DC-DC converter with floating ground

[electrodacus]

I'm just curious if anyone knows a way the circuit in the diagram below can either have a constant V2 larger than V1 or constant output current different from input current.
To not be accused of setting up a trap
I think that is not possible to have a constant 30V output see question A)
,but it is possible to have a V2 fluctuation between 5V and 30V so Yes for question B) 



The green box can contain any circuit inside no necessarily just a DC-DC converter. It can not contain any energy when connected to the circuit.
It has just two wires connecting to that box so box is in series with the load (100Ohm resistor as example) and the 10V battery



A) Is it possible for V2 to be a constant 30V
B) Is it possible for V2 to peak to 30V multiple times while V2 never drops below 5V ?

[magic]

Input and output currents surely are equal because all current entering through the input must exit through the output since there are no other external connections...

Maybe you meant to connect the negative terminals to ground? If so, this looks like an ordinary switching boost converter - obviously they are possible and there is lots of them, and there is lots of information how they work.

[electrodacus]

Input and output currents surely are equal because all current entering through the input must exit through the output since there are no other external connections...

Maybe you meant to connect the negative terminals to ground?

I agree with you but looking to see if this is a common opinion.
I did not meant to connect the negative terminals to ground. The ground on that DC-DC converter is floating.

[SiliconWizard]

I'm more used to having the signal flow drawn from left to right, but whatever.

The OP's title explicitely says 'with floating ground', so I'm assuming the drawing matches that and is not a mistake.
Such an arrangement would normally just yield Vout = Vin - some drop, typically a diode drop in the boost converter.

[electrodacus]

I'm more used to having the signal flow drawn from left to right, but whatever.

The OP's title explicitely says 'with floating ground', so I'm assuming the drawing matches that and is not a mistake.
Such an arrangement would normally just yield Vout = Vin - some drop, typically a diode drop in the boost converter.

You are allowed to make any modifications you want inside that DC-DC converter short of adding a battery.

[magic]

Since output current equals input current, output voltage can't be higher than input voltage. Otherwise output power would be higher than input power, which is impossible without some energy source is the converter. It doesn't have to be a battery, though - a hamster wheel generator would work just as well

[electrodacus]

Since output current equals input current, output voltage can't be higher than input voltage. Otherwise output power would be higher than input power, which is impossible without some energy source is the converter. It doesn't have to be a battery, though - a hamster wheel generator would work just as well

no hamsters allowed.
What about an inductor or capacitor ? There are already plenty of those in a DC-DC converter.

[magic]

A charged capacitor connected between the input and the output could provide a voltage boost, for a short time.

Otherwise, it's an obvious consequence of conservation of charge and conservation of energy that such circuit cannot do anything useful.

Is this a free energy thread?

[electrodacus]

A charged capacitor connected between the input and the output could provide a voltage boost, for a short time.

Otherwise, it's an obvious consequence of conservation of charge and conservation of energy that such circuit cannot do anything useful.

Is this a free energy thread?

Yes you are correct. A circuit can be setup so that internally a capacitor is charged then connected across the input and output.
No it is not a free energy thread.

[SiliconWizard]

By transforming a quadripole into a dipole, it will be hard to get anything else than |V2| <= |V1|.
Maybe we can try leveraging some quantum effect.

[electrodacus]

By transforming a quadripole into a dipole, it will be hard to get anything else than |V2| <= |V1|.
Maybe we can try leveraging some quantum effect.

So you are saying that there is no way to get V2 above V1 even for a short period of time ?
The DC-DC converter (jut a name can be any configuration you want i side) can supply itself internally so instead of just wasting that energy as heat it can charge a capacitor that then can connect in series with the battery.
magic already mentioned that.

[SiliconWizard]

By transforming a quadripole into a dipole, it will be hard to get anything else than |V2| <= |V1|.
Maybe we can try leveraging some quantum effect.

So you are saying that there is no way to get V2 above V1 even for a short period of time ?

No, for a transient, the question is obvious and answered already by magic. Just put anything in series which has a charge. It will just be extra energy that has been stored previously though.
But once this charge is depleted, game over. So what does that bring to the table?

[electrodacus]

No, for a transient, the question is obvious and answered already by magic. Just put anything in series which has a charge. It will just be extra energy that has been stored previously though.
But once this charge is depleted, game over. So what does that bring to the table?

Say you have 10V battery and load is just a simple resistor in parallel with a low value capacitor.
The box called DC-DC converter has an actual DC DC converter inside that charges a capacitor to say 20V
Input to DC-DC converter is the battery positive directly and GND is trough the Load.
After the capacitor is charged it is disconnected from the internal DC-DC converter and connected across the input and output of the green box. So you have 10V battery in series with a 20V capacitor and then with the load that is a resistor in parallel with a small value capacitor.
The logic can be done by a micro controller and moving the capacitor can be done with mosfets.

Of course this cycle can be repeated say once every few seconds or whatever is needed to charge and discharge that capacitor and during the charge period the Load voltage V2 will be quite a bit lower than V1 but then for some other period V2 can be larger than V1.

[SiliconWizard]

You can't do that if your only return path is through the load. As shown in the diagram.

[electrodacus]

You can't do that if your only return path is through the load. As shown in the diagram.

You mentioned something about a diode voltage drop in your initial posts.
So if there is a voltage drop between input and output and a current flow will that not result in energy being dissipated as heat inside the DC-DC converter box mostly on that diode ?
Can you not just take that energy and charge a capacitor instead of wasting it as heat ?
The return path is trough the load.

I think you are imagining that the box is just a DC-DC converter. I mentioned that you can have anything you want in that green box just not a battery or a hamster.

[electrodacus]

You can't do that if your only return path is through the load. As shown in the diagram.

Here I made a diagram that is maybe better than a description.
There is a DC-DC converter inside the box with positive input from battery+ and battery- trough the 100Ohm load.
This can me a max power point type that can say maintain 5V at the input and so with 5V drop on 100Ohm 50mA are available for DC-DC converter times 5V that is a respectable 0.25W witch can be used to charge the capacitor to 20V
Then when that gets to 20V the DC DC converter is disconnected and the capacitor is connected thus for a moment V2 will be 30V

[Andy Chee]

I believe you are fundamentally describing a variant of a Cuk regulator or charge pump:

https://www.analog.com/en/technical-articles/differences-between-the-uk-converter-and-the-inverting-charge-pump-converter.html

It may be possible to float the switch control circuitry (the blue line in your diagram), but given you have common grounds anyway (the grey line in your diagram), why would you even bother going through the hassle?

If this is just a thought experiment, then yes, charging a capacitor in series, then flipping its direction, would indeed boost the output voltage.  The control circuitry can be made floating and it would still work.

