Author Topic: MH's ideal coil and voltage question (Read 487957 times)

tinman · « **Reply #1365 on:** June 28, 2016, 03:00:59 PM »

Quote from: poynt99 on June 28, 2016, 02:42:10 PM

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There is no "potential difference" with a resistor yet current flows.

If there is no potential difference across a resistor,current will not flow through it--or did i miss something here?

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Brad, IK is not the same as cemf or bemf.

IK is BackEMF.
All three are the result of the same action--a magnetic fields change over time,where the value is determined by the !rate! of change over time--among other things.

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With standard feedback in amplifiers, even with 100% feedback the signal passes through at unity gain. Something to think about perhaps.

As the coil is ideal,then any energy put into that coil,should be returned,as an ideal coil cannot dissipate energy. So for example,if we put 100 joules of energy into that coil,we should get 100 joules of energy back out of that coil--one way,by IK collection.
Dose that sound right?

Brad

tinman · « **Reply #1366 on:** June 28, 2016, 03:10:47 PM »

Some quotes by MarkE

1-That is correct. BEMF resists changes in the current that any applied voltage attempts to drive.

2-Eddy currents induce voltage that opposes the driving voltage. They are about resisting current changes.

3-Generator BEMF is not inductive kickback. But inductive kickback is definitely BEMF.

4-BEMF is very simple: It is the EMF that results from changing magnetic flux crossing perpendicular to a conductor that acts in opposition to the applied voltage across the length of the conductor.

So from what i understand,is IK BackEMF,and BackEMF/CEMF are only set apart as IK is a separate function to that of motor BackEMF and CEMF,but although the function is separate,the function that gave rise to the IK is still the same--a magnetic fields change in time.

Brad

poynt99 · « **Reply #1367 on:** June 28, 2016, 03:21:14 PM »

Quote from: tinman on June 28, 2016, 03:00:59 PM

If there is no potential difference across a resistor,current will not flow through it--or did i miss something here?

I presumed, perhaps incorrectly, that when you referred to "potential difference" you meant the difference between the emf and cemf. Was I wrong? If not, then replace the inductor with a resistor, now you have the same voltage across the voltage source as the resistor, yet current flows.

Quote

IK is BackEMF.
All three are the result of the same action--a magnetic fields change over time,where the value is determined by the !rate! of change over time--among other things.

IK generally refers to the terminal voltage when an energized inductor goes open.

Quote

As the coil is ideal,then any energy put into that coil,should be returned,as an ideal coil cannot dissipate energy. So for example,if we put 100 joules of energy into that coil,we should get 100 joules of energy back out of that coil--one way,by IK collection.
Dose that sound right?

Sure.

tinman · « **Reply #1368 on:** June 28, 2016, 03:47:27 PM »

Quote from: Erfinder on June 28, 2016, 03:26:35 PM

A good net search reveals all of this...we have been up and down this road for years....whats next....Is this leading up to something? Doesn't feel like it is....

When is the perspective going to change so that folk can begin discussing something that leads to something practical?

Regards

Something is all screwed up Erfinder.
Seems now there is to much debating what is and what is not,and far less time building and research on the bench.

It seems the more you learn,the less things make sense.

Think im going back to the !old days!,and let my bench do the talking.
I dont even know why im here,talking about a situation that dose not exist

I went further forward just building and testing,that i ever have discussing that which dose not exist.

Anyway, a few new exciting devices coming up,and a couple that i have already started on-->and one big one hiding in the background,near completion

In saying that,do you know how to cut Tungsten carbide ?

Brad

partzman · « **Reply #1369 on:** June 28, 2016, 04:26:23 PM »

Regarding the Emf (applied voltage) equaling the Cemf (induced counter-emf) in an inductor with the analogy of an amplifier with feedback raises some questions IMO. Having designed, built, and manufactured many types of amplifiers with various types of feedback loops thru the years, they all have a common summing point somewhere in the circuitry. This summing point is actually a current node where the input and inverted output voltages are converted to opposing currents in ratios to yield an overall amplifier gain that one desires. The higher the open loop gain, the more accurate the designed gain.

This all seems to fit our Emf=Cemf situation at first glance but I would like to know "where is the summing point" in a single inductor? If it exists it should be able to be shown mathematically.

