Winding Question: does STAR make more torque than DELTA, even with adjusted phase current?

[Hummina Shadeeba]

And wouldn’t it be much more efficient in the motor when at low speeds and a low kv it would be less amps to get a needed torque than if higher kv.

Nope. Regardless of winding or termination, to make more torque means more heat, and making the same torque makes the same heat, and making less torque makes less heat.

Same heat produced for a given torque output regardless of kv but what of efficiency in the motor: with a lower kv (higher kt) motor needing less amps to produce that same torque that’s less amps so greater efficiency no? Maybe I’m skipping the higher voltage.

 

[John in CR]

And wouldn’t it be much more efficient in the motor when at low speeds and a low kv it would be less amps to get a needed torque than if higher kv.

Nope. Regardless of winding or termination, to make more torque means more heat, and making the same torque makes the same heat, and making less torque makes less heat.

Same heat produced for a given torque output regardless of kv but what of efficiency in the motor: with a lower kv (higher kt) motor needing less amps to produce that same torque that’s less amps so greater efficiency no? Maybe I’m skipping the higher voltage.

It assumes same speed of course, so battery voltage doesn't really matter that much as speed is determined by the "apparent voltage" to the motor so as long as battery voltage is sufficiently high. That means you could run same pack voltage and partial throttle on the higher Kv motor for identical results. Same heat means same efficiency.

 

[Hummina Shadeeba]

And wouldn’t it be much more efficient in the motor when at low speeds and a low kv it would be less amps to get a needed torque than if higher kv.

Nope. Regardless of winding or termination, to make more torque means more heat, and making the same torque makes the same heat, and making less torque makes less heat.

Same heat produced for a given torque output regardless of kv but what of efficiency in the motor: with a lower kv (higher kt) motor needing less amps to produce that same torque that’s less amps so greater efficiency no? Maybe I’m skipping the higher voltage.

It assumes same speed of course, so battery voltage doesn't really matter that much as speed is determined by the "apparent voltage" to the motor so as long as battery voltage is sufficiently high. That means you could run same pack voltage and partial throttle on the higher Kv motor for identical results. Same heat means same efficiency.

To get to the same speed with different kv motors and same battery the apparent voltage would need be higher on the low kv motor right?
to produce the same torque the low kv would need less amps
So to calculate total wattage is it more voltage with the lower kv motor but less amps and vis versa with the high kv..and end up equal?

Because with a motor at half the kv it needs half the amps to produce a desired torque and I’m imagining amp hours in a pack and it’s using half as much for the same torque

 

[madin88]

Are you running Hall sensors or sensorless? Isn't there a phase difference between wye and Delta? If running sensors, you might have to re set the phase advances, otherwise you'll push huge current and just generate a useless field.

I know nothing of this nucular controller and how it works... Just hypothesising.

Yes the motor has Hall sensors. The difference between Delta and Star are 30 electrical degrees.
Nucular controllers do have auto tuning and will set hall offset if needed automatically, but when using controllers without such function it will be an issue.

Why would a motor only work with one or the other termination?

Aside from hall signals another reason are circulating currents within the winding when wired in Delta. These currents can occur because of the parallel wired coils and they will create additional drag and heat.

---

This thread is about torque differences when changing the winding which i would like to discuss, and since nobody did proof me wrong, can we say that it is certainly true? Or is there anything i missed?

Here another example with very simple math, lets take a motor with 1Ohm phase to phase resistance, first wired in Delta with 100V between two phases:
Then we get by using I = U/R:
Iw1 = 100A, Iw2 and Iw3 = 50A

Motor in Star needs 1,73 of the voltage and 1/1,73 of the current so we get:
Iw1 and Iw2 = 86,5A, Iw3 = 0A

100A in Delta vs. 86,5A in Star per coil means less abuse of the steel in Star, and more potential for torque.

 

[madin88]

It assumes same speed of course, so battery voltage doesn't really matter that much as speed is determined by the "apparent voltage" to the motor so as long as battery voltage is sufficiently high. That means you could run same pack voltage and partial throttle on the higher Kv motor for identical results. Same heat means same efficiency.

Partial throttle on a high kV motor always means higher losses compared to WOT on a lower kV motor because the controller has to work as a step down converter all the time with around 3 times lower inductance (if wired from Star to Delta).