In practice, there's no point in floating a capacitor based converter.  OTOH floating transformer based converters happens all the time.

[electrodacus]

I believe you are fundamentally describing a variant of a Cuk regulator or charge pump:

https://www.analog.com/en/technical-articles/differences-between-the-uk-converter-and-the-inverting-charge-pump-converter.html

It may be possible to float the switch control circuitry (the blue line in your diagram), but given you have common grounds anyway (the grey line in your diagram), why would you even bother going through the hassle?

Are you attempting to extend this principle to a multi kilovolt generator or similar that would require such isolation?

This circuit is not useful and is just an exercise to understand that V2 can be larger than V1 (intermittently) but not continues if the ground is floating.
It is the analog of a mechanical setup that I was trying to explain but I think the electrical setup is easier to understand first.
There where people in the past that disagreed with the fact that V2 can not be continuously higher than V1 despite the floating GND.

[Andy Chee]

It is the analog of a mechanical setup that I was trying to explain but I think the electrical setup is easier to understand first.

Electrical analog of the hydraulic ram?

https://en.wikipedia.org/wiki/Hydraulic_ram

[electrodacus]

Electrical analog of the hydraulic ram?

Yes it will be the electrical equivalent of that also.
Energy storage is used in that case to be able to pump water at higher elevation.
People seem to always remember about the frictional losses and the fact that you can not get rid of them in real world but fail to think about the energy storage witch is I will say as important.

[Andy Chee]

As a crude switched capacitor booster demonstration, you can use a mechanical DPDT switch in order to flip the capacitor forwards and backwards.  Make the capacitor large enough so that your fingers can move at reasonable speed.  Hey presto!  No electrical ground for the controller!

[electrodacus]

As a crude switched capacitor booster demonstration, you can use a mechanical DPDT switch in order to flip the capacitor forwards and backwards.  Make the capacitor large enough so that your fingers can move at reasonable speed.  Hey presto!  No electrical ground for the controller!

Yes reversing the capacitor polarity works also if you do not want V2 to be 3x V1
I'm not interested in building this just want people to understand that a "magic box" with just two wires connected, in series with the load and no connection to ground can start with zero energy and can have V2 >> V1 for short periods of time proportional with the amount stored in the capacitor and inversely proportional with the load current.
I should have added also a small capacitor in parallel with the 100Ohm to show that V2 increases to somewhere close to 30V then decreases well below 10V before another cycle can be repeated and it is not possible to maintain 20 or 30V at V2
 

[Andy Chee]

I'm not interested in building this just want people to understand that a "magic box" with just two wires connected

I'm confused.  In your opening post, you said you wanted to build and test it:

If you think it is possible I will like to know how in order to build and test it if it makes sense.

[electrodacus]

I'm not interested in building this just want people to understand that a "magic box" with just two wires connected

I'm confused.  In your opening post, you said you wanted to build and test it:

If you think it is possible I will like to know how in order to build and test it if it makes sense.

What I mentioned is what you quote exactly and the question was V2 continuously higher than V1 not intermittent.
Someone mentioned that not even intermittent is possible and thus why the discussion got here.

[EPAIII]

Yes, it may be possible to charge the series capacitor that you show and then reverse it's connections to place it is series with the battery to get a higher Voltage across the load FOR A SHORT TIME. But here's what the load will see as these processes happen, using YOUR second drawing, which I would modify.

1. First, starting with the capacitor discharged, it will look like a DC short so the load will see the full battery Voltage or +10 V.

2. As the capacitor charges that charge will show as an increasing Voltage that is bucking the battery so the load will see V2 as decreasing from +10 V down to zero or some lower number depending on how long the capacitor charging is allowed to continue.

3. Then the capacitor is reversed so it's +10 V is in series with the battery and the load sees +20 V for an instant.

4. Then the capacitor starts to discharge through the load and battery so it's Voltage decreases and the load sees a Voltage that decreases from the +20 V value back down to the +10 V value.

5. And it repeats. V2 looks something like this:

{My attachment should be HERE! But this stupid BB software does not seem to want that. I give up. It is below}

I would alter your drawing by having the battery Voltage connected to the load while the capacitor is charging. That way the half of my drawing below 10 V would become a straight line at 10 V.

But even that is NOT how I would double the Voltage to the load. I do not see any reason why a floating ground is needed in this circuit. I would use the battery Voltage to run a DC to AC converter which produces an AC Voltage. Then I would use a conventional Voltage doubler circuit (diodes and capacitors) to boost the Voltage to the 20 VDC level. This would include A GROUND connection to the battery minus terminal.

Or, BETTER YET, just use a DC to DC converter in boost mode WITH A GROUND connection to the battery minus terminal.

The "floating ground" buys you nothing in a battery powered circuit. The whole thing can float. It can float on top of 100,000 VDC if you wish and it will still work. Just turn the 100,000 VDC OFF before changing the battery.

By transforming a quadripole into a dipole, it will be hard to get anything else than |V2| <= |V1|.
Maybe we can try leveraging some quantum effect.

So you are saying that there is no way to get V2 above V1 even for a short period of time ?

No, for a transient, the question is obvious and answered already by magic. Just put anything in series which has a charge. It will just be extra energy that has been stored previously though.
But once this charge is depleted, game over. So what does that bring to the table?

[EPAIII]

Good grief! Why are you wasting our time with such things if you are not trying to actually use them?

Can a "magic box" be built to do this? Probably.

Can it be built according to your second diagram. Perhaps, but not as you show it.

Would anybody actually build it that way? Likely NOT! Except, perhaps yourself.

If you want a CONSTANT higher Voltage at V2, you are going to have to have something in there that is connected to the battery minus terminal. Notice, I did not call it GROUND. So, you are correct, you do not need an actual ground. But you do need a capacitor or something that is connected to the battery's negative terminal to hold that higher Voltage while the circuit, whatever circuit, goes about the job of creating it.

Your V2 is the Voltage across the load and the load is connected to the battery negative terminal. So that Voltage, by your own drawings, is referenced to that point. That point MUST be included in any circuitry that creates the higher Voltage you want. Otherwise that created Voltage is NOT referenced to YOUR own reference point and can not be said to have ANY VALUE with respect to it.

All Voltages are measured across TWO points, one of which is usually considered a common. And that common is often called "ground". That's just the way we talk and think about things.

I have wasted enough time on this.

As a crude switched capacitor booster demonstration, you can use a mechanical DPDT switch in order to flip the capacitor forwards and backwards.  Make the capacitor large enough so that your fingers can move at reasonable speed.  Hey presto!  No electrical ground for the controller!

