larsb
1 MW
The 5V of BEMF is caused by the RPM of the shaft which is performing work on something. That work requires power which it is receiving from the shaft. Power at the shaft is voltage times current. There is 15V of voltage headroom left to add more power to the motor.
Nope, that’s still not clear. BEMF is not performing shaft work, and it’s not doing electrical work as it’s not driving current. Compare this to mechanical work that needs both force and distance to do work.
maybe you mean it lowers available voltage (that is part of the input power) that might do work. But this cannot be true as max efficiency could never exceed 50% in this case. Plenty of motors do 80-90% on the dyno so the way it works has to be like this :wink:
(calculation does not take losses into account so we can only go lower than the calculated efficiency in real life)
Never thought about how BEMF comes into the power equations before.
An example:
kV 10 (just number to play with)
Voltage 10 (just number to play with)
Max rpm 100 (calculated)
Rpm at max power (half of the full rpm)= 100/2= 50
Voltage needed to reach 50 rpm: 5V (remember controller doesn’t need to apply full voltage to reach this rpm and a BLDC controller with PWM is not a short circuit)
Stall torque 10 Nm (just number to play with)
Torque is half the stall torque at max power = 10/2=5Nm
P=M*n/9.55 (torque-power equation)
Max power is then 5*50/9.55= 26.2W
Effective voltage to drive current=5V
Torque constant is 9.55/kV Nm/A so 5Nm needs 5/(9.55/10)= 5.2A input current
Eff=pout/pin= 26.2/(5*5.2)=100%-ish due to rounding error