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Author Topic: MH's ideal coil and voltage question  (Read 477894 times)

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #135 on: May 09, 2016, 10:21:11 PM »
For Brad:

Don't get put off with what may look like a bunch of Engineering mumbo-jumbo talk.  I never intended to go in this direction.  All of the mumbo-jumbo talk can be expressed in ordinary simple English that anybody that wants to learn can truly learn.  I am glad that you are all fired up about this subject.

So I am going to ask you again to rethink your stance.  Accept the fact that at 3 seconds the current has climbed to 2.4 amps in a perfectly straight line.  Assuming that you do accept this fact, then the next step for you and your peers is to figure out how and why, and how you can apply this to real life.

I am still hoping that you and your peers will be able to answer the question and explain why.  At this point the thread has been littered with clues.  The best-case scenario would be for you to leverage off of this information and brainstorm with your friends and answer the simple question about the current waveform and demonstrate that you understand the how and why of what's going on.

MileHigh

minnie

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Re: MH's ideal coil and voltage question
« Reply #136 on: May 09, 2016, 10:53:23 PM »



   Oh, when I do 4/1.66666 I get something like 2.4A. Amazing!
                John.

Magneticitist

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Re: MH's ideal coil and voltage question
« Reply #137 on: May 09, 2016, 10:53:39 PM »
The Tau business is fairly simple to explain.

When you transition from a finite Tau to an infinite Tau the current waveform goes from an inverse exponential curve to a straight line.  Note that it is a straight line with a constant slope of V/L.

So Tau at infinity just means the current trace is a perfectly straight line.  Since it is a straight line the concept of "reaching 63% of the maximum value" does not apply anymore because that concept does not exist when the current waveform is a perfectly straight line.  i.e.; "There is no time constant."

So Tau being infinity does not mean stopping current flow, it means linearly increasing current flow.

Since we are discussing limits, the only possible way for the current to flat-line at zero "forever" would be for the inductance to be infinity.  Then you have a "more real" Tau = infinity because this time L/R becomes infinity/R.

So when Tau = infinity/R that gives you the horizontal current trace stuck at zero with a slope of zero (V/infinity), whereas when Tau = L/0, you get a current trace that is a straight line with with a slope of V/L.


is it really that simple to you? how can this not be construed as utterly confusing and even contradictory? I suppose I just have it completely backwards, but isn't our ability to calculate the current at a given time based upon the constant? but yet we remove the constant and un-tether it to infinity like what would happen to any normal inductor without a time constraint with which to calculate it? how do we know the rate of charge would follow a straight line incline, isn't the magnetic field supposed to be able to follow an equally opposing one making it just a flatline forever?

if you were to 'zoom in' at roughly the 99.5% point of the inductors charge you may see that you have only just begun charging it. it would almost be like looking at this examples straight incline with an infinite or 'no time constraint'.  the idea of 'infinity' or 'zero' seem to introduce some natural 'stalemate' as another user put it. we are expected to look at 0 as a balance of opposing forces, or a neutral. an at-rest. yet in this case the 0 is just supposed to represent a lack thereof as if there is a difference. it's not something I'm tying to argue to death it's just very counter intuitive in my opinion sorry of that offends you.

EMJunkie

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Re: MH's ideal coil and voltage question
« Reply #138 on: May 09, 2016, 11:10:51 PM »
Ok,im going to need help understanding this Poynt

The zero current meaning no load is no problem,but having an infinite load(a load of infinite proportions),would this not require a current flow of infinite magnitude?,and we have no current flow.


Brad


Personally I think the term: "infinite load" is just silly.

I would prefer the term: "infinite resistance" as Current (I) is directly proportional to the Resistance (R) of the load and the Voltage (V) applied accross the Load. Ohms Law Exquation: V/R = Current (I)

With: "infinite resistance", no current can flow.

I know this is what Poynt meant, but this mix in words is confusing for others that dont know.