Yes reversing the capacitor polarity works also if you do not want V2 to be 3x V1
I'm not interested in building this just want people to understand that a "magic box" with just two wires connected, in series with the load and no connection to ground can start with zero energy and can have V2 >> V1 for short periods of time proportional with the amount stored in the capacitor and inversely proportional with the load current.
I should have added also a small capacitor in parallel with the 100Ohm to show that V2 increases to somewhere close to 30V then decreases well below 10V before another cycle can be repeated and it is not possible to maintain 20 or 30V at V2

[magic]

You can make a magic box which will store small amount of energy internally and make V2 sometimes lower and sometimes higher than V1.
You cant' even make it sometimes higher and sometimes equal.

There are switched capacitor voltage doublers, aka charge pumps. They need ground connection to produce constant V2>V1 and their input current is higher than output current.

[electrodacus]

Yes, it may be possible to charge the series capacitor that you show and then reverse it's connections to place it is series with the battery to get a higher Voltage across the load FOR A SHORT TIME. But here's what the load will see as these processes happen, using YOUR second drawing, which I would modify.

1. First, starting with the capacitor discharged, it will look like a DC short so the load will see the full battery Voltage or +10 V.

2. As the capacitor charges that charge will show as an increasing Voltage that is bucking the battery so the load will see V2 as decreasing from +10 V down to zero or some lower number depending on how long the capacitor charging is allowed to continue.

3. Then the capacitor is reversed so it's +10 V is in series with the battery and the load sees +20 V for an instant.

4. Then the capacitor starts to discharge through the load and battery so it's Voltage decreases and the load sees a Voltage that decreases from the +20 V value back down to the +10 V value.

5. And it repeats. V2 looks something like this:

{My attachment should be HERE! But this stupid BB software does not seem to want that. I give up. It is below}

I would alter your drawing by having the battery Voltage connected to the load while the capacitor is charging. That way the half of my drawing below 10 V would become a straight line at 10 V.

But even that is NOT how I would double the Voltage to the load. I do not see any reason why a floating ground is needed in this circuit. I would use the battery Voltage to run a DC to AC converter which produces an AC Voltage. Then I would use a conventional Voltage doubler circuit (diodes and capacitors) to boost the Voltage to the 20 VDC level. This would include A GROUND connection to the battery minus terminal.

Or, BETTER YET, just use a DC to DC converter in boost mode WITH A GROUND connection to the battery minus terminal.

The "floating ground" buys you nothing in a battery powered circuit. The whole thing can float. It can float on top of 100,000 VDC if you wish and it will still work. Just turn the 100,000 VDC OFF before changing the battery.

You are correct for the case where you reverse the connection of that capacitor you can get a peak of 20V but when it will discharge it will drop to zero not to only +10V as the capacitor is reverse charged.

My question was if there is any possibility to get V2 > V1 continuously not intermittent.
And the reason I ask this is because people claim that the mechanical analog of this circuit  can have V2 > V1 continually.
I was even banned on this forum for arguing that will not be possible.

The way I drawn the example with an MPPT DC-DC converter it is possible to  have V2 in the +5V and +30V range never get to zero but still it requires that V2 drops below 10V (below V1). It can be done even as +9V to +30V but it will be impossible to keep it  above +10V (above V1) continually.
 

[electrodacus]

You cant' even make it sometimes higher and sometimes equal.

This is exactly what I want to hear and I agree fully with you.
When I claimed the same about the mechanical analog of this circuit nobody will agree with me and I was even banned on this forum.

It is strange that people understand the electrical version but not the mechanical version of this circuit.
During that discussion I even showed this electrical circuit and some people there claimed you can have a higher V2 continually but that is not the case this time.

That is why I mentioned that I will build this if someone has a version they think can show V2 > V1 continually.
I spent thousand's of $ to build the mechanical analog and proved that  V2 decreases and people that have seen that say that my experiment is flawed as they expect V2 will never decrease below V1

[Zero999]

You can't do that if your only return path is through the load. As shown in the diagram.

Here I made a diagram that is maybe better than a description.
There is a DC-DC converter inside the box with positive input from battery+ and battery- trough the 100Ohm load.
This can me a max power point type that can say maintain 5V at the input and so with 5V drop on 100Ohm 50mA are available for DC-DC converter times 5V that is a respectable 0.25W witch can be used to charge the capacitor to 20V
Then when that gets to 20V the DC DC converter is disconnected and the capacitor is connected thus for a moment V2 will be 30V

No, it's not possible to do this. A voltage regulator which goes inline with the input like that, must have a common i.e. 0V connection. There is no way to design a regulator which goes series with a power supply like that, be linear, or switched mode and regulates the output voltage. All voltages are relative, so common connection is required to monitor the output potential difference.

[electrodacus]

No, it's not possible to do this. A voltage regulator which goes inline with the input like that, must have a common i.e. 0V connection. There is no way to design a regulator which goes series with a power supply like that, be linear, or switched mode and regulates the output voltage. All voltages are relative, so common connection is required to monitor the output potential difference.

When you say not possible. Are you referring to not possible to have V2 > V1 continually or not possible to even have V2 for any amount of time above V1?
In the diagram you quoted it is possible to have V2 up to 3x V1 (30V) for short periods of time before V2 decreases below V1 in order to recharge the capacitor back to 20V.

[Andy Chee]

When you say not possible.

I believe he's talking about the voltage feedback to the switching controller.  At minimum, the feedback network MUST be referenced to the output voltage ground in order to measure the output voltage.

OTOH, if this regulator is just a thought experiment, then you don't really need feedback and just leave it open loop.  However, without feedback you can't really call it a "regulator" can you?

[electrodacus]

I believe he's talking about the voltage feedback to the switching controller.  At minimum, the feedback network MUST be referenced to the output voltage ground in order to measure the output voltage.

OTOH, if this regulator is just a thought experiment, then you don't really need feedback and just leave it open loop.  However, without feedback you can't really call it a "regulator" can you?

You can have an internal microcontroller running say at 3.3V from a separate capacitor where you have a voltage reference or just use the 3.3V supply to fairly accurately charge the capacitor to 20V.
I do not want for this to get complicated the point I'm trying to make is that is possible to have V2 higher than V1 but only intermittent and V2 needs to drop below V1 else the capacitor can not be charged.

[Zero999]

No, it's not possible to do this. A voltage regulator which goes inline with the input like that, must have a common i.e. 0V connection. There is no way to design a regulator which goes series with a power supply like that, be linear, or switched mode and regulates the output voltage. All voltages are relative, so common connection is required to monitor the output potential difference.

When you say not possible. Are you referring to not possible to have V2 > V1 continually or not possible to even have V2 for any amount of time above V1?
In the diagram you quoted it is possible to have V2 up to 3x V1 (30V) for short periods of time before V2 decreases below V1 in order to recharge the capacitor back to 20V.

