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Re: Tube Data

Hi Malcolm,
                   my comments:
> Original Poster: "Malcolm Watts" <malcolm.watts-at-wnp.ac.nz>
>
> Hi Nick, all,
>
> > Original Poster: "Megavolt Nick" <tesla-at-fieldfamily.prontoserve.co.uk>
> >
> > Hi All,
> >           One of the main characteristics of an electron emission device
> > like a vacuum tube is that it will have a comparatively high output
> > impedance. The tube I mention below can supply 400ma at 5kV -
> >
> > Z = V/I
> >
> > Z = 5000/0.4
> >
> > Z = 12500
>
> I am aware of that - I do design valve audio equipment. However, the
> only part the primary of the transformer plays as far as its
> inductance goes is to present a sufficiently high shunt impedance at
> the lowest frequency of interest, the impedance match actually
> being effected by considering the turns ratio, coupling constant
> (nearly 1 in this situation so is generally disregarded) and the actual
> load impedance.
Yes

>
> > This means that to avoid damaging the tube the load connected to it must
> > have an impeadance of at least 12.5 kiloOhms.  Therefore it must use a
> > primary coil with a large inductance to provide this impedance.
>
> What happens if you shunt this impedance with a considerably
> lower load impedance?
You have to bother to do all the maths ;-)

>
> > For a typical disruptive TC operating at 300kHz with a 0.1µF primary
> > capacitor you would need about 28µH of primary inductance to tune, which
has
> > an impedance of only 5 Ohms; whereas a vacuum tube design would use a
much
> > larger primary to provide the large impedance, about 6.6mH would be
needed.
>
> Comparing disruptive discharge and CW is an apples and oranges
> situation isn't it? In one case, you are simply transferring a block of
> energy - in the other, you are supplying energy on a continuous
> basis.
Yes - the primary impedance of the disruptive tc is really calculated as a
PFN calculation rather than a simple impedance.

>
> > This arrangement would suit a magnifier system as the large primary
needed
> > could be wound co-axially with the secondary for most of its length
giving
> > the very high K factor a magnifier needs.
>
> I am inclined to think that you are really looking at a 1/4 wave
> transformer situation for a CW magnifier. Quantifying the load
> impedance is the interesting bit. Things would get more interesting
> for effecting a match if you include loose link coupling.
>        Again, does anyone bother working this out properly or is the
> general rule to take a suck it and see approach?

The reason that my reply was so quickly shot down by anyone who knows what
they're doing (malcolm :-) is that to actually calculate the load impedance
of a vttc requires rather a lot of maths I'd rather not do - you have to
take a sec. backwards approach and calculate the charateristic impedance for
the topload, then work out the sec. impedance based on the load the sec is
seeing (the topload) and the sec.'s own charateristics, then work out the
presented impedance through the pri. from this, factor in the pri
characteristics and you're there.  Definitely not before breakfast ;-)

The real problem is that there are two large unknowns - the k factor and the
topload impedance.  These would require some fairly concerted effort to
measure to any reliable accuracy.
Fairly sharp calculus would be a good idea as well :-)

Regards
Nick Field

>
> Regards,
> Malcolm
>
> <snip>
>
>
>