OverUnity Research

One of the most annoying things about participating in these free energy forums is the way it seems to be accepted as fact by almost everyone
that the discharge of energy from the magnetic field of a coil is (back emf).

My understanding is that it is regular emf because it is in the same direction as the applied emf, the back emf is only there to oppose the charging of the coil,
and the discharging of the coil if through another inductance.

So what is everyone else's opinions ? Is the emf produced from the release of energy from a magnetic field of a coil (emf) or is it (Back emf) ?

I'm interested in the reason it would be consistently considered back emf, and not emf as it should be in my opinion.

Cheers

I think it is this misconception that fuels a lot of mistaken claims, people see the coil discharge referred to as back emf and think the coils magnetic field is comprised
of back emf. I think this leads some to believe that the coil discharge is free energy from the environment, And maybe that back emf consumes energy from the supply
or something. People also seem to think that reactive power is back emf. Same misconception i think.

I think this confuses newer experimenters to no end and ruins or retards their understanding right from the beginning of their learning.

Hardly anyone says anything to rectify the situation and it seems to me when no one says any different the newer experimenters assume it must be fact.

I'm kind of bemused by this and disappointed.

Cheers

@Farmhand

The reason many don't respond to these things is because we have already done so too often. Not long after you help someone understand these things another piper comes along and erases the correct understanding.

'Back emf' is constantly misused. This term has more to do with motors than anything else. Counter emf is the term used for networks of inductance. Both are used interchangeably and incorrectly by almost everyone.

The emf created by the collapse of a magnetic field around a coil is a voltage (emf). emf doesn't even define energy.

It doesn't matter because there is no more free energy in B/Cemf than there is when you release a compressed spring.

The greatest mistake is to rely upon the Wiki's. In general, the information is good. The problem is that any jerk-off can edit the information. With the Wiki folks overloaded, it could be forever before they correct the edited information.

Yes exactly emf doesn't define energy, I'm very much still a learner and will be always. But I don't understand the mindset of some people.

As soon as Carroll (Citfa) explained it to me it made complete sense (it was logically correct as far as I could tell ) and I did some quick research to confirm it.

And yes Counter emf is a better term when not talking of motors, that makes sense too, point taken.  O0

I don't want to try to get too technical because i may confuse myself. But I think this is an important part of the basics.

When an unqualified person like me tries to explain these simple things we often get swamped with all kinds of BS and are expected to then explain everything from
tuned circuits to black holes. hahahaha

People with the qualifications need to state these things every now and then clearly and simply, with a simple logical explanation of why it is so.

Thanks ! I need to check occasionally to make sure I'm not having a nightmare or going mad. :D


Cheers

Quote from: Farmhand
But I don't understand the mindset of some people.

Aye, it is an ages old phenomenon.  The meanings of
certain words or expressions within any language
change over the course of time.

In many cases the changes come about unwittingly
caused by those who lack education or experience.
And once the new meanings are accepted by those
within the "following" it is already too late to stop
the corruption.  The change may not make it into
the dictionaries and reference works as a permanent
change but in the minds of those convinced it will
be handed down from generation to generation to
live on within the "following."

In the earliest stages of our technical educations we
pass through that embryonic phase where we
"know just enough to be dangerous."  It is while
we are there that we tend to use terms incorrectly.

As we progress on to higher education and understanding
most of us will "self-correct" our wrong usage.  But
always present is a certain "rebel" element who delight
in warped exposition and make it their goal to change the
way things are said...

While we're on the subject of misnomers, is anyone able
to make sense of this?

Quote from: Dumped on 2012.12.04, 04:59:26
While we're on the subject of misnomers, is anyone able
to make sense of this (http://www.energeticforum.com/renewable-energy/10610-3-battery-generating-system-51.html#post217163)?

I can't make any sense of it, but i think what they are seeing is a combination of effects. One being the good batteries natural behavior of rising terminal voltage when loaded
along with the switched inductor (motor) causing some shock release of energy from the "dead" but still contains charge battery.

Several misconceptions going on there I think. The dead battery can only put out a negative voltage if the voltage is measured "upside down". Regardless it is still just a difference of potential.

I think a similar effect could be had from a switched inductor of similar inductance to the motor and a capacitor and resistor with similar capacitance and resistance as the battery.

Couldn't a bad battery be modeled with a capacitor and a resistor, maybe several capacitors in series with different resistors across only some. Like some bad cells worse than others and maybe one good cell.
Possibly with a low voltage battery cell in parallel with an unresistored capacitor/cell.

Cheers

Quote from: Farmhand on 2012.12.03, 23:31:47
...
So what is everyone else's opinions ? Is the emf produced from the release of energy from a magnetic field of a coil (emf) or is it (Back emf) ?
...

Back emf is emf. What's the problem? It is well named. Emf is not specifically linked  to the magnetic field. When you power a circuit with a battery, the battery provides an electromotive force acting onto the electrons. When this circuit is a coil, a magnetic field is created by the electrons moved by the electromotive force from the battery. If you open the circuit, the magnetic field collapse becomes the source of a new electromotive force through the induction phenomenon, and so, we get "back" emf.

There is a complete parallel between coil and capacitor, magnetic field and electric field, current and voltage, open a circuit and short cut a circuit.

When you charge a capacitor from a battery, you store energy in the electric field thanks to the electromotive force from the battery that moves the electrons from one plate to the other one. When you short the capacitor, you collapse the electric field and recover the energy from the pulse.
When you power a coil from a battery, you store energy in the magnetic field thanks to the electromotive force from the battery that put the electrons in motion, creating a current. When you open the coil, you collapse the magnetic field and recover the energy from the pulse.

Although the question of a capacitor seems not problematic, "back emf" which is the coil equivalence of shorting a capacitor, seems to be a magic formula in the free energy field, while there is strictly no difference of principle.
The only difference is practical: our capacitors are almost perfect because the dielectric resistance is so high that it can be considered as infinite, while our coils should have no resistance, i.e be superconductors, to be equivalent to our near perfect capacitors. This is why we always need a generator to maintain a current in a coil after having established it while we don't need a generator for maintaining the voltage of a capacitor after it is charged, at least for a significant time. But once again, to open a coil or short a capacitor is the same thing: to recover "back" the emf that has been used to create respectively the magnetic or electric field.

When i say "Back emf" I mean (counter emf), "counter emf" or "back emf" being the force resisting the applied emf.

The collapsing of the magnetic field of a coil does not produce "counter emf" it produces emf, counter emf is in opposition to applied emf,
the emf produced when the magnetic field of a coil collapses does not oppose the applied emf, it emulates it or aids it.

Energy returning to the supply is not back emf either it is reactive power. That's how I see it.

Back emf or counter emf can only be the force that opposes emf while the emf is applied.

Reverse emf is not Back emf or Counter emf either.

If you wanted to apply a reverse emf to a circuit supplied by a battery as the source of emf then just reverse the polarity of the battery connections to the circuit and a reverse emf would be applied.
Reverse emf would be opposed by counter emf also just like the forward emf.

Cheers

Quote from: Farmhand on 2012.12.04, 10:12:03
When i say "Back emf" I mean (counter emf), "counter emf" or "back emf" being the force resisting the applied emf.

The collapsing of the magnetic field of a coil does not produce "counter emf" it produces emf, counter emf is in opposition to applied emf,
the emf produced when the magnetic field of a coil collapses does not oppose the applied emf, it emulates it or aids it.

I would say that you already knew what Back emf was.

Now you can fight the good fight and educate the others  C.C

Good Luck  O0

There cannot be EMF without BEMF.  It is an action reaction pair and cannot be separate under any circumstances.  I just take it from my view that back emf is taken from mechanic of a reaction force.  For the case of friction and static force, there is still a reaction force though not dynamic.  We'll just stick to inductor case.

In mechanic, we have F = ma, more specific, F = m dv/dt .  In electric we have V= L di/dt. 

When you push a heavy ball, you will feel a force exerted back with the magnitude of  m dv/dt .  This is the reaction force.  Now what if you push a ball of paper?  If you can match any speed, you will still experience the same magnitude because it has more acceleration now with less mass.   

What if you try to stop a ball that in motion?  You will exert a force V=L di/dt. 

You cannot separate V=L di/dt.   It's better to see V as EMF, L di/dt as a reaction force and not mix both of them together as one kind. 

Quote from: Farmhand on 2012.12.04, 05:30:17
I can't make any sense of it, but i think what they are seeing is a combination of effects. One being the good batteries natural behavior of rising terminal voltage when loaded
along with the switched inductor (motor) causing some shock release of energy from the "dead" but still contains charge battery.

Several misconceptions going on there I think. The dead battery can only put out a negative voltage if the voltage is measured "upside down". Regardless it is still just a difference of potential.

I think a similar effect could be had from a switched inductor of similar inductance to the motor and a capacitor and resistor with similar capacitance and resistance as the battery.

Couldn't a bad battery be modeled with a capacitor and a resistor, maybe several capacitors in series with different resistors across only some. Like some bad cells worse than others and maybe one good cell.
Possibly with a low voltage battery cell in parallel with an unresistored capacitor/cell.

Cheers

Special meter, mysterious part, Creative commons license, patent. Such foolishness...

Quote from: Farmhand on 2012.12.04, 10:12:03
...
The collapsing of the magnetic field of a coil does not produce "counter emf" it produces emf, counter emf is in opposition to applied emf,
the emf produced when the magnetic field of a coil collapses does not oppose the applied emf, it emulates it or aids it.

When the field is collapsing, the current in the coil tends to prevent the collapse. For this purpose the current "tries" to continue to circulate as before the collapse, thus opposing the field decrease. That's why there is no current reverse at the moment of the coil opening, and that's why the voltage increases under the effect of the current which tries to circulate and escape at the coil end. Back emf is a counter emf figthing the collapse of the field.
I believe that I now understand what you mean, you think "counter emf = emf against emf" while I think "emf against field variation", with paying a great attention to what is the cause and the effect to distinguish between direct and counter effect.

The following is a good example of how the meanings
of technical terms are changed over time.

When both Counter e.m.f. and Inductive Kickback were
first discovered and named the meanings were very
clear.

Today, however, "back" as in Back e.m.f. can mean
either "pushing back in opposition during charge" or
it can mean "getting back by discharge."

As long as we take it upon ourselves to assign new
meanings to technical terms the confusion will persist.

Quote from: exnihiloest on 2012.12.04, 09:34:56
Back emf is emf. What's the problem? It is well named. Emf is not specifically linked  to the magnetic field. When you power a circuit with a battery, the battery provides an electromotive force acting onto the electrons. When this circuit is a coil, a magnetic field is created by the electrons moved by the electromotive force from the battery. If you open the circuit, the magnetic field collapse becomes the source of a new electromotive force through the induction phenomenon, and so, we get "back" emf.

There is a complete parallel between coil and capacitor, magnetic field and electric field, current and voltage, open a circuit and short cut a circuit.

When you charge a capacitor from a battery, you store energy in the electric field thanks to the electromotive force from the battery that moves the electrons from one plate to the other one. When you short the capacitor, you collapse the electric field and recover the energy from the pulse.
When you power a coil from a battery, you store energy in the magnetic field thanks to the electromotive force from the battery that put the electrons in motion, creating a current. When you open the coil, you collapse the magnetic field and recover the energy from the pulse.

Although the question of a capacitor seems not problematic, "back emf" which is the coil equivalence of shorting a capacitor, seems to be a magic formula in the free energy field, while there is strictly no difference of principle.
The only difference is practical: our capacitors are almost perfect because the dielectric resistance is so high that it can be considered as infinite, while our coils should have no resistance, i.e be superconductors, to be equivalent to our near perfect capacitors. This is why we always need a generator to maintain a current in a coil after having established it while we don't need a generator for maintaining the voltage of a capacitor after it is charged, at least for a significant time. But once again, to open a coil or short a capacitor is the same thing: to recover "back" the emf that has been used to create respectively the magnetic or electric field.

Quote from: GibbsHelmholtz on 2012.12.04, 14:04:43
There cannot be EMF without BEMF.  It is an action reaction pair and cannot be separate under any circumstances.  I just take it from my view that back emf is taken from mechanic of a reaction force.  For the case of friction and static force, there is still a reaction force though not dynamic.  We'll just stick to inductor case.

In mechanic, we have F = ma, more specific, F = m dv/dt .  In electric we have V= L di/dt. 

When you push a heavy ball, you will feel a force exerted back with the magnitude of  m dv/dt .  This is the reaction force.  Now what if you push a ball of paper?  If you can match any speed, you will still experience the same magnitude because it has more acceleration now with less mass.   

What if you try to stop a ball that in motion?  You will exert a force V=L di/dt. 

You cannot separate V=L di/dt.   It's better to see V as EMF, L di/dt as a reaction force and not mix both of them together as one kind. 

I agree. Nevertheless there is a cause and an effect. When you increase a current, the field increases with a delay that can't be less than d/c, d being the distance from the flowing electrons, where the field is observed. The current is the cause, the field the effect. Inversely when you move a magnet near a coil, the change of the magnetic field can't be felt by the coil before a d/c delay and so the current is the effect of the moving field which is the cause. So the reaction force is a bit delayed but it doesn't change anything because the energy that is not "fighted" by the reaction force during the time d/c, is used to build the field, the whole phenomenon still obeying the energy conservation.

Quote from: Dumped on 2012.12.04, 21:09:08
...
Today, however, "back" as in Back e.m.f. can mean
either "pushing back in opposition during charge" or
it can mean "getting back by discharge."
...

That's why math is so important in physics. The need of measurements first initiated by Galileo who studied the fall of bodies along inclined planes and the need to avoid the amguities of the natural language have convinced the scholars after the Renaissance to adopt math as the best choice language.

I've been here before, so here's my two-penneth just to clear up a minor point.

Relating to Bemf/Cemf. Some people seem to think that Cemf and Bemf are totally different things, referring to different phenomenons: They are not, they are the same thing. Back emf = Counter emf.

It simply comes down to this. In the UK we use the term Back emf, in the USA they use the term Counter emf. Just as we say 'anti-clockwise' and they say 'counter clockwise'. Simply two different ways of saying exactly the same thing.  :)

It may also depend upon the industry specific terminology used.

Since rotating machinery is a fairly large part of my business I follow the terminology used in my industry.

For inductors, CEMF is analogous to the counter force in a spring when it is being compressed and BEMF is analogous to that same stored energy from the compression of the spring being released by the spring when it is released.

For motors, CEMF is not commonly used but clearly understood because the windings of a motors are inductors and BEMF relates to the motor slip angle or the rotational delay between driving energy and the actual following angular position of the rotor. BEMF is the term used when slip angle and angular position/velocity are needed.

Either term is a force - be it energy being stored or released. One can't exist without the other otherwise tank circuits wouldn't have any ring.

I never understood the UK usage of 'anti-' in such cases. In the U.S. (and every other country I work with except the U.K. and India) 'anti' would be applicable to comparing sine to cosine. Example: a resonant tank circuit ..... a series L/C tank will have a maximum of current when anti-resonant(resonant for a series L/C) and a parallel tank will have a minimum of current when resonant.

Because I deal with these things every day I must speak the same lingo as the rest of the folks. Should I proclaim that BEMF and CEMF were the same action/reaction .... well, it wouldn't be a good thing. They might cancel my IEEE subscription  :'(

I have come to realize that modern learning institutions are globbing concepts together, perhaps to streamline since most folks won't see the differences. Sad, IMHO.

To me counter emf and back emf are one and the same thing, both only exist to oppose emf and neither has anything to do with regained energy or the emf produced when a magnetic field collapses.

Both are the same to me. And both can only exist to oppose emf and are always present when emf is applied either directly by us or indirectly applied by the coil.

For me the emf produced by the collapse of the magnetic field of a coil will always be just emf, and the emf produced by the coil is in fact opposed by counter emf while the emf is applied.

It can only be one or the other, either the emf produced by the collapsing magnetic field is emf or it is counter emf. Counter emf opposes the applied emf so it can only be regular emf.

That's logic. If the coil was to produce counter emf as a result of the magnetic field collapse then how could it be opposed by more counter emf in the discharge current path ?

Because when the coil discharges and current flows in the discharge path counter emf is present to oppose the force that created the current. Is that not so ?

Cheers

P.S. There is a similar misuse of the term "reactive power", people think they can power loads with reactive power. Which is impossible. Loads can only be powered by "real" power that dissipates energy in the load,
as soon as that happens the power is real power, reactive power can only be the unused part of the apparent power which is returning to the supply or being stored, if reactive power released from the coil of a motor
is stored in a capacitor then the power supplied by the capacitor back to the load must be considered as apparent the power, if returned to the capacitor it is reactive again the power dissipated in the load is real power.



..

Quote from: Collins Dictionary
back emf

Definitions

noun

(electrical engineering) an electromagnetic force appearing
in an inductive circuit in such a direction as to oppose any
change of current in the circuit.

The above concise definition, while quite good, covers every
possible scenario.  In order to add precision to any application
which focuses on the concept it would be necessary to define
the circumstance which produces the emf.

Many experimenter discussions are characterized by a paucity
of explanatory verbiage and numerous abbreviations.

With a little more effort any potential ambiguities could be
prevented or minimized.  In this era of "text speak" and
weird spellings that is probably too much to hope for...

The key word is "change."

That is a massive simplification in my opinion. In my opinion the "key" word there is "appearing".
Going by that definition the emf produced from the collapsing field is back emf and that back emf
is opposed by back emf because it is causing a change in the current in the discharge path, particularly
if the discharge path for the current from the coil discharge is a different path to the current path for charging the coil.
In the latter case it would involve back emf opposed by back emf. The coils field collapsing is in my opinion a direct result of the lack of emf to maintain it
and so it cannot remain and must collapse, I don't think it collapses just because there is a rule that says it must resist a change in current. I think it collapses because
of the lack of emf to maintain it, the emf it produces is a result of that, the result is it "appears" to resist a change in the flow of current. But In reality it doesn't collapse
for the purpose of resisting a change in current, it collapses because the emf that built and maintains it is removed. Just like a balloon the balloon doesn't deflate to maintain
air current it deflates because the force that inflated it and keeps it inflated is removed. If there was truly momentum involved, then when the applied emf was removed the magnetic field would continue to "build"
or increase for some time after the emf was removed.Is that the case ? I think it is more like pressure that is released when the force that created it is removed.
Truth is I think that the energy will go whichever way it can go easiest. If it is easier for the energy from the field collapse to escape by reversing the flow of current I think it will do that, but I couldn't be sure unless i tested it.

