Can a DC Capacitor Transformer Transform Direct Current Voltage?

[Low-Q]

DC "transformer"

I just had an idea not using induction as in regular AC transformers, but using the electric field in a capacitor using DC in order to transform DC voltage directly without advanced electronics. The idea is using a special capacitorl. The capacitor have a primary pair of foils, while there is a secondary pair of foils that is "picking up" the electric field from the primary foils.

Is it possible to transform DC voltage from one specific voltage and up or down to another specific voltage using the method above? If there is anything that must be explained in a better way, please let me know.

Vidar

 

[mfb]

With capacitors, that is not possible. Just imagine how a current flow would look like - there is no way for the current to flow continuously. Voltage dividers exist, but they need resistors and waste a significant fraction of the voltage as heat, so they are not real "transformers" in that sense.
There are DC-DC-Converters, but they all need some active electronics.

 

[jim hardy]

If there is anything that must be explained in a better way, please let me know.

Magnetic Induction implies relative motion between the conductor and the field, even if that 'motion' is just the field's intensifying, or expanding if that's easier to visualize.

... to transform DC voltage...

What are some formulas things like voltage and current and capacitance ?
DC implies to me things are steady.
What are you planning to move?

 

[Baluncore]

Some of the most precise DC voltage dividers and multipliers are built using capacitors.
But they do not use the principle you suggest of a three electrode capacitor.
Take a look at page 8 of the LTC1043 data sheet at; http://www.linear.com/product/LTC1043

 

[Carl Pugh]

Yes it is possible to transform a voltage up or down using just a capacitor.
If the plates of a capacitor have a charge and the plates are moved closer together or farther apart, the voltage on the plates will change.

Capacitor
Voltage=Charge/Capacitance
So if Capacitance is halved, Voltage is doubled

However this is probably not a practical way to change the voltage.

 

[analogdesign]

Yes it is possible to transform a voltage up or down using just a capacitor.
If the plates of a capacitor have a charge and the plates are moved closer together or farther apart, the voltage on the plates will change.

Capacitor
Voltage=Charge/Capacitance
So if Capacitance is halved, Voltage is doubled

However this is probably not a practical way to change the voltage.

In practice this is most certainly done in integrated circuits at least. It is the very basis of switched-capacitor analog signal processing. You can't move the capacitor plates in an IC of course, but you can do things like rearrange capacitors from serial to parallel and switch in additional capacitors after charging some, for example. Typically you need op amps to enforce virtual grounds so you can avoid charge sharing issues.

These types of circuits are truly ubiquitous, but most people aren't aware of them because they are buried deep into the electronic products we use.

Besides analog signal processing, on-chip capacitive DC-DC converters are a hot research topic right now to deal with the crisis in I/O pins. They typically use a parallel-to-serial structure, although many of them are quite innovative. They are used in every Flash memory oxide charging circuit, for instance, and are poised to be used almost everywhere soon.

 

[meBigGuy]

Capacitive DC to DC converters are common, and available as stand alone chips.
http://www.linear.com/product/LT1054

Also explained well here
http://www.maximintegrated.com/app-notes/index.mvp/id/725

As to the original question, I don't see any way to transform voltages without something changing, field wise. Moving the plates is an interesting concept.

 

[Baluncore]

Moving the plates is an interesting concept.

That is called a parametric amplifier.
The parameter being varied is the capacitance. If charge is fixed then; dV = Q / dC.

 

[sophiecentaur]

I am sure someone has tried to use the effect in the design of a microphone.

 

[Low-Q]

Thanks for all feedbacks. Maybe I should make a 4-foil capacitor just to figure out what happens to the charge of third and forth foil (secondary foils) while applying charge to the first and second foils (primary foils), because it does not look like your answers agree with each other :-)
I will post the results here.

Vidar

 

[meBigGuy]

I think the answers here all agree, but you are not interpreting them accurately. Assuming the foils are stationary there will be charges developed on all foils as you apply an initial charge (as in series capacitors), and then all will stablize and nothing will change. If you then remove charge, things will change also as they do in series capacitors.

The two concepts presented here are that you can physically move the foils containing constant charge which will change the voltage, or you can transfer charge with switches.

 

[sophiecentaur]

I think a diagram would help here. I am not sure we all have the same arrangement in mind. An isolate foil, placed parallel to the outer foils would not 'acquire' any charge, it would just polarise and behave as if it were not there. You would, in fact have two capacitors, each of greater value (twice if it's dead centre) in series, giving the same net capacitance for the main capacitor.
The situation with two foils, placed in between would depend on whether and how they were connected.

The Cockroft Walton voltage multiplier works using spark gaps to alter the connection between a set of parallel capacitors to a series connection - giving several times more kV. But that relies on non-linearity.

 

[jim hardy]

I am sure someone has tried to use the effect in the design of a microphone.

