Calculating KV for BLDC Motor Re-Winding with Different Wire Materials

[metastable]

I'm looking at the formulas for re-winding BLDC motors to change the motor constants KV (max rpm/v) and KT.

I've read that the formula is as follows (assuming wye termination):

Original Turns Per Tooth # * Original KV = New Turns Per Tooth * New KV

which can be rearranged:

(Original Turns Per Tooth # * Original KV) / New Turns Per Tooth = New KV

which implies for a given stator/rotor:

(10 Original Turns Per Tooth # * 100 Original KV) / 5 New Turns Per Tooth = 200 New KV

my question is as follows:

Assuming 10 turns gives 100kv originally with copper wire, If I change the wire material to a different conductive material such as aluminum (same 10 turns), will the KV still be exactly 100kv? If the answer is no, what formula do I use to estimate the new KV?

 

[anorlunda]

Assuming 10 turns gives 100kv originally with copper wire, If I change the wire material to a different conductive material such as aluminum (same 10 turns), will the KV still be exactly 100kv? If the answer is no, what formula do I use to estimate the new KV?

Changing the material gives different resistive losses. But in any case, losses depend on load and they are often neglected. So, other than resistance, changing materials has no effect on KV.

 

[metastable]

Changing the material gives different resistive losses. But in any case, losses depend on load and they are often neglected. So, other than resistance, changing materials has no effect on KV.

Thank you for this reply. I wanted to write the kv/turns formula using only SI variables and came up with these:

A = Change Factor of KV (rpm/v)
B = Change Factor of Conductor Resistivity (ohm-meters)
C = Change Factor of Conductor Volume (meters^3)
D = Change Factor of Conductor Resistance (ohm)

A=sqrt(B/(C*D))
B=D*C*A^2
C=B/(D*A^2)
D=B/(C*A^2)

For example in the previous example assuming same gauge copper wire for both old & new winding:

(10 Original Turns Per Tooth # * 100 Original KV) / 5 New Turns Per Tooth = 200 New KV

Becomes:

Change Factor of KV (rpm/v) = A = sqrt(B/(C*D))

A = sqrt(1/(0.5*0.5))

A = 2

So the new kv is the original kv (100kv) times A=(2)... giving 200kv.

---------------------

Or suppose we keep the same turns and same thickness but switch to a material with double the resistivity...

Change Factor of KV (rpm/v) = A = sqrt(B/(C*D))

A = sqrt(2/(1*2))

A = 1

So the new kv is the original kv (100kv) times A=(1)... giving 100kv.

---------------------

Or suppose we keep the same material, same # turns but use half the wire cross section...

Change Factor of KV (rpm/v) = A = sqrt(B/(C*D))

A = sqrt(1/(0.5*2))

A = 1

So the new kv is the original kv (100kv) times A=(1)... giving 100kv.

----------------------

My question becomes:

Is it acceptable to rewrite the turns/KV formula in the following manner?

(Original Turns Per Tooth # * Original KV) / New Turns Per Tooth = New KV

Becomes:

A = Change Factor of KV (rpm/v)
B = Change Factor of Conductor Resistivity (ohm-meters)
C = Change Factor of Conductor Volume (meters^3)
D = Change Factor of Conductor Resistance (ohm)

A=sqrt(B/(C*D))
B=D*C*A^2
C=B/(D*A^2)
D=B/(C*A^2)

 

[anorlunda]

Why do you want to put resistivity in there at all?

Edit: I did not try to read or understand your change factor equations.

 

[metastable]

I included resistivity because if I keep the same conductor volume and # turns, but I switch from copper to aluminum, as you say the KV stays the same but the resistivity and resistance changes. So if resistivity isn't included in the formula, I don't believe I can calculate the new KV from volume and resistance changes alone because if the resistivity of the material changes, I believe the same conductor volume and resistance implies a different number of turns and therefore a different KV.

 

[anorlunda]

I believe the same conductor volume and resistance implies a different number of turns and therefore a different KV.

Do you mean how many turns will fit in a given volume? Otherwise, you count turns by counting them one-at-a-time. Three turns of wire is three turns, regardless of material, volume or resistance.

So I'm not understanding what you are saying.

 

[metastable]

Suppose I unwind the three copper turns from the original motor,

Next I make an identical shape aluminum wire,

Next I measure the new resistance & it is higher,

In order to make the new wire have the same volume and resistance as the first wire, I have to shorten the wire and increase the cross section slightly,

Shortening the wire changes the # turns when the new wire of the same volume and resistance is re-wound to the stator.

Changing the # turns changes the KV of the motor with the new wire, although the volume and resistance are unchanged.

