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Re: NST power rating -- another perspective
Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br>
Tesla list wrote:
>
> Original poster: "John H. Couture" <couturejh-at-mgte-dot-com>
>
> Gerry -
>
> It appears I misunderstood what you were trying to measure, something I am
> prone to do. You mention maximum power transfer and the Thevenin theorem. I
> have used the Thevenin theorem only for bridge problems. What does (^^2)
> mean? I cannot follow your Thevenin equations but maybe Antonio, Paul N, or
> some other coiler can help.
"^^" should be "^", exponentiation. Let's see:
> This could very well be for a pure resistive load, but for me this is not an
> interesting case. The voltage transfer function for an L R RL circuit (where
> L and R represent thevenin parameters of the NST and RL is the load
> resister) is:
> VL = Vsec(oc) [ RL/(sL + R + RL)]
In Laplace transforms, ok.
> the current in RL is:
> IL = Vsec(oc) / (sL + R + RL)
Ok too.
> the power across RL is:
> PL = Vsec(oc)^^2 x RL / [(sL)^^2 + sL(R + RL) + (R + RL)^^2]
Power is not calculable so directly in this way. A product in the time
domain is not a product of transforms.
A possible method is to assume sinusoidal steady state conditions
at a frequency of w rads/s and then see that the current magnitude in
RL is: IL=Vsec/sqrt((w*L)^2+(R+RL)^2)
And the power in RL is, assuming Vsec as rms value: PL=RL*IL^2
Solving for the RL that results in maximum PL: RL=sqrt((w*L)^2+R^2))
Or: The optimum resistive load has the same magnitude of the
Thevenin output impedance.
The maximum output power is then:
PLmax=Vsec^2/(2(sqrt((w*L)*2+R^2)+R)
Note that the short-circuit current of the transformer is:
Isc=Vsec/sqrt((w*L)^2+R^2)
and the output VA rating is VA=Vsec^2/sqrt((w*L)^2+R^2)
PLmax is then always smaller than one half of the the output
VA rating, varying between VA/2 when w*L>>R to VA/4 when R>>w*L
Note that this is only valid with sinusoidal signals.
Antonio Carlos M. de Queiroz