Re: NEC-LIST: 50 versus 77 (75) ohm transmission line

Dear All,

I have been tracing the equations for the (why coax. are 50
ohms issue, but I could not track the Minimum Inner-conductor
Temperature Rise condition. I would highly appreciate if someone
could help me: how is it derived ?

    I thought :

    (Total Loss) = (Inner-cond.Loss) + (Outer-cond.Loss)
    when Outer-cond.Dia. is given(constant), then (Outer-cond.Loss) is
    constant.

    So, (minimum Total Loss condition) = (minimum Inner-cond.Loss condition)

    And also = (Minimum Inner-conductor Temperature Rise condition)

    So, the Zo should be 77ohm/sqrt(epsilon), instead of
    36ohm//sqrt(epsilon).

    Where is the error ?

I found it very informative to draw graphs with Excel, and learn that
the curves are very shallow, meaning deviation from optimum condition
or Zo yields little degradation.

I note below some equations I found in textbooks and some that I
solved. I received request from some people, and others may be
interested too.

Hideho.

Coax. Formulae:
    b : radius of outer-conductor
    a : radius of inner-conductor
    eq. in programming like format.
    ln: log.natural.
    IS units.
(a) Characteristic Impedance Zo:
    Zo=60/sqrt(epsilon) * ln(b/a) [ohm]
(b) Electric Field Intensity E
    E=V/(a*ln(b/a)) [V/m] on inner-conductor surface (max.E).
    E=V/(r*ln(b/a)) elsewhere.
(c) Attenuation alpha (approx.):
    alpha= R/2*Zo + G*Zo/2 [Neper/m]
    alpha= R/2*Zo (for conductor loss only)
(d) Conductor Resistance of one-conductor-material coax :
    R=rho*( 1/(2*pi*delta*b) + 1/(2*pi*delta*a) )
    rho : resistivity of conductor [ohm*m]
(e) skin depth :
    delta = sqrt(2*rho/omega/mu) [m]
    omega = 2*pi*f
    mu = 4*pi*1e-7, permeability

(1) Maximum Standing Voltage condition :
from eq.(b) :
    V=E * a*ln(b/a)
for b=constant, putting the derivative to 0:
    dV/da =E * ln(b/a) -1 = 0
    hence ln(b/a)=1 or b/a=e
so :
    Zo = 60/sqrt(epsilon) [ohm]

(2) Maximum Transmitted Power condition :
    P=V^2/Zo=sqrt(epsilon)/60*E^2*a^2*ln(b/a)
for b=constant, putting the derivative to 0:
    d(a^2*ln(b/a))/da = a*(2*ln(b/a) -1) = 0
    hence ln(b/a)=1/2 or b/a=sqrt(e)
so,
    Zo = 30/sqrt(epsilon) [ohm]

(3) Minimum Attenuation condition :
for conductor loss only,
    alpha = rho*sqrt(epsilon)/(240*pi) * (1/a+1/b) / ln(b/a) [Neper/m]
for b=constant, putting the derivative to 0:
    d ((1/a+1/b) / ln(b/a))/da
      = 1/(a*ln(b/a))^2*(1 + b/a - ln(b/a)) = 0
    hence 1 + b/a - ln(b/a) = 0
    from numerical computation,
    b/a = 3.59112148
    ln(b/a) = 1.278464544
so,
    Zo=76.7078726/sqrt(epsilon) [ohm]

==================================================
Hideho YAMAMURA,
Senior-Engineer,
Hardware Technology Engineering Department,
Advanced Product and Technology Development Operation,
Enterprise Server Division,
Hitachi, Ltd.
==================================================
Received on Sun Mar 25 2001 - 03:03:36 EST