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Simple DC motors Current, Torque, EMF and Faraday's Law

  1. Jun 14, 2013 #1
    Ok, so I'm doing an experimental write up for an experiment on adjusting voltage on the speed of a simple DC motor, and changing the number of coils.

    I thought that I understood what I was saying but I've recently become a little confused with the concepts of torque vs speed specifically when trying to talk about a torque/speed graph.

    I know that as torque increases, speed decreases, but I've also been talking about EMF and Faraday's law todo with how as the motor increases in speed the current drops until the torque is quite small (Due to the equation τ=BAIN cosθ, τ=torque, B = magnetic field, A = area of coil, I = current, N = number of turns) and then the forces are equal as the small torque remaining is balanced due to friction, and also the magnetic field working against the motor that has been produced as it cuts the magnetic field (this is faradays law correct?)

    3. The attempt at a solution
    So what I'm not sure about is, do I say that as the number of coils increases, the torque increases (due to the A and N in the above equation) and as such the speed is reduced, or Is it more technically that as the number of coils increases, the generated back emf is larger, and also the magnetic field produced by this back emf is larger, which also causes it to be slowed.

    Sorry if this does not make sense.. the motor we used is similar to this: http://www.miniscience.com/projects/Magnet_Motor_kit/Magnet_Motor_LL.jpg
     
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  3. Jun 14, 2013 #2

    Andrew Mason

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    With an internal combustion engine, torque increases as speed increases. But not with a DC electric motor. The torque is a function of the strength of the magnetic fields in the armature and field windings, which is a function of the current passing through these windings. In a serial motor (such as a starter motor in your car) both the field windings and armature are drawing enormous current when it just starts so that is when maximum torque occurs.

    As the motor speeds up, back EMF is generated in the windings which reduces the current and the torque.

    AM
     
    Last edited: Jun 15, 2013
  4. Jun 14, 2013 #3
    So the Torque is always there, but it's just being negated by the back EMF?

    But in the experimentation, we're expecting the motor speed to decrease as we increase coil size, despite having more torque, is this just due to more windings causing more back EMF and then the extra weight is the reason for the slower speed?
    Or is Faraday's law in effect causing a magnetic field that opposes the one that moves the motor?
     
  5. Jun 15, 2013 #4

    CWatters

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    An idea lossless DC permanant magnet motor running flat out with no load draws no current and generates no torque. If there was torque it would keep accelerating.

    When you apply power to one of these motors it accelerates until the back emf matches the supply voltage. The back emf is proportional to the number of turns so the more turns the lower the rpm at which the back emf is equal to the supply voltage. This is why more turns => slower motor.
     
  6. Jun 15, 2013 #5

    CWatters

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    This page and graph might help...

    http://lancet.mit.edu/motors/motors3.html

    This is the torque vs speed for constant voltage and constant number of turns...

    colorTS1.jpg

    This sort of plot is produced by putting the motor on a variable load/brake. eg The speed is being forced to change and the torque measured.

    Note how at max speed the torque is zero and when stalled there is a max torque.

    If you increase the number of turns the slope gets steeper.

    PS When thinking about Torque vs Speed it matters what is causing the speed to change. eg is it the load, the supply voltage or the number of turns you are changing. The effect on the plot for torque vs speed is different in each case.
     
  7. Jun 15, 2013 #6
    In my case however, we didn't change the load on the motor, the only thing we were doing was increasing the number of coils, and expecting a linear relationship between the speed of the motor and the number of turns, decreasing speed as turns increase. So This increases torque, but from my understanding the torque is just the force that turns the motor, so it has more torque with more turns, but is going slower. And I can't see why that is unless the back EMF is cancelling out the current producing the torque at a slower rotational velocity?
     
  8. Jun 15, 2013 #7

    CWatters

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    That's what I would expect. More turns means the required backemf is achieved at a lower speed.

    Ah I see the confusion. You are wondering why increasing the torque doesn't increase speed.

    The torque produced by a motor is not a constant it's a curve (Edit: Well a straight line). It's missleading to say that increasing turns increases torque. You can only say that it changes the torque curve. The actual torque produced at any one time depends on the load.