    Chris Sykes
        hyiq.org


MileHigh

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Re: MH's ideal coil and voltage question
« Reply #139 on: May 10, 2016, 12:20:29 AM »

is it really that simple to you? how can this not be construed as utterly confusing and even contradictory? I suppose I just have it completely backwards, but isn't our ability to calculate the current at a given time based upon the constant? but yet we remove the constant and un-tether it to infinity like what would happen to any normal inductor without a time constraint with which to calculate it? how do we know the rate of charge would follow a straight line incline, isn't the magnetic field supposed to be able to follow an equally opposing one making it just a flatline forever?

if you were to 'zoom in' at roughly the 99.5% point of the inductors charge you may see that you have only just begun charging it. it would almost be like looking at this examples straight incline with an infinite or 'no time constraint'.  the idea of 'infinity' or 'zero' seem to introduce some natural 'stalemate' as another user put it. we are expected to look at 0 as a balance of opposing forces, or a neutral. an at-rest. yet in this case the 0 is just supposed to represent a lack thereof as if there is a difference. it's not something I'm tying to argue to death it's just very counter intuitive in my opinion sorry of that offends you.

For starters, just forget about all of the talk about an inductor being an infinite value.  The take away from this is as follows:  As the resistance gets lower and lower, the time constant gets longer and longer, and the inverse exponential curve for the current waveform gets flatter and flatter.  When the resistance drops to zero it becomes a straight line and then the concept of a time constant becomes invalid.

So, how much of a mind bender is it really to go from a sloped current waveform that is nearly perfectly flat and you only start to detect a slight curve in the waveform after five hours and a sloped current waveform that is perfectly straight?

What you really want to do if you are interested is work with your peers and focus on trying to understand what the circuit is doing and and answering the question.  Focus on doing the research and trying to answer the question.  If you know what you are talking about, this is a very simple problem with a very simple explanation.

Also as a side note, if you are serious, you have to work on your language and start using commonly accepted terms and phrases to describe electronic circuits.  You need to reign in the meaningless expressions and the mixing up of variables and concepts and use the proper term for the proper concept.

Magneticitist

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Re: MH's ideal coil and voltage question
« Reply #140 on: May 10, 2016, 12:47:20 AM »
Well this is what I wanted to do from the start, but not with this imaginary test circuit.
How about we do that with the original circuit in question, with your input included?

no offense but I didn't come here to simulate how your test circuit may possibly function.
I don't really want to argue what 0 means in the real world either.

if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged, is that going to show us what's firing the transistor? according to what it seems like you were originally alluding to, then no. in order to really make use of this lesson on inductors and how we can understand what may be happening in let's say, our joule thief circuits, we need to add more information into the scenario than inductance and voltage in ideal terms. there are a lot of different ways inductors can act under varying circumstances and to assume some simple ideal DC over an ideal inductor of 5H scenario really paints some kind of picture as to 'how we should proceed' fine tuning our joule thief seems like a lesson tailored more toward spiteful cerebral-measuring than genuine intentions of shedding light on the topic (that was) at hand.

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #141 on: May 10, 2016, 12:55:09 AM »
Brad:

I am going to do a preemptive strike to get you off of your fixation so that you start to move forward and both answer the question but more importantly understand the answer and the underlying concepts.

For the first part of the question the cat is out of the bag and the formula for the current is i = 0.8*t.

It seems crazy doesn't it?  All that arguing and anguish and wild imaginary theories and the answer for the first three seconds is i = 0.8*t.

Okay, so after one second the current is 0.8 amperes.

I attached the formula from Hyperphysics for a voltage source energizing an RL circuit.  You can see the red trace for the current and the formula.  You can see the "V/R" in the formula and that might tempt you to say that that's proof that the current goes to infinity but that would be dead wrong.  The red formula does not work when the resistance becomes zero because the term "V/0" is invalid and undefined.

So now I am going to prove to you that the first part of the answer given to you is correct.  We are going to solve the formula for t=1 second, L=5 Henrys and R=0.001 ohms.   Sound fair enough?  If I am correct the two answers for the current value at one second should be very close to each other.

From the red formula:

i = 4/0.001 * (1 - 2.7182818^(-0.001/5))

i = 4000 * (1 - 2.7182818^-0.0002)

i = 4000 * (1 - 0.99980001999866673333)

i = 4000 * 0.000199980013332666693332

i = 0.799920005333 amperes.

