Timeline for Confusion about the meaning of floating
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2017 at 22:21 | comment | added | Bimpelrekkie | Yes diff. signals are indeed used to be able to suppress common mode disturbances and noise which can be in the ground connections but also be picked up by the signal lines. But it will be picked up the same by both signal lines so if we only measure the difference voltage then the disturbance is eliminated! Read here: electronics.stackexchange.com/questions/92813/… | |
| Dec 16, 2017 at 22:10 | comment | added | user1245 | @Bimpelrekkie I see. When grounds are shared that 1000V common mode is dangerous as you say. Voltmeter does not have this issue as you also mentioned. So can we say that the reason that differential amplifiers are used in daq systems for diff. signals with connecting source and daq analog grounds is because to reject the noise with respect to a common ground? If they were designed as how voltmeters measured, would they have problem to reject common mode noise since there is no common ground? I hope I understand what you mean. | |
| Dec 16, 2017 at 21:52 | comment | added | Bimpelrekkie | But that differential amplifier does see that +1000V because it is grounded. It will blow up if you measure +999V and +1001V. That 1000V is the common mode voltage and all amplifiers have a limited common mode input voltage range. So a differential amplifier does not behave in the same way as a hand-held voltmeter. | |
| Dec 16, 2017 at 21:52 | comment | added | Bimpelrekkie | @user134429 You should realize that "ground" is just a reference point for us, humans, or for simulators. Inside that voltmeter, the ground will be local and only connected to the COM input of the meter. Indeed there is no other connection. So that voltmeter is floating, if you measure between -1V and +1 V or +999V and +1001V, the voltmeter will show 2 V in both cases. It does not "see" that 1000V difference to ground. | |
| Dec 16, 2017 at 19:04 | comment | added | user1245 | @Bimpelrekkie A voltmeter has only two input terminals. Imagine a device has two diff. outputs -Vin=-1VDC and +Vin=+1VDC and this device has a circuit ground call it AGND1. And when one hooks up +Vin and -Vin to the voltmeter probes, the voltmeter will show 2VDC. And the voltmeter is a circuit and it has also a ground lets call AGND2. Now AGND1 and AGND2 are definitely not connected. Isnt it how things are normally done? Where am I wrong? | |
| Dec 16, 2017 at 18:43 | comment | added | Bimpelrekkie | @user134429 You seen to think that AGND1 and AGND2 are not connected, well usually they are connected. Usually as in a star grounding scheme (Google: star ground to learn more). So AGND1 and AGND2 are local grounds and they are connected on a higher level. If they would not be connected then you can expect all sorts of problems. Your lack of knowledge comes from lack of experience, I advise you to see how things are normally done and try to understand why they are done that way. | |
| Dec 16, 2017 at 18:06 | comment | added | user1245 | @Bimpelrekkie Please see my Edit 2, I would be glad to have your comment on this. | |
| Dec 16, 2017 at 12:31 | history | edited | Bimpelrekkie | CC BY-SA 3.0 | added 644 characters in body |
| Dec 16, 2017 at 12:24 | comment | added | Bimpelrekkie | @user134429 I added a section to my answer. | |
| Dec 15, 2017 at 22:58 | history | edited | Bimpelrekkie | CC BY-SA 3.0 | added 960 characters in body |
| Dec 15, 2017 at 19:27 | comment | added | user1245 | Please also see my edit where I expanded my question to a more confusing case. | |
| Dec 15, 2017 at 15:32 | comment | added | Bimpelrekkie | Yes static charges are ESD and we all know what that can do to semiconductors. But not PCB traces. Yes I think the power of a static discharge is too low to burn a PCB trace. To burn a PCB trace with a static discharge you'd need a very big device to hold your charge. A capacitor maybe? But then there's another plate, perhaps with a path to ground, and that's your loop so not a static discharge anymore. | |
| Dec 15, 2017 at 15:15 | comment | added | Eugene Sh. | That cannot happen from charges equalizing I'd say. - why not? You think the power would be too low? Static discharges known to be an enemy of electronics. Yet maybe the PCB traces should not be such sensitive... | |
| Dec 15, 2017 at 15:14 | comment | added | Bimpelrekkie | It even happened to burn the ground trace when connected two devices with RS-232 That cannot happen from charges equalizing I'd say. So there was some other connection as well (to close the current loop). That path probably included mains voltage or some other power source and then what you describe is possible. In that case the circuits were not really floating! | |
| Dec 15, 2017 at 15:11 | history | edited | Bimpelrekkie | CC BY-SA 3.0 | added 254 characters in body |
| Dec 15, 2017 at 15:07 | comment | added | Eugene Sh. | Good question... I would think about some static charge accumulated over time. But from experience, I have encountered some pretty high differences. It even happened to burn the ground trace when connected two devices with RS-232. | |
| Dec 15, 2017 at 15:06 | history | edited | Bimpelrekkie | CC BY-SA 3.0 | added 151 characters in body |
| Dec 15, 2017 at 15:04 | comment | added | Bimpelrekkie | Sure but how is that potential difference created? Usually it is charge-buildup or capacitive coupling. If the circuits are really properly floating that voltage difference will be eliminated when you measure it as the charges are equalized (assuming you're using a voltmeter with a finite impedance). | |
| Dec 15, 2017 at 14:55 | comment | added | Eugene Sh. | When you have two floating devices the potential difference between their grounds might be high, but not the voltage in each one of them. Unless you directly measure it you can't even know... | |
| Dec 15, 2017 at 14:52 | comment | added | Bimpelrekkie | Yes but then you're using too much voltage which damages the components and makes the circuits instantly non-floating anymore. You should have use a more sane voltage. | |
| Dec 15, 2017 at 14:50 | history | edited | Bimpelrekkie | CC BY-SA 3.0 | added 99 characters in body |
| Dec 15, 2017 at 14:49 | comment | added | Eugene Sh. | I believe this is a dangerous definition. When you connect two floating device, there is a chance the current will flow and pretty high one (yet instantaneous) and burn things right away | |
| Dec 15, 2017 at 14:47 | history | answered | Bimpelrekkie | CC BY-SA 3.0 |