RE: The PING Test

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

Original poster: "John H. Couture by way of Terry Fritz <twftesla-at-qwest-dot-net>" <couturejh-at-mgte-dot-com>


The Ping Test is the most fundamental of all Tesla coil tests. The TC
operates using a dampened wave and the Ping test also uses a dampened wave
unlike other tests. A robust pulse generator is needed for good results. I
used a HP 214A which can produce  pulses of 100 volts into a 50 ohm load.
These pulse generators are still available on the surplus market at
resonable prices.

In the Ping test the TC secondary coil is pinged with a pulse from the
generator and the ring down of the dampened wave noted on a standard
oscilloscope. The test setup is very simple and a storage or digital scope
is not necessary because the pings are repetitive. The decrement or
decreasing of the waveform can then be used to give the TC researcher
important information regarding how his TC system will perform.

A crucial and difficult to determine TC parameter of the secondary coil is
the Q factor of the coil. The Q factor can be used to estimate the secondary
voltage of the TC. The equation is
      Vs = Q x Vp   where Vs = secondary voltage
      Vp = primary voltage

The Q factor can also be found by the equation
      Q = 6.283 F L/R  where F is the frequency,
L is the inductance of the secondary and R is the resistance.
The problem with this equation is that both Q and R are unknowns. However,
if we can find the Q factor with the Ping test we can then find the R
parameter.

In order to illustrate how the Ping test can be used to determine the Q
factor I tested a 4" x 15" secondary coil with 1000 turns of #28AWG. The
coil inductance was 30.6 mh and the frequency was 91.84 KHz. When the coil
was pinged the scope showed the dampened waveform decreased to 10% in 27
cycles. This gave a Q factor of
      Q = pi/log dec = 3.1416/(ln(10))
      Q = 1.364 x cycles = 1.3644 x 27
      Q = 36.82

If the TC primary voltage is 15 KV the estimated secondary TC voltage would
be
      Vs = Vp x 1.4 x Q x eff
      Vs = 15 x 1.4 x 36.82 x .50
      Vs = 386.61 KV
with an overall efficiency of 50%. The efficiency would vary depending on
the quality of construction, the wire insulation, etc. The overall
efficiency can be found with the lamp tests mentioned in earlier posts.

To find the effective resistance "R" when the coil is in operation and 35.47
is the Q factor.
      R = 6.283 x F x L/Q
      R = 6.283 x 91840 x 0.0306/36.82
      R = 479.55 ohms
The DC resistance of the coil was 76.9 ohms. The effective AC/DC operating
resistance ratio of the coil is
      AC/DC Ratio = 479.55/76.9 = 6.24

This resistance ratio with the same H/D can be changed using less turns and
larger wire or more turns and smaller wire. Should the  secondary turns be
increased or decreased to increase the secondary voltage and the spark
length with the same H/D coil?

John Couture

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