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?