1
\$\begingroup\$

I know that it's usually perfectly possible to connect two grounds, and that voltage is measured against it's ground, but this got me wondering about batteries. I'm assuming somewhere in my train of thoughts I'm missing something, but I'm having difficulty figuring out where. So, let's go with an example of what I think should happen.

  1. If I plug a battery (say my laptop) in at home, charge it and unplug, from what I get the ground of the battery remains at the level at which the ground in my house was.
  2. I take my laptop to work, whole different city where the ground is probably at a different level from at home. I would guess that this difference would lay anywhere between -10 and 10 volt, but might be enough to fry a laptop. I don't really know.
  3. I plug in my laptop. The moment I plug it in, the ground in my battery equalizes with the ground at the office. This means that a voltage of about 10V shortly gets short-circuited, and my laptop is busted.

I know this doesn't happen, so I clearly got something wrong, but what exactly? And if moving a battery from one ground to the other isn't a risk, then why is connecting two grounds ever a risk?

\$\endgroup\$
3
  • 6
    \$\begingroup\$ Once you unplug it, the concept of level ceases to exist. (some static electricity may remain, but that's negligible) \$\endgroup\$
    – Agent_L
    Oct 20, 2016 at 10:50
  • 3
    \$\begingroup\$ (1) is fantasy. (2) most likely the voltage is smaller, but the current available might be huge. (3) When you plug in, you only create one side of the circuit. Where does this convoluted logic emerge from - I'm intrigued... \$\endgroup\$ Oct 20, 2016 at 12:52
  • \$\begingroup\$ You can imagine a really tiny capacitor between your laptop and the Earth. When you plug it in this capacitor discharges, and because it's really tiny, not much happens. \$\endgroup\$
    – user253751
    Oct 21, 2016 at 5:25

4 Answers 4

19
\$\begingroup\$

Voltages do not have a "universal" reference frame. A voltage is also known as a "potential difference", meaning that a voltage is just the difference between two points. So they can reference to any other voltage.

Engineers can work on a 400,000V cable if they don't connect to ground; indeed they are sometimes in a cage hanging from the overhead cable, which was put there by helicopter.

So when you connect a 10V battery to a ground point, it says, "okay, I'm now at 10V relative to this ground", and you disconnect it and then connect it to some other ground point, it says, "now I'm 10V relative to this ground" and so on. It has forgotten the first ground. The battery cannot remember the old voltage reference, indeed it never "knew" it, it just carries on generating the same potential difference relative to whatever one of its terminals happens to be connected to. You can take the same battery, charge it to +10 on its positive terminal relative to ground, then disconnect it, connect the positive terminal to ground, and the other one is now -10.

Which is a bit like magic, but it's really just boring old physics.

\$\endgroup\$
6
  • 3
    \$\begingroup\$ The pedant in me wants to challenge your initial assertion somehow. Clearly everything can be referenced to everything else in a static field. \$\endgroup\$ Oct 20, 2016 at 12:57
  • 1
    \$\begingroup\$ @SeanHoulihane hard to challenge it as it currently stands IMO; there ain't a single universal reference frame in contemporary physics. Since the entire concept of potential energy in a field requires a well-defined field in the first place; you'd have to define your "everything" first, losing the "universality" by then. \$\endgroup\$
    – user20088
    Oct 20, 2016 at 14:15
  • 1
    \$\begingroup\$ @vaxquis Agreed, partly there is some truth there, but equally in every reference, all points have a potential. So its not wrong, just not the whole story. \$\endgroup\$ Oct 20, 2016 at 14:54
  • 1
    \$\begingroup\$ What Ian is saying is that you cannot just say "here there are 10V". What you must say is "here there are 10V with respect to there". And 'there' is arbitrarily. \$\endgroup\$ Oct 20, 2016 at 19:38
  • 1
    \$\begingroup\$ @SeanHoulihane I wasn't trying to tell the whole story, I was trying to provide a useful way of understanding in a few paragraphs. We often leave out elements of truth to enhance utility. For instance, every wire is a transmission line to some degree, but in lots of cases you can ignore that, and introducing that would just increase complexity and confusion. I was just trying to get across that a voltage is always a difference between two points, and you can move them around. Hence for instance, the apparent magic of charge pump circuits. \$\endgroup\$
    – Ian Bland
    Oct 20, 2016 at 20:42
14
\$\begingroup\$

As soon as you disconnect your laptop from one ground, its ground potential can go anywhere, not just +/- 10v.

