Minimum limit for current and maximum limit for voltage while calculating power

Question

I have a brushless Motor then i am using in conjunction with a ESC, powered by a Lithium Polymer Battery(3S 1300mah). I just got a new Lithium Polymer battery (6S 15000mah) from a friend and am planning on using it to power my motor.

From the motor details shown by the manufacturer here:- http://www.emaxmodel.com/views.asp?hw_id=15 , it shows the motor being rated at 3S 17A MAX. If i were to calculate it's wattage, it would be (4.2 * 3) * 17 = 219.3W, assuming a voltage of 4.2V per LiPo Cell. However, as 17A is the max rated current, and that the max continuous current is not given, i would be using 15A as an estimate, which would give me a wattage of 189 Watts.

While the motor is shown to be using 3S LiPo batteries, would i be able to power it with a 6S LiPo Battery with a lower current draw?( I'm assuming that the motor will draw less current since more voltage is supplied to it).

Am I right in thinking that i should be able to power the motor with a variety of voltages, as long as the power(voltage * current) does not exceed the max rated wattage of the motor?

If so, given an extreme example, would i be able to power the motor at 100V @ 1.89A(not that it would be feasible or would i ever do it), or is there a max limit to the voltage and/or a min limit to what the current supplied must be for the motor to work?

All thoughts, opinions and answers would be greatly appreciated.

Answer 1

If so, given an extreme example, would i be able to power the motor at 100V @ 1.89A(not that it would be feasible or would i ever do it), or is there a max limit to the voltage and/or a min limit to what the current supplied must be for the motor to work?

Power limit for the motor is 204 watts - that's stated in the improved link.

Also note that the speed of the motor is dependent on voltage (\$K_V\$) and that is 1270 rpm per volt applied. The more applied voltage the faster it spins and, for a fixed mechanical load, the power will increase proportionally with RPM squared. So, for a given mechanical load, power quadruples for a doubling of voltage.

This cannot be avoided because the motor spins faster with more voltage applied. Output power is \$2\pi n T\$ where n is revs per second and T is torque to the load and for a fixed mechanical load, a doubling in n also means a doubling in T hence power output to the load quadruples with a doubling of voltage.

At 100 volts applied (with only a light mechanical load), the speed of the shaft would be 127,000 RPM and the motor will shake to bits well before that point.