I mean it's impossible to regulate the output voltage, without a common connection to the input.

You schematic is also drawn contrary to the convention of having the input on the left and output on the right.

Here it is flipped the right way round, with the common/0V connection added.

When you say not possible.

I believe he's talking about the voltage feedback to the switching controller.  At minimum, the feedback network MUST be referenced to the output voltage ground in order to measure the output voltage.

OTOH, if this regulator is just a thought experiment, then you don't really need feedback and just leave it open loop.  However, without feedback you can't really call it a "regulator" can you?

No, it still won't work because of KCL.

[electrodacus]

I mean it's impossible to regulate the output voltage, without a common connection to the input.

You schematic is also drawn contrary to the convention of having the input on the left and output on the right.

Here it is flipped the right way round, with the common/0V connection added.

I do not think there is any convention regarding where the input needs to be but if that makes it easier to visualize I have no problem with that.
I do not want that connection added. That connection missing is the entire point of the exercise.

With the floating 0V negative connection a circuit inside the green box can still get the V2 = 3*V1 as peak voltage.
So I can have a V1 = 10V constant the internal circuit will be able to calculate that it has 10V at the input or 10.1V using a micro controller with cleaver algorithm and current and voltage measurements.

But I do not want to get the circuit and explanation to complex. I just want for everyone to agree that V2 can be three times V1 for short periods of time and V2 needs to decrease below V1 in order to be able to recharge that internal capacitor back to 20V and so increase the V2 again.
 

No, it still won't work because of KCL.

Kirchhoff current law will not prevent V2 to be higher than V1 temporarily with no other connection on that green box other than Battery+ and Load+ (the 100Ohm resistor).
The box can contain capacitors that are energy storage devices.
Those capacitors can be charged from battery trough the 100Ohm load resistor so V2 < V1 then after those are charged they can be connected in series with the battery in order to have V2 >> V1 for some limited amount of time as capacitor will of course discharge.
You can have another capacitor internally that supplies a low power micro controller and be able to accurately calculate the battery voltage and even the resistor value as you can have an internal voltage reference say a 1.2V to be able to compare against.
So you can accurately fluctuate V2 to be between say 5V and 30V
 

[Someone]

With the floating 0V negative connection a circuit inside the green box can still get the V2 = 3*V1 as peak voltage.

Your arguments are (intentionally?) impossible to follow as you swap freely between several poorly defined questions.

There is no 3x multiplier proven/shown/known. If you had a device which can increase the voltage by any amount then just add more of them in series to increase the voltage to arbitrarily high multipliers.

So, your argument is faulty to begin with. Start again with a clear and consistent point.

[HuronKing]

So, your argument is faulty to begin with. Start again with a clear and consistent point.

If it helps, I'm extremely skeptical he is trying to talk about DC-DC converters at all. Consider his post history - he has talked about this before.

Any minute now we will begin, hehe, jousting at windmills, literally.

[electrodacus]

Your arguments are (intentionally?) impossible to follow as you swap freely between several poorly defined questions.

There is no 3x multiplier proven/shown/known. If you had a device which can increase the voltage by any amount then just add more of them in series to increase the voltage to arbitrarily high multipliers.

So, your argument is faulty to begin with. Start again with a clear and consistent point.

a) V1 is always 10V (battery voltage).
b) V2 can be made to fluctuate between 5V and 30V using just that green box with only two wires connected to it.  There is no battery in the box but there are multiple capacitors, inductors, mosfets and even a micro controller.

Is your claim that is not possible ?

[electrodacus]

If it helps, I'm extremely skeptical he is trying to talk about DC-DC converters at all. Consider his post history - he has talked about this before.

Any minute now we will begin, hehe, jousting at windmills, literally.

I only talk about this electrical circuits. This should be in the area of expertise for most people in this forum.

The claim is that green box has only two wires connected to it. It is irrelevant what is inside as long as box contains no electrical energy when connected to the circuit.

Is it possible to have the V2 fluctuating between 5V and 30V ?
If not why ?

[PlainName]

So, your argument is faulty to begin with. Start again with a clear and consistent point.

If it helps, I'm extremely skeptical he is trying to talk about DC-DC converters at all. Consider his post history - he has talked about this before.

Any minute now we will begin, hehe, jousting at windmills, literally.

Exactly my thought after the first post. I have money on where this is leading to 

[electrodacus]

Exactly my thought after the first post. I have money on where this is leading to 

It is not a trick question and it will not lead to anything else on this forum.
It seems there are multiple opinions on this problem so what is yours?
Is A) or B) possible ?  (see original post as it was edited recently and the two questions where added).

[PlainName]

I am trying very hard not to have an opinion, simply because I know where doing so will lead. Be I see that we have some fresh users, so... 

[electrodacus]

I am trying very hard not to have an opinion, simply because I know where doing so will lead. Be I see that we have some fresh users, so... 

Are you afraid that having an opinion will get you banned or that you can be wrong ?
There already a variety of opinions and there is no consensus.
I expressed my opinion.

[PlainName]

I am scared of the rabbit holes.

[Someone]

Your arguments are (intentionally?) impossible to follow as you swap freely between several poorly defined questions.

There is no 3x multiplier proven/shown/known. If you had a device which can increase the voltage by any amount then just add more of them in series to increase the voltage to arbitrarily high multipliers.

So, your argument is faulty to begin with. Start again with a clear and consistent point.

a) V1 is always 10V (battery voltage).
b) V2 can be made to fluctuate between 5V and 30V using just that green box with only two wires connected to it.  There is no battery in the box but there are multiple capacitors, inductors, mosfets and even a micro controller.

Is your claim that is not possible ?

Where do I claim that is not possible? I claim you are adding additional information (30V) which is irrelevant. V2 could be made to oscillate/transient between any arbitrary voltages +/- whatever. So you are adding fake argument and confusion where there should be none (this entire thread).

Should you wish to argue things, it has to be from logic and rational basis. Instead we will just point out how your discussion is faulty and only making things worse not better.

[Someone]

Is A) or B) possible ?  (see original post as it was edited recently and the two questions where added).

... and there we have it. Changing the post to make it look like people are disagreeing with you. There was no need to edit the original post when you can just add your new argument inline. Now we'll have to aggressively quote everything you say to keep the discussion coherent.

[bdunham7]

The claim is that green box has only two wires connected to it. It is irrelevant what is inside as long as box contains no electrical energy when connected to the circuit.

To be crystal clear, do you mean that the box contains no electrical energy (or presumably other energy) at the moment it is connected to the circuit or that it never contains (stores) any energy during the time that it is connected to the circuit?

[bdunham7]

This is exactly what I want to hear and I agree fully with you.
When I claimed the same about the mechanical analog of this circuit nobody will agree with me and I was even banned on this forum.

