Thread: Winding coils
View Single Post
  #72   Report Post  
Old December 7th 03, 09:30 PM
John Devereux
 
Posts: n/a
Default

Bill Turner writes:

On Sun, 07 Dec 2003 20:00:37 GMT, John Popelish
wrote:

The inductive component of the impedance remains
essentially constant through resonance. What is non ideal about the
inductor is that it does not exhibit just inductance, but a parallel
combination if inductance and capacitance. Ignoring the capacitance
and calling the effect variable inductance is just not as accurate a
way to describe what is going on.


__________________________________________________ _______

Your point is well taken, but look at it this way:

Say I give you a black box containing an inductor with two terminals on
the box. If I have you measure the inductance at one and only one
frequency, there is no way for you to know whether it is an inductor
operating well below its self-resonance point, or an inductor operating
near its self-resonance point. To the outside world, at ONE frequency,
they appear identical; same reactance, same inductance.


No, you are neglecting the phase. The two cases would have very
different phase shifts (the current would be out of phase with the
applied voltage, by different amounts), depending on whether you were
below, at, or above resonance.

And yet, at some other (lower) frequency, they will measure quite
differently. This is the basis for my observation that inductance does
indeed vary with frequency, based on the parasitic capacitance present
in all inductors.

And yes, if you can factor out the self-capacitance, then the inductance
would indeed be constant with frequency. The problem is, no one has
ever figured out how to do that with an actual coil. It can't be done.


Yes it can. This is what a network analyser or impedance bridge does,
(as I understand it, I've never actually had to use either!).

At low frequencies the black box would be inductive. The current would
lag the voltage. At resonance the voltage would be in phase with the
current (the black box would appear resistive). At high frequencies
the current would lead the voltage. It would appear capacitive.


--

John Devereux