 

[anorlunda]

In order to make the new wire have the same volume and resistance as the first wire,

So you're trying to change the motor "constants" while holding the volume and resistance of the windings constant. If you over-constrain, there is no solution. To get new "constants", something has to change.

The logical way to proceed is to determine how many turns you need to achieve the new motor constants. Then choose a wire to make those turns within the volume constraints. Finally, check the heat load I²R losses for all the turns the diameter and material of the wire selected, and see if that heat load is within limits. It will not be simple to find out how much heat the motor can get rid of. But it will be unlikely that the heat is the same as the original motor.

We need help from from someone with motor winding experience.
@jim hardy , @Baluncore @Tom.G ?

Edit: BLDC motors also have digital controllers. Might the controllers need to be reprogrammed if the motor is rewound with different number of turns?

 

[jrmichler]

In any motor, winding resistance causes I2R heating, and is a significant source of motor efficiency losses. To that end, motors have been built with superconducting windings. The goal of a motor designer is always to minimize efficiency losses - winding resistance, stator steel hysteresis, windage, bearing friction, cooling fan, etc.

Motor current is primarily controlled by BEMF (back EMF), not winding resistance.

 

[jim hardy]

You're asking about an electromagnetic phenomenon.
So think simple :: Volts = blv = Flux Density X Length(of Conductor) X Velocity
No resistance term there^ ..

In my machinery class we measured K_V by spinning the motor at known RPM with known excitation and measuring voltage it produced at zero current
so wire size didn't matter only length of it being swept through the field's magnetic flux

Volts per turn is a measure of flux
since you're not changing the iron you're really not changing the flux
so as @anorlunda said 'why are you even considering resistance ?'
Force on charges inside the conductors is QVcross B and there's no resistance term there.

Going back to the original question

my question is as follows:

Assuming 10 turns gives 100kv originally with copper wire, If I change the wire material to a different conductive material such as aluminum (same 10 turns), will the KV still be exactly 100kv?

I can't think of any mechanism by which changing from copper wire to aluminum would affect .QVcross B enough to measure
so K_V and K_T won't be affected either
and my answer is "Yes, for all pracrtical purposes".

In the spirit of attention to detail i must mention copper is slighty diamagnetic
see circled volume susceptibility numbers here
https://en.wikipedia.org/wiki/Magnetic_susceptibility

so it'll affect reluctance of the magnetic path hence B out around the tenth digit
which is well beyond the ability to measure with everyday test equipment

What motor formulas do include armature resistance ?

In a squirrel cage motor , rotor resistance does affect torque and slip
But your BLDC probably has a permanent magnet rotor .

Keep your thinking simple.

old jim

 

[metastable]

So you're trying to change the motor "constants" while holding the volume and resistance of the windings constant. If you over-constrain, there is no solution. To get new "constants", something has to change.

Ok I believe you are suggesting that I can't change the KV "constant" while leaving the volume and resistance of the winding unchanged...

But let's assume the original motor is wye, copper, 10 turns, 100kv, 0.06ohm, 1 unit cross section

copper -> aluminum (same geometry wire) gives:

copper = 1.68*10^(-8) ohm meters resistivity
aluminum = 2.65*10^(-8) ohm meters resistivity

2.65/1.68 = 1.57738x greater resistance

therefore w/:

wye, aluminum, 10 turns, 100kv gives, 1 unit cross section gives:

0.0946428ohm new resistance = 0.06ohm (original copper) * 1.57738 (change factor resistivity w/ aluminum)

giving:

wye, aluminum, 10 turns, 100kv, 0.0946428ohm, 1 cross section

I now modify the aluminum wire to have the same volume and resistance as the original copper wire:

sqrt(1.57738 change factor resistivity w/ aluminum) = 1.255937896553806517075

1 unit original cross section * 1.255937896553806517075 = 1.255937896553806517075 new cross section

10 turns / 1.255937896553806517075 = 7.962177132674476138125 new turns

new resistance = 0.06ohm = (0.0946428ohm aluminum / 1.255937896553806517075 new cross section) / 1.255937896553806517075 new turns

125.5937896553806517075 new kv = (10 original turns * 100 original kv) / 7.962177132674476138125 new turns

---------------------------

Conclusions

I began with wye, copper, 10 turns, 100kv, 0.06ohm, 1 unit cross section.

I switched to wye, aluminum, 7.962177132674476138125 turns, 125.5937896553806517075kv, 0.06ohm, 1.255937896553806517075 unit cross section, same winding volume.

The aluminum motor now has the same winding volume, the same resistance, and according to the "traditional" kv formula, greater KV.

