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Back emf motor

  1. Aug 11, 2008 #1
    If i have a series DC motor and at say X rpm it has a back emf of 11.5v and supply voltage of 12v and i increased the supply voltage (keeping rpm the same) would the back emf stay the same? and if not would the diffrence between back emf and supply increase?

    Thank you.
     
  2. jcsd
  3. Aug 11, 2008 #2

    NoTime

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    To keep the rpm the same you need to put a load on the motor.
    What happens to the back emf as you increase the load?

    Edit: I missed the word series.
    It may actually be possible to design one where the rpm goes down as supply voltage increases.
    Not sure.
    With the interaction of the field coils and armature the answer is probably unique to the particular motor.
     
    Last edited: Aug 11, 2008
  4. Aug 17, 2008 #3
    Would i be right in thinking that back emf is indapendant from supply voltage and is just proportional to RPM?
     
  5. Aug 17, 2008 #4

    NoTime

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    Back emf is always equal to the supply voltage.
    Or attempts to become so, which can cause some pretty spectacular, if unwanted, results.
     
  6. Aug 17, 2008 #5

    Redbelly98

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    Isn't back-emf necessarily less than the supply voltage? Otherwise no current would flow?

    I'm thinking of a simple, DC permanent-magnet motor, like those used for computer cooling fans.
     
  7. Aug 17, 2008 #6

    NoTime

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    When states are changing back emf won't be equal.
    IIRC you need the equal condition to be achieved or current will increase/decrease until it is.
    Disregarding some resistive or inductive losses.
    Like a resistor where the voltage drop equals the terminal voltage.
     
  8. Aug 17, 2008 #7

    Redbelly98

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    It should also depend on how much mechanical load there is. There would be significant air resistance in the case of a fan. For negligible load or other losses, yes back emf should equal applied voltage to get negligible current and torque.
     
  9. Aug 18, 2008 #8
    I know that as rpm increases back emf increase and current drops which reduces the torque. So what I’m trying to work out is this:
    If I had a motor spinning at 3000rpm and had a load of 10NM a supply voltage of 12v and back emf of 5v. If I then doubled the voltage to say 24v and increased the load so the rpm stayed the same then the difference between the back emf and supply voltage should increase? Because the net voltage needs to be higher for enough current to flow to provide torque right?
     
  10. Aug 18, 2008 #9
    But if i double the voltage then the current should double at a set rpm? So if I had a 12v motor that at 3000rpm had a back emf of 5v and a current of say 20A if doubled the voltage the current should rise to 40A? To do this the net voltage needs to be double so at 24v the back emf needs to be 10v and the net would become 14v twice the original net voltage.

    Or would the back emf stay at 5v and the net voltage rise to 19v in which case current would more than double?
     
  11. Aug 18, 2008 #10

    Redbelly98

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    Assuming we're talking about simple DC permanent-magnet motors, then yes the difference would increase to 24-5 or 19V. The current would increase to 19V/R, where R is the motor's resistance.

    That's an increase in current from the original 7V/R, so the current will more than double: 19/7=2.7
     
  12. Oct 17, 2008 #11
    efficent storage of back emf

    I've been recording some very large back emf spikes with my magnetic motor. So far I haven't found an efficient way to store the current. I think something like a mosfet and a capacitor bank might be the solution but I'm a novice. My input power is 9-24volts DC output spikes are in the range of 30-150Volts AC

    If anyone has an idea or a circuit for this application send me a message. I'm aware of the standad Newman replication crcuit but think its missing much of the potential.

    I have some videos of my motor on youtube if you want a look.

    http://www.youtube.com/user/kevininbc

    Cheers,
    Kevin
     
  13. Oct 17, 2008 #12

    uart

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    No you were correct the first time, for a series motor (if you make the simplifying assumption of a linear iron circuit) then you do indeed end up with both the current and the back EMF proportional to the applied voltage if the RPM is kept constant.
     
  14. Oct 17, 2008 #13

    rbj

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    to model this completely you have to consider the different components of the DC motor (the rotor and the stator) separately, then hook them up in series and see what happens.

    the back e.m.f. of the rotor is proportional to both (or the product of) the rotor speed and the current in the stator. i think there is also a back e.m.f. of the stator which is proportional to the rotor speed and the current in the rotor. if they are in series, the currents are identical and the voltages add up. i think that the torque the rotor produces is proportional to the product of the two currents. the resulting speed (given the torque) is a function of the mechanical system that the motor is coupled to; i.e. it's different if the motor is running free which results in very low torque and very low winding currents (but the back e.m.f.s nearly equal the winding voltages) than if it's a blocked rotor that has no back e.m.f. but has lots of current (and torque).
     
  15. Oct 17, 2008 #14

    uart

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    No rjb you dont get a back emf in the stator because it's stationary wrt the field (the rotor turns but it's field remain stationary due to the action of the commutator).

    In a series motor if you make the simplifying assumption of a linear iron circuit then the back EMF is proportional to the product of rotational speed and current, so E=kIw (denoting the rotational speed as w).

    So if R is the total resistance of the brushes and both windings, then in steady state I = (V-E)/R, which after plugging in the above and simplifying gives :

    I = V/(R + kw)

    E = kIw = kwV/(R+kw)

    Thus, as I stated previously, both E and I are proportional to V if w is fixed.
     
    Last edited: Oct 17, 2008
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