We know that Emf (applied voltage to a coil) = L*dI/dt as we can easily prove this with experimentation. We can rearrange the formula to solve for any unknown variable with the remaining known factors, so we are again assured this formula is correct.

We also know that Lenz dictates that with an "induced" current, Cemf = -L*dI/dt and this again can be proven experimentally.

What is difficult for me to accept and understand is to say the the Emf=Cemf in a single inductor with a voltage applied for if they are equal mathematically, they cancel mathematically and no current flows. I have to agree with TM on this. I would be happy to see mathematical proof to the contrary.

The current in an inductor is in phase with the applied voltage to the inductor which fits the Emf equation above and this would seem to indicate that the EMF wins in the production of output current over the Cemf. If the two are in a feedback interaction as the amplifier analogy implies, what is the ratio that would produce these results? What magnitudes or Emf and Cemf would have to exist to satisfy the EMF equation? IMO, Cemf would have to equal zero or the equation is invalid!

So, IMO Cemf does not exist in a single inductor but Cmmf does as I posted earlier.

pm

Grumage · « **Reply #1370 on:** June 28, 2016, 05:32:24 PM »

" In saying that,do you know how to cut Tungsten carbide ? "

Dear Brad.

You can use any Diamond edged cutting disc and plenty of water.

One of those motor powered ceramic tile cutters should do the trick.

Cheers Grum.

MileHigh · « **Reply #1371 on:** June 28, 2016, 06:48:09 PM »

During the course of the rioting more than 1000 oscilloscopes were destroyed.

picowatt · « **Reply #1372 on:** June 28, 2016, 07:28:26 PM »

Quote from: partzman on June 28, 2016, 04:26:23 PM

Regarding the Emf (applied voltage) equaling the Cemf (induced counter-emf) in an inductor with the analogy of an amplifier with feedback raises some questions IMO. Having designed, built, and manufactured many types of amplifiers with various types of feedback loops thru the years, they all have a common summing point somewhere in the circuitry. This summing point is actually a current node where the input and inverted output voltages are converted to opposing currents in ratios to yield an overall amplifier gain that one desires. The higher the open loop gain, the more accurate the designed gain.

This all seems to fit our Emf=Cemf situation at first glance but I would like to know "where is the summing point" in a single inductor? If it exists it should be able to be shown mathematically.

We know that Emf (applied voltage to a coil) = L*dI/dt as we can easily prove this with experimentation. We can rearrange the formula to solve for any unknown variable with the remaining known factors, so we are again assured this formula is correct.

We also know that Lenz dictates that with an "induced" current, Cemf = -L*dI/dt and this again can be proven experimentally.

What is difficult for me to accept and understand is to say the the Emf=Cemf in a single inductor with a voltage applied for if they are equal mathematically, they cancel mathematically and no current flows. I have to agree with TM on this. I would be happy to see mathematical proof to the contrary.

Partzman,

I would have thought anyone familiar with the concept of negative feedback would realize that my step wise description of the action of an inductor's CEMF was actually a smooth and continuous, self-regulating process.

Looking at slices in time, I could as well describe the action of the feedback resistor of an amplifier, emitter degeneration, or various other electronic, chemical, or mechanical feedback mechanisms in a step wise fashion, even though they may actually be a smooth and continuous process.

Regarding where the "summing junction" is will depend on your measurement reference point. The rate of current change dependent voltage that is the inductor's CEMF, appears across the inductor's terminals and is "summed with" the EMF applied across those terminals.

The two formulae you posted exactly describe the feedback mechanism related to CEMF. As well, both you and Tinman are indeed correct when you state that if the CEMF equals the applied EMF, no current will flow.

However, (in the 4V across 5H example discussed) the only time the CEMF is equal to the applied EMF of 4 volts is when the RATE OF CHANGE of the current flowing thru the inductor is .8 amps per second.

If, for example, the rate of current change were to become even slightly less than .8 amps per second, the generated CEMF would also become less than 4 volts, which would allow the current flow to increase until the .8 amps per second rate of change is again achieved. It is this feedback mechanism that regulates the rate of change of the current flowing thru the inductor to be .8 amps per second.

Although discussed in a rather step-wise manner, like the negative feedback in an amplifier or the degeneration of an emitter, it is a smooth and continuous process.

