In article , Bill Turner
writes:

On Tue, 09 Dec 2003 04:59:29 GMT, wrote:

I maintain it does. Otherwise the formula X=2piFL is invalid.

NO! In the above equation, X varies when F varies. The equation
does NOT mean that L varies as F varies.

_________________________________________________ ________

Perhaps an example will make it clear.

Suppose you have a coil which measures 1 uH at 1 MHz. It is known to
have a self-resonant (parallel) frequency of 100 MHz.

OK, it has a distributed capacity of 2.533 pFd.

The circuit being measured is composed of a pure inductance
of 1.000 uHy and pure capacitance of 2.533 pFd in parallel.

We can neglect the losses in each one of those components
for the sake of illustration.

You measure its reactance at 1 MHz using the formula X=2*pi*F and find
it to be 6.28 ohms.

The inductive reactance is 6.28319 Ohms at 1 MHz and the
capacitive reactance is 62.8326 KOhms at 1 MHz.

In terms of susceptance, the B_sub_L is 0.159153 and
15.9153x10^-6 mhos, respectively. Total susceptance is then
0.159137 mhos or 6.28389 Ohms. [reactance meter probably isn't
calibrated that close to show the slight change due to distributed
capacity]

At 2 MHz you find it to be 12.56 ohms.

At 2 MHz, the inductive reactance is 12.5664 Ohms or 0.0795775
mhos while the distributed capacitance has 31.8306 mhos. The
total susceptance is 0.0795456 mhos or 12.5714 Ohms. That is
within 0.0907% of 12.56 Ohms and darn few reactance measuring
gizmos are calibrated that close...

At 10 MHz you find it to be 62.8 ohms.

Okay, at 10 MHz, the inductive susceptance is 0.0159155 mhos
and the capacitive susceptance is 159.153 x 10^-6 mhos, the
total being 0.0157563 mhos or 63.4665 Ohms reactance. That's
an error of 1.061% from 62.8 and still fairly reasonable for the error
of a reactance meter or whatever.

So far the reactance is changing linearly with respect to frequency.
(Actually it is not perfectly linear, but the difference at these
frequencies is small and probably would not be observed with run of the
mill test equipment.)

Okay, that's progress. We are agreed that test equipment can have
errors...he said with a grin having worked in metrology and a
standards lab for over 2 years in the past... :-)

But, as you approach 100 MHz, you find the change is obviously no longer
linear.

Ah, but you are measuring TWO things at the same time, the
parallel of the true inductor and its distributed capacity. Once you
are into measuring multiple elements, you need a test setup to
try to get a handle on the individual components. That is why I
brought up the "true inductance" two-frequency test on a Q meter
that has a calibrated tuning capacitor. That WILL establish the
equivalent pure capacitor due to distributed winding capacity in the
coil (the physical inductor form).

Once you KNOW the distributed capacity, its just a matter of some
button-pushing on a good scientific handheld calculator to derive
true inductance from the reactance readings of both inductance and
distributed capacity. [I recommend an HP 32 S II as an RPN fan]

The parallel capacitance will definitely exist as more picoFarads in
a circuit such as a FET gate which has a very high parallel resistance
(or very low conductance if you can think in terms of admittance).
That FET input capacitance will change the higher frequency resonance
even lower.

Offhand, I'd say that 2.533 pFd distributed capacity is rather high and
probably is around 1.0 pFd (solenoidal type, no core)...but a FET
gate input and its PCB traces to ground plane can be an additional
2.0 pFd. That's 3.0 pFd total and the self-resonance of that circuit is
now 91.888 MHz.

This is not an illusion. If you have an inductance meter which uses 99
MHz as a test frequency, it WILL MEASURE 80.4 uH.

That "inductance meter" is still measuring TWO THINGS AT THE
SAME TIME. The physical coil still has two components, the
pure inductance in parallel with a pure capacitance representing
the distributed capacity of the windings. Those are inseperable
unless you do something like the "true inductance" test at octave
separation frequencies or equivalent.

A Q Meter of any kind made today, last year, or back in the pre-
history before 1947, MEASURES THREE THINGS AT THE SAME
TIME! Yet the Q Meter is still accurate enough to derive the
equivalent parallel resistance, parallel inductance, and parallel
capacitance of the physical coil's windings. [it actually measures
conductance and susceptance as a total magnitude and relates
that to the Q or loss factor while the calibrated frequency setting
and calibrated variable capacitor allow separate "inductance"
measurement even though the Q Meter is "looking" at both L and
C_sub_d in parallel]

ANYONE using test equipment SHOULD be aware of what their
equipment does, how it works (in general), and what it really
measures. Since inductance does NOT change in a passive coil
(that isn't otherwise influenced by magnetic fields), what anyone
measures on a particular coil is THREE THINGS: The conductance
due to losses and the susceptance due to BOTH parallel inductance
and parallel capacitance. Conductance will change with frequency
depending on a lot of different factors (coil form, coil core, wire used,
shield used (if any), dielectric of the former material, core permittivity,
etc.). Susceptance will change with frequency because of the TWO
components...BUT THE INDUCTIVE COMPONENT DOES NOT
CHANGE.

