# Field weakening & efficiency



## riwe (Nov 17, 2011)

Hi EV-lovers. 

I trying to get my head around the field weakening function when used with a permanent magnet motor. 
This article (http://www.appliancemagazine.com/ae/editorial.php?article=2297) talks about the constant torque region in which no field weakening is done. The reason for this is that the voltage generated by the spinning motor is lower than or equal to the voltage coming from the controller.

Within the constant torque RPM range we have full torque. If we want to increase the RPM we need to apply a current, *Id*, to “fight” the magnetic field created by the permanent magnets in the motor. Since the magnetic “power” from the magnets are related to the torque we will see a decrease in torque when increasing *Id*. Potentially one could demagnetize the magnets by applying to much *Id.*

This picture is showing what I am trying to describe:










Now to my question. Most motor manufacturers, provide a dyno graph that shows the efficiency of the motor. Normally you will find the highest efficiency close to the rated max RPM of the motor. 

Based on the information regarding *Id* and RPM above I would say that the highest efficiency of the motor and controller can be found within the constant torque region were we don’t have to apply an extra current to fight the magnetic field. 

How does this work? Where is the “power” needed to create *Id* taken from? 

Best regards

Rikard


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## major (Apr 4, 2008)

riwe said:


> Hi EV-lovers.
> 
> I trying to get my head around the field weakening function when used with a permanent magnet motor.
> This article (http://www.appliancemagazine.com/ae/editorial.php?article=2297) talks about the constant torque region in which no field weakening is done. The reason for this is that the voltage generated by the spinning motor is lower than or equal to the voltage coming from the controller.
> ...





> Where is the “power” needed to create *Id* taken from?


From the motor's power source obviously. Although there are other methods to weaken the field which do not require Id, like physical displacement.

I would not say most show highest efficiency at or near maximum RPM. Most I think peak in the area of base speed. Sort of like this:


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## riwe (Nov 17, 2011)

major said:


> From the motor's power source obviously. Although there are other methods to weaken the field which do not require Id, like physical displacement.
> 
> I would not say most show highest efficiency at or near maximum RPM. Most I think peak in the area of base speed. Sort of like this:



Thank you Major, 

When looking at that graph I would say that the dark red area would be around the motors constant torque region, right?

I am using the Sevcon controller and for all I know the only way to get higher RPM is using Id.

I did a test with my Sevcon Gen4 and the me1304 from motenergy. at 48volt and 180A of motor current the gen4 needed to apply -200A (RMS) in order to keep the PRM above 2000. 

8,64Kw in to the controller and the output from the me1304 shaft was approx 5.1Kw which tells me that 3,54Kw was transformed to heat. That gives me an efficiency of approx 59%. 

Does that sound about right?

Best regards

Rikard


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## major (Apr 4, 2008)

Hi Rikard,

That is terrible efficiency. I saw where they claim 92%, If your number is correct, get help from your dealer or your money back.

Other types of field weakening require special constructed motor. 

How do you measure shaft output power?

major


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## riwe (Nov 17, 2011)

major said:


> Hi Rikard,
> 
> That is terrible efficiency. I saw where they claim 92%, If your number is correct, get help from your dealer or your money back.
> 
> ...


Hi, I've seen the 92% one as well but that is only around max rpm. If I run the me1304 with a 96volt controller I will be able to reach hi rpm without applying I'd and lowering torque. This would give me a higher efficiency, right?

I think that applying hi demagnetizing current will lower overall efficiency, but I can't really back that up.

The output was measures by a dyno. 

best regards

rikard


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## major (Apr 4, 2008)

riwe said:


> Hi, I've seen the 92% one as well but that is only around max rpm. If I run the me1304 with a 96volt controller I will be able to reach hi rpm without applying I'd and lowering torque. This would give me a higher efficiency, right?
> 
> I think that applying hi demagnetizing current will lower overall efficiency, but I can't really back that up.
> 
> ...


You're measuring at rated load and 2000RPM, right? You should be within a few percent of their efficiency statement or they are just lying to you about it. No excuse for 59% 

And the Id component of the current is highly reactive. It should only contribute to loss by a fraction of I²R in the winding, like a percent or two of input power.


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## riwe (Nov 17, 2011)

major said:


> You're measuring at rated load and 2000RPM, right? You should be within a few percent of their efficiency statement or they are just lying to you about it. No excuse for 59%
> 
> 
> And the Id component of the current is highly reactive. It should only contribute to loss by a fraction of I²R in the winding, like a percent or two of input power.


