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From: Phil Allison on 2 Jun 2010 21:42

"Jeroni Paul"

> "Frank Zabcar"
>
> The current in an ideal capacitor is given by ...
>
> I = C x dV/dt
>
> Can you use this to estimate the current draw?
>

I tried this taking only the fall ramp:
dV = 0,781 V
dt = 4,72 us
C = 1000 uF

It results in 165 A. I may have done something wrong but I think the
capacitor ESR may play a role here as it causes an increase of dV not
reflected at I.

** Sure does !!

An **ideal** 1000uF cap has an impedance of under 1 milliohm at 200kHz -
in reality no such beast exists so the simple formula above fails
hopelessly.

A standard grade 16 volt, 1000uF cap has an ESR of around 200 milliohms ( at
high frequencies) while a low ESR type has about 20 - 30 milliohms and both
have series inductance that amounts to 20 milliohms at 200kHz.

Your 1000uF cap is getting hot cos it has way too much ESR !!!

Get a low ESR type, rated to handle 2 amps or more of ripple.

The Panasonic FM series 16V, 1000uF cap has a rated ESR of 19 milliohms and
a ripple current rating of 2.2 amps.

Note:

The ESR of electros goes DOWN when the cap gets hot - by as much as a
factor of 5.

Means cheap, high ESR electros can be made to do the job in a SMPS at the
expense of having a very short life.


..... Phil




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