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I know that back-EMF can be considered as a voltage source in series with the motor which is proportional to speed. This is the common understanding, and I totally get it. Before I understood this, I developed an alternate explanation on my own, and I wonder if it has any validity.

Think of this: an inductor resists change in current. A bigger inductor resists it more. A stalled motor resists change in current. A spinning motor resists it more.

A small inductor at a given current has some stored energy. A bigger inductor at the same current has more stored energy. A stalled motor at a given current has some stored energy. A spinning motor at the same current has more stored energy.

Hopefully you can see what a student might intuitively hypothesize: a motor's windings exhibit an inductance that increases with the motor's speed. Not because it's magically growing more turns of wire of course, but perhaps it's a sort of mechanical inductor, storing energy in the motor's momentum, rather than in a magnetic field. My intuitive understanding of an inductor is, after all, a flywheel. Maybe this is an inductor that actually is a flywheel.

Can this analogy be stretched further? In a resistive and inductive load, AC current lags behind AC voltage. Add more inductance, and current lags more. In a motor, current lags behind voltage. If the motor is spinning faster, does it lag more?

And if that much is true, can it be shown that back-EMF is equivalent to an inductance that increases with motor speed?

And if not, why? Intuitive examples would be appreciated first, then the math. I never seem to understand when presented in the opposite order.

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Interesting. The back-emf (modeled as a voltage source proportional to speed) is not equivalent to an inductance that depends on speed. Furthermore, there is no possible L(w) you can come up with that will make that assertion true.

I will describe a simple experiment, but in essence I'll be saying that they can't be equivalent because upon a motor load change, an inductor dependent on speed L(w) will not affect the stationary state current (torque after all transients have died down, becoming a contradiction), while a voltage source dependent on speed v(w) will (which makes sense).

Assuming a DC motor, a simple proof is to imagine that the load on the motor gets reduced. Because there is less load, the motor speeds up. Also imagine we wait for some time so that all transients go away (t=inf.). Now let's see what happens with both models:

With the back-emf modeled as a voltage source, its voltage increases because speed increased. This means that the current decreases, because the difference between the power voltage source and the back-emf voltage got smaller. This means torque decreased, which makes sense because we reduced the load on the motor.

On the other hand, no matter what inductance value you give to the "back-emf inductor", the current on the motor would remain the same, because inductors are short-circuits in dc. But this does not make sense, because torque is proportional to current and if the current remains the same, torque remains the same, but we started this analysis saying that we reduced the load on the motor.

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  • \$\begingroup\$ This made me think of maybe a simpler disproof of my alternate model: there's no way to limit the current of a motor with inductance alone. Even an ideal motor with zero resistance will run at a finite speed, and also zero current if there is no torque, but with only inductance in the model, the current would always increase. \$\endgroup\$
    – Phil Frost
    Jan 12, 2013 at 5:39
  • \$\begingroup\$ Exactly, I thought of the same simplification after I wrote it. \$\endgroup\$ Jan 12, 2013 at 5:48
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An ideal motor may be modeled as a "transmission" between the electrical and mechanical sides, with a "gear ratio" of "k volt seconds per revolution" for some constant k. Just as a mechanical transmission bidirectionally commutes changes in one side's torque or rotational speed to changes in the other side's torque and rotational speed, so too with the motor. A normal transmission scales by a dimensionless quantity, but that doesn't pose a problem. I can't figure out how to make Google's dimensional analysis to work with torque, but one assumes that a motor drives something some particular distance from its shaft, one can then change the formula to use meters instead of revolutions.

If one assumes k equals pi, then applying one amp to the motor will yield (1 amp * (1 volt second per meter)), which is to say one newton of force. Applying one volt to the motor will cause motor's output to move at a rate of (1 amp / (1 volt second per meter)), which is to say one meter per second. Moving the output at a rate of one revolution per second will cause the voltage to be one volt; applying one newton of force will make the motor draw one amp. Just as with an ideal mechanical transmission, the motor establishes an instantaneous correspondence between what's happening on both sides.

Of course, real motors don't behave quite like ideal motors, but most real motors may be modeled as an ideal motor with a series inductor and resistor on the electrical side, and with an attached mass and some friction on the mechanical side. Commutation issues may cause behaviors to vary somewhat from that simplified model, but in many cases it works well enough to be useful. Because of commutation issues, a motor's inductance may vary slightly depending upon its exact mechanical position. Nonetheless, a motor's inductance is relatively independent of speed--the faster a motor is turning, the faster the inductance will vary between the values it has at different positions, but for the most part it will behave like a relatively-constant inductance.