It is strange that people understand the electrical version but not the mechanical version of this circuit.

OK, there it is--somehow I knew I'd find it in there somewhere.  Your circuit is not an analog of the mechanical device that you just can't get yourself to understand.  And I'm sure it will be impossible to get you to understand that as well.  Congratulations, you just got me to waste another 5 minutes.

[electrodacus]

Is A) or B) possible ?  (see original post as it was edited recently and the two questions where added).

... and there we have it. Changing the post to make it look like people are disagreeing with you. There was no need to edit the original post when you can just add your new argument inline. Now we'll have to aggressively quote everything you say to keep the discussion coherent.

The post change can not make people agree or disagree with me. It is to make it simpler to answer with Yes or NO to A) and B)
Those that answered before the two questions where added never mentioned A) or B)
The question was always the same and yes people where not all in agreement with me or with others before or after I added the questions to original post.
It is a problem and based on feedback I can improve the formulation to make thing more clear.
So what is your answer ?

[electrodacus]

To be crystal clear, do you mean that the box contains no electrical energy (or presumably other energy) at the moment it is connected to the circuit or that it never contains (stores) any energy during the time that it is connected to the circuit?

Yes I was clear that there is no electrical energy in the box when box is connected to the circuit.
Of course it can store energy after it is connected as it contains instructors and capacitors and both are energy storage devices. Even if it contained a simple copper wire that also has inductance and capacitance so it is still an energy storage device.
You can not get rid of energy storage in a real world application of any type electrical or mechanical.

OK, there it is--somehow I knew I'd find it in there somewhere.  Your circuit is not an analog of the mechanical device that you just can't get yourself to understand.  And I'm sure it will be impossible to get you to understand that as well.  Congratulations, you just got me to waste another 5 minutes.

It is your opinion.
Do you have an answer to this problem ?
Simple questions A) and B) in the original post.

[magic]

a) V1 is always 10V (battery voltage).
b) V2 can be made to fluctuate between 5V and 30V using just that green box with only two wires connected to it.  There is no battery in the box but there are multiple capacitors, inductors, mosfets and even a micro controller.

Is your claim that is not possible ?

Is it possible to have the V2 fluctuating between 5V and 30V ?
If not why ?

Do you have an answer to this problem ?

Is it possible that you have a brain and could start using the answers people posted to solve your trivial problems?

Or is this some sort of democracy where truth is established by a majority vote?

[electrodacus]

Is it possible that you have a brain and could start using the answers people posted to solve your trivial problems?

Or is this some sort of democracy where truth is established by a majority vote?

It is a physics problem and this is a sort of survey.
You say the problem is trivial and maybe it is for you but based on other comments it does not seem to be trivial.

I will love to know your answer. I gave mine in the original post.

[boB]

How about an LTspice simulation of one of the answers ?

[electrodacus]

How about an LTspice simulation of one of the answers ?

Is this a question directed to me ?
There are two questions A) and B) in the original post with my answers in bold No and Yes.

A) Is it possible for V2 to be a constant 30V?  No
B) Is it possible for V2 to peak to 30V multiple times while V2 never drops below 5V ?  Yes

So do you not agree with one of my answers ? Or with both ?

If you say Yes to question A) the it will be you that will need to prove that as I can just use an empty box or a wire between the two points and made my case.
If you say No to second question I can describe to you how it will work but the diagram I offered as an example should be sufficient.

The capacitor will be disconnected from the two box terminals and connected to the output of a DC-DC boost converter.
The boost converter can be a max power point so one that maintain a constant voltage at the input and takes what is available.
So say DC-DC MPPT is set to 5V then the other 5V will drop across the 100Ohm resistor meaning 50mA will flow trough the circuit.
5V * 50mA = 0.25W are available with say a 70% efficient DC-DC converter 0.175W are available to charge a large electrolytic capacitor connected at the output of this DC-DC converter.
When this capacitor gets to 20V some analog trip circuit can disconnect the DC-DC and connect the capacitor between the two connection points on the green box.
Thus for a short moment V2 = 10V + 20V = 30V
It can get way more complicated than this with a microcontroller and current measurement to calculate exactly what the resistance value is and what the battery voltage is but this are unnecessary complications just to say that it can be done.

[Andy Chee]

So do you not agree with one of my answers ? Or with both ?

In order to resolve disagreement, we need proof.  Ltspice simulation can be used as proof.

[electrodacus]

So do you not agree with one of my answers ? Or with both ?

In order to resolve disagreement, we need proof.  Ltspice simulation can be used as proof.

As I mentioned above for question A) it will be you that needs to provide the proof.
The question B) I explained in details so it should be sufficient proof but it can be arranged to be simulated in LTspice tho it will take time as it is a relatively complex simulation due to switch logic involved.
They should likely already have some DC-DC boost converters with max power point tracking so that part should be easy.

I provided spice simulation as proof for the Veritasium video on how electricity works and it was of no help. If people do not understand the basics a simulation or even an real experimental demonstration will not help with anything and cost me time.

[Someone]

A) Is it possible for V2 to be a constant 30V?  No
B) Is it possible for V2 to peak to 30V multiple times while V2 never drops below 5V ?  Yes

So do you not agree with one of my answers ? Or with both ?

Now that you have stated a somewhat clear question who has disagreed with you? No-one! So why immediately ask if every poster agrees/disagrees?

[soldar]

I'm more used to having the signal flow drawn from left to right, but whatever.

Me too. Power, voltage, signal flow from left to right and from top to bottom. I often redraw schematics that way because it helps me analyze and understand the circuit.

When I reverse engineer a circuit I start out with a schematic which resembles the PCB but I rearrange it in steps until it is easier to conceptualize.

[tggzzz]

Is A) or B) possible ?  (see original post as it was edited recently and the two questions where added).

... and there we have it. Changing the post to make it look like people are disagreeing with you. There was no need to edit the original post when you can just add your new argument inline. Now we'll have to aggressively quote everything you say to keep the discussion coherent.

The post change can not make people agree or disagree with me. It is to make it simpler to answer with Yes or NO to A) and B)
Those that answered before the two questions where added never mentioned A) or B)
The question was always the same and yes people where not all in agreement with me or with others before or after I added the questions to original post.
It is a problem and based on feedback I can improve the formulation to make thing more clear.
So what is your answer ?

The acceptable way to change your opinion is:

• add a new post indicating your new opinion
• optionally changing the early post, using strikethrough to indicate what you no longer wish other people to consider, and adding "EDIT: ..." to show your new opinion.

Doing that makes you appear to be thoughtful and considerate.

"Silently" changing the early post makes you look like an insecure troll.