    The no load (no torque) speed will reduce. The max torque (no speed) will increase. You might not have been changing the load but what was the load?
     
    Last edited: Jun 15, 2013
  9. Jun 15, 2013 #8
    the load was just the friction and any energy loss in the motor from heat I'd say.

    So to explain why increasing number of turns reduces the maximum speed, I just say that the back-emf is achieved at a lower speed? My teacher said that I should be refering to a torque vs speed graph, But i'm finding it really hard to explain how.. This is my understanding at the moment, is this accurate? Torque increases when number of turns increases, according to the equation t=BAIN cos(theta) as N and A would increase for more turns (area and number of turns) therefore there is a higher torque. However, this higher torque is accompanied by a loss of speed due to a greater back-emf being generated, reducing the current to zero at a lower speed. meaning that despite the higher torque, the motor spins slower.

    So technically it's not that higher torque causes lower speed, just that the things causing the higher torque, are also causing a lower maximum speed through back-emf?

    Edit: Sorry to be so annoying about this :S It's 1/5th of my overall mark and we switched teachers halfway through the assessment so i'm a bit stressed.
     
  10. Jun 15, 2013 #9

    CWatters

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    Ok so we can probably say thats close to the "no load" situation where torque = 0.

    Yes although perhaps elaborate a bit.

    I would save a copy of that plot I posted above. Then add two lines showing what happens when you increase or reduce the number of turns. eg work out what happens to the no load speed and stall torque and join with a straight line. Need not be 100 to scale.

    I'm in a rush to go out for lunch but...

    When you draw the chart you will see that the lines cross so as you change the number of turns at some loads (some torque) the speed is higher and at others it is lower.
     
  11. Jun 16, 2013 #10
    Ok, one final question because I forgot to mention it, for coils with more turns, e.g when comparing one with 5 turns to one with 7, the 7 turns coil, in our findings has less gradient when increasing voltage, when plotting speed, so as the voltage increases for both, they both increase in speed but the one with more turns increases less per voltage.

    Is that what's supposed to happen? I don't see why it would? Unless more coils means it's running less efficiently?
     
  12. Jun 16, 2013 #11

    CWatters

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    That's exactly what you would expect but it's not due to a loss of efficiency...

    As we said earlier, increasing the number of turns reduces the speed at which the back EMF matches the supply voltage. That applies at any voltage.

    The relationship between voltage and rpm is...

    RPM = Motor Constant (rpm/volt) * Supply voltage (volts)

    If you plot rpm against voltage you should get a straight line where the gradient (aka slope) is equal to the motor constant.

    Increasing the number of turns reduces the motor constant and hence the gradient of the slope. That's the primary effect.

    There is a secondary effect in that increasing the number of turns does increase the resistance of the windings so more is lost as heat. However the main cause of reducing gradient is the change in motor constant.
     
  13. Jun 16, 2013 #12
    I have a formula similar to that, because I had seen that before, but couldnt find any references.

    I have back emf = motor constant * motor speed

    I can see this is related but I can't quite link it.

    And the motor constant is reducing for more turns because, of a greater back emf?

    Oh! I think I get it..
    When you increase the number of turns, the back emf is achieved quicker. So the motor constant, is less because the back emf is generated quicker per volt, causing less rpm/volt?

    I'm so sorry for being slow with all this :(
     
  14. Jun 16, 2013 #13

    CWatters

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    Yup that's it.

    However the motor constant in..

    back emf (V) = motor constant (volts/rpm) * motor speed (rpm)
    and
    RPM = Motor Constant (rpm/volt) * Supply voltage (volts)

    aren't identical although they are roughly the reciprocal of each other.

    http://en.wikipedia.org/wiki/Motor_constants
     
  15. Jun 16, 2013 #14
    Do you know of a place where I can reference that equation for RPM = motorconstant * supply voltage? I can't seem to find it anywhere outside of wikipedia..

    Thank you so much for the help!
     
  16. Jun 16, 2013 #15

    CWatters

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