Charge a coil, then completely open the negative end and see if it causes a current back to the supply so as to discharge the field energy. I think that could be considered as reactive power.



Cheers

Inductors and their electrical characteristics can be
difficult to comprehend.  For some reason capacitors
are a whole lot easier for most of us.

T=RC is easily intuitive.

T=L/R requires a lot of thought.

I  can still fondly remember my own early experiences
with inductors.  I got my hands on a power transformer
from an old radio and used it for many experiments with
a 6 Volt lantern battery.  I was fascinated by the large
flamelike sparks that would be produced when breaking
contact with the battery energizing the transformer low
voltage output winding. 

And the occasional nearly half inch long sparks that could
be created from the high voltage primary winding when
making and breaking the connections to the battery and
the low voltage winding.

Not to mention the power of the shocks I'd get when
careless!

I was hooked on electricity and couldn't rest until I
knew how that stuff worked...

Later on I was able to obtain a Model T ignition coil
which seemed to me worth its weight in gold.  Used it
to make my first Jacob's Ladder and to ignite pieces of
paper in the very hot sparks.  And then a very rudimentary
spark gap radio transmitter.

Ah yes, those were the days!  The smell of Ozone brings
back many memories.

Quote from: Farmhand on 2012.12.05, 02:30:06
...The coils field collapsing is in my opinion a direct result of the lack of emf to maintain it and so it cannot remain and must collapse
...

If there was a lack of emf, the field would collapse instantly.
The field must collapse but don't collapse instantly because each attempt of collapsing is partially compensated by an emf that prolongs the field. An emf always opposes its source, because the energy must be taken somewhere, and the question becomes a simple question of viewpoint about which source is that one to be preserved.
If you are using a generator to establish a field in a coil, the emf appears when the field is increasing (self-induction) and opposes the source, so you will say "it's back emf or it's counter emf, it is depleting my source". If you open the coil circuit, the field collapses and the coil itself is the source. You may say it's emf, not counter emf: it's an emf generated by the coil which is the source, a real magnetic generator as a dynamo providing current from a changing field. But you may also say that it's a counter emf: it opposes the field collapse and thus it opposes the current variation which should lead to stop it.
Question of viewpoint while the phenomenon at work is always the same, emf from field variation or field variation from emf, depending on which is the cause and which is the effect. "Back emf", "counter emf" are just words from professional jargon of electricians to qualify this emf relatively to the useful generator.

Quote from: Farmhand
In my opinion the "key" word there is "appearing".

Could it be that "change" and "appearance" are
simultaneous?  Whenever current through an
inductor is non-changing there is no induced
emf.  Or, whenever the magnetic field strength
within an inductor is non-changing there is no
induced emf.

In both cases, absent change, the appearance
of the induced emf does not manifest.

Quote from: WaveWatcher on 2012.12.05, 00:33:19
It may also depend upon the industry specific terminology used.

Maybe, but I'm of the opinion that you won't see 'counter emf' mentioned in a uk produced text book, only 'back emf'. Hence I'm also rather inclined to think the opposite would be true of an American text book.  I certainly never came across the term Cemf on the course I did, but the term Bemf I remember from inductor theory.

Happy to be proved wrong though!

Quote from: Farmhand on 2012.12.05, 01:43:15


P.S. There is a similar misuse of the term "reactive power", people think they can power loads with reactive power. Which is impossible. Loads can only be powered by "real" power that dissipates energy in the load,
as soon as that happens the power is real power, reactive power can only be the unused part of the apparent power which is returning to the supply or being stored, if reactive power released from the coil of a motor
is stored in a capacitor then the power supplied by the capacitor back to the load must be considered as apparent the power, if returned to the capacitor it is reactive again the power dissipated in the load is real power.



..

If you have a primary and secondary.  Secondary has a resistive load.  The power giving out by the resistive load is real power.  The resistive  load is powered by the primary. 

http://www.energeticforum.com/renewable-energy/3806-back-emf-vs-collapsing-magnetic-field-spike.html

Yes I do agree inductors can be complicated, I also agree that it very much depends on the perspective and how it is looked at.
I do appreciate everyone's opinions, impressive responses, and for me there is always something to learn from looking at things from another's perspective.
I don't wish to appear argumentative for the sake of argument, after all most of you guys have put inductors to many good uses for a long time
so I will not be arbitrarily dismissing or ignoring any comments.

We all know that counter emf does not actually consume from the supply. Don't we ? In a transformer it simply restricts input current and in a generator it represents the energy transferred
from shaft power (mechanical force) to electrical power in my opinion. All the so called Lenz mitigating devices still output relative to the effect of Lenz.

In a transformer no energy is dissipated due to counter emf. In a motor the effect of back emf does the same thing I think
and in the generator it takes the form of Lenz effect kind of in an inverted way, I think.

I don't think there are any losses involved in the restriction of current in a motor or transformer due to the effect of back or counter emf.

The way most folks refer to back emf which I disagree with most is the recovery of the energy from the magnetic field of a coil using a diode,
the coil is not charged with counter emf and it acts a source of emf to charge the battery or run the load, the way the charge battery is usually
in series with the supply battery presents a counter emf to the value of the charge battery voltage, meaning because of the 12 volts above the
circuit positive rail voltage when the coil discharges it faces 12 volts (counter emf) automatically, if two 12 volt batteries are charged in series, and
those two are in series with the supply battery then the emf produced by the coil faces 24 volts (counter emf) automatically.
if there is any inductance in the discharge path then the emf of the discharging coil is opposed by counter emf produced by that as well because it is causing a change in the current.

In the case of AC sine wave then the emf is constantly changing and so on the decline of the wave there is a constant discharge of magnetic
field energy which appears as emf right along with and aiding the applied emf. After all the energy discharged from the coil was from the applied emf.

Anyway it is an interesting subject. Thanks for the replies so far.


EDIT: (some edits made for clarity, I hope  :) )

Hi Wings, yes that thread does explain some things, that thread needs to be linked in the Don Smith thread on that other forum.  ;D
Thanks for digging that up. I'm not sure anyone here requires tuition from Aaron though, meaning no disrespect, to anyone.


..

I've been doing some research and simulation on transformer loading. 

When we load the secondary, the magnetic field interaction between two winding cause more input.  We can simulate secondary loading by input artificial current through the secondary.  Better yet, we can use magnet to simulate secondary loading.  So far on the sim, I cannot see how this would not work.

What if we have a watt meter connect to a transformer.  Instead of loading the secondary, place a strong magnet to the transformer core to simulate loading.  What would be observed on the watt meter? 

Quote from: GibbsHelmholtz on 2012.12.06, 14:01:16
I've been doing some research and simulation on transformer loading. 

When we load the secondary, the magnetic field interaction between two winding cause more input.  We can simulate secondary loading by input artificial current through the secondary.  Better yet, we can use magnet to simulate secondary loading.  So far on the sim, I cannot see how this would not work.

What if we have a watt meter connect to a transformer.  Instead of loading the secondary, place a strong magnet to the transformer core to simulate loading.  What would be observed on the watt meter? 

Not much would happen with the Watt meter because there is still no load, if no energy is taken out of the transformer
then only losses would be replaced as far as energy goes, the Watt meter might show some extra Watts due to reactive power there could be more losses as well.
Main point is there would not be a significant continuous input unless significant energy is taken out.

Cheers

Quote from: GibbsHelmholtz on 2012.12.06, 14:01:16

What if we have a watt meter connect to a transformer.  Instead of loading the secondary, place a strong magnet to the transformer core to simulate loading.  What would be observed on the watt meter? 

The watt meter would show increased energy usage. How much depends upon how strongly your applied magnet can saturate the core of the transformer. Keep in-mind that all you are doing is decreasing the inductive reactance of the transformer and therefore creating more current through the primary. The energy will be lost as heat.

A more controllable method is to connect the secondary to the primary reversed polarity - then insert a suitable variable resistor between a primary and secondary wire so you can control the effects. This can be dangerous when using a transformer not rated to work as a reactive load element.

Almost all reactive load elements you can buy have the losses controlled by fixed spacing gaps in the core lamination.

These are typically used to add to resistive loadbanks to bring the power factor of the test load closer to real-world loads.

Google 'Simplex' or 'Avtron' load banks. Both should have a technical paper covering the types of load banks. I know Avtron did have one. I participated in the writing of it  ;)

A real pro,

That's what I have in mind.  This is also what I see when the secondary is loaded.  It must has a similar effect causing input to increase.  I would not say secondary loading saturate the core, but primary and secondary current making the domain reluctant to move.  Though the material does not saturate itself, it is no differ from saturation phenomena. 

Saturation should cause inductance to decrease, so inductive reactance decreased. 
Maybe we should rethink transformer operation.   

Quote from: GibbsHelmholtz on 2012.12.06, 14:01:16
I've been doing some research and simulation on transformer loading.  

When we load the secondary, the magnetic field interaction between two winding cause more input.  We can simulate secondary loading by input artificial current through the secondary.  Better yet, we can use magnet to simulate secondary loading.  So far on the sim, I cannot see how this would not work.

What if we have a watt meter connect to a transformer.  Instead of loading the secondary, place a strong magnet to the transformer core to simulate loading.  What would be observed on the watt meter?  

A transformer presents an input impedance R+L in series, where R is the resistance of the primary coil and L its inductance. In a good transformer, L must represent a high impedance at the working frequency. Therefore only a weak leakage current flows in the primary (if the secondary is open).
If you put a magnet that saturates the transformer core, L decreases, the impedance becomes low, and consequently a current that can be very strong flows in the primary.

Note that if you put a magnet, you don't really simulate a loaded transformer, because the secondary plays no role, no current flows in the secondary, what is not the case when a real load is connected to the secondary and its resistance is also concerned.

Quote from: GibbsHelmholtz on 2012.12.06, 14:01:16


What if we have a watt meter connect to a transformer.  Instead of loading the secondary, place a strong magnet to the transformer core to simulate loading.  What would be observed on the watt meter?  

The Watt meter will show the reactive power if that is what it is set to, if it is set to real power, as some can be only the real power will be shown.
I was considering real power when Watts was stated. The real power shown would only be representing the losses I think. Which should be not that much.

With respect to simulated loading, real power is the value to take into account. And with no load actually drawn the real power can only represent the losses.

Cheers

Quote from: GibbsHelmholtz on 2012.12.07, 05:43:34
A real pro,

That's what I have in mind.  This is also what I see when the secondary is loaded.  It must has a similar effect causing input to increase.  I would not say secondary loading saturate the core, but primary and secondary current making the domain reluctant to move.  Though the material does not saturate itself, it is no differ from saturation phenomena.  

Saturation should cause inductance to decrease, so inductive reactance decreased.  
Maybe we should rethink transformer operation.  

My mistake. You are correct. I should have said the primary current increases since the primary inductive reactance decreases. I was thinking in terms of the reactive load increasing (inductive reactance decreasing).
When the secondary is not connected the transformer is no longer a transformer. It is simply an inductor. Transformer theory no longer applies.
----So often we don't know when to stop applying rules and theories to something when it looks like it hasn't changed  :D

Remember that current lags voltage when a load has inductance. The inductor absorbs energy from the circuit and then equally returns energy to the circuit, so a pure reactance doesn't dissipate power. You will not have a pure reactance, so you will dissipate some power. How much depends upon how good you are at introducing losses.

Control core saturation curves and inject core losses:
   a) use other windings as control windings via the same methods used in mag-amps
   b) replace some lamination with air or aluminum to create eddy current losses.
    c) apply magnets to the core - your magnet best be very large and/or powerful  :)

These losses will appear on the watt meter just as surely as the losses in the transformer secondary loads.

Quote from: exnihiloest on 2012.12.07, 09:05:58


Note that if you put a magnet, you don't really simulate a loaded transformer, because the secondary plays no role, no current flows in the secondary, what is not the case when a real load is connected to the secondary and its resistance is also concerned.

This is what I want to know.  What exactly is the difference between two cases.  They both seems to increase input, they both interact via magnetic field.  The magnet has current that seems to represent secondary current. 

Quote from: Farmhand on 2012.12.07, 10:31:17
The Watt meter will show the reactive power if that is what it is set to, if it is set to real power, as some can be only the real power will be shown.
I was considering real power when Watts was stated. The real power shown would only be representing the losses I think. Which should be not that much.

With respect to simulated loading, real power is the value to take into account. And with no load actually drawn the real power can only represent the losses.

Cheers

The Watts meter is set to real power.  When you put a magnet to the core, more current flow through the primary and real power reading increase. 

Quote from: GibbsHelmholtz on 2012.12.07, 14:12:35
This is what I want to know.  What exactly is the difference between two cases.  They both seems to increase input, they both interact via magnetic field.  The magnet has current that seems to represent secondary current. 

These cases are completely different. With the magnet you have a R-L circuit, not a transformer, the open secondary plays no role.
The losses are not the same, the U/I phase are not the same and so on...

Gibbs,

Quote from: exnihiloest on 2012.12.07, 14:53:03
These cases are completely different. With the magnet you have a R-L circuit, not a transformer, the open secondary plays no role.
The losses are not the same, the U/I phase are not the same and so on...

And I stated earlier:

Quote
When the secondary is not connected the transformer is no longer a transformer. It is simply an inductor. Transformer theory no longer applies.

The magnet isn't creating losses exactly the same way as a loaded secondary. The losses created by applying a magnet create transformer core losses by changing the saturation curve.

Mag-amps use the same principle to control loads by changing the saturation curves and/or X_L of the Mag-amp in series between the power source and load with a small DC control signal.

Quote from: WaveWatcher on 2012.12.08, 00:08:14

And I stated earlier:

The magnet isn't creating losses exactly the same way as a loaded secondary. The losses created by applying a magnet create transformer core losses by changing the saturation curve.

Thanks WaveW,

I thought by having the magnet, core losses becomes less?  Besides, current increase means more losses to wire resistance, not core? 

Yes, I understand that Exn. also say they are differ, but can it be more specific? 
For example. 
When the secondary is loaded, what happens to primary current? 
When secondary is loaded, what happens to phase?

When magnet is biased, what happens to primary current?
When magnet is biased, what happens to phase? 

Quote from: GibbsHelmholtz on 2012.12.08, 01:22:04
...
Yes, I understand that Exn. also say they are differ, but can it be more specific?  
For example.  
When the secondary is loaded, what happens to primary current?  
When secondary is loaded, what happens to phase?
...

Like to compare potatoes and carots. Not the same color, not the same shape, not the same genetics...
Take the equivalent transformer schematic and do the math in both cases.
(http://www.overunityresearch.com/index.php?action=dlattach;topic=1585.0;attach=8795)

Quote from: exnihiloest on 2012.12.08, 08:43:16
Like to compare potatoes and carots. Not the same color, not the same shape, not the same genetics...
Take the equivalent transformer schematic and do the math in both cases.

Well... I take it that loading secondary cause primary current to increase and making it more in phase. lol  So does biasing the magnet.  The only differ that would be in question is that there is no stored field in transformer while there is stored field in coil. 

@Gibbs

QuoteWell... I take it that loading secondary cause primary current to increase and making it more in phase. lol  So does biasing the magnet.  The only differ that would be in question is that there is no stored field in transformer while there is stored field in coil.  

I think the best way to really learn is to experiment and prove the matter for yourself. Consider the fact that what you seek has been considered to be impossible therefore no matter how smart the critics may consider themselves to be at best they can only tell you their opinion of what they think doesn't work. A non-working unit is pointless.

On another note, the storing of energy in a magnetic field is dependent on the dynamics of the field and if a core is saturated then how could we expect it to store anything more. I dug up one example for you, the picture below is one channel  from an Arduino based hall effect magnetometer I built to prove matters for myself. It is measuring only the magnetic field intensity from a laminated open inductor in series with a switched resistive load. Note this is not a coil sensor measuring induced voltage it is an independent sensor measuring only the field intensity a known distance from the core.
Now how does this measure of only the magnetic field relate to the current flow in the windings?, why is the peak falling with each current impulse?. Now if the initial current flow is impeded by the self-inductance of the coil then why is the field strength maximum when the current is lowest?. This can easily be explained by what is common knowledge however it is best that you answer these questions for yourself because that is the way we really learn and understand things.

AC

Thanks AC,

I have my own questions and do things on my own but mostly to verify.  So far I found what established is pretty solid.  In my opinion, the problem with free energy isn't something technical but rather something overlooked.  Getting too deep may get ourselves lost. 

My general conclusion for abundant energy is relative motion and magnetic field.  Relative motion is everywhere since the universe is dynamics.  Magnetic field can be created strong at low cost.  A strong field can be create with resonance.  The question is could the magnetic field turn random motion into something useful.  I'm stuck at this for now.

Ah, Good old "back EMF", let's see, what heretical thing can I say about it?

Well, if we can cancel this effect we have to first cancel or invalidate Faradays law of induction, which would mean a changing magnetic field would no longer produce or induce an electric field.  Bye bye electro DYNAMICS!  All will be left with is statics, how boring!

No, we don't want to cancel the back EMF, which is the induced voltage from the changing current's own magnetic field and has a polarity that acts "back" against the current, no we don't want to do that we can't, instead we can be clever and work at fempto seconds before the back emf effect reacts against the whole coil.  I dont think those MIT brain heads thought of that yet!   >:-)

EM

 :D Nobody really answered what is the back EMF. Clearly when I think about it this effect do not occur when the secondary of transformer is opened. This tell me it's related to reflection of energy wave and most particularly to the current flow ! That also means Tom Bearden is probably right about the relaxation time and the most desired experiemnt is to confirm it with long relaxation time material.....

While experimenting, I stumble upon this result. 