And therefore, I maintain that inductance DOES vary with frequency.

How can it be otherwise?

The baseline taught in all textbooks (where I learned it first) and in
classes (where I learned it second) all agree that one MUST
separate the components into their "pure" form and THEN derive the
component parts by different tests. That is how it is perceived by
most other folks based on a lot of first-principle demonstration.

Inductance of a coil DOES NOT CHANGE WITH FREQUENCY.
Basic definition. First-principle stuff by definition.

You CAN say that APPARENT inductance changes if you are just
doing one kind of test. "Apparent" isn't going to work well when this
coil is dropped into a circuit thinking that "inductance changes with
frequency" and the circuit contains a lot of other sneaky little
components that can shoot the "apparent" reading way off. No one
successfully works with L-C and active-device networks using this
"apparent" reading. One separates the component parts first, then
combines them into manageable parallel-equivalent or series-equivalent
circuits.

The ILLUSION is from looking at an impedance- or admittance-measuring
instrument such as a Q Meter and thinking its calibrated inductance
dial "measures inductance." It doesn't...but it comes very close. That is
just the calibrated variable capacitor tuning to resonance at specific
frequencies...as a convenience to the user. The capacitance markings
will be accurate but any external coil that has significant parallel
capacitance from its windings will add to the calibrated capacity on the
dial. Some Q Meters allow variable frequency settings to do things
like the octave-separation-of-frequency measurement of the external
test parallel capacity.

An impedance or admittance bridge type of instrument can yield
different "errors" and "illusions" depending on their type/kind.

Len Anderson
retired (from regular hours) electronic engineer person

In article ,
(Avery Fineman) writes:

At 2 MHz, the inductive reactance is 12.5664 Ohms or 0.0795775
mhos while the distributed capacitance has 31.8306 mhos. The
total susceptance is 0.0795456 mhos or 12.5714 Ohms. That is
within 0.0907% of 12.56 Ohms and darn few reactance measuring
gizmos are calibrated that close...

Make that 31.8306 x 10^(-6) mhos in second line. :-(

Len Anderson
retired (from regular hours) electronic engineer person

In article ,
(Avery Fineman) writes:

At 2 MHz, the inductive reactance is 12.5664 Ohms or 0.0795775
mhos while the distributed capacitance has 31.8306 mhos. The
total susceptance is 0.0795456 mhos or 12.5714 Ohms. That is
within 0.0907% of 12.56 Ohms and darn few reactance measuring
gizmos are calibrated that close...

Make that 31.8306 x 10^(-6) mhos in second line. :-(

Len Anderson
retired (from regular hours) electronic engineer person

:

The inductance is not changing. What you are measuring is not pure
inductance but the coil has a stray capacitance. That is what is making
the
coil seof resonate.

__________________________________________________ _______

I am well aware of that, but you are tap dancing around the relevance of
the formula X=2*pi*F*L.

Just answer this: If I have a coil of very high Q (no appreciable
resistance), and I apply 100 volts of 100 MHz AC to it, and measure a
current of 2 milliamps through it, then:

1. What is its reactance?
2. What is its inductance?

--
Bill, W6WRT

I am not tap dancing around that formula. To be a self-resonance , the coil
must have some capacitance somewhere. You are leaving that out by just
using one formula. You also have to use the Xc=/(2pi*f*C) to get the
capacitance and then to find the final results you have to use Xo=Xl-Xc.

As to what you are asking for , it could be almost anything for the
inductance. It all depends on the stray capacitance. ?Remember you into
this self resonance so there has to be some capacitance. There is an
infinante number of solutions to the inductance of this problem.

The simple circuits that are most often found to measuer inductance do not
take the capacitance that may be there into concideration.

Just as if you are using a simple SWR meter and have a SWR of 2:1. You do
not know if you have a 25 ohm resistor or a 100 ohm resistor hooked to it.
Either will show 2:1. Also if there is any reactance , either inductive or
capacitive the simple SWR meter will not show that either. Again there is
an infiniante number of capacitors and inductor and resistors that will show
up as a 2:1 swr. You need a beter instrument.

