Re: High current discharge from capacitor

From: Jonathan Kirwan (jkirwan_at_easystreet.com)
Date: 08/07/04 

Date: Sat, 07 Aug 2004 10:09:33 GMT

On Fri, 6 Aug 2004 20:12:20 +0100, "markp" <map.nospam@f2s.com> wrote:

>> Sounds kind of like an "all-fire specification." Could it be a 'squib' to
>> ignite something?
>>
>
>Not ignite, melt. It's a gas release plug.

So, if you delivered the energy over a longer time period, conduction would
prevent it from always reaching the required temperature?

...

Capacitive discharge into an R isn't going to supply a constant 28V, you know.
The V will vary as energy is delivered. And the total energy as a function of
time is:

              _ t
         1 /
  E(t) = - * | [V(t)*V(t)*dt]
         R _/
           0

Substituting and solving for t yields,

  t = (1/2) * R*C * ln( C*V^2 / (C*V^2 - 2*E) )

If you try this equation with E=84Joules, V=28Volts, and C=540,000uF, and
R=2Ohms (rounded up) you'll see that it won't deliver 84 Joules in .2 seconds.
Even with over 210 Joules sitting on it, it cannot pump out 84 Joules into 2
Ohms in under .2 seconds. Upping the C value, even to 4F, won't help either.

Upping the voltage to 45V and using C=100,000uF, it takes .177 seconds to
deliver the 84 Joules into 2 Ohms. Can you accept 45V?

Sizing the cap on the basis of specifying that there be no more than 20% drop in
.2 seconds, doesn't mean that it will actually operate that way. The 20% drop
figure is made up and the .2 seconds is a requirement, but these don't have much
to do with what the capacitor and heater *will* do. So sizing a capacitor on
these assumptions, using C=I*dt/dV, won't calculate something that can actually
deliver the energy in the time you specified.

Note in the time equation above that C is both inside the ln() function as well
as outside, while V is only inside the ln() function. This means that if you
reduce C, but increase V to compensate so that you have the same stored energy,
the time to deliver the energy will actually be faster -- even though the stored
energy is just the same either way. For delivery speed of a given amount of
energy, you want to store it on a smaller C and with a larger V.

I decided to try out the energy calculations because I just didn't feel
comfortable with the glib 530,000uF to 540,000uF calculations.

Jon

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