Ah! Ok , we've found the problem!
If for now we pretend our motor is a normal brushed motor. And we're running "forward", so the B-leg is connected to ground and remains that way. Now we start PWM-ing the A-leg at 50%. During the on-time, the current through the motor ramps up according to U = L DI/dt. Here, U is the voltage across the inductor. When the motor is actually running at about 50% of its rated speed (at the battery voltage), then U = U - U = 1/2 U.
Then during the off-time, we turn on the A lowside FET. Or we let the body diode do its thing, the difference is 0.6V, or very small....
Now the current ramps down according to the same formula, and because we're at exactly 50%, the resulting number is also the same....
So the current through the motor is a triangle, ramping up and down depending on the inductance, battery voltage and the PWM period. But the current taken from the "bus" is ONLY taken during the "on-time". So the bus sees a current, with triangular topside, but with zero in the times that the motor current would be dropping. Depending on the battery and the length of the leads, some of the ripple may be taken by the battery. But when I did the math, the output resistance of the battery was so significant that the battery+leads was essentially a current source. So almost no change in current flowing from the battery during the whole PWM period.
As the PWM percentage was set to 50%, the battery will be providing about 50% of the motor current. During the on-time the other half is provided by the capacitor and during the off-time that current recharges the capacitor.
So your article seems to somehow assume that the motor current continues to flow through the bus capacitor during the off-time, while in fact it doesn't. It is possible that they have a different situation as with VESC, so that their calculation is valid for their situation, but not ours.
Discuss hardware related to the VESC such as the NRF nunchuk.
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Hi...i am a new user here. As per my knowledge you can run your VESC at min 12V. This is for the FET drivers to be properly powered. You would get enormous RPMs if you apply the full voltage to the motor, but that is not a problem.
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