This is the dyno from the me1114 which should be the air-cooled equivalent of m1304. Actually the efficiency at 2000RPM is 40%, or am I reading it wrong?









What do you mean whit "And the Id component of the current is highly reactive"? 

Best regards

Rikard


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## major (Apr 4, 2008)

riwe said:


> This is the dyno from the me1114 which should be the air-cooled equivalent of m1304. Actually the efficiency at 2000RPM is 40%, or am I reading it wrong?
> 
> 
> 
> ...


Yes, you are reading that incorrectly. It does not show any 2000 RPM point.

And reactive current is out of phase with voltage on an AC line which means it does not calculate directly to real power. 

I have to go out of town and can't explain now. But it appears you are not measuring your motor just reading from a published graph?

Regards,

major


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## riwe (Nov 17, 2011)

Take another look at the grap in the dyno above, I've update it with a 2000RPM Point for efficiency. 

I will do some more studying on the Id.

Thank you!

Rikard


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## dougingraham (Jul 26, 2011)

Yes, you are reading it wrong. At no point does the RPM go below 4000. It starts at about 4200 with no load applied and at the max load of the test it has dropped to just below 4000 rpm. The RPM is the top line which is solid black.

The efficiency at the left side is mostly a factor of the no load current which has to do with iron losses. At the right side it is the copper losses that predominate. I would say that over the normal usage range this motor has an efficiency in the upper 70% range.

The X-axis is clearly torque. What is the X-axis scale? Is it foot - lbs or newton - meters. Or something else?


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## sergiu tofanel (Jan 13, 2014)

Regarding the appliance magazine article, the author should not be allowed to teach this stuff. He may be a good engineer, but he's lousy at explaining. That being said, I'll take a crack at it.

From what I gather from the article, he is talking about a 3 phase BLDC motor, where the phase commutation is done electronically. With that in mind, the motor speed is limited by the commutation frequency. There are two ways to make an AC or BLDC motor go fast: increase the number of poles in the stator winding, or increase the driving frequency. Since adding extra poles to the motor is expensive, all we have left is increasing the frequency. 

The coils inside a motor are limited by their impedance. There are two components to this: resistance, which is the opposition to the flow of electric current (V = IR), and inductance, which reduces the current flow as a result of changes in current (V = L*dI/dt). In other words, an inductor will "oppose" any increase (or decrease) in the flow of current by creating an opposing voltage. The faster we try to change the current flowing through a coil, the more opposition we get, which is seen as an increase in resistance. 

Now, back to the motor. The power supply has a fixed maximum voltage, which cannot be increased without a booster. So as we try to increase the rotating speed of the motor, we must increase the driving frequency. In other words, we must force the current to slosh through the sine wave at a faster and faster rate. A fast change in current results in more and more opposition from the coils, which is seen as a higher "resistance". The field weakening the article was describing is not intentional. It is a result of the higher "resistance" that the coils exhibit when driven at higher and higher frequencies. So given a limited supply voltage, one must concede that there is less and less available current to fight higher and higher EMF's.


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## dougingraham (Jul 26, 2011)

riwe said:


> Take another look at the grap in the dyno above, I've update it with a 2000RPM Point for efficiency.


The point you have indicated is where the torque is about 12 whatever units are being expressed on the x-axis. The RPM at that point is 4200, the voltage is 72, the current is about 20 amps, the watts out is about 600 and the efficiency is about 40%. Input watts being 1440 and output watts being 600 would give an efficiency of 42%.


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## gunnarhs (Apr 24, 2012)

riwe said:


> Take another look at the grap in the dyno above, I've update it with a 2000RPM Point for efficiency.
> 
> I will do some more studying on the Id.
> 
> ...