 

[Hummina Shadeeba]

Are you running Hall sensors or sensorless? Isn't there a phase difference between wye and Delta? If running sensors, you might have to re set the phase advances, otherwise you'll push huge current and just generate a useless field.

I know nothing of this nucular controller and how it works... Just hypothesising.

Yes the motor has Hall sensors. The difference between Delta and Star are 30 electrical degrees.
Nucular controllers do have auto tuning and will set hall offset if needed automatically, but when using controllers without such function it will be an issue.

Why would a motor only work with one or the other termination?

Aside from hall signals another reason are circulating currents within the winding when wired in Delta. These currents can occur because of the parallel wired coils and they will create additional drag and heat.

---

This thread is about torque differences when changing the winding which i would like to discuss, and since nobody did proof me wrong, can we say that it is certainly true? Or is there anything i missed?

Here another example with very simple math, lets take a motor with 1Ohm phase to phase resistance, first wired in Delta with 100V between two phases:
Then we get by using I = U/R:
Iw1 = 100A, Iw2 and Iw3 = 50A

Motor in Star needs 1,73 of the voltage and 1/1,73 of the current so we get:
Iw1 and Iw2 = 86,5A, Iw3 = 0A

100A in Delta vs. 86,5A in Star per coil means less abuse of the steel in Star, and more potential for torque.

with delta utilizing all three phases at once while star is only using 2 (right?) wouldnt the delta allow higher max torque as it can distribute the magnetic field through more teeth and therefore less likely to saturate?

ive heard that all three phases are powered and producing torque when running foc program, is that so and how?

so possible for a motor to run in star but not delta if the phases (or even separate teeth windings?) are imbalanced and a voltage across the two parallel paths will happen... but not the inverse where the motor would run in delta but not star.

 

[John in CR]

Nope. Regardless of winding or termination, to make more torque means more heat, and making the same torque makes the same heat, and making less torque makes less heat.

Same heat produced for a given torque output regardless of kv but what of efficiency in the motor: with a lower kv (higher kt) motor needing less amps to produce that same torque that’s less amps so greater efficiency no? Maybe I’m skipping the higher voltage.

It assumes same speed of course, so battery voltage doesn't really matter that much as speed is determined by the "apparent voltage" to the motor so as long as battery voltage is sufficiently high. That means you could run same pack voltage and partial throttle on the higher Kv motor for identical results. Same heat means same efficiency.

To get to the same speed with different kv motors and same battery the apparent voltage would need be higher on the low kv motor right?
to produce the same torque the low kv would need less amps
So to calculate total wattage is it more voltage with the lower kv motor but less amps and vis versa with the high kv..and end up equal?

Because with a motor at half the kv it needs half the amps to produce a desired torque and I’m imagining amp hours in a pack and it’s using half as much for the same torque

Yes to all, but you have to look at a pack in terms of watt hours, and keep in mind that the stuff going down the phase wires is different, so it would be the same sucked out of the battery for same rpm and torque for the different wind motors...except that the lower apparent voltage higher phase current of the fast wind motor at partial throttle to match performance with the slow wind motor would have higher losses in the phase wires.

 

[John in CR]

Are you running Hall sensors or sensorless? Isn't there a phase difference between wye and Delta? If running sensors, you might have to re set the phase advances, otherwise you'll push huge current and just generate a useless field.

I know nothing of this nucular controller and how it works... Just hypothesising.

Yes the motor has Hall sensors. The difference between Delta and Star are 30 electrical degrees.
Nucular controllers do have auto tuning and will set hall offset if needed automatically, but when using controllers without such function it will be an issue.

Why would a motor only work with one or the other termination?

Aside from hall signals another reason are circulating currents within the winding when wired in Delta. These currents can occur because of the parallel wired coils and they will create additional drag and heat.

---

This thread is about torque differences when changing the winding which i would like to discuss, and since nobody did proof me wrong, can we say that it is certainly true? Or is there anything i missed?

