Lenz's Law & DC motors

[AngelTyrael]

Hey there, hope this is located in the right thread!

I've been searching online for a quite a decent amount of time trying to wrap my head around this concept.

We've been asked to research the link between Lenz's law and DC motors, and for some reason I was given the impression that back emf was a positive to DC motors.

After some online research, all that I've concluded is that back emf is produced when the voltage is increased, which leads to the motor operating at its maximum capacity.

What is the most obvious effect of Lenz's law and the way in which DC motors function?

Thanks in advance.

 

[ericgibbs]

What is the most obvious effect of Lenz's law and the way in which DC motors function?

Thanks in advance.

hi,
Without the back emf from the motor winding the motor would draw a current equal to the applied motor voltage divided by its winding resistance

The back emf in the motor winding is in the opposite sense to the applied voltage [Lenz's Law] so the actual motor current is roughly [Vapp- Vemf]/ Rmtr.

Does that help.?

 

[AngelTyrael]

Is this essentially what you are getting at?:

'Because the back emf opposes the supply emf, it acts to limit the
current through the armature. This means that the motor will build
speed until the back emf has reduced the current through the motor to
a value just able to provide the load torque. Thus a motor running at
low rotational speed needs a high magnetic field strength but, at higher
speed, the same motor needs a smaller magnetic field strength to
maintain the critical current.'

What I'm having trouble understanding is how the back emf 'reduces the current... to a value just able to provide the load torque'

 

[ericgibbs]

Is this essentially what you are getting at?:

'Because the back emf opposes the supply emf, it acts to limit the
current through the armature. This means that the motor will build
speed until the back emf has reduced the current through the motor to
a value just able to provide the load torque. Thus a motor running at
low rotational speed needs a high magnetic field strength but, at higher
speed, the same motor needs a smaller magnetic field strength to
maintain the critical current.'

What I'm having trouble understanding is how the back emf 'reduces the current... to a value just able to provide the load torque'

Lenzs Law Back emf = - di/dt...... its in the opposite sense to the voltage applied to the motor terminals.

So [Vapplied -Vbemf]/Rmtr

Assume 200Vdc is Vapp and the Vbemf = -190V and the motor has a dc resistance of say 1 ohm

Lenzs Law == [200 -190]/1 = 10Amps....

Without the Vbemf it would be 200/1 =200A

EDIT: these figures are for an unloaded motor.

Found a link that explains in more detail:
**broken link removed**

 

[AngelTyrael]

Okay that's cleared some up for me, thanks!


Just an add-on, to make sure I've got this cemented:

Different countries, different interpretations, ultimately,
Lenzs Law Back emf = - di/dt is ΔΦB/Δt?

Also, either I haven't been exposed to the formula, or once again it's lost in translation..

[Vapplied -Vbemf]/Rmtr is derived from where? Is Vbemf regarded as a positive value, thus why it is subtracted and not added?

These queries seem awefully simple for these forums but the help is much appreciated.

 

[ericgibbs]

Okay that's cleared some up for me, thanks!


Just an add-on, to make sure I've got this cemented:

Different countries, different interpretations, ultimately,
Lenzs Law Back emf = - di/dt is ΔΦB/Δt?

The di/dt is the same as Δi/Δt. its the rate of change of current with respect to time, whereas ΔΦB/Δt is the rate of change of magnetic flux with respect to time.

Also, either I haven't been exposed to the formula, or once again it's lost in translation..

[Vapplied -Vbemf]/Rmtr is derived from where? Is Vbemf regarded as a positive value, thus why it is subtracted and not added?
Its not a positive value, its in the opposite polarity to the applied voltage.


These queries seem awefully simple for these forums but the help is much appreciated.

hi,
Remember for a fixed supply voltage the motor speed will fall when a load is applied to the motor shaft,
as it slows the Vbemf drops so the current goes up. The torque load line is not flat.

I hope this has helped.

EDIT: a study of Faraday's Law will be helpful.
http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/farlaw.html