At 12:45 PM 02/11/2011, you wrote:

>Hi all,
>a 2.2uF capacitor charged to 400V discharges in 20uS: what is the formula
>to calculate the average discharge current? Huge precision is not necessar=
y,
>it's just to size appropriately a MOSFET/BJT switch.

I think you will need to look at peak current and safe operating
area of the switch. The peak current is not determined by the
capacitor value, but by any series inductance and resistance and
the MOSFET characteristics (they are essentially a constant current
sink when driven on).

Take, for example, the FDP20N50F (a 20A 500V MOSFET).

http://www.fairchildsemi.com/ds/FD/FDP20N50F.pdf


With 8V drive at 25=B0C it will _typically_ self-limit at only 60A.
Worst case could be quite a bit higher, but then the capacitor ESR
and inductance could come into play.

It can withstand 60A at 400V for 10usec. That will be enough to
discharge 1.5uF completely in 10usec..

(C =3D 60A * 10E-6 / 400VDC)

... might be okay with 2.2uF, but
you also might want to limit the current (perhaps with 10-20uH of
inductance) or look for something a bit beefier if reliability is important=
..

If you can measure what's actually happening.. an actual peak
current in the 20A range will be a lot safer.. the SOA (safe
operating area) curves allow for 100usec at that current.

Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the rewar=
d"
speff@interlog.com             Info for manufacturers: http://www.trexon.co=
m
Embedded software/hardware/analog  Info for designers:  http://www.speff.co=
m




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