Timeline for Oscilloscope probe noise
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Oct 18, 2020 at 18:00 | history | tweeted | twitter.com/StackElectronix/status/1317888187558223874 | ||
| Oct 7, 2020 at 2:27 | comment | added | frr | Oscilloscope inputs tend to have a native input impedance of low megaOhms real + like 2-3 picofarad imaginary. Your 10 GSps 'scope possibly has the imaginary part lower (under a picofarad?) - I'd expect your 'scope to have analog bandwidth in the low GHz. When a probe is attached to the circuit under test, that impedance combines with the circuit under test, Kirchhoff laws apply etc. More precisely: the probe cable has its own contribution to 'scope input impedance, the 1:10 divider is resistive too, without a divider the probe's coax cable gives wavelength effects etc. In case you look at RF. | |
| Oct 6, 2020 at 19:10 | history | became hot network question | |||
| Oct 6, 2020 at 15:32 | vote | accept | CommunityBot | ||
| Oct 6, 2020 at 14:03 | answer | added | analogsystemsrf | timeline score: 1 | |
| Oct 6, 2020 at 13:23 | comment | added | Kyle B | If you connect your probe tip directly to probe ground, does this phantom signal go away? It should. After you note that effect, Google "Faraday's Law". The amount of signal you'll capture depends on loop area. THe reason for this may become more clear, and once you understand the reason, you're better able to avoid it! | |
| Oct 6, 2020 at 12:20 | history | edited | JRE | CC BY-SA 4.0 | edited body; edited title |
| Oct 6, 2020 at 12:16 | answer | added | Andy aka | timeline score: 8 | |
| Oct 6, 2020 at 12:07 | comment | added | user220456 | Could you please provide the formula you used to calculate the amplitude | |
| Oct 6, 2020 at 12:06 | comment | added | user220456 | Ok. So, there is not actual physical coupling right? Just coupling through air? | |
| Oct 6, 2020 at 11:56 | comment | added | Andy aka | The coupling is capacitive from the wiring in your walls. It might be 1 or 2 pF. As you get closer it might be tens or hundreds of pF Work out the impedance at 50 Hz and see that it forms a potential divider with the unconnected probe input impedance. That's it basically. | |
| Oct 6, 2020 at 11:45 | comment | added | user220456 | @Andyaka,Thank you. Just a question. Could you please provide me an image of the coupling as an answer and how the scope impedance (1Mohm in this case or 50ohms in other case) is affecting the coupling amplitude and frequency? | |
| Oct 6, 2020 at 11:41 | comment | added | user220456 | @Bimpelrekkie, could you please explain how a scope put in high impedance, will capture the coupling? Could you provide a image of how the coupling is happening here please? I have confused so much regarding this coupling and from where it is coupling | |
| Oct 6, 2020 at 11:41 | comment | added | Andy aka | How close you are to the source and how low the input impedance of the probe is. | |
| Oct 6, 2020 at 11:40 | comment | added | user220456 | @Andyaka, so, the probe is picking up the electric field form the air? What determines the amplitude of the 50Hz signal? | |
| Oct 6, 2020 at 11:38 | history | edited | user220456 | CC BY-SA 4.0 | deleted 144 characters in body |
| Oct 6, 2020 at 11:33 | comment | added | Ruperto | also it might help: The signal ground of the scope has to be clipped/connected closest to the signal ground of the DC signal. I hope it helps | |
| Oct 6, 2020 at 11:20 | comment | added | Bimpelrekkie | as I am measuring only DC signal with the oscilloscope, I could use the AC Coupling mode to remove this unwanted noise, ripple Uhm, no! AC coupling mode blocks the DC so that means if you measure a DC voltage with a small ripple: in AC coupling mode, you would only see the ripple. Looks like that ripple is 50 Hz, is the mains frequency in your country also 50 Hz perhaps? Also remember that a scope in put is high impedance so even the slightest coupling to mains wires will show up in the trace. | |
| Oct 6, 2020 at 11:19 | comment | added | Andy aka | What you are seeing is the electric field in the air around you caused by AC power wiring in your building. I estimate that your country uses 50 Hz. | |
| Oct 6, 2020 at 11:09 | history | asked | user220456 | CC BY-SA 4.0 |