Here another example with very simple math, lets take a motor with 1Ohm phase to phase resistance, first wired in Delta with 100V between two phases:
Then we get by using I = U/R:
Iw1 = 100A, Iw2 and Iw3 = 50A

Motor in Star needs 1,73 of the voltage and 1/1,73 of the current so we get:
Iw1 and Iw2 = 86,5A, Iw3 = 0A

100A in Delta vs. 86,5A in Star per coil means less abuse of the steel in Star, and more potential for torque.

with delta utilizing all three phases at once while star is only using 2 (right?) wouldnt the delta allow higher max torque as it can distribute the magnetic field through more teeth and therefore less likely to saturate?

ive heard that all three phases are powered and producing torque when running foc program, is that so and how?

so possible for a motor to run in star but not delta if the phases (or even separate teeth windings?) are imbalanced and a voltage across the two parallel paths will happen... but not the inverse where the motor would run in delta but not star.

I don't know exactly how the electricity flows with delta vs WYE termination, but all it does is change the Kv, so there is no gain or difference in torque potential. It's just higher or lower max rpm difference with the same supply voltage. Our motors are just torque machines, so if one did have a torque advantage then of course all motors would be terminated that way.

Forget windings or delta/star. If you want more torque potential use a bigger motor or lower your gearing. I climb hills all the time due to our mountainous terrain. Plus the total load of me and my bikes is much higher than most. The majority of the motors I use have a Kv of over 18rpm/volt and the lowest is over 14rpm/volt. My motors aren't drastically bigger or heavier, yet I enjoy acceleration not even possible on the short wheelbase and high CG of typical ebikes, and my bikes have a top speed higher than any other hubmotored ebike. What makes that all possible is my use of small drive wheels combined with high efficiency motors that are capable of far higher rpm without iron core losses getting out of hand than other hubmotors. The only thing preventing me from surpassing the 200kph mark is that I've been unwilling to pony up the $$$ for a pair of controllers capable of significantly higher voltage than 128V and battery current of at least 150A each. With a low slung bike plus some aero touches I could probably get right at 200kph using the HubMonster on my EEB frame because it's in a 22.5" OD 16" moto wheel compared to the 19.25" OD of the wheels on my other HubMonsters, but when I break thru 200, I want to blow past it using a much higher voltage than on my MadAss beast's 31s pack that gets me past 182kph.

 

[Hummina Shadeeba]

There is the possibility of a voltage between the two parallel paths in the delta though when not wound balanced, and even with the right amount of turns on each tooth and all those turns very neatly done I’ll see slight variations in inductance per phase

And wye allows better thermal transfer as it uses less turns of wire which can trap heat.

And in my experience you can fit a slightly bigger stator in the motor case with wye as the wye bundle uses less space than needed to deal with the connection to the phase wires with delta


In my obsessive quest to get the most copper in a motor winding LRK with wye showed best when just looking at how much useful wire can be gotten in and that’s likely the biggest benefit

 

[John in CR]

In my obsessive quest to get the most copper in a motor winding LRK with wye showed best when just looking at how much useful wire can be gotten in and that’s likely the biggest benefit

Fitting maximum copper is definitely the best goal.

 

[mxlemming]

This thread is about torque differences when changing the winding which i would like to discuss, and since nobody did proof me wrong, can we say that it is certainly true? Or is there anything i missed?

Here another example with very simple math, lets take a motor with 1Ohm phase to phase resistance, first wired in Delta with 100V between two phases:
Then we get by using I = U/R: Iw1 = 100A, Iw2 and Iw3 = 50A

Motor in Star needs 1,73 of the voltage and 1/1,73 of the current so we get:
Iw1 and Iw2 = 86,5A, Iw3 = 0A

100A in Delta vs. 86,5A in Star per coil means less abuse of the steel in Star, and more potential for torque.

Madin,

Looking to the fundamentals of an AC synchronous motor, the torque is the cross product of the magnetic flux vector in the stator (produced by the current flowing in the coil, the number of turns and the radius...) and the magnetic flux in the rotor. If you re-configure the windings in star vs delta, you don't change the mechanics/physics of the situation, the iron still saturates at the same flux etc..., and therefore you cannot gain any more or less torque for the same current in the physical copper. The star delta transformation means the back EMF goes up, while the current down, as you have pointed out, but there is no magic improvement - only the prospect of the controller not controlling it right and wasting energy into the non torque producing flux.

Regarding your thoughts about unused phase possibility and the delta having two paths which can create a loop of current... etc... this may be a consideration in a badly controlled BLDC 6 step mode, where one set of transistors is switched off, but is not the case in sinusoidal commutation where there is always a recirculation path through the 3 phase bridge.

