Pushing The Limits of Motor Amps

@SimosMCmuffin you are a master with words. :hugs::hugs::hugs:

Thanks, I actually had a good conversation about this particular phenomenon with a customer at work, who develop and manufacture BMS’ for industrial use. We’re developing a product for them.

The scenario we were talking about was, what happens if you’re regen braking and your battery pack gets disconnected for a reason or another.

Two things can happen, if your motor controller is fast enough it can detect the dangerously high DC bus voltage (VESC high voltage cut-off) and stop the regen braking, at which point you can’t slow down anymore, because you can’t dumb the energy anywhere, but at least you didn’t blow up your motor controller. Or your controller isn’t fast enough, it will boost the DC bus voltage too high and blow itself up, at which point again, you can’t slow down anymore and you blew up your motor controller :smile:

The bleed resistor could prevent this, as talked in the http://www.electric-skateboard.builders/t/diy-break-chopper-protection-against-overvoltage-no-cutoff/49823/34 thread, but depending on it’s location (battery, BMS, motor controller) and the electronic/mechanical fault that happens, it might still not be able to save the day. The best place for it would be at the motor controller, because then even if the battery and the BMS get disconnected, you can dump the energy right at the motor controller.

5 Likes

Dude, I had to read that like 4 times but I definitely gained some insight there; thank you for the knowledge bomb.

So, I got the bulk of this, but to some up (well I am in no place to sum up the eloquence)… the biggest risk is not having the battery connected more than merely the board not being on as there is nowhere to dump the excess BEMF voltage? If the motor controllers are switched off but the battery is connected, will the BEMF voltage still dump the excess into the battery or can it only do that function if it is switched on?

Depends on if your using an antispark with fets or a loop key style… As one can be back powered one cannot.

1 Like

Of course, I get it…Still took me too long to figure out why it mattered… duh and lol. Gracias.

1 Like

Even the anti-spark/BMS might allow current to flow back to the battery from the “load side”, if it’s not a fully blocking FET setup. Loop key is indifferent to the direction of current, as one could guess.

The Vedder anti-spark would not stop this, as the current is able to go through the mosfet’s body diode in the anti-spark itself + through the body diodes in the inverter bridge, in the case that it’s just the purely the motor generating a high BEMF voltage. Not 100% sure what’s the conduction path during regen braking in the mosfets, but regardless vedder anti-spark would still allow the regen braking to happen.

A fully blocking switch setup usually has 2 N-channel mosfets with their Drains connected together, as shown below from the BQ76200’s datasheet. The current would be able to go through the first mosfet’s body diode, but then would be blocked by the second one. The second mosfet would need to be in conducting mode, for the current to go to the battery.

1 Like

I’m revisiting this, as I made a mistake in some of my wordings in regards to top speed.

I’ve been gearing for the top speeds I want to achieve at dead battery. So as the battery dies I won’t really ever notice as it is outside of my normal riding range.

2 Likes

@SimosMCmuffin what is it that makes the motor less efficient when it is producing torque while further from it’s no-load speed?

when further from the no-load speed the back emf will be less but why would that make the motor less efficient?

similarly why would a 100kv motor doing 20mph be more efficient than a 200kv motor doing 20mph? Are motor amps more inefficient than battery amps for some reason?

Well, the further the motor is being run from it’s no-load speed, the more amps you have to give it (hence it’s further from it’s no-load speed), which directly equate to more losses just in the windings due to their resistance. That is a pretty simple fact that should always be true. I think there are then more dynamic losses which are speed dependent.

Then there is the mechanical efficiency perspective, with the electric power IN and mechanical power OUT. for example if we are not driving any load, but we are spinning the motor, then our mechanical efficiency is 0%, because there is no useful mechanical work being done, but we are still using electrical power to just spin the motor.


It’s mainly because of the higher currents with the higher Kv motors. Not so much in the motor itself, but everywhere else. I mean all things being equal in the motor, same stator size, same amount of copper with only the amount of winding turns changing, hence the different Kv values. They should still be able to run identically at different battery voltages, but with the higher Kv motor, the current everywhere else in the electrical system will be higher, so more resistive losses in the wiring and in the transistor stage.

Or so my understanding goes.

