Revision as of 08:13, 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

E = k/(1+k)

for a given desired efficiency keep

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)

Maximum Power

Maximum generator power as a function of omega.

For a generator / motor:

omega = Ks * Vemf    (Vemf is back emf)
or Vemf = omega/Ks

Vm = Vemf - I * Rm  (Vm is voltage at terminals)

I = Vm / Rl          
let Rl = Rm for max output power

then Vm = Vemf - (Vm/Rm)*Rm
or
Vm = omega/Ks - Vm

so at max output power
Vm = omega/2Ks

output power is V^2/R so

Pmax = Vm^2 / Rm 
or
Pmax = omega^2 / 4 Rm Ks^2

lets define the power constant as 
 
Pk = 1/4RmKs^2   

so Pmax = Pk * omega^2