---------------------------

Now I test the proposed formula:

A = Change Factor of KV (rpm/v)
B = Change Factor of Conductor Resistivity (ohm-meters)
C = Change Factor of Conductor Volume (meters^3)
D = Change Factor of Conductor Resistance (ohm)

A=sqrt(B/(C*D))
B=D*C*A^2
C=B/(D*A^2)
D=B/(C*A^2)

----------------

A=sqrt(B/(C*D))

A=sqrt(1.57738/(1*1))

A = 1.255937896553806517075

100kv original * 1.255937896553806517075 = 1.255937896553806517075 new KV

^same value given by the "traditional" formula

----------------

So my question is: is it appropriate to write the SI based formula and then use it to calculate the KV instead of the traditional formula?

 

[jim hardy]

In order to make the new wire have the same volume and resistance as the first wire,

why would you want to do that ?

So my question is: is it appropriate to write the SI based formula and then use it to calculate the KV instead of the traditional formula?

Of what formulas do you speak ?
You've not shown them,
and i do not understand
(1) this obsession with resistance
(2) how you propose to wind 7.962177132674476138125 turns

old jim

 

[Tom.G]

Offhand, I see no advantage to changing to Al wire from the existing Cu. Just cut a few turns off the existing windings and you're done. Can you state a reason for trying to match the existing DC winding resistance? If so, can you use a smaller gauge Cu wire? (By the way, Al is rather difficult to solder)

Realize that with fewer turns you will need more motor current to reach the same torque.

Cheers,
Tom

 

[metastable]

why would you want to do that ?

to summarize:

I've read that the formula is as follows (assuming wye termination):

Original Turns Per Tooth # * Original KV = New Turns Per Tooth * New KV

which can be rearranged:

(Original Turns Per Tooth # * Original KV) / New Turns Per Tooth = New KV

I wanted to write the kv/turns formula using only SI variables and came up with these:

A = Change Factor of KV (rpm/v)
B = Change Factor of Conductor Resistivity (ohm-meters)
C = Change Factor of Conductor Volume (meters^3)
D = Change Factor of Conductor Resistance (ohm)

A=sqrt(B/(C*D))

Why do you want to put resistivity in there at all?

so as @anorlunda said 'why are you even considering resistance ?'

Conclusions

I began with wye, copper, 10 turns, 100kv, 0.06ohm, 1 unit cross section.

I switched to wye, aluminum, 7.962177132674476138125 turns, 125.5937896553806517075kv, 0.06ohm, 1.255937896553806517075 unit cross section, same winding volume.

The aluminum motor now has the same winding volume, the same resistance, and according to the "traditional" kv formula, greater KV.

125.5937896553806517075 new kv = (10 original turns * 100 original kv) / 7.962177132674476138125 new turns

Now I test the proposed formula:

A = Change Factor of KV (rpm/v)
B = Change Factor of Conductor Resistivity (ohm-meters)
C = Change Factor of Conductor Volume (meters^3)
D = Change Factor of Conductor Resistance (ohm)

A=sqrt(B/(C*D))
B=D*C*A^2
C=B/(D*A^2)
D=B/(C*A^2)

----------------

A=sqrt(B/(C*D))

A=sqrt(1.57738/(1*1))

A = 1.255937896553806517075

125.5937896553806517075 new KV = 100kv original * A

125.5937896553806517075 new KV = 100kv original * 1.255937896553806517075

In order to make the new wire have the same volume and resistance as the first wire,

why would you want to do that ?

I make the volume and resistance of the copper and aluminum wires the same in order to answer @anorlunda and your ( @jim hardy ) question about why the resistivity and resistance terms are needed in the proposed formula to predict the KV changes from changes to resistance, resistivity and volume instead of using changes to "turns".

Of what formulas do you speak ?
You've not shown them,
and i do not understand
(1) this obsession with resistance

These are the formulas I speak of:

I wanted to write the kv/turns formula using only SI variables and came up with these:

A = Change Factor of KV (rpm/v)
B = Change Factor of Conductor Resistivity (ohm-meters)
C = Change Factor of Conductor Volume (meters^3)
D = Change Factor of Conductor Resistance (ohm)

A=sqrt(B/(C*D))
B=D*C*A^2
C=B/(D*A^2)
D=B/(C*A^2)

(2) how you propose to wind 7.962177132674476138125 turns

Good point. If we require the turns to be whole numbers, what is the closest to the original resistance we can get with aluminum wire?