EDIT: wiring wise the difference between high and low Kv systems, is the wire insulation breakdown voltage and wire resistance. This goes for the whole electrical system overall. With very high voltages we wouldn’t need as much copper to carry the same power, because low current, but we would need more insulation strength/thickness to stop the insulation from breaking down due to the higher voltage. Opposite of that with low voltages we would need more copper with less insulation to be able to carry the higher current.

I think this makes sense, if you look at something like an electric car with the battery voltages in the multiple hundreds of Volts. They still have pretty thick wires with thick insulation on them, but imagine how much copper (which is expensive) they would need if their system was a low voltage one. I think it’s more economical to have higher voltage, use less expensive copper and more cheap insulation on wires, because the electronics can handle the higher voltages these days.

1 Like

Copper losses should be relatively linear when ignoring skin effect of the windings. However the core losses increase exponentially with higher switching frequencies and flux densities. The magnetic field of any electromagnet tries to resist change thereby introducing a hysteresis in the system and causing losses. At even higher frequencies eddy currents induced in the stator taken an even stronger effect.

Shouldn’t copper losses be on an exponential curve in relation to current? P = I^2 * R ?

1 Like

Yeah true, but different Kv motors the same size should also have the same amount of copper. So the one with the lower Kv should have a higher resistance and vice versa. Therefore the copper loss shouldn’t be responsible for a large difference in efficiency.

1 Like

assume the motor amp limit is constant… therefore copper losses and torque are constant. torque in newton meters multiplied by the angular speed of the motor in radians per second is the mechanical power produced. as the motor spins faster (closer to no load), more and more mechanical power is produced for the same constant copper losses (constant motor amp limit), and more mechanical power for the same losses means greater efficiency.

with the same km, pack voltage, battery amps and motor amps, the 100kv motor is less efficient, even when assuming the 200kv motor has double the Io & iron loss.

suppose these 2 motors are geared such that for both the present ground speed is 20mph… both are the same KM… which is greater efficiency?

calculator source: bavaria-direct.co.za

^even with double the assumed iron losses & Io with the higher kv motor (io input doubled in this example) & the same km, the higher kv motor has greater efficiency, not less… 98% motor efficiency & 97.1% system efficiency for the 200kv compared to 95.5% motor efficiency & 94.6% system efficiency for the 100kv.


^in this example, for the same km but higher kv motor to draw the same current at the same effective voltage at the same ground speed, the higher kv motor must spin at more than double the angular speed of the lower kv motor (thanks to gearing), but the higher kv motor at same current has exactly half the torque as the lower kv motor, but half the torque & more than double the angular speed means the higher kv motor (same km) is actually generating more mechanical power with lower losses for the same input power, and therefore the higher kv motor is more efficient for the same km, when properly geared for the mechanical load.

1 Like

@professor_shartsis (professor of …)

the values you plugged in you have the winding resistance as a fourth of the value of the other not half which I think would make sense…double the turns double the resistance. and the Io should be the same I believe if it’s taken at the same rpm

http://powerditto.de/indexditto.html

https://translate.google.com/translate?sl=auto&tl=en&js=y&prev=_t&hl=en&ie=UTF-8&u=http%3A%2F%2Fpowerditto.de%2Findexditto.html&edit-text=&act=url

heres a page I recently got that has a lot of links. pages and pages. it does explain how a higher kv motor will have more power similar to your explanation somewhere in one of the links. speed is a cheap addition to the motor with less losses even if the same copper volume in the motor.

don’t think we have to worry about skin effect at our frequencies though right?

@SimosMCmuffin

this makes sense as a reason why greater torque is more inefficient following I2r

yes, half the turns half the resistance… double the cross section area of the wire (for same copper volume) and half the resistance again for the same km. 1/2 * 1/2 = 0.25

Io is measured at the maximum unloaded angular speed of the motor.

makes sense with the 1/4 resistance and halving twice thanks

ive seen the Io taken at different speeds but if it’s at the maximum speed of the motor it will be more than double for the double kv as the motor would be going double the speed and the eddy current being exponential based on speed

I currently run motormax 60A and battery 30A. While there be any benefit raising motormax to 80A? It’s a dual 6355 190kv setup (torque motors)

~%33.3 greater torque until you reach the 30a battery amps limit at higher rpms.

1 Like

As @professor_shartsis said more :rocket::rocket::rocket: sauce in low rpms until you reach your battery amp limit. Basically better low end acceleration.