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Why does speed of DC motor increase when flux is reduced?

  1. Mar 19, 2015 #1
    I understand that from KVL
    e = v -IaRa


    But physically what causes the speed to increase?

    What force causes the rotor to accelerate?

    In fact physically, the rotor is moving because of field flux interacting with armature current's flux.
    So physically if I reduce one flux, the speed must reduce, since i have reduced the cause of motion.
  2. jcsd
  3. Mar 19, 2015 #2


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    Well I've looked at this for a while and I can't see what's wrong with your algebra. So the only thing I can offer is,
    When the flux decreases, e reduces, so I increases (I = (V-e)/R ) so the torque MAY be altered, because T = K.I.F
    The torque would have to change if the motor were going to change speed.
    If reducing the flux were going to increase the speed, then the torque would have to increase, at least temporarily.
    But if there were no flux, there could be no torque and the motor must slow down. So ultimately I would expect that reducing flux must slow the motor.

    So far my calculations do not support that! I must struggle some more with the maths. I think it is essentially a maths problem.

    Edit -
    I think you are right, but although you have reduced one factor causing motion, you have increased another (current)

    I wonder if the issue is that the equations take no account of friction? They presumably apply to a motor whether or not it is in equilibrium (with friction, load, etc balancing torque) so I can see why they do not mention it.
    Last edited: Mar 19, 2015
  4. Mar 19, 2015 #3

    jim hardy

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    what happens to e ? What then happens to Ia ?
  5. Mar 19, 2015 #4
    Emf (voltage) of ideal DC machine is:
    ε = k⋅Φ⋅n​
    k=p⋅z/60a...constant dependent of DC machine construction
    n... speed of machine in rpm

    So if you keep voltage fixed it is evident that relation between Φ and n must be inverse proportional. Just as simple as that.
  6. Mar 19, 2015 #5
    I understand mathematically that if the flux is reduced then speed must increase.
    E = V - IaRa

    E is proportional to speed and flux.

    So if V and IaRa are constant, then Speed and Flux are inversely proportional.

    But this is mathematically. The machine doesn't look at the equation n changes its speed. What is the physical reason for the speed to increase if I reduce the flux?

    What extra force is provided to push the rotor and accelerate?

    I was thinking that DC field is constant. If we reduce we are giving the armature a change of flux. Maybe this change of flux adds an extra 'e' that causes. This extra e pushes the rotor to a new equilibrium speed and this speed is now the steady speed.

    If we follow mathematical laws, Speed is inversely proportional to Flux. So if I reduce the flux to zero suddenly, field circuit trips, then would the rotor accelerate to dangerous speeds?

    I think once the motor starts rotating the flux acts as obstruction to speed, like it tries to maintain the rotor in steady position, so if i reduce the flux, i'm reducing the obstruction.

    Analogy: a cheetah needs to eat in order to run. The food is flux while its legs are armature. If the cheetah its food it will run, but if it eats a lot, it will be huge n the weight wont allow it to run. If it becomes slim it can run fast, but if it is empty then it has no energy(torque) to run.
  7. Mar 20, 2015 #6

    jim hardy

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    counter emf decreases. So armature current goes up, look at your equations. That gives torque to accelerate the machine.

    [QUOTE ]So if I reduce the flux to zero suddenly, field circuit trips, then would the rotor accelerate to dangerous speeds?[/QUOTE]
    If you could reduce flux to zero there'd be no torque and no counter emf
    so armature current would go to Vapplied/Rarmature , a very large number of amps.

    In actuality you can only reduce field current to zero and you are left with residual flux of the field structure.
    So RPM goes to a very large number unless load can hold it back.
    An unloaded motor that loses field current will overspeed and wreck itself .. The armature wires get slung by centrifugal force out of their slots into the airgap and crushed against stator..


    Stick to analogies that'll lead you to same equations figured out by our predecessors. The imagination can believe anything so we must lead not follow it.

    see also https://www.physicsforums.com/threads/speed-control-by-flux-weakening.706529/#post-4478852
  8. Mar 20, 2015 #7
    Thanks the thread was very helpful...

    "force F on a charge Q with velocity V in a magnetic field B : F = QV cross B (vector cross product) .
    That force F is responsible for counter-EMF.
    Counter EMF is what opposes inflow of armature current.
    Since counter EMF is proportional to product of velocity V (think RPM) and flux B (think field),,,

    IF you reduce either RPM or field
    THEN counter emf decreases so more current will flow into the armature.
    That increases electrical power input to the motor.
    So the motor will accelerate until counter emf once again is sufficient to balance applied voltage.

    Always keep in mind your two basic DC motor formulas:
    Counter EMF = (K X Flux) X RPM, where K is a constant for that particular motor
    and Torque = 7.04 X (K X Flux, both same as above) X Armature Amps."
    --by Jim Hardy

    Isn't torque on rotor proportional to both Field Flux and Armature Flux(or armature current)...

    Torque on rotor(armature) is caused due to interaction of flux by field and flux by armature current at armature conductors.

