Proof of common sense when maths is lacking concept

*Ratch*

milesyoung,

Give it a second read-through

I did, several times.

in the simplest case there are two causes of changing magnetic flux in the winding. One is a change in current in the circuit

OK, I believe you are referring to the change caused by the alternating line voltage.

the other is a change in the angular position of the rotor. The latter is what limits the current.

Are you sure you don't mean the angular velocity of the rotor?

The latter is what limits the current

A big inductance coil does not have a rotor, and yet it limits current. In any case, anything that causes a conductor to cross a magnetic flux in a perpendicular manner will generate a voltage in the conductor.

Ratch

milesyoung

OK, I believe you are referring to the change caused by the alternating line voltage.

I'm talking about a change in instantaneous current.

Are you sure you don't mean the angular velocity of the rotor?

No (edit: As in, yes I'm sure :). A change in rotor position will change the magnetic flux through the winding and induce a voltage. A change in velocity will chance the rate of change of magnetic flux through the winding.

A big inductance coil does not have a rotor, and yet it limits current. ...

Yes but this isn't what limits the current as the angular velocity of the rotor increases. How could it? This voltage isn't a function of anything that has to do with the rotor. It's only a function of the rate of change of current in the winding, considering the self-inductance of the winding to be constant.

How else would you explain an increasing opposing voltage in a DC motor? For a constant current the inductive voltage drop associated with the self-inductance is clearly not present.

*Ratch*

milesyoung,

I'm talking about a change in instantaneous current.

Caused by?

change in rotor position will change the magnetic flux through the winding and induce a voltage. A change in velocity will chance the rate of change of magnetic flux through the winding.

A change of position is velocity, a change of velocity is acceleration.

Yes but this isn't what limits the current as the angular velocity of the rotor increases. How could it?

Partly by inductive reactance, whether the rotor is turning fast or slow.

This voltage isn't a function of anything that has to do with the rotor.

Are we still talking about the back-voltage?

It's only a function of the rate of change of current in the winding, considering the self-inductance of the winding to be constant.

Stator winding or rotor winding?

How else would you explain an increasing opposing voltage in a DC motor?

Permanent magnet DC motor? OK, then rotor winding. So both the rotation of the rotor and the switching of the commutator can change the current.

How else would you explain an increasing opposing voltage in a DC motor?

Yes, changing current with constant inductance produces voltage proportional to current change. Now, what all can change the current? Rotation and the commutator?

For a constant current the inductive voltage drop associated with the self-inductance is clearly not present

The current may be constant from the source, but the commutator changes its direction in the rotor.

Ratch

milesyoung

Caused by?

Why does it matter?

A change of position is velocity, a change of velocity is acceleration.

I wrote "change in the angular position of the rotor". If I replace "position" with "velocity", which I would argue your response would ask me to consider, the statement is false.

Partly by inductive reactance, whether the rotor is turning fast or slow.

Yes _partly_ if the instantaneous current is changing - which it isn't necessarily in a DC motor. The effect that's limiting current in your "big inductance coil" might also be present, but this is not what is the limiting factor that increases with the angular velocity of the rotor and it's not usually what would be called "back-EMF" in motors.

Are we still talking about the back-voltage?

Back-EMF is a confusing term because its used by different people to describe different things. Usually it means the part of the "back-voltage" that's dependant on the angular velocity of the rotor but some will lump the voltage produced by a change in the instantanous current in with it.

Stator winding or rotor winding?

Stator winding. Why would it have something to do with the rotor winding?

Permanent magnet DC motor? OK, then rotor winding. So both the rotation of the rotor and the switching of the commutator can change the current.

It can certainly not change the current in the stator winding (that would make it AC), which is where the back-EMF appears.

Yes, changing current with constant inductance produces voltage proportional to current change. Now, what all can change the current? Rotation and the commutator?

Again, I have no idea why you would bring the commutator into this - we're naturally talking the stator winding.

Edit:
Ah sorry I completely missed the "Permanent magnet DC motor" part. In that case I'm referring to the part of the rotor windings on the stator side of the brushes (which sees no alternating current).

sophiecentaur

Quote by milesyoung

Again, I have no idea why you would bring the commutator into this - we're naturally talking the stator winding.

