Re: NST power rating -- another perspective

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net> 


 > Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br>
 >
 >  >
 > "^^" should be "^", exponentiation. Let's see:

I think ^^ is a C language notation because ^ is a bitwise operator for
exclusive or.  But in anycase the important thing is we understand the
meaning to be exponentiation. I'm only glad to adapt to this convention (of
course, I would rather not be constrained by ascii characters).


 >  > 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.

My VL, IL, and PL should have been expressed as VL(s), IL(s), and PL(s) and
Vsec(oc) is assumed to be the transformed Vsec(oc).  I meant to keep the
expressions in frequency domain. The power in frequency domain, I believe,
will be the product of VL(s) and IL(s).  One can then take the inverse
Laplace transform to convert back to time domain (for the general
case).....or, one can do what you suggest below.

 > 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)

Agreed.  Evaluate the polynomials at s=jw, w=2pi x f, and f = 50 or 60
Hertz.  Then find the magnitude of current and voltage across RL by
combining the imaginary and real parts using SQRT(imaginary^2 + real^2).
Its then easy to compute the power.

I did go thru a simple case and evaluated the magnitude of the vectors in
the way you suggested.  I found the same results [ie, RL = magnitude of (R +
XL) for maximum power transfered into RL].  Here are two very simple cases
that might shed light on the 25% of VA rating:

CASE 1 with resistive source:

     source voltage = 1.   source resistance = 1,  Isc = 1,  so VA rating = 1

     Add a  load resistance of 1, IL = 1/2,  VL = 1/2 and PL = 1/4 (or 25% of
VA)


CASE 2 with inductive source:

      source voltage = 1,  source impedance (XL) = 1,  Isc = 1, so VA rating
= 1

     Add a load resistance of 1, total current limiting impedance is sqrt(1^2
+ 1^2) or 1.414.  IL = .707, VL = .707,  and PL = 1/2  (or 50% of VA)


 > 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

Agree,

 >
 > Note that this is only valid with sinusoidal signals.
 >
 > Antonio Carlos M. de Queiroz

Thanks for your response,

Gerry R.
Ft Collins, CO