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Power and Efficiency
Efficiency is a ratio of output to input. It can be in any reasonable
units. One method to determine efficiency for Tesla coils is shown in
the Tesla Coil Notebook. This method uses spark length for the output
and the power transformer wattage for the input. In order to use this
method certain conditions must be established so overall efficiencies
are never over 100%. It is assumed that coils under 250 watts per foot
of spark (WPFS) are all about 80% efficient (small coils).
For example, the small coil shown in the T.C. Construction Guide uses
18 watts and produces a one inch spark (.083 ft). This is
18/.083 = 217 WPFS. As this is under the 250 WPFS the efficiency is
assumed to be about 80%.
For larger coils the WPFS is always over 250 WPFS. For example, a
15 KVA coil producing a 12 foot spark would be rated at
15000/12 = 1250 WPFS. To convert this to efficiency the following
equation is used:
% Efficiency = 250 x 80%/WPFS = 20000/1250 = 16%
A graph showing this relationship appears in the T.C. Notebook as
Fig 7. So far this graph seems to be working. However, if enough data
is collected from many more coils that proves otherwise, the graph
can be easily changed. It is obvious that large coils can be as low
as 10% efficient (2000 WPFS). The graph is for classical coils only.
I do not have enough information available to make a graph for
magnifiers, tube coils, etc.
Try this method with your coil and let me know how you make out.
Does sombody have a better method? It should be based on
measurable units.
Comments are welcome.
Jack Couture