What Are the Effects of Counter-EMF on Electric Motors?

[Momento]

Hallo everyone,

I know as a motor speeds up! It acts like a generator and due to that fact. It produces counter-EMF. However, I'd like to know a few things about this effect!

Would counter-EMF cause the motor to draw more current?
Would it effect the motor negatively?

Finally,(I doubt this is possibile) If counter-EMF was greater! Then input what would happen?(THEORETICALLY).

Thanks all,

 

[mdjensen22]

why would you suspect a higher current draw from a counter EMF?
Counter EMF is what causes the motor current to drop from the inrush levels once the motor starts spinning.

Counter EMF is inherent to a motor, so I think you'd be hard-pressed to argue that it has some adverse effect on the motor.

What do you think would happen if counter EMF was higher?

http://en.wikipedia.org/wiki/Counter-electromotive_force

 

[uart]

Finally,(I doubt this is possible) If counter-EMF was greater! Then input what would happen?(THEORETICALLY).

If by "input" you meant the return of energy to the source (also known as generation or regeneration) then yes it is very possible, and something that is very commonly done. In this case however the mechanical "load" must provide the driving torque while it's regenerating.

 

[Momento]

I knew what COUNTER-EMF is... But I knew it wouldn't be higher then input.

But Theoretically what would happen? Just imagine. Will it turn off the motor or what do you all think?

 

[Momento]

My theory is since the counter EMF (THEORETICALLY) is greater then input, the motor would not operates anymore because that force is countering the input?

 

[NascentOxygen]

My theory is since the counter EMF (THEORETICALLY) is greater then input, the motor would not operates anymore because that force is countering the input?

If you power a motor from a battery, and conditions cause the motor to produce a higher counter emf than the battery voltage, then nett energy is being delivered to the battery.

On steep ascents, e.g., slopes of foothills or escarpments, it is sometimes necessary to timetable electric locomotives so that at the time one heavily-laden train is ready to climb the steep incline, another on a parallel track is descending and thus generating extra voltage for the overhead electricity cable.

 

[Momento]

If you power a motor from a battery, and conditions cause the motor to produce a higher counter emf than the battery voltage, then nett energy is being delivered to the battery.

On steep ascents, e.g., slopes of foothills or escarpments, it is sometimes necessary to timetable electric locomotives so that at the time one heavily-laden train is ready to climb the steep incline, another on a parallel track is descending and thus generating extra voltage for the overhead electricity cable.

I noticed from this thread and reading about this matter. That its actually a good thing and not something to worry about! The greater the C-EMF the less input current to be drawn!

Overall I thought C-EMF was a bad effect... However, it really it isn't.

(Please correct me If I'm wrong... At the end we all are here to learn.)

Thanks everyone!

 

[Momento]

Let me use an example to illustrate my understanding: I have an electric motor and a heavy wheel acting as the load.

Now when the electric motor is turned on without the wheel attached as a load, it will instantly run easily and goes fast because there is no mechanical resistance of the load. So the motor would eventually speed up and not need more current because of C-EMF. We can agree that the hardest part was only the beginning for the motor to start.

Now when I attach a physical load load on the motors shaft/rotor. In our case that's the wheel... The motor starts to draw more and more current for torque! So there is a force for it to start and rotate, and the RPM significantly drops and the C-EMF would be less generated.However, if the "LOAD" was not a major "resistance" to the motor... In a sense the motor can "handle" the wheel as a "load". It will need more current at the start up with the load because it needs a stronger rotational force to "move" the object from rest. Eventually it will speed up and speed up and becoming less of a load, then the C-EMF would generally reduce its motor input current and then the only thing needed between both RPM & Torque is really the higher RPM and low torque since the wheel is already in rotational motion. Or should we say: Rotational inertia.
Is this a good example clarifying my understanding? Hope it is.

Thank you!
Mo,

 

[Momento]

If you power a motor from a battery, and conditions cause the motor to produce a higher counter emf than the battery voltage, then nett energy is being delivered to the battery.

That means the motor would either stop or counter rotate. Because I'm trying to imagine what you're saying its difficult o see this happen.
The rotor itself would become a generator. So in a sense when CEMF is higher it can provide power back to the battery.However, the motor would stop its motion I think or counter rotate because of this change... Any opinions?

 

[NascentOxygen]

That means the motor would either stop or counter rotate. Because I'm trying to imagine what you're saying its difficult o see this happen.
The rotor itself would become a generator. So in a sense when CEMF is higher it can provide power back to the battery.However, the motor would stop its motion I think or counter rotate because of this change... Any opinions?

If an electric locomotive is going down a steep slope, gravity speeds it up faster than the motor would drive it at that speed setting. So the back emf is greater than the battery voltage — a condition where current is now forced into the battery terminal instead of coming out of it. There is no reason for the motor to stop or reverse direction. It's a case of the vehicle wheel's turning the motor, and in these circumstances it's called a generator.