Hi, as stated from members before , the dyno tests starts at 70V /4000 RPM with very low load. At this point efficiency is very low (30-40%)
When more load is applied (holding voltage, and rpm "constant" ) amperage, torque and efficiency increases, efficiency up to 80%.
The highest efficiency should be near base-speed of motor (which is??) and under full load. For a PM-AC this should normally be about 90%.
So I assume that either
1) The motor is not fully loaded
2) The base speed is not 4000 RPM

And now to the FOC - algorithm which holds the voltage/rpm "constant" under different load. Assuming magnets are in the rotor you apply three AC phases to the stator which are typically 120 degree shifted. 
When adding the three components (either in rotor or stator frame) you get 2 currents Id and Iq (and two corresponding magnetic fields induced by this currents) which are 90 degree shifted. 
One magnetic component is parallel to the magnetic rotor field (induced by Id), the other to the stator coil field (induced by Iq). Together they produce the torque required for the rotation of the load in addition to the magnetic field of the PM present:

T = k1*Iq X (k2*Id + Bpm),

k1 and k2 are constants based on stator inductance,Bpm is the PM-field
For PM-machine the present PM-field in the rotor is usually strong so you try to keep the current Id as little as possible to simplify control.
That leaves Iq as the torque determining component (Id -> 0):

T = k1*Iq X Bpm, with Bpm the constant field from the PM. 

As load increases, Iq is increased, this works up to base speed.
Over base-speed Id must be set negative to produce a field opposing Bpm.


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## riwe (Nov 17, 2011)

Thank you Sergui for the clarification!


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## riwe (Nov 17, 2011)

dougingraham said:


> The point you have indicated is where the torque is about 12 whatever units are being expressed on the x-axis. The RPM at that point is 4200, the voltage is 72, the current is about 20 amps, the watts out is about 600 and the efficiency is about 40%. Input watts being 1440 and output watts being 600 would give an efficiency of 42%.


Yes you, and Major, are right! I was reading it wrong. Thank you!


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## riwe (Nov 17, 2011)

gunnarhs said:


> Hi, as stated from members before , the dyno tests starts at 70V /4000 RPM with very low load. At this point efficiency is very low (30-40%)
> When more load is applied (holding voltage, and rpm "constant" ) amperage, torque and efficiency increases, efficiency up to 80%.
> The highest efficiency should be near base-speed of motor (which is??) and under full load. For a PM-AC this should normally be about 90%.
> So I assume that either
> ...


During testing we took some logs as well. Both Iq and Id was monitored and we saw that the highest motor rpm without applied Id is approx. 1650 rpm. This is using a 48V controller. 








The picture above shows the motor running at approx. 2400rpm. Id applied is at the moment -138 A. We can also see that Id is not present until approx. 1650rpm.

As I was told the base speed is dependent on the voltage of the controller. If I would use a 72 Volt controller the base speed would be higher because the controller can handle a higher voltage from the motor before applying Id.
Id in motenergy motors are “normally” (I’ve seen this value in 3 different configurations from Sevcon) set to -200A and we hit that value during our tests. If we increased the Id we got a higher RPM. 

I would think that even if I dont run the motor at 72volt the efficiency would be higher and as Major said motenergy claim up to 92% between 24-72V.

I guess there is something wrong with either my config or the way we measure. 

To be continued...

Thank you all!

Rikard


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## Tesseract (Sep 27, 2008)

riwe said:


> What do you mean whit "And the Id component of the current is highly reactive"?


Reactive current is that current which is out of phase with respect to the applied voltage. Look up "power factor" for more detail.

As you already know, you need to supply "negative" magnetizing current (Id - which is read as "I_sub_d) to get a PM motor above base speed. Below and up to base speed, the stator current and voltage waveforms applied to a PMAC motor are in phase; above base speed, the inverter needs to adjust the PWM waveform so that the stator current appears to lead the stator voltage (thus it is analogous to brush advance in a DC motor). This reactive Id current is stored as magnetic flux in the motor for half of the fundamental AC period then returned to the inverter and stored as electric charge in the DC link capacitance for the other half of the period. If all components were lossless and ideal then no net energy is required to supply the reactive portion of the stator current waveform: it just sloshes back and forth between motor and inverter.

There are, of course, numerous losses in the inverter and motor which the battery pack must supply extra power to overcome: I²R losses in the wiring and windings as well as capacitor ESR, plus magnetic hysteresis and eddy losses in the stator laminations and rotor magnets. That last item is key: field-weakening causes the rotor magnets to heat up.

