Is a larger Kv equivalent to more voltage?

Question

I've been a fan of first person view drone Racing for years now and, in thinking of building a new quadcopter, I have a question I'm unable to resolve myself.

Please correct me if I'm wrong:

As i understand, a brushless motor has a maximum torque and power (Watts), if you push it beyond its limits it will start heating and eventually burn.

A brushless motor will always try to match the speed required drawing as much amps as is needed so increasing the power used.

Any brushless motor has a Kv value related to the coils windings that determines the rpm it will get without load per volt applied.

Then, can I say, for example, 1000 Kv at 1 volt will be equivalent to 500 Kv at 2 volts in terms of power, rpm and torque? So could I say it will generate the same lift?

Of course more voltage will reduce the amps but that's not related to the question, I think.

If this is true, I can't understand why actual drones do not run on bigger Kv and fewer cells since cells add weight and Kv not.

EDIT:

As a example of my question the DRL RacerX has broke the speed world record using T-motor f80 wich is a 2408 at 2500kv the manufacturer says it should be run at 16V achieving 680W but the 1900kv says 22V around 1000W.

The DRL used it at 42V (the 2500kv version) still with a 5 inch prop (Sure the motor will not survive for a long time), they just seemed to increase the amp rating of the controllers and just hope the motor will last long enought.

Why did they use such a high voltaje (2 x 5s) since that will not only increase the total weight but that will also increase the amps needed (as its increasing the load).

Im sure they had done their research and made the best they could but i want to understand why is that the best way of achieving it, the main question that comes to my mind is why not a larger prop? it should be much more efficient right? and still that drone is for a max speed record maneuverability is not needed

Answer 1

OK so lets start with Please correct me if I'm wrong

As I understand, a brushless motor has a maximum torque and power (Watts), if you push it beyond its limits it will start heating and eventually burn.

Yes(mostly).

• A motor has a maximum torque called the "stall torque" that occurs when the motor is kept from moving (stalled). The current that flows in that condition is called the "stall current". In this case, all the energy that is going into the motor is producing heat and if the motor cannot get rid of it, the temperature will rise until something bad (and often irreversible) happens.
• When a motor is operating it can also overheat; but, often in an aircraft the prop wash will help to cool the motor to reduce the risk of that.
• Another failure mode can happen as you push to higher RPMs. Every motor has a breaking point. Here is a video of a really cool 3D printed motor that the author was able to get to 30,000 RPM!!! Note what happened at max RPM - it didn't burn up.

A brushless motor will always try to match the speed required drawing as much amps as is needed so increasing the power used.

Again, Yes(mostly).

The nature of an open loop DC Motor is that the motor output power (Power = RPM*Torque) will rise to the point that it matches the power available to it (input power - losses). RPM is not the only thing that limits this. An example of this is when you grab the chuck of an electric drill and it slows down because the fixed power available forces the RPM to drop to compensate for the increased torque required.

Any brushless motor has a Kv value related to the coils windings that determines the rpm it will get without load per volt applied.

Yes Kv is a pretty good model of what drives the RPM on an unloaded motor. What is happening is that a motor also acts as a generator. As the RPM increases, it generates a Voltage (called the Back EMF) that increases pretty linearly with the RPM. This voltage works against the source voltage. At some point the EMF will rise to match the source voltage and no more current will be able to flow into the motor. If $V_{emf} = K_e \omega \\$ then $K_v = 1 / K_e$. This is where the $K_v$ comes from.

Then, can I say, for example, 1000 Kv at 1 volt will be equivalent to 500 Kv at 2 volts in terms of power, rpm and torque? So could I say it will generate the same lift?

NO Generating lift means torque (load) and $K_v$ only applies in the theoretical world of no load.

Since lift = work and work = power, then (for a lossless motor) the same amount of power will be required to produce the same lift and if you are using the same propeller, the relationship between lift/torque and RPM will be the same; so, the required RPM and torque will be the same regardless of the $K_v$ of the motor.

In the real world there is another parameter $K_i$ and this parameter establishes the relation between current and Torque. And, due to some interesting (and insidious) math (which I will get to in a bit) $K_i \propto 1 / K_v$. So, as you get a bigger $K_v$ and get more RPM for the same voltage, you will get a smaller $K_i$ and get less torque for the same current. The same power will be required and still no one will get a free lunch.