[IanB]

There were people in the past that disagreed with the fact that V2 can not be continuously higher than V1 despite the floating GND.

You would need to provide a context reference for that statement, by quoting instances where "people in the past" said this.

[Andy Chee]

There were people in the past that disagreed with the fact that V2 can not be continuously higher than V1 despite the floating GND.

You would need to provide a context for that statement, by quoting instances where "people in the past" said this.

I can provide context.

The DC-DC floating converter problem is supposed to be an electrical analogy to describe the energy interactions behind Veritasium's/Xyla's "downwind faster than wind" vehicle.

"people in the past" is a reference to people discussing the vehicle physics, as opposed to the DC-DC converter problem.  But the OP believes they are directly equivalent.  Go figure.

[IanB]

I fully understand that context. I knew what this was about from the very first post in the thread, since we've been here before.

But I want electrodacus to provide a direct quote where someone said the DC-DC converter could provide a continuous step up in voltage without a third electrical connection. Otherwise he is just being disingenuous and misrepresenting people.

Because I really don't believe anyone on this forum would be likely to say that about the "DC-DC converter" picture in the first post.

[Zero999]

How about an LTspice simulation of one of the answers ?

Is this a question directed to me ?
There are two questions A) and B) in the original post with my answers in bold No and Yes.

A) Is it possible for V2 to be a constant 30V?  No
B) Is it possible for V2 to peak to 30V multiple times while V2 never drops below 5V ?  Yes

So do you not agree with one of my answers ? Or with both ?

If you say Yes to question A) the it will be you that will need to prove that as I can just use an empty box or a wire between the two points and made my case.
If you say No to second question I can describe to you how it will work but the diagram I offered as an example should be sufficient.

The capacitor will be disconnected from the two box terminals and connected to the output of a DC-DC boost converter.
The boost converter can be a max power point so one that maintain a constant voltage at the input and takes what is available.
So say DC-DC MPPT is set to 5V then the other 5V will drop across the 100Ohm resistor meaning 50mA will flow trough the circuit.
5V * 50mA = 0.25W are available with say a 70% efficient DC-DC converter 0.175W are available to charge a large electrolytic capacitor connected at the output of this DC-DC converter.
When this capacitor gets to 20V some analog trip circuit can disconnect the DC-DC and connect the capacitor between the two connection points on the green box.
Thus for a short moment V2 = 10V + 20V = 30V
It can get way more complicated than this with a microcontroller and current measurement to calculate exactly what the resistance value is and what the battery voltage is but this are unnecessary complications just to say that it can be done.

Higher voltage output pulses are possible but not a steady voltage.

How about an LTspice simulation of one of the answers ?

Here's a simulation. A capacitor is first connected in series with the supply and the load, which charges it up. It is then connected the other way round, negative plate to V1 and positive to the output. The capacitor first discharges, adding its voltage to V1, before charging in in the opposite direction. Higher voltages are possible by connecting more capacitors in parallel and then in series for the discharge cycle.

This circuit isn't practical application. It's just a theoretical demonstration. It's possible to use solid state relays connected to a microcontroller for the timing. The microcontroller could be powered via another capacitor which is charged in series with the load resistor.

[electrodacus]

A) Is it possible for V2 to be a constant 30V?  No
B) Is it possible for V2 to peak to 30V multiple times while V2 never drops below 5V ?  Yes

So do you not agree with one of my answers ? Or with both ?

Now that you have stated a somewhat clear question who has disagreed with you? No-one! So why immediately ask if every poster agrees/disagrees?

Just the comment before yours Andy Chee asked for a LTspice simulation as proof for those answers.
Should I understand you do not disagree with my answers to those two questions?
As there will be people asking the opposite of what you are asking.

[electrodacus]

I'm more used to having the signal flow drawn from left to right, but whatever.

Me too. Power, voltage, signal flow from left to right and from top to bottom. I often redraw schematics that way because it helps me analyze and understand the circuit.

When I reverse engineer a circuit I start out with a schematic which resembles the PCB but I rearrange it in steps until it is easier to conceptualize.

I agree and sorry for having the diagrams drawn in a non standard way.  I was thinking that is a simple enough diagram that it will not affect anyone's ability to understand.

[electrodacus]

The acceptable way to change your opinion is:

• add a new post indicating your new opinion
• optionally changing the early post, using strikethrough to indicate what you no longer wish other people to consider, and adding "EDIT: ..." to show your new opinion.

Doing that makes you appear to be thoughtful and considerate.

"Silently" changing the early post makes you look like an insecure troll.

I did not changed my opinion. I made the original post more clear and easier to answer by adding an example of what can be inside the green box and asked two questions that can be answer with a simple Yes or No.

[coromonadalix]

again  an useless thread  wholl endup shutted down ... pffffffffff


and seeing your past threads    "electrodacus"   you seem to love making some ruckus 

im not a mod, but you'd be on a warning

[electrodacus]

I fully understand that context. I knew what this was about from the very first post in the thread, since we've been here before.

But I want electrodacus to provide a direct quote where someone said the DC-DC converter could provide a continuous step up in voltage without a third electrical connection. Otherwise he is just being disingenuous and misrepresenting people.

Because I really don't believe anyone on this forum would be likely to say that about the "DC-DC converter" picture in the first post.

I will try to make a search but it was over a year ago and it will be some work to search trough all those comments.
So should I understand you agree with my answers to both of those questions A) and B)  ?

[coromonadalix]

i agree with nothing, you seriously need to relearn electronics / applications  and a few other things, instead of starting thread and sucking life out of many members here

all you answers can be found elsewhere

its a derailing thread  thats all       



im beginning to wonder if some people want click counts or post / thread counts   the way i see things

in the fpga or micro processor thread  one guy  started 4 thread of the same part with slight differences, and before it would have been blocked,mad a complaint and its still going   

[electrodacus]

Here's a simulation. A capacitor is first connected in series with the supply and the load, which charges it up. It is then connected the other way round, negative plate to V1 and positive to the output. The capacitor first discharges, adding its voltage to V1, before charging in in the opposite direction. Higher voltages are possible by connecting more capacitors in parallel and then in series for the discharge cycle.

This circuit isn't practical application. It's just a theoretical demonstration. It's possible to use solid state relays connected to a microcontroller for the timing. The microcontroller could be powered via another capacitor which is charged in series with the load resistor.

Thanks for doing the simulation and thanks for answering my questions clearly.
I think there are at least 2 or 3 people excluding me that agree with you so I'm looking for those that did not provided a straight answer but doubted that is the case some asking for simulation proof.
 