When the coil is moving back and forth with just one magnet, the signal reading is high. (illustrated in Indo1)
When the coil is moving back and forth with both magnets, the signal reading is low. (illustrated in Indo2)
Why is this? 

Bravo Gibbs


This is the solution  O0

Quote from: GibbsHelmholtz on 2012.12.10, 13:31:07
While experimenting, I stumble upon this result. 

When the coil is moving back and forth with just one magnet, the signal reading is high. (illustrated in Indo1)
When the coil is moving back and forth with both magnets, the signal reading is low. (illustrated in Indo2)
Why is this? 

Because the two magnets have field density between them that has less change while moving.

If you wish to see stronger signal, align the two magnets so they repel each other (assuming the same distance and velocity of movement.)

Quote from: WaveWatcher on 2012.12.11, 00:45:45
Because the two magnets have field density between them that has less change while moving.

If you wish to see stronger signal, align the two magnets so they repel each other (assuming the same distance and velocity of movement.)

Thanks WaveW,

Can we establish a field that would show constant flux between?
If the magnets move closer together, will the flux change? 

The flux density will never be perfectly uniform between attracting magnets. If the coil is the same diameter as the magnets you should find a distance between magnets where the density is almost uniform and therefore create little to no induction in the movements shown.

If you move them closer the density increases.

No matter what orientation you choose a change in flux density occurs and will induce an emf in the coil.

What is the goal of these experiments?

years ago I read of an interesting way to cancel the backEMF  or maybe just generate extra energy,  or whatever,  the point was that they were claiming it worked and they had photos of this special coil and the theory was quite interesting.   I'll try to describe what I remember:

The basic principle was to electrostaticaly charge a coil, or perhaps place it into a high E-field,  and the coil if tuned with a capacitor will begin to oscillate all by itself.

The principle was that it becomes a negative resistance, and the inventors were describing the charges on the wire and the currents and how all of this comes about etc...

Anybody have any recollection?  It was one of those "we give this to the world" type of disclosures.

EM

Quote from: WaveWatcher on 2012.12.11, 02:30:22
The flux density will never be perfectly uniform between attracting magnets. If the coil is the same diameter as the magnets you should find a distance between magnets where the density is almost uniform and therefore create little to no induction in the movements shown.

If you move them closer the density increases.

No matter what orientation you choose a change in flux density occurs and will induce an emf in the coil.

What is the goal of these experiments?

Thanks again,

I thought so.  I guess it's similar to the electric field between parallel plate volts/meter.  You can make a uniform field but if the magnet gets closer, flux density increase.  If density does not increase, then we would have free energy. 

I have a thought experiment that induction with PMM could have COP 2. 

Suppose we have two superconductor rings.  Inject two equal current into the rings.  Now let them attract to each other like PMM.  As they speed up toward each other, EMF induces in both rings and current energy convert into motion.  Overall, we converted all input from current into motion energy.   Now if one ring is a permanent magnet, we only need to inject current into one ring.  The same thing happens, but this time only 1 current used up with the same motion energy.

Interesting thoughts.

However, it is my understanding that a superconductor completely rejects external magnetic fields. It should be impossible to induce an emf into a superconducting ring.
Also, since superconductors reject all external magnetic fields there is no magnetic attraction like associated between two magnetic materials. There is 'field pinning' but that results in something more like station keeping than attraction or repelling.

Looking back at some threads tonight,i happened across this one.

I hope those here today have a better understanding of BEMF than those of times past.

Quote: When i say "Back emf" I mean (counter emf), "counter emf" or "back emf" being the force resisting the applied emf.

The collapsing of the magnetic field of a coil does not produce "counter emf" it produces emf, counter emf is in opposition to applied emf,
the emf produced when the magnetic field of a coil collapses does not oppose the applied emf, it emulates it or aids it.

Energy returning to the supply is not back emf either it is reactive power. That's how I see it.

Back emf or counter emf can only be the force that opposes emf while the emf is applied.

Reverse emf is not Back emf or Counter emf either.

If you wanted to apply a reverse emf to a circuit supplied by a battery as the source of emf then just reverse the polarity of the battery connections to the circuit and a reverse emf would be applied.
Reverse emf would be opposed by counter emf also just like the forward emf.

I hope this member now knows better,along with  everyone else here.
If not,then maybe the mistakes above should be corrected.


Brad

Just to clarify a little more,  CEMF happens on the rise of current and opposes the direction of flow.  BEMF occurs on the reduction of current and aids the direction of current flow, what is so hard about this, it is first year electronics.  They are distinctly two different things, not the same at all, and I think easy to see.

Room

Could it be that "Back EMF" is in reality a shortening of
KickBack EMF?

CEMF occurs during 'charging' of the inductor.

KickBack occurs during the 'discharging' of the inductor.

Yes, it really is all very clear and easy to see in both DC and
AC applications.

Except, of course, to 'Newbies" and "Embryonics,' but even
they shall see it in due course as they continue learning.

Quote from: Room3327 on 2018.08.16, 17:38:41
Just to clarify a little more,  CEMF happens on the rise of current and opposes the direction of flow.  BEMF occurs on the reduction of current and aids the direction of current flow, what is so hard about this, it is first year electronics.  They are distinctly two different things, not the same at all, and I think easy to see.

Room

Well no actually.
CEMF dose not appose the direction of current flow-it reduces current flow by decreasing the potential difference between the source EMF and the self induced EMF.
It could be seen as an impedance,but not an opposition.

The same applies with BEMF, which is a self generated EMF by motoring action,where the motor also acts as a generator.

Inductive kickback or flyback is a result of the magnetic field reversing direction through the coil,where the windings of the coil are the reference point.
This causes the current to continue to flow through the coil in the same direction it was before the source was disconnected,but where the voltage across the coil inverts.

The load impedance value across the coil during this flyback period determines as to how fast the field around the coil collapses.


Brad

In Electronics Education of the '60s Counter Electromotive Force
included not only Inductive Action but also the Counter EMF
developed in a motor by generator action.  Counter EMF was
defined to be any opposition to Source EMF and Current.

When AC is applied to any Inductor we see opposition to
current flow while the AC current is increasing during each
half cycle.  We also see inductive discharge aiding source
current while AC current is decreasing during each half cycle.
'ELI the ICE man' was drilled into our heads to aid our
remembering that Inductive Current Lags the Applied
Voltage.

The term 'Back EMF' was not part of the Electrical/Electronics
technical vocabulary back then.  I wonder when it entered into
the jargon?

The meaning of 'counter emf' is intuitively quite clear.  The
meaning of 'back emf' is a bit ambiguous.  Is Back EMF intended
to mean 'giving back' or 'pushing back?'

author=muDped link=topic=1585.msg69249#msg69249 date=1534479858]

QuoteWhen AC is applied to any Inductor we see opposition to
current flow while the AC current is increasing during each
half cycle.

I do not agree with this muDped,as there is never any opposition to current flow.
There is (what could be called)an impedance to current flow,due to the low potential difference between the source EMF and the self induced EMF(CEMF)
This dose not only apply to AC,as the same will be seen when pulsed DC is applied.

  We also see inductive discharge aiding source
current while AC current is decreasing during each half cycle.

I have seen this with pulsed DC,but never with AC.
I do not believe that an AC frequency high enough could be produced that would exceed the speed of a collapsing magnetic field.

QuoteThe term 'Back EMF' was not part of the Electrical/Electronics
technical vocabulary back then.  I wonder when it entered into
the jargon?

When they realised that there was a difference between a self induced EMF of an inductor connected to a source,and a magnetically generated EMF like that of a PM DC motor.

QuoteThe
meaning of 'back emf' is a bit ambiguous.  Is Back EMF intended
to mean 'giving back' or 'pushing back?'

Pretty much. BEMF is power generated by the actions of the rotor windings of the motor cutting through the magnetic fields of the stator,thus producing a current flow that is in the same direction to that of the source.
The current value the source provides is source voltage divided by winding resistance minus the self generated current. So the faster the motor spins,the more self generated current it make's,and the less current the source has to supply.

With inductors,there is no change in speed,where the frequency remains the same,and so we have CEMF.

So as you can see,the two are different.


Brad

Quote from: TinMan on 2018.08.17, 10:45:56
author=muDped link=topic=1585.msg69249#msg69249 date=1534479858]





I do not agree with this muDped,as there is never any opposition to current flow.
There is (what could be called)an impedance to current flow,due to the low potential difference between the source EMF and the self induced EMF(CEMF)
This dose not only apply to AC,as the same will be seen when pulsed DC is applied.

  We also see inductive discharge aiding source
current while AC current is decreasing during each half cycle.

I have seen this with pulsed DC,but never with AC.
I do not believe that an AC frequency high enough could be produced that would exceed the speed of a collapsing magnetic field.

When they realised that there was a difference between a self induced EMF of an inductor connected to a source,and a magnetically generated EMF like that of a PM DC motor.

Pretty much. BEMF is power generated by the actions of the rotor windings of the motor cutting through the magnetic fields of the stator,thus producing a current flow that is in the same direction to that of the source.
The current value the source provides is source voltage divided by winding resistance minus the self generated current. So the faster the motor spins,the more self generated current it make's,and the less current the source has to supply.

With inductors,there is no change in speed,where the frequency remains the same,and so we have CEMF.

So as you can see,the two are different.


Brad

Yes let's confuse the issue again, we don't want anyone to understand this stuff, let's fight over symantics and how one person views something different over another.  Oppose subtract call it something else,  the fact is when current is rising in a coil it self induces a reverse or counter EMF in the same coil. That voltage is reducing the applied EMF by some factor, call it opposing or call it subtracting, from the applied emf or COUNTERING it either way it does the same thing. When the applied emf on a coil is removed or reduced the stored energy in the coil is released and induces an emf in the coil that aids or adds to the applied emf of that coil, this is called Back EMF because it is giving BACK some of the input energy.  Is that correct enough for you? Sorry if I didn't use the words you wanted me to.

For those who are mystified by BEMF manifesting itself as high voltage spikes this is what causes that phenomena.

A coil is charged and has manifested a magnetic field around itself, then power is cut off and that field collapses.  When it collapses it self induces into itself a pulse of energy.  In that pulse is all the energy that it took to manifest the magnetic field minus a small amount for losses.  If that pulse of energy is going into a load it will supply to that load the same amount of energy (VA) that made the field, it might be (example) 12 V and 3 A or 36 W. using some of the BEMF energy as amps reduces the voltage (example) you use 200 ma.  36 W /.2 you will have 180 Volts. Now if you take that same pulse of energy and apply it to no load or very high impedance load (meter) that energy has to go somewhere and with very little current flow it manifests as voltage (example) 36,000 V and 1ma. also equals 36 W.  That's how you can get 100's and even thousands of volts from a BEMF pulse, but that pulse will never return you more energy then went into it because there are losses, so you can never get more out then you put in with BEMF, yes you can get a lot more voltage but never more energy.

You will also notice that the higher the voltage spike the shorter the pulse length and the higher the current the longer the pulse length.

Room

 O0 O0 O0 O0

I agree completely with everything you just posted.  And since the time it takes to discharge the coil is dependent on the circuit impedance the discharge is going through, then it should be obvious that you can indeed have a frequency that is faster than the discharge of the coil.  Ham Radio operators are well aware of this because they have to tune their circuits to match the frequency being used.  Or more commonly called tuning to resonance.

Respectfully,
Carroll

The earliest experiments with electricity that I was able to do
as a young lad were with 6 Volt lantern batteries and transformers
salvaged from old discarded radios.  The flaming arcs one is able
to make when momentarily connecting the primary of the
transformer to the battery was quite amazing.  The arcs made a
slight 'poofing' sound as they developed.  Naturally, at some
point in playing with them one gets both fingers across the
wires where the arc develops and one receives a potent shock.
Truly amazing and magical stuff.  Even small audio transformers
of the day were able to be used as high voltage spark generators
at the secondary when the primary was driven with an auto radio
vibrator.

Ah yes, semantics. The breakfast of idiots.Sorry to waste server space.  The reason why specific words are so important to some is because all they have is the tongue and no mind. This is the larger congregation of souls. This is also why idealists are crucified. One can find more bullets in the adventurer's back than arrows in the front.And we must not forget: blather, spew and repeat. But only in the codified language of the coddled masses...

And how does one get line breaks correct in the thread server?

 author=Room3327 link=topic=1585.msg69256#msg69256 date=1534528380]
     

QuoteYes let's confuse the issue again, we don't want anyone to understand this stuff, let's fight over symantics and how one person views something different over another.

I am not trying to confuse the issue,i am trying to clarify it.
If we are to get anywhere,then clarity is needed,and it appears that clarity is just not here in this simple subject.

Quotethe fact is when current is rising in a coil it self induces a reverse or counter EMF in the same coil. That voltage is reducing the applied EMF by some factor

And this is what i am talking about.
To say

Quoteinduces a reverse or counter EMF

to me sounds like a voltage of the opposite polarity is being produced by the coil. When you reverse something,that means opposite direction,or opposite polarity--as in-reverse the polarity of a circuit.
And to say that the CEMF reduces the applied EMF is just not right. The applied EMF will remain the same across the coil at all times,regardless of the CEMF value,because the CEMFs polarity is the same as,but always lower than the applied EMF.

QuoteWhen the applied emf on a coil is removed or reduced the stored energy in the coil is released and induces an emf in the coil that aids or adds to the applied emf of that coil,

As EMF denotes voltage(not current),then when the applied EMF is removed,the voltage across the coil is of the opposite polarity to that of the applied EMF. I fail to see how this adds to or aids the applied voltage.

Quotethis is called Back EMF because it is giving BACK some of the input energy.  Is that correct enough for you?

If i am to be honest,then no,it's not correct enough for me.
When an applied EMF is removed from a coil,you get what is well known as inductive kickback-or flyback. This is not BackEMF. Inductive kickback/flyback produces an EMF across the coil that is of the opposite polarity to that of what the applied EMF was. BackEMF produces a voltage that is the same polarity to that of the applied EMF.

Even wikipedia has it wrong

https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Counter-electromotive_force.html

Quote: For example, the voltage drop across an inductor is due to the induced magnetic field inside the coil.[1][2] The voltage's polarity is at every moment the reverse of the input voltage.

The voltage's polarity is not the reverse of the input voltage at all.
The CEMFs polarity across the coil is the same as the applied EMF.
The only time the voltage polarity is reverse.opposite to that of the applied EMF,is when the source EMF is interrupted/broken/disconnected-->inductive kickback.


No wonder so many get confused with this subject.

Quote Giantkiller
Ah yes, semantics. The breakfast of idiots.Sorry to waste server space.  The reason why specific words are so important to some is because all they have is the tongue and no mind.

Mock me if you will,but in this game,things like this should be well defined.


Brad

Quote from: TinMan on 2018.08.18, 00:53:35
author=Room3327 link=topic=1585.msg69256#msg69256 date=1534528380]
     


I am not trying to confuse the issue,i am trying to clarify it.
If we are to get anywhere,then clarity is needed,and it appears that clarity is just not here in this simple subject.

And this is what i am talking about.
To say  to me sounds like a voltage of the opposite polarity is being produced by the coil. When you reverse something,that means opposite direction,or opposite polarity--as in-reverse the polarity of a circuit.
And to say that the CEMF reduces the applied EMF is just not right. The applied EMF will remain the same across the coil at all times,regardless of the CEMF value,because the CEMFs polarity is the same as,but always lower than the applied EMF.

As EMF denotes voltage(not current),then when the applied EMF is removed,the voltage across the coil is of the opposite polarity to that of the applied EMF. I fail to see how this adds to or aids the applied voltage.

If i am to be honest,then no,it's not correct enough for me.
When an applied EMF is removed from a coil,you get what is well known as inductive kickback-or flyback. This is not BackEMF. Inductive kickback/flyback produces an EMF across the coil that is of the opposite polarity to that of what the applied EMF was. BackEMF produces a voltage that is the same polarity to that of the applied EMF.

Even wikipedia has it wrong

https://ipfs.io/ipfs/QmXoypizjW3WknFiJnKLwHCnL72vedxjQkDDP1mXWo6uco/wiki/Counter-electromotive_force.html

Quote: For example, the voltage drop across an inductor is due to the induced magnetic field inside the coil.[1][2] The voltage's polarity is at every moment the reverse of the input voltage.

The voltage's polarity is not the reverse of the input voltage at all.
The CEMFs polarity across the coil is the same as the applied EMF.
The only time the voltage polarity is reverse.opposite to that of the applied EMF,is when the source EMF is interrupted/broken/disconnected-->inductive kickback.


No wonder so many get confused with this subject.

Quote Giantkiller
Ah yes, semantics. The breakfast of idiots.Sorry to waste server space.  The reason why specific words are so important to some is because all they have is the tongue and no mind.

Mock me if you will,but in this game,things like this should be well defined.


Brad

I'm not mocking you Brad, I am merely trying to educate some of those that do not understand this stuff without confusing them even more, for the most part you are right, I wasn't disagreeing with you, I was trying to simplify it for others.  Yes the emf across a coil reverses with BEMF and the current continues in the same direction it was going.  The counter EMF is in reverse to the applied EMF, the applied EMF is delayed from reaching it's maximum value across the inductor until the CEMF is over or the rise in current has stopped as far as I know, and the CEMF is the reverse polarity of the applied EMF, I think your wrong there.
Room

Hi Folks,

My take on this topic is this (let's say we DC pulse a coil): when current starts flowing in a coil, a magnetic field starts also building up. Here is the "problem": the increasing magnetic field (in fact it is an electromagnetic field) induces an EMF (a potential, a voltage) in the same coil (as per Faraday induction) just because the coil surely embedded in its own changing magnetic field. This induced EMF always works against the input voltage that started to drive the input current into the coil (normal Lenz law).

I imagine this as if there are small variable voltage sources created in each turn of the coil: they immediately appear when input current starts flowing and as this current increases so does the amplitude of these internally induced voltages too. When the current reaches its maximum value possible (the 5L/R time just lapsed) and becomes steady state, then these internally induced voltages reduce to zero volt because the increasing magnetic field stops increasing any further, hence no more induction can happen in the coil. 

