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I was going through a lecture regarding oscilloscope grounding methods and saw this image.

Assume I have the isolated widget powered by an external battery. Now, I am going to measure the signals using an oscilloscope.

Question : Why does the current not flow between the "Power In" rail in my widget and the "Mains earth ground" when I connect my oscilloscope alligator clip ground to the "Power In" Rail of the widget - as shown in red in my above image?

I believe that the Current follows through the path of least resistance. When the scope ground (which is connected to the earth's main) is connected to the +V in my widget, does the +V not have a least resistance path to ground (Earth in this case - assuming earth is lower potential that +V rail)?

I am having this confusion and not able to understand clearly why the current does not flow. Can someone please explain in simple terms on how to understand this?

I understand the when I connect the scope ground to any point on my widget, that point will be "Earth" referenced. But, I am not getting how the current does not flow between the +V and Earth when the above connection is made?

P.S. the lectures says that since the widget is isolated and has its own +V and negative, "there are no loops formed when the oscilloscope ground is connected to the +V and that's why current does not flow from the +V to the scope ground" - I am still not getting this. Any analogy or illustrations with simple terms explanation will greatly help me understand.

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    \$\begingroup\$ Because current must flow in a loop from its source and back to its source. There isn't really much more to say than this at this level...at least not without a lot more words to not describe very much without knowing more about your understanding of what things are. Do you also wonder why just connecting the + terminal of a battery to a circuit does not power it? Because if you are, then you should ask about that instead of this since that is closer to the true problem. If it does not confuse you then the case you posted about should not confuse you either since the answer is identical. \$\endgroup\$ – DKNguyen Sep 26 '20 at 6:25
  • \$\begingroup\$ I understand what you said. I am only not getting the fact that why does current not flow through the path of least resistance. +V connected to ground (Earth over here) is the path of least resistance. Why does it not flow in this way? If your statement - Current must flow from source and back to source again - is reasonably logical , then my question is also reasonably logical - "Why does the current from +V not flow to the Earth connected by the probe ground when it is been provided a least path of resistance to the ground." \$\endgroup\$ – Newbie Sep 26 '20 at 7:02
  • \$\begingroup\$ @Newbie its because the V+ in the battery powered circuit is only positive relative to the battery GND. To the rest of the world it's floating/undefined until you connect the oscilloscope gnd. Now V+ of the Battery battery powered circuit is equal to the oscilloscope gnd. Still no current flows. you just have defined voltage potentials now. The Battery battery powered circuit gnd is not equal to oscilloscope gnd but negative (oscilloscope gnd minus V+). So if you now measure at the Battery battery powered circuit voltages will be shown as negative values. \$\endgroup\$ – schnedan Sep 26 '20 at 7:10
  • \$\begingroup\$ Thank you. I am starting to get some understanding from your comment @schnedan. Could you please provide it as an answer briefly in little more simpler terms with the number example, please? I am still finding some difficulty as to why current does not flow. I got that we defined the voltage potentials. But current flow, not able to understand \$\endgroup\$ – Newbie Sep 26 '20 at 7:37
  • \$\begingroup\$ And also, assume +V is 9V and as I connect the V+ to the oscilloscope ground and remove after sometime, does the V+ still have the same potential (9V) as before? \$\endgroup\$ – Newbie Sep 26 '20 at 7:43
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its because the V+ in the battery powered circuit is only positive relative to the battery GND. To the rest of the world it's floating/undefined until you connect the oscilloscope gnd. Now V+ of the Battery battery powered circuit is equal to the oscilloscope gnd. Still no current flows. you just have defined voltage potentials now. The Battery battery powered circuit gnd is not equal to oscilloscope gnd but negative (oscilloscope gnd minus V+). So if you now measure at the Battery battery powered circuit voltages will be shown as negative values

So first have a look to the oscilloscope - its galvanic isolated (but neutral and secondary side GND might be coupled). If GND is floating, wherever you connect it, it is tied to the potentials of the device under test.

