I bought a motor from Alienpowersystems.com, the motor is rated to 8s (30v) and 2200w max.
I want to run the motor with 12s (45v) for better efficiency and lower temps. Will the motor run fine if I use 12s and limit the current draw in the vesc to 45A so the maximum output doesn’t exceed 2200w? 2200w/45v=48A
I cannot see why it shouldn’t be possible because higher voltage equals to less current, and that equals to thinner wires since the heat generation gets lower.
May not be wise to over volt the motor…have you asked Alien?
Yeah I would recommend asking Bruno (APS-guy) by email
Yhea, this is the reply I got: “They are 8s max rated. i know people use them on 10s, but honestly i can not say to use on 12s. you can try but it is on your own risk.”
What is it that determines the max voltatage? My perception is that the manufacturer sets a maximun voltage to prevent user to draw more than the maximum output from the motor, in this case 2200w. With a VESC I’m able to strangle the current and the output won’t exceed 2200w.
Not sure what factors limit voltage but I would assume putting up to 12s would put a huge strain on its wiring …it’s a big jump from 8s…10s I can see …but 12s is gonna fry something
Hmm alright, thanks for your response. I haven’t got any theoretical answer to this question why it’s not possible so I’ll stick with my theory and try it. I will post my results here, in the future!
Thank you!
“More Watts means more heat in the windings too lf course and the final determinant is that you have to limit either the current or duration to avoid overheating.”
“Conventional wisdom suggests that high voltage, low amps should be marginally better than the reverse…waste heat comes from I^R”
“If you increase voltage and decrease the current (given power) generally you could be closer to peakeff running. (but there is an optimum couple)”
Nothing says it shouldn’t be possible.
I say burn up the motor for science.
Haha… But I think(and hopes) that the motor won’t burn up! 
be careful about this “high voltage equals low current” thing because it is not that simple. if you just up the voltage it will draw more current! the high voltage low current “rule” if you can call it that, only applies if you also change the motor to a lower kv motor.
As I said, one of the ingrediens is to make this work is to limit the current draw in the vesc to 45A. The motor will actually draw less current for the same wattage. It can be calculated with this formula: P=UxI.
But you are right, if the current wasn’t possible to limit, the wattage would exceed 2200w and the motor would burn.
I read that the real motor max voltage is related to both the winding enamel and the bearing max rpm. The enamel likely can do much can more. The high kv of a little motor will mean high rpm so the bearings are a limit. Really the max voltage is more so there to limit the amps which are much more available with the high voltage. With the amp limit you should be fine. Maybe see what rpm the bearings can spin
I’m thinking that there may be some physical constraints that will keep it from running 12s at an extended period of time…and as you mentioned, bearings may not hold up to higher rpm…
Will be interesting to see what can be concluded from testing an 8s motor at 12s.
I agree with LHB, Just do it for Science, then let us know what happened.
Trial and Error is the purest form of Science!
Hail Skatan, may his bearings deliver us from eternal road rashiness.
So I have read quite a bit in the subject. What’s creating the magnetic field is the current, not the voltage. Of course, the voltage must be there becuase 0v =0A. The higher the voltage is the higher the current gets. It can be calculated with Ohms Law U=IxR.
As mentioned before; If the motor has higher voltage it needs lower current for the same wattage. But if the current is lower, the magnetic field must be lower becuase the current is lower, right? And if the magnetic field is weaker the motor will have less torque but it will be possible to reach higher RPM:s. That’s why high voltage motors has lower kv, the high voltage doesn’t create much torque as the ones running on lower current, so they need to have lower RPM per volt to compensate.
This is just a theory by me, but I think it seems legit. I could run the motor with 12s but it would weaker (slower acceleration) but the top speed would be higher. Some of you mentions that the bearing will not hold up at higher RPM:s, and you are surely right. But it’s not hard to replace bearings on brushless motors if it breaks.
@longhairedboy @Namasaki @Michaelinvegas @Hummie @lowGuido @Haimindo
I’ve been endlessly trying to get to the bottom of voltsge in a motor as well
Given the same motor with the same winding it makes sense that it would have less torque running more volts because wattage is amps and volts multiplied and amps make inductance. But I’ve heard it different ways and on here, forget his name but he’s just a half a head avatar, he wrote some formula which I forget explaining a decrease in temp with voltage in the motor. I’m still questioning if amps alone create inductance or if an amp is not simply and amp and if it’s pushed with voltage it’s different
@tor totally get what your saying…bearings are an obvious choice for failure point because of the higher rpm, and yes you can replace them with high spec bearings…but that’s just one point of possible failure of course … Windings…epoxy not being able to take the heat and increase speed…Gees what else is in a motor? IMO anything that gets hot on 8s will get hotter on 12s weather is electrical resistance or physical resistance…which will shorten the lifespan of the motor…so will agree that it will work…but not as long as it would at 8s
We just don’t want you to squeeze the “magic smoke” (like @longhairedboy likes to call it) out of a perfectly good motor 