PW

minnie · « **Reply #1373 on:** June 28, 2016, 08:05:38 PM »

It looks as if by what PW is saying that for the 5hy inductor with
4 volts applied what MH is saying is true. Result!!!!
John.

partzman · « **Reply #1374 on:** June 28, 2016, 08:24:47 PM »

Quote from: picowatt on June 28, 2016, 07:28:26 PM

Partzman,

I would have thought anyone familiar with the concept of negative feedback would realize that my step wise description of the action of an inductor's CEMF was actually a smooth and continuous, self-regulating process.

Looking at slices in time, I could as well describe the action of the feedback resistor of an amplifier, emitter degeneration, or various other electronic, chemical, or mechanical feedback mechanisms in a step wise fashion, even though they may actually be a smooth and continuous process.

Yes I am familiar with the dynamics of negative feedback loops and agree with what you are saying here.

Quote

Regarding where the "summing junction" is will depend on your measurement reference point. The rate of current change dependent voltage that is the inductor's CEMF, appears across the inductor's terminals and is "summed with" the EMF applied across those terminals.

This is where I have a problem when we "sum" a positive applied voltage or Emf with any value of - voltage or Cemf other than zero and still maintain the fixed applied Emf. Are we saying that +4 + (-4) = 4?

Quote

The two formulae you posted exactly describe the feedback mechanism related to CEMF. As well, both you and Tinman are indeed correct when you state that if the CEMF equals the applied EMF, no current will flow.

However, (in the 4V across 5H example discussed) the only time the CEMF is equal to the applied EMF of 4 volts is when the RATE OF CHANGE of the current flowing thru the inductor is .8 amps per second.

Respectively this seems contradictory to me. How can we have any dI or rate of change of current if Emf=Cemf? Forgive me but I just can not wrap my head around that. If this is true, what mathematical expression will support this condition?

Quote

If, for example, the rate of current change were to become even slightly slower than .8 amps per second, the generated CEMF would also become less than 4 volts, which would allow the current flow to increase until the .8 amps per second rate of change is again achieved. It is this feedback mechanism that regulates the rate of change of the current flowing thru the inductor to be .8 amps per second.

Although discussed in a a rather step-wise manner, like the negative feedback in an amplifier or the degeneration of an emitter, it is a smooth and continuous process.

PW

To respond to this, I will repeat my last paragraph of my previous post-

"The current in an inductor is in phase with the applied voltage to the inductor which fits the Emf equation above and this would seem to indicate that the EMF wins in the production of output current over the Cemf. If the two are in a feedback interaction as the amplifier analogy implies, what is the ratio that would produce these results? What magnitudes or Emf and Cemf would have to exist to satisfy the EMF equation? IMO, Cemf would have to equal zero or the equation is invalid!

So, IMO Cemf does not exist in a single inductor but Cmmf does as I posted earlier."

I might add here that IMO the nexus or feedback summing point in a single inductor is at the physical interface between turns involving the bucking or cancellation of the H field or flux field.

pm

picowatt · « **Reply #1375 on:** June 28, 2016, 08:27:06 PM »

Quote from: minnie on June 28, 2016, 08:05:38 PM

It looks as if by what PW is saying that for the 5hy inductor with
4 volts applied what MH is saying is true. Result!!!!
John.

There was never any dispute that the voltage across the inductor was 4 volts when it was connected across an ideal voltage source of 4 volts.

However, the induced voltage associated with the CEMF of the inductor, as connected, can only be observed indirectly by measuring its effect on the current flowing thru the inductor and noting that the current increases at a specific rate.

The dispute was more so with regard to also referring to the voltage across a resistor as CEMF. Although I understand what MH was saying, I believe "CEMF" should be used only as defined with regard to inductors.

PW

picowatt · « **Reply #1376 on:** June 28, 2016, 08:54:16 PM »

Quote from: partzman on June 28, 2016, 08:24:47 PM

Yes I am familiar with the dynamics of negative feedback loops and agree with what you are saying here.

This is where I have a problem when we "sum" a positive applied voltage or Emf with any value of - voltage or Cemf other than zero and still maintain the fixed applied Emf. Are we saying that +4 + (-4) = 4?

