PopularFX
Home Help Search Login Register
Welcome,Guest. Please login or register.
2020-11-25, 13:04:34
News: Check out the Benches; a place for people to moderate their own thread and document their builds and data.
If you would like your own Bench, please PM an Admin.
Most Benches are visible only to members.

Pages: [1]
Author Topic: Coil Antics  (Read 536 times)

Group: Elite Experimentalist
Hero Member
*****

Posts: 3681


Buy me some coffee
Carry over from unrelated thread.

A thought experiment of a simple circuit--shown below.
If you build the circuit to find the answer,can you keep it to your self until such time as the thought experiment has been concluded. We then can all build the circuit,and share our finding's and understandings of the inductors behavior in this circumstance.

First up,lets talk LEDs.
We have say a standard 10mm white LED,and this LED conducts fully at 3 volts.
Now,we know when we have a 3 volt potential across the LED of the correct polarity a current must flow,and the current flow is in the direction indicated by the diode arrow of the LED.

So we first look at diagram 1 (V1).
This is pretty straight forward .
During the on time,current flows through D1,L1,and R1 as indicated by the blue arrow.
When the source is switched off L1 then becomes the source,and current flows through R1 in the direction indicated by the red arrow,and through L1 in the same direction as it was when the FG was the source.

Now we look at diagram 2 (V2)
We have the same coil,wound in the same direction,but we have tapped into the center of the windings,and added two LEDs between the windings.
When the FG is on,we once again see the current flow through D1,L1,R1,and now LED2.
LED2 see's a potential of 3 volts across it,and it starts to conduct,and current flows through L1 as in the first test.
But what happens this time when the FG switches off,and the magnetic field starts to collapse ?.
As you state,we see a reversal of voltage across the coil.
We now see in excess of the needed 3 volts across LED1,of the right polarity for the LED to conduct-->but the current flow is in the wrong direction !apparently! for that LED to work  ???
So we have the correct potential now the FG has switched off,and the voltage across L1 has inverted,and enough voltage for LED1 to light,but the diode part of LED1 will not allow current to flow in that direction,as the current is suppose to continue to flow in the same direction once the source is disconnected-->right  ;D

So,will Led1 light up when the source(FG) is disconnected,and the field starts to collapse?
Or will the current continue to flow in the same direction through LED2 when the field collapses ?.


Brad
« Last Edit: 2020-11-05, 08:53:39 by TinMan »


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 3681


Buy me some coffee
Gyula's reply

Hi Brad,

I have redrawn the 2nd circuit a little, see attachment.  I think the LED diodes separate L1 coil into two parts at the tap point, I use L1a and L1b symbols.
I indicated the voltage polarities the collapsing field would create across the two 'half' coils. 
So if the 'half' coils act as generators and their voltages are in series as shown then LED 2 will continue to light up as long as the collapsing field maintain current.

So my answer is that LED 2 will remain lighted up.    Agree with this?

Notice: I edited this post to make some correction in the drawing and in the text-

Gyula


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 3681


Buy me some coffee
muDped's reply

I believe Gyula has arrived at the correct analysis.

The external resistor (R1) will experience Polarity Reversals of both Current and Voltage.

The LED configuration (essentially within the Inductor) will NOT experience Polarity Reversal.

The external resister sees the Inductor as a Boost Configuration.

The internal LED pair will see the Inductor as a Buck Configuration.


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 3681


Buy me some coffee
Gyula

I have redrawn circuit 1 so as to represent what i believe you are saying.

So it seems to me you are saying that the center tapped end's of the coil at the LEDs will retain the same polarity throughout each cycle,but the two end's of the coil(or R1) will see a polarity inversion when the input pulse shuts off.
See 1st diagram.

But now we look at it in a different way.
We now remove LED2,and we replace that with a 1 ohm resistor.
See diagram 2
Gyula,dose your theory still stand now?.
Will the polarity at R2 remain the same throughout each cycle with R2 now replacing LED2 ?


