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DC Motor Winding Questions 7

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NoonesMan

Electrical
Aug 10, 2006
4
Hello all, I am a 23 year old RadioShack Store Manager. I have an knack for electronics and physics and I do most of my research on the internet. This is my first post and when I joined it mentioned nothing of a introduction procedure so I assumed that I should just ask my questions ^_^

My questions concern small brush-type DC motors with a 3-pole arm, referred to as 130-sized. These size motors are very popular among the growing communities of people racing 1/24 scale R/C controlled performance cars.
There are some basics that experimentation has uncovered, i.e. Fewer turns of wire makes the arm spin faster with sacrificed torque where many turns of wire creates more torque with less RPM, Neodymium magnets increase torque and reduce RPM, and weaker ferrite magnets can increase RPM with reduced torque (where the better trade off is with neo mags).
My main concern is the lack of sources willing to share this information, or willing to go beyond the basics of motors. The people I've found couldn't explain the different effects of different wire turns, or heavier/lighter gauge wire. None of these people could determine how to compensate the coils for increased permanent magnet strength, or explain any advantages to weaker permanent magnets.
Don't get me wrong, there are plenty of sources willing to sell a motor built to specs, but they don't make money if I wanted to build my own. its been tough finding info on simple DC motors.

I hope this post isn't too much for anyone willing to help. I'll thank anyone in advance ^_^ This information will go to serve the communities I frequent in their own desires to build their motors, or at the very least understand them fully.
 
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Permanent magnet DC motors follow the physics performance of classic DC motor theory. “Interaction between current carrying conductors and electromagnetic field”.
Motor action is based on Ampere’s law, for the force F on a conductor with length (L), carrying current (i) while immersed into a magnetic field with density B:

F = B*L*i

The torque will be the sum of forces on each current carrying conductor times the radius.
B will depend on the magnet strength and air gap. The current “i” will depend on the voltage applied, resistance of the conductor and “cemf” induced on the rotating winding. The current is not steady until the rotor reaches a steady speed.

The counter electromotive force “cemf” is a voltage induced on the rotating conductor function of the speed “rpm” , the magnetic field density B and the number of turns, length, span etc all included in a constant Kw .

cemf = Kw*B*rpm

The voltage applied to the motor terminals Va (armature) is balanced by the cemf, the conductors resistance Ra, and the current Ia.

Va = cemf + Ia*Ra = Kw*B*rpm + Ia*Ra

Solving the equation above for “rpm”

Rpm = (Va-Ia*Ra)/ (Kw*B)

It can be seen that stronger magnetic field will reduce the speed. The change of turns is not that obvious, but a reduction of turns reduces the resistance “Ra” and the “cemf”, increasing the current (torque).

All these parameters must be keep not exceeding the temperature limit of the insulation materials.
 
Hello NoonesMan,

There are several sites that explain why DC motors behave like that. But I shall try to write up a short summary:

There are a couple of physics laws that govern the behaviour of motors and generators. There is the induction law that says that voltage = magnetic flux times wire length times wire speed (E=B*l*v) and there is the law of magnetic force that says that force = magnetic flux times wire length times current (F=B*l*I). These formulae are valid in the SI system of units. You need to add constants if you use other systems with units like feet, pounds etcetera.

You have already observed that torque goes down when you use a ferrite instead of a stronger magnet. That, as you can see, follows form force equation (F goes down as B is reduced) and since torque is torque arm (which is constant) times force, it is clear that torque goes down with flux.

The speed of a DC motor is determined by the applied voltage. As long as the induced counter-EMF (the voltage generated in the motor winding) is less than applied voltage, the motor accelerates until applied voltage is balanced by counter-EMF.

So, if the magnetic field is reduced, the voltage formula (E=B*l*v) says that induced counter-EMF also goes down. That means that the applied voltage forces more current to flow and the motor accelerates.

This fact is used in larger motors with a separate field winding (not permanent magnet) so that you can increase speed above rated speed (aka base speed) at rated voltage by reducing field current.

There are a few secondary effects that you should know about (effect of internal armature voltage drop - IR drop - and armature reaction - a kind of parasitic field weakening) but I think that you should start with these basic formulae and find out more from your own experiments, which I find very recommendable.

Get yourself a simple DMM, a variable power supply and, perhaps, a stroboscope or tach for speed measurement. You will soon be on the way to becoming a motor expert with these simple tools.


