ESC and KV

[Steve Jeffery]

Being quite new to this hobby I have a question:-

As I understand it an ESC is basically a device that takes DC and converts it to a variable frequency AC. Motor speed is proportional to frequency. Therefore how can there be a fixed KV for a motor. There must be some limiting factor as increasing the frequency from the ESC and the motor will go faster and KV will go up (Is there some intelligence in the ESC that senses the motor speed and carries on increasing the motor speed until it will not go any faster ?).

Writing this has got the grey cells working and made me realize that I should have taken more notice of 3 phase motor theory when at college!!

Steve

[GrahamC]

Posted by Steve Jeffery on 18/10/2013 09:00:46:

(Is there some intelligence in the ESC that senses the motor speed and carries on increasing the motor speed until it will not go any faster ?).

Steve

Yep that's right. You are spot on. The ESC 'listens' to the feed back from the motor in order to detect where the rotor is in its rotation. I believe that the phenomenon that it detects is 'back EMF' the pulse produced in the coil as it moves through the magnetic field.

I'm sure the experts will be along in a bit, and I'll find out if I've understood it correctly!

Edited By GrahamC on 18/10/2013 09:19:07

[Peter Beeney]

Steve, - Forget frequency for a moment, the motor speed is controlled by the applied voltage and the size of the load. The motor actually drives the ESC, not the other way round, it’s only acting as an electronic commutator, plus as a speed controller, but that’s a separate issue. The kV is actually the generated voltage, but in our case it’s morphed sideways into a unit of motor performance. There is a crude drawing in my pictures somewhere of motor commutation, which I did in the past for another thread, when there was some discussion about ‘is it AC or DC?’. I’m off flying now, but I’m back later and will explain all, if you so wish.

PB

[Simon Chaddock]

Steve

Please don't think these ESC converts DC into AC. It does no such thing.

It is a variable speed switch that takes over the function of the commutator in a brushed motor. As Peter described the clever bit is the fact that this switching is controlled by the ESC sensing the position of the rotating part of the motor.

It only works if the motor is actually going round so to start it just gives a random pulse to one coil. If this starts the rotation in the correct direction fine, if not, it does it again until it does.

If you look carefully you can sometimes actually see the rotor jump a couple of times before it starts spinning.

The speed control is a completely different function where the ESC switches the current on and off at a much higher frequency than the commutation. The effective voltage (and hence the speed) that the motor sees is controlled by the ratio of time on to off. This method is used on both brushed and brushless speed controllers.

Edited By Simon Chaddock on 18/10/2013 20:15:58

[Shaunie]

Maybe I'll put the cat amongst the pigeons here or maybe not:-

To my mind whether the ESC converts DC to AC or whether it provides commutation is purely a matter of visualisation because I believe both to be true, just a matter of viewpoint really. It's DC at the battery but AC as far as each winding is concerned.

The visualisation that the motor commands the ESC is useful in that the ESC's task is to keep in time with the motor which demands this happens.

The important part for the OP's question is the back emf characteristic. As the motor starts up there is no back emf and a large current is drawn, the motor starts to rotate so a back emf is generated (generator effect) which causes current to fall because it opposes the battery voltage, at some point as the motor speeds up the back emf approaches the battery voltage and current will then fall almost to zero. Obviously this will not happen as power output will be zero so there is always some difference between ideal Kv and actual speed achieved:-

(I=(Vbatt-Vbackemf)/R) (I think that would be right anyway)

It is this that prevents the motor speed becoming infinite! Same as a brushed DC motor really.

This brings the question that if the motor was driven beyond this speed by an external source would it regenerate and start to charge the battery? Just a thought.

Shaunie.

[Peter Beeney]

Simon, - This particular little discussion has been going on for a long time time now, and I would have to say there is no doubt that the ESC does indeed convert DC to AC. If we define AC, (Alternating Current), this is simply DC, (Direct Current), travelling first one way through a conductor and then changing around and travelling in the opposite direction. A moments reflection on the operation of the standard mechanical commutator of a brushed motor will show this to be so. For the motor to operate the segments of the comm. switch the direction of the current every 180 degrees or half revolution. Thus the current reverses direction once every complete turn. To get the whole and complete picture you need to really start right at the beginning, with the reasons why the motor runs at all.

I’d consider that the ESC is just a very necessary part of the motor. The motor controlling the switching speed of the ESC, and the ESC then in turn supplying the motor with current to make it turn, this is definitely AC, and I’d personally not see this at all to be visualisation, so to speak. All motors are generators and will all generate a voltage, this is also most definitely an alternating voltage, and it’s amplitude is determined by the speed at which it’s being turned.

A permanent magnet motor won’t runaway, or it’s most unlikely to, anyway, but a series wound motor such as a car starter motor runs up to infinity unloaded, (in theory), and in certain circumstances other motors can reach uncontrolled speeds, possibly capable of doing some damage to themselves.

We’re rather putting the cart way in front of the horse here, in terms of the principles as to why and how motors run, but the unloaded speed always takes a finite current to drive it. As the speed increases the internal mechanical resistances and so on get more difficult to overcome and thus the useful torque gets less. The motor runs proportionally slower and thus more current flows. If we carried this idea far enough, by steadily increasing the applied voltage, eventually the torque would absorb so much current, and the speed so proportionally slow, that we would be unable to put a load on it. If we now consider our load to be a propeller we can instantly see a little clue as to the reason why when we change from say a 3 cell to a 4 cell pack we have to reduce the propeller size to prevent overloading the motor.