One other important caveat is that the inverter can be destroyed if it messes up the timing of the Id waveform, or stops producing it altogether (say from losing the encoder signal or because a fault input causes the inverter to shutdown). This is because the PMAC motor acts as a generator any time the rotor is spinning with a terminal voltage that is proportional to RPM and inversely proportional to negative Id. If negative Id goes away or is mis-timed then the back EMF from the motor will suddenly rise - at least until the anti-parallel diodes across the inverter's switches start conducting (acting now as a 3 phase full wave rectifier). This will result in a current flowing into the battery pack proportional to the difference in motor and battery voltage divided by the total circuit resistance. If the motor is spinning at twice base speed then it will produce an unloaded terminal voltage that is twice the battery pack voltage, and given that total battery, inverter and motor circuit resistance is usually less than 0.1 ohm, it is quite possible for this current to exceed 1000A in even a ~100V system. Destruction of the inverter soon follows, and don't forget that the motor will be exerting a tremendous braking force at the same time as well.




sergiu tofanel said:


> ...There are two ways to make an AC or BLDC motor go fast: *increase the number of poles in the stator winding*, or increase the driving frequency. Since adding extra poles to the motor is expensive, all we have left is increasing the frequency.


Bolded text is incorrect - you need to _decrease_ the number of poles to increase synchronous speed.


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## riwe (Nov 17, 2011)

Wow! I acctually think I got all of that J This is so much fun! Thank you.
So to answer my first question, it is NOT *Id* that is causing my loss of efficiency. 

Motenergy claims minimum 85% at 24-72V and that what I should get. I’m thinking that the dyno might be inaccurate. I will go over my Sevcon logs again to see if I can find anything that looks funky.

One question on that. Lets say we are doing regen with a PMAC that has the same profile as my motor. No Id needed up to 1650RPM. Lets say we are spinning the motor at 3000RPM and we are taking out 20A of regen current via a 48V controller. At 3000RPM the voltage from the motor is 76 volts. The controller is configured to keep the regen current at 20A until the DC link voltage reaches 58 volt. After that the controller cuts back the regen current until the DC-link upper limit is reached and the controlled will most likely trigger a high voltage alarm. If we keep spinning the motor at this point it’s going to get smelly, right? 

So my question is, what happens to the energy that is “removed” from regen current during cutback phase? I saw in the lab that the current got as low as 0,8A during regen. What happens to the other 19,2A? 

In order to really control this I would need a break resistor or a clutch to disconnect the motor before EMK hits the roof.


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## gunnarhs (Apr 24, 2012)

riwe said:


> During testing we took some logs as well. Both Iq and Id was monitored and we saw that the highest motor rpm without applied Id is approx. 1650 rpm. This is using a 48V controller.
> 
> The picture above shows the motor running at approx. 2400rpm. Id applied is at the moment -138 A. We can also see that Id is not present until approx. 1650rpm.
> 
> ...


Ok, I looked up the motor specs here http://www.motenergy.com/me1114.html.
According to this your last test with a 48V controller (and voltage?) I would expect the best efficiency at 48V 1650 RPM. What efficiency did you measure at this point? 
The best efficiency is most dependent on the motor structure and usually all motors have ONE sweet spot.
In this sweet spot the efficiency should be 90% (under full load) but I would not expect it over a long range of RPM
I myself have only used PM-machines as loads in dynos and as motors in small vespas. 
In the dyno-case we had to be careful not to get out of the recommended specs, meaning changing poles/gearing according to the test motors speed.
For the vespas I could not see this (aircooled) motors doing something magic except being smaller than an equivalent power brushed DC.
I do not believe the myth of the 90% + efficiency of PM machines over long voltage /RPM range and independency of load. And as Tesseract points out you seem to have to apply extra cooling during field weakening.


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## riwe (Nov 17, 2011)

The motor I am using is actually the me1304, but the only thing that I found that differs from the me1114 is the motor inductance, me1114=82uH & me1304=66uH.
1650 is, as shown in the previous picture, max rpm without applying *Id*. Sadly I don’t have a efficiency value from the RPM. I will try and get a new dyno test done here in Stockholm. 
This is by the way a grap taken under load with the following config:
*Id = -200A
DC voltage = 50V
Rated stator current = 180A (RMS)*








During this test we saw that *Id* was the only thing limiting the RPM. When *Id* was increased, RPM increased. BUT -200A sounds pretty hi to me….

Anyways, I plan to use this motor in a boat I will see pretty linear resistance, no downhill in the Baltic ocean and it’s not a planing hull. It sounds to me that if I want 1200 rpm on the propeller shaft and I am using a 1:2,85 transmission, I should not use a 48V system since that would put me in that field weakening RPM-range. 

If I go for a 72V system I will have a full torque range from 0 to approx. 3000 rpm and I won’t be doing a lot of weakening. 

Does that make sense?

