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Old Jan 04, 2012, 07:51 AM
Aka: Tom Jenkins
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So a brushless motor is a 3 phase motor?
Does the ESC feed AC or DC to the motor?
thanks
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Old Jan 04, 2012, 11:56 AM
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A brushless motor is a brushed motor with electronic commutation instead (ESC) of mechanical (brushes) commutation. Motor tells the controller when to switch, the controller senses the BEMF signal to detect zero-crossing.

Both motortypes share the same (simplified) motor model:
rpm = Kv x (U- (Im-I0) x Rm)

For pictures of signals and a link see post #8 en #9.


Gelukkig Nieuw Jaar Ron
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Old Jan 04, 2012, 04:42 PM
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Originally Posted by ApexAero View Post
So a brushless motor is a 3 phase motor?
Does the ESC feed AC or DC to the motor?
thanks
Both brushed and brushless motors have three sets of coils which are alternately energized and de-energized by the commutator. The commutator is energized by a DC voltage. The amplitude of that voltage controls the speed at which the motor turns. The only difference between the two types of motors is that the brushed motor has an internal mechanical commutator built into the motor and the brushless motor has an external electronic commutator built into the ESC. In both cases the commutator, whether it be physically located in the motor or in the ESC, determines which set of coils is energized at any given time and which direction the current flows through the energized coils. As noted in an earlier post, if you look at the signal being applied from the mechanical commutator of a brushed motor to the motor's coils and the signal applied by the electronic commutator in a burshless motor's ESC to the motor's coils, those signals will be exactly the same.

So, is a brushless motor a 3 phase AC motor? No. A brushless motor is just a modified version of the brushed motor the speed of which is still controlled by the amplitude of a single DC voltage applied to the input. A true 3 phase AC motor has 3 separate AC voltages applied to the input and the frequency of those voltages determine the motor's RPM.

Larry
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Old Jan 04, 2012, 04:58 PM
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Quote:
Originally Posted by Ron van Sommeren View Post

Both motortypes share the same (simplified) motor model:
rpm = Kv x (U- (Im-I0) x Rm)
Why Im-I0 ?
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Old Jan 04, 2012, 11:39 PM
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Io is the no-load current. It basically doesn't contribute to the magnetic field (back EMF) of the motor.

For brushed motors it is relatively constant over a wide range of applied voltages.

For brushless motors it is roughly proportional to input voltage.

The model below is a standard model for DC motors.

By using Ohm's law, it is possible to characterize any motor if the Kv, Io, Rm are known.

Kv should be measured using the motor as a generator so that there is no voltage drop in the armature.

When Kv is measured by applying a voltage under no load and then measuring RPM, the presence of the no-load current Io permits only an approximate (but close) value for Kv.
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Old Jan 05, 2012, 02:19 AM
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How can a motor run with no load if there is no amps contributing to the magnetic field?
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Old Jan 05, 2012, 07:50 AM
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Originally Posted by yomgui View Post
How can a motor run with no load if there is no amps contributing to the magnetic field?
A good question, but not exactly what I said.

I stated that the no load current didn't contribute to the back EMF (Eg in the drawing).

Apply a voltage to a motor. Current will be drawn, a magnetic field and torque will be created and the motor will rotate. Remove the voltage source and the motor will still rotate. No current will be drawn. Why is the motor still rotating with no applied voltage and no current drawn? The energy from the initial startup current is stored in the armature in the form of rotational energy. There is no torque on the motor and therefore no current drawn or generated.

When voltage is first applied to the motor, current flows which creates a magnetic field in the armature. The motor turns and acts as a generator which generates a back voltage which reduces the current.

Under no load, the small remaining current does not contribute to the back EMF and is used to overcome friction, windage and magnetic losses. Under no load conditions this is defined as the no-load current Io.

A perfect motor with 100% efficiency under no-load conditions after starting up would rotate with applied voltage at the RPM defined by Kv and no current draw because all the initial energy applied from the voltage source is stored in the armature and doesn't dissipate.

If a load were applied to the 100% efficient motor, the current would rise to exactly match the power required by the load and all the energy drawn from the voltage source would be applied to the load. No extra current would be be required for the armature's magnetic field because in a 100% efficient motor, the magnetic field created from the startup current and stored in the armature in the form of rotational motion wouldn't dissipate.

True Kv is measured under no current conditions by using the motor as a generator.
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Old Jan 05, 2012, 11:51 AM
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Quote:
Originally Posted by Martyn McKinney View Post
Io is the no-load current. It basically doesn't contribute to the magnetic field (back EMF) of the motor.
No motor current, be it load current or no-load current, contributes to the back EMF. Back EMF is created by the motor's windings moving through the magnetic fields of the permanent magnets. That is what creates the voltage in generator mode.