To model this internal voltage source, I imagine its effect as if you connect many variable 'batteries' in series with each turn of the coil, see the drawing, the switch is ON).
Now the question is the polarity direction of these internal voltage sources. We know the induced voltage opposes input current and we know the induced voltage in the coil cannot be higher than the applied voltage that started the coil current because the input current is still in the increasing state in the closed circuit. If we sum up each such voltage source, then the summed amplitude should approach that of the outside voltage source that started the input current flow but never reach it, the difference is always the I²R loss coming from coil wire resistance (+some other losses like radiation and core loss etc).

Maybe this approach helps better understand the process during the 5L/R time duration that happens in a coil when current starts flowing in it.

Gyula

Quote from: gyula on 2018.08.18, 14:09:29
Hi Folks,

My take on this topic is this (let's say we DC pulse a coil): when current starts flowing in a coil, a magnetic field starts also building up. Here is the "problem": the increasing magnetic field (in fact it is an electromagnetic field) induces an EMF (a potential, a voltage) in the same coil (as per Faraday induction) just because the coil surely embedded in its own changing magnetic field. This induced EMF always works against the input voltage that started to drive the input current into the coil (normal Lenz law).

I imagine this as if there are small variable voltage sources created in each turn of the coil: they immediately appear when input current starts flowing and as this current increases so does the amplitude of these internally induced voltages too. When the current reaches its maximum value possible (the 5L/R time just lapsed) and becomes steady state, then these internally induced voltages reduce to zero volt because the increasing magnetic field stops increasing any further, hence no more induction can happen in the coil. 

To model this internal voltage source, I imagine its effect as if you connect many variable 'batteries' in series with each turn of the coil, see the drawing, the switch is ON).
Now the question is the polarity direction of these internal voltage sources. We know the induced voltage opposes input current and we know the induced voltage in the coil cannot be higher than the applied voltage that started the coil current because the input current is still in the increasing state in the closed circuit. If we sum up each such voltage source, then the summed amplitude should approach that of the outside voltage source that started the input current flow but never reach it, the difference is always the I²R loss coming from coil wire resistance (+some other losses like radiation and core loss etc).

Maybe this approach helps better understand the process during the 5L/R time duration that happens in a coil when current starts flowing in it.

Gyula

Yes, I think this does help people to see how CEMF works, thanks Gyula.
and thank you CITFTA for the thumbs up.
Room

 author=Room3327 link=topic=1585.msg69263#msg69263 date=1534555614]
     

QuoteI'm not mocking you Brad,

I was not referring to you room.

Quotefor the most part you are right,

Which part is wrong?.

QuoteI was trying to simplify it for others.  Yes the emf across a coil reverses with BEMF and the current continues in the same direction it was going.

No,no it dose not.
BEMF is of the same polarity to that of the applied EMF
BEMF is the self generated EMF within an electric motor.
Once again,BEMF is being mixed up with inductive kickback.

QuoteThe counter EMF is in reverse to the applied EMF,

Once again,that is wrong.
The CEMF is of the same polarity to that of the applied EMF--just like BEMF is of the same polarity to that of the applied EMF in an electric motor.

Quote, and the CEMF is the reverse polarity of the applied EMF, I think your wrong there.

I am not wrong.
Both BEMF and CEMF are of the same polarity to that of the applied EMF.

When you drop a voltage across an inductor,what happens?
Why is there very little current flow through that inductor at T=0 to the first time constant?.
If the polarity produced by the CEMF or BEMF was opposite to that of the applied EMF,what would be the value of current flow at T=0 to the first time constant?
Get your self two 12 volt batteries,and hook them up reverse polarity to each other,and see what happens-how much current flow is there at the connection points?.
Then get 1 battery with say 12.6volts across it(the EMF),and one with say 12.2 volts across it(the CEMF),hook them up so as the polarity is the same(parallel),and then measure what the current flow is between them?.
My guess is you will get smoke in one test,and very little current flow in the other test.

Quotethe applied EMF is delayed from reaching it's maximum value across the inductor until the CEMF is over or the rise in current has stopped as far as I know

EMF denotes voltage,where voltage is the value of that EMF.
When you drop a voltage across an inductor,there is no delay to that voltage reaching it's maximum value,as you just dropped that voltage across the inductor.
The voltage across the inductor will always be the value you dropped across it,providing the source and connecting wires are able to handle the maximum current flow through that inductor without voltage drop.
The value of the CEMF has nothing to do with the voltage across the inductor,only the current flowing through the inductor.

If the connecting wires or source cannot maintain the voltage across the coil as the current rises,then the opposite to what you stated above is true,where the maximum voltage will appear across the coil at T=0,and then drop off as the CEMF reduces-->current rises. But this is only due to either the source or connecting wires being unable to handle the maximum current flow through the coil.

QuoteI am merely trying to educate some of those that do not understand this stuff without confusing them even more,

It would seem that the teachers need to go back to school them self,as those teachers seem to be the cause of the confusion.

QuoteI wasn't disagreeing with you,

It would seem that you are.


Brad

Quote from: gyula on 2018.08.18, 14:09:29
Hi Folks,

My take on this topic is this (let's say we DC pulse a coil): when current starts flowing in a coil, a magnetic field starts also building up. Here is the "problem": the increasing magnetic field (in fact it is an electromagnetic field) induces an EMF (a potential, a voltage) in the same coil (as per Faraday induction) just because the coil surely embedded in its own changing magnetic field. This induced EMF always works against the input voltage that started to drive the input current into the coil (normal Lenz law).

I imagine this as if there are small variable voltage sources created in each turn of the coil: they immediately appear when input current starts flowing and as this current increases so does the amplitude of these internally induced voltages too. When the current reaches its maximum value possible (the 5L/R time just lapsed) and becomes steady state, then these internally induced voltages reduce to zero volt because the increasing magnetic field stops increasing any further, hence no more induction can happen in the coil. 

To model this internal voltage source, I imagine its effect as if you connect many variable 'batteries' in series with each turn of the coil, see the drawing, the switch is ON).
Now the question is the polarity direction of these internal voltage sources. We know the induced voltage opposes input current and we know the induced voltage in the coil cannot be higher than the applied voltage that started the coil current because the input current is still in the increasing state in the closed circuit. If we sum up each such voltage source, then the summed amplitude should approach that of the outside voltage source that started the input current flow but never reach it, the difference is always the I²R loss coming from coil wire resistance (+some other losses like radiation and core loss etc).

Maybe this approach helps better understand the process during the 5L/R time duration that happens in a coil when current starts flowing in it.

Gyula

Thank you Gyula.

As you can see,the CEMF is of the same polarity to that of the applied EMF.

We also now can work out why very little current flows at from T=0 to the first time constant-->the potential difference between the EMF and CEMF is very low. As time go's by,the rate of change of the magnetic field reduces,and in turn the value of the CEMF reduces,and the potential difference between the applied EMF and CEMF increases,causing a greater current flow.
This is why i asked Room to carry out the two tests in my last reply.


Brad

Quote from: TinMan on 2018.08.19, 00:42:30
author=Room3327 link=topic=1585.msg69263#msg69263 date=1534555614]
     


I was not referring to you room.

Which part is wrong?.

No,no it dose not.
BEMF is of the same polarity to that of the applied EMF
BEMF is the self generated EMF within an electric motor.
Once again,BEMF is being mixed up with inductive kickback.

Once again,that is wrong.
The CEMF is of the same polarity to that of the applied EMF--just like BEMF is of the same polarity to that of the applied EMF in an electric motor.

I am not wrong.
Both BEMF and CEMF are of the same polarity to that of the applied EMF.

When you drop a voltage across an inductor,what happens?
Why is there very little current flow through that inductor at T=0 to the first time constant?.
If the polarity produced by the CEMF or BEMF was opposite to that of the applied EMF,what would be the value of current flow at T=0 to the first time constant?
Get your self two 12 volt batteries,and hook them up reverse polarity to each other,and see what happens-how much current flow is there at the connection points?.
Then get 1 battery with say 12.6volts across it(the EMF),and one with say 12.2 volts across it(the CEMF),hook them up so as the polarity is the same(parallel),and then measure what the current flow is between them?.
My guess is you will get smoke in one test,and very little current flow in the other test.

EMF denotes voltage,where voltage is the value of that EMF.
When you drop a voltage across an inductor,there is no delay to that voltage reaching it's maximum value,as you just dropped that voltage across the inductor.
The voltage across the inductor will always be the value you dropped across it,providing the source and connecting wires are able to handle the maximum current flow through that inductor without voltage drop.
The value of the CEMF has nothing to do with the voltage across the inductor,only the current flowing through the inductor.

If the connecting wires or source cannot maintain the voltage across the coil as the current rises,then the opposite to what you stated above is true,where the maximum voltage will appear across the coil at T=0,and then drop off as the CEMF reduces-->current rises. But this is only due to either the source or connecting wires being unable to handle the maximum current flow through the coil.


It would seem that the teachers need to go back to school them self,as those teachers seem to be the cause of the confusion.

It would seem that you are.


Brad

Brad I am not going to continue the argument, and it would seem that one of us needs to go back to school.  As no one has taken your side in this, I would reconsider WHO needs to go back to school.  I don't know where you got your erroneous information, but I like you, so I tried to be diplomatic with you on correcting your erroneous ideas but that time has passed. Get your act together and quit confusing everyone on this subject.
Room

I guess I'm not done I just spotted this. In Blue my response.

Quote from: TinMan on 2018.08.19, 00:42:30
author=Room3327 link=topic=1585.msg69263#msg69263 date=1534555614]
     


I was not referring to you room.

Which part is wrong?.

No,no it dose not.
BEMF is of the same polarity to that of the applied EMF
BEMF is the self generated EMF within an electric motor.
Once again,BEMF is being mixed up with inductive kickback.

Once again,that is wrong.
The CEMF is of the same polarity to that of the applied EMF--just like BEMF is of the same polarity to that of the applied EMF in an electric motor.

I am not wrong.
Both BEMF and CEMF are of the same polarity to that of the applied EMF.

When you drop a voltage across an inductor,what happens?
Why is there very little current flow through that inductor at T=0 to the first time constant?.
If the polarity produced by the CEMF or BEMF was opposite to that of the applied EMF,what would be the value of current flow at T=0 to the first time constant?
Get your self two 12 volt batteries,and hook them up reverse polarity to each other,and see what happens-how much current flow is there at the connection points?.
Then get 1 battery with say 12.6volts across it(the EMF),and one with say 12.2 volts across it(the CEMF),hook them up so as the polarity is the same(parallel),and then measure what the current flow is between them?.
My guess is you will get smoke in one test,and very little current flow in the other test.

EMF denotes voltage,where voltage is the value of that EMF.
When you drop a voltage across an inductor,there is no delay to that voltage reaching it's maximum value,as you just dropped that voltage across the inductor.

That is if you look at it as being in parallel with the CEMF, but in reality the CEMF is in series with the source not parallel, could this be your problem in seeing this correctly?

The voltage across the inductor will always be the value you dropped across it,providing the source and connecting wires are able to handle the maximum current flow through that inductor without voltage drop.
The value of the CEMF has nothing to do with the voltage across the inductor,only the current flowing through the inductor.

If the connecting wires or source cannot maintain the voltage across the coil as the current rises,then the opposite to what you stated above is true,where the maximum voltage will appear across the coil at T=0,and then drop off as the CEMF reduces-->current rises. But this is only due to either the source or connecting wires being unable to handle the maximum current flow through the coil.


It would seem that the teachers need to go back to school them self,as those teachers seem to be the cause of the confusion.

It would seem that you are.


Brad

Room

Dear Room,

I chime in if you do not mind with a question on one of your sentences you wrote in Reply #65 to Brad, I quote:

"Yes the emf across a coil reverses with BEMF and the current continues in the same direction it was going. The counter EMF is in reverse to the applied EMF, the applied EMF is delayed from reaching it's maximum value across the inductor until the CEMF is over or the rise in current has stopped as far as I know, and the CEMF is the reverse polarity of the applied EMF, I think your wrong there."

Please specify the circuit you had thought of when you wrote your 1st sentence. Is it a closed circuit that include a voltage source like a battery and a coil and a switch that was on? I need to know in which circuit you meant the "emf across a coil reverses with BEMF and the current continues in the same direction it was going".

If you had thought of an already open circuit where the current in a coil is just switched off, then I agree: the polarity of the appearing voltage spike across the coil (due to the sudden collapse of the field) will have a flipped polarity with respect to the input supply voltage that charged up the coil. 


You accepted my post #66 how CEMF works (at least that is how I think it works). In the drawing I had attached there, the polarity of the small CEMF 'batteries' for each turn of the coil is identical to that of the input voltage source. Of course this is a closed circuit and we need to consider the 5L/R time duration during which we examine the CEMF.

Why does Brad say this in his Reply #69:

"We also now can work out why very little current flows at from T=0 to the first time constant-->the potential difference between the EMF and CEMF is very low. As time go's by,the rate of change of the magnetic field reduces,and in turn the value of the CEMF reduces,and the potential difference between the applied EMF and CEMF increases,causing a greater current flow."

My answer is: because the amplitude of the CEMF is the highest at the t=0 and this highest value means an equal voltage value to the input supply voltage and as time starts passing from the t=0 moment, the value of CEMF will exponantially decrease. But we cannot measure CEMF during these moments (it is impossible I think), we can only measure a manifestation of it in the coil current as an exponential current flow.

I found a simulation for this here:
http://www.learnabout-electronics.org/ac_theory/images/Fig-4-5-1.swf (http://www.learnabout-electronics.org/ac_theory/images/Fig-4-5-1.swf)  (Java applet needs to be allowed to run in your computer).  The simulation is embedded into this link:
http://www.learnabout-electronics.org/ac_theory/dc_ccts45.php (http://www.learnabout-electronics.org/ac_theory/dc_ccts45.php) 

Observe the exponential curve: it has the steepest rate of change when it just starts from zero.  Mathematically, the tangent line to the curve is quasi vertical near t=0 and becomes quasi horizontal as the curve flats out i.e. the rate of change (of the current hence magnetic flux) diminishes towards a minimum at t=5L/R .

In this link they call it back EMF (bemf, BEMF) which is created during the 5L/R time duration (I would prefer counter EMF, CEMF but English is my 2nd language) so no problem for me which of them is used as long as everybody knows what circuit and under what conditions they are used in.

You may ponder why the amplitude of the CEMF or BEMF is the highest at the moment the supply voltage is switched across a coil?
What is more: why the CEMF inside a coil has the same polarity the input supply has (as shown in my drawing)?   

Because this is the only explanation which describes reality at t=0 moment: the coil current is zero and starts building up from that value. This zero current is possible if you connect two quasi identical batteries in parallel, positive to positive and negative to negative: between them there is very little, quasi zero current (there are no identical batteries to the last fraction of mVs in practice, that is).

To put this CEMF "phenomena" otherwise: the moment a voltage source appears across a coil, the coil immediately "creates" the same voltage amplitude internally and first blocks current just because the created voltage polarity is the same as that of the supply voltage.

I am not for or against anybody here, I am for correct science whenever it is possible. 

Gyula

Hi gyula, I will try and answer your questions.

Quote from: gyula on 2018.08.19, 22:09:48
Dear Room,

I chime in if you do not mind with a question on one of your sentences you wrote in Reply #65 to Brad, I quote:

"Yes the emf across a coil reverses with BEMF and the current continues in the same direction it was going. The counter EMF is in reverse to the applied EMF, the applied EMF is delayed from reaching it's maximum value across the inductor until the CEMF is over or the rise in current has stopped as far as I know, and the CEMF is the reverse polarity of the applied EMF, I think your wrong there."

Please specify the circuit you had thought of when you wrote your 1st sentence. Is it a closed circuit that include a voltage source like a battery and a coil and a switch that was on? I need to know in which circuit you meant the "emf across a coil reverses with BEMF and the current continues in the same direction it was going".

If you had thought of an already open circuit where the current in a coil is just switched off, then I agree: the polarity of the appearing voltage spike across the coil (due to the sudden collapse of the field) will have a flipped polarity with respect to the input supply voltage that charged up the coil. 

In BEMF before the EMF is removed (example) current is flowing left to right through a coil from negative to positive.  So the left end of the coil is neg. and the right end is pos. ( its connected to source Positive current flow is neg. to pos.) so now the source is removed from the coil and we start the bemf cycle, in order for the BEMF to aid or give back its energy the coil must release it in the same direction it was flowing. Which means the coil now becomes neg. on the right and pos. on the left --It reverses In order to keep current flowing in the same direction neg. to pos.

You accepted my post #66 how CEMF works (at least that is how I think it works). In the drawing I had attached there, the polarity of the small CEMF 'batteries' for each turn of the coil is identical to that of the input voltage source. Of course this is a closed circuit and we need to consider the 5L/R time duration during which we examine the CEMF.

Why does Brad say this in his Reply #69:

"We also now can work out why very little current flows at from T=0 to the first time constant-->the potential difference between the EMF and CEMF is very low. As time go's by,the rate of change of the magnetic field reduces,and in turn the value of the CEMF reduces,and the potential difference between the applied EMF and CEMF increases,causing a greater current flow."

My answer is: because the amplitude of the CEMF is the highest at the t=0 and this highest value means an equal voltage value to the input supply voltage and as time starts passing from the t=0 moment, the value of CEMF will exponantially decrease. But we cannot measure CEMF during these moments (it is impossible I think), we can only measure a manifestation of it in the coil current as an exponential current flow.

To be honest I am not certain what the question here is, at T=0 the current is almost a vertical line meaning it is increasing very rapidly as it increases more and more the CEMF is continually growing and by T=2 it is slowing down the rise in current till T=5. does this help?


I found a simulation for this here:
http://www.learnabout-electronics.org/ac_theory/images/Fig-4-5-1.swf (http://www.learnabout-electronics.org/ac_theory/images/Fig-4-5-1.swf)  (Java applet needs to be allowed to run in your computer).  The simulation is embedded into this link:
http://www.learnabout-electronics.org/ac_theory/dc_ccts45.php (http://www.learnabout-electronics.org/ac_theory/dc_ccts45.php) 

Observe the exponential curve: it has the steepest rate of change when it just starts from zero.  Mathematically, the tangent line to the curve is quasi vertical near t=0 and becomes quasi horizontal as the curve flats out i.e. the rate of change (of the current hence magnetic flux) diminishes towards a minimum at t=5L/R .

