Can Back EMF in Inductors and Motors be Eliminated or Reduced?

[nemesiswes]

Well I have been researching a lot about inductors and motors and so on and I just had a question that I can't seem to answer. Is it possible to get rid of or reduce the back emf generated in say a inductor or motor. I know you can decrease the inductance of a inductor in many ways, but is there a way to directly get rid of or decrease the Back emf's effect on the inductor. Like one idea I had was to oppose the back-emf generated on the inductor by just applying another voltage in opposition to the back emf plus the original applied voltage.

So yeah is there and what ways can you if it is possible?

 

[sophiecentaur]

Why did you want to do this? Had you a particular application in mind?
Did you want to increase the current into a motor when it's running at high speed, for instance?

 

[nemesiswes]

I guess that could be an application, but I mainly just wanted to know out of curiosity. plus i always see those over unity sites talking about it too, so I just wanted to know if it is even possible and how you would do it.

 

[nemesiswes]

Sooo does nobody have an answer to this? Is it that it is just impossible and I am dumb for thinking other wise, lol. I am sure it is possible though, even I have an idea of how you might and that is to just apply a opposite voltage to the back emf, basically increase the source voltage. If it works or not, I do not know.


So is there any other ways you might reduce or get rid of the back-emf?

 

[sophiecentaur]

If you take an inductor then it will produce a 'back emf' when you apply an AC voltage across it, limiting the current through it. It has a so-called Reactance (as opposed to the Resistance of a wire / resistor / lump of carbon) and this reactance is frequency dependent. A Capacitor also has a reactance, which is frequency dependent but has the opposite Sign. It is possible to add a capacitor in series with an inductor which will have equal and opposite reactance at a particular, 'resonant' frequency. The effective back emf of the combination is zero so, apart from some residual resistance, ti will be a short circuit at that frequency.
Is that what you wanted?

 

[nemesiswes]

huh, interesting, so does that mean that if you wanted to decrease the rise time of the current on a inductor, regardless of the inductance value, as long as you had the appropriate capacitor in series with that inductor, you could decrease the rise time needed for the current to reach a specific value?

or is there something I am missing (not understanding)?

 

[nemesiswes]

I was messing with multi-sim and it seems that what I said last was true. I figured it was since no reactance or at least very little means little to no opposition to the current and thus it's rise time can be much much quicker. However the Capacitor and Inductor do have high voltages across themselves, I am pretty sure this is because of there individual reactances effect on the current. Faster current rise time would equal higher voltage across the capacitor and inductor.

 

[sophiecentaur]

I was messing with multi-sim and it seems that what I said last was true. I figured it was since no reactance or at least very little means little to no opposition to the current and thus it's rise time can be much much quicker. However the Capacitor and Inductor do have high voltages across themselves, I am pretty sure this is because of there individual reactances effect on the current. Faster current rise time would equal higher voltage across the capacitor and inductor.

Your rise time will shorten and produce 'ringing' (overshoots at the resonant frequency) on the edges of a square wave signal. This can be 'good' or 'bad', depending on your requirement,

 

[sophiecentaur]

However the Capacitor and Inductor do have high voltages across themselves, I am pretty sure this is because of there individual reactances effect on the current.

Yes - the reactances cancel out so loads of current can flow between the ends of the series pair.
If you take the same L and C and connect them in parallel, you will get a voltage maximum across the pair at the resonant frequency when the effect is that the two Susceptances (=1/Reactance) are equal and opposite. Hence the Impedance goes very high and no net current will flow into the circuit (although a lot of current is 'sloshing about' between L and C during each cycle of oscillation).

 

[nemesiswes]

I was wondering if it is possible to design a circuit where the current (voltage ) only flows in one direction but still has the canceling of reactances? I thought one way is to have a pulse source pulse at specific points, so it would act like the capacitor except in only one half the wave. this way you could have a square wave powering the inductor, just with no negative current, only positive and still have the reactance canceling effects.

Is this possible, if it is then how would the circuit work? I am not sure how the to build such a circuit in multisim.

 

[sophiecentaur]

The nearest thing to what you want would be a 'current source'. You want the current to follow your signal volts independently of the load.
A good current source can be achieved over a limited range of volts using a feedback circuit. But the collector of a transistor is quite a good current source. What you find is that the volts you can see on an oscilloscope are nothing like what you'd expect when feeding an inductor but the current is 'right'.

 

[FOIWATER]

Research interpole, or compensating winding.

This winding is added in series with the armature winding, and is placed on the brush axis of your motor, if your brushes shift out of the neutral plane due to armature reaction, there will be a voltage induced into them due to the fact they now reside within the magnetic field, and not the neutral plane... this voltage is obviously in the opposite of the voltage induced by the interpole.. it does in fact, negate counter emf.

As a result, you have less sparking of your brushes when armature reaction occurs... since they short circuit the armature coils through the commutator.. and there is no voltage at that point if applied voltage + counter EMF = 0.

It is much more interesting than I wrote here, and can be explained in much more detail, obviously.

I am sure you must know about how CEMF is induced, so I won't bother humoring you.. but about the neutral plane..

if you know that the purpose of your brushes is to conduct current onto the commutator, and ultimately split the current in your armature coils to produce two constant current directions (and therefore two constant torque directions, due to the switching effect of the rotating commutator) Then it is easy to understand that, There must be no opposition to the applied voltage at that point of commutation (commutation, literally means a reversal, in this application a reversal of current) There must be no opposition, no CEMF induced in the armature coils that the brushes/commutator reverse the current in. The point at which the coils directly under your brushes do not exist within the magnetic field of the motor as it rotates (therefore no CEMF induced because faradays law conditions are not met) is called the motor neutral plane. This neutral plane though, can be shifted as motors draw large currents. The magnetic field of the motor can be influenced by the strong magnetic (rotating) field of the armature. Meaning that, the brushes may now exist within the magnetic field, and may now have CEMF developed within them. That's why brushes start sparking under large loads (meaning slower speeds, which means less cemf opposing source voltage, which means more current drawn, which means a LARGER rotating magnetic field that influences the position of the stator field).
So, with an INTERPOLE or COMPENSATING winding connected in series electrically with the armature, but strategically placed physically so that it will exist within the desired neutral plane, we can induce a voltage into the brushes which is of opposite direction as the CEMF, and lowers it to an acceptable value, as if we were inside the neutral plane. so, in summary, interpoles offset counter emf effects that are caused by neutral plane shift (armature reaction) in DC series and shunt wound motors.

Hope it helped, but maybe way off what you wanted

 

[NascentOxygen]

Sooo does nobody have an answer to this? Is it that it is just impossible and I am dumb for thinking other wise, lol.

Back emf is a property of the inductor. It is as integral to the inductor as sodium is to sea salt. Without the back emf due to changing currents, there can be no inductance.

v=L·di/dt is the equation that relates back emf to inductance, for a given rate-of-change-of-current. If you want to see no back emf, then you must maintain the current constant. That's all find and dandy until the day your neighbourhood loses its electricity due to a thunderstorm, and your inductor current suddenly drops to zero. So you must be prepared for that eventuality, too, and its resultant back emf! There are ways to reduce the back emf, but they all involve simultaneously reducing di/dt. Unfortunately (and fortunately!), the laws of physics are inescapable!