But let's assume the original motor is wye, copper, 10 turns, 100kv, 0.06ohm, 1 unit cross section

copper -> aluminum (same geometry wire) gives:

copper = 1.68*10^(-8) ohm meters resistivity
aluminum = 2.65*10^(-8) ohm meters resistivity

2.65/1.68 = 1.57738x greater resistance

therefore w/:

wye, aluminum, 10 turns, 100kv gives, 1 unit cross section gives:

0.0946428ohm new resistance = 0.06ohm (original copper) * 1.57738 (change factor resistivity w/ aluminum)

giving:

wye, aluminum, 10 turns, 100kv, 0.0946428ohm, 1 cross section

sqrt(1.57738 change factor resistivity w/ aluminum) = 1.255937896553806517075

1 unit original cross section * 1.255937896553806517075 = 1.255937896553806517075 new cross section

10 turns / 1.255937896553806517075 = 7.962177132674476138125 new turns

^7.96... is not whole / integer

8 turns is the closest whole number of turns

8 new turns / 10 turns = 0.8 change factor of volume

1 / 0.8 change factor of volume = 1.25 change factor of cross section to retain the same volume

giving:

wye, aluminum, 8 turns, 1.25 cross section

0.060571392ohm = (0.0946428ohm aluminum * (8 new turns / 10 turns)) / 1.25 change factor of cross section

giving:

wye, aluminum, 8 turns, 1.25 cross section, 0.060571392ohm

125 new kv = (10 original turns * 100 original kv) / 8 new turns

giving:

wye, aluminum, 8 turns, 1.25 cross section, 0.060571392ohm, 125kv, same volume

--------------------------

Comparision

wye, copper, 10 turns, 100kv, 0.06ohm, 1 unit cross section

wye, aluminum, 8 turns, 125kv, 0.060571392ohm, 1.25 unit cross section, same volume

--------------------------

Test of proposed formula

change factor of resistance = 1.0095232 = 0.060571392ohm aluminum / 0.06ohm copper

A = Change Factor of KV (rpm/v)
B = Change Factor of Conductor Resistivity (ohm-meters)
C = Change Factor of Conductor Volume (meters^3)
D = Change Factor of Conductor Resistance (ohm)

A=sqrt(B/(C*D))

A=sqrt(1.57738/(1* 1.0095232))

A = 1.25

100kv original kv * A = new kv

100kv original kv * 1.25 = 125kv new kv

comparison:

(10 original turns * 100 original kv) / 8 new turns = 125 new kv

^both formulas give same results.

-------------------------

Conclusions

If we require the turns to be whole numbers, what is the closest to the original 0.06ohm resistance we can get with aluminum wire? What is the new KV?

0.060571392ohm = closest to the original resistance we can get with aluminum wire

125kv new kv = new kv requiring whole number turns and aluminum wire with the same volume and closest possible resistance to the original copper wire

Both the "traditional" kv formula and the proposed formula based on resistance, resistivity and volume changes predict the same outcome.

------------------------

Offhand, I see no advantage to changing to Al wire from the existing Cu. Just cut a few turns off the existing windings and you're done. Can you state a reason for trying to match the existing DC winding resistance? If so, can you use a smaller gauge Cu wire? (By the way, Al is rather difficult to solder)

Correct, I don't expect any benefit with aluminum wire though silver wire could have an advantage in terms of resistance. I wanted to use aluminum as the example because any differences between aluminum and copper would be more exaggerated than the comparison of silver and copper.

------------------------

So I want to know if it is appropriate to use the following formula to calculate the expected kv of the new winding, if silver or aluminum wire is used instead of copper?

A = Change Factor of KV (rpm/v)
B = Change Factor of Conductor Resistivity (ohm-meters)
C = Change Factor of Conductor Volume (meters^3)
D = Change Factor of Conductor Resistance (ohm)

A=sqrt(B/(C*D))

 

[jim hardy]

So I want to know if it is appropriate to use the following formula to calculate the expected kv of the new winding, if silver or aluminum wire is used instead of copper?

I think it is NOT appropriate.
Reason i think that is
Your kv changed because you changed your number of turns from ten to eight ..
Your kv did NOT change because of your change from copper to aluminum..

Why do you think it is appropriate ?

 

[metastable]

You're asking about an electromagnetic phenomenon.
So think simple :: Volts = blv = Flux Density X Length(of Conductor) X Velocity
No resistance term there^ ..

So if the flux density, length of conductor, velocity and voltage are known then are you saying it would be impossible to calculate the conductor resistance and resistivity values?

 

[anorlunda]

Our answers in this thread do not seem to be helping the OP. We are going in circles.

@metastable , you have misconceptions that we can't fix here. You need to go back to basics about wires, coils, inductors, fields, forces, circuits, and motors. We can't provide that here.

Thread closed.