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Here they define being floating as:

enter image description here

They mention Ungrounded = Floating.

But in another forum someone wrote:

The signal is consider floating when it does not have the same ground with your device. Earth has nothing to do with it. Earth is just another ground.

I'm a bit confused with the meaning of floating. Is the source floating in the below system?:

enter image description here

If it is not floating can you give an example of a system where the source ground is floating?

EDIT:

A floating source is connected to a differential amplifier. If I add a ground where the red arrow points the simulation circuit amplifies this signal very well. But if I don't use a ground the simulation corrupts.

In real do we really need a ground at that point or is this only needed in SPICE simulation? Because if I add a ground it is not floating anymore in the diagram. This is really confusing.

enter image description here

EDIT 2:

Even more confusion.

I always encounter such circuit topology for differential amplifiers:

enter image description here

Please notice that, above the input diff signals i.e the source and the diff. amplifier again share the same ground.

But when I look at the input terminals for a voltmeter or a diff. ended data acquisition board, there is no extra ground. There are inputs for -Vin and +Vin, but not GND.

Imagine now that I have a device which has an analog ground called AGND1 and this device has two differential outputs say 2V and -2V relative to its own AGND1. Now if I hook up its differential outputs to the voltmeter or a diff. ended DAQ board which has its own ground call it AGND2, we are facing a situation where AGND1 and AGND2 are not connected. But still these systems work as below:

enter image description here

As you see in a typical voltmeter or diff ended. DAQ board connection we don't connect two sytems grounds AGND1 and AGND2.

So the diff. amplifier topology I encounter uses grounds common but in real the grounds are not connected.

This is also very confusing since I dont know where my lack of knowledge comes from.

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    \$\begingroup\$ Any battery-powered device has floating grounds/signals. And if you want, for instance connect it so some wired communication channel with another device, you will have to establish a common ground (as long as there is no galvanic isolation on the transceivers, such as optical coupling). \$\endgroup\$ – Eugene Sh. Dec 15 '17 at 14:27
  • \$\begingroup\$ Lets say Source in my figure is a 9V battery. As you see one of the battery terminals(GND1) is then connected to AIGND. So one of the battery terminal is connected directly to the AGND of the measuring device. Are you sure we still call this floating? \$\endgroup\$ – atmnt Dec 15 '17 at 14:32
  • \$\begingroup\$ It should be noted that "floating" is a sort of colloquial term in electrical engineering and is not well defined. You'll hear it used in different ways with slightly different meanings as some of the answers point out. \$\endgroup\$ – kjgregory Dec 15 '17 at 20:20
  • \$\begingroup\$ @kjgregory See my question in edit with the example circuit. I ask: "In real do we really need a ground at that point or is this only needed in SPICE simulation?" Whats your opinion? \$\endgroup\$ – atmnt Dec 15 '17 at 20:39
  • \$\begingroup\$ In SPICE you probably need it to satisfy the simulator. In a practical system, it depends on many things. Like what IS the floating source? What are your design concerns? What is the environment for the circuit, etc. \$\endgroup\$ – kjgregory Dec 15 '17 at 20:51
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Floating is a voltage term and, like any voltage, it must have a reference.

That is: "Object A can be floating with respect to object B."

If your shown circuit, both grounds are wired together, so the source, V1, is NOT floating with respect to the amplifier.

However, if this was a battery operated widget, with no other connection, the whole thing is floating with respect to the ground under your feet.

schematic

simulate this circuit – Schematic created using CircuitLab

The following schematic on the other hand has a floating source.

schematic

simulate this circuit

BTW: Just to confuse you further, there is a whole other meaning of floating.

In the schematic below the two inputs A, and B are unconnected and we call that floating. In this case they are actually tied to ground through the pull-downs, but the left end is still considered as floating whether the pull-downs are there or not.

schematic

simulate this circuit

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  • \$\begingroup\$ As for ground, I think it is considered floating if not physically connected to Earth (well, that's the Wiki definition). Signal is floating is when does not have common ground (not necessarily non-floating). \$\endgroup\$ – Eugene Sh. Dec 15 '17 at 14:41
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    \$\begingroup\$ @EugeneSh. common reference would be more appropriate me thinks. GRound just confuses everyone. \$\endgroup\$ – Trevor_G Dec 15 '17 at 14:43
  • \$\begingroup\$ @Trevor Thanks, could you tell me in your "second" circuit example you said the source V1 is floating. What if that source were differential signalling, would it still be floating? Can you also give a circuit example with that, would be very glad! \$\endgroup\$ – atmnt Dec 15 '17 at 18:20
  • \$\begingroup\$ Please also see my edit where I expanded my question to a more confusing case. \$\endgroup\$ – atmnt Dec 15 '17 at 19:27
  • \$\begingroup\$ @user134429 in that instance the source is floating, but you also have no feedback around the op-amp so it is just a really bad comparator. \$\endgroup\$ – Trevor_G Dec 15 '17 at 19:37
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In my definition a circuit is "floating" if no current flows when I connect it to my ground or any other voltage relative to my ground, using one wire.

A circuit is not floating when I can make a current flow.

OK, I can apply 1 Million Volts and a current will flow. I'm talking about applying a voltage difference which will not damage any components or break isolation etc.

In your first picture the right source is indeed floating, if I connect one wire to it from my ground or any point in my circuit (the grounded source on the left) then no current will flow. There would only be the connection I just made so no current can flow.

In your second picture there are 2 connections between the source on the left and the amplifier at the right. This means these circuits are not floating in relation to each other.

I think your confusion comes from the statement Ungrounded = floating.

"Earth is indeed just a ground (reference). Imagine circuits A and B which are floating in relation to each other, they cannot share a ground (or any other connection).