 

[NascentOxygen]

Let me use an example to illustrate my understanding:


I have an electric motor and a heavy wheel acting as the load.

Now when the electric motor is turned on without the wheel attached as a load, it will instantly run easily and goes fast because there is no mechanical resistance of the load. So the motor would eventually speed up and not need [strike]more[/color][/strike] much[/color] current because of C-EMF. We can agree that the hardest part was only the beginning for the motor to start.

Now when I attach a physical load load on the motors shaft/rotor. In our case that's the wheel... The motor starts to draw more and more current for torque! So there is a force for it to start and rotate, and the RPM significantly drops and the C-EMF would be less generated.However, if the "LOAD" was not a major "resistance" to the motor... In a sense the motor can "handle" the wheel as a "load". It will need more current at the start up with the load because it needs a stronger rotational force to "move" the object from rest. Eventually it will speed up and speed up and becoming less of a load, then the C-EMF would generally reduce its motor input current and then

Right so far. (see minor correction)

the only thing needed between both RPM & Torque is really the higher RPM and low torque since the wheel is already in rotational motion. Or should we say: Rotational inertia.

I don't understand that.

 

[Momento]

If an electric locomotive is going down a steep slope, gravity speeds it up faster than the motor would drive it at that speed setting. So the back emf is greater than the battery voltage — a condition where current is now forced into the battery terminal instead of coming out of it. There is no reason for the motor to stop or reverse direction. It's a case of the vehicle wheel's turning the motor, and in these circumstances it's called a generator.

Ok, is there a way to control/decrease/re-direct C-EMF? Under the same speeds its in. I was thinking if the C-EMF was 20 V and my input was 10, I could possibly used a step up transformer to make it 21 V. So the C-EMF would not be the HIGHER potential. The motor starts to rotate very very fast reaches it potential! It would not draw any current a all. So we could apply a higher EMF then the produced C-EMF and still the motor would not consume any current.

 

[Momento]

I don't understand that.

Sorry didn't describe it properly. I ment that the load, would only require torque only in the beginning of motion. When it moves faster and faster and faster. It somewhat become's easier to rotate. Not much torque is then needed.

 

[NascentOxygen]

Sorry didn't describe it properly. I ment that the load, would only require torque only in the beginning of motion. When it moves faster and faster and faster. It somewhat become's easier to rotate. Not much torque is then needed.

That's right, with no load the motor has to develop only sufficient torque to spin its own rotor, overcoming bearing friction and air resistance around the rotor windings. You maybe could encase the motor in a vacuum if you wanted to minimize air resistance losses.

 

[NascentOxygen]

Ok, is there a way to control/decrease/re-direct C-EMF? Under the same speeds its in. I was thinking if the C-EMF was 20 V and my input was 10, I could possibly used a step up transformer to make it 21 V. So the C-EMF would not be the HIGHER potential. The motor starts to rotate very very fast reaches it potential! It would not draw any current a all. So we could apply a higher EMF then the produced C-EMF and still the motor would not consume any current.

Back emf is usually a desirable thing! By putting power back into the battery, it allows the driver of an electric car to recoup some energy when coasting downhill. Charging the battery exerts a braking effect, so this slows the car without needing to wear out brake linings. In electric trains, they something switch a big resistor across the motor-generator to waste that generated power when slowing or going downhill, by doing this it still affords the braking effect. If you don't want braking, then just disconnect the motor and it has nothing to absorb the current it is trying to generate. It freewheels and the locomotive rockets faster and faster downhill without braking.

 

[I_am_learning]

In phyiscs you can't say this thing is desirable, and this thing isn't desirable. Physics is all related together. If you accept the Faraday's Law of elctromagnetic induction or you think it is a good thing, then back-emf is just its consequence. If there was no back emf, (well it can't be), then nothing would work. Because, not having back emf means, physics theories not working, which means, we know nothing.

On more technical terms, The Work Output of a DC motor = Back emf * Current
If there was no back emf (which occurs when the motor is stalled), all the input power is lost in the heating.

 

[Momento]

Back emf is usually a desirable thing! By putting power back into the battery, it allows the driver of an electric car to recoup some energy when coasting downhill. Charging the battery exerts a braking effect, so this slows the car without needing to wear out brake linings. In electric trains, they something switch a big resistor across the motor-generator to waste that generated power when slowing or going downhill, by doing this it still affords the braking effect. If you don't want braking, then just disconnect the motor and it has nothing to absorb the current it is trying to generate. It freewheels and the locomotive rockets faster and faster downhill without braking.

Im studying about all the effects of C-EMF. I know its desirable. However, in some causes you must know how to control it.