Thanks

Rikard


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## gunnarhs (Apr 24, 2012)

riwe said:


> In order to really control this I would need a break resistor or a clutch to disconnect the motor before EMK hits the roof
> .


Yes you should have a large breaking resistance when testing 
(you can lower it later when adding battery which absorbs the regen current). 
There should also be an over-current-protection in the controller.



> It sounds to me that if I want 1200 rpm on the propeller shaft and I am using a 1:2,85 transmission, I should not use a 48V system since that would put me in that field weakening RPM-range.
> 
> If I go for a 72V system I will have a full torque range from 0 to approx. 3000 rpm and I won’t be doing a lot of weakening.
> 
> Does that make sense?


Ok assuming you use continuous motor power of 9kW 
(Motor has continuous power of 10 kW, but you should not exceed 125A) 
DC voltage at battery would be at least 80V 
We assume 90% RMS-factor, I assume quadratic pulses (not sinus)
You would drive the motor at 72V AC (rms) / 125A (rms) .
Then according to specs you get 18 Nm effective motor torque at 3900 RPM (80% efficiency assumed).
At the propeller you would have then 51 Nm (2,85* 18 Nm) , 
at max 1368 RPM (3900/2,85).
Would that be enough?

I would think like you that this is better than having to do field weakening but as I say I am no PM-motor expert


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## riwe (Nov 17, 2011)

gunnarhs said:


> Yes you should have a large breaking resistance when testing
> (you can lower it later when adding battery which absorbs the regen current).
> There should also be an over-current-protection in the controller.
> 
> ...


 Thank you Gunnar!

Just to recap on your calculations. The _*Nm per Amp*_ value differs between 0,14125 and 0,15 depending on who you ask. But lets use 0,145. At 125A*0,145 I would get 18,1 Nm. Just like you said. The controller would use a sinus pulse, not a quadratic, if that makes any difference. 
My guess is that the rpm must be lower than 3900 in order to stay away from the *Id*. How did you get to 3900?
3900 on the motor gives us 1368rpm on the shaft with 51,3 Nm of torque.
Power (kW) = Torque (Nm) * 2 pi * rotational speed (rpm) / 60000
51,3*2pi*1368/60000 = 7,349Kw 
7,349*1,36 = 9,99hp
So approx. 10Hp on the propeller shaft. That sounds good.


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## gunnarhs (Apr 24, 2012)

riwe said:


> My guess is that the rpm must be lower than 3900 in order to stay away from the *Id*. How did you get to 3900?


From the chart you posted for 72V bench test.. 
I back-checked in calculation if it would theoretically fit.
In practice this could be lower (or higher?), dependent on load I assume.
When using sinus then you have to use higher DC bus voltage to remain 72V RMS which would give you this high RPM. If 3900 is to high you could reduce DC-bus voltage


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## riwe (Nov 17, 2011)

Hey! 

I made some progress regarding the *Id* issue, turns out the gen4 uses the *Ke *(motor voltage constant) when calculating the field weakening current. 

So a higher *Ke *will make the controller apply *Id *at a lower RPM. 

How do I calculate *Id*? 

This is the formula I use:

Ke= V line-to-line(rms) / rad/s

But then I found this:
 With a BLDC motor use an ac voltmeter to measure the voltage between any 2 wires of​ the 3 motor wires and then convert the line-to-line voltage to the phase voltage value by​dividing the line-to-line voltage by _3 _=1.73.
http://support.ctc-control.com/customer/elearning/younkin/motorParameters.pdf

Im confused regarding the PMAC/BLDC terms. Can anyone clarify what Ke formula I should use when calculating for a motenergy brushless DC motor?

Best regards

Rikard


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## major (Apr 4, 2008)

riwe said:


> Hey!
> 
> I made some progress regarding the *Id* issue, turns out the gen4 uses the *Ke *(motor voltage constant) when calculating the field weakening current.
> 
> ...


I would think you need to confirm with Sevcon because of all the different unit systems and conventions which exists. The motor manufacturer likely has specified such a value in his sales literature. It actually is just a method for defining the base speed for the particular source voltage so the controller can determine when to start field weakening.


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## riwe (Nov 17, 2011)

major said:


> I would think you need to confirm with Sevcon because of all the different unit systems and conventions which exists. The motor manufacturer likely has specified such a value in his sales literature. It actually is just a method for defining the base speed for the particular source voltage so the controller can determine when to start field weakening.


Hey Major! 

Ok, Sevcon it is. 

Thank you!


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