Re: Tube Data

To: tesla-at-pupman-dot-com
Subject: Re: Tube Data
From: "Malcolm Watts" <malcolm.watts-at-wnp.ac.nz> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)
Date: Sun, 27 Feb 2000 22:37:26 -0700
Approved: twftesla-at-uswest-dot-net
Delivered-To: fixup-tesla-at-pupman-dot-com-at-fixme

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.
 
> 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?
 
> For a typical disruptive TC operating at 300kHz with a 0.1湩 primary
> capacitor you would need about 28焙 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.

> 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?

Regards,
Malcolm

<snip>