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[TCML] VSWR, Q factor, Loads and Impedances



From another thread:
> Low VSWR can assure more radiation than a high VSWR which
> insures large sparks.

> I'm not sure that the concept of VSWR is even applicable
> to Tesla coils.

VSWR doesn't have any great meaning here.

It would start to apply if you broke the connection between coil
base and ground and inserted an RF source.   Then the VSWR would
be the ratio of top volts to base volts once the steady state
condition has been reached - ie once the resonator has 'rung up' and
the continuous power coming in from the source balances the power
dissipated in the resonator's losses and load.

Consider a loss-free resonator with some load Zload applied.  If
the resonator has characteristic impedance Zc, then the input
(base) impedance seen by the source is

  Zbase = Zc^2 / Zload

This is the impedance 'inversion' effect of a quarter-wave resonator.

If the power passing from source to load is P, and all the
impedances are purely resistive, then

   P = Vbase^2 / Zbase
and
   P = Vtop^2/ Zload

(since no power is lost in the loss-free resonator)

Equating these, we get

   Vtop/Vbase = sqrt( Zload / Zbase) = VSWR

and pulling in the 'inversion' equation,

   VSWR = Vtop/Vbase = Zload / Zc = Zc / Zbase

Suppose now another case, this time a lossy unloaded resonator
having Q-factor Q.  In steady state we have

   Zbase = Zc / Q
and
   Vtop = Zc * Ibase
and
   Vbase = Zbase * Ibase

Putting these together gives VSWR = Vtop/Vbase = Zc / Zbase = Q

So the VSWR here has the same value as the Q factor, although they're
distinct quantities with different definitions.

When our RF source is removed and the base is grounded again, we
can no longer use the ratio Vtop/Vbase since Vbase is now clamped
to zero.  The VSWR becomes undefined and we must use instead the
Q factor.

--
Paul Nicholson
--

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