An example of all this might be the car dynamo of yesteryear. If you hooked it up to a 12 volt battery it would run as a motor, at one given and fairly slow speed, the segments on the commutator changing the DC to AC to enable the motor to run. If it’s then driven by the engine, using a fan belt, this because there was also a fan for cooling the radiator bolted on as well, the same segments would convert the generated AC to DC. Admittedly a bit lumpy, but ok for charging the battery.

Hope this is all reasonably clear……

PB

[Shaunie]

"As the speed increases the internal mechanical resistances and so on get more difficult to overcome and thus the useful torque gets less. The motor runs proportionally slower and thus more current flows. If we carried this idea far enough, by steadily increasing the applied voltage, eventually the torque would absorb so much current, and the speed so proportionally slow"

Sorry Peter but you seem to be saying here that its the mechanical losses stopping speed increasing, this is not the case at all. When the back EMF (directly proportional to motor speed) becomes equal to the supply voltage then the motor current would fall to zero and a perfect motor would stop accelerating, an imperfect motor would stop accelerating at some point before this when power consumed is equal to the losses (or load if one is applied).

The internal resistance of the motor is very low relatively speaking. The motor current follows the rule I=V/R, it's just that V is the differential between the two voltages not applied voltage. Due to the low resistance the motor speed does not need to be held very far below it's target speed for the current to rise dramatically and burn out occurs.

Series wound motors are a special case, when run unloaded the current is low as a result the field magnets weaken and therefore the speed continues to increase, as you say often catastrophically.

Shaunie.

[Peter Beeney]

The way I might consider the the facts could be like this. Although it’s not by any means necessarily correct, of course, however at the moment I can’t think of any other answers. This applies to permanent magnet motors, such as the brushless type.

When a motor accelerates with no load, the self-generated back emf increases proportionally to the increasing speed to oppose the applied voltage. If the speed were increased to the point where the back emf equalled the applied voltage no current would flow, therefore no magnetic field would be created around the conductor, which would mean there could be no interaction between the two magnetic fields and so the motor would stop. Even a perfect motor couldn’t run in this state; but this is rather an impossible situation anyway. However, a practical motor will run up to a speed such that the back emf generated is sufficiently high enough to reduce the current to a level where the motor starts to slow down; immediately the back emf is then reduced, the current increases again, and the motor speeds up once more. The current is then again reduced, the motor again starts to slow down…… …the nett result of all this shilly-shallying is that the motor quickly reaches a steady state equilibrium speed, with exactly enough current flowing to develop sufficient torque to overcome the rotor inertia and frictional losses etc. The power output of the motor is twofold, a mechanical rotary output, useful, which can be measured in watts, and a heat output, not so useful, also measured in watts; but negligible at this stage. The no-load current is sometimes given in the spec.

If we now add a load, a propeller, the motor speed will fall. Consequentially the back emf value will also fall. Thus the current will now rise, increasing the magnetic field strength and therefore the torque to compensate for the reduced revs. The mechanical output power is the torque x the rpm. But, of course, the increased current has also resulted in the unwanted heating effect also increasing. Straightway the ratio of useful power to not useful power is falling. As we increase the load, the ratio decreases still further. Taking this on to an end point, such as a crashed model with the motor stalled say, there is no output power at all because the revs are zero, but the heating current, because there is no back emf, is at a maximum. This may be indicated by the smoke signals now conveniently rising from the tangled pile indicating it’s whereabouts and saying please close the throttle!

Now if we double the input voltage the whole thing starts again. But, of course, the motor is going to rev twice as fast; therefore the mechanical internal resistances and losses will require proportionally more power to drive them; resulting in that the rotor's unloaded equilibrium speed will be slower with respect to the applied voltage, the unloaded current will inexorably increase, and along with it our unwanted heating effect. Also the internal electrical resistance of the motor doesn’t change, so that if the voltage is doubled the current flowing through this resistance will also double, and so the heating effect will increase, too. Adding a load will again slow the motor, and from what I’ve just said, the current will be increased considerably; the heating effect even more so. The torque too will now be greater, but at some point the heating disadvantage will outweigh the power advantage, and then it all gets too hot. Therefore it’s important to keep the revs as high as possible.

Again taking the unloaded motor toward an end point by continuing to increase the voltage the motor will continue to run ever faster. At some point the motor frictional losses, which by now are increasing at a rate perhaps twice as fast as the voltage will become a sufficient enough load to limit the no-load speed to one that is proportionally slower than if a lower voltage were applied; and again the electrical resistance hasn’t changed, therefore because of the higher voltage a higher current rate will flow. The speed would only need to decrease slightly to increase the current greatly. Now we are approaching the point where all the power is being used to drive the motor, adding a load is only going to increase the losses.

The speed in the first instance is also going to be governed by the physical characteristics of the design, but again this is not going to change simply by altering the input voltage.

This is just a very general ramble, I’m sure it would take some considerable field testing to establish some actual test results; I’m surmising a bit here anyway, and I don’t have any figures to go on. All this is only really to indicate a possible reason for the fact that if you increase the battery voltage you may have to reduce the prop size to allow the revs to increase. But that might not necessarily be decreasing the power output.

PB