volts/amps question

[skellyo]

Originally Posted by VintageMA

To follow up on what suicideneil said - if you let it the motor will pull as many amps as needed for the demand and blow itself up in the process if you let it. There is nothing to "limit" the current.

Like he said for Neu - the max amps is the amount of current the motor can handle for 30 seconds and the max rated voltage.

For example (and this doesn't correlate to any specific motor):

Rating: 3000 watts max for 30 seconds, max voltage 30 volts - max amps = 3000/30 = 100 Amps for 30 seconds.

A few scenarios running with lipo packs:

4S = 14.8 volts; 3000 / 14.8 = 202.7 Amps @ 30 sec.
5S = 18.5 volts; 3000 / 18.5 = 162.2 Amps @ 30 sec.
6S = 22.2 volts; 3000 / 22.2 = 135.1 Amps @ 30 sec.

It's really just a measure of heat dissipation, power and the max RPMs of the motor (to calculate the max voltage a motor can handle)

Your numbers aren't really true. The max power rating of the motor is based on a specific input voltage. The current does not increase with a lower voltage to achieve the same power as the motor is not a constant power device.

[zeropointbug]

No, it's not a constant power device... but, that's what the motor can handle (before hitting specific temp) at different voltages... when running a lower voltage, it can handle more current passing through it's windings, this has to do with efficiency curves of motors at various voltages/rpms. It's kinda hard to explain, but you get it don't you?

[skellyo]

Originally Posted by zeropointbug

No, it's not a constant power device... but, that's what the motor can handle (before hitting specific temp) at different voltages... when running a lower voltage, it can handle more current passing through it's windings, this has to do with efficiency curves of motors at various voltages/rpms. It's kinda hard to explain, but you get it don't you?

I don't believe that's true either. Higher current in the motor will produce more heat, period, no matter the voltage applied to it. If it were designed to handle 100A for 30S, it's not going to be able to handle 200A for 30S just because you half the voltage. This has nothing to do with efficiency curves either, it has to do with I^2R losses in the motor windings.