[major]

Why do some motors have a small and others large wire for winding size(dia), and a high or low number of comutator segments? Second what does the number of winding segments mean? High v/s low.

Hi Dink,

The basic parameter (for the armature) is the number of series turns per pole which then determines the voltage speed relationship for a given magnetic flux per pole. One must also know the armature winding pattern to know the number of parallel circuits.

For a given flux and speed, the more turns, the higher the generated voltage. For a motor running on a fixed voltage, this means the more turns, the slower it turns. But also, the more turns, the higher the torque for a given current.

There is a finite space in the armature for the wire (turns), so when you increase the number of turns, you have to make the wire smaller. And the smaller the wire and the higher the turns, the higher the resistance, and lower the current capability.

So, few turns, large wire, high speed, high current. Many turns, small wire, low speed, low current.

The number of commutator segments (or bars) is chosen by the designer with various considerations, voltage being a primary factor relating to acceptable level of sparking. Often the number of comm bars is set equal to the number of slots in the laminated armature core. How that choice is determined gets into the magnetic structure and manufacturing process and cost benefit analysis which I think is beyond discussion.

Generally speaking, higher voltage motors have higher comm bar counts.

I hope that addressed your questions.

major

 

[major]

But I'm interested to know one more thing about DC serie motor.

At a fix given current, maybe 1000A, what determines the capacity of a motor to sustain his max torque over the increasing rpm?

Hi Yab,

At a fixed given current, the series motor will always have the same torque. Torque = Kt * Ia * Flux. Kt is a machine constant and is proportional to the number or turns in the armature.

In your plot, the torque is constant at 240 lb.ft. from 0 to 2500 RPM. This is obviously at the current limit of the controller, a constant current value. So, in the equation, Ia is constant, and Flux is constant, so the torque is constant.

Above 2500 RPM the torque falls off because the current decreases. And for a series motor, as Ia decreases, so does the Flux. Ia = (Vm - Eg) / Req. Vm = applied motor voltage, Eg = generated voltage in the armature, and Req is the equivalent series resistance of the motor.

Eg = Kt * Rad/s * Flux. Kt is the same constant as in the torque equation, numerically if metric units are used.

If you wrap your head around all these equations at the same time, you will see that as the motor speed increases, Eg increases, and unless Vm can increase, Ia will decrease and hence torque will decrease.

I read someting about resitivity of motor in ohm (mΩ) vs voltage given to the motor vs voltage produce by the motor, but it's not clear for me.

It is the resistance of the motor. And for these EV motors, is in the milliohm range. I tried to show how that plays into it with Req in the equations.

Also, the number of commutator segments can play a role?

For these EV motors (series wound and capable of 1000A), they have bar wound armatures instead of round wire wound armatures. Meaning that the armature conductors are rectangular copper ribbon. This makes for a robust motor having the conductor brazed or welded to the commutator segment, low resistance winding and mechanically secure assembly. In most cases, these bar wound armatures have a single turn coil and the number of armature coils equal to the number of commutator segments (bars).

So the armature constant, Kt, relates directly to the number of comm bars. This is generally true for the EV suitable motors you will encounter, but there are some winding tricks occasionally used which make that statement not universal

Hope that confused everyone

major

 

[major]

If I understand correctly , for the same voltage, the same current and same quantity of bar in the armature, a motor with less resitivity will be capable to sustain his max torque to a higher RPM.

I think you are placing too much concern on resistance (not resistivity, which is a property of copper). Motors of same size and basic design (comm bar count) will have resistance near enough not to make appreciable difference.

So, because the the copper has a very little resistivity, the most resistive part of a serie DC motor is probably the brush and the brush contact area on collector.
If it is true, a motor with very large brush can tolerate more current and can also give his max torque at higher RPM.

Yes, this does play a part in it and for simplification I just included this in the _Req_ for those equations. However, when doing detailed motor calculations, the brush drop (Voltage drop including brush and surface contact) is considered separately because it is a nonlinear function of current and surface speed. It is also dependent on the temperature, brush material, spring force, seating, humidity and sunspot activity.

Generally speaking, the larger the total cross sectional area of the brushes, the higher current capability. At higher currents, like 1000 A, this might account for a volt or two difference between one motor to another of similar size. So in a hundred volt system, will mean little as to being able to carry maximum torque to high speed.