:

The inductance is not changing. What you are measuring is not pure
inductance but the coil has a stray capacitance. That is what is making
the
coil seof resonate.

__________________________________________________ _______

I am well aware of that, but you are tap dancing around the relevance of
the formula X=2*pi*F*L.

Just answer this: If I have a coil of very high Q (no appreciable
resistance), and I apply 100 volts of 100 MHz AC to it, and measure a
current of 2 milliamps through it, then:

1. What is its reactance?
2. What is its inductance?

--
Bill, W6WRT

I am not tap dancing around that formula. To be a self-resonance , the coil
must have some capacitance somewhere. You are leaving that out by just
using one formula. You also have to use the Xc=/(2pi*f*C) to get the
capacitance and then to find the final results you have to use Xo=Xl-Xc.

As to what you are asking for , it could be almost anything for the
inductance. It all depends on the stray capacitance. ?Remember you into
this self resonance so there has to be some capacitance. There is an
infinante number of solutions to the inductance of this problem.

The simple circuits that are most often found to measuer inductance do not
take the capacitance that may be there into concideration.

Just as if you are using a simple SWR meter and have a SWR of 2:1. You do
not know if you have a 25 ohm resistor or a 100 ohm resistor hooked to it.
Either will show 2:1. Also if there is any reactance , either inductive or
capacitive the simple SWR meter will not show that either. Again there is
an infiniante number of capacitors and inductor and resistors that will show
up as a 2:1 swr. You need a beter instrument.

On Tue, 09 Dec 2003 09:19:26 -0800, Bill Turner
wrote:

On Tue, 09 Dec 2003 09:59:03 +0200, Paul Keinanen
wrote:

L = Xl / (2 pi f) applies only to _pure_inductive Xl

It does _not_ apply to L = X / (2 pi f) in which X is some combination
of Xl and Xc !

_________________________________________________ ________

And what is the basis for the second statement above? Network analysis
normally combines reactances of opposite signs and treats them as a
single reactance, provided the frequency is constant of course.

Yes, that is true.

For a series resonant circuit X = Xl + Xc and since Xl and Xc have
opposite signs, the total reactance is less than either of them. At
resonance, their magnitude is the same and they cancel each other,
thus, the reactance is zero at resonance.

The situation is a bit more complicated with parallel resonant
circuits (hopefully I got the formula right)

X = 1 / ( 1 / Xl + 1 / Xc)

At resonance 1/Xl and 1/Xc have equal magnitude but opposite sign and
the divisor goes to zero and X to infinity. At resonance X also
changes sign.

Most test equipment simply measure the total reactance
X = 1 / ( 1 / Xl + 1 / Xc) at a specified frequency f.

Using the notations

Xl = 2 pi f L
Xc = -1/(2 pi f C)

Please show _mathematically_ how you from a single measurement of the
total reactance X at frequency f can solve both the inductance L and
capacitance C.

If you can not do it, then the single measurement of the total
reactance X is not sufficient for solving both L and C :-).

You could also start with the simpler series resonance (X = Xl + Xc)
case in which the measured reactance X is the end to end reactance.

And please, do not try to invent your own formulas such as
L = X / ( 2 pi f) when the components of X are unknown. That formula
is only valid in a special case when X is known to be a pure inductive
reactance Xl.

Paul OH3LWR

On Tue, 09 Dec 2003 09:19:26 -0800, Bill Turner
wrote:

On Tue, 09 Dec 2003 09:59:03 +0200, Paul Keinanen
wrote:

L = Xl / (2 pi f) applies only to _pure_inductive Xl

It does _not_ apply to L = X / (2 pi f) in which X is some combination
of Xl and Xc !

_________________________________________________ ________

And what is the basis for the second statement above? Network analysis
normally combines reactances of opposite signs and treats them as a
single reactance, provided the frequency is constant of course.

Yes, that is true.

For a series resonant circuit X = Xl + Xc and since Xl and Xc have
opposite signs, the total reactance is less than either of them. At
resonance, their magnitude is the same and they cancel each other,
thus, the reactance is zero at resonance.

The situation is a bit more complicated with parallel resonant
circuits (hopefully I got the formula right)

X = 1 / ( 1 / Xl + 1 / Xc)

At resonance 1/Xl and 1/Xc have equal magnitude but opposite sign and
the divisor goes to zero and X to infinity. At resonance X also
changes sign.

Most test equipment simply measure the total reactance
X = 1 / ( 1 / Xl + 1 / Xc) at a specified frequency f.

Using the notations

Xl = 2 pi f L
Xc = -1/(2 pi f C)

Please show _mathematically_ how