The only thing controlling the voltage with respect to any other ground of the isolated common terminal (aka laptop 'ground') is its capacitance and leakage path to the other ground. This capacitance can be very low, 10s or 100s of pF is common, so very little charge is needed to get large voltage differences. If the leakage is poor, say it's a dry day, then it is very easy to get and keep kV or even 10s of kV (yes, I do mean thousands of volts) between your isolated laptop and another ground, just by walking across a carpet.

It's this large potential difference that sometimes gives you a 'belt' when you touch a filing cabinet or a car, after walking across to it, on a dry day. 10v difference is unlikely to fry your laptop, but kV might, depending on where it hits.

If the spark goes to the laptop ground, then all will be well, it can handle it. If it goes to USB conductors, then you will probably be OK. These have been 'hardened' against ESD (ElectroStatic Discharge) as people are expected to plug in things that have been charged up by walking across a carpet and so are at different ground potentials. The RJ45 port should be OK as well, this is transformer isolated.

As a matter of course, I always try to use myself to ground anything I'm carrying to the local conductors before plugging in. If I get a belt, then I feel the effort was worthwhile.

\$\endgroup\$
2
  • \$\begingroup\$ +1 for mentioning capacitances (which are easy to forget), and for mentioning that the real reason the laptop doesn't fry is because engineers planned it that way =) \$\endgroup\$
    – Cort Ammon
    Oct 20, 2016 at 18:52
  • \$\begingroup\$ Talking of static, I was thinking about a discussion about it here some days ago, and how it really doesn't seem to be as much of a problem as in e.g. the 1970s when it seemed you could kill CMOS just by looking at it. Then I thought back to that time, when I was a child, and how for a while there was apparently no textile but nylon; clothes, sheets, furnishings, carpets... \$\endgroup\$
    – Ian Bland
    Oct 20, 2016 at 20:47
5
\$\begingroup\$

why is connecting two grounds ever a risk?

This can be a risk - making a galvanic connection between ground A and ground B can pass serious levels of current but, this is usually referred to as ground fault currents and is typical in factories with big AC machines having significant ground currents - this ground current will find its way to the lowest potential in the factory and, if you help this flow of current by connecting parallel grounds, some of that current will pass through your equipment and potentially cause problems.

Regarding the battery comparison, chargers usually produce galvanically isolated outputs so there isn't really a ground connection. However, due to leakage capacitance in the charger, voltages up to several tens of volts may be present on the battery 0 volt point relative to real ground/earth of the building. This voltage will be AC and rising and falling as per the AC mains frequency.

So, you disconnect the laptop and, providing there is no discharge path to "real earth" through your fingers/body, the internal 0 volts of the laptop may be left at a significant potential - this potential is fairly random to predict - at what point in the AC cycle the disconnect happened is somewhat arbitrary of course.

However, once disconnected, the laptop can be reconnected to ground in another building (for instance) and the only thing you would notice (if you were armed with a perfect means of looking at the current that flows), is a surge lasting a few microseconds as the laptop acclimatized to the new ground potential.

\$\endgroup\$
2
\$\begingroup\$

Voltages only make sense in the context of an electric circuit. When you disconnect your laptop from the charger at home, that circuit ceases to exist (well, if you want to get technical about it, there is a tiny capacitance between the metal parts of the charger and the metal parts of the laptop, which quickly tends toward zero as the distance increases).

Once you plug your laptop at the office, the inverse process takes place: first the capacitance between your laptop and the charger grows, charged to some initial voltage. As you bring metal parts closer, that capacitance increases, and the voltage it was charged to decreases proportionally. When metal parts finally come in contact, the capacitance undergoes a discharge (that's why most connectors are designed so that grounds come into contact first).

At this point you have a new circuit composed of your laptop and your office charger, both at the same potential. The voltage between the laptop and your first charger is completely irrelevant now: the capacitance between the two is reduced to nothing, and they are separated by many meters of air (the shortest path being from charger 1 to Earth and from Earth to charger 2), so no sane amount of voltage could result in any measurable current. From practical point of view, there is simply no electrical circuit from your home to the laptop in your office.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.