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Group: Tech Wizard
Hero Member
*****

Posts: 601

Hi Brad,

I think yes,  the polarity of the voltage across R2 will remain the same throughout each cycle. 
Interesting thought experiments for sure. 

Thanks
Gyula
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 3681


Buy me some coffee
Hi Brad,

I think yes,  the polarity of the voltage across R2 will remain the same throughout each cycle. 
Interesting thought experiments for sure. 

Thanks
Gyula

Yes,it gets the mind thinking Gyula.

However,i do not agree that the polarity across the resistor will remain the same during the inductive kickback part of the cycle. If the coils voltage inverts during the inductive kickback part of the cycle (which we all know it does),then the voltage across R2 must also invert,but the current will continue to flow in the same direction throughout the cycle as per normal.

If i am correct,and R2 is once again replaced by the LED,then LED 2 will not conduct during the kickback part of the cycle,as the polarity is incorrect for LED 2. However,if i am correct,then the polarity is correct for LED 1,but the current flow direction is incorrect !apparently!  ???

Perhaps it is time for you to build the circuit Gyula,and see what happens  O0 ,and PM me your findings.
This way,anyone else that wants to have a stab at the thought experiment part of this thread can continue to do so without knowing the outcome just yet.


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Hero Member
*****

Posts: 1461
Tinman
Quote
If i am correct,and R2 is once again replaced by the LED,then LED 2 will not conduct during the kickback part of the cycle,as the polarity is incorrect for LED 2. However,if i am correct,then the polarity is correct for LED 1,but the current flow direction is incorrect !apparently!

I did similar experiments countless times and can say as a fact LED 2 will conduct when the inductor field collapses. What really surprises me is that so few here seem to have taken the time to prove it for themselves. Decades ago, when I first read that the voltage alternates but not the current in a pulsed inductor the first thing I did was start doing experiments to understand exactly why.

Here is another interesting fact, if we induced the coil with an external coil the voltage and current would alternate lighting LED 1 and 2 but the current will not alternate when the coil is excited by conduction as depicted and only LED 2 lights. Why do you suppose that is?.

It's really interesting stuff because in the inductor circuit, when we charge the inductor with a voltage/current the source current is opposed by the coils expanding magnetic field which we call self-induction. That is on charging the magnetic field direction, expanding, wants to induce a voltage in the coil in the opposite direction to the applied voltage which limits the source current. However on discharge once the source is removed the field collapses, contracting, and there is no opposition because there is no source or voltage to oppose it.

What many have missed is that the magnetic field direction (expanding or contracting) always determines the induced voltage in the coil which can oppose or aid a source voltage/current. For example if you did not cut the source current but simply reduced it the contracting magnetic field would aid the source current. This is why inductors or chokes are often used to smooth or limit the source current. The inductor adds current when it's falling and limits it when it's rising due to the magnetic field direction.

As well, this is kind of important because it does not matter why the magnetic field changes to induce a voltage/current only that it does. Do you understand?, there could be any number of ways to do it and all are valid so long as the magnetic field changes in direction, expanding/contracting with respect to the conductors.

Regards
AC





---------------------------
I don't like morning people... or mornings or people

“Progress is impossible without change, and those who cannot change their minds cannot change anything.” George Bernard Shaw
   

Group: Tinkerer
Hero Member
*****

Posts: 2287
In Pulsed DC Applications with Switching Power Supplies employing just a single
Inductor, the inductor has been oft referred to as a DC Transformer.

In the Typical Buck Configuration where the Load is Series connected to the
Inductor, the Load will not see any change in Polarity of Load Current.

While in the Boost Configuration where a Load may be connected Across an
Inductor, the Load will see that the Inductor is capable of producing a Change
in Polarity with respect to the Source Voltage and a consequent Load Current of
Opposite Polarity to Source Current.

For some Noobs it is a difficult concept to grasp and can cause confusion until
the Full and Complete Comprehension of the Switching Power Supply is attained.

It is for that reason that long ago when I was actually teaching Advanced Electronics
that I focused on the Switching Power Supply in all of its manifestations as being the
Ideal Tool to teach how the Inductor does its Magic.

