Back EMF ```Name: Michael Status: Other Grade: Other Location: Outside U.S. Country: Australia Date: Spring 2010 ``` Question: My question concerns Back EMF in DC brushed motors. I was told that the max velocity of a motor occurs when the Back EMF is equal to the voltage across the motor. I am confused as to what this voltage is? Is it simply the input voltage? . . . or is it the voltage in the armatures which is calculated by multiplying the armature current by the resistance of the wire coils? I also have two equations I am hoping will find the input voltage I need to power the motor: Voltage input = armature current x terminal resistance + Back EMF (what exactly is the terminal resistance, my understanding is that it is the resistance due to the coils of wire on the armatures?) And Back EMF = Flux density of passing field x Armature current x length (also what is the length described here? Is it one length of the armature of the side that is perpendicular to the field? . . . or is it the total length perpendicular, which would be one length multiplied by how many coils? and maybe multiplied by 2 as well as there is two of those lengths perpendicular per coil?) If someone could clarify what those values are that I have asked about, thank you! And if someone can explain the steps I need to do in order to find the input voltage required? (I know my torque and angular velocity that I need my motor to achieve.) Replies: Hi Michael, I think you are greatly over thinking this! First of all, Back EMF can never equal the voltage impressed across the motor, because if it did, no current could flow into the motor. However, a typical motor running without load will achieve a speed where Back EMF will come close to equaling the impressed voltage. There must be slightly more impressed voltage than Back EMF, to allow enough current to flow, to overcome frictional losses. Once a load is added, all this goes out the window, because the motor slows down, reducing the Back EMF, and allowing greater current to flow, and greater power developed to drive the load. In short, the current a motor consumes, is proportional to the applied voltage minus the induced Back EMF. Back EMF is simply a voltage induced in a motor's armature, as a result of it spinning in its own magnetic field ...the same way a generator works. This voltage is induced in the opposite polarity of the operating voltage applied to the motor, and thus tends to impede the current flow caused by the applied voltage. This is why the "stall current" (the motor current that occurs when the armature is prevented from turning) is far higher than running current. The faster the armature turns, the greater the Back EMF, and the more the Back EMF, the more the motor's operating current is opposed and reduced. The above action explains why a typical motor has its maximum torque at zero RPM (where no BACK EMF can exist, and thus maximum operating current will be drawn), and that torque linearly falls as speed increases (because Back EMF linearly increases as RPM increases, thus reducing operating current). If a motor had absolutely no frictional losses at all, then your statement that "the max velocity of the motor occurs when the Back EMF is equal to the voltage across the motor", would be true. In the real world, however, motors have friction to overcome, and loads to drive. So in reality, a motor will increase its speed, increasing its Back EMF and reducing its current drawn as its speed goes up, until a balance is reached and the current it uses generates just enough power to overcome friction and load. At that point, the motor no longer accelerates and its speed remains constant. Regards, Bob Wilson Click here to return to the Engineering Archives

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