0
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ Have you got access to a proper pdf data sheet? The detail on the web page is very poor. For a start 17 amps can only be taken for less than 1 minute and no other current value is offered. This makes estimating normal running maximum power difficult. I will add that increasing the voltage doesn't mean current going down but up. A motor has no idea what power it is taking. \$\endgroup\$ – Andy aka Nov 12 '14 at 13:45
  • \$\begingroup\$ Hi andy, i don't have the pdf data sheet for the motor, the best info i have is here :- modelaccessories.co.uk/emax-brushless-motor-specifications.html under GT 2210/13 where the no load current and power is shown. Am i wrong in thinking that the motor will only draw as much power as is needed? If i increase the throttle, it should draw more power and vice versa if i reduce the throttle. If so, if the votlage supplied is increased, shouldnt the current draw drop? \$\endgroup\$ – Kenneth .J Nov 12 '14 at 13:51
  • \$\begingroup\$ The way to run a motor off twice the voltage, taking half the current, is ... rewind it with twice as many turns. \$\endgroup\$ – Brian Drummond Nov 12 '14 at 14:53
1
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.