In this link they call it back EMF (bemf, BEMF) which is created during the 5L/R time duration (I would prefer counter EMF, CEMF but English is my 2nd language) so no problem for me which of them is used as long as everybody knows what circuit and under what conditions they are used in.

Many people do not know the difference between CEMF and BEMF and think they can interchange them whenever they want.

You may ponder why the amplitude of the CEMF or BEMF is the highest at the moment the supply voltage is switched across a coil?
What is more: why the CEMF inside a coil has the same polarity the input supply has (as shown in my drawing)?   

It does not have the same polarity as I just explained to Brad Because they are not in parallel they are in series with the source. The CEMF is not at it's highest at the moment the switch is closed It is at it's lowest.

this is the only explanation which describes reality at t=0 moment: the coil current is zero and starts building up from that value. This zero current is possible if you connect two quasi identical batteries in parallel, positive to positive and negative to negative: between them there is very little, quasi zero current (there are no identical batteries to the last fraction of mVs in practice, that is).

To put this CEMF "phenomena" otherwise: the moment a voltage source appears across a coil, the coil immediately "creates" the same voltage amplitude internally and first blocks current just because the created voltage polarity is the same as that of the supply voltage.

It doesn't first block current it first allows the current and then as the CEMF develops it slows it down untill the coil is fully charged, any more energy added at that point is basically wasted depending on what you want to do.

I am not for or against anybody here, I am for correct science whenever it is possible. 

I agree I am not for or against anyone here, I too am for correct science.  Room

Gyula

Hi Room,

From what you answered last night: "In BEMF before the EMF is removed (example) current is flowing left to right through a coil from negative to positive.  So the left end of the coil is neg. and the right end is pos. ( its connected to source Positive current flow is neg. to pos.) so now the source is removed from the coil and we start the bemf cycle, in order for the BEMF to aid or give back its energy the coil must release it in the same direction it was flowing. Which means the coil now becomes neg. on the right and pos. on the left --It reverses In order to keep current flowing in the same direction neg. to pos."

it comes that you considered first the charge up of a coil and then you continued with switching the current off (so now the source is removed from the coil) and that the coil polarity reverses.  So this is the whole current on and off process for a coil.

But so far in my earlier two posts I have discussed only the charge up process of a coil from the moment a voltage source is switched across it till the moment the 5L/R time duration lapses. It is important to separate this initial switch-on transient event from the switch-off event and it is the latter event you mainly include in your answers to Brad and to me (at least it sounds to me like that).

Regarding your answer: "To be honest I am not certain what the question here is, at T=0 the current is almost a vertical line meaning it is increasing very rapidly as it increases more and more the CEMF is continually growing and by T=2 it is slowing down the rise in current till T=5. does this help?"

The question was from Brad to me, in fact it was not a question but an explaining sentence, I quote the revelant part he wrote to me: "We also now can work out why very little current flows at from T=0 to the first time constant..."

Well, notice in the simulation (I linked in my earlier post) the coil current starts from a zero value and then increases rapidly, almost vertically indeed. But it still has low values after the t=0 moments just after a voltage source appears across the coil. What you wrote in the second half of your answer: "as it increases more and more the CEMF is continually growing" is what we need to discuss.

In the simulation it is the coil current which is included on the vertical axis of the graph. 
I would like to understand why do you say the CEMF is continually growing?
The CEMF is a potential, a voltage appearing within the 5L/R time duration inside the coil and it cannot be measured directly, only a manifestation of it can be measured if we measure coil current. 

So why do you say the CEMF is increasing in this situation? (i.e. when there is a closed circuit for a coil and a voltage source)

I will continue later on tonight with commenting the rest of your answers (I will be busy till then) but I think as a first step this question needs to be clarified and we can go on from that point.

Thanks, 
Gyula

Quote from: Room3327 on 2018.08.19, 16:49:01
Brad I am not going to continue the argument, and it would seem that one of us needs to go back to school.  As no one has taken your side in this, I would reconsider WHO needs to go back to school.  I don't know where you got your erroneous information, but I like you, so I tried to be diplomatic with you on correcting your erroneous ideas but that time has passed. Get your act together and quit confusing everyone on this subject.
Room

Room

I have read your last few posts,and i can tell you now--it is not me that should look more into this.

Reading one of your positions on this,i see you believe that a coil placed across a battery is a series connection,and that is where i am making my mistake.

I can assure you that it is not me making the mistake here,as a coil placed across a battery is a parallel connection. To help you understand this,first you must understand the difference between series and parallel connections.

In a series connection,the components are connected so as they all have the same current flowing through them.
Series connected circuits normally require a minimum of two components plus the source--in this case a battery.
A parallel connection is where the circuit components are connected to the source so as they all have the same potential difference across them as the source. An inductor placed across a source-such as a battery,will at all times have the same potential difference across it as that of the battery.

If you place a resistor across a battery,it is a parallel connection,as the potential difference across the resistor will be the same as that of the battery
If you place a capacitor across a battery,it is a parallel connection,as the potential difference across the capacitor will be the same as that across the battery--providing no ohmic losses through the connecting wires.
And the same applies for an inductor placed across a battery--parallel connection  O0

Quote: Many people do not know the difference between CEMF and BEMF and think they can interchange them whenever they want.

There is not a lot of difference.
CEMF is normally used in relation to AC or solid state devices,such as a voltage being dropped across an inductor.
BEMF is normally used in reference to the generating action of an electric motor-such as a PM DC motor-->but not limited to.
Both relate to magnetic fields cutting through conducting loops(coils),and producing a voltage that is of the same polarity to that which created it.

Quote: The CEMF is not at it's highest at the moment the switch is closed It is at it's lowest.

Im sorry Room,but that is also incorrect.
At T=0 to the first time constant,is when the CEMF is at it's highest value,which is why the current flow through the coil during the first time constant is at it's lowest.
The instant a voltage is dropped across a coil (T=0),is when the rate of change of the magnetic field is at it's greatest--this is why the CEMFs value is at it's maximum at this time,and why the current flow is at it's minimum--the potential difference between the applied EMF and the CEMF is at it's smallest value.

Quote: It doesn't first block current it first allows the current and then as the CEMF develops it slows it down untill the coil is fully charged,

Once again,that is incorrect,and in actual fact completely opposite to what happens.
The CEMF at T=0 is at it's highest value,and so the potential difference between the applied EMF and CEMF is at it's lowest value,and so very little current can flow.
As the magnetic field being built by the inductor starts to close in on it's maximum value,and rate of change reduces,the current flow through the coil starts to increase,due to a lower CEMF value.

Quote: In BEMF before the EMF is removed (example) current is flowing left to right through a coil from negative to positive.  So the left end of the coil is neg. and the right end is pos. ( its connected to source Positive current flow is neg. to pos.) so now the source is removed from the coil and we start the bemf cycle, in order for the BEMF to aid or give back its energy the coil must release it in the same direction it was flowing. Which means the coil now becomes neg. on the right and pos. on the left --It reverses In order to keep current flowing in the same direction neg. to pos.

Once again,BEMF and inductive kickback are being mixed up.

Once again-BEMF is not inductive kickback.
An example of BEMF--
A PM DC motor produces BEMF.
BEMF is the self generated EMF within the motor.
Once again,the BEMF produced is of the same polarity to that of the source.

Inductive kickback--
When a voltage is dropped across a coil,the magnetic field being built starts at the core(if it has one),cuts through the conducting loop's,and then into free space.
When the source is removed,the magnetic field collapses from the surrounding free space,through the conducting loop's,and back into the core. During this process,the coil now becomes the source. The magnetic field is now collapsing from free space,through the conducting loops,back into the core. This is why the voltage inverts across the coil--the magnetic field is cutting through the conductive loops in the opposite direction. It is also why the current continues through the coil in the same direction.

Inductive kickback has nothing to do with CEMF or BEMF.

I hope that helps explain where you are making your mistakes.


Brad

Quote from: Room3327 on 2018.08.19, 16:49:01
Brad I am not going to continue the argument, and it would seem that one of us needs to go back to school.  As no one has taken your side in this, I would reconsider WHO needs to go back to school.  I don't know where you got your erroneous information, but I like you, so I tried to be diplomatic with you on correcting your erroneous ideas but that time has passed. Get your act together and quit confusing everyone on this subject.
Room

Well some of us are confused,that is for sure.
This is the very reason i booted this thread up again.

I can assure you that my ideas are not erroneous,and i can prove every statement i have made on the bench without fail-i have been doing it for many years now Room.

I can show you in real experiments on the bench the value of the BEMF produced within a DC motor,and give you an exact percentage of that BEMF to that of the applied EMF.

I can show you the very same with pulsed coils,where the value of the CEMF throughout the 5 time constants can be calculated exactly,and a percentage given of the CEMF in relation to that of the applied EMF.

Can you do the same?


Brad

Quote from: TinMan on 2018.08.20, 14:22:27
Room

I have read your last few posts,and i can tell you now--it is not me that should look more into this.

Reading one of your positions on this,i see you believe that a coil placed across a battery is a series connection,and that is where i am making my mistake.

I can assure you that it is not me making the mistake here,as a coil placed across a battery is a parallel connection. To help you understand this,first you must understand the difference between series and parallel connections.

In a series connection,the components are connected so as they all have the same current flowing through them.
Series connected circuits normally require a minimum of two components plus the source--in this case a battery.
A parallel connection is where the circuit components are connected to the source so as they all have the same potential difference across them as the source. An inductor placed across a source-such as a battery,will at all times have the same potential difference across it as that of the battery.

If you place a resistor across a battery,it is a parallel connection,as the potential difference across the resistor will be the same as that of the battery
If you place a capacitor across a battery,it is a parallel connection,as the potential difference across the capacitor will be the same as that across the battery--providing no ohmic losses through the connecting wires.
And the same applies for an inductor placed across a battery--parallel connection  O0

What I am saying is THE CEMF is in SERIES with the source not the inductor there is a difference.

Quote: Many people do not know the difference between CEMF and BEMF and think they can interchange them whenever they want.

There is not a lot of difference.
CEMF is normally used in relation to AC or solid state devices,such as a voltage being dropped across an inductor.
BEMF is normally used in reference to the generating action of an electric motor-such as a PM DC motor-->but not limited to.
Both relate to magnetic fields cutting through conducting loops(coils),and producing a voltage that is of the same polarity to that which created it.

There is a tremendous difference between the two, CEMF occurs when the current is rising and opposes the rise and BEMF occurs when the current is falling, neither do the two work the same or cause the same phenomena.  Have you ever seen a high voltage spike occur with CEMF can you grab the CEMF and bottle it up in a capacitor as you can with BEMF.  They are, again, two distinctly different things.

Quote: The CEMF is not at it's highest at the moment the switch is closed It is at it's lowest.

Im sorry Room,but that is also incorrect.
At T=0 to the first time constant,is when the CEMF is at it's highest value,which is why the current flow through the coil during the first time constant is at it's lowest.
The instant a voltage is dropped across a coil (T=0),is when the rate of change of the magnetic field is at it's greatest--this is why the CEMFs value is at it's maximum at this time,and why the current flow is at it's minimum--the potential difference between the applied EMF and the CEMF is at it's smallest value.

Before the switch is thrown there is no potential there anywhere in the circuit HOW can CEMF be at it's maximum when it only occurs when current flows and the CEMF is created by that current flow.

Quote: It doesn't first block current it first allows the current and then as the CEMF develops it slows it down untill the coil is fully charged,

Once again,that is incorrect,and in actual fact completely opposite to what happens.
The CEMF at T=0 is at it's highest value,and so the potential difference between the applied EMF and CEMF is at it's lowest value,and so very little current can flow.
As the magnetic field being built by the inductor starts to close in on it's maximum value,and rate of change reduces,the current flow through the coil starts to increase,due to a lower CEMF value.

At T=0 there is no current flow and NO CEMF, where would the CEMF come from if there was some at T=0 (grab it quick sounds like free energy to me)

Quote: In BEMF before the EMF is removed (example) current is flowing left to right through a coil from negative to positive.  So the left end of the coil is neg. and the right end is pos. ( its connected to source Positive current flow is neg. to pos.) so now the source is removed from the coil and we start the bemf cycle, in order for the BEMF to aid or give back its energy the coil must release it in the same direction it was flowing. Which means the coil now becomes neg. on the right and pos. on the left --It reverses In order to keep current flowing in the same direction neg. to pos.

Once again,BEMF and inductive kickback are being mixed up.

Once again-BEMF is not inductive kickback.
An example of BEMF--
A PM DC motor produces BEMF.
BEMF is the self generated EMF within the motor.
Once again,the BEMF produced is of the same polarity to that of the source.

Inductive kickback--
When a voltage is dropped across a coil,the magnetic field being built starts at the core(if it has one),cuts through the conducting loop's,and then into free space.
When the source is removed,the magnetic field collapses from the surrounding free space,through the conducting loop's,and back into the core. During this process,the coil now becomes the source. The magnetic field is now collapsing from free space,through the conducting loops,back into the core. This is why the voltage inverts across the coil--the magnetic field is cutting through the conductive loops in the opposite direction. It is also why the current continues through the coil in the same direction.

Inductive kickback has nothing to do with CEMF or BEMF.

Inductive kickback is the SAME thing as BEMF these two are interchangeable.

I hope that helps explain where you are making your mistakes.

I hope that helps explain where you are making your mistakes.

Brad

We mustn't overlook the importance of the
electro-magnetic field and how it generates
by induction whatever voltage is developed
within the inductor to resist or oppose changes
in current flow.  That is the 'magic' of the
Inductor and its amazing properties.

The electro-magnetic field represents stored
energy.  As the intensity of the magnetic field
changes a force is generated to oppose either
an increase in the level of current flow or a
force to sustain that level of current flow when
it tends to decrease.

The abrupt cessation of current flow is the
foundational characteristic of the DC Transformer
at the heart of the Switch Mode Power Supply,
which is able to either Buck or Boost as desired.

It is unfortunate that the language and terminology
of the technical world can become corrupted by
uneducated "builders" who create their own lingo
to explain their often incorrect theories.  This lingo
catches on within the "builder" community who
proudly demonstrate their disdain for 'conventional
theory,'  and so it goes...

Those who don't know the language create their own.

What I can say? I think mistake is to consider EMF when you have to watch how current and fields interract. But you are more experienced od course

https://en.wikipedia.org/wiki/Counter-electromotive_force

These are the accepted definitions used by those trained in electronics.

Quote from: CITFTA on 2018.08.20, 18:44:00

https://en.wikipedia.org/wiki/Counter-electromotive_force

These are the accepted definitions used by those trained in electronics.

I disagree with what it says in line one, It is saying Back EMF "is the electromotive force or voltage that OPPOSES the change in current which induced it". Back EMF does not oppose the current which induced it, it AIDS it but if you believe CEMF and BEMF are the same thing, then I am talking about Inductive kickback..  I think Wikipedia is wrong as they so often are but I can work with that too.
Now if I'm wrong and I could be and CEMF and BEMF (Why would there be two names for the same thing  in electronics?)are the same exact thing fine then call, what I call BEMF, Inductive kickback.  It still doesn't change what I have been saying all along about how the two work.  CEMF is NOT at it's maximum when or before the switch is thrown. it's at it's minimum, the CEMF is in series with the source not parallel etc. If I am wrong, I will admit it and apologize, if I am right I will expect the same from those who disagree with me.

Room

HI Room,

A simple thought experiment clearly shows you are correct about CEMF being zero at the instant a voltage is applied.  Just ask what causes the CEMF to be generated?  It is generated by the current flowing through the turns of the coil.  So until there is some current flowing there can be NO CEMF.  So at the instant voltage is applied the CEMF is zero.

In the language of experimenters BEMF can mean anything.  For those of us actually trained in electronics it is the opposing voltage generated in the motor windings because those windings are moving through a magnetic field.  However you will find many people that actually work in the field that also call this force CEMF.

No one that I am aware of that has actually worked or been trained in electronics calls the inductive kick-back BEMF.  As some of you have agreed on it is not the same thing at all.

I am basing my answers on over 50 years of actually working in electronics in an industrial  environment and actually having earned a degree in electronics.

Everyone is entitled to their own opinion as to what they want to call things but it only adds confusion to those trying to actually learn the proper terms.  It also caused some major misunderstandings about what someone may be trying to say.

Respectfully,
Carroll

Quote from: CITFTA on 2018.08.20, 20:17:11
HI Room,

A simple thought experiment clearly shows you are correct about CEMF being zero at the instant a voltage is applied.  Just ask what causes the CEMF to be generated?  It is generated by the current flowing through the turns of the coil.  So until there is some current flowing there can be NO CEMF.  So at the instant voltage is applied the CEMF is zero.

In the language of experimenters BEMF can mean anything.  For those of us actually trained in electronics it is the opposing voltage generated in the motor windings because those windings are moving through a magnetic field.  However you will find many people that actually work in the field that also call this force CEMF.

No one that I am aware of that has actually worked or been trained in electronics calls the inductive kick-back BEMF.  As some of you have agreed on it is not the same thing at all.

I am basing my answers on over 50 years of actually working in electronics in an industrial  environment and actually having earned a degree in electronics.

Everyone is entitled to their own opinion as to what they want to call things but it only adds confusion to those trying to actually learn the proper terms.  It also caused some major misunderstandings about what someone may be trying to say.

Respectfully,
Carroll

Agreed, Thank you. I have worked in electronics for 56 years.
Room

Hi gentlemen,

Someone wrote the text below that I quote and I agree with it.

