Timeline for Using an oscilloscope to measure the gain of an op-amp circuit (with capacitors)
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2022 at 11:51 | comment | added | Andy aka | 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. | |
| Jan 11, 2022 at 11:44 | comment | added | Kalamakra | 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? | |
| Jan 11, 2022 at 11:20 | comment | added | Andy aka | .... 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. | |
| Jan 11, 2022 at 11:19 | comment | added | Andy aka | 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..... | |
| Jan 11, 2022 at 11:16 | vote | accept | Kalamakra | ||
| Jan 11, 2022 at 11:14 | comment | added | Kalamakra | 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? | |
| Jan 11, 2022 at 11:01 | history | edited | Andy aka | CC BY-SA 4.0 |
added 496 characters in body
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| Jan 11, 2022 at 10:55 | history | answered | Andy aka | CC BY-SA 4.0 |