I thought of an idea few hours ago... How about using a "DIODE"? To block/redirect the C-EMF? Is that possibile?

I understand its a good thing but not in all causes. So using a "DIODE" could possibile force the C-EMF to go with the flow of the original EMF?

 

[Momento]

Because, not having back emf means, physics theories not working, which means, we know nothing.

Not necessarily.
Maybe in some severe-circumstances something could differ.

 

[jim hardy]

What's emf? It's a force on a charge, be it magnetic chemical or electrostatic in origin..
What's current? Charge in motion.

back emf is kinda unavoidable. it comes from the same thing that makes the motor turn, namely the force on a charge that's moving in a magnetic field.

Charge traveling forward inside the conductor feels a force perpendicular to both the magnetic field and the direction of its motion. That mutual perpendicularity is called a "Vector Cross Product", though the name isn't really important. It pushes the charge sideways which makes the motor try to turn.

When the motor starts to turn, the wire itself now has motion relative to the magnetic field, and THAT motion creates another "vector cross product" which pushes the charge backward against its forward motion along the conductor. That opposes the applied voltage, trying to reverse current flow. So it's called 'back emf' or counter emf.

And it's really quite a nice thing , it makes motors and transformers practical.

 

[Momento]

What's emf? It's a force on a charge, be it magnetic chemical or electrostatic in origin..
What's current? Charge in motion.

back emf is kinda unavoidable. it comes from the same thing that makes the motor turn, namely the force on a charge that's moving in a magnetic field.

Charge traveling forward inside the conductor feels a force perpendicular to both the magnetic field and the direction of its motion. That mutual perpendicularity is called a "Vector Cross Product", though the name isn't really important. It pushes the charge sideways which makes the motor try to turn.

When the motor starts to turn, the wire itself now has motion relative to the magnetic field, and THAT motion creates another "vector cross product" which pushes the charge backward against its forward motion along the conductor. That opposes the applied voltage, trying to reverse current flow. So it's called 'back emf' or counter emf.

And it's really quite a nice thing , it makes motors and transformers practical.

It truly is a wonderful effect. For some reason I though it was... A burden. However, a very useful effect.

Now my previous question is: Is it possibile to use a "Diode" to force only the input EMF? And re-direct C-EMF to another position?

 

[NascentOxygen]

I thought of an idea few hours ago... How about using a "DIODE"? To block/redirect the C-EMF? Is that possibile?

I understand its a good thing but not in all causes. So using a "DIODE" could possibile force the C-EMF to go with the flow of the original EMF?

Yes, a diode would block reverse current. The back emf will still be generated, since emf is a property of the motor and its rotational speed, but the diode would block it externally.

 

[shahed7744]

Just remember two things, when driving a motor the input power is V_in*I and output power is V_backemf*I=Tw(w is the velocity and T torque), in any system you can not generate output power greater than the input power,so whenever the C-EMF is greater than V_in, its the mechanical load that is giving power through its rotational energy,so it will eventually slowdown until the C-EMF is less than the V_in unless there is another source of energy which will maintain the velocity of the load, in that case it is simply operating as a generator.

 

[Momento]

Yes, a diode would block reverse current. The back emf will still be generated, since emf is a property of the motor and its rotational speed, but the diode would block it externally.

Yes! I do not want to block it, but rather control it or reverse its direction

Just remember two things, when driving a motor the input power is V_in*I and output power is V_backemf*I=Tw(w is the velocity and T torque), in any system you can not generate output power greater than the input power,so whenever the C-EMF is greater than V_in, its the mechanical load that is giving power through its rotational energy,so it will eventually slowdown until the C-EMF is less than the V_in unless there is another source of energy which will maintain the velocity of the load, in that case it is simply operating as a generator.

Thanks for that useful information!

 

[jim hardy]

Now my previous question is: Is it possibile to use a "Diode" to force only the input EMF? And re-direct C-EMF to another position?

i don't see how it could be possible inside of a wire. The counter emf acts on the exact same same charge that is making the motor turn.

look up "right hand rule" , QV cross B
http://physicsed.buffalostate.edu/SeatExpts/resource/rhr/rhr.htm

 

[Momento]

i don't see how it could be possible inside of a wire. The counter emf acts on the exact same same charge that is making the motor turn.

look up "right hand rule" , QV cross B
http://physicsed.buffalostate.edu/SeatExpts/resource/rhr/rhr.htm

What? Counter-EMF is the cause of the motor's motion? I don't think it is... Maybe its the cause that a motor would not feed more input current. That I agree, however, being the source of motion that I disagree.

The only thing I wanted to do is re-direct/control the C-EMF. That is possibile by simply adding a diode to that wire. It somewhat... Controls the C-EMF.
Now how can I apply the RHR on the C-EMF? I tried and found it the force acted on it is the same direction as input EMF is that possibile?Cheers,

 

[Momento]

I just realized its pointless to block C-EMF... Even if there is a diode, the induced EMF will be canceled by the diode and the C-EMF, if the C-EMF is greater then the EMF, it will cancel it and the motor would just act as a generator.