Quote:
Originally Posted by Martyn McKinney View Post
Kv should be measured using the motor as a generator so that there is no voltage drop in the armature.
If there is no voltage drop in the armature where does the voltage output of the generator come from?

Quote:
Originally Posted by Martyn McKinney View Post
When Kv is measured by applying a voltage under no load and then measuring RPM, the presence of the no-load current Io permits only an approximate (but close) value for Kv.
As long as Rm and Io are known kv can be calculated with the equation kv = RPM / (V - (Rm * Io)). That is just the RPM equation re-arranged to solve for kv. In this case Io is Rm when the motor has no load. That in itself shows that Io should not be subtracted from Im when calculating RPM.

Larry
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Old Jan 05, 2012, 12:14 PM
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Originally Posted by Martyn McKinney View Post
...Why is the motor still rotating with no applied voltage and no current drawn? The energy from the initial startup current is stored in the armature in the form of rotational energy. There is no torque on the motor and therefore no current drawn or generated.
I'm confused here (a more or less permanent state for me). Could you please define "rotational energy". And what do you mean by "there is no torque on the motor"? I thought the motor created torque, not accepted it.

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Originally Posted by Martyn McKinney View Post
Under no load, the small remaining current does not contribute to the back EMF and is used to overcome friction, windage and magnetic losses.
Where does the back EMF come from then?

Larry
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Old Jan 05, 2012, 12:24 PM
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Originally Posted by Lnagel View Post
No motor current, be it load current or no-load current, contributes to the back EMF. Back EMF is created by the motor's windings moving through the magnetic fields of the permanent magnets. That is what creates the voltage in generator mode.
I don't disagree with your statement. The point I was trying to make was that the no-load current doesn't contribute to the back EMF because the armature does not rotate at the RPM defined by Kv X Applied Voltage because of the voltage drop from Io.



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Originally Posted by Lnagel View Post
If there is no voltage drop in the armature where does the voltage output of the generator come from?
When used as a generator (which is how Kv should be measured without knowing Io), there is no current flow and therefore no voltage drop in the resistance of the armature.

Quote:
Originally Posted by Lnagel View Post
As long as Rm and Io are known kv can be calculated with the equation kv = RPM / (V - (Rm * Io)). That is just the RPM equation re-arranged to solve for kv. In this case Io is Rm when the motor has no load. That in itself shows that Io should not be subtracted from Im when calculating RPM.

Larry
I see your point and am wondering whether language or convention has come into play here. I always use Ia for total armature current. I now realize that I misinterpreted the significance of yomgui's initial question.

Ia has 2 components, Io and the load current. RPM is determined by the voltage at the armature which is lessened by the sum of these two components. The question then arises in the original poster's equation, what is the definition of Im?

For a motor with no load, RPM would be equal to Kv x (U-IoRm).

For a motor with a load, RPM would be equal to
KV X (U - (IoXRm + IloadXRm))=Kv x (U-(Iload+Io)Rm)

The original poster said: rpm = Kv x (U- (Im-Io)Rm

But Im = Iload +Io

therefore rpm = Kv x (U-(IloadRm) which ignores Io and is therefore not correct.

Both you and yomgui are correct!


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In this case Io is Rm when the motor has no load.
I don't understand this sentence.
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Old Jan 05, 2012, 12:58 PM
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Originally Posted by Lnagel View Post
I'm confused here (a more or less permanent state for me). Could you please define "rotational energy". And what do you mean by "there is no torque on the motor"? I thought the motor created torque, not accepted it.
When the armature is rotating, it has stored the electrical energy drawn from the applied voltage. Remove the applied voltage and the armature still rotates. Apply a resistive load to the terminals of the rotating motor and the motor will act as a generator with the load absorbing the rotational energy in the form of current until the armature stops rotating.

When the motor is acting as a motor it can generate torque to a load.

When a motor is acting as a generator, torque can be applied to the motor.



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Originally Posted by Lnagel View Post
Where does the back EMF come from then?

Larry
Faraday's law. When a conductor (the armature, even with no current) is moving through a magnetic field (the magnets), a voltage is generated.
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Old Jan 05, 2012, 02:29 PM
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Originally Posted by Martyn McKinney View Post
When the armature is rotating, it has stored the electrical energy drawn from the applied voltage. Remove the applied voltage and the armature still rotates. Apply a resistive load to the terminals of the rotating motor and the motor will act as a generator with the load absorbing the rotational energy in the form of current until the armature stops rotating.
Sorry but I still don't quite get it. Where is this electrical energy stored? Is it the inertia of the spinning rotor? And when the applied voltage is removed are you saying the motor will keep spinning indefinitely until a resistive load is applied?