In summary, there is no way you get the massive boost in performance you are talking about, only the possibility that one combination is not being controlled right and causing large field currents and much poorer performance.

 

[madin88]

with delta utilizing all three phases at once while star is only using 2 (right?) wouldnt the delta allow higher max torque as it can distribute the magnetic field through more teeth and therefore less likely to saturate?

There is always the situation, no matter if delta or star, where one winding is doing nothing for torque. This is at the moment when the magnets moving over the corresponding phase are changing the polarity from north to south there, which is right at the moment when BEMF of this coil is crossing the zero line.

so possible for a motor to run in star but not delta if the phases (or even separate teeth windings?) are imbalanced and a voltage across the two parallel paths will happen... but not the inverse where the motor would run in delta but not star.

People which wind motors say that this imbalanced current flow comes from BEMF and harmonics, and the more sinewave it looks the more efficient the motor will run in delta.
The Neumotor is not pure sinewave so maybe this has some effects, that a given % of current is just circulating around making heat and not beeing used for torque.
Designing a motor to be pure sinewave is quite hard to achieve so this might be the reason why most EV motors are wound in Star, and RC motors usually have delta winding because manufacturers wanna keep assembling process simple and not having another solder joint within the motor which also would take up space. Most DIY people then take the effort and go with Y winding to improve ETA.

 

[madin88]

Thanks for your input. Ive seen that you work on your own VESC controller so i am happy that you share your knowledge here :thumb:

Looking to the fundamentals of an AC synchronous motor, the torque is the cross product of the magnetic flux vector in the stator (produced by the current flowing in the coil, the number of turns and the radius...) and the magnetic flux in the rotor. If you re-configure the windings in star vs delta, you don't change the mechanics/physics of the situation, the iron still saturates at the same flux etc..., and therefore you cannot gain any more or less torque for the same current in the physical copper.

Where i am doing hard is to understand that peak current each winding does see is equal at given torque betweeen Delta and Star (when everything adjusted by factor 1,73)? The math says it is not, but maybe mine was a bit too simple..
I also try to understand the effect of the phase shift of 30° and that hall sensors need be changed what this is doing with the current.

In summary, there is no way you get the massive boost in performance you are talking about, only the possibility that one combination is not being controlled right and causing large field currents and much poorer performance.

I am not saying there is a massive boost in performance, only that this motor did make more torque when i had it runnning in Star even when i was pushing more then twice of the phase current now in Delta. Therefore i want to find a reason.
And to be mentiond, maybe i forgot to say, the motor is already deep in saturion at those currents so we are in an area where motor constants are changing with kT beeing lower. If not in saturation, then it's totally clear that there is no performance difference.

 

[madin88]

i like those simulations

https://people.ece.umn.edu/users/riaz/animations/sinvec.gif

http://people.ece.umn.edu/users/riaz/animations/vecmovieplus.gif

 

[Hummina Shadeeba]

I read somewhere that with foc all three phases are somehow powered. Is that so?

 

[larsb]

Yes. that's what the GIFs above show

 

[John in CR]

...the motor is already deep in saturion at those currents so we are in an area where motor constants are changing with kT beeing lower.

What motor and how are you not running into heat problems?

 

[mxlemming]

I am not saying there is a massive boost in performance, only that this motor did make more torque when i had it runnning in Star even when i was pushing more then twice of the phase current now in Delta. Therefore i want to find a reason.
And to be mentiond, maybe i forgot to say, the motor is already deep in saturion at those currents so we are in an area where motor constants are changing with kT beeing lower. If not in saturation, then it's totally clear that there is no performance difference.

I'm afraid that if you're deep into saturation, I'm kind of out of my depth. My work hasn't got to running in saturation yet. I'm starting to think about the edge cases, auto tuning PIDs, modifying the transforms to avoid low pwm off etc... Probably have to consider saturation at some point... now I've got FOC actually running.

Are you running foc or bldc? If you said already, I missed it. It is important for considering what happens.

 

[madin88]

What motor and how are you not running into heat problems?

Hi John, it the Neumotor 8057-75kv
There is a thread about my project with this motor and a few others here have been using it too.
https://endless-sphere.com/forums/viewtopic.php?f=30&t=91144&start=125#p1486864
Yes it heats up in no time when i set phase amps too high, probably also because of it's low weight (sub 3kg) and that it is an outrunner.

