(I hope I haven't duped this message).
I think you are right and wrong at the same time.
The original discussion did relate to the topic above, but your
statement was an absolute one unrelated to the topic.
In the example below (presumably an arc has been struck and the cap bank
is busy discharging), it would not be correct to regard the "impedance"
of the arc as 50 ohms, it could be closer to zero.
Caps, like batteries, do have an internal resistance which limits the
charge flow.
At the end of the day, the impedance of the wiring out of the cap bank
into the arc will limit the charge flow.
What everyone has ignored in this thread so far, is that the whole
process is being refuelled by the power supply (if the d.c. path is
available), which probably hasn't smoked yet, or the continued RF
output, and if an arc is being discussed, maintaining the arc once it
gets going needs only a tiny part of the charge needed to "strike" it in
the first place. The vapourised conductors or semiconductors maintain
the "burn" until there is just about nothing left. This is not a long
period of time, probably less than milliseconds.
Therefore in discussing the arcing of resistors, chokes, blocking caps,
or tuning caps, they surely have to be considered as being hot (very
hot) prior to the time to the arc/parasitic, call it what you like,
occurs.
Having said this, I wouldn't be surprised if the normal (i.e.,
explosive) failure mode in tube PAs is simply overheated, or
dirty/contaminated, high voltage components letting go.
Do you agree?
Ian ZS6BTE
Jon Ogden wrote:
>
> on 3/10/00 12:29 AM, Ian Roberts at itr@nanoteq.co.za wrote:
>
> > Jon Ogden wrote:
> >> Capacitors discharge at an exponential rate and cannot deliver an
> >> instantaenous spike.
> >
> > Speaking from memory, I recall that a 1 micro size cap can provide a
> > peak current of 20 amps, therefore 32 micros can provide...?
>
> It's still not an INSTANTENOUS function. A cap doesn't discharge all
of
> it's energy at once. And 20 amps from 1 uF sounds like an awful lot.
>
> Let's see....
>
> Let's say we have 32 uF at 4000 volts as in the example all along.
>
> The stored energy is:
>
> J = (32E-6 * 16,000,000)/2
> J = 512/2
>
> J = 256 Joules
>
>
> Now, if 1 uF can deliver 20 amps then at 32 uF, we have 640 Amps peak
> capability.
>
> So at 4000 Volts and 640 Amps, we get 2,560,000 Watts.
>
> Is the typical ham amplifier power supply really capable of delivering
a
> peak power of 2.56 MegaWatts?
>
> Knowing that we have 256 Joules of available energy and that a Joule is
a
> Watt second, this means that the cap would be discharged in 0.39 uSec.
>
> Let's say we have connected the supply output to a 50 Ohm resistor
connected
> to ground.
>
> We know that capacitor discharge is expressed by the formula:
>
> V = E *(e^(-t/RC))
>
> where E is capacitor voltage and e is the natural logarithmic base
(2.718).
>
> Let's see how much voltage our 32 uF capacitor bank has left after 0.39
> uSec.
>
> V = 4000*(e^(-0.39E-6/(50*32E-6)))
>
> V = 3999.025
>
> So....I don't think a 1 uF cap can provide a peak current of 20 Amps.
>
> The amount of time for the capacitor bank to discharge to 0.5 volts
(since
> you can't mathematically take the log of 0) is:
>
> 0.5 =4000* e^(-t/RC)
>
> 1.25E-4 = e^(-t/RC)
>
> -8.987 = -t/RC
>
> 8.987 = t/(50*32E-6)
>
> t = 14.3792 milliseconds.
>
> There for we have approximately 256 Joules of energy released in
14.3792
> mSec. This would be 17,803 Watts delivered to the resistor in that
period
> of time.
>
> A lot less than 2.56 MW.
>
> We also know the resistor can handle 64 kW at 5 msec. At 15 msec
(close to
> the 14.3792 millisecond number) the power handling is 64kW/3 or 21.333
KW.
>
> The resistors will NOT be destroyed.
>
> There!
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