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Re: Reflections in Power Transfer
Subject: Re: Reflections in Power Transfer
Date: Wed, 7 May 1997 13:19:12 +1200
From: "Malcolm Watts" <MALCOLM-at-directorate.wnp.ac.nz>
Organization: Wellington Polytechnic, NZ
To: tesla-at-pupman-dot-com
Hi all,
here is the post I promised on power transfer/reflection. It
also relates to the recent question on ground wiring. The relevant
part will be highlighted.
In any power transfer situation, there is a maximum power that
can be transferred when source and sink impedances match resistively
and must be conjugate if other impedances are involved (e.g. if
there is a choke in the circuit, there must be a capacitor also so
that total reactances = 0.
Situation 1: A transformer hooked to the mains. No load = no power
transfer. Reactive energy is returned to the mains minus some loss
in winding resistance and mains wiring.
Short-circuit load = no power transfer. In this case, power is
dissipated in any circuit resistances e.g. windings, mains wiring
since these in effect become the load.
Resistive load = power transfer.
Situation 2: The Coil we all know and love. No load (secondary spark)
= no power transfer to the spark. Energy is dissipated in circuit
resistances, most notably the gap and RF ground. Robert Stephens has
an excellent video showing how dramatic the difference in gap
brilliance is between the loaded and unloaded situations.
Short circuit across secondary = no power transfer (zero voltage
across the "load" = zero current through it). This also gives the
gaps a wake-up call.
Either of these two situations causes a prolonged energy trade
unless the quenching is pretty stunning. I haven't yet successfully
quenched a gap when a significant amount of energy remains in the
primary or secondary.
You can see this effect for yourself if you run a coil at below
breakout power, then slowly move a grounded discharge rod toward the
terminal. You should get a bright-dull-bright pattern in the gap as
the rod is moved in to touch the terminal. Running a TC short or open
like this hammers the primary components.
Situation 3: Secondary alone. As shown by the Corums this can be
viewed as a trnasmission line section terminated by a capacitance at
the top end and a near short-circuit at the primary end. The higher
the top capacitance, the shorter the line. With no added capacitance
it is approx 1/4 wavelength long electrically at the resonant
frequency. In fact the effective length of the line determines Fr, and
the effective length is itself determined by the distributed L and C
components.
There is an equation relating the characteristic impedance of
a 1/4 wave transformer such as this and its terminating impedances.
Zo = SQRT(L/C) if series resistance is low and shunt conductance is
low.
In terms of the terminations, Zo = SQRT(Ztop x Zbottom) where
Ztop and Zbottom are the terminating impedances. In any instance
where Zo is not the same as a terminating impedance, a standing wave
appears on the line as a result of boundary reflections. Since
Zbottom is near zero Ohms, Ztop is an extremely high impedance (no
spark). If a spark results such that Ztop = Zo, there is no
reflection from that end of the line and VSWR drops to around 1.
If Ztop = a short circuit, it is not = Zo and a reflection once
again takes place. In this case, the secondary is now terminated by
two short circuits and the dominant resonant mode is half-wave at two
times the open circuit resonant frequency or thereabouts. The
reflected wave from the top is completely out of phase with the
primary and appears to the primary as a high impedance which the
primary cannot transfer power to.
So how come the primary can transfer power to the secondary if
the secondary is unterminated at the top end? It is because of the
phase relationships in the circuit. In this case, the reflections are
in-phase with the primary oscillations by the time they get back to
the bottom of the coil and bounce off the bottom end short circuit.
This only works if the primary is situated near the short circuited
end. Anywhere else and the phase relationships are less than optimum.
In brief: No load on the top end causes a reflection from the
top that is in-phase with the wavefront arriving at the top. A short
circuit load causes a reflection that is 180 degrees out of phase
with a wavefront arriving at the top. An optimum load causes no
reflection at all.
Regarding grounding, the equation relating the various
impedances clearly shows that less than an ideal ground causes output
impedance to be lower than it could be (= less voltage rise because
of increased ground path losses = lower Q).
That'll do. I could write a technical treatise on it but the Corums
have done a pretty good job already. Hope this isn't too
simplified (or complex depending on your point of view).
Malcolm