higher gain than expected, of the order of 2 dB. Maybe it is an effect of the capacitors. To generate the signal they used a function generator GW INSTEK model GFG-8215A and an oscilloscope BK Precision 2190E (rated for max BW 100MHz) fitted with oscillocope probes. The input voltage was around 1.3 V

LvW

Joined Jun 13, 2013
1,621
Juan Martinez - I can confirm your calculations.
Surprisingly - if I am not wrong - up to now, I did not see here any correct calculation for the midband gain.
Your circuit is the classical MFB-topology with 2 resistors and 2 capacitors only.
This circuit can be found in many relevant publications - as well as all the typical equations/formulas.
The midband gain can be found as
Am=(R2/R1)/(1+C2/C1).
When we insert your parts values we get
Am=-1.513 identical to
Am=3.59dB

Last edited:

MrAl

Joined Jun 17, 2014
9,757
Hello, thank you for your answer. Yes, I think the students made a mistake in the scale of the first graph (can't check cos they are in recess now), but limiting the analisis to the following, which is a plot of their experimental data with a theoretical curve I made that matches the LTSpice simulations:
View attachment 281205
I get a similar thing, a higher gain than expected, of the order of 2 dB. Maybe it is an effect of the capacitors. To generate the signal they used a function generator GW INSTEK model GFG-8215A and an oscilloscope BK Precision 2190E (rated for max BW 100MHz) fitted with oscillocope probes. The input voltage was around 1.3 V
Hello again,

Well if they used 1.3 volts input and plotted the output as is then it would read 30 percent higher than theoretical AC analysis where the assumption is 1.0 volts AC input. You could double check if that may have been a big problem, and also yes the capacitor values could affect this a lot.