Re: Capacitor Selection

["Andrew Kilgore" <amkilgore@xxxxxxx>]

Hi,

Well, I really know very little about this...
(In fact, if you've read any of my other posts, you'll soon realise that I
know very little about most things :o))

My instincts would tell me that current leakage with regards to a capacitor
is the "leakage" of charge across the plate.
In an ideal world you would be able to connect a capacitor to a circuit,
charge it, remove it from the circuit and it would hold it's charge
indefinitely. However, this ain't an ideal world. In the above example when
you remove the capacitor from the circuit, it would maintain it's charge,
but this would gradually decrease over time due to "leakage"
across the
plates.

I know that different types of capacitors have different types of
dielectrics (the material used between the plates) and this will effect the
leakage across the plates.

Anyway, that's enough guessing for now.
I think it's about time some of the electronic buffs on here put us all out
of our misery...?!?

Thanks,
Andy.


----- Original Message -----
From: "Hawes,Timothy Edward (GEG)" <haweste@xxxxxxx>
To: <ukha_d@xxxxxxx>
Sent: Monday, August 04, 2003 11:27 AM
Subject: RE: [ukha_d] Capacitor Selection


James, Andy,

Thanks. I'm still a bit confused though as the "CV" was related
to leakage
current i.e. leakage current: 0.01CV or 4µA (whichever is greater).

Any thoughts ?

Tim H.
(wishing I'd paid more attention at school now too . . .)

-----Original Message-----
From: James Hoye

> I think I know the equation you're talking about: Q = CV.
> As far as I remember, Q is the charge is Columbs, C is the capacitance
in
> Farads and V is the voltage (in Volts) across the plates.
>
> Not sure if this helps at all in this instance.

Ooops... I was way out :$

Yes it was Q = CV.

Funny how you spend months doing things like AC theory and never actually
seem to use them once you've done the exam.

James H

+


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