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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