Re: Sync Gap Timing (phase angle considerations)

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

Original poster: "Dr.Resonance by way of Terry Fritz <twftesla-at-qwest-dot-net>" <resonance-at-jvlnet-dot-com>


You are speaking essentially of the potential and current across the
capacitor as it charges.  The charging current does lag but that is not
important as you simply wait for a long time period until the cap can fully
charge and then you fire it. There is a short build time for the current in
the inductor to hit a max value but this time period is very short with
respect to the initial charging current on the capacitor.  Unless you have a
very poor quality cap of high inductance this will not be a problem as it
should reach full discharge current in well under 1 microsecond.

You time the spark gap to fire as the max available potential across the
capacitor and be sure to use a scope to directly measure this potential ---
don't just try to measure the AC transformer supply potential.

Best regards,

Dr. Resonance




----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Sunday, November 17, 2002 5:30 PM
Subject: Re: Sync Gap Timing (phase angle considerations)


 > Original poster: "harvey norris by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <harvich-at-yahoo-dot-com>
 >
 >
 > --- Tesla list <tesla-at-pupman-dot-com> wrote:
 >
 >  > >This has been asked before, but it keeps coming up,
 >  > and always bugs me:
 >  > >Do you want the gap to fire at the AC peaks and
 >  > troughs, or at the zero
 >  > >crossings?
 > Look at the components of a tesla primary. It consists
 > of a relatively low inductive reactance L, where X(L)=
 > 2pi*F*L and a very high capacitive reactance in C.
 > Capacitive reactance is made by the formula
 > X(C)=1/(2pi*F*C) Reactance is measured in ohms. These
 > equations use henries for L and Farads for C. Since C
 > is typically a very low number as expressed in Farads,
 > it being in the denominator means that X(C) will be a
 > very high number value in ohms of reactance, and since
 > X(C)>> X(L) -at- 60 hz, the fact that L and C are in
 > series implies that the total series arrangement is
 > predominantly capacitive reactance at 60 hz, so we can
 > treat it simply as a capacity.
 >
 > With an inductive reactance it is commonly stated that
 > the resultant amperage lags the impressed voltage by
 > 90 degrees on the AC cycle. This may not be entirely
 > true as it should lag the impressed voltage by the
 > phase angle that inductor makes by plotting out its
 > resistance and its inductive reactance with y and x
 > axis', but here we are only concerned with how the
 > amperage in time from the displacement current on the
 > capacity will act with respect to its voltage input.
 > It does the opposite effect as the inductor, so it is
 > said that the voltage across the cap will LEAD the
 > impressed voltage by ninety degrees. We can ask the
 > question, when in  the time of the AC current through
 > a capacity  does the capacity contain the greatest
 > voltage? Obviously in this case the greatest voltage
 > across the cap concurrently occurs with the point when
 > greatest amperage is occuring across the cap. But IT
 > IS THAT AMPERAGE WITH RESPECT TO THE INPUT VOLTAGE
 > THAT IS ITSELF 90 degrees out of phase with the input.
 >
 > Jackson in "Introduction to ELECTRIC CIRCUITS" notes
 > the following; pg 334..
 >
 > the alternating current "through" a pure capacitance
 > leads the voltage across it by 90 degrees.
 >
 > Now for this case a special further statement can be
 > made. Since we are equating  in time the voltage
 > stored on the capacitor with the amperage "through"
 > the capacitor, and the amperage through that capacitor
 > is also 90 degrees out of phase with the input
 > voltage, we can also state the obvious; The voltage
 > across the capacitor is 90 degrees out of phase with
 > the source AC voltage.
 >
 > What this means is that when the arc gaps are brought
 > together, the voltage on the capacitor plates will
 > naturally wish to fire when they contain the most
 > voltage, and with respect to the source AC voltage
 > cycle that will be near its zero crossing point, since
 > these relationships are already 90 degrees out of
 > phase.
 >
 > All of this can also easily be realized by simple
 > argument, whereby it should be realized that in the
