Reactive to resistive ballast ratio of sqrt(3):1 ? (fwd)
---------- Forwarded message ----------
Date: Wed, 13 May 98 22:41:34 EDT
From: Jim Monte <JDM95003-at-UCONNVM.UCONN.EDU>
Subject: Reactive to resistive ballast ratio of sqrt(3):1 ?
I have been thinking about the advantages of including resistive ballast
in the charging circuit, and I stumbled on something a little intersting.
I found that an arc is most stable with the reactive and resistive
components in the ratio of sqrt(3):1. How does this compare with
values that people find to work well in practice?
Below is the justification for my stability claim:
Consider the circuit below representing the charging portion of a
primary when the gap is firing.
- Ia ->
+--- Z = R + jX -------+
V is the secondary voltage of the HV transformer, Z is the ballast
impedance (transformed via the turns ratio of the transformer if on
the primary side, but this leaves the X/R ratio unchanged anyhow), and
Ra is a crude approximation of the spark gap as a resistance which will
later approach zero.
I defined stability as the partial derivative of the magnitude of
the arc current with respect to the arc resistance --
d|Ia|/dRa, found its minimum (actually I found the minumum of its square,
d|Ia|**2/dRa, which was a bit easier to do and has the same minimum
location), and then differentiated again to show that the extremum
was in fact a minumum.
Current magnitude squared is
|Ia|**2 = V**2/((R+Ra)**2+Z**2)
Defining stability as the first partial with respect to Ra,
d|Ia|**2/dRa = -2*V**2*(R+Ra)/((R+Ra)**2+Z**2)**2
Taking the second partial,
d2|Ia|**2/dRa2 = 2*V**2*(3*(R+Ra)**2-X**2)/((R+Ra)**2+X**2)**3
Setting to zero for min, X**2 = 3*(R+Ra)**2.
Letting Ra -> 0, X = sqrt(3)*R
Taking the third partial,
d3|Ia|**2/dRa3 = 12*V**2*(R+Ra)*(X**2-2*(R+Ra)**2)/((R+Ra)**2+X**2)**4
Evaluating at X**2 = 3*(R+Ra)**2,
d3|Ia|**2/dRa3 = 12*V**2*(R+Ra)**3/((R+Ra)**2+X**2)**4 > 0,
so this is a minimum.
Comments? Does this seem to be a reasonable approach?