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  • \$\begingroup\$ I don't know if this directly address my question, but it is interesting information anyway. I've never thought of motors quite like this. Perhaps it's interesting to note that a major factor in motors deviating from this ideal model is the winding resistance; if it were zero, any attempt to slow the motor by increasing the mechanical load would result in more (perhaps infinite) current to be drawn until the back-emf was equal to the supply voltage. Also, decreasing the supply voltage would allow the back-emf to drive an infinite current to instantly stop the motor. \$\endgroup\$
    – Phil Frost
    Jan 29, 2013 at 2:58
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    \$\begingroup\$ @PhilFrost: It's worthwhile to note that if a motor which is running is held shorted, it will stop quickly; the motor's resistance is the main factor that prevents the stop from being instantaneous. More interestingly, if the motor is rapidly switched between being shorted and being connected to the supply, it will rapidly slow down to a fraction of its original speed, and any excess speed will cause current to be driven back into the supply. \$\endgroup\$
    – supercat
    Jan 29, 2013 at 4:24
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No, they're not at all equivalent. Back EMF is, as you say, a voltage source. The voltage depends on the speed of the motor and nothing else. Any current that flows as a result of that voltage depends only the external impedance connected to the motor.

On the other hand, the energy stored in an inductor is essentially a current source, and it will (attempt to) produce whatever voltage is required to get that current to flow in the external circuit, which is what gives rise to the "inductive kick" effect. Of course, the magnitude of the current in question is modified over time by the terminal voltage of the inductor.

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    \$\begingroup\$ Simple proof of this, (works with a motor that doesn't need power to generate the stator field, e.g. permanent magnet DC motor, BLDC motor, stepper motor) ... rotate the motor WITHOUT applying a voltage. Now it's not back EMF, it's just EMF! \$\endgroup\$
    – user16324
    Jan 12, 2013 at 10:16
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OK. Back to "Back EMF." As for the original question: "Is it valid to consider back EMF in a motor equivalent to increased inductance?" The answer is NO. An inductor gives you back the energy you apply against the Back EMF-to build the magnetic field-as electrical energy. A motor CONVERTS the energy you apply against the Back EMF-into mechanical energy.

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I don't think we can directly consider the back emf equivalent to an inductance because equivalence means a condition is necessary and sufficient for the result and the result is also necessary and sufficient for the condition which in this case is not true as the back emf is a voltage which has Volts as units and is not an intrinsic characteristic of a circuit but it needs to be created by some means whereas an inductance is an intrinsic characteristic of a circuit that can be calculated from the characteristics of the circuit , however if we go back to the basics we can clearly see that the back emf expression holds some inductance in its formula and we can find that in the term K of the back emf =KW The origin of this is the theory saying a change in flux through a surface attached to a loop creates an electric potential difference in the loop opposing the flux change now if we apply that using the definitions (the flux through a surface, the variation ) we find = back emf= - d(BSurfacecostheta)/dt When there is no rotational movement and B is parallel to the surface vector or perpendicular to the surface so the flux is maximum cos theta disappears and this is from where we have the flux =LI and then the voltage equal to -LdI/dt, where L is coming from the conversion of the magnetic field into it`s value using the current from the formula giving the magnetic field versus the current which normally says integral of B over a closed loop is mUZEROI so L generally depends on the geometry of the loop which defines the form of the loop Now if the system is rotating the change will be on BCOS theta rather then just B given that the surface is constant and now you can see from where omega W is coming in the expression of the back emf and we can still express the flux as proportional to something through a constant this time proportional to omega something like L1W but this time L1 is not exactly the same as L in the expression of -LdI/dt but it will necessary include the equivalent max current generating the max magnetic field opposing the original magnetic field so where L is henry L1 is in henryamps and you can see that the formula -LdI/dt is in henryamps/sec and same thing L1W is henryamps/sec because L1 is now in henry*amps so in conclusion if you want to understand any questions relative to back emf and electromagnetism interactions and i would say even generally any question going back to the basics and the root definitions helps a lot rather than jus trying to use established knowledge to find solutions , in this context i find the lessons of professor Lewin on youtube https://www.youtube.com/@lecturesbywalterlewin.they9259 invaluable , i promise you will never get bored , you will always look for more and will help you understanding a lot of things and certainly ask more questions

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