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DC-DC converter with floating ground
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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.




--- Quote from: electrodacus on December 28, 2023, 02:33:34 am ---
--- Quote from: Andy Chee on December 28, 2023, 02:27:01 am ---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!

--- End quote ---

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

--- End quote ---
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:

--- Quote from: EPAIII on December 28, 2023, 06:32:39 am ---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.



--- End quote ---

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:

--- Quote from: magic on December 28, 2023, 08:01:08 am ---You cant' even make it sometimes higher and sometimes equal.

--- End quote ---

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:

--- Quote from: electrodacus on December 28, 2023, 12:58:26 am ---
--- Quote from: SiliconWizard on December 27, 2023, 11:43:51 pm ---You can't do that if your only return path is through the load. As shown in the diagram.

--- End quote ---

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



--- End quote ---
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.
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