MH's ideal coil and voltage question

[tinman]

I'll venture a guess.

I see 6 separate test conditions.

1.  Locked rotor, no PM, no pole pieces.
     Armature acts only as an inductor with CEMF, no BEMF, max I determined by R_DC

2.  Locked rotor, no PM but with pole pieces installed.
     Similar to above but with larger inductance, CEMF only, no BEMF, max I determined by R_DC

3.  Locked rotor, with PM and with pole pieces installed.
     As above but with larger inductance or shift in BH curve (saturation), CEMF only, no BEMF, max I determined by R_DC

4.  Spinning rotor, no PM, no pole pieces (no magnetic or conductive material in proximity to rotor, spun via external means)
     Rotor acts as an inductor with modulated inductance and R_DC, some noise, CEMF only, no BEMF, max I determined by R_DC

5 

6.   


PW   

Only 5 and 6 are applicable for this test.
All we want to know is how much of the current draw reduction is due to the PMs being in place when the motor is spinning.

Spinning rotor, no PM but with pole pieces installed, rotor spins as attraction motor
     Rotor still has inductance but those effects are swamped by the BEMF now present.  Max I determined by BEMF.

If the PMs are removed,what BackEMF are you referring to ?.
The motor used is the same as the one on the right side of the pic below. When the magnets are removed,so too is the steel cylinder that housed them. The only thing left is the two bolts to hold the Ali bearing carriers together-see pic below.

Spinning rotor, with PM and pole pieces installed, rotor spins as normal PM motor.
     As above, max I determined by BEMF, more torque (or RPM) available for a given BEMF (current draw)

So the first thing to do,was find the power draw of the motor free spinning(no load),as in standard trim-motor complete.
The second part of the test was to simply remove the magnets and the steel tube that housed them,so as we are left with the bare rotor. I then spun that rotor by way of the second motor seen in the pic below,at the same RPM(or very close to)as the motor was spinning in the first part of the test. This is so we know that the coils were switching at the same rate as they were in the first part of the test.

I was going to do various RPM test,but we will have to stick with just the one in this test

So now this is where it starts to get confusing PW,with you wanting to use BackEMF to denote the generating effects in a DC PM motor,and CEMF for pulsed/AC fed inductors.

You say in answer 5--> Rotor still has inductance but those effects are swamped by the BEMF now present.
If the magnets are now removed,where is this generated BackEMF coming from?


Brad

[tinman]

Tinman,

Are you stating that you believe that the only amount of current that can flow thru an inductor with zero DC resistance is the amount of current that flows at T=0?

PW

The CEMF induced current is determined by the rate of change of the magnetic field in time--yes?
If as Poynt says,that the current will rise at a steady linear rate of 800mA/s,then it must also be true that the rate of change of the magnetic field in time will also change at that steady linear rate.That being the case,then the value of the CEMF,and there for the self induced current,must also rise at that steady linear rate--the joys of having no exponential current curve--no time constant.

As there is no time constant--the process is infinite,then the determined action/reaction is determined at T=0. The coil is also ideal,and there for dose not dissipate any energy,and so none is lost such as it is in a coil with resistance.

The answer given for MHs question is not correct,as the fact that there is no time constant ,was simply ignored--and it should not have been.
Having no time constant results in a steady linear climb in EMF induced current,and there for must also result in a steady climb in self induced current that is in opposition to that which created it.


Brad

[picowatt]

Only 5 and 6 are applicable for this test.
All we want to know is how much of the current draw reduction is due to the PMs being in place when the motor is spinning.

If the PMs are removed,what BackEMF are you referring to ?.
The motor used is the same as the one on the right side of the pic below. When the magnets are removed,so too is the steel cylinder that housed them. The only thing left is the two bolts to hold the Ali bearing carriers together-see pic below.

Your image looks like my test #4 conditions (although the two bolts remain in proximity to the armature).

The wording of the my test #4 conditions is with regard to elimination of induced eddy currents and their subsequent magnetic field.

Not knowing whether or not you would have pole pieces remaining, test condition #5 is without the PM's but with pole pieces remaining.

It looks as if you are comparing my test conditions #6 with something close to my test #4.

PW

[picowatt]

The CEMF induced current is determined by the rate of change of the magnetic field in time--yes?
If as Poynt says,that the current will rise at a steady linear rate of 800mA/s,then it must also be true that the rate of change of the magnetic field in time will also change at that steady linear rate.That being the case,then the value of the CEMF,and there for the self induced current,must also rise at that steady linear rate--the joys of having no exponential current curve--no time constant.

As there is no time constant--the process is infinite,then the determined action/reaction is determined at T=0. The coil is also ideal,and there for dose not dissipate any energy,and so none is lost such as it is in a coil with resistance.

The answer given for MHs question is not correct,as the fact that there is no time constant ,was simply ignored--and it should not have been.
Having no time constant results in a steady linear climb in EMF induced current,and there for must also result in a steady climb in self induced current that is in opposition to that which created it.


Brad

Again, to be clear, are you stating that you believe that the only amount of current that can flow thru an inductor with zero DC resistance is the amount of current that flows at T=0?

A simple yes or no would be most helpful.

PW

[poynt99]

The CEMF induced current is determined by the rate of change of the magnetic field in time--yes?
If as Poynt says,that the current will rise at a steady linear rate of 800mA/s,then it must also be true that the rate of change of the magnetic field in time will also change at that steady linear rate.That being the case,then the value of the CEMF,and there for the self induced current,must also rise at that steady linear rate--the joys of having no exponential current curve--no time constant.

As long as the current is rising at a steady rate of 0.8A/s, the cemf will be a steady 4V, it will not rise at a steady linear rate.

As there is no time constant--the process is infinite,then the determined action/reaction is determined at T=0. The coil is also ideal,and there for dose not dissipate any energy,and so none is lost such as it is in a coil with resistance.

The answer given for MHs question is not correct,as the fact that there is no time constant ,was simply ignored--and it should not have been.
Having no time constant results in a steady linear climb in EMF induced current,and there for must also result in a steady climb in self induced current that is in opposition to that which created it.

It is very unlikely anyone agrees with you. How are you going to prove this?