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Originally Posted by Martyn McKinney View Post
When used as a generator (which is how Kv should be measured without knowing Io), there is no current flow and therefore no voltage drop in the resistance of the armature.


Faraday's law. When a conductor (the armature, even with no current) is moving through a magnetic field (the magnets), a voltage is generated.
These two statements seem to be at odds with each other.

Larry
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Old Jan 05, 2012, 02:51 PM
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Originally Posted by Lnagel View Post
Sorry but I still don't quite get it. Where is this electrical energy stored? Is it the inertia of the spinning rotor?
The original question was:
Quote:
Originally Posted by yomgui View Post
How can a motor run with no load if there is no amps contributing to the magnetic field?
The answer depends on the definition of the word run and whether one is dealing with a perfect motor or a real motor.

If run means that the armature is turning, no current needs to be flowing when the armature is turning.

When voltage is first applied, the armature will rotate and accelerate. This rotation will generate a back EMF which in a perfect motor will match the input voltage and make the no load current Io go to zero. No current will be drawn under no load conditions in a perfect motor. The initial energy (current) supplied from the voltage source has gone into rotational energy in the armature. The same thing may be accomplished by mechanically rotating the armature, in which case no electrical energy has gone into the armature, only mechanical energy.

In a real motor, the current drawn under no load conditions is compensating for friction, windage and magnetic losses.

Remove the voltage source from a perfect motor and the motor will continue rotating forever. In a real motor it will slow down due to windage and friction.

Remove the voltage source from a perfect motor which is rotating and apply a resistive load to the terminals. What will happen?

The motor will generate current, slow down and ultimately stop when it has generated the same amount of energy that was original put into its rotational motion by the applied voltage or mechanical energy. The current that is generated is a result of the existing rotational motion of the armature.

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Originally Posted by Lnagel View Post
And when the applied voltage is removed are you saying the motor will keep spinning indefinitely until a resistive load is applied?
In a perfect motor, yes. In a real motor, no, because of friction, windage and magnetic losses that occur in a real motor.

Originally Posted by Martyn McKinney View Post
When used as a generator (which is how Kv should be measured without knowing Io), there is no current flow and therefore no voltage drop in the resistance of the armature.


Faraday's law. When a conductor (the armature, even with no current) is moving through a magnetic field (the magnets), a voltage is generated.

Quote:
Originally Posted by Lnagel View Post
These two statements seem to be at odds with each other.
Sorry, I don't understand what you mean by this. The only thing I can think of is your interpretation of the words "voltage drop".

My understanding is that voltage drop occurs only across resistive or reactive elements when there is current flowing. In a generator, with no load, there is no current flowing and therefore no voltage drop. The output voltage and the RPM may then be used to calculate Kv.
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Old Jan 05, 2012, 03:13 PM
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Originally Posted by Martyn McKinney View Post
Sorry, I don't understand what you mean by this.
Here you are saying that there can be no armature voltage without current.

When used as a generator (which is how Kv should be measured without knowing Io), there is no current flow and therefore no voltage drop in the resistance of the armature.

Here you are saying there can be armature voltage without current.

Faraday's law. When a conductor (the armature, even with no current) is moving through a magnetic field (the magnets), a voltage is generated.

Larry
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Old Jan 05, 2012, 03:38 PM
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Originally Posted by Lnagel View Post
Here you are saying that there can be no armature voltage without current.

When used as a generator (which is how Kv should be measured without knowing Io), there is no current flow and therefore no voltage drop in the resistance of the armature.
We seem to be having a communication problem. I have been on these threads many years and recognize both you and yomgui as posters who "get it".

I am NOT saying there can be no armature voltage without current.
I said that in a generator there can be no VOLTAGE DROP without current.
Voltage drop is the voltage ACROSS a resistive or reactive element caused by current passing through a resistive or reactive element, such as the armature resistance. Voltage drops (which are usually losses) are subtracted from voltage sources.

In a unloaded generator (which should be considered a voltage source) there is generated OUTPUT voltage but there is NO VOLTAGE DROP, because there is no current passing through the armature resistance.

Quote:
Originally Posted by Lnagel View Post
Here you are saying there can be armature voltage without current.

Faraday's law. When a conductor (the armature, even with no current) is moving through a magnetic field (the magnets), a voltage is generated.

Larry
Of course there can be armature voltage without current.
Every unloaded generator generates a voltage without any current (as long as the armature is rotating). Faraday's law.
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