If this is true, I can't understand why actual drones do not run on bigger Kv and fewer cells since cells add weight and Kv not.

So from the above discussion, $K_v$ really has not affect on the number of batteries that are required for the same performance. If you increase $K_v$, sure you need less voltage; but, you will draw more current either requiring more cells to reach the required current or to match the battery life. The power required by the motor times the run time will result in the energy required in Joules which also will be the battery capacity in Ah times its voltage (which also is Joules).

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Now, regarding the T-motor f80 2408 you mentioned, Mini Quad did a pretty detailed Bench Test. Here is some of the data. It looks pretty sweet to me. Note that for the 5" prop it hits about 28,000 RPM at 16V (28000/16=1750$K_v$???) - NO - this is loaded not unloaded. Also note how much current was required - 30A Average with 162A peak. Max nominal discharge for a battery is normally about 25C. That would mean you would need to use a 16V, 1200mAh battery pack. The real limit, however, will likely be peak current. I believe a LIPO battery probably can't do more than 50C discharge rate; so, 162/50 means you will probably need at least a 16V 3200mAh battery which will NOT be small.

There is also going to be some magic in the Motor Controller to be able to get the full power out of this motor by optimizing the voltage on each motor phase and also managing the currents.

I hope this helping you understand that there is A LOT more to getting the maximum power to weight than just having a large $K_v$.

Answer 2

It is helpful if you think of this in terms of overall power, rather than just voltage.

Imagine you have a machine which requires 100W per motor for your desired flight performance characteristics. This 100W can be delivered to the motor using any voltage/current combination you wish.

For instance, you could deliver it by supplying 100V at 1A, or 1V at 100A, or any other combination.

Your proposed approach, using higher KV motors, would mean using lower voltage but higher current.

The electrical engineering problems that you experience with higher current are harder to solve than those you experience at higher voltages (at least in this range of voltage and current). Switching 100V at 1A is much easier than switching 1V at 100A.

Answer 3

Power is power is power.

As you mention in your post,

Of course more voltage will reduce the amps but that's not related to the question, I think.

Well, it is related to the question, because

$$ P = IV \\ $$

If you have constant power output, then current decreases as voltage increases because you can rearrange that as $I = P/V$. In this expression, you can clearly see that a bigger denominator (bigger voltage) means smaller current, again for a constant power output.

As I mentioned, power is power; you seem to have a good feeling for the electrical power, but consider also that there is a mechanical power:

$$ P = \tau \omega \\ $$

That is, power is torque ($\tau$) times speed ($\omega$). Now you can hopefully see where I'm going. Just like the constant power expression for electrical power shows that current declines for increasing voltage, mechanical power can be rearranged to show:

$$ \boxed{\tau = \frac{P}{\omega}} \\ $$

Torque declines with increasing speed for a constant power!

Torque is important because it affects your ability to accelerate:

$$ \tau = I \alpha \\ $$

Where $I$ is the moment of inertia of the load and $\alpha$ is the angular acceleration.

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Now, with some of the math explained, consider the performance of a racing vehicle. Depending on the course, you may want a vehicle with a higher top speed, or you may want a vehicle with more acceleration.

Assume for a moment that the maximum power output of the drone is fixed by the "class" or performance tier for the competition. For example, single A racing might have a 10 watt limit, but triple A racing might have a 250 watt limit. This is hypothetical, but a good argument for the need for a constant power assumption.

You, as the designer and/or crew chief, would obviously want a motor that performs at the maximum allowable power output for your class. Now, if you have a track that requires a lot of turns, you want to be able to change the speed quickly. Changing speed requires an acceleration, and again, your acceleration is a linear function of your available torque.

If you want a more maneuverable, agile drone, you would pick a motor with a lower Kv constant. This reduces the top speed of the drone in exchange for more torque, meaning you are able to change the speed of the props (and thus change directions) with more authority.

If you want a drone with a higher top speed, then you would pick a motor with a higher Kv constant. This sacrifices torque (and thus maneuverability) for a higher top speed. This would be used for a drone equivalent of drag racing.

Ultimately it is up to the crew chief, pilot, designer, etc. - whoever makes the hardware decisions - to pick the motors that provide the best performance for a given track. As mentioned at the top of the post, it all has to do with the trade-off imposed by the constant power assumption.