second is your example in 3 steps starting with no connection to full connection with GND and probe:

example

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  • \$\begingroup\$ Oscilloscope ground IS connected to earth (both in normal oscilloscopes and in the OP's picture) and you are saying it is galvanically isolated. That is incorrect. \$\endgroup\$ – Andy aka Sep 26 '20 at 10:10
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    \$\begingroup\$ That is common to most bench oscilloscopes, but there are PC scopes with just a DC jack, battery powered mobile scopes... You cant say its general true for all scopes. Also it does not change anything how the oscilloscope would display the voltage nor that the GND of the battery powered circuit would be minus to the oscilloscope gnd \$\endgroup\$ – schnedan Sep 26 '20 at 10:35
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    \$\begingroup\$ But the op's picture shows it is earthed. \$\endgroup\$ – Andy aka Sep 26 '20 at 10:51
  • \$\begingroup\$ @Andy aka See Transistors comment after his answer; There is a conflict between the image and text (OP stated that the lecture asserted that they ARE isolated, as given conditions) The ground symbols are different from each other, adding uncertainty to the diagram. Ground symbols are not always consistent (I've seen vary by manufacturer, country, or vintage, 'draftsman', etc. I've even seen inconsistencies on the same print from one page to another, or one revision to another). The OP wants clarification, If the lecture statement is correct, then the grounds are different, not connected. \$\endgroup\$ – troubleshooter Sep 28 '20 at 15:05
  • \$\begingroup\$ Ahem @troubleshooter what are you talking about? It certainly isn't what I'm talking about. \$\endgroup\$ – Andy aka Sep 28 '20 at 16:32
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enter image description here

Figure 1. Note that the author of the schematic has (correctly) used two different symbols for (1) EARTH and (2) GND.

Since (1) is not connected to (2) there is no closed loop for current to flow.

(I would argue that the symbols should be swapped as the origin of symbol (2) represents metal plates buried in the Earth and symbol (1) represents connection to a chassis.)

enter image description here

Figure 2. Various earth / ground symbols. Source: Ground - Earth - chassis.

"Ground" is often used as a reference point from which other voltages are measured. When various parts of the circuit are isolated from each other best practice is to use a different ground symbol for each section to make this clear.

Finally, if the circuit common is connected to earth (as in The Earth) we take that as a real zero volts, in much the same way as the surveyor might use sea-level or some local reference as an absolute reference.

Building rocket

Figure 3. The building on the ground and shot off into space. In the space situation (electrically isolated) we can call any floor the 'ground' floor.

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    \$\begingroup\$ The answer to your question hinges on whether the ground symbols in your diagram are eventually connected to the same grounds somehow (try ohming from the one ground to the other!). If they are connected, then connecting the scope ground to the + WILL short the battery (depending on the ground quality/resistance). If not, then it will 'float' as others have stated. There is confusion on what your diagram actually is 'calling' grounds and chassis ground return. \$\endgroup\$ – troubleshooter Sep 27 '20 at 16:18
  • \$\begingroup\$ Yes and no, the OP's question says, "P.S. the lectures says that since the widget is isolated ..." so there can't be a connection between the two. I'm also being careful to use the term "earth" to differentiate between a real ground and an arbitrary ground on the isolated circuit. \$\endgroup\$ – Transistor Sep 27 '20 at 16:43
  • \$\begingroup\$ I agree with your comment too. But it looks like what the 'lecture states' and what the lecture image shows, contradict each other. If the material showed the image as the OP posted it, with the green lines and the word "Earth", and the symbol on the 'widget' being symbol most often used to indicate an 'earth' ground, then we could take the statement as correct (as you point out), and the image in error, or at least misleading... \$\endgroup\$ – troubleshooter Sep 28 '20 at 5:52
  • \$\begingroup\$ We could speculate that the author (of the lecture) was encouraging students to look at written documentation to define the conditions, etc; and not rely on what they 'think' something is saying or showing. (or maybe so they don't just 'look at the pictures'). But I suspect the reality was just a poorly illustrated image, or they just are using their ground symbols differently than customary. \$\endgroup\$ – troubleshooter Sep 28 '20 at 6:10
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Take a regular 1.5 volt battery that is unconnected to to anything. You'd expect it to hold its charge and not supply any current - because there is an open-circuit across it. Now, if you earthed the negative terminal of the battery you'd expect the positive terminal to be at +1.5 volts relative to earth. And, there's still no closed-circuit for current to flow and, the battery retains its charge and still produces 1.5 volts.

However, if instead of earthing the negative terminal you earthed the positive terminal. Now the negative terminal would be at -1.5 volts relative to earth. But still there is no current flowing because there is no load on the battery.