[electrodacus]

again  an useless thread  wholl endup shutted down ... pffffffffff


and seeing your past threads    "electrodacus"   you seem to love making some ruckus 

im not a mod, but you'd be on a warning

Why will the thread be "shutted down" ?
Not quite sure why you think I "love making some ruckus".
This is a question about an electrical circuit with floating GND.
I do expect much more people here know the correct answers to this but it will not be all.
For those that did not know the answer to this even if they do not comment just read this it is a learning experience.
I do not think I said anything controversial for most people. 

[electrodacus]

i agree with nothing, you seriously need to relearn electronics / applications  and a few other things, instead of starting thread and sucking life out of many members here

I do not understand.
What do you agree or disagree with ?
I'm guessing you are refereeing to me when you say I need to relearn electronics. That sounds like you disagree with my answers to those two questions. Or maybe it is with something else I said ?

It is hard for me to understand why anyone will think I'm a troll or I enjoy this. I do not.

[Zero999]

Thanks for doing the simulation and thanks for answering my questions clearly.
I think there are at least 2 or 3 people excluding me that agree with you so I'm looking for those that did not provided a straight answer but doubted that is the case some asking for simulation proof.

No one said that the output voltage can never exceed the input, just that the output voltage can't be continuous. The original post was poorly worded and the fact you appear to have gone back and changed it hasn't done you any favours.

[PlainName]

I did not changed my opinion.

Then why are you here making yet another thread on the same thing? You're not here to learn; you are here solely to get everyone else to agree with your erroneous but fixed opinion. You won't stop unless every last one says "Oh yes, you're right".

That's it. The thread serves no other purpose except for you to spout off and get agreement that you are right. But you are wrong, and since you will never accept that you will just forever be creating this very same thread again. And again. And again.

The only reason you're doing it here, now, is because you've been banned from every other respectable forum for doing the exact same thing there.

[electrodacus]

No one said that the output voltage can never exceed the input, just that the output voltage can't be continuous. The original post was poorly worded and the fact you appear to have gone back and changed it hasn't done you any favours.

I changed it because it was poorly worded based on the feedback from others. Sorry if you think is still poorly worded. Can you point out what part so I can make it more clear ?

[coromonadalix]

good point  in all of it  PlainName

 

and others do complain,    the mod(s) may eventually close it down


but until then     Electrodacus will continue and never stop   

[electrodacus]

I did not changed my opinion.

Then why are you here making yet another thread on the same thing? You're not here to learn; you are here solely to get everyone else to agree with your erroneous but fixed opinion. You won't stop unless every last one says "Oh yes, you're right".

That's it. The thread serves no other purpose except for you to spout off and get agreement that you are right. But you are wrong, and since you will never accept that you will just forever be creating this very same thread again. And again. And again.

The only reason you're doing it here, now, is because you've been banned from every other respectable forum for doing the exact same thing there.

Did I made a tread on this floating GND problem ?
Yes you are correct that I'm not here specifically to learn but I do not mind if that happens in the process.
What other forum was I ever banned from ?
I do not get your problem with me. Do you agree or not with my answers to those two simple questions ?
If you do then I do not see the problem and if you do not then please state so and provide some evidence. 

[coromonadalix]

man i think we deal with a machine  loll   an AI  bot is here

"not here specifically to learn"

you talk like an advocate

im done,  you're black listed, sadly   this forums doesnt seem to provide this functionality

[electrodacus]

man i think we deal with a machine  loll   an AI  bot is here

"not here specifically to learn"

you talk like an advocate

im done,  you're black listed, sadly   this forums doesnt seem to provide this functionality

I'm being honest not an advocate.
Care to answer the two questions before you leave ?

[Simon]

Input and output currents surely are equal because all current entering through the input must exit through the output since there are no other external connections...

Maybe you meant to connect the negative terminals to ground?

I agree with you but looking to see if this is a common opinion.
I did not meant to connect the negative terminals to ground. The ground on that DC-DC converter is floating.

Then you know enough already to tell everyone the solution and quit the wind up!

[electrodacus]

Then you know enough already to tell everyone the solution and quit the wind up!

I'm terrible at expressing myself.

The original post contains a diagram that has a battery in series with a green box (anything can be in that box as long as it is not containing energy when connected) and a resistor as the load.

There are two questions
A) Is it possible for V2 to be a constant 30V
B) Is it possible for V2 to peak to 30V multiple times while V2 never drops below 5V ?

And I provided my answer to those questions
A) No
B) Yes

A lot of people seems to agree with me and some do not agree or do not want to provide an answer but ask for some proof my answers are correct.

Here you can see Zero999 agreed with me and even provided an LTspice example (simpler example than mine) still it shows V2 >> V1 intermittently.
https://www.eevblog.com/forum/chat/dc-dc-converter-with-floating-ground/msg5247495/#msg5247495
Also there you can see a description of how a more complex circuit can work but the idea is the same as in the simpler example provided by Zero999 and EPAIII

[Zero999]

My criticism wasn't that you edited your post, but lacked transparency in doing so.

Why did you ask the question if you already knew the answer? It it was a brain teaser, then you should have been upfront about it. Going from your posting history, I think many people here thought it was some sort of free energy scheme.

I'm not the smartest person here. Far from it. I'm sure many others are capable of figuring it out. Indeed many probably did, but didn't post a solution due to lack of will or time.

[IanB]

I'm not the smartest person here. Far from it. I'm sure many others are capable of figuring it out. Indeed many probably did, but didn't post a solution due to lack of will or time.

The reason most people are not responding is because he is not asking the question in good faith. It is, as Simon said, a wind up.

This thread, along with other previous ones, is just a big waste of time.

[electrodacus]

My criticism wasn't that you edited your post, but lacked transparency in doing so.

Why did you ask the question if you already knew the answer? It it was a brain teaser, then you should have been upfront about it. Going from your posting history, I think many people here thought it was some sort of free energy scheme.

I'm not the smartest person here. Far from it. I'm sure many others are capable of figuring it out. Indeed many probably did, but didn't post a solution due to lack of will or time.

Sorry for not being clear. Yes it was a brain teaser and I always mentioned that I know the answer to the question I was asking.
I think you are modest. You are sure fairly smart.
I did not asked for a solution if you agreed with my answers.
I asked for a solution if you disagreed with me on question A)

[electrodacus]

The reason most people are not responding is because he is not asking the question in good faith. It is, as Simon said, a wind up.

This thread, along with other previous ones, is just a big waste of time.

Using an analogy to prove a point is in bad faith ?

[bdunham7]

Using an analogy to prove a point is in bad faith ?

