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I'm struggling with the concept of floating voltages.

For a grounded supply I can connect one side to an effectively infinite ground and get full power. It may vary depending on its construction, but if the grounded supply provides X volts then it is necessarily true that:

[A] (V+ to G) - (V- to G) = X, correct?

Now suppose I want to "ground" a floating supply. My understanding is that the supply's V+ and V- separated behavior with respect to some "ground" of practically infinite capacitance must be characterized by the source of the voltage:

  1. A floating driven power supply cannot send power through a path that does not directly connect its V+ and V- terminals, as roughly explained here.
  2. A chemical power supply, like most conventional batteries, cannot send power that doesn't connect its V+ and V- terminals, because the driving chemical reaction will not occur without electron transport exactly matching the needs on both terminals simultaneously.
  3. But a capacitor should be able to send power from either its V+ or V- terminals, right? It depends on the charge of the G, but not only does the relationship [A] above hold, but also the capacitor can dump all of its power into an arbitrary ground if V+ and V- are independently and separately connected to that ground. I.e., there is no such thing as a "floating" capacitor?

(Evidently I'm assuming that a "floating" supply is defined to be one in which power only flows between V+ and V-. Is that accurate? And are there any other "floating" power supply categories than #1 and #2.)

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  • \$\begingroup\$ The ground reference section of this answer may help. \$\endgroup\$ – Transistor May 14 '16 at 15:59
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To add to Barry's answer I'd also make these points:

A floating driven power supply cannot send power through a path that does not directly connect its V+ and V- terminals, ... A chemical power supply, like most conventional batteries, cannot send power that doesn't connect its V+ and V- terminals,

Within lumped circuit analysis, no device can deliver power to any other device that isn't part of a circuit connected to two of its terminals.

But a capacitor should be able to send power from either its V+ or V- terminals, right?

Current always flows in complete circuits. This means that whenever there is current flowing in one terminal of a capacitor, an equal current is flowing out of the other terminal. Because of this you can't say "the V+ terminal delivered power" or "the V- terminal delivered power".

The capacitor delivers power to the rest of the circuit when current flows out of its more positively charged plate and in the more negatively charged plate. The capacitor absorbs power from the rest of the circuit when current flows in to the more positively charged plate and out of the more negatively charged plate.

It doesn't matter which terminal is labeled "+" or "-", it matters which one is at a more positive potential and which way the current is flowing at the instant when you are measuring the power flow.

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  • \$\begingroup\$ To clarify: Charge a capacitor to xV. It has positive charge on one plate and negative on another. Now connect V+ through load to a "ground." Current flows until charge on V+ equals ground charge. That's "power from V+" right? Capacitor will now show voltage of x/2. Now connect V- through load to the same "ground." Current flows until V- charge equals ground charge. We've "taken power from V-." Right? If not, why not? \$\endgroup\$ – feetwet May 14 '16 at 16:39
  • \$\begingroup\$ When you wrote "Now connect V+ through load to a ground..." you are not analyzing it correctly. No current will flow in that situation unless the V- terminal is also connected and there is complete circuit path for the current to flow through. Also, "Current flows until charge on V+ equals ground charge" is incorrect. The V+ potential will equalize with the ground potential, but this does not necessarily involve any current flowing (if V- is not connected). The ground charge is considered infinite, so there is never a possibility of equalizing charge. \$\endgroup\$ – The Photon May 14 '16 at 16:41
  • \$\begingroup\$ I think I'm getting close to an understanding: A capacitor stores a charge differential between V+ and V-. These can literally be counted as a difference in electrons between the two plates. Whatever G's electrostatic value, it can't equal the charge on both plates. It's a third virtual plate with infinite electron capacity. So if connected to G one at a time either V+ or V- (likely both) will independently send or receive electrons to match the charge on G, right? That's a current. Are you saying that's true but that the current carries no power if done separately from each terminal? \$\endgroup\$ – feetwet May 14 '16 at 16:49
  • \$\begingroup\$ You are so confused about this that your question doesn't even really make sense. Maybe bring this question to chat? \$\endgroup\$ – The Photon May 14 '16 at 16:51
  • \$\begingroup\$ Conclusion from chat: I had forgotten that you can't "push or pull" electrons on one side of a capacitor; it will resist such action unless offsetting electrons are pulled or pushed on the opposite terminal. Thanks! \$\endgroup\$ – feetwet May 14 '16 at 17:21
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The term "floating", be it a power supply, battery or other circuit component, simply means that neither of the component's two leads are grounded. To draw power from a floating device does require that the load be connected to both leads of the device. This is true whether the device is floating or not. The amount of current (power) that can be drawn from a power supply does not depend on whether the supply if floating or not. It only depends on the load connected to the supply's 2 output leads. Grounding floating supplies is done for safety reasons and/or circuit requirements but not to allow power to be drawn from the supply.

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Voltage is relative (to some potential). Usually for simplicity we pick certain potential and call it Ground. It's possible, that in system several grounds will be present. And they are not necessarily connected. If they are not connected, they are called floating.

The term itself refers to anything that is not "constrained" to any known potential in the system. For example floating input is one that it's voltage may be anything and you can't guess. This is why sometimes you will see pull down or up resistors, so instead floating node you will get known voltage.

To your question: capacitor indeed may be a floating supply. Actually, there is at least one very common case when it is used this way: boost capacitor in switching systems is exactly that.

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