So as you can see, when we introduced a resistor of 0.001 ohms, after one second the current was just a tiny tiny smidgen less than the 0.8 amperes you get when there is no resistor.   You note that the calculation for when there is no resistor is 10 times easier as compared to when there is a resistor.

So there is your proof in the pudding.  It's time for you to move on and try to answer the question and more importantly understand the how and the why.

MileHigh

tinman

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Re: MH's ideal coil and voltage question
« Reply #142 on: May 10, 2016, 01:12:19 AM »
The Tau business is fairly simple to explain.

When you transition from a finite Tau to an infinite Tau the current waveform goes from an inverse exponential curve to a straight line.  Note that it is a straight line with a constant slope of V/L.

So Tau at infinity just means the current trace is a perfectly straight line.  Since it is a straight line the concept of "reaching 63% of the maximum value" does not apply anymore because that concept does not exist when the current waveform is a perfectly straight line.  i.e.; "There is no time constant."

So Tau being infinity does not mean stopping current flow, it means linearly increasing current flow.

Since we are discussing limits, the only possible way for the current to flat-line at zero "forever" would be for the inductance to be infinity.  Then you have a "more real" Tau = infinity because this time L/R becomes infinity/R.

So when Tau = infinity/R that gives you the horizontal current trace stuck at zero with a slope of zero (V/infinity), whereas when Tau = L/0, you get a current trace that is a straight line with with a slope of V/L.

Im sorry MH,but that dose not work either,as the V has no slope ,due to that fact that the voltage is ideal.

I will once again post the definition of an ideal voltage.

Quote: An ideal voltage source is a two-terminal device that maintains a fixed voltage drop across its terminals. It is often used as a mathematical abstraction that simplifies the analysis of real electric circuits.

This is the definition that is used to define behaviours in circuits such as yours.
The voltage of 4 volts from your ideal voltage supply,will not alter during the 3 seconds after T=0.

Regardless of whether it is L/0 or R/0,Tau is always infinite,meaning that the current will not rise in the case of an ideal inductor.
You are trying to use real world calculations to define outcomes in an ideal situation that dose not exist,and then you are surprised by the outcome.

Poynts simulation shows the results expected,and if he ran that simulation for !lets say! 100 years,we would see the current rise expected for such a low value of resistance--there is nothing wrong with his sim.


Brad

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #143 on: May 10, 2016, 01:19:38 AM »
Well this is what I wanted to do from the start, but not with this imaginary test circuit.
How about we do that with the original circuit in question, with your input included?

no offense but I didn't come here to simulate how your test circuit may possibly function.
I don't really want to argue what 0 means in the real world either.

if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged, is that going to show us what's firing the transistor? according to what it seems like you were originally alluding to, then no. in order to really make use of this lesson on inductors and how we can understand what may be happening in let's say, our joule thief circuits, we need to add more information into the scenario than inductance and voltage in ideal terms. there are a lot of different ways inductors can act under varying circumstances and to assume some simple ideal DC over an ideal inductor of 5H scenario really paints some kind of picture as to 'how we should proceed' fine tuning our joule thief seems like a lesson tailored more toward spiteful cerebral-measuring than genuine intentions of shedding light on the topic (that was) at hand.

I will repeat to you, the circuit is only "imaginary" if you have a bad attitude and refuse to open up your mind and learn something new and you refuse to apply your knowledge.  I will repeat to you, that there is almost no difference between an inductor with zero ohms resistance and an inductor with 0.001 ohms resistance and I just proved it to you and Brad in my previous posting where I did the "real" calculation.

If you can't understand how closely an ideal inductor matches a real inductor and how much can be learned from trying to understand how an ideal inductor works then read over the thread again.  I am not here to argue that anymore.

<<<  if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged  >>>

The quoted sentence above does not really make any sense and that is the whole point of this exercise.  Understand how an inductor works so you can talk about using one sensibly and also understand how a Joule Thief works and then build better Joule Thieves.

