The_UnforgiveN

Hi all..
I've confused

Now, we have a brush DC motor. We show that in a circuit as RLE load, am I right? R-> internal resistance, L -> inductance due to armature coils, E -> back emf due to rotation.

OK, lets take a look to the process.

Assume that I've powered the motor. At that instant, current is at its max because of zero speed of armature. During the speed up, back emf occurs and opposes the main supply voltage. At this time current begns do decrease. First question; when current decreases what happens to inductor, due to dI/dt change? Does inductor forms another back emf that opposes the first created back emf? Because it tries to maintain current?

Later at the instant we cut the power off, some says that very large back emf occurs and damages the switch if we have not any flyback diode. Another question; what's the origin of that huge back emf? In my mind, if we cut the power off, main supply voltage becomes zero and only voltage occurs due to still rotating armature. And this cannot be larger than main supply voltage.

If you say that, this huge emf forms because of the inductor (dI/dt), why there's no emf like this at the begining of the procedure (when powering the system)

whatever, can you enlight me a bit I need roles of inductor and armature speed in creation of the back emf during normal process and at the instant power cut off.

Thank you very much

ericgibbs

hi,

Look at www.penzar.com download the demo,,,, then goto

http://www.ecircuitcenter.com/Circui...otor_model.htm

You should be able to set up a demo motor

Regards

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Eric
"Good enough is Perfect"

PIC tutorials:
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The_UnforgiveN

I don't know how to build a circuit with SPICE. Can you answer my above questions a bit? Thanks

ericgibbs

hi,

For a description of the workings of a dc motor, Google for:

'dc motor back emf'

There a lots of articles that will answer your question, there is no point in me repeating information thats already available.

If you have a problem with a particular part, come back to us.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's: www.winpicprog.co.uk/
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

Hero999

If it's a permanent magnet motor the back EMF is dependant of the motor speed more than the inductance, of course there'll be a short inductive spike when the power is disconnected but after that there'll be a a sag in voltage untill the rotot slows down.

Connect the motor to a large weight, run it up to full speed, disconnect the power and measure the voltage. It will first drop to within 80% to 60% of the supple then gradually decay to 0V.

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The_UnforgiveN

ericgibbs, my questions were already about particular parts, just about back emf, not whole dc motor operation. I've googled so much. But they do not distinguish the back emf due to speed and due to inductance. I know how a dc motor operates. And lately if you don't wanna answer then, don't.

Hero999, your answer is much more clear, thank you.

ericgibbs

hi,
If you had explained at the start you had googled already, but you didnt get the answers you needed, then I could have, possibly, given you the answers myself.

We get so many OP's asking questions, who have not taken the time to look thru the web pages and search for the answers.

How are we supposed to know, if you don't tell us?

Anyway, as you say, Hero has been good enough to answer your question.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's: www.winpicprog.co.uk/
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

The_UnforgiveN

Hi,
If you will say "google bla bla" to whoever asks smt, then why does this forum exist? Maybe because of this behaviour I couldn't find anything related to my topic, at this forum too.

But you're right at that point, I should mention when I google before I ask smt. By the way, if you have a good link for this topic, can you post please?

Thanks again

ericgibbs

Hi,

>> Now, we have a brush DC motor. We show that in a circuit as RLE load, am I right?
>> R-> internal resistance, L -> inductance due to armature coils, E -> back emf due to rotation.
R= Resistance of the windings, L= Inductance of the windings, E= - Emf generated due to the angular rotation

>> Assume that I've powered the motor. At that instant, current is at its max because of zero speed of armature.
The current is not a max at that instant, due to the inductance of the winding, it quickly rises to a maximum.
The current is creating the initial magnetic field within the motor.

>> During the speed up, back emf occurs and opposes the main supply voltage.
>> The back EMF is proportional to the speed of rotation. At this time current begns do decrease.
>> First question; when current decreases what happens to inductor, due to dI/dt change?
Nothing happens to the inductor, its a function of the windings and the motor core.

>> Does inductor forms another back emf that opposes the first created back emf? Because it tries to maintain current?
No.

>> Later at the instant we cut the power off, some says that very large back emf occurs and damages the switch if we have not any flyback diode.

>> Another question; what's the origin of that huge back emf? In my mind, if we cut the power off, main supply voltage becomes zero and only voltage occurs due to still rotating armature. And this cannot be larger than main supply voltage.
Lenzs Law states that EMF = -L * di/dt, as the inductance is a function of the winding and core, its the rate of change of current with respect to time that generates this back EMF.

So if the motor is running at say full load and you switch OFF the supply voltage, in say, 10mSec, assuming the inductance to be 1 H and the current was say, 10 amps, then you would get 1 * 10/0.01 = - 1000 volts.
Its important to remember this back emf will generate a current in the opposite sense to the original applied voltage and it will cause regenerative breaking of the motor.

The flyback diode provides a path for the current due to the back EMF and effectively clamps it to a safe value.

>> If you say that, this huge emf forms because of the inductor (dI/dt), why there's no emf like this at the begining of the >> procedure (when powering the system)

A back EMF is generated whenever a current flowing in a magnetic circuit is suddenly switched off.

__________________
Eric
"Good enough is Perfect"

PIC tutorials:
Nigel's: www.winpicprog.co.uk/
Gramo's: www.digital-diy.net/
Bill's: www.blueroomelectronics.com/

The_UnforgiveN

Awesome.. That was what I've asked. Thank you very much for your reply ericgibbs