    I understand that if field flux is reduced then back emf reduces n armature increases, so torque increases. but doesn't decrease in field flux compensate for the increase in armature current, for the torque to be constant.
  9. Mar 20, 2015 #8

    jim hardy

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    nope. Armature resistance is intentionally low so that the motor can have good torque.

    One could build such a motor as you describe but its performance would be disappointing for most uses.

  10. Mar 20, 2015 #9
    Thanks... so it's
    Flux reduces
    Back emf reduces
    Ia increases n provides for the additional T that pushes the rotor to a higher speed
  11. Mar 20, 2015 #10


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    And, when you consider that an electric motor is there in order to do some work on a mechanical load, the 'runaway' situation that you were starting off with will not necessarily apply. The load will limit the top speed - and either that speed will be less as the Field reduces or the winding current will get too great.
  12. Mar 27, 2015 #11


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    If lower field flux increases torque and speed, why do we not design motors with lower field flux? (Although most references here seem to be motors with field windings, I am thinking of permanent magnet motors, since the field can be controlled independently of armature/rotor current.)
    Would not the ideal motor then have no field flux at all? This seems a paradox, since I'd expect such a motor to get warm, but not move at all!

    And I don't understand Jim's comment,
    I'd have thought, you can independently control the field in a shunt wound motor, by simply not connecting the field winding to the same supply. If you were trying to do what OP asked, surely you could apply the required reverse current to reduce the field (including residual flux) to exactly zero? It would require a bipolar supply, but I'm pretty sure I've seen these around.

    As everyone says, the equations seem to imply that speed increases when flux falls and that this is mediated by increased armature/rotor current, but it also seem to me that despite the increased current, torque must fall, as it depends on the product of flux and current. (Can the fall in back emf increase the current sufficiently to increase the torque? It seems to me that, at best, I is inversely proportional to e, so the torque would remain constant.)
  13. Mar 27, 2015 #12

    jim hardy

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    Yes one can do what you propose. One has then made an essentially a fancy resistor of very few ohms.

    i think I vs e is direct but i get your point.

    Are you varying voltage applied to both armature and field, or just to field?

    What fraction of applied voltage is consumed by Counter-EMF, and what fraction by Armature Resistance? In a motor you want the lion's share going to Counter-EMF because that's what makes the mechanical work. Armature resistance just makes heat.
    What's slope of X/(1-X) over range X=0 to X=0.99 ?
  14. Mar 29, 2015 #13


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    The thing about this one is that it is easy to ignore the consequence of reducing the field current / field. The temptation is to think that you're getting something for nothing and it just isn't like that.
    If you stall any motor the torque will be proportional to the product of the field and the armature current - and the current will just be limited by the winding resistance. (Smoke and flames, in many cases). How are you powering this thing? An assumed voltage source (very low source resistance) will allow as much current as you want - and waste as much power as you care to demand. The torque can get as high as you like if the resistance is small enough. You don't 'run out of' anything, despite what your intuition tells you and the motor just gets less and less efficient. Intuition has a lot to answer for in these cases.
  15. Mar 29, 2015 #14

    jim hardy

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    My basic motors course was, well, intended to teach the basics.
    Lab was a lot of fun. We had 7.5hp machines and dynamometers , huge rheostats, hundreds of amps available, and fine old wood cased electrical meters that were some years later donated by university to the Smthsonian as historical artifacts.

    Let's try an example with numbers
    then the formulas will be easy to derive


    Let's assume nice round numbers for a starting point
    Vapplied = 100VDC
    Rarmature = 0.1 ohm
    Rfield = 100
    flux Φ = 1 weber per amp
    torque = 7.04 ft lbs
    We'll assume this motor is connected to some constant torque load like an elevator, so as to only change one variable at a time.

    Ifield = Vapplied/Rfield = 100V/100Ω = 1 amp
    so Φ = 1 weber
    (Since this is a thought experiment i arbitrarily set K for this machine = 1, ordinarily you'd determine it from open circuit test on a dynamometer)

    Since Torque = 7.04 K Φ Iarmature ,,, Iarmature must be 1 amp

    Counter emf must be Vapplied - IarmatureRarmature = 99.9 volts

    Since counter emf = K Φ RPM , RPM = (counter emf)/KΦ = 99.9RPM

    Power = 2Π X torque X RPM/33000 = 0.134 hp

    Now we'll halve field current
    so Φ = 1/2 weber
    Torque is still 7.04 ft-lbs , = 7.04 K Φ/2 Iarmature
    so Iarmature doubles to 2 amps
    Counter emf drops by increased voltage across Rarmature, now 2 amps X 0.1 ohm , to new value of 99.8 volts,
    Counter emf = KΦRPM so RPM = 99.8 = RPM X 1/2 , RPM = 199.6

    Power = 2Π X Torque X RPM = 0.276 hp, note it and RPM both nearly doubled.

    In real life the load is probably not constant torque but something like a fan or vehicle where torque goes up with speed.
    So it would settle at somewhat higher speed and somewhat higher torque.

    It's counterintuitive until one works some problems.