Gunnar Englund
 
So, if I am to understand correctly, the simple fact of the motor spinning induces voltage in the coils. When that induced voltage balances the applied voltage the motor has reached its max RPM. (usually only with no load?)

When the stationary magnetic field is reduced, it induces less voltage for the same RPM allowing for more RPM until the induced voltage matches the applied voltage. Stronger magnetic fields induce more voltage forcing the RPM to stay lower.

In the case where the stationary magnetic strength and applied voltage remains fixed with no limit on current supplied, if the coil length of wire is reduced, it induces less voltage which allows increased RPM. More length in the coil induces greater voltage and supresses RPM. The differences in coil length then determine available torque to maintain RPM?

So imagining this motor spinning at max RPM, I apply resistance to it's motion. This drops RPM, thus dropping induced voltage, allowing for more current to flow increasing torque to maintain RPM. I could continue until the magnet wire has reached it's maximum current handleing (also dissapating heat) and the motor cannot continue to compensate with enough current to generate a magnetic force enough to overpower the resistance of the mechanical load.


If you all see no flaws in the ways that I've interpreted your responces, know you've done your deeds well!
 
Um..

Par1:
Correct.

Par2:
Correct.

Par3:
Number of loops changes the voltage not length.
Length does affect torque.

Par4:
Correct.



Keith Cress
Flamin Systems, Inc.-
 
Does this context work better?

In the case where the stationary magnetic strength and applied voltage remains fixed with no limit on current supplied, if the turns in the coils are reduced, it induces less voltage which allows increased RPM. More turns in the coils induce greater voltage and supress RPM. The number of turns of the coils then determine available torque to maintain RPM?

One last question: Two motors, with constant voltage and the only difference being the coil's gauge of wire, would perform the same under (theoretical) no load?
 
Yes.
Yes.

Last Q:
No. (Theoretically) The one with larger wire would have less IxR losses so supply voltage would not be lost as resistive heating thereby requiring the cemf to be slightly larger. (spin faster)

Keith Cress
Flamin Systems, Inc.-
 
I can see that you sorted it out quite well. Just one thing (I know you are a practical man, Keith). Noone put an important "(theoretical)" in his last question and that means that there is no current in the motors. Then (but only then) do the motors spin with the same speeds.

Otherwise you have got it perfectly right.

Gunnar Englund
 
You're saying that the motor would speed the same 'no load' speed? With different size wire? No current in the motors?? (Not quite following you skogs.) The unloaded motor still has parasitic loads; bearings, windage.

Keith Cress
Flamin Systems, Inc.-
 
Hi folks;
Consider a motor with a slightly overhauling load. Possibly a transit vehicle on a slight downgrade.
A point will be reached where the applied voltage will equal the back EMF, and the losses will be supplied by the overhauling load.
Will the practical coincide with the theoretical at this point?
Another interesting point. With a wound field, the field strength depends more on the wire gage than on the number of turns. With a given wire size, if you double the number of turns, you approximately double the resistance and so half the current. Net result,- approximately the same number of amp-turns.
respectfully
 
Wow, I love the responces guys very helpfull.

I did sort it out well because I like to think I have a strong understanding of most electronic principles (I had an electronic kit at 7 and it went up from there :p ), but having these facts laid out in front of me with all the (somewhat) hidden factors (cemf) explained in plain detail really painted the whole picture for me.

Well, I could have laid out my parameters a little better with that last question *cough* and really, after I mulled it over, I could have answered it myself. The motors would be pulling current in that question, the absence of load I was referring to was physical load. (i.e. perfect bearings, no air resistance, no brush friction, and possibly a head start (to overcome mass differences))

Anyway I can answer my own question: Mass & Electrical Resistance seem to be the only factors left, and they directly relate to the guage of wire. Even with the restraints in my question the performance varies (with the above factors) when you begin to add physical resistance again. Put in the physical properties again and the issues point to where the best resistance/mass tradeoff is.

I would assume that the resistance lost allows overcompensation of the gained mass allowing in better performance, in almost every aspect. The exception being undersupplied power...which is another ballpark altogether.

Anywho, Thank you all. You have shown me how simple simple DC motors actually are. I'm going to share this info with the forum I mentioned earlier with credit to you all ^_^
 
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