It would be better to consider a permanent magnet motor first, surely - there is no stator winding. A brush motor with a permanent field magnet will behave just like a dynamo. By Lenz's law, the EMF generated by such a dynamo will be 'such as to oppose its cause' i.e. it will be against the polarity of the supply volts. The commutator is necessary for this to happen and it is, of course, necessary for the motor to be turning in the first place.
If you want a stator winding then, as this passes DC (series or shunt connection) will that affect or produce any 'back EMF'? I don't think so.
The load on the motor is what limits the speed of the motor and that governs the difference between supply voltage and back EMF, which determines the current that can flow through the resistance of the rotor. The resistance that the supply 'sees' would be supply volts / rotor current.

For a series wound motor, the field current will drop as the speed builds so the speed regulation is not so good.

milesyoung

You're absolutely right, consider the simple DC-motor model which probably needs no introduction:

http://www.library.cmu.edu/ctms/ctms...r/motorsim.htm

Let the motor be loaded so the rotor has a constant angular velocity and let the current in the circuit be constant in magnitude. Would you say the back-EMF in this case has anything to do with the self-inductance L?

sophiecentaur

If you were to increase the supply volts, the current would increase significantly until the revs built up, a reducing the current to slightly more than before. That looks, to me, mighty like a big parallel capacitor, relating to the MI of rotor and load.

milesyoung

I'm not quite sure what you mean. I can't see how you could argue that the induced voltage from flux cutting the rotor windings could be attributed to self-inductance.

toneboy1

Quote by sophiecentaur

Sorry for trying to teach my Grandmother to suck eggs!

Reading this thread reminds me how similar electric motors and radio antennae behave (and even loudspeakers). The up-front resistance is usually just not relevant to their performance - a dipole is an 'open circuit' but it can look like a useful 70Ω at the right frequency. Electrical resistance turns up where you'd least expect it but Energy Conservation requires it.
The Back EMF is only a 'way of looking at it', I think. There must be an alternative way of looking at it involving self inductance of a rotating machine.

heh, it wasn't quite like that. Thanks for the interesting input.

Quote by Ratch

TB1,

Just to follow up on your previous question about that math expression. If you look at equation (6) of the attachment, you can find an alternative form of that expression. If fact, there are many different forms. Once you have the denominator rationalized, you can use partial fraction expansions. Then the number of different expressions is only limited by your imagination.

Ratch

So you're saying I could force it into that form after partial fractions, interesting.
THOUGH it would take ages to even get to (1) if you didn't have a computer to aid in expanding and simplifying the fraction, wouldn't it?
Also, I dont quite see where H came from when you introduced it.

Thanks

*Ratch*

TB1,

So you're saying I could force it into that form after partial fractions, interesting.
THOUGH it would take ages to even get to (1) if you didn't have a computer to aid in expanding and simplifying the fraction, wouldn't it?

Yep, lots of work without a computer. But you can see that eq(6) is the same equation as eq(1).

Also, I dont quite see where H came from when you introduced it.

After rationalizing the left side, I multiplied the numerator by what it took to make it look like the right side. The multiplier was "H". Of course, I also had to multiply the denominator by the same multiplier to balance the fraction out.

Ratch

sophiecentaur

Quote by milesyoung

I'm not quite sure what you mean. I can't see how you could argue that the induced voltage from flux cutting the rotor windings could be attributed to self-inductance.

'Equivalent Circuits' don't always make physical sense, I know, but when you start a motor, a lot of energy goes in. The rate of current reduction will be exponential with a time constant. That can be modelled with a CR series circuit across a parallel resistor. It's a simplification but it even works as the supply volts are reduced.

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milesyoung	You're absolutely right, consider the simple DC-motor model which probably needs no introduction: http://www.library.cmu.edu/ctms/ctms...r/motorsim.htm Let the motor be loaded so the rotor has a constant angular velocity and let the current in the circuit be constant in magnitude. Would you say the back-EMF in this case has anything to do with the self-inductance L?

sophiecentaur Recognitions: Gold Member Science Advisor	If you were to increase the supply volts, the current would increase significantly until the revs built up, a reducing the current to slightly more than before. That looks, to me, mighty like a big parallel capacitor, relating to the MI of rotor and load.