# Using an oscilloscope to measure the gain of an op-amp circuit (with capacitors)

I have an upcoming laboratory exam where we have to construct an op-amp circuit (from our experiments the circuits were always varied and probably will be a bit more difficult on our exam.)

We are always asked to measure the circuit's gain at certain frequencies to find its bode plot. I know how to construct the circuit and plug the oscilloscope to it.

The problem is I do not know which values of frequencies I should be testing, or at which frequency value I should know that the gain does not change.

I heard something about dividing the gain found at the low and high frequencies by sqrt(2) to determine the bode plot. Can I get an explanation on what to do or the thing about dividing by sqrt(2)?

Edit: A friend of mine told me something about choosing voltage values at which distortion does not occur at the output signal. Also would appreciate a comment about this.

• The problem is I do not know which values of frequencies I should be testing, or at which frequency value I should know that the gain does not change.  - this is where you test it and find out. OK, a pro would be able to estimate where these points in the spectrum are but, given that it's an experiment, you should try it at different frequencies and discover those points. Commented Jan 11, 2022 at 10:24
• So, I should be trying frequencies until Vout/Vin starts to settle to a certain value. Also, could you explain dividing by sqrt(2) and distortion at the output. Thank you for commenting. Commented Jan 11, 2022 at 10:30
• I don't follow what would you even divide by sqrt(2) and what would it accomplish with your task. Have you read the course material what you should do and how instead of asked friends? Commented Jan 11, 2022 at 10:34
• There is no course material, we are suppose to attend the exam with the knowledge we gained from the experiments. So, neither I or my friends know anything about what kind of circuit we will be asked to build. Also, I found out that the gain value at the low and high cut off frequencies should equal to Av(max)/sqrt(2) or 0.707*Av(max), that's what the dividing by sqrt(2) was about. Commented Jan 11, 2022 at 10:40

The problem is I do not know which values of frequencies I should be testing, or at which frequency value I should know that the gain does not change.

This is where you test it and find out. OK, a pro would be able to estimate where these points in the spectrum are but, given that it's an experiment, you should try it at different frequencies and discover those points. The whole idea about an experiment (your words) is that it is a voyage of discovery.

As for the dividing by $$\\sqrt2\$$, this relates to the half power points of the spectrum that you uncover in your experiment. Here's a simple magnitude bode plot of a single-order low-pass filter: -

Picture from here.

A signal that falls to 3 dB below its nominal maximum level level is said to have reduced in power by 2. Consider this; if the original signal was 10 volts at its nominal maximum level then, the power associated with that signal into a 1 Ω load would be: -

$$V^2/R = 100/1 = 100\text{ watts}$$

Half the power level would be 50 watts and that would be a voltage of: -

$$V = \sqrt{50\text{ watts}\times 1 \text{ ohm}} = 7.071 \text{ volts}$$

Or, put another way 10 volts divided by $$\\sqrt2\$$

And, in decibel terms that's 3 dB down on 10 volts because 20 log (7.071/10) = 3.01 dB. Yes, 3.01 dB is strictly speaking the half power point but, as engineers we refer to it as the 3 dB point.

A friend of mine told me something about choosing voltage values at which distortion does not occur at the output signal. Also would appreciate a comment about this.

You should use your oscilloscope to ensure that the op-amp output does not visibly distort when making measurements in your experiment. Distortion is not your friend here (unless you are a guitarist) and, it can skew the 3 dB points a little if you are not careful about managing things correctly.

Here's another 2 bode plots (of a single-order low-pass and high-pass filter) with a little more information: -

Image from here. The point of showing the above is that for many op-amp circuits, there can be two 3 dB points; one at a low frequency and one at a higher frequency. Think audio processing circuits for example. So, you might end up with an experiment that produces a magnitude bode plot like this: -

Image from here.

• In conclusion, can we say that before even trying to find the cut-off frequencies, we should adjust the voltage value so that there is no distortion. Then, we should try the frequency values to find the gain at which it does not change (which makes it our Av(max)) then we can multiply Av(max) by 0.707 (1/sqrt(2)) to find the corresponding gain value for our cut off values. Now, one last question. Should we choose the max voltage value at which the output signal is not distorted, or does any voltage value work under the distortion threshold? Commented Jan 11, 2022 at 11:14
• Adjust the input voltage so that when the output is at it's maximum, there is no visible distortion. Of course you can keep well away from using maximum signals but, if you go too small in value, background noise will start to erode at the accuracy of measurements you take. I'd probably run at a level that is half the maximum signal that shows some sign of distortion. But distortion can be fickle; slew rate limitations can be subtle and tricky to see..... Commented Jan 11, 2022 at 11:19
• .... They affect the higher frequencies so, maybe check at the apparent 3 dB point of your highest frequency signal and reduce the input amplitude to keep the output signal at a quarter full-scale level to avoid anomalies due to slew rate limitations. You'll soon get a feel for it. Commented Jan 11, 2022 at 11:20
• Thank you so much for your answers. Wouldn't changing the frequency also contribute to the distortion at the output, is this why I should use the half of the max voltage where there is no distortion. Also, would the cut-off frequency values change if the input voltage is changed, or would it stay the same for all voltage values? Commented Jan 11, 2022 at 11:44
• Once you have minimized any significant level of output distortion (by reducing the input level to a moderate value) then, lowering the input level even further will not make any change to the cut-off frequency. Changing the frequency is just changing the frequency; providing you do not visibly distort the output then there is nothing to worry about. Commented Jan 11, 2022 at 11:51