I am not a physics guy, so not even the basic concept of a DC motor is easy for me. My question is as follows:

How do these parts of a motor affect its RPM and Torque?

I had my research a while ago so I filled out some of it; please correct if there is something wrong.

More turns - less RPM, more torque, less battery consumption
Less turns - more RPM, less torque, more battery consumption
More winds (number of wires) - less RPM, more torque, ?
Less winds (number of wires) - more RPM, less torque, ?
Thicker wire - ?, ?, ?
Thinner wire - ?, ?, ?
Stronger magnets - less RPM, more torque, ?
Weaker magnets - more RPM, less torque, ?
Bigger commutator - ?, ?, ?
Smaller commutator - ?, ?, ?

Also, if you know any tutorials for a guy who has the brain of a six-year old boy regarding a DC motor, please let me know. Thanks and apologies for my English!


I'm working with constant voltage 2.4v (2 Ni-MH AA Batteries)

  • $\begingroup$ how are you distinguishing winds with turns?? $\endgroup$ – Vineet Menon Mar 13 '12 at 7:03
  • $\begingroup$ I'm sorry if it confuses you, what I mean in "winds" is the number of wires to use... $\endgroup$ – domanokz Mar 13 '12 at 7:11
  • $\begingroup$ The number of wires attached to the commutator... $\endgroup$ – domanokz Mar 13 '12 at 7:16
  • $\begingroup$ got a schematic? $\endgroup$ – Timtam Mar 13 '12 at 7:52
  • $\begingroup$ No I don't, I'm just trying to understand the concept... $\endgroup$ – domanokz Mar 13 '12 at 8:16

The RPM is restricted by frictional losses in the engine and the tendancy of the engine to explode if you rotate it too fast. In that sense the RPM limit is down to how well the engine is made and what it's made of rather than any fundamental EM properties.

The torque is dependant on how large a magnet field the engine generates, and this is mostly dependant on the current it draws and the number of turns in the coil. So the thickness of the wire has no effect except if it affects the resistance and therefore the current drawn. Likewise the commutator won't have any effect.

Have you had a look at http://en.wikipedia.org/wiki/Electric_motor#Performance. This gives a pretty thorough description and isn't too technical.


This is in addition to @JohnRennie's excellent answer. First thing, the "more this less that" depends upon if the motor is being operated under constant voltage or current. I'm talking about torque only, RPM will increase if torque increases generally, but other friction forces and self-induction make it a headache to solve.

Condition           Const Voltage       Const Current
More turns -        No change           Increases
More winds          No change           Increases
Thicker wire -      Increases           No change
Stronger magnets -  Increases           Increases
Bigger commutator - Affects friction only

FYI, I used the equations $V=IR$ (Ohm's law), $R=\frac{\rho\ell}{a}$ (definition of resistivity), and $\tau=BI|\sin(\omega t)|$ (a formula I derived for a DC motor) to analyse this.

  • $\begingroup$ Yes I'm working on constant voltage 2.4v (2 AA Ni-MH BAtteries). $\endgroup$ – domanokz Mar 14 '12 at 2:43

The physical quantity of iron and copper tells you everything about the speed/torque properties of the motor. The way you wind the copper is very minor: you get the same motor performance with thousands of tiny windings or with dozens of fat windings, except you need to operate with higher or lower voltage depending. That's assuming you wind it for basic two-pole operation. You could wind it like an AC motor for multi-pole operation, but I don't think there's any reason to do that with a DC motor, because you only lose power whereas in AC it is desirable to wind motors for different speed, which is not a problem in DC.

Having wound your motor, the basic fact of DC motors is: voltage is speed, and current is torque. Double the voltage, double the speed. That's the armature voltage: the field voltage should be kept constant. It works because the motor has to spin fast enough that the field voltage generates a back-voltage in the armature equal to the supply voltage, and the less you apply at the aramature, the slower it needs to spin to generate that quantity. The motor will draw as much current as it needs to get the torque necessary to rev up to the speed demanded by the voltage.

The torque is limited by the overheating of the wires to the point of melting the insulation. The maximum speed, at full excitation, is limited by the maximum armature voltage which I haven't thought of before, but is PROBABLY limited by the arcing of the commutators. You can actually increase the speed even more by REDUCING the field excitation, but at the cost of much reduced power.


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