EMF and Back EMF

Hi.
Can somebody tell me what exactly the differences is between EMF and BACK EMF in DC Motors?
Thank you.

Back EMF is the voltage of the motor itself acting as a generator, the back EMF being reverse polarity, for example if you had a PM DC motor and applied 10 volts, it might turn 1000 RPM, if you disconnect the power and drive the motor with an external means at 1000 RPM in the same direction it would generate close to 10 volts at the opposite polarity.

Think of it like this, say you have a simple DC motor, with an NPN transistor to sink current thru the motor:
EMF is created when you pump current thru the motor. Current flow creates a magnetic field in the motor coil which interacts with the permanent magnets in the motor to create movement.
When you Stop pumping current thru the motor, the magnetic field in the coil collapses, and creates a current (Back EMF) that wants to keep flowing in the motor. A diode across the motor coils dissipates that current thru the motor. Without a diode, the current hits says the NPN transistor which is now off and looks like a high value resistor. Current x resistance = voltage, without the diode that voltage could damage the transistor.

Thanks.
Which one is the factor of Electric Torque in DC Motor.(EMF or Back EMF)?Or maybe I am wrong.
Very thank you.

Somebody has any animation or movie in this case?I couldn't understand very well.Thanks.
I've searched google but nothing very good.

leoncorleone:
Hi.
Can somebody tell me what exactly the differences is between EMF and BACK EMF in DC Motors exactly?
Thank you.

emf is electromotive force..... it is nothing more than the plain words called 'driving voltage'. emf is really just a term for voltage developed by any kind of voltage generating device..... to distinguish it from usual voltage drops across passive components like resistors.

When we're dealing with a motor system .... the 'emf' term when used by itself is nothing more than the power supply voltage. If we want to move the motor with electricity.... we need to apply a voltage.

You know that if you use your hand to manually turn a motor shaft..... this usually produces a voltage across the motor coil (due to magnetic and electric effects learned in physics etc). In any case..... if a motor is turning..... it develops a voltage.

So if we apply a power source (aka voltage source) to the motor system..... the motor will have the usual voltage drops through it's own resistive and inductive components.... but will also have an additional generator voltage.

While supply voltage is trying to force current into the motor.... the motor's generator voltage is trying to force current into the power source.

The overall effect is supply voltage up against motor's generator voltage.

ie.... blue corner of the ring is power supply voltage.... up against red corner generator voltage. blue corner is emf. red corner is back emf.

Whoever is higher in voltage will determine which direction the current (or real power) flows.... toward source... or toward motor.

So the voltage difference is supply voltage minus generator voltage. Aka emf minus backemf.

A spinning motor still generates some amount of its own voltage.... so it generates some amount of 'emf'. But since it goes against the grain of the power supply's voltage.... they got to make up a name for the motor's internally generated voltage.... hence the 'back emf' term.

leoncorleone:
Thanks.
Which one is the factor of Electric Torque in DC Motor.(EMF or Back EMF)?Or maybe I am wrong.
Very thank you.

Neither have anything to do with torque on their own. Torque is produced by force, and force is due to current
and magnetic field only.

The difference between applied EMF and back EMF is what remains to push current through the
winding's resistance (ohm's law), so the current will depend on the difference between applied EMF
and back EMF.

In an unloaded motor the difference is small, the current is small and the torque is small.

If you short the terminals the difference is proportional to the negative of the back EMF, so the
torque is large and reverse (hard braking).

Or to put it another way :

Consider the motor with the rotor shaft locked. When you apply voltage (EMF) the stalled motor draws current dependant on the rotor and brush resistance only.

If the motor shaft is free to rotate as you apply voltage (EMF) you will find the motor current reduces. This is because the rotor, rotating in the magnetic field of the stator, generates a voltage (Back EMF) which opposes the drive voltage. The current drawn is now a function of (EMF - Back EMF)

Theoretically if motor run at infinity speed (an impossibility) the Back EMF could equal the drive voltage (EMF) and the motor current would be zero since there is no differential voltage to produce current flow.

Motor torque is determined by motor current so the higher the Back EMF with respect to EMF then the less torque is produced. At starting conditions (zero speed) motor current is a maximum and Back EMF is a minimum so torque is at a maximum, but speed is at a minimum and since power output is a function of the product of torque and speed then shaft power is a minimum. As speed increases, torque reduces due to decreasing current and at some point (the maximum efficiency point) the product of rising speed and reducing current (torque) reaches a maximum, after which the product then tails off. This produces the classic "hump" shaped curve of motor efficiency.

jackrae:
Or to put it another way :

Consider the motor with the rotor shaft locked. When you apply voltage (EMF) the stalled motor draws current dependant on the rotor and brush resistance only.