And in this case, with same controller, a bigger motor can probably give his max torque at higher rpm than a comparable smaller motor. (Example: Warp 11" vs Warp 9")

Not necessarily. Larger motors generally produce more torque. But generally run slower. And may not have higher power rating. There are just too many variables. It is unlikely you can find two motors of different sizes with all other design features similar to even make a comparison. You could do analytically, but I'm not up to it tonight

For example, I think the Warp9 and Warp11 have the same size and number of brushes. The W11 would have a larger comm and therefore likely a slight advantage in short time base current overload capability. Other difference is, I think, the W9 has 49 slots and bars and the W11 has 25 slots and 75 bars. So the W11 has a higher Kt. Even though the W11 is larger diameter, I think the W9 has a longer core. But the W11 has a higher flux at saturation.

Some speculation on my part there. I have never seen the Warp11 up close and personal. I have seen the W9, but not used it much.

If your goal is to take the torque at current limit higher in motor RPM, increase battery voltage. And figure out a way not to zorch it

major

 

[major]

I have been reviewing this post and previous ones I have posted,and need some help understanding. I understand what the commutator is and it's segments, but the windings confuse me.

Hi Dink,

The commutator (which is made of segments or bars) only conducts current from the stationary world to the rotating armature and switches that current in direction at the appropriate time and position. The commutator actually adds nothing to torque or power production in the motor, and in fact loses torque and power in friction and voltage drop.

The windings in the armature are what do the work. More specifically, the portion of the winding in the steel core (magnetic path). The windings are the coils and are connected to the commutator, each end of the coil to a comm segment or bar. The armature coils (windings) can be loops (turns) of round wire inserted into slots in the steel core of armature, or, as in the picture above, can be single turns (called hairpins) of copper rectangular conductors (sometimes also called bars) inserted into the armature core slots.

Hope that helps,

major

 

[major]

Not exactly.............

GE 11": 49 bars and 4 big brush (4 poles).

Warp11: 25 bars and 8 brush (4 poles).

I suspect it has 25 slots with a 49 bar comm. 2 coils per slot.

 

[major]

Major,
how do you have an armature design that has a different number of Commutator bars and a different number of armature slots?
How does this work?

2 coils per slot. 1 coil per comm bar. 25 slots * 2 coils/slot = 50 coils. Minus one dead (unconnected) coil = 49 coils for 49 comm bars. Dead coil keeps spacing and balance. Odd number of comm bars needed for 4 pole wave armature winding.

I had thought (never having seen one) that it was 25 slots and 75 bars, using 3 coils/slot. I think GE also makes an 11 incher using 65 slots and bars. And maybe other designs as well.

what advatages does this have over the same number bars and same number slots? like 49 and 49 in the GE 11?

Some pros and cons for each way. Existing tooling plays a big part in the choice. For high volume production, equal bars and slots are probably best, for both design and process.

major

 

[major]

You can see.....

Warp 11 have 25 slot in armature, but 49 bar comm.

I would like to know the difference rpm and torque of each motor at same voltage and current!..

Hi Yab,

So both motors have 49 bar comms. They both use single turn armature coils and wave winding. So they would have equal Kt values for those equations in my post #9.

But, both the speed and torque equations include the term _Flux_. And we do not know the Flux values for these motors, so it is impossible to draw an accurate comparison of performance from the information at hand.

Both motors use steel although we do not know the exact type. But we could assume that the saturated (maximum) flux density would be the same. To get a ballpark comparison, measure the steel armature core diameter and stack length. From this, calculate the cylindrical surface area. For a comparative ratio between the two, just use ratio of D*L, one to the other. The larger D*L will have more Flux at saturation, meaning more torque/amp on overload.

Back down at non-overload conditions, it is much harder to draw such conclusions. The field coils turns (mmf) come into play and a host of other design features like air gap length (distance between pole face and armature surface), pole pitch, etc.

Gross estimation: Same comm bar count, same size = about the same speed torque current voltage relationships.

Regards,

major

 

[major]

Is it a big disavantage to have one big brush hold with two spring compare at 2 brush with one spring on each?
I assume the same area of brush contact (2x1 = 1x2).

Hi Yab,

In a perfect world it does not matter; the contact area is equal. But In our world, the 2 brushes are better because they can react separately to imperfections in the commutator. This improves the life or service interval and lessens the chance of problems. And the 2 brushes may each seat a bit more quickly. But it is a bit more costly to use 2 twice as many brushes.

major

 

[major]

Small fortune = $1.00

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