Yes there are different ways of using inductors in circuits.  A good start is right here.  This example applies directly to the energizing phase of a Joule Thief.  If you understand the basics then moving to regular inductors is easier.  And like I told you in many circuits real inductors for all practical intents and purposes act exactly like ideal inductors.

"Spiteful cerebral-measuring" is you just being defensive because you are in the realm of real electronics here, not the usual throwing around of meaningless catch phrases.

The best thing you could do for yourself is have the same spark as Brad and follow it through and learn something real and substantial about electronics.  I will remind you, you read the question and you did not have the slightest idea what to do.  I am hoping that you and Brad arrive at a successful conclusion to this debate.

I am not going to touch any Joule Thief thread, sorry.

tinman

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Re: MH's ideal coil and voltage question
« Reply #144 on: May 10, 2016, 01:20:28 AM »
Brad:

I am going to do a preemptive strike to get you off of your fixation so that you start to move forward and both answer the question but more importantly understand the answer and the underlying concepts.

For the first part of the question the cat is out of the bag and the formula for the current is i = 0.8*t.

It seems crazy doesn't it?  All that arguing and anguish and wild imaginary theories and the answer for the first three seconds is i = 0.8*t.

Okay, so after one second the current is 0.8 amperes.

I attached the formula from Hyperphysics for a voltage source energizing an RL circuit.  You can see the red trace for the current and the formula.  You can see the "V/R" in the formula and that might tempt you to say that that's proof that the current goes to infinity but that would be dead wrong.  The red formula does not work when the resistance becomes zero because the term "V/0" is invalid and undefined.

So now I am going to prove to you that the first part of the answer given to you is correct.  We are going to solve the formula for t=1 second, L=5 Henrys and R=0.001 ohms.   Sound fair enough?  If I am correct the two answers for the current value at one second should be very close to each other.

From the red formula:

i = 4/0.001 * (1 - 2.7182818^(-0.001/5))

i = 4000 * (1 - 2.7182818^-0.0002)

i = 4000 * (1 - 0.99980001999866673333)

i = 4000 * 0.000199980013332666693332

i = 0.799920005333 amperes.

So as you can see, when we introduced a resistor of 0.001 ohms, after one second the current was just a tiny tiny smidgen less than the 0.8 amperes you get when there is no resistor.   You note that the calculation for when there is no resistor is 10 times easier as compared to when there is a resistor.

So there is your proof in the pudding.  It's time for you to move on and try to answer the question and more importantly understand the how and the why.

MileHigh

I have filled in the blanks in your diagram,and also noted that there will be no voltage curve when the voltage is ideal,and set at 4 volts for 3 seconds-as per your original question.

I am pretty sure Poynt has not yet finished looking into this,and i think you will find that i am correct when i say that your ideal voltage of 4 volts across the ideal inductor for the first 3 seconds from T=0 ,remains 4 volts,as an ideal voltage dose not change in value over time-regardless of the load,and so there is no such voltage curve as you have depicted in your diagram .

Brad

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #145 on: May 10, 2016, 01:31:08 AM »
<<< Im sorry MH,but that dose not work either,as the V has no slope ,due to that fact that the voltage is ideal. >>>

There is no slope associated with the voltage so I don't know what you are talking about.   Forget the Tau business for now, it's not relevant.

Read again, Poynt stated that his sim did not run right the first time and when he tweaked the R value it ran as expected and he is in agreement with me now.

The best thing you could do for yourself is follow-though and answer the complete question and demonstrate competence in the subject matter.  You have been given a very generous start, now it's up to you and your peers to follow through.  You still have a long way to go.

The question still has not been answered....

Magneticitist

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Re: MH's ideal coil and voltage question
« Reply #146 on: May 10, 2016, 01:31:42 AM »
I will repeat to you, the circuit is only "imaginary" if you have a bad attitude and refuse to open up your mind and learn something new and you refuse to apply your knowledge.  I will repeat to you, that there is almost no difference between an inductor with zero ohms resistance and an inductor with 0.001 ohms resistance and I just proved it to you and Brad in my previous posting where I did the "real" calculation.