If the motor shaft is free to rotate as you apply voltage (EMF) you will find the motor current reduces. This is because the rotor, rotating in the magnetic field of the stator, generates a voltage (Back EMF) which opposes the drive voltage. The current drawn is now a function of (EMF - Back EMF)

Theoretically if motor run at infinity speed (an impossibility) the Back EMF could equal the drive voltage (EMF) and the motor current would be zero since there is no differential voltage to produce current flow.

No, you don't need infinite speed - at no load speed plus a smidgeon there will be zero current (no load
isn't actually no load, there is friction and air-resistance to be overcome). Faster than that and the motor
is a generator and pushing current back the other way. When back-EMF > applied EMF you have a generator.

So now you would have have a perpetual motion machine - I think not

There is NO way a motor powered only by electricity applied to its terminals and freely rotating can draw no current - or even generate a voltage which is greater than the voltage which is driving it.

Only if mechanical power is input to the motor shaft as mechanical energy can a motor generate an EMF greater than that being used to electrically power it. ie a regenerative braking condition

However I'm willing to be proven wrong

The condition of zero current is not based on speed as speed itself does not remove energy from a system.
its the friction/heat/acoustic losses relative to speed. A zero loss system (impossible) is the condition for zero current ( if current was removed it would stay spinning forever with no friction! )

Infinite speed would have infinite friction losses too, infinite energy would still be required to drive the motor (gets a bit hard to calculate).

A motor could achieve infinite speed given a zero friction no load condition. At no point though would it be zero current as it would always be accelerating.

Edit: Although I think this would need an infinite voltage potential too! And there is going to be a limit due to the speed of electrons through the wires.

A motor would achieve infinite speed given a zero friction no load condition. At no point though would it be zero current as it would always be accelerating.

So it would be running at infinity + revs ?? I thought there was nothing greater than infinity - unless you're Buzz Lightyear who used to go "To infinity and beyond"

But we're losing the point, ie "what is EMF, what is Back EMF and which determines torque". Question has I hope been answered to the OP's satisfaction

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Hehe.. you are right though, it's impossible.. it would take an infinite time to accelerate to infinity too so it would never actually get there. Makes my head hurt just thinking about it.

All that's needed for back-EMF to be greater than applied EMF is simply for external mechanical input energy to be greater than whatever losses there are at that speed.

If speed is zero, any mechanical energy added would create back emf as long as it moves (overcomes friction ). Otherwise its as MarkT said "no load speed plus a smidgeon".

alka:
If speed is zero, any mechanical energy added would create back emf as long as it moves (overcomes friction ). Otherwise its as MarkT said "no load speed plus a smidgeon".

'fraid neither of those conditions above make sense

By definition :-

If speed is "zero" then it cannot move to create back emf

If a motor is being powered then it runs at load speed and no faster

I am talking about just about the conditions to create zero-current. When back EMF becomes the same as applied. This does need something physically turning the motor faster than its normal speed at that power.

I guess its not "back-EMF" if its just being turned by something else with no power applied.. its just a generator.

One thing worth mentioning here is the motor constant (Kv) - this can be given in units of
rpm/volt or inversely as volts / (radian/second) [ Vs/rad ], the constant of proportionality of
back-EMF w.r.t. angular velocity. It is often quoted for motors (usually as rpm/volt)

If expressed as Vs/rad, the constant is the same as the torque constant for the motor,
if expressed as Nm/A. (This identity is a simple consequence of conservation of energy).

These constants are really just encoding the length of the winding, radius of the rotor and
strength of the magnets in the motor, if you know the fundamental physics of a current
in a magnetic field you'll recognize:

F = BIl, (force on current I of length l in field B)
V = Blv (induced voltage on conductor of length l moving at velocity v)

which can be combined with
T = Fr (torque in terms of force and radius)
w = v/r (angular velocity given the tangential speed and radius)

to give V = Blr w
and T = Blr A

So V/w = T/A = Blr

l is the combined length of all the winding turns, ie l = 2Na where a is the length of the rotor
and N is the turn count.

So Kv = 2BNra, ie for larger motors you need fewer turns (lower N) for the same motor constant,
and for really large motors that actually forces you to use higher voltages since N cannot be less than
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