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'[EE] Capacitor discharge current formula?'
2011\11\02@124727 by

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 necessary,
it's just to size appropriately a MOSFET/BJT switch.

Thank You very much,
Mario

PS: I wanna play in the game too! Here's my attempt, maybe correct, maybe not:

I'm using two formulas. The first is:

thus I get 0.00088 Coulomb inside that charged capacitor.

Second formula:

Ampere = Coulomb / second

thus, if I'm not wrong, the right answer should be:

44 Ampere

Am I right or wrong?

If I'm right, then (since this is a pulse application) I need a switch which
can handle peak currents of more than 44A (say 60A to be safe).

Did I do my calculation correctly?

I'm 100% self-taught *blush* :o

Cheers,
Mario
Hi Mario,

Your calculation is indeed correct. However, I'm not sure that average
current is the best metric for you.

When a capacitor charges or discharges through a resistance (i.e, the
resistance is what limits the current), then the energy transferred
into or out of the capacitor is equal to the energy dissipated in the
resistor.

So, if your FET or BJT is the device which limits the current during
the discharge, then during each discharge pulse, the capacitor energy
will be dissipated in the FET/BJT.

Taking the average current and then doing an I^2*R calculation on it
will give you, I think, an unrealistically-high average power.

Sean

On Wed, Nov 2, 2011 at 12:45 PM, Electron <electron2k4infinito.it> wrote:
{Quote hidden}

>
i = C*dV/dt. A passable estimate, therefore, is
2.2u*400/20u = 44A.

Mike H

On Wed, Nov 2, 2011 at 10:45 AM, Electron <electron2k4infinito.it> wrote:

{Quote hidden}

>
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 necessary,
>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°C 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 = 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 reward"
speffinterlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

Hi,
thank You all for your replies.

My 20uS figure is actually due to the load (a motorbike coil whose
DC resistance is around 0.1ohm) which acts (at least on my 'scope)
quite like a constant current sink. I've made real world measurements
on it.

The capacitor I'm using is pulse-rated and can withstand that dI/dt
rate.

Of course the DC resistance of the coil doesn't say much, as it's
just the primary of a pulse transformer, then we have the secondary,
a resistor and a spark plug with a built in second resistor.

But if I pretend to be a transistor and ground it, on my 'scope I can
see that the capacitor discharges in 20uS, hence the calculation of
the pulse current that the switch has to widthstand. True, I could
have hacked together a current probe and measure it, but I just
thought why not calculate it.

If You think I better measure it, due to the uncertainty of the coil
inductance, capacitor ESR, etc.. I will definitely do it!

Thanks again.

With kind regards,
Mario
Hi Mario:

You can use a ferrite core to make a current transformer, see:

http://www.prc68.com/I/JouleThief.shtml#CC

Have Fun,

Brooke Clarke
http://www.PRC68.com

On Nov 2, 2011, at 10:32 AM, Mike Hord wrote:

> i = C*dV/dt. A passable estimate, therefore, is
> 2.2u*400/20u = 44A.

Heh.  I did t = RC to get an "equivalent" discharge resistance, and  V=IR to get the resulting current, hoping that "average" and the 1/e  factor in the time constant end up balancing each other "close enough."

This also results in an answer of 44A...

BillW

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