Revision as of 08:12, 20 April 2007

Electrical Efficiency

Efficiency as a fraction input mechanical power that is available at the output terminals:

E = Pout / (Pout + Ploss) 
.
Pout = I^2 * Rl and  Ploss = I^2 * Rm thus
.
E = I^2 * (Rl / (Rl+Rm))
.
let Rl = k * Rm then
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E = k/(1+k)
.
for a given desired efficiency keep
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Rl >= Rm * E/(1-E)
.
some results:
k = 4 -> 80% 
k = 9 -> 90%

Power Efficiency

For a load that is a multiple of the generator resistance, a fraction of the maximum output power is obtained. If the load equals the generator resistance, then the maximum output power is obtained.

Let the power efficiency be:

Pe = Pout / Pmax

some math...

Pe = 4k/(1+k)^2

64% -> k = 4   (80% electrical efficiency)
36% -> k = 9   (90% electrical efficiency)