    If one halves voltage applied to both armature and field , same torque...

    Ifield = Vapplied/Rfield = 50V/100Ω = 1 amp
    so Φ = 1/2 weber
    Since Torque = 7.04 K Φ Iarmature , Iarmature must be 2 amp

    Counter emf must be Vapplied - IarmatureRarmature = 50 - 2X 0.1 = 49.8 volts

    Since counter emf = K Φ RPM , RPM = (counter emf)/KΦ = 49.8/0.5 = 99.6 RPM

    I usually think of torque as a property of the load.
    One could repeat the above exercise inserting for torque some f(RPM) ?

    old jim
    Last edited: Mar 29, 2015
  16. Mar 29, 2015 #15
    Why is torque still 7.04 when the field is reduced to half its original value?

    Isn't it the motor that runs the elevator? Why would the motor torque remain constant? shouldn't the motor torque reduce n not be able to raise/lower the elevator?

    it's the motor that is running the elevator n not the other way around.

    something like: if the elevator needs 7.04 torque, then if the motor can give it, then it will give but if the motor cannot supply it then the elevator will simply not function.

    sorry, i'm having trouble understanding the concept.

    if i reduce the field, does the I(input) increase. Because i'm reducing the load on the system, since back emf acts as load for Vsupply, so if I reduce field, then back emf is reduced, n I(input) is increased.
  17. Mar 29, 2015 #16

    jim hardy

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    elevator is a fixed weight.

    So it takes the same torque to lift it regardless of speed.
    That's why induction motors for elevator service have high starting torque.

    Read the rest of the post - a motor driving a fan would require a different torque.
    When studying interactions one should change only one variable at a time.

    Usually one is selecting a motor to a load and he is given a torque-speed curve for the load.
    It's the load that sets the torque. Motor will come to a speed where it delivers that torque.

    If you wish, repeat the exercise with load torque as some function of RPM.

    The point was to show interaction of power source, counter emf and flux.

    Those two formulas characterize a DC motor in steady state.

    With 100 volts applied at standstill, that motor would have counter-emf of zero
    so Iarmature would be 100V/0.1Ω = 1000 amps
    with KΦ still = 1
    torque would be 7.04 X 1 X 1000 = 7040 ft-lbs
    which would really accelerate that elevator.
    So you'd want to start the motor at reduced voltage.
    Last edited: Mar 30, 2015
  18. Mar 30, 2015 #17


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    I think that is because you are ignoring the 'limitless' supply of current that is available. Whatever the torque per Amp that this motor can supply (less and less as the field is reduced), there is an amount of current that will produce that torque. Having a weak field makes it a very poor motor (inefficient) but doesn't mean it can't go faster.
    There is a clue in the words 'counter-intuitive', that keep turning up in this thread. :smile:
  19. Mar 30, 2015 #18

    jim hardy

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  20. Mar 30, 2015 #19


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    It has to be said that you could build a motor that would slow down as the field was reduced. Such a motor would have a high armature (or supply) series resistance. Fact is that such motors are not made and they are not the sort of 'semi-ideal' motor (system) that we are discussing here. I think it is this sort of motor that our 'nay sayers' are sub-consciously considering in their minds.
  21. Mar 30, 2015 #20
    by load I meant electrical load...back emf would act as electrical load on system

    and 'n' means and. i understand it can be confusing here with 'speed'

    the 'limitless supply of current' part helps. thanks.
  22. Mar 31, 2015 #21

    jim hardy

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    Thanks for clarifying ...

    It might help you to figure what would be speed of that hypothetical motor at zero torque (hint counter emf = applied voltage, armature current is zero)
    and what would be its torque at zero speed( hint counter emf = zero, and we worked an example above )
    and what would be its torque at halfway between zero and no load speed.
    Plot them and you'd have a primitive speed-torque curve for the motor.

    Then change Rarmature to one ohm and repeat.

    Then change Rarmature to ten ohms and repeat.

    That'd give you a feel for what Sophie is saying about motor characteristics dependent on armature resistance.

    Realize that any load has a speed-torque curve also, and a system of motor&load will settle out where the two curves cross.


    your very first post asked this question:
    Lorentz force QV cross B on charges moving in the armature conductors.

    It falls out from Right Hand Rule

    old jim
    Last edited by a moderator: Apr 18, 2017
  23. Mar 31, 2015 #22
    Thanks for the detailed reply.

    I knew that force is caused by interaction of flux caused by armature n field current. I am aware of the left hand rule n lorentz force, but i didn't know where the additional force came from to accelerate the rotor.

    Now i understand it's from the additional armature current. Thanks for that:)
  24. Mar 31, 2015 #23


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    There could be a problem with definition here. An electrical load means the Power consumed whereas a mechanical 'load' is usually a Force or Torque. The two need to be treated as distinct from one another.
  25. Mar 31, 2015 #24

    jim hardy

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    Thank you for sticking with us. It is rewarding to see
    "the light come on".

    Thanks for the feedback .
  26. Oct 19, 2015 #25
    i want to increase the speed of stick ....what should i do? can any1 explain me
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