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RE: Primary Heating



Original poster: "Lau, Gary by way of Terry Fritz <twftesla-at-qwest-dot-net>" <Gary.Lau-at-compaq-dot-com>

My innermost primary connection is by means of a very beefy copper strap,
securely soldered to the end of the tubing, and the connection to the strap
~ 2" distant by means of a 1/4" bolt.  The outermost primary connection is
the movable tap, and this by means of two fuseholder clips.  This is where
I'd expect to see ohmic heating, but do not.

Gary Lau
MA, USA

Original poster: "Paul Nicholson by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <paul-at-abelian.demon.co.uk>

David Rieben wrote:

> most of the heating is confined to the 2 innermost turns

and others have reported similarly.  Wonder why.  We can exclude
a localised current max in the primary at the operating frequency
because the self-cap of the primary is very small compared with
the added parallel cap of the primary tank, eg 100pF/82nF < 1%, so
that the primary current is virtually uniform [*].

David, do you have a good low-resistance connection at the inner
end of the primary?  Perhaps you can run some 60Hz AC current through
the primary and see if the localised heating still occurs.

If a 60Hz current doesn't reveal the same localised hot spot, then
we might have to look for sources of HF. 

> the current max always occurs at the grounded end as opposed to

We expect uniform current in the primary at Fres.  But if higher
frequency energy is being developed in the coil anywhere (parasitic
resonance involving the gap, secondary arcs redistributing mode 
energy, etc) then it could be interesting.

> (V and I are running 90* out of phase, I think).  

Yes, I lags V by 90 degrees.

First, we should eliminate the simplest explanation - lossy connection
to primary inner, or heat from the gap conducting through to the
primary.

[*] providing we ignore the capacitance between primary and
secondary.  This can be quite high for the inner-most turns, and
the resulting displacement current adds a non-uniform component to
the primary current.  But this component can never exceed the coil
base current (if sec base is properly grounded), so therefore this 
non-uniform component of Ipri must be less than the main primary
current by at least the factor sqrt(Csec/Cpri).
--
Paul Nicholson
--