If circuit A is connected to "earth" then circuit B cannot be connected to "earth" in any way. If circuit B was connected then it would not be floating anymore in relation to A.

Both circuits A and B can have a ground but they cannot share it or share any other connection.

My battery or solar powered calculator called circuit C is floating in relation to both circuit A and circuit B as it has no connections whatsoever to A nor B.

A simple trick to check if a circuit is floating is to draw a (dotted) line to separate the two circuits. The dotted line cannot cross any wires!

Like so:

enter image description here

Beware that a ground symbol could be used in more than one place and then it really is a connection as well although there's no visible wire.

I am unable to draw a dotted line to separate the source and amplifier in your 2nd picture. Therefore they are not floating in relation to eachother.

Edit

Confusion about this circuit:

enter image description here

Really, it is not that confusing!

This is only one circuit so it could float with respect to ground but does not have to. It really makes no difference as ground is just a reference point. The ground between the 2 9V batteries is a good point.

There is no need for any other ground symbols unless you want those to have a direct connection to that same ground (between the batteries).

If you add a ground to the - terminal of V1 you short it to the ground and disrupt the working of the circuit.

So no, there should be no ground added not in the simulator and also not in the real world!

But this circuit will not work well because there is no path for the base currents of the transistors. You have to set a commonmode voltage using resistors which will also supply that base current.

To solve that do this:

schematic

simulate this circuit – Schematic created using CircuitLab

DC voltage source V2 must be a voltage in the common-mode range that the amplifier can handle. You can also make V2 zero and remove it.

This solution preserves the diffential nature of the signals. You could also ground (or apply a DC voltage) on one side (see Trevor's answer) and that works but then the signal is not differential anymore.

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    \$\begingroup\$ 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 \$\endgroup\$ – Eugene Sh. Dec 15 '17 at 14:49
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    \$\begingroup\$ 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). \$\endgroup\$ – Bimpelrekkie Dec 15 '17 at 15:04
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    \$\begingroup\$ 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! \$\endgroup\$ – Bimpelrekkie Dec 15 '17 at 15:14
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    \$\begingroup\$ 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... \$\endgroup\$ – Eugene Sh. Dec 15 '17 at 15:15
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    \$\begingroup\$ 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. \$\endgroup\$ – Bimpelrekkie Dec 15 '17 at 15:32
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Current travels in loops. When one system is floating relative to the other, it means the loops are not on communcation (not onnected).

Consider a New York subway car. The big loop is from substation, to third rail, to car propulsion system, to running rails and back to substation. There is no way to insulate the wheels from the chassis of rhe car, so the chassis is part of the big loop. Sometimes a car will lose contact with the running rails due to snow, ice, rust etc. If there were any ground jumpers between the cars, the propulsion current would try to return via that ground jumper to a car with good contact.

There is also a control system that lets the motorman control each car's propulsion system, detect blocked-open doors, announcements, conductor's intercom, etc. etc. You really don't want propulsion current returning through the control wires. So this system is isolated, or "floating", from the propulsion current.


In your case, the other system is not isolated from yours, because it is tied by Q3 and Q4. This will pull the other system to about the potential of your system. Or vice versa, all a matter of perspective.

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  • \$\begingroup\$ Please also see my edit where I expanded my question to a more confusing case. \$\endgroup\$ – atmnt Dec 15 '17 at 19:28
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Idealy, you don't want ground there. If anything, you want to split your vsin into two separate additive inputs and put a ground in the middle of that. If you put a ground on either side of it as it stands, you'll end up with an amplifier that's not working optimally. This is because you're pinning one side of your inputs to the a single voltage. Most op-amps work better with differential inputs (one signal goes up while the other goes down). The vsin split into two with a ground in the middle of them is the correct way to simulate this.

schematic

simulate this circuit – Schematic created using CircuitLab

The reason spice is having troubles without you putting a reference ground in place is because it's seeing your op-amp as a simplified block diagram and isn't comprehending the internals of the op-amp. Through the op-amp, you are actually connected to ground but spice would never know because it's using a simplified model.

In the real world, you don't need a dual/split sine wave as ground is just a reference to measure voltage from. A single sine wave input into a BJT op-amp is probably fine without any sort of reference outside of the op-amp. If it were a MOSFET op-amp, I would most certainly recommend putting bleed-off resistors between the inputs and ground to prevent any floating signals from creating too high of a voltage on the op-amp inputs. Even on a BJT op-amp, I wouldn't be opposed to bleed-off resistors to further prevent unexpected or catastrophic occurrences.

To answer Edit 2:
While this may work. They may still be giving you a simplified diagram of what goes on in the voltmeter or DAQ. There should be some safety circuity in place to prevent extreme potential differences between devices that don't share grounds. This may be in the form of high resistance bleed off resistors or zener diodes on the DAQ or voltmeter. Without some kind of circuit protection, there's a good chance ESD would destroy the device.

The other thing to keep in mind here is that even though the devices aren't externally connected to the same ground, they are still connected between those two wires to eachother's grounds indirectly. Depending upon the transistor technology, this may be sufficient in real devices to prevent any sort of floating voltage issues.

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  • \$\begingroup\$ Please see my EDIT 2 on this matter. \$\endgroup\$ – atmnt Dec 16 '17 at 18:17
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Stop using the word ground and you'll be off to a better start. Refer to it as a common reference point. Blue is only blue by agreement. Same is true for electric circuits; i.e. ground is only ground by agreement. Floating, in short, is like schrodinger's cat; it's both postive and negative until you measure it but only at the TIME you measure it. Occasionally postive and occasionally negative and such is this post.

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