Inductors react against a change in current.  Applying a voltage to its terminals will produce a current that will then create a magnetic field like any other wire. An inductor is called as such because its magnetic coupling is high. In other words, the magnetic field it creates will have an effect on the element itself. If the magnectic field varies with time (which is the case when you just applied a voltage to the inductor terminals), an electric field will be induced within the wire the counter the change in magnetic field. This E-field will tend to "stop" the current that would normally be present without the inductance effect. Thus current "lags" behind voltage. Too quick of change in current creates an e-field that decreases dI/dt.
You can see an inductor like a wheel that is free to spin due to the flow of a river. If the flow changes too quickly, the wheel, due to its inertia, will tend to keep spinning for a moment. The same is true if the water of the river goes from steady to flowing in a short instant."

Reference https://www.physicsforums.com/threads/why-does-current-lag-behind-voltage-in-inductor.479918/ (https://www.physicsforums.com/threads/why-does-current-lag-behind-voltage-in-inductor.479918/)

To me the appearing E-field sounds to be the CEMF and it must be at its maximum value when a voltage is just applied across a coil because otherwise the current would not be near to zero just just pico or nano seconds later the t=0 moment.  And the CEMF must be reducing to zero within roughly the 5L/R time duration when the coil current reaches it maximum value, defined by Ohm's law. 

Gyula

PS Room, I wrote to in Reply #74...

Since I do have a copy of The Radio Amateur's Handbook
(1970) I looked into the terminology in their article about
Inductance.  Interestingly, the terms used there are induced e.m.f.
and back e.m.f.  No mention at all of Counter emf.

I also have a copy of Electricity and Magnetism by Norman
E. Gilbert (3rd Edition 1950) where the terms used are
Induced emf and Back emf.  Back emf in this book includes
self induced inductor emf and motor induced emf.  In his
description of how Induction Coils work Gilbert used the
term Induced emf to describe what we now call
'kickback' or 'flyback.'

The Allied Dictionary of Electronic Terms published by
Allied Radio, Chicago, Illinois (6th Edition 1962) defines
Counter emf and Back emf to include normal inductor
action as well as what we now know as kickback.  The
voltage developed within a motor is called Induced emf.

I'm unable to locate my Navy training manuals but I recollect
that we used Counter emf extensively.

It seems that whatever terminology we learned was very much
dependent upon where we studied.

Quote from: CITFTA on 2018.08.20, 20:17:11
HI Room,

A simple thought experiment clearly shows you are correct about CEMF being zero at the instant a voltage is applied.  Just ask what causes the CEMF to be generated?  It is generated by the current flowing through the turns of the coil.  So until there is some current flowing there can be NO CEMF.  So at the instant voltage is applied the CEMF is zero.

In the language of experimenters BEMF can mean anything.  For those of us actually trained in electronics it is the opposing voltage generated in the motor windings because those windings are moving through a magnetic field.  However you will find many people that actually work in the field that also call this force CEMF.

No one that I am aware of that has actually worked or been trained in electronics calls the inductive kick-back BEMF.  As some of you have agreed on it is not the same thing at all.

I am basing my answers on over 50 years of actually working in electronics in an industrial  environment and actually having earned a degree in electronics.

Everyone is entitled to their own opinion as to what they want to call things but it only adds confusion to those trying to actually learn the proper terms.  It also caused some major misunderstandings about what someone may be trying to say.

Respectfully,
Carroll

Carroll

Here is another thought experiment

When dose an electric motor draw the least amount of current?
1-when it starts and RPM is at it's lowest,hence BEMF is at it's lowest value--or
2-when RPMs have reached there maximum,and the BEMF is at it's maximum.

3-is the greatest rate of change of the magnetic field taking place at 1 or 2 ?.

4-is the polarity of this BEMF the same polarity to that of the source-or of the opposite polarity?


Brad

Quote from: gyula on 2018.08.20, 21:51:36
Hi gentlemen,

Someone wrote the text below that I quote and I agree with it.

Inductors react against a change in current.  Applying a voltage to its terminals will produce a current that will then create a magnetic field like any other wire. An inductor is called as such because its magnetic coupling is high. In other words, the magnetic field it creates will have an effect on the element itself. If the magnectic field varies with time (which is the case when you just applied a voltage to the inductor terminals), an electric field will be induced within the wire the counter the change in magnetic field. This E-field will tend to "stop" the current that would normally be present without the inductance effect. Thus current "lags" behind voltage. Too quick of change in current creates an e-field that decreases dI/dt.
You can see an inductor like a wheel that is free to spin due to the flow of a river. If the flow changes too quickly, the wheel, due to its inertia, will tend to keep spinning for a moment. The same is true if the water of the river goes from steady to flowing in a short instant."

Reference https://www.physicsforums.com/threads/why-does-current-lag-behind-voltage-in-inductor.479918/ (https://www.physicsforums.com/threads/why-does-current-lag-behind-voltage-in-inductor.479918/)

To me the appearing E-field sounds to be the CEMF and it must be at its maximum value when a voltage is just applied across a coil because otherwise the current would not be near to zero just just pico or nano seconds later the t=0 moment.  And the CEMF must be reducing to zero within roughly the 5L/R time duration when the coil current reaches it maximum value, defined by Ohm's law. 

Gyula

PS Room, I wrote to in Reply #74...

Very correct Gyula.

Some believe it is the current flowing throgh the coil that creates the magnetic field,when in fact it is the increasing electric field that creates the magnetic field. You simply cannot have a moving/expanding electric field without a magnetic field.

1-A voltage is dropped across a coil
2-An electric field starts expanding through the coil.
3-The expanding electric field produces a magnetic field.
4-The magnetic field cuts through the winding,producing a CEMF.
5-This CEMF,which is of the !!same!! polarity to that of the source,reduces the potential difference between the coil and source,thus by ohms law,reduces the amount of current that can flow through the resistance of the coil it's  self.

This can all be shown on the bench to be true,and the numbers add up to be exact.


Brad

Quote from: CITFTA on 2018.08.20, 20:17:11
HI Room,

A simple thought experiment clearly shows you are correct about CEMF being zero at the instant a voltage is applied.  Just ask what causes the CEMF to be generated?  It is generated by the current flowing through the turns of the coil.  So until there is some current flowing there can be NO CEMF.  So at the instant voltage is applied the CEMF is zero.


Respectfully,
Carroll

And that is where the mistake is being made.
It is the electric field that is producing the CEMF,not the current flowing through the coil.
This is why voltage leads current in an inductor.
This is why voltage across a secondary in a transformer is alway in phase with the voltage across the primary of that transformer.

The EMF across the secondary is produced by the electric field of the primary-not the current flowing through it.

Some ask why i booted this thrrad up again?.
Well,by now,im guessing you can see why.


Brad

My Navy manuals are in storage somewhere.
Here's a copy of one that is very similar.

Quote from: muDped on 2018.08.21, 03:28:35
My Navy manuals are in storage somewhere.
Here's a copy of one that is very similar.

Thanks for those papers muDped

I only hope those here will take the time to read and understand them.


Brad.

Quote from: Room3327 on 2018.08.20, 20:24:11
Agreed, Thank you. I have worked in electronics for 56 years.
Room

Many smart men once said heavier than air machines could never fly.

That statement was often made by the same people that watched birds flying over head.

I have been dabbling in electronics for maybe 10 years on a hobbiest basis.
Dose your 56 years mean you know better than me?,or maybe you have been looking at it all wrong ?

I am mostly self taught,and if something dose not sound right,i search for the truth on my bench--and i find the correct answers.

To you i may be an amature,but i stand by what i say.


Brad

 author=muDped link=topic=1585.msg69278#msg69278 date=1534786174]

QuoteIt is unfortunate that the language and terminology
of the technical world can become corrupted by
uneducated "builders" who create their own lingo
to explain their often incorrect theories.  This lingo
catches on within the "builder" community who
proudly demonstrate their disdain for 'conventional
theory,'  and so it goes...

Those who don't know the language create their own.

I see in the thread below muDped,that you also use the term flyback.

http://www.overunityresearch.com/index.php?topic=1580.0

Quote: If the forward voltage characteristic of the diode is greater than the
supply voltage (1.5 Volts max in the example) then there will only
be a small leakage current.  But, your point is well taken;  I would
much prefer that all output measurements be made across the
coil winding directly and isolated from the supply rather than as
its shown which places the coil during flyback in series aiding with
the source cell so that whatever voltage appears across the load is
the sum of Vsource and Vflyback.

Quote from: CITFTA on 2018.08.20, 20:17:11



No one that I am aware of that has actually worked or been trained in electronics calls the inductive kick-back BEMF.  As some of you have agreed on it is not the same thing at all.



Respectfully,
Carroll

That is correct Carroll

Inductive kickback is caused by an abrupt interruption of current flow from the source.
BEMF is produced while the coil is connected to the source.
The two are totally different.

Brad

 author=Room3327 link=topic=1585.msg69277#msg69277 date=1534784323]

QuoteWhat I am saying is THE CEMF is in SERIES with the source not the inductor there is a difference.

No,the CEMF is in parallel with the source EMF,as the CEMF has the same polarity across the coil as the source EMF.
This is why it is said that the CEMF apposes that of the EMF-->positive of the CEMF to positive of the applied EMF.

QuoteCEMF occurs when the current is rising and opposes the rise and BEMF occurs when the current is falling,

CEMF is the self induced EMF within inductors.
The CEMF is at it's greatest from T=0,and falls as the rate of change of the magnetic field decreases.

BEMF is the self generated EMF within electric motors.
The value of the BEMF is at it's lowest when the motor is first switched on,and at it's highest value when the motor reaches running RPM. The BEMF is produced due to the coils of the rotor cutting through the fields of the permanent magnets--referring to a PM DC motor here.

Quoteneither do the two work the same or cause the same phenomena.

They both work the same,in that they both reduce the current due to lowering the potential difference between the applied EMF and BEMF/CEMF
The greater the rate of change of the magnetic field in relation to the coils,the greater the value of the CEMF/BEMF.

QuoteHave you ever seen a high voltage spike occur with CEMF can you grab the CEMF and bottle it up in a capacitor as you can with BEMF.  They are, again, two distinctly different things.

Once again,you have mixed up BEMF with inductive flyback/kickback.
They are not the same thing--not even close.

https://www.precisionmicrodrives.com/content/ab-021-measuring-rpm-from-back-emf/
Quote: Back electromotive force (back EMF, BEMF) is a voltage that appears in the opposite direction to current flow as a result of the motor's coils moving relative to a magnetic field. It is this voltage that serves as the principle of operation for a generator.

https://www.sealevel.com/support/what-is-flyback-voltage-and-what-should-i-do-about-it/
What is flyback : A "Flyback Voltage Spike" occurs when the supply current to an inductive load such as a solenoid is interrupted.

QuoteBefore the switch is thrown there is no potential there anywhere in the circuit HOW can CEMF be at it's maximum when it only occurs when current flows and the CEMF is created by that current flow.

Well of course nothing will happen until the switch is closed.
T=0 represents the very instance the switch is closed.

Current flow dose not create the CEMF.
The CEMF is created by the electric field produced by the applied EMF.
Get your self a transformer,and drop an AC voltage across the primary.
Now,will the secondaries voltage be in phase with the primaries voltage or current?
As most of us know,the secondaries voltage wave form will be in phase with the primaries voltage wave form,and the primaries current wave form will be 90* behind.
This clearly shows you that it is the primaries electric field that is inducing the EMF across the secondary coil--not the primaries current flow.

We also know that a moving or changing electric field is accompanied by a magnetic field--you cant have one without the other. It is this magnetic field that cuts through the coil windings,creating the CEMF--not current flow.

QuoteAt T=0 there is no current flow and NO CEMF, where would the CEMF come from if there was some at T=0 (grab it quick sounds like free energy to me)

One would think that after 56 years of working electronics(as you claim)you would know your statement to be wrong.
It is as i stated above.
At T=0,the CEMF is at it's greatest value,and this value is only slightly less than that of the applied EMF value. This is why very little current flows at T=0--the potential difference between the applied EMF and CEMF is at it's lowest.
The CEMF value is high,because at T=0,the  rate of change of the magnetic field is at it's greatest. This is the same as an electric motor spinning at it's maximum RPM.

QuoteInductive kickback is the SAME thing as BEMF these two are interchangeable.

As i have stated many times before,and provided links that show you the difference,the two are very different.

QuoteI hope that helps explain where you are making your mistakes.

As i stated before,it is not me making the mistakes here--regardless of your 56 years in electronics,verses my 10 years of dabling on the bench.


Brad

You only need very fast scope 4-channel and a coil to check all this and end this topic.  O0

Just got my PC working again after moving house and I note the disagreement between two individuals here.  I see them making opposing statements that are completely opposite yet they are both correct from their different perspectives or viewpoints.  I'll elaborate on that in a later post but I will say that it took me a long time to work out why you don't get a 180 degree phase shift across a transformer, and it all comes down to perspective.  We say that there is a voltage drop across a resistor but we don't talk of a voltage drop across an inductor.  We are taught that V=-L*di/dt (note the minus sign) but also that V=omega*L*i (note the absence of a minus sign) and that second V is the AC voltage drop across the reactance.  So voltage drop is a sort of back emf and that also applies to resistors.  Apply a voltage to a resistor and clearly the polarity of the voltage across the resistor is the same as the that of the supply as seen by an external viewer.  But now take an internal view going around the circuit and what is the polarity as seen by a moving electron?  In the battery the polarity is in the same direction as the current, in the resistor it is in the opposite direction.  So we could say that the resistor creates that opposing polarity, but we don't, we call it voltage drop.
Smudge

After just reviewing texts and manuals from the past
(1945 onward) it seems that there is no standardization
in the definitions of the various terms we've been
discussing.  They are used interchangeably by the
various authors and authorities to help explain the
concept of induced voltage by a changing magnetic
field which always opposes the change.

Perhaps the concept is more important than the
necessity to label the various aspects with rigid
terminology.

'Kickback' and 'flyback (a TV term)' seem to be
more recent additions to the group of terms.

It seems there is freedom to utilize whichever term
suits ones' fancy to describe each aspect of 'induced
voltage.'

Well, perhaps we've been taught wrong all these years! :o  For a different view, look at Miles Mathis' theory on photons being the field particle rather than electrons.

http://milesmathis.com/updates.html

Regards,
Pm

Quote from: Smudge on 2018.08.21, 19:19:06
Just got my PC working again after moving house and I note the disagreement between two individuals here.  I see them making opposing statements that are completely opposite yet they are both correct from their different perspectives or viewpoints.  I'll elaborate on that in a later post but I will say that it took me a long time to work out why you don't get a 180 degree phase shift across a transformer, and it all comes down to perspective.  We say that there is a voltage drop across a resistor but we don't talk of a voltage drop across an inductor.  We are taught that V=-L*di/dt (note the minus sign) but also that V=omega*L*i (note the absence of a minus sign) and that second V is the AC voltage drop across the reactance.  So voltage drop is a sort of back emf and that also applies to resistors.  Apply a voltage to a resistor and clearly the polarity of the voltage across the resistor is the same as the that of the supply as seen by an external viewer.  But now take an internal view going around the circuit and what is the polarity as seen by a moving electron?  In the battery the polarity is in the same direction as the current, in the resistor it is in the opposite direction.  So we could say that the resistor creates that opposing polarity, but we don't, we call it voltage drop.
Smudge

That's not a quote above in red I expected to hear, but I am listening.  My knowledge of this matter was taught to me by engineers in the R&D department that I worked for at a transformer manufacturing company.  We designed and wound all types of transformers for the OEM market and our own products, I can only guess that motors and generators, which I have not worked with, may have some differences from ordinary transformers and inductors that I don't know about, but it is hard to imagine Brad and I are seeing things from opposite ends and yet we are both right?
Thank you for your reply.
Room

I thank everyone who posted on this maybe one day we can all be on the same page with this topic if we keep at it.

Quote from: Smudge on 2018.08.21, 19:19:06
Just got my PC working again after moving house and I note the disagreement between two individuals here.  I see them making opposing statements that are completely opposite yet they are both correct from their different perspectives or viewpoints.  I'll elaborate on that in a later post but I will say that it took me a long time to work out why you don't get a 180 degree phase shift across a transformer, and it all comes down to perspective.  We say that there is a voltage drop across a resistor but we don't talk of a voltage drop across an inductor.  We are taught that V=-L*di/dt (note the minus sign) but also that V=omega*L*i (note the absence of a minus sign) and that second V is the AC voltage drop across the reactance.  So voltage drop is a sort of back emf and that also applies to resistors.  Apply a voltage to a resistor and clearly the polarity of the voltage across the resistor is the same as the that of the supply as seen by an external viewer.  But now take an internal view going around the circuit and what is the polarity as seen by a moving electron?  In the battery the polarity is in the same direction as the current, in the resistor it is in the opposite direction.  So we could say that the resistor creates that opposing polarity, but we don't, we call it voltage drop.
Smudge

The phase relationship of the seconary to that of the primary will depend on what configuration your transformer is.
Take a toroid transformer for instance,where primary and secondary are wound around the core in the same direction together. The phase relationship will be in phase.
If we take a MOT for example,the phase relationship between primary to secondary will be 180* out-but that is to be expected,as the end of the primary is inducing the start of the secondary,and so the primaries output is inverted to that of the seconday. Simply swap the scope probes over on either,and now we have an in phase relationship.

But the question here is--is the polarity of the CEMF across the coil the same as that of the applied EMF

And-is the value of the CEMF at it's greatest just after T=0, or when the current flow has reached its maximum value?.

Simple really.


Brad

Quote from: Room3327 on 2018.08.21, 20:40:00

I thank everyone who posted on this maybe one day we can all be on the same page with this topic if we keep at it.

Yes.lets stick with it ,and work it all out.

I have been watching many video's on youtube,and read many electronics web pages on the subject,and all i can say is-WTF.
How can all these !so called! ex-spurts have it so wrong.

Anyway,im sure Smudge will join in here.

In the mean time,here is a video i found,that is actually correct,and shows the polarity of the CEMF to that of the applied EMF from T=0

As you can see,the polarity of the induced CEMF across the coil,and the applied EMF from the source is the same polarity.

You will also note that the instant the switch is closed,(T=0),is when the current flow is at it's minimum,and the CEMF value is at it's maximum.

https://www.youtube.com/watch?v=4LuwwIFZD_A


Brad

Here is a video showing BEMF.