Thanks TinMan for this most excellent challenge and thanks too to all who responded
with their explanations.  Well Done!


---------------------------
Many people, especially ignorant people, want to punish you for speaking the truth, for being correct. Never apologize for being correct, or for being years ahead of your time. If you're right and you know it, speak your mind. Even if you are a minority of one, the truth is still the truth - Ghandi
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 3681


Buy me some coffee
 author=Allcanadian link=topic=3983.msg85243#msg85243 date=1604595529]
Tinman

Quote
Here is another interesting fact, if we induced the coil with an external coil the voltage and current would alternate lighting LED 1 and 2 but the current will not alternate when the coil is excited by conduction as depicted and only LED 2 lights. Why do you suppose that is?.

Well thats easy.
When the coil is induced by an external magnetic field,then the coils back EMF is what we see across the coil.
But if the coil is induced by an EMF,then that EMF is always higher than the back EMF of the coil,and so we see the induced EMF across the coil during the induction phase,not the back EMF.


Quote
I did similar experiments countless times and can say as a fact LED 2 will conduct when the inductor field collapses. What really surprises me is that so few here seem to have taken the time to prove it for themselves. Decades ago, when I first read that the voltage alternates but not the current in a pulsed inductor the first thing I did was start doing experiments to understand exactly why.

The question is-->have you done this very experiment ?.

Lets look at each turn of the inductor,and the voltage across each turn during both the on and off time of the source.

For this part of the thought experiment we are going to use a 10 turn coil.
Our source is going to apply 10 volts across the coil when switched on.
This means that during the on time of our source,each turn of the coil will have 1 volt across it-see 1st diagram below.
Now,you would have to agree that the only reason we can see 1 volt across each turn is because the coil wire has resistance. So,if we measured the voltage across 1/2 a turn,we would see .5 volts-and so on.
So we can measure a voltage difference across any resistance within the coils wire.

In the 1st diagram,we see the 1 volt measured across each turn when the switch is closed-->take note of prob orientation. And lets just say that during the off time,when the source switch is opened,we get a 3 volt spike across each turn. As noted and known,the voltage across the coil invert's when the field collapses-->once again,take note of probe orientation.
Do we agree so far AC ?.

Ok,onto the second diagram.
We have now cut the coil winding at turn 5,and bridged that cut with a resistor. For the sake of this experiment the resistors resistance value is the same as 1 turn of wire. !Remember,we can measure a voltage across each turn due to the resistance in the wire,and R2 has the same resistance value as 1 turn of wire.
 So AC,can you please place (what you believe to be)both the voltage value across R2 in each case,and the polarity of that voltage across R2 in each case.

Everyone else is free to have a go as well.
But here is the question to be answered
Will you see the polarity stay the same,as though you are measuring the current flow through that resistor,which is suppose to remain in the same direction throughout the cycle.
Or will you see a polarity flip as we do in each turn of the coils windings?.


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 3681


Buy me some coffee
Please answer previous post before taking on this one.

Ok,time to make you think even harder.

I know some,well maybe most here think they have it all sorted when it comes to coils and inductors.
In most cases that would be true. But lets continue on  O0

As you will see in the diagram below,our R2 resistor is now replaced with a nichrome resistor-same wire they make wire wound resistors out of. This resistor replaces 1 turn of wire on the coil.

As we know,during the inductive kickback part of the cycle,the current will flow through R1 as indicated by the red arrow,as the voltage across the coil inverts. But what will R2 now show us?
Will it show the direction of current flow during the inductive kickback part of the cycle,and maintain it's original polarity as during the on time from the source,or will it show a polarity inversion like every other turn of the coil will during the inductive kickback part of the cycle?  ??? . If the voltage dose invert across our nichrome resistor,why would it not invert across our R2 resistor in the previous setup ?.


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Tinkerer
Hero Member
*****

Posts: 2287
Is it important to know in advance the resistance of the Nichrome Wire Turn?

It might affect the "answer" to the question?


