confused about Dielectric

[Bluga]

i'm reading the following paragraph but i'm confused at what its saying(bold).

My question is: if i have a capacitor initally not charged and is connected to a voltage source thorugh a switch (open),
now if i insert a dielectric material and closes the switch, doesn't this charge up the capacitor??

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In measuring the dielectric constant, a basic understanding of capacitance theory is beneficial. By storing charges,
capacitors have found applications in memory storage devices where the stored charges can represent digital "ones"
and "zeros." Problems arise because of the switching times of a capacitor to charge and discharge this stored charge.
To compensate this, capacitors are being incorporated directly on the substrates. But in order to store an appreciable
amount of charge, the surface area of the conductor plates must be increased,
which decreases the number of capacitors which can be incorporated on a given substrate.

To counter this, materials have been inserted in-between the conductor plates. These materials, or dielectrics,
act as insulators to increase the charge storage capabilities of the capacitor. This results because the dielectrics
contain charged molecules which are randomly oriented. When an external field is applied, by dropping a potential
across the two plates, the charged molecules align themselves with the electric fieldThis alignment of charges
produces dipoles where the positive charges of each molecule are in the direction of the applied field and the
negative charges oppose the field. An internal electric field, which is opposite in direction
of the external electric field, will result. Consequently a reduction of the overall electric field and the overall potential occurs.
Referring again to the definition of capacitance, if the potential across the two plates is reduced, the capacitance is increased.

 

[rimshaker]

When the cap is connected to a DC voltage source (switch closed), in effect it 'charges' up to whatever the voltage source is. There is now a built-in potential (voltage) across the capacitor. It's already 'charged' up, that might be the confusing part. Inserting a dielectric will simply reduce this potential, due to that parasitic dipole. Inserting the dielectric before of after the switch is closed doesn't matter, the voltage will still be less than the voltage source. Once it's removed, the voltage will rise up to whatever the DC source is.

 

[Bluga]

Thank you, my question is:

if i have a capacitor initally connected to a voltage source (12V) thorugh a switch (switch is OPEN, so capacitor is NOT CHARGED),

now, if i insert a dielectric material (k=3.4) and CLOSES the switch at same time, it charges up the capacitor right? will it be bigger than 12v or smaller?

Furthur if i supply addtional power to the capacitor, how will it affect the capacitor?

 

[CTho9305]

regardless of dielectric, if you put 12V dc across a capacitor, it will charge to 12 volts. now, the difference is that with better dielectrics, you will have more charge stored. That means you can draw more milliamp-seconds from it.

I still dont understand how/why a dielectric does what it does though.

 

[Eskimo]

Dielectrics can serve two important properties in a capacitor. The first is the addition of physical strength to a system. As you know capacitance is inversely proportional to the seperation between the two terminals. This fact leads designers to want to place the terminals as close to eachother as possible. Unfortunately at high values of capacitance the electric field set up inside the capacitor can physically deform the capacitor plates unless a physical medium exists to support the plates.

Additionally dielectric materials provide an increase in dielectric strength. Defined as the maximum electric field you can apply before the electrons are literally ripped across the region seperating the plates. Air only has a dielectric strenght of about 0.3MegaVolt/cm. After that value there is an ionization and charge will spark across the void. By contrast Silicon Dioxide has a dielectric strength of about 5-10MV/cm. Of course we use Si02 as our main dielectric of choice in semiconductors and while 10MV/cm sounds like a lot when you consider we have film thicknesses of 20-25Angstroms that's not that much.

 

[Menelaos]

Bluga, I think I see your problem.

The voltage applied to the capacitor is the potential difference corresponding with the external electric field, mentioned in the paragraph you quoted.

Take a cap without dielectric and charge it to eg. 12 V. Now put the dielectric in between the plates. The dielectric will polarize (transl.?), as described in the paragraph, and the voltage will drop below 12 V. The cap starts charging until the voltage is again 12 V.

So the overall potential with dielectric is lower, using the same amount of charge; or
the stored charge is bigger, using the same overall potential.

Hope this helps,

Menelx.