Once again,you will see that the BEMF voltage polarity across the motor,is the same polarity to that of the source.

https://www.youtube.com/watch?v=qwtiX3pFCIE&t=26s


Brad

Quote from: TinMan on 2018.08.22, 09:57:26
As you can see,the polarity of the induced CEMF across the coil,and the applied EMF from the source is the same polarity.

Of course it is as measured from the common ground.  But a physicist who is interested in the forces on an electron will look t it differently.  He will look at the polarity, not as measured relative to ground, but as seen relative to the velocity direction of the moving electron.  Then the battery polarity is in the opposite direction to the battery's electron velocity while the CEMF polarity is in the same direction as the coil's electron velocity.  He looks at CEMF as being the reverse polarity to the forward EMF, one is driving the electrons and the other is retarding the electrons.  And that is where the minus sign comes in for V=-L*di/dt and for V=-N*dPhi/dt.  You are correct from your perspective, and so is the physicist from his perspective.  It's a pity we are not taught that V=-R*i for resistors, instead we we are given V=R*i and told to consider it as a voltage drop.  But clearly the voltage across the resistor is retarding the electron flow.
Smudge

Hi Smudge,

Okay, I understand what you wrote and would like to make a notice as an addition.

I took the attached picture from the video TinMan linked to ( https://www.youtube.com/watch?v=qwtiX3pFCIE&t=26s (https://www.youtube.com/watch?v=qwtiX3pFCIE&t=26s) ) and indicated the polarity of the V_Back virtual voltage source (the CEMF source) inside the motor and indicated the polarity of the 100V battery driving the motor.

So it shows that the two positives of them are common and their negatives are common, this is the only way to have the two current values calculated at start-up (10 A) and then at the full and unloaded rotor speed (1.5 A).
So the polarity of the CEMF and that of the input voltage is the same.   This has been one of the points TinMan (and I) wanted to clarify.

Thanks, Gyula

PS Edited for more clarity.

Additionally, the resistor is dissipating power as it
'resists' electron flow.  Inductors and Capacitors
dissipate very little power as they go about their 'magic.'
L and C both have a small resistive characteristic
which is a bit of a nuisance.

Quote from: Smudge on 2018.08.22, 18:27:57
Of course it is as measured from the common ground.  But a physicist who is interested in the forces on an electron will look t it differently.  He will look at the polarity, not as measured relative to ground, but as seen relative to the velocity direction of the moving electron.  Then the battery polarity is in the opposite direction to the battery's electron velocity while the CEMF polarity is in the same direction as the coil's electron velocity.  He looks at CEMF as being the reverse polarity to the forward EMF, one is driving the electrons and the other is retarding the electrons.  And that is where the minus sign comes in for V=-L*di/dt and for V=-N*dPhi/dt.  You are correct from your perspective, and so is the physicist from his perspective.  It's a pity we are not taught that V=-R*i for resistors, instead we we are given V=R*i and told to consider it as a voltage drop.  But clearly the voltage across the resistor is retarding the electron flow.
Smudge

Smudge

We are looking at the polarity of the CEMF to that of the applied EMF,not the flow of electrons.
But,even looking at the flow of electrons,we can clearly see ,although the direction of flow is opposite,the polarity across the source,and the CEMF across the coil is the same.

This is hard to show with an inductor,but very easy to show with say a PM DC motor,where we can increase the value beyond that of the applied EMF.

I believe some here are mixing up BEMF with inductive flyback,in which case the voltage across the inductor dose indeed invert.


Brad

The polarity reversal of the Induced Voltage is essential
to the operation of a Buck Converter.  The Induced
Voltage while the Inductor is charging reverses as the
Inductor goes into discharge.  Charge and Discharge
Current move in the same direction.

While a Capacitor doesn't reverse polarity while
charging or discharging, the direction of the current
does reverse.  Charge Current and Discharge Current
move in opposite directions.

Quote from: TinMan on 2018.08.23, 00:14:15
Smudge

We are looking at the polarity of the CEMF to that of the applied EMF,not the flow of electrons.
But,even looking at the flow of electrons,we can clearly see ,although the direction of flow is opposite,the polarity across the source,and the CEMF across the coil is the same.

Let's look at a dual carriageway road where there is a strong wind blowing parallel to the road.  Clearly the wind is in the same direction at both carriageways.  The polarity of the wind is the same.  But that is from the perspective of a stationary observer looking down on the road.  If you are a driver, on one carriageway you have a headwind slowing you down and on the other carriageway you have a tail wind speeding you up.  So what perspective is the most important?  You say we are not looking at the flow of electrons, yet the so called "experts" that you think have got it wrong are like the car drivers, they are interested in what is slowing the electrons down or what is speeding them up.  You are like the guy looking down onto the road, you see the wind/voltage polarity from that perspective.  Both you and the "experts" are right, so why continue arguing about it?
Smudge

Quote from: Smudge on 2018.08.23, 08:22:03
Let's look at a dual carriageway road where there is a strong wind blowing parallel to the road.  Clearly the wind is in the same direction at both carriageways.  The polarity of the wind is the same.  But that is from the perspective of a stationary observer looking down on the road.  If you are a driver, on one carriageway you have a headwind slowing you down and on the other carriageway you have a tail wind speeding you up.  So what perspective is the most important?  You say we are not looking at the flow of electrons, yet the so called "experts" that you think have got it wrong are like the car drivers, they are interested in what is slowing the electrons down or what is speeding them up.  You are like the guy looking down onto the road, you see the wind/voltage polarity from that perspective.  Both you and the "experts" are right, so why continue arguing about it?
Smudge

We were just trying to be clear as to the polarity of the CEMF,thats all.

Now-next
Which is correct ?
1-The CEMF is at it's greatest value at T=0 (moment the voltage is dropped across the coil)
2- The CEMF is NOT at it's maximum when  the switch is thrown. it's at it's minimum,


Brad

Quote from: forest on 2018.08.21, 16:29:43
You only need very fast scope 4-channel and a coil to check all this and end this topic.  O0

i tried to, but its not that easy, or be more specific on how.

see below circuit, 12V SLA, switch (mr hand) and a 5.9mH coil.

The screenshot shows the:

battery voltage (blue)
coil voltage (yellow)
coil current (green).

I don't think the yellow (coil voltage) is the Cemf, its just the voltage across he coil when switch is active.
It follows the sagging battery voltage (blue) trace.

Green is the current following the normal inductor current build up curve.

Itsu

Hi Itsu,

Would you mind measuring the DC resistance of the coil?  Just to learn about coil time constant.  (of course the resistance could be calculated from the measured current)

Thanks,, Gyula

Gyula,

the coil measures 0.6 Ohm.

Itsu

Quote from: Itsu on 2018.08.23, 15:02:58
Gyula,

the coil measures 0.6 Ohm.

Itsu

I calculate that at the point where you have 1 amp of current flowing,the CEMF will be 600mV lower than that of the source EMF

Looking at your scope shot,at the point of 1 amp of current flow,the source EMF is at 12.6v.
This means that your CEMF has a value of 12v.


Brad

Thanks Itsu.

The peak current is about 14 A and the battery voltage drops to about 9 V from the unloaded 12.6 V. The voltage drop across the internal battery resistance is 12.6-9 = 3.6 V so the battery resistance is about 3.5/14 = 0.25 Ohm. This should be added to the coil DC resistance: 0.6 + 0.25 = 0.85 Ohm so time constant, L/R is 0.0059/0.85 = 0.0069 second i.e. 6.9 ms. I indicated the switch-on moment in the scope shot as t=0 and drew the 1xL/R (6.9ms) and the 5xL/R (34.5ms) vertical lines to cut the exponential curve of the coil current.

I agree with Brad CEMF calculation at the i = 1A coil current moment.  I indicated the 1A coil current in a horizontal pink line and projected it up to cut the battery voltage.   The measured and calculated values are well within the ballpark.

Gyula

Quote from: TinMan on 2018.08.23, 13:30:17
We were just trying to be clear as to the polarity of the CEMF,thats all.

Now-next
Which is correct ?
1-The CEMF is at it's greatest value at T=0 (moment the voltage is dropped across the coil)
2- The CEMF is NOT at it's maximum when  the switch is thrown. it's at it's minimum,
Brad

1 is correct.  The moment the voltage is applied to the coil at switch closing the current (and hence the flux created by the current) starts to rise and the value of that flux rise (dPhi/dt) is at its greatest.  The dPhi/dt always creates a CEMF of a value such that the sum (you must always sum the voltages as you go round a circuit) of the forward EMF from the battery and the CEMF from the coil equates to the voltage drop across the series resistance wherever that might be (could be in the battery, the coil or as an added resistor or all three in series).  So at T=0 the initial current is 0, there is no voltage drop across the resistance and the CEMF equates in value to the supplied EMF.  Thereafter the current rises in value while its rate of change (di/dt) and hence also the flux dPhi/dt and the CEMF fall, leading to the well known exponential waveforms.  As T tends to infinity the current plateaus at a fixed value determined by the resistance, dPhi/dt tends to zero hence CEMF also goes to zero.
Smudge

OK Guys,

i have zoomed in on the switch close point and increased the vertical amplitude / div. value for yellow/blue.
I also changed the current probe to 1A/div. and moved it out of the way of the yellow trace botom line.

The blue cursors are the unloaded battery voltage versus the coil voltage at 1A time (480mV delta).

Itsu

I have added the battery resistance and the coil resistance to Itsu's circuit.  I also show the emf polarity arrows.  V₁ is battery voltage, V₂ and V₃ are the voltage drops aross the two resistors and V₄ is the CEMF.  As it's a closed circuit the voltages around the circuit must sum to zero.  Thus V1-V2-V3-V4=0 (note the minus signs!).  At any point in time the CEMF V₄ is easily calculated from Itsu's measurements where he scopes V3+V4 and also the current.  At any point in time the magnitude of the CEMF V₄ is equal to L*di/dt and also equal to N*dPhi/dt (Phi is flux, not flux-linkage) but it is normal practice to show the counter polarity by including a minus sign.
Smudge

Quote from: Smudge on 2018.08.23, 16:05:02
1 is correct.  The moment the voltage is applied to the coil at switch closing the current (and hence the flux created by the current) starts to rise and the value of that flux rise (dPhi/dt) is at its greatest.  The dPhi/dt always creates a CEMF of a value such that the sum (you must always sum the voltages as you go round a circuit) of the forward EMF from the battery and the CEMF from the coil equates to the voltage drop across the series resistance wherever that might be (could be in the battery, the coil or as an added resistor or all three in series).  So at T=0 the initial current is 0, there is no voltage drop across the resistance and the CEMF equates in value to the supplied EMF.  Thereafter the current rises in value while its rate of change (di/dt) and hence also the flux dPhi/dt and the CEMF fall, leading to the well known exponential waveforms.  As T tends to infinity the current plateaus at a fixed value determined by the resistance, dPhi/dt tends to zero hence CEMF also goes to zero.
Smudge

Thank you Smudge.

Ok guys I cannot write much because I have here only Polish language and it continously correct je while typing. What I thought is the coil with pulse square wave input current probe between input and the coil and the measure of voltage across first turn of the coil. Third and fourth channels used to measure voltage across second turn od he same coil and a shunt resistor and measure current induced. In second turn there should be the current from the source minus the induced current. Maybe we can lewej something from this setup

Smudge,  Thank you for your input on this, so just to be clear on this, TinMan and I were both right and we were both wrong.  I was wrong on the minimum and maximum CEMF/BEMF, had it backwards in my old memory and TinMan was wrong about the CEMF/BEMF being in parallel  with the source, it is actually in series with the source.  Is this correct or am I still wrong?  I really would like to come to a consensus on this matter for everyone to understand this stuff.

I am still not feeling right on the naming convention CEMF/BEMF because it happens on both current rise and current fall with AC and DC.  One reduces and one adds to the AC signal.  I think?  Should they both be called the same name, I believe it is rather confusing? I don't know the answers, but again we should nail this stuff down.

Respectfully
Room

Quote from: Room3327 on 2018.08.25, 16:58:52
Smudge,  Thank you for your input on this, so just to be clear on this, TinMan and I were both right and we were both wrong.  I was wrong on the minimum and maximum CEMF/BEMF, had it backwards in my old memory and TinMan was wrong about the CEMF/BEMF being in parallel  with the source, it is actually in series with the source.  Is this correct or am I still wrong?  I really would like to come to a consensus on this matter for everyone to understand this stuff.

I am still not feeling right on the naming convention CEMF/BEMF because it happens on both current rise and current fall with AC and DC.  One reduces and one adds to the AC signal.  I think?  Should they both be called the same name, I believe it is rather confusing? I don't know the answers, but again we should nail this stuff down.

Respectfully
Room

How close do you wish to look Room?.

First i must say the value of the EMF and CEMF are measured in volts,and volts shows us a polarity,where we have 0,ground,negative as a reference point to measure the value of the EMF.
So,the CEMF has the same polarity across the inductor as that of the applied EMF,  + to +,and - to -.

Now,in relation to Smudges comment,and most accepted ,is this-->So at T=0 the initial current is 0

Depending on how close you wish to look,and how accurate you wish to be,the stated in blue is actually incorrect.

At T=0,there is an instantaneous rise in current which come before or at the rise of voltage across the inductor.
The value of this current,and it's time duration depends on the configuration of the inductor it self. The value of this current at T=0 can be greater than half that of the steady state current through the coil after the 5th time constant. This instant rise in current is the self capacitance of the inductor being charged. This has a side effect,that being the CEMF across the coil rises to a greater value than that of the source EMF. Just after the capacitance has been charged,the resultant CEMF that now has a higher value than that of the applied EMF makes the inductor now act as a negative resistance,where the inductor becomes the source for a short time.
So now,at this point in time,our inductor is sending energy back to the source.

This is the area i am researching ATM in another thread,where there is a tight capacitive coupling between the primary and secondary of a transformer.

But like i said,it depends on who you wish to believe--those with many years of experience,or a hack(me) that messes around on the bench in his spare time.


Brad

Quote from: Room3327 on 2018.08.25, 16:58:52
Smudge,  Thank you for your input on this, so just to be clear on this, TinMan and I were both right and we were both wrong.  I was wrong on the minimum and maximum CEMF/BEMF, had it backwards in my old memory and TinMan was wrong about the CEMF/BEMF being in parallel  with the source, it is actually in series with the source.  Is this correct or am I still wrong?  I really would like to come to a consensus on this matter for everyone to understand this stuff.

The CEMF/BEMF is both in series and in parallel so you are both right and wrong!  Now that might sound daft but as always it depends on how you look at it and what you see.  We usually draw circuits with the source on the left and the load on the right both standing up from the common ground along the bottom and connected across the top.  Then you see them in parallel.  But you could also draw it with both the source and the load across the top in series, with the LH and and RH end of this series both connected down to ground.  Then you see them in series.  From the parallel viewpoint the polarity arrows both point in the same direction (say upwards) and that is Tinman's view.  From the series viewpoint the polarity arrows point in opposite directions.  For solving a closed circuit where you wish to determine the current produced by the various EMF's you must follow the series viewpoint around the circuit and sum the various EMFs using those polarity arrows. 

Quote
I am still not feeling right on the naming convention CEMF/BEMF because it happens on both current rise and current fall with AC and DC.  One reduces and one adds to the AC signal.  I think?  Should they both be called the same name, I believe it is rather confusing? I don't know the answers, but again we should nail this stuff down.

Personally I consider counter EMF and back EMF to mean the same thing, but Tinman has used references to suggest that they apply to different circumstances.  I have suggested that simple resistance produces a "counter" or "back" EMF proportional to current in order to illustrate the offending series/parallel viewpoints.  However the accepted wisdom is that these EMFs are proportional to the rate-of change of magnetic flux.  (As an aside Mr Lenz doesn't help when he talks about induced current and not induced voltage.)  Perhaps it would be better if we used a common term like MIEMF (magnetically induced emf) for all these situations, then we would not be wasting energy on arguing which term is correct.  I don't think you can say that one reduces and one adds to the AC signal without specifying the circuit you are referring to.  It is common in most AC circuits for both addition and reduction to happen at different parts of the cycle.
Smudge

Quote from: TinMan on 2018.08.26, 03:35:40
How close do you wish to look Room?.

First i must say the value of the EMF and CEMF are measured in volts,and volts shows us a polarity,where we have 0,ground,negative as a reference point to measure the value of the EMF.
So,the CEMF has the same polarity across the inductor as that of the applied EMF,  + to +,and - to -.

Now,in relation to Smudges comment,and most accepted ,is this-->So at T=0 the initial current is 0

Depending on how close you wish to look,and how accurate you wish to be,the stated in blue is actually incorrect.

At T=0,there is an instantaneous rise in current which come before or at the rise of voltage across the inductor.
The value of this current,and it's time duration depends on the configuration of the inductor it self. The value of this current at T=0 can be greater than half that of the steady state current through the coil after the 5th time constant. This instant rise in current is the self capacitance of the inductor being charged. This has a side effect,that being the CEMF across the coil rises to a greater value than that of the source EMF. Just after the capacitance has been charged,the resultant CEMF that now has a higher value than that of the applied EMF makes the inductor now act as a negative resistance,where the inductor becomes the source for a short time.
So now,at this point in time,our inductor is sending energy back to the source.

This is the area i am researching ATM in another thread,where there is a tight capacitive coupling between the primary and secondary of a transformer.

But like i said,it depends on who you wish to believe--those with many years of experience,or a hack(me) that messes around on the bench in his spare time.


Brad

Brad Thank you for your reply, yes I have a lot of years experience but that does not make me an expert on everything, I have never claimed to be an expert on anything except Islam.  You rebooted this thread to get to the bottom of CEMF/BEMF and I was trying to help, I also would like to know the bottom line on this stuff, it's OK I can be wrong, but I bet many have a better handle on this now then they did, I'm not here for me, I am just trying to help. I meant no disrespect to you and you are an excellent experimenter.
Room

Quote from: TinMan on 2018.08.26, 03:35:40
How close do you wish to look Room?.