 > first half of the source frequency AC cycle, (when it
 > is now making its zero crossing point), the cap has
 > been charged in one polarity , and if a gap is then
 > available for disharge, it would happen in that time
 > period. If no gap were available it is in the second
 > half of the source frequency AC cycle where the stored
 > energy is returned to the source. Jackson also notes;
 >
 > inductance and capacitance alternately store energy
 > during the charging quarter cycle, only to return this
 > energy to the circuit during the next (discharging)
 > quarter cycle.
 >
 > The potential confusion that may develope with that
 > statement is that the quarter cycles being refered to
 > are NOT the quarter cycles of the source voltage AC,
 > but the ACTUAL AC on the component itself acting in 90
 > degree phased time with respect to the source. So it
 > is very obvious that (without the complication of
 > having a gap and consequent quenching considerations)
 > after the zero crossing point of the input source AC,
 > the capacitor starts returning energy as the input
 > polarity has changed, but with respect to the AC
 > voltage itself on that capacitor, this is the portion
 > of time between the first and second quarter cycle.
 >
 > Let us hope the textbook has cleared up any potential
 > confusion on this matter...
 >
 > However there is one more important point to be dealt
 > with where Jackson ALSO states...
 > the current through a PURE inductor lags the voltage
 > across it by 90 degrees.
 > The very evident problem with this statement is that
 > "pure" inductances in the real world dont exist,
 > unless we are dealing with a superconductor of no
 > resistance. We have to actually form a phase angle
 > using the resistance of the inductor to find out how
 > far the amperage will lag the voltage source. What
 > this actually means is that in the real world with
 > most values of inductance at 60 hz, if the inductance
 > is not a very high value, the phase angle lag would
 > actually be MUCH smaller than 90 degrees. This is
 > quoted from;
 > HW Jackson on AC Power/Reactive vs Real Phase Angles.
 > http://groups.yahoo-dot-com/group/teslafy/message/4
 >
 > Jackson now describes the condition of power in a
 > circuit containing (equal)resistance and
 > (inductive)reactance.
 > "If a circuit consists of equal resistance and
 > inductive reactance in series, the current lags the
 > the applied emf by 45 degrees."
 >
 > "Now in this circumstance the Q of the coils, or the
 > ratio of X(L)/R seems to be the parameter for
 > determining the true phase angle."
 >
 > This article also shows the procedure for taking the
 > arctan of Q to find the phase angle involved with an
 > inductor.
 >
 > "Thus here in this example we can see the larger the Q
 > the greater the phase angle of the reactive currents.
 > Speaking of reactive currents being 90 degrees out of
 > phase with the voltage is then strictly a misnomer,
 > Imagined as the condition of a described IDEAL
 > component of zero resistance and not a REAL component
 > having resistance, as these calculations have shown at
 > 60 hz it takes a VERY LARGE inductance to even become
 > close to a REAL 45 degree phase angle in its reactive
 > condition."
 >
 > Actually that statement only applies for inductors of
 > smaller gauge wire. A quick example here can be made
 > here for a 500 ft spool of 14 gauge wire having 11 mh
 > and 1.2 ohms.
 > X(L) = 2 pi*F*L = 6.28*60*.011= 4.1448 ohms
 > Q = X(L)/R =4.1448/1.2 = 3.454
 >
 > Now in trigometry the Y coordinate [X(L)] and the X
 > coordinate R gives a tangent as Y/X
 >
 > What we need to know to find that phase angle is the
 > degrees necessary to have a tangent of 3.454
 >
 > To easily find this on a calculator we can instead
 > employ the "inverse tangent" trigonometric function.
 >
 > This is commonly shown as arctan, or tan^-1 on
 > calculators where tan^-1 (3.454) = 1.289 radians
 >
 > Trigonometric functions on calculators always usually
 > deal in radians, where there are 2 pi radians/ 360
 > degrees.  Since the answer is specified in radians we
 > can divide by the conversion factor, which is the same
 > thing as multiplying by 180 degrees/pi radians* 1.289
 > radians = (180*1.289)/3.14 = 73.89 degree phase angle.
 >
 > If the same inductor had more resistance, this phase
 > angle would not be so high.
 >
 > Sincerely HDN
 >
 >