---------------------------
Many people, especially ignorant people, want to punish you for speaking the truth, for being correct. Never apologize for being correct, or for being years ahead of your time. If you're right and you know it, speak your mind. Even if you are a minority of one, the truth is still the truth - Ghandi
   

Group: Elite Experimentalist
Hero Member
*****

Posts: 3681


Buy me some coffee
Is it important to know in advance the resistance of the Nichrome Wire Turn?

It might affect the "answer" to the question?

Im not sure how it will effect the answer,as we are just asking as to what polarity the voltage will be across the nichrome resistor,not the value of that voltage.
But for arguments sake,let's say it's value is 1 ohm.   


Brad


---------------------------
Never let your schooling get in the way of your education.
   

Group: Tinkerer
Hero Member
*****

Posts: 2287
As at the instant of Inductive Kickback the Nichrome Wire Turn becomes equivalent to a
Source with Substantial Internal Resistance in Series with Similar Sources with smaller
Internal Resistances, it would appear that...

At some point within the Inductive Discharge the Nichrome Turn could transition from being
a Source to being a Load.

The Time Constant of the Nichrome Turn would be significantly smaller than the Time Constants
of the Copper Wire Turns which could result in its Discharge earlier than the other Copper Coils.

Similar to what can happen in a series string of Electrical Cells where one of the Cells has
Discharged more than the others and in its weakness can be overpowered by the others
to experience a Polarity Reversal.

Yep, that's my story and I'm stickin' to it. ;)

Assuming of course that the circuit operates as a Dis-Continuous Switching arrangement.
« Last Edit: 2020-11-07, 04:34:04 by muDped »


---------------------------
Many people, especially ignorant people, want to punish you for speaking the truth, for being correct. Never apologize for being correct, or for being years ahead of your time. If you're right and you know it, speak your mind. Even if you are a minority of one, the truth is still the truth - Ghandi
   
Group: Tech Wizard
Hero Member
*****

Posts: 601
Hi Brad,

I have sent you a PM.

Gyula
   
Hero Member
*****

Posts: 1461
Tinman
Quote
I know some,well maybe most here think they have it all sorted when it comes to coils and inductors.
In most cases that would be true. But lets continue on

I would agree, because it was never the coil in itself but it's energy state.

As Tesla implied, a simple coil could cook us internally if it was excited and radiated EM waves in the microwave range. It could slowly kill us in some frequency ranges causing cellular degeneration like cancer or scramble our DNA in others. Thus the coil in itself is obviously not a cause of anything in itself but a response to some external force applied to it.

In fact, a coil is only 1% material moving near the speed of light and 99% free space full of EM wave energy so what is it ... exactly?. As Tesla implied, I impart a force and the material becomes radiant, particulate, carrying a charge with it through some distance. It was just a coil, but now it is something else which carries energy with it, so what is it?. The fact remains that most peoples supposition of what something is with respect to a material thing has no basis in reality. It is only 1% material which means there perspective is 99% flawed based on the reality we already know as scientific facts.

The fact remains that nuclear energy, aka atomic bomb, is based upon the velocity of very small particles in material things moving at horrific velocities liberated within a small time frame. Now we know all matter contains this energy thus at some point one would think we could liberate it in other ways. So if somehow we could liberate even a fraction of a fraction of 1% of this atomic energy in a coil in a circuit the energy would be enormous by any standard.

So... it's a brain fart at best because the scientific facts we know are infinitely stranger than what most supposedly normal people can even imagine. Thus this supposition of normal is more ignorance to the scientific facts we should already know from a high school education in my opinion. I mean we all went to school didn't we?, the atom, molecules, stuff moving near the speed of light... what part do people not understand?.

Regards
AC
« Last Edit: 2020-11-11, 18:31:08 by Allcanadian »


---------------------------
I don't like morning people... or mornings or people

“Progress is impossible without change, and those who cannot change their minds cannot change anything.” George Bernard Shaw
   
Pages: [1]
« previous next »


 

Home Help Search Login Register
Theme © PopularFX | Based on PFX Ideas! | Scripts from iScript4u 2020-11-25, 13:04:34