 

[madin88]

Probably have to consider saturation at some point... now I've got FOC actually running.

I can imagine that BEMF will change and that there will be more torque ripple since the torque from the peak currents will be cut off somewhere.
If you look here in ES to those using Hubdrive than almost everyone is running phase amp limits which lead more or less to saturation

Are you running foc or bldc? If you said already, I missed it. It is important for considering what happens.

FOC, with Nucular 24F

 

[mxlemming]

Madin,

Sorry to dredge this from the depths... I'm close to buying a Neumotor 8038. Just doing battle with them to actually send me a price for the extended shaft and actually sell me one. It is looking promising.

Did you ever resolve this and work out why/if star made more than Delta on your Neumotor?

After a few more months of controller development (very slowly I'll add... Coding has somewhat stopped now I've got auto tune, crude field weakening and have been poking 100+amps at an 80100 motor,I'm more tempted to spend my hours zooming round in the sun) I have no further insight. As far as I can tell it should be equal once current accounted for as previously discussed.

I wonder if this is one we could potentially get Vasili to answer, since
1) he might know since he clearly "gets it" and
2) it might be a function of nucular controller more than motor

 

[larsb]

Yes, this is really interesting. Difference in inductance might between delta/star might be hard for the nucular to handle?

 

[mxlemming]

I cheekily asked Vasili on telegram and he replied very fast... But unfortunately that he wasn't sure but suggested

Vasiliy:
"
What i can guess is that on delta much less inductance, with certain motors it may produce some less torque because of current ripple
"

Without knowing how the nucular controller works, or even it's switching frequency I can only guess.

Maybe one day I'll have a motor that i can easily reconfigure and I'll work it out for sure, but until then I think I'll write it off as "something has gone wrong for madin".

 

[bmwupgrade]

I don't think you guys have looked at the comparing it fairly

I would rather evaluate motor winding as per potential. Lets say we have:
motor winded 9 threads 7 turns as a star = it gets kv 40 for example on 60v = top speed we like and its ok

no we switch for same motor but with triangle(delta) winding, we only need 4 turns to get to same 40kv (4*1.73=6.92)
- now our resistance on motor has dropped from 14 turns (7 turn 2 teeth) down to only 4 turns= resistance is only 0.2857 of what it used to be)

lets calculate heat up with composition for 1.73 phase current increase
we had: (100a^2)*1ohm*1sec=10k of heat
now we have (173^2)*0.2857ohm*1sec=8.55k of heat

Result, we already reduced heat production of the motor = we can add more phase current

STEP 2
we have had 9x7=63 strands before but after we reduced it to 9x4=36 strands.
--- this means we can add more parallel strands, at least up to 63. we ca do 16x4 now!

16/9 is 1.777 gain in capacity, or rather 1-9/16= 43% reduction in resistance again.

Lets re calculate now what we can do
we had: (100a^2)*1ohm*1sec=10k of heat
now we have (173^2)*(0.2857ohm*(1-0.43))*1sec=4.87k of heat

Result is we dropped the heat by over 52% for same torque and kv!

------ were you guys said you saw "loss of torque" in delta vs star, you were right because it is essentially HIGHER kV, and the higher the kV the lower the low end torque. Phase current of 1.73 does not compensate for that.----

now that motor with 16x4 delta and 173a phase would have equal torque of 9x7 100a star, but would also heat up over 52% less. Correct me if I'm wrong some where.

Now about the potential I was speaking about
now we have (173^2)*(0.2857ohm*(1-0.43))*1sec=4.87k of heat
lets adjust to get 10k heat as in original motor (355^2)*(0.2857ohm*(1-0.43))*1sec=10.0k of heat

so we can now get 355a instead of 173a where we were equal in torque (according to theory, when also compensate for kv) = that is DOUBLE ? the phase torque ?

 

[-larva]

I'm not sure I follow your logic but I think you need more turns in delta to get same kv, not less(if everything else in motor stays same)

Edit: And when you adjust winding turns in motor to get same kv, you would also get same kt.
So 7 turns in star vs. 7*1.73=12 turns in delta.
63/12=~5strands if you then calculate resistance for both windings they would be same