Quote:
Originally Posted by Qbranch
As with electrical PWM, though there is less power output (due to lower duty cycle) there is no loss, since instead of doing resistive (linear) control of the output, we're just engaging/disengaging from the power source in varying ratios... weather the power source be an input torque or an input voltage source.
-q
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We may be describing the same thing, but in different ways. The analogy to PWM control is an excellent one. If a speed control is set up so that a motor draws 100 watts, then the load on the power source is 100 watts plus whatever the speed controller gives off in heat. If the duty cycle is increased so that the motor draws 200 watts, then the load on the power supply is just over 200 watts.
Electric watts (voltage x current) are the same as mechanical watts (rpm x torque), and power in must equal power out (plus heat loss due to inefficiencies).
So when driving our robot motors (we'll assume a constant voltage situation), when you use PWM to chop the average current to the motor you get a lower motor speed... but you also get a lower current draw on the battery. In this mechanism since you are lowering the speed of the output shaft, relative to the input shaft, it must also increase the torque on the output shaft relative to the input shaft in order to keep the power equations balanced.
Like I say... I think we're both kind of getting at the same thing from different ways, rather than debating a point here, and the PWM model is an excellent way of looking at this system for people familiar with speed controls.
Jason