Respectively this seems contradictory to me. How can we have any dI or rate of change of current if Emf=Cemf? Forgive me but I just can not wrap my head around that. If this is true, what mathematical expression will support this condition?

To respond to this, I will repeat my last paragraph of my previous post-

"The current in an inductor is in phase with the applied voltage to the inductor which fits the Emf equation above and this would seem to indicate that the EMF wins in the production of output current over the Cemf. If the two are in a feedback interaction as the amplifier analogy implies, what is the ratio that would produce these results? What magnitudes or Emf and Cemf would have to exist to satisfy the EMF equation? IMO, Cemf would have to equal zero or the equation is invalid!

So, IMO Cemf does not exist in a single inductor but Cmmf does as I posted earlier."

I might add here that IMO the nexus or feedback summing point in a single inductor is at the physical interface between turns involving the bucking or cancellation of the H field or flux field.

pm

My apologies Partzman, but I have done about the best I can to describe CEMF in terms of being a rate of current change dependent feedback mechanism.

Perhaps someone else can do better, I'll just be repeating myself...

PW

poynt99 · « **Reply #1377 on:** June 28, 2016, 10:46:49 PM »

PW understood when I mentioned negative feedback, that it was a similar process, not a process reflecting exactly what happens in an amplifier. I even stated that it was "similar". PW got it, others apparently took it literally, current summing nodes and all.

CEMF appears to behave as a negative feedback process. How is the nfg applied? It is applied via reverse induction caused by the circulating E field, which drives the inductor current in a direction opposing the current caused by the applied Vin.

And yes, it does appear that current can and does flow when the cemf=Vin.

Current flows if the inductor is replaced with a resistor. I think we all agree with that. The voltage across the resistor is a voltage drop. It is both a voltage drop and an induced emf in the case of the inductor however. The two values happen to be the same. It is not the induced emf that is driving the opposing current though. Simple proof supporting this is if the inductor was replaced with a 4V supply, there would be no current flow, period. The mechanism limiting the current therefore must be via an induced opposing current. The feedback mechanism is via the magnetic and electric fields, which relate directly to the current in the inductor. The two are inter-linked. If the current rises, the B field rises.

Clearly cemf is not the most suitable term for this scenario, it only confuses the issue. Even the official definitions don't make sense; "Increasing current in a coil of wire will generate a counter emf which opposes the current." How does an emf oppose a current? An emf can only oppose another emf. Unless we are misinterpreting what "they" are saying. Opposing may not mean what many of us first and may still be assuming; that oppose as to actively "DRIVE" the circuit. Perhaps what they really mean is that the voltages oppose, just as the case with a voltage source and a resistor; i.e. it is a "passive" voltage drop only. The end result of the inductor's behaviour is that it behaves exactly like a dynamic resistor.

picowatt · « **Reply #1378 on:** June 28, 2016, 11:11:32 PM »

Quote from: poynt99 on June 28, 2016, 10:46:49 PM

The end result of the inductor's behaviour is that it behaves exactly like a dynamic resistor.

I am a bit uncomfortable with this statement.

The CEMF is a generated effect due to induction. A better visualization for the inductor's CEMF might be as a variable voltage source in series with a conductor. The voltage of the variable voltage source is constantly adjusted as necessary to maintain the .8 amps per second rate of change to the current flowing thru the conductor (with regard MH's applied 4 volts and 5H inductor).

PW

(Added: I am uncomfortable with the use of "dynamic resistor" as it seems to imply a dissipative mechanism)

tinman · « **Reply #1379 on:** June 29, 2016, 01:47:27 AM »

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The end result of the inductor's behaviour is that it behaves exactly like a dynamic resistor.

I think we have to leave resistors of any kind out of this,as there is no such thing as an ideal resistor.
If a resistor did not dissipate energy,then it would have a resistive value of 0 ohms,and so,you have no resistor.

A resistor has no inductance,dose not store energy,and has resistance.
An ideal coil has inductance,dose store energy,and has no resistance.

The applied EMF see's the number of turns in the coil. The CEMF also see's the same number of turns in the coil.
As soon as the EMF induced current starts to flow,an equal and opposite CEMF produced current will start to flow,due to the fact that there are no losses associated to R.
The only reason the self induced current is of a lesser value to that of the EMF induced current in a real world inductor,is due to the losses associated to R in that inductor.

Brad

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