Generator Efficiency

Here several useful generator equations are derived.

Electrical Efficiency

Electrical efficiency as a fraction input mechanical power that is available as electricity the output terminals. Of course the voltage of the avaialbe power might not be optimal - but thats what switching power supplies are for.

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 possible generator output power is obtained. If the load equals the internal generator resistance, then the maximum output power (Pmax) 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

Its also possible to size a generator based on two fundamental properties of the generator. Here we derive the maximum generator power as a function of rotational velocity 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

For example:
A generator might produce 70 volts at 2500 rpm.  The internal resistance is 7.7 ohms.

ks = 2500/70 (rpm per volt)

Pk  = 25e-6   units are v^2/(ohm rpm^2)
at 1000 rpm maximum power is 25.5 watts
at 2000 rpm 101.8 watts.