Understanding Back EMF in DC Motors

[Harrison G]

Homework Statement

Hello, i decided to sign up here to ask one question: A dc motor with load has verry small resistance. We can almost say the resistance is only that of the copper. Without load the motor is accelerating as much as it can per given voltage and it generates strong back emf that resists the flow of current, hence the rrsistance is big. So can we say for the sake of calculations that: I=V-Vemf/Rwire. Then we just divide that current on the battery Voltage and find the overal resistance: Re=V/I. In other words we express that back emf as a resistance

Homework Equations

The Attempt at a Solution

 

[CWatters]

So can we say for the sake of calculations that: I=V-Vemf/Rwire. Then we just divide that current on the battery Voltage and find the overal resistance: Re=V/I. In other words we express that back emf as a resistance

More typically you would write..

Vsupply = (I*Rwire) + Vbemf

which is the same equation rearranged.

But it's not usual to express back emf as a resistance.

To maximise efficiency you want I²R losses in the windings to be small so Rwire to be as small as possible. I have a brushless motor with just 2 (Two) turns on the windings.

 

[andrevdh]

Strickly speaking this is not a back emf. It is an induced emf. Or am I wrong?

 

[Harrison G]

More typically you would write..

Vsupply = (I*Rwire) + Vbemf

which is the same equation rearranged.

But it's not usual to express back emf as a resistance.

To maximise efficiency you want I²R losses in the windings to be small so Rwire to be as small as possible. I have a brushless motor with just 2 (Two) turns on the windings.

But if the motor plays the role of emitter resistance its ok, right?

 

[CWatters]

What do you mean by "emitter resistance"?

 

[Harrison G]

Strickly speaking this is not a back emf. It is an induced emf. Or am I wrong?

When When

What do you mean by "emitter resistance"?

I mean when you connect the motor to the emitter pin of an npn bjt. In order to calculate Rb you need to know what the resistance of the motor is.

 

[Harrison G]

When When

I mean when you connect the motor to the emitter pin of an npn bjt. In order to calculate Rb you need to know what the resistance of the motor is.

I know that its best to connect it to collector, but thing are different with H bridges

 

[Harrison G]

Strickly speaking this is not a back emf. It is an induced emf. Or am I wrong?

The voltage across the motor couses current. That current creates magnetic field. That field then interacts with that of the stator and the rotor spins. But as it is spining inside magnetic field electromagnetic induction happens that induces an emf into the coil. By law that emf opposes the voltage that coused it. So that emf creates a current in oposite direction of the battery current and both electron flows began opposing each other.Thats why its called ''back'' or ''counter'' emf- Becouse that electromotive force opposes the original electromotive force

 

[Harrison G]

Mr CWatters, i mean that in the charcteristics of dc motors is given that: Dc motor current with no load by 3V- 150mA Dc motor current when max load is applied by 3V- 2100mA And i assume that u could say that with no load when 3V is given, the resistance then is 3:150mA=20ohms And with maximum load there is almost no spinning so the resistance of the motor will be almost only that of the coil: 3V:2100=1,42 ohms. And now knowing that we can use it to bias a transistor that is used in H bridge, and use the max load resistance( that resistance since i assume that the motor will have hard time driving something) to set a base resistor. In that case the motors resistance is like an emitter resistance

 

[CWatters]

I'm struggling to understand your notation. I don't understand what "3V- 2100mA" means because you can't subtract current from voltage.

Suppose the stall current of your motor was 2A and your transistor has a gain of at least 100 then the base current needs to be at least 2/100 = 20mA. Does that help?

 

[Harrison G]

I'm struggling to understand your notation. I don't understand what "3V- 2100mA" means because you can't subtract current from voltage.

Suppose the stall current of your motor was 2A and your transistor has a gain of at least 100 then the base current needs to be at least 2/100 = 20mA. Does that help?

Yeah it does, thank you

 

[andrevdh]

The voltage across the motor couses current. That current creates magnetic field. That field then interacts with that of the stator and the rotor spins. But as it is spining inside magnetic field electromagnetic induction happens that induces an emf into the coil. By law that emf opposes the voltage that coused it. So that emf creates a current in oposite direction of the battery current and both electron flows began opposing each other.Thats why its called ''back'' or ''counter'' emf- Becouse that electromotive force opposes the original electromotive force

I am happy with what you are saying up to the point where the text is underlined. What you are referring to is Lenz's law. It says that the induced emf will be such that the magnetic field of the coil will try and keep the changing magnetic flux (that it is from the stator in this case due to the spinning motion) contant. Thus the induced emf is due to the changing magnetic flux it is experiencing from the stator and not due to its own voltage in the coil. Still, I am open to be convinced otherwise.

 

[Grim Arrow]

I am happy with what you are saying up to the point where the text is underlined. What you are referring to is Lenz's law. It says that the induced emf will be such that the magnetic field of the coil will try and keep the changing magnetic flux (that it is from the stator in this case due to the spinning motion) contant. Thus the induced emf is due to the changing magnetic flux it is experiencing from the stator and not due to its own voltage in the coil. Still, I am open to be convinced otherwise.

