DC-DC converter with floating ground

[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.