Using an analogy to explain something to someone who is learning from you--not debating you--is sometimes quite effective.  Using an analogy in an argument is like....well it doesn't work as well because you have to defend both the correctness of the example you are using and the validity of the relationship between that example and the point you are actually trying to 'prove'.  In this case your circuit is not analogous to the issue that is really on your mind.  But that doesn't matter because you will maintain that it is and now we have an additional thing to disagree over--and you've gotten nowhere.  So while not "bad faith" in all cases (although arguments using analogies certainly can be and are very commonly presented in bad faith in the legal profession IMO) they are almost always a bad idea when arguing contested matters.

[tggzzz]

My criticism wasn't that you edited your post, but lacked transparency in doing so.

Why did you ask the question if you already knew the answer? It it was a brain teaser, then you should have been upfront about it. Going from your posting history, I think many people here thought it was some sort of free energy scheme.

I'm not the smartest person here. Far from it. I'm sure many others are capable of figuring it out. Indeed many probably did, but didn't post a solution due to lack of will or time.

Sorry for not being clear. Yes it was a brain teaser and I always mentioned that I know the answer to the question I was asking.
I think you are modest. You are sure fairly smart.
I did not asked for a solution if you agreed with my answers.
I asked for a solution if you disagreed with me on question A)

Let's record that, so he can't rewrite history again.

Completely disreputable behaviour.

[electrodacus]

Using an analogy to explain something to someone who is learning from you--not debating you--is sometimes quite effective.  Using an analogy in an argument is like....well it doesn't work as well because you have to defend both the correctness of the example you are using and the validity of the relationship between that example and the point you are actually trying to 'prove'.  In this case your circuit is not analogous to the issue that is really on your mind.  But that doesn't matter because you will maintain that it is and now we have an additional thing to disagree over--and you've gotten nowhere.  So while not "bad faith" in all cases (although arguments using analogies certainly can be and are very commonly presented in bad faith in the legal profession IMO) they are almost always a bad idea when arguing contested matters.

Please just answer questions A) and B)

[electrodacus]

My criticism wasn't that you edited your post, but lacked transparency in doing so.

Why did you ask the question if you already knew the answer? It it was a brain teaser, then you should have been upfront about it. Going from your posting history, I think many people here thought it was some sort of free energy scheme.

I'm not the smartest person here. Far from it. I'm sure many others are capable of figuring it out. Indeed many probably did, but didn't post a solution due to lack of will or time.

Sorry for not being clear. Yes it was a brain teaser and I always mentioned that I know the answer to the question I was asking.
I think you are modest. You are sure fairly smart.
I did not asked for a solution if you agreed with my answers.
I asked for a solution if you disagreed with me on question A)

Let's record that, so he can't rewrite history again.

Completely disreputable behaviour.

Can I know what are you talking about ?  What will I want to rewrite there ?
Can you please answer A) and B) so that it is recorded.

[PlainName]

Can you please answer A) and B) so that it is recorded.

Recorded for what purpose? What are you hoping to achieve?

[electrodacus]

Recorded for what purpose? What are you hoping to achieve?

It was as a comedic response to him recording my answers
Whatever you say is recorded anyway by the forum.

[Andy Chee]

Please just answer questions A) and B)

Your analogy is flawed and incorrect. That's my answer.

[electrodacus]

Your analogy is flawed and incorrect. That's my answer.

The questions are not about any sort of analogy.
So you are refusing to answer this questions because you are not sure about the correct answers ?
I'm just assuming that if you where sure you will had no problem answering whatever you think this is analog to some other type of problem or not.

[EPAIII]

Yes, it may be possible to charge the series capacitor that you show and then reverse it's connections to place it is series with the battery to get a higher Voltage across the load FOR A SHORT TIME. But here's what the load will see as these processes happen, using YOUR second drawing, which I would modify.

1. First, starting with the capacitor discharged, it will look like a DC short so the load will see the full battery Voltage or +10 V.

2. As the capacitor charges that charge will show as an increasing Voltage that is bucking the battery so the load will see V2 as decreasing from +10 V down to zero or some lower number depending on how long the capacitor charging is allowed to continue.

3. Then the capacitor is reversed so it's +10 V is in series with the battery and the load sees +20 V for an instant.

4. Then the capacitor starts to discharge through the load and battery so it's Voltage decreases and the load sees a Voltage that decreases from the +20 V value back down to the +10 V value.

5. And it repeats. V2 looks something like this:

{My attachment should be HERE! But this stupid BB software does not seem to want that. I give up. It is below}

I would alter your drawing by having the battery Voltage connected to the load while the capacitor is charging. That way the half of my drawing below 10 V would become a straight line at 10 V.

But even that is NOT how I would double the Voltage to the load. I do not see any reason why a floating ground is needed in this circuit. I would use the battery Voltage to run a DC to AC converter which produces an AC Voltage. Then I would use a conventional Voltage doubler circuit (diodes and capacitors) to boost the Voltage to the 20 VDC level. This would include A GROUND connection to the battery minus terminal.

Or, BETTER YET, just use a DC to DC converter in boost mode WITH A GROUND connection to the battery minus terminal.

The "floating ground" buys you nothing in a battery powered circuit. The whole thing can float. It can float on top of 100,000 VDC if you wish and it will still work. Just turn the 100,000 VDC OFF before changing the battery.

You are correct for the case where you reverse the connection of that capacitor you can get a peak of 20V but when it will discharge it will drop to zero not to only +10V as the capacitor is reverse charged.

My question was if there is any possibility to get V2 > V1 continuously not intermittent.
And the reason I ask this is because people claim that the mechanical analog of this circuit  can have V2 > V1 continually.
I was even banned on this forum for arguing that will not be possible.

The way I drawn the example with an MPPT DC-DC converter it is possible to  have V2 in the +5V and +30V range never get to zero but still it requires that V2 drops below 10V (below V1). It can be done even as +9V to +30V but it will be impossible to keep it  above +10V (above V1) continually.

You are neglecting the battery Voltage.

And you were not clear as to the duration of the boost to +20 V. You talked about both continuously and for a brief time.

I only talked about what your second drawing suggested. I did not intend to completely rule out the possibility of a different circuit accomplishing "the impossible". I will not be so brash as to say that it is totally impossible. Just because one possible circuit does not do it, does not mean that none can.

You may want to take a course in logic.

[electrodacus]

I only talked about what your second drawing suggested. I did not intend to completely rule out the possibility of a different circuit accomplishing "the impossible". I will not be so brash as to say that it is totally impossible. Just because one possible circuit does not do it, does not mean that none can.

You may want to take a course in logic.

I specifically mentioned that you can have any circuit you want in the green box as long as there is no energy in the box (so no batteries or hamster wheels in the box).

I will actually be so brash as to say that is totally impossible.
As far as I'm aware nobody has demonstrated that conservation of energy can be broken. So it will be logic to claim that any circuit inside the box will not be able to accomplish "the impossible".  Magic is not real despite multiple claimed evidence of magic smoke