Perfect ! To OP - try a search on 'electret'

 

[Baluncore]

The Cockroft Walton voltage multiplier works using spark gaps to alter the connection between a set of parallel capacitors to a series connection …

Actually the Cockcroft Walton generator uses diodes and capacitors to pump charge, there is no capacitance variation or parallel to series conversion.

A Marx impulse generator uses spark gaps to change charged capacitors from a parallel charge to a series voltage sum.

The Wimshurst machine has been described a variable capacitance voltage multiplier.

 

[sophiecentaur]

Actually the Cockcroft Walton generator uses diodes and capacitors to pump charge, there is no capacitance variation or parallel to series conversion.

A Marx impulse generator uses spark gaps to change charged capacitors from a parallel charge to a series voltage sum.

The Wimshurst machine has been described a variable capacitance voltage multiplier.

Oh yes - you are right about the 'Cocky Wally' etc.. My memory!

I had always understood the Wimshurst in terms of Inductive Charging but I guess it's only a different way of looking at the same thing.

 

[Low-Q]

Hi,

Made a rough drawing of the principle. However it does not seem that Windows 7 is capable of doing the simple trick to make file transfer into an FTP site possible (Easy on XP, but in 7, no). It is not possible to upload pictures directly to this forum so I must try to describe the basic principle.

Imagine 4 conductive plates oriented in parallell with a small space between them. Supply voltage to the two inner plates, and wait till they are fully charged, and keep the voltage supply connected. Now measure the voltage over the two outer plates that has no physical contact to the inner plates. Will there be any measurable voltage over the two outer plates?

Vidar

 

[sophiecentaur]

Hi,

Made a rough drawing of the principle. However it does not seem that Windows 7 is capable of doing the simple trick to make file transfer into an FTP site possible (Easy on XP, but in 7, no). It is not possible to upload pictures directly to this forum so I must try to describe the basic principle.

Imagine 4 conductive plates oriented in parallell with a small space between them. Supply voltage to the two inner plates, and wait till they are fully charged, and keep the voltage supply connected. Now measure the voltage over the two outer plates that has no physical contact to the inner plates. Will there be any measurable voltage over the two outer plates?

Vidar

Are you having a problem with this? If you go to 'advanced' there is a 'manage attachments' button below the editing window. It will handle most common formats and presents them as clickable thumbnails.

 

[Low-Q]

"Attachment" ofcourse - not "Insert image". Thanks! Here it is.
The lighter blue and red are the "output" negative and positive charge.

 

[sophiecentaur]

As the two extra plates are insulated, there can be no change in the charge on them. They will polarise by induction and the fields around the capacitor will change. But I can't think of any reason for any change of potential

 

[mfb]

Imagine 4 conductive plates oriented in parallell with a small space between them. Supply voltage to the two inner plates, and wait till they are fully charged, and keep the voltage supply connected. Now measure the voltage over the two outer plates that has no physical contact to the inner plates. Will there be any measurable voltage over the two outer plates?

If the plates are insulated and all distances are small compared to the plate sizes, roughly the voltage you applied to the inner foils.

As soon as you connect anything conducting (like most voltmeters), the voltage will drop quickly.@sophiecentaur: A potential change does not need a changed charge. You can draw the system as 3 capacitors in series, where the middle one (going from the outer side of one plate to the outer side of the others) is tiny compared to the other capacitors. This tiny capacitance will get nearly the full voltage.

 

[meBigGuy]

Imagine 4 conductive plates oriented in parallell with a small space between them. Supply voltage to the two inner plates, and wait till they are fully charged, and keep the voltage supply connected. Now measure the voltage over the two outer plates that has no physical contact to the inner plates. Will there be any measurable voltage over the two outer plates?

Vidar

Not really. Just whatever charge was capacitively coupled from the inner plates when the voltage was applied. And no more will appear when you drain them. (except you will capacitively drain a little from the first two plates). No additional energy will ever be transferred into the outer plates from the inner plates unless the inner plate voltage is changed. You will be able to pick up more energy on the outer plates from your local radio station than from the inner plates.

 

[sophiecentaur]

Not really. Just whatever charge was capacitively coupled from the inner plates when the voltage was applied. And no more will appear when you drain them. (except you will capacitively drain a little from the first two plates). No additional energy will ever be transferred into the outer plates from the inner plates unless the inner plate voltage is changed. You will be able to pick up more energy on the outer plates from your local radio station than from the inner plates.

But bringing the plates into position can involve work being transferred, I think, because of the induced charges on a finite width plate. How doe this square with what we've been discussing?

 

[meBigGuy]

But bringing the plates into position can involve work being transferred, I think, because of the induced charges on a finite width plate. How doe this square with what we've been discussing?

My comments were for stationary plates.

 

[sophiecentaur]

But whether the plates are stationary and the supply is turned on or the plates are brought into position, there could still be work involved or energy transferred. Either way, there is a possible difference between the energy situations with and without the extra plates. However, if the plates have no resistance, then there would be no potential across them (i.e. across the metal of each plate and not 'between' them) so my worry may be unfounded.