 

[jim hardy]

What? Counter-EMF is the cause of the motor's motion? I don't think it is...

you are correct. Counter EMF and motion arise from the same cause, but one does not cause the other.

QV cross B is the cause of both torque(which in turn causes motion) and counter EMF.

It's "Thought Experiment time":

Place a paper in front of you.
Draw a straight horizontal line representing current in wire, let's just say it moves left. That's the direction of index finger above.
Next stand a pencil pointing up for magnetic flux , that's direction of middle finger above.
Next observe force on the moving charge in the wire pushes the wire away from you, in direction of thumb. That's the torque that tries to rotate the motor.

Now - assume the wire has started to move away from you as the motor begins rotation.
There's a second force exerted on the charge in the wire due to that motion.
What direction is the force due to rotation of motor??

When i do the diagram, my index finger pints in direction of wire's motion , away from me
my middle finger points up in same direction as pencil
and my thumb points to right, opposite direction of current flow.

AND THAT is why counter emf opposes current flow and is proportional to speed
and is the reason your next statement is exactly correct

Maybe its the cause that a motor would not feed more input current.

Observe that effect is in addition to the torque effect. They're different effects not cause and effect.

On next statement i think we agree but I'm not exactly certain

That I agree, however, being the source of motion that I disagree.

QV cross B is cause of both motion and counter EMF.

i hope you will practice the above exercise until it's second nature. It was for me the key to understanding electric machinery. I have never forgot the day it burst into my understanding - i was sitting in freshman Physics class trying to figure out what my freshman EE professor had gone over the previous day in machinery class.

I suffer a mild autism and have to make my mental model at the microscopic level agree with the equations or i will not accept the equations. I experimented with a bar magnet, battery and small wire until i could see the mutual perpendiculariy of the vector cross product. The right hand rule is a memory aid for me and apparently other people too. Make sure you become fluent in vector manipulations, they are fundamental in EE.

I have enjoyed seeing your understanding of this deepen, so thank you EDIT in fact thank you ALL for letting [STRIKE]us[/STRIKE] me play in your sandbox. I learn from everyone. EDIT (and i shouldn't presume to speak for everyone jh)

Best wishes for your studies and a successful career.

old jim

 

[jim hardy]

if the C-EMF is greater then the EMF, it will cancel it and the motor would just act as a generator.

exactly. An oversped motor is a generator. Counter emf succeeds in reversing current (hence power) flow.

 

[Momento]

exactly. An oversped motor is a generator. Counter emf succeeds in reversing current (hence power) flow.

Yea just relaized after long hours thinking about it... That once a motor goes over its potential speed from its source, C-EMF would be greater then INPUT and will stop IN.EMF, so eventually in this point the motor becomes a good old generator.

 

[Momento]

Question though, C-EMF is only generated in the part of a conductor that is exposed to a magnetic field that is constantly changing right? The more conductor exposed the more C-EMF generated, the lesser the exposition the less C-EMF generated right?

 

[jim hardy]

right

one of the formulas you can derive from that basic understanding is

e = Blv

e = voltage
l = length of conductor
v = velocity of conductor relative to field
B = magnetic flux density

"""after long hours thinking about it...""" bravo !
cross check your microscopic understanding against the formulas in your book and make them meld. Once you have the basic physics imprinted you can derive the formulas with ease, which beats cramming for exams.

old jim

 

[Momento]

right

one of the formulas you can derive from that basic understanding is

e = Blv

e = voltage
l = length of conductor
v = velocity of conductor relative to field
B = magnetic flux density

"""after long hours thinking about it...""" bravo !
cross check your microscopic understanding against the formulas in your book and make them meld. Once you have the basic physics imprinted you can derive the formulas with ease, which beats cramming for exams.

old jim

Makes sense!

Because a conductor with many turn's will generate a significantly powerful C-EMF, if there is a small portion exposed to the magnetic field it would generate a weak C-EMF compared to INPUT-EMF,

Thanks Jim! I appreciate all you're efforts everyone thanks!

 

[CGOLDING]

When C-EMF is produced - is it true that some of the energy is dissipated as heat - if so is there an equation which can be used to find out how much is lost in heat.

 

[NascentOxygen]

When C-EMF is produced - is it true that some of the energy is dissipated as heat - if so is there an equation which can be used to find out how much is lost in heat.

The energy losses as heat arise from friction and I².R losses in conductors (that includes eddy currents in metal). Primarily, it's the I².R copper losses that are load-dependent; if there is no armature current, these losses are a minimum. The counter emf determines the current, this in turn determines the Ohmic losses.