Joule Thief 101

[tinman]

Faradays Law predicts no such thing!!!

Again, E.M.F = -N dϕ/dt

The (-) sign is 180 Degrees, not 90 Degrees. He infered your Current and Voltage were the same.

See attached Image:

Mate, I am only standing for this because its wrong, its not right we let others learning, see these errors and believe them to be fact.

PW, Brad, others, I am sorry but its true, errors should be corrected!

   Chris Sykes
       hyiq.org

Brad, this quote is wrong:

Brad, your scope shot shows the expected 90 degree phase difference as predicted by Faraday's Law:

Actually,that quote is correct in regards to the scope shot i provided with the schematic and scope probe placements.
According to Faraday's law of electromagnetic induction, rate of change of flux linkages is equal to the induced emf,so the induced EMF will be greatest when the rate of change of flux is greatest.
If we look at the schematic below,with the attached scope shot,we can see that the EMF is indeed greatest when the rate of change of flux is greatest,and the EMF is 0 when the rate of change of magnetic flux is 0.

The blue trace is our current,and from that current trace we can see when the rate of change of magnetic flux will be greatest,which is when the currents wave form is passing through the 0 volt line on the scope,at which point we can see that the induced EMF is at it's highest value.. When the current is at it's peak,then the rate of change of magnetic flux is at it's lowest-even though the magnetic flux will be at it's greatest,at which point we can see the induced EMF is 0.

Remember-the question was asked based around the schematic and associated scope shot supplied. If a resistive load is placed on the L2 coil,then the current trace on L1,and the voltage trace(our EMF trace)on L2 will then become very close to 180* to each other. This is once again assuming that we are operating the transformer at it's designed frequency,load,and the load is resistive.

I think maybe that you have either-
1-seen the current trace as showing the peak as being the greatest rate of change of the magnetic flux
Or 2- missing the fact that L2 (the secondary coil) has no load on it,where as if it did,then the EMF of L2 would shift a further 90*(there about's) to line up with the current now flowing through it,which would then show a 180* (or close to) phase difference to that of L1s current trace.

So to sum up-
TK is correct,and Faradays law of induction holds true for the scope shot given with the attached schematic.
Only when a load is attached to L2(our secondary) dose -N dϕ/dt come into play,as L2 will only produce it's own CEMF when a current is flowing through it,and this CEMF(as we know)will be 180*(or close to) the EMF that induced it.

Hope that clears that up.


Brad

[EMJunkie]

Brad, your scope shot shows the expected 90 degree phase difference as predicted by Faraday's Law:

Actually,that quote is correct in regards to the scope shot i provided with the schematic and scope probe placements.
According to Faraday's law of electromagnetic induction, rate of change of flux linkages is equal to the induced emf,so the induced EMF will be greatest when the rate of change of flux is greatest.
If we look at the schematic below,with the attached scope shot,we can see that the EMF is indeed greatest when the rate of change of flux is greatest,and the EMF is 0 when the rate of change of magnetic flux is 0.

The blue trace is our current,and from that current trace we can see when the rate of change of magnetic flux will be greatest,which is when the currents wave form is passing through the 0 volt line on the scope,at which point we can see that the induced EMF is at it's highest value.. When the current is at it's peak,then the rate of change of magnetic flux is at it's lowest-even though the magnetic flux will be at it's greatest,at which point we can see the induced EMF is 0.

Remember-the question was asked based around the schematic and associated scope shot supplied. If a resistive load is placed on the L2 coil,then the current trace on L1,and the voltage trace(our EMF trace)on L2 will then become very close to 180* to each other. This is once again assuming that we are operating the transformer at it's designed frequency,load,and the load is resistive.

I think maybe that you have either-
1-seen the current trace as showing the peak as being the greatest rate of change of the magnetic flux
Or 2- missing the fact that L2 (the secondary coil) has no load on it,where as if it did,then the EMF of L2 would shift a further 90*(there about's) to line up with the current now flowing through it,which would then show a 180* (or close to) phase difference to that of L1s current trace.

