Question about capacitors ?!

Thread Starter

greenpeace

Joined Sep 22, 2007
30
hello everybody ,

before I introduce the problem , I have to tell you that I'm not an electrical engineer ,
but I have a problem related to your major & I hope someone can help me !

suppose that we have to insert a capacitor in a circuit

but for that capacitor = instead of consisting of to adjacent plates

it consists of many plates ( say 8 or 10 or any other )

( I don't know if it is called multi-capacitor or not :confused: )
The question is / how these plates will be connected ?
& what is the benefits of using many several plates instead of just 2 ?


hope someone can help me & give me as more information as possible
 

Gadget

Joined Jan 10, 2006
614
-|-|-|-|-|-|-|-|-|-|----------- terminal 1
--|-|-|-|-|-|-|-|-|-|---------- terminal 2

By interleaving the plates a greater plate surface area is achieved in a small space meaning more capacitance. Commonly seen in Tuning gangs.
 

Thread Starter

greenpeace

Joined Sep 22, 2007
30
thanks < Mr . Gadget :)

but can you explain how the plates are connected more please :(

( keep in mind that my knowledge in this field is very poor )

I'm mechanical engineer :D

just explain it in words
& if you can / explain that illustration about terminals !
 

Papabravo

Joined Feb 24, 2006
14,691
The metalic plates are soldered or welded to a metalic conductor. One set of plates is connected to one of the two terminals. The other set of plates is connected to the other set of terminals. In this configuration there is no way for a DC current to flow. Only an AC current can flow through a capacitor. That concept may be beyond your understanding at the moment, but you can take it on faith for the time being until you have a chance to investigate the matter further.
 

Thread Starter

greenpeace

Joined Sep 22, 2007
30
-|-|-|-|-|-|-|-|-|-|----------- terminal 1
--|-|-|-|-|-|-|-|-|-|---------- terminal 2

By interleaving the plates a greater plate surface area is achieved in a small space meaning more capacitance. Commonly seen in Tuning gangs.

I didn't understand how each plate is connected to the plate next to it ??
 

Thread Starter

greenpeace

Joined Sep 22, 2007
30
The metalic plates are soldered or welded to a metalic conductor. One set of plates is connected to one of the two terminals. The other set of plates is connected to the other set of terminals. In this configuration there is no way for a DC current to flow. Only an AC current can flow through a capacitor. That concept may be beyond your understanding at the moment, but you can take it on faith for the time being until you have a chance to investigate the matter further.
I will try to illustrate my imagination of that capacitor
and you tell me which configuration is the right one !
 

Thread Starter

greenpeace

Joined Sep 22, 2007
30
I want your answers to consider :

* the way of connecting each plate to the plate next to it .

* charge distribution in the plates .

* the separation between the plates ( whether the plates are equally separated , or there is separation between two sets of plates < beside the separation between each plate > )
 

thingmaker3

Joined May 16, 2005
5,084
I have to tell you that I'm not an electrical engineer, but I have a problem related to your major & I hope someone can help me !
My major was Theater Arts, but I'll answer your question anyway.;)

Your drawings are capacitances in series. They are solved using 1/C1 + 1/C2 + 1/C3... + 1/Cn = 1/Ct where Ct is the equivalent capacitance. Ergo, if you have eight plates instead of two, and they are all the same size and same distance apart with the same dielectric, you have 1/8 the Farads you would have with a "normal" capacitor of the same specification.

There is no reason to do this, of course. Much less costly would be to simply make a smaller capacitor.
 

Thread Starter

greenpeace

Joined Sep 22, 2007
30
My major was Theater Arts, but I'll answer your question anyway.;)

Your drawings are capacitances in series. They are solved using 1/C1 + 1/C2 + 1/C3... + 1/Cn = 1/Ct where Ct is the equivalent capacitance. Ergo, if you have eight plates instead of two, and they are all the same size and same distance apart with the same dielectric, you have 1/8 the Farads you would have with a "normal" capacitor of the same specification.

There is no reason to do this, of course. Much less costly would be to simply make a smaller capacitor.
nice to shear knowledge :)

by this configuration we increase the cross-sectional area
and therfore increasing the capacitance .

but the problem is , how each plate is connected to the plate next to it ?
and what about the charge ditribution in each plate ?
 

thingmaker3

Joined May 16, 2005
5,084
nice to shear knowledge :)
I've often been accused of "cutting up.":D

by this configuration we increase the cross-sectional area and therfore increasing the capacitance .
Such is not what your drawings show. Was this your intent? To show what others in this thread have described? The scheme in your drawings decreases capacitance.

but the problem is , how each plate is connected to the plate next to it ?
By any method cost-effective to the manufacturer. The plates would not be connected to the next plate, though, rather to the one beyond. See the "variable tuning capacitor" in this reference: http://encyclopedia2.thefreedictionary.com/capacitor
and what about the charge ditribution in each plate ?
Even, equal, and pretty much uniform.
 

Thread Starter

greenpeace

Joined Sep 22, 2007
30
Such is not what your drawings show. Was this your intent? To show what others in this thread have described? The scheme in your drawings decreases capacitance.
I told you before that i cannot imagine its shape
that's why i draw it , so you can tell me if I'm right or wrong :D


By any method cost-effective to the manufacturer. The plates would not be connected to the next plate, though, rather to the one beyond. See the "variable tuning capacitor" in this reference: http://encyclopedia2.thefreedictionary.com/capacitor
thanks alot for the link

see the attachement and answer my question :)
 

Attachments

thingmaker3

Joined May 16, 2005
5,084
The gray plates are connected electrically as one plate. The white plates are connected electrically as the other plate. When the gray plates rotate into the spaces between the white plates, capacitance is increased.
 

Thread Starter

greenpeace

Joined Sep 22, 2007
30
The gray plates are connected electrically as one plate. The white plates are connected electrically as the other plate. When the gray plates rotate into the spaces between the white plates, capacitance is increased.
can this be done without rotation ?

I mean is there any type of capacitors with many plates
where the two sets are stationary ?
 
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