If you can't understand how closely an ideal inductor matches a real inductor and how much can be learned from trying to understand how an ideal inductor works then read over the thread again.  I am not here to argue that anymore.

<<<  if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until
we can see how many seconds it takes to become what we can consider fully charged  >>>

The quoted sentence above does not really make any sense and that is the whole point of this exercise.  Understand how an inductor works so you can talk about using one sensibly and also understand how a Joule Thief works and then build better Joule Thieves.

Yes there are different ways of using inductors in circuits.  A good start is right here.  This example applies directly to the energizing phase of a Joule Thief.  If you understand the basics then moving to regular inductors is easier.  And like I told you in many circuits real inductors for all practical intents and purposes act exactly like ideal inductors.

"Spiteful cerebral-measuring" is you just being defensive because you are in the realm of real electronics here, not the usual throwing around of meaningless catch phrases.

The best thing you could do for yourself is have the same spark as Brad and follow it through and learn something real and substantial about electronics.  I will remind you, you read the question and you did not have the slightest idea what to do.  I am hoping that you and Brad arrive at a successful conclusion to this debate.

I am not going to touch any Joule Thief thread, sorry.


why do you always do that? is this sentence REALLY that hard for you to understand?
"<<<  if we were to let's say, graph out the joule thiefs inductor voltage to current ratio over time until we can see how many seconds it takes to become what we can consider fully charged  >>>"

is that really such a laymans way of stating it that you simply refuse to attempt comprehending it?

What are we doing in this exercise? Are we not basically graphing an inductor charging according to a time constant? (which in this case doesn't exist) you just happen to use a 5H inductor, nowhere near what the JT inductor is in Brads circuit, and a 0 resistance which has caused a conundrum you are still not understanding the nature of. And yet and still you are claiming to know how an inductor with 0 resistance will act.

I'm literally stating what we just did in practice by analyzing this exercise, and am comparing it to how we can use this same circuit theory analysis to gain an understanding about what is going on
with the joule thief circuit you no longer wish to shed light on.

I'm not going to keep telling you that I'm not attacking your reasoning for using such theoretical models.. It's common practice for even the greenest of EE students. I keep telling you it's not helping us understand the confusion with Brads circuit. You could have easily generated a different question that was more tailored to that specific circuit if you had some revelation you wished to portray in lesson form.


tinman

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Re: MH's ideal coil and voltage question
« Reply #147 on: May 10, 2016, 01:36:00 AM »
Below i have added the correct voltage and current traces as would be seen using an ideal voltage,and an ideal inductor.

You will note that it is also the same as what Poynt's sim showed during his 100 second run time.


Brad

P.S
I have incorrectly worded current trace !across! inductor,when it should be current through inductor-as im sure you are well aware of.

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #148 on: May 10, 2016, 01:39:21 AM »
I have filled in the blanks in your diagram,and also noted that there will be no voltage curve when the voltage is ideal,and set at 4 volts for 3 seconds-as per your original question.

I am pretty sure Poynt has not yet finished looking into this,and i think you will find that i am correct when i say that your ideal voltage of 4 volts across the ideal inductor for the first 3 seconds from T=0 ,remains 4 volts,as an ideal voltage dose not change in value over time-regardless of the load,and so there is no such voltage curve as you have depicted in your diagram .

Brad

Read my posting again and understand it, I solved for R=0.001 ohm so I was not discussing an ideal inductor there.

MileHigh

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Re: MH's ideal coil and voltage question
« Reply #149 on: May 10, 2016, 01:47:54 AM »
Magneticitist:

Sorry for being picky but I am not going to discuss Joule Thieves on this thread.  For the main coil in a Joule Thief, there is no such thing as a "inductor voltage to current ratio" and I am making a point to you about communicating effectively.  How can you expect to talk shop about Joule Thieves if the person you are talking to is a beginner that would not have a clue what you are saying when you say "inductor voltage to current ratio" which itself doesn't really make sense.  I won't go after you anymore about this but now you are fully aware.

Note, this thread is all about the energizing of an inductor.  My advice to you is to forget about Joule Thieves on this thread and focus on trying to answer the question and understand what is going on.

MileHigh