So to sum up-
TK is correct,and Faradays law of induction holds true for the scope shot given with the attached schematic.
Only when a load is attached to L2(our secondary) dose -N dϕ/dt come into play,as L2 will only produce it's own CEMF when a current is flowing through it.

Hope that clears that up.


Brad

Hey Brad, I think we will have to agree to disagree on this one.

EMF is not Current. EMF is "Columb's of Charge" measured in Volts. Which is a quantity that can be measured on a "Generators" Terminals, when No Current is Flowing.

It is Apples and Oranges we are comparing when fitting your Scope Shots to behaviours.

Faraday's law of induction is a basic law of electromagnetism predicting how a magnetic field will interact with an electric circuit to produce an electromotive force (EMF)

We both know, and have discussed on many Ocasions, The Magnetic Field is Current, they are one in the same things. The Magnetic Field is Not Voltage - the mesaurement of EMF!

   Chris Sykes
       hyiq.org

[EMJunkie]

Also, the Phase Angle can change, as both you and TK have shown, at any time, through adding a Load, or change in Frequency, and others reasons. E.G: Adding a DC Bias to the device.

By definition, Faradays Law of Electromagnetic Induction, including Heinrich Lenz's contribution, predicts EMF. Which is by definition the (-) of the Source, which is mathematically equal to 180 degrees, or Anti Phase.

Electrical Science for Technicians

Ref: Electrical Science for Technicians: Page 198

   Chris Sykes
       hyiq.org

[MileHigh]

I wouldn't be surprised if there is a lot of moaning and groaning and shameless outright lying when it comes to the wine glass questions.  Or there will be more strutting little fake-ass peacocks flashing their phony colours and saying, "Oh, I already knew that."

This is a no-strutting-peacock no-outright-lying zone when it comes to the wine glass questions.  So if you want to answer them now then give it a shot, don't dare say after the fact that you could answer the questions.  Show that you have some moral fiber and character.

Here are the questions, only short, simple answers will qualify as correct answers.  Don't even think about a copy-paste.

How does a wine glass resonate, what is the mechanism?
How is the resonant frequency of a wine glass determined?

Let's see if any of the resonance fetishists can answer the questions successfully.

[tinman]

Hey Brad, I think we will have to agree to disagree on this one.



It is Apples and Oranges we are comparing when fitting your Scope Shots to behaviours.

 

   Chris Sykes
       hyiq.org

EMF is not Current. EMF is "Columb's of Charge" measured in Volts.

That is correct,and the yellow trace in the scope shot is measuring the columb's of charge by way of showing us a voltage trace.

Which is a quantity that can be measured on a "Generators" Terminals, when No Current is Flowing.

That is also correct. And if we use a PM generator,the EMF produced from the stator coil will be 0 when the PM is directly at the center of the stator coil's core-where the induced magnetic field into that core is at it's greatest,but the rate of change of that magnetic flux through the core is 0,and so the EMF is also 0.

The Magnetic Field is Current, they are one in the same things.

The magnetic field is the result of current flow through a conductor.
As i said,and showed above with the generator example,the magnetic flux through the core will be at maximum when the EMF is at 0,and so the current flow will be at maximum when the rate of change of magnetic flux is 0,and the EMF will also be at 0 when the rate of change of magnetic flux is at 0.
See pic below.

The Magnetic Field is Not Voltage - the mesaurement of EMF!

That is correct.
But a changing magnetic field through a conductor produces an EMF in that conductor.

We both know, and have discussed on many Ocasions,

Yes,but it has been with a load placed on the secondary,which will then put the secondaries CEMF 180* to that of the primaries EMF,as the secondary now will have current flowing through it,and will produce it's own CEMF which is 180* from the EMF of the primary that produced it.
The primary has an EMF placed across it,and the secondary produces an EMF that apposes it-(a counter EMF-our 180* phase relationship) - but only when current flows through it.
But if the secondary is open(as it was in my schematic),then the EMF across the secondary is inline with the rate of change of the magnetic flux through the core,not the strength of the magnetic flux in the core-where maximum field strength is indicated by peak current,and maximum rate of change is indicated at the 0 volt line of the current trace.

Brad