Yes, he means indeed Lenzs law. Ya see, the reason for induced emf in the verry basic physical way is (atleast as i understand it) is that when magnetic lines cross the coil, they kinda zap or push the electrons from the atoms there in a certain dirrection that depends on the direction that the coil or the stator spins. And so that direction is such, that when the rotor spins, the magnetic lines of the stator induce emf that couse a current which is opposite to the one that coused tje rotor to spin. Or that emf gives a current that tries to stop the spinning

 

[andrevdh]

Yes, the induced emf will be of such polarity that it results in that the motion of the rotor is opposed (we can discuss this if you want to if you were to look at the diagrams on Lenz's Law on the HyperPhysics website: http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html). Problem is it is not the only source so the resultant current is in most instances determined by the power source (unless the induced emf is stronger than that of the power souce - not sure this is possible?). I am more of the thought that the polarity of the induced emf will change as the rotor spins, so it would sometimes add to that of the power source and sometimes it would oppose it. This happens because the pole of the rotor approaches the pole of the stator on the one side, but leaves it on the other side, so the magnetic flux first increases then decreases resulting in the induced emf changing its polarity as the rotor spins around. I would therefore think that it would create a ripple on the DC.

 

[Grim Arrow]

Yes, the induced emf will be of such polarity that it results in that the motion of the rotor is opposed (we can discuss this if you want to if you were to look at the diagrams on Lenz's Law on the HyperPhysics website: http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html). Problem is it is not the only source so the resultant current is in most instances determined by the power source (unless the induced emf is stronger than that of the power souce - not sure this is possible?). I am more of the thought that the polarity of the induced emf will change as the rotor spins, so it would sometimes add to that of the power source and sometimes it would oppose it. This happens because the pole of the rotor approaches the pole of the stator on the one side, but leaves it on the other side, so the magnetic flux first increases then decreases resulting in the induced emf changing its polarity as the rotor spins around. I would therefore think that it would create a ripple on the DC.

That would be true if the magnetic fields of the coils were constant. Then once the emf would support the source V, once they would oppose its current.
Lets say it like this: In generators the magnetic field that the stator creates is constant(it always points in one dirrection). When in our case the rotor is driven into motion by some outer force, it spins in one dirrection. The metal frame attached to the rotor crosses the magnetic field lines of the stator in a certain way, and in it a vurrent is beign induced, which dirrection deppends on the way the frame spins in the stator. And when u connect a load ressistance to the frame conductor, in such a way that one lead of the resistor always is connected to the same lead of the coil and vice versa, it will appear that current changes dirrection. Now let's made a motor from that generator by applying voltage to its leads. But we apply that voltage in such way that as before it also produces a current flow that always flows in one way through the frame. In that case, the induced current once will oppose the source one and once will support it. But in the ordinary dc motor, in order to prevent the rotor from beign stuck between the magnet poles, the coils are connected to the supply voltage through a comutator. When the rotor is spinning, the comutator between two brushes(these are the two ends of the battery). These brushes can't move, but the commutator is free to spin between them, alternating the dirrection of current through the coils in such ways that coincidentaly the indused current always happen to oppose the Voltage source current. You can see more about that commutator thing in youtube.

 

[andrevdh]

Thank you for your patience I am beginning to get the picture or what you are getting at. The motor seems to be designed that as its pole approach the pole of the stator the current or voltage of the source is such that it is attracted towards the pole of the stator and as it leaves this pole the direction of the current or applied voltage is changed around so that it is repelled, that way the spinning motion of the rotor of the motor is always supported due to the interaction between the rotor and stator. This would mean that the induced emf is opposite to that of the applied voltage as you say. In Physics speak back emf is normally reseved for another phenomena - an induced emf in a coil due to a change in its own magnetic field, not as in this case the magnetic field of an external source, but I can understand why you use the term back emf in this context.

Lenz's Law: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c2

 

[Grim Arrow]

Thank you for your patience I am beginning to get the picture or what you are getting at. The motor seems to be designed that as its pole approach the pole of the stator the current or voltage of the source is such that it is attracted towards the pole of the stator and as it leaves this pole the direction of the current or applied voltage is changed around so that it is repelled, that way the spinning motion of the rotor of the motor is always supported due to the interaction between the rotor and stator. This would mean that the induced emf is opposite to that of the applied voltage as you say. In Physics speak back emf is normally reseved for another phenomena - an induced emf in a coil due to a change in its own magnetic field, not as in this case the magnetic field of an external source, but I can understand why you use the term back emf in this context.

Lenz's Law: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html#c2

Thats right, in inductors back emf is coused by Self induction, which is another type of Electromagnetic induction and it appears only when the crossed by its own magnetic lines. However here we are talking about coils moving in a magnetic stator which is the verry basic type of electromagnetic induction.