So, repeat this thought experiment with a light load of 10 kohm across the battery - is there any difference to what I wrote above?

Now repeat with a more complex circuit directly attached across the battery. Providing that the more complex circuit doesn't have any extra connection to something that might be earthed, should you expect anything to change? Will current flow into the earth?

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  • \$\begingroup\$ I still don't get it. Why doesn't current flow to the earth ground - which is at a Lower Potential than +V (1.5V) and low resistance than its actual widget GND? I am connecting the ground when the board is drawing power from the battery \$\endgroup\$ – Newbie Sep 26 '20 at 8:28
  • \$\begingroup\$ The negative terminal of the battery is at a lower potential than the positive terminal but current only flows when you connect a conducting circuit across the battery. There is nothing special about earth other than it is at a potential that is loosely shared amongst the local population. What if the battery were on the international space station? Earth would be the chassis of the spacecraft and much more conductive than actual earth. \$\endgroup\$ – Andy aka Sep 26 '20 at 8:45
  • \$\begingroup\$ @Newbie Because the oscilloscope ground is not the same "ground" as the ground of the widget, which is actually the v-. You can't empty a battery by connecting v+ to earth. (Unless v- is also connected to earth) \$\endgroup\$ – Fredled Sep 28 '20 at 20:05
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OK, lets simplify this; If the circuit was in a metal box, say a transistor radio, on a plastic table, (no scope connected).

If you took a voltmeter, measured voltage from any part of the radio's circuit, to let's say - a water pipe (or anything in the room but not on the radio), you would read nothing. The battery, and circuit, does not 'see' the outside world, or earth ground. NOW, if you connect the scope ground to ANY part of the radio, lets just say--- the positive battery post. NOW you have introduced a connection to ground (of the outside world, and your AC powered test equipment most importantly). The outside world is tied at that one point now.

Everything in that radio will now have a voltage with respect to earth ground, (or anything electrically powered from AC or grounded in some way).

That point is now 'earth ground voltage'. Everything else connected in that circuitry will now have some sort of reading. The readings to each other will still be the same within the circuit, but you will also be able to get readings of how much it is 'above' or 'below' the point you have now grounded; in this case, everything will read 'below' earth ground (negative).

In this example, with scope grounded to the positive, no current flows if it is the ONLY ground point connected. The negative battery power cannot get to the positive, except through the radio's circuit as normal. ONLY if you ground the radio chassis in some way; say you accidentally put a second scope probe ground on the battery NEGATIVE, or if you put the box on a metal grounded table (or touched the radio chassis and the scope case at the same time); THEN the battery power has a 'ground' route back to the positive; you have a short (or ground loop). [think 'antenna' or cable TV feed that is grounded!]

So in this example you connected your scope probe to battery '+', then the battery NEGATIVE will NOT read '0'volts, but will read minus x volts (the batteries voltage). And everything in between will read somewhere between 0V and negative battery's voltage. (Kind of upside down, so to speak).

If you had connected the scope ground to the battery negative, then everything would read 'right side up' as you are accustom to, and also to readings taken with no grounds attached as with a voltmeter and radio alone.

If you grounded your scope 'in between' somewhere in the circuitry; then your scopes readings would be some 'above' and some 'below' ground level (DC-wise). Remember, AC waveforms are all going to be the same. Technically this would be referred to as DC-offset considerations.

SUMMARY: You won't have any shorts (or ground-loops) as long as you ONLY have ONE defining ground connection (and chassis is NOT grounded, that would be TWO ground sources in this example). As long as there is only one connection point* (between the two 'worlds'), or no connection, there will be no short circuit. *Obviously, you can have more than one ground, as long as they ALL connect to the same common place electrically speaking.

Note; the metal box could be connected to battery negative (typical in traditional circuitry for shielding and current return), battery positive (I repair vintage Germanium transistor circuits where all the voltages are reversed, uses negative 'hot', and positive 'ground'), or box could be connected to nothing. [think of a vintage Dodge with positive ground battery; it didn't short out because it's bumper touched a parking pole or if parked next to a negative grounded car! Only if you tied an external ground into their circuits, or tied their circuits together, would you get fireworks. PS, You CAN connect the Dodge's negative 'HOT' to the negative of a 'negative grounded' car, and it's positive 'ground' to positive; just don't let the bumpers touch!]

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