First i must say the value of the EMF and CEMF are measured in volts,and volts shows us a polarity,where we have 0,ground,negative as a reference point to measure the value of the EMF.
So,the CEMF has the same polarity across the inductor as that of the applied EMF,  + to +,and - to -.

From your viewpoint yes, but not from the viewpoint of electrons travelling around the circuit.  It is from that other viewpoint that the words "back" and "counter" were used to describe the polarity.

Quote
Now,in relation to Smudges comment,and most accepted ,is this-->So at T=0 the initial current is 0

Depending on how close you wish to look,and how accurate you wish to be,the stated in blue is actually incorrect.

It is correct for a pure inductor and the rate of rise from zero is given by di/dt=V/L

Quote
At T=0,there is an instantaneous rise in current which come before or at the rise of voltage across the inductor.
The value of this current,and it's time duration depends on the configuration of the inductor it self. The value of this current at T=0 can be greater than half that of the steady state current through the coil after the 5th time constant. This instant rise in current is the self capacitance of the inductor being charged.

Or any other shunt capacitance that might be there.  It comes from the rate of rise of voltage and is given by i=C*dV/dt.

QuoteThis has a side effect,that being the CEMF across the coil rises to a greater value than that of the source EMF. Just after the capacitance has been charged,the resultant CEMF that now has a higher value than that of the applied EMF makes the inductor now act as a negative resistance,where the inductor becomes the source for a short time.
So now,at this point in time,our inductor is sending energy back to the source.

In any LC circuit energy can slosh back and forth

Quote
This is the area i am researching ATM in another thread,where there is a tight capacitive coupling between the primary and secondary of a transformer.

But like i said,it depends on who you wish to believe--those with many years of experience,or a hack(me) that messes around on the bench in his spare time.

It's hacks like you that will find the odd behaviour that is often hidden behind the theories.  Good luck with that.
Smudge

Quote from: Smudge on 2018.08.26, 15:02:35
The CEMF/BEMF is both in series and in parallel so you are both right and wrong!  Now that might sound daft but as always it depends on how you look at it and what you see.  We usually draw circuits with the source on the left and the load on the right both standing up from the common ground along the bottom and connected across the top.  Then you see them in parallel.  But you could also draw it with both the source and the load across the top in series, with the LH and and RH end of this series both connected down to ground.  Then you see them in series.  From the parallel viewpoint the polarity arrows both point in the same direction (say upwards) and that is Tinman's view.  From the series viewpoint the polarity arrows point in opposite directions.  For solving a closed circuit where you wish to determine the current produced by the various EMF's you must follow the series viewpoint around the circuit and sum the various EMFs using those polarity arrows. 


Personally I consider counter EMF and back EMF to mean the same thing, but Tinman has used references to suggest that they apply to different circumstances.  I have suggested that simple resistance produces a "counter" or "back" EMF proportional to current in order to illustrate the offending series/parallel viewpoints.  However the accepted wisdom is that these EMFs are proportional to the rate-of change of magnetic flux.  (As an aside Mr Lenz doesn't help when he talks about induced current and not induced voltage.)  Perhaps it would be better if we used a common term like MIEMF (magnetically induced emf) for all these situations, then we would not be wasting energy on arguing which term is correct.  I don't think you can say that one reduces and one adds to the AC signal without specifying the circuit you are referring to.  It is common in most AC circuits for both addition and reduction to happen at different parts of the cycle.
Smudge

Smudge,  Thank you for your most informative posts.  As I said above to Brad I am not a so called expert, I don't profess to be an expert in anything,  I just have a lot of knowledge in many things and probably not all correct. I do defer to your knowledge as I do look at you as the expert here.  If I have offended you or Brad in any way I sincerely apologize that is not my intention, yes I am a bit gruff, sorry, as an old biker I tend to talk a bit ruff and yes at my age I still ride my harley when ever I can. I hope this has helped people that don't have a handle on it and for me it has been fun.  Have a good day and again Thank you.
Room

Quote from: Room3327 on 2018.08.26, 18:08:34
Smudge,  Thank you for your most informative posts.  As I said above to Brad I am not a so called expert, I don't profess to be an expert in anything,  I just have a lot of knowledge in many things and probably not all correct. I do defer to your knowledge as I do look at you as the expert here.  If I have offended you or Brad in any way I sincerely apologize that is not my intention, yes I am a bit gruff, sorry, as an old biker I tend to talk a bit ruff and yes at my age I still ride my harley when ever I can. I hope this has helped people that don't have a handle on it and for me it has been fun.  Have a good day and again Thank you.
Room

Hi Room

Dont know if you are interested or not,but i have provided a close up(time expanded) scope shot view of T=0

You will note that once the coils self capacitance charges,and the current begins to fall,the CEMF value is higher than that of the applied EMF,and we now see current flowing back into the source(area between green lines).

But as i said,it depends on how accurate you want to be,and how close you want to look.


Brad

Quote from: TinMan on 2018.08.27, 12:39:22
Hi Room

Dont know if you are interested or not,but i have provided a close up(time expanded) scope shot view of T=0

You will note that once the coils self capacitance charges,and the current begins to fall,the CEMF value is higher than that of the applied EMF,and we now see current flowing back into the source(area between green lines).

But as i said,it depends on how accurate you want to be,and how close you want to look.


Brad

Hi Brad,
Just so I am on the same page, this is a scope shot of just T=0 not the entire time period is that correct.  You are saying the first peak is the self capacitance of the coil charging after which the rest of the total time period T1-T5 is then the normal looking curve.?  So what this means is there is a spike of current charging the internal capacitance before the normal curve is started if one looks close enough.  OK I can see how that can occur I just can't say I have ever seen it before, but I am not saying you are wrong.  I would like to see the entire curve T0- T5 as well as the expanded T=0 screen shot of this coil just to observe it all, interesting coil or is this from a motor?

And yes it does look like the voltage is higher then source at that point, So are you saying that the internal capacitance charging current is creating the initial maximum CEMF/BEMF at T=0 ?
Room

Quote from: Room3327 on 2018.08.27, 17:17:16
Hi Brad,
Just so I am on the same page, this is a scope shot of just T=0 not the entire time period is that correct.  You are saying the first peak is the self capacitance of the coil charging after which the rest of the total time period T1-T5 is then the normal looking curve.?  So what this means is there is a spike of current charging the internal capacitance before the normal curve is started if one looks close enough.  OK I can see how that can occur I just can't say I have ever seen it before, but I am not saying you are wrong.  I would like to see the entire curve T0- T5 as well as the expanded T=0 screen shot of this coil just to observe it all, interesting coil or is this from a motor?

And yes it does look like the voltage is higher then source at that point, So are you saying that the internal capacitance charging current is creating the initial maximum CEMF/BEMF at T=0 ?
Room

Hi Room

Yes,that is an expanded view at T=0,and the normal curve you see starts after the initial charge up of the coils self capacitance.

If you look at the current and voltage trace closely when viewing the normal curve on a scope,you can actually see the spikes at T=0.

I will get you a series of scope shots,where i expand the time durations a bit more each screen shot of the same event/pulse of the inductor when i get home from work today.

I will see if i can find a coil with a higher value self capacitance,and clean up the waveforms a bit.


Brad

Quote from: TinMan on 2018.08.28, 00:12:44
Hi Room

Yes,that is an expanded view at T=0,and the normal curve you see starts after the initial charge up of the coils self capacitance.

If you look at the current and voltage trace closely when viewing the normal curve on a scope,you can actually see the spikes at T=0.

I will get you a series of scope shots,where i expand the time durations a bit more each screen shot of the same event/pulse of the inductor when i get home from work today.

I will see if i can find a coil with a higher value self capacitance,and clean up the waveforms a bit.


Brad

Hi Brad,
Great, we can look at one piece at a time.
Room

Quote from: Room3327 on 2018.08.28, 17:30:15
Hi Brad,
Great, we can look at one piece at a time.
Room

Hi Room

I didnt make it to the bench last night,had some mechanical repairs to do on the wife's car.
Will try again tonight.


Brad

Quote from: Room3327 on 2018.08.28, 17:30:15
Hi Brad,
Great, we can look at one piece at a time.
Room

Hi Room

Below is a scope shot from T=0.
For this test i use a MOT,as the inductance value is high,and gives a better indication of what i am showing.

In the first shot,you see the normal current curve(blue trace),and the voltage trace (yellow)
If you look carefully at the first(non expanded)pic,you will notice a thin blue trace at T=0. It stands out quite well when using a MOT.
So keep an eye on the time per division setting on each scope shot,and the voltage per division on the last scope shot,where i dropped it down to 100mV P/D

So the scope shots below are all of the same scope shot,but where i froze the scope shot,and expanded the time on each.

The first is at 10uS
The second at 1uS
The third at 200nS
The forth at 50nS
On the forth shot,the V/PD was raised to 100mV,and that is over 10 ohms.
So within the first 50nS from T=0,we have a current of 40mA.

This is my area of current research.
Something very interesting happens in this area when using a bifilar wound coil,and that i will show tomorrow night.

Brad

Quote from: TinMan on 2018.08.30, 13:17:02
Hi Room

Below is a scope shot from T=0.
For this test i use a MOT,as the inductance value is high,and gives a better indication of what i am showing.

In the first shot,you see the normal current curve(blue trace),and the voltage trace (yellow)
If you look carefully at the first(non expanded)pic,you will notice a thin blue trace at T=0. It stands out quite well when using a MOT.
So keep an eye on the time per division setting on each scope shot,and the voltage per division on the last scope shot,where i dropped it down to 100mV P/D

So the scope shots below are all of the same scope shot,but where i froze the scope shot,and expanded the time on each.

The first is at 10uS
The second at 1uS
The third at 200nS
The forth at 50nS
On the forth shot,the V/PD was raised to 100mV,and that is over 10 ohms.
So within the first 50nS from T=0,we have a current of 40mA.

This is my area of current research.
Something very interesting happens in this area when using a bifilar wound coil,and that i will show tomorrow night.

Brad

Brad,

Nice!  O0

Pm

Quote from: TinMan on 2018.08.30, 13:17:02
Hi Room

Below is a scope shot from T=0.
For this test i use a MOT,as the inductance value is high,and gives a better indication of what i am showing.

In the first shot,you see the normal current curve(blue trace),and the voltage trace (yellow)
If you look carefully at the first(non expanded)pic,you will notice a thin blue trace at T=0. It stands out quite well when using a MOT.
So keep an eye on the time per division setting on each scope shot,and the voltage per division on the last scope shot,where i dropped it down to 100mV P/D

So the scope shots below are all of the same scope shot,but where i froze the scope shot,and expanded the time on each.

The first is at 10uS
The second at 1uS
The third at 200nS
The forth at 50nS
On the forth shot,the V/PD was raised to 100mV,and that is over 10 ohms.
So within the first 50nS from T=0,we have a current of 40mA.

This is my area of current research.
Something very interesting happens in this area when using a bifilar wound coil,and that i will show tomorrow night.

Brad

I agree with partzman very nice,  and Yes I think you are right and that large current peak at T=0 must be self or internal coil capacitance, What else could it be?  Is it SM's 'kick' ? I don't know but very interesting, are you working on something that keeps the 'KICK' and stops the rest of the cycle, it makes me wonder about ELI voltage leading current in an inductor although you can see it in the rest of the cycle but like you say right at T=0 must be internal capacitance, so what is this project you are working on I am very curious?
Room

Good work Brad!  Well illustrates that there is not
a perfect inductor or capacitor in the Real World.

The instant that power is applied to the inductor
a very sharp transition occurs which has very high
frequency composition.  At those frequencies, which
are far above the self-resonant frequency of the
inductor, the internal capacitance of the inductor
will dominate.

You've given us Proof Positive.

I suspect the transient response of the Capacitor is
quite similar.  At the first instant of transition it
could be expected that the capacitor's internal
inductance would dominate. But only very briefly.

The questions of "When does an inductor look like
a capacitor?" and "When does a capacitor look like
an inductor?" have been answered.

Hi Brad,

Isn't this the problem Tesla worked on early in his career,  At the moment of switch closure on motors and generators they were zapping people and Tesla fixed it for Edison and never got paid.  How did he fix it I forget?
Room

Simply stunning! so close to solution for many devices including TPU  O0

To all here.
ATM i am exploring this area,and carrying out hours of research in this area of capacitive coupling between primary and secondary coils within transformers,where the two winding are wound either one atop the other,or bifilar type.

Tonight i will post some scopeshots of a bifilar transformer,with a 1 to 1 ratio being pulsed with a 1% duty cycle.

Some of you may not believe what you see,but it is real,and may just be the secret to the TPU,or something we have been overlooking all these years.

I believe i have found a way to make electrons flow through the secondary (current in other words)without current flowing through the primary--so no input power as we know it,but output power from the secondary.

ATM i am working alongside another person,who's research is the basis of my indepth research. Once we have solid evidence of a COP+ device,im sure it will be presented. But as it stands ATM,we are still in the primary stages of research.

Brad


Brad

Wow looking forward to seeing it Brad. Is it tonight yet?

Quote from: TinMan on 2018.08.30, 13:17:02
Hi Room

Below is a scope shot from T=0.
For this test i use a MOT,as the inductance value is high,and gives a better indication of what i am showing.

In the first shot,you see the normal current curve(blue trace),and the voltage trace (yellow)
If you look carefully at the first(non expanded)pic,you will notice a thin blue trace at T=0. It stands out quite well when using a MOT.
So keep an eye on the time per division setting on each scope shot,and the voltage per division on the last scope shot,where i dropped it down to 100mV P/D

So the scope shots below are all of the same scope shot,but where i froze the scope shot,and expanded the time on each.

The first is at 10uS
The second at 1uS
The third at 200nS
The forth at 50nS
On the forth shot,the V/PD was raised to 100mV,and that is over 10 ohms.
So within the first 50nS from T=0,we have a current of 40mA.

This is my area of current research.
Something very interesting happens in this area when using a bifilar wound coil,and that i will show tomorrow night.

Brad

I used the same 50ns setting at TM, but due to my lower selfcapacitance of my coil the current charge peak
is much lower (20mA or so), see green trace below,  yellow is the voltage (12V SLA) across the coil.

But i fail to see what this has to do with the topic at hand in this thread being the CEMF, BEMF, Polarity etc.

I think as mentioned earlier here, this is a normal capactance charge current of a capacitor, in this case the
self capacitance of a coil.


Itsu

Quote from: Room3327 on 2018.08.30, 19:58:48
Hi Brad,

Isn't this the problem Tesla worked on early in his career,  At the moment of switch closure on motors and generators they were zapping people and Tesla fixed it for Edison and never got paid.  How did he fix it I forget?
Room

That always reminded of a champagne bottle I was holding, while a mate was tapping a knife between the cork and bottle as the cork had snapped off. A tiny bit of gas escaped, he moved the knife which shut it of and the the bottle exploded. I was still holding the base but he ended with 4 hours of micro surgery and rehab.

Quote from: Itsu on 2018.08.31, 10:55:58

I used the same 50ns setting at TM, but due to my lower selfcapacitance of my coil the current charge peak
is much lower (20mA or so), see green trace below,  yellow is the voltage (12V SLA) across the coil.

But i fail to see what this has to do with the topic at hand in this thread being the CEMF, BEMF, Polarity etc.

I think as mentioned earlier here, this is a normal capactance charge current of a capacitor, in this case the
self capacitance of a coil.


Itsu

If anyone else would like me to discontinue with my findings and research in this thread,please let me know here in this thread.

I will hold off for a couple of days ,until those contributing to this thread let me know.


Brad

Sounds like it deserves its own thread but happy to see it here first :) If you're seeing data that helps me better understand these concepts I'd love to see it. Sounds related to me.

Quote from: TinMan on 2018.08.31, 12:52:52
If anyone else would like me to discontinue with my findings and research in this thread,please let me know here in this thread.

I will hold off for a couple of days ,until those contributing to this thread let me know.


Brad

Hmmmm,    did i say that you should discontinue to put your findings and research in this thread?
I don't think so, i just pointed out that the noticed cap charge effect has to my knowledge nothing
to do with the Cemf/Bemf discussion, nothing more.

Itsu

Quote from: Itsu on 2018.08.31, 15:17:09
Hmmmm,    did i say that you should discontinue to put your findings and research in this thread?
I don't think so, i just pointed out that the noticed cap charge effect has to my knowledge nothing
to do with the Cemf/Bemf discussion, nothing more.

Itsu

But i fail to see what this has to do with the topic at hand in this thread being the CEMF, BEMF, Polarity etc.

To me it sounds like you are saying that what i was showing has nothing to do with the topic of this thread--so why is it here?.

Anyway,a new topic has been started,so as to keep this one on track.

Thread name is T=0


Brad

Since you asked, its there because i think it has nothing to do with the topic in this thread, simple.

That is what i was saying, NOT that you should discontinue to put your findings and research in this thread, those are your words.

Itsu

Seems some rocket scientist bumped this old thread at OU.com,so i had a read through it.

I found a post made by Poynt99 in this thread relating to BEMF and inductive kickback.

Hope you dont mind me posting your explanation here Poynt?

Quote: Perhaps a simple way to distinguish between bemf and inductive kickback (IK) is to ask yourself, "does the effect take place simultaneously, or in sequence?"

bemf is a simultaneous phenomenon, whereas the IK effect takes place in a certain sequence of events. In a motor, the instant you apply power there is very little if any bemf generated, so the input current is high, but once the motor is running at speed, the applied emf, and bemf are in opposition simultaneously. When you are switching inductors, there is a definite sequence of events; i.e. you energize a coil, then remove the applied emf, then the coil tries to continue the current flow and as such reverses its output voltage, most often in a short high voltage spike. But that is determined by the load seen by the IK spike. THAT is IK.

So generally speaking, if you are switching/pulsing inductors, you are generating/collecting IK. If you are powering motors, the bemf effect is taking place, limiting the current into the motor. Without bemf, motors would draw huge amounts of current all the time.

Hope that clears things up a bit.

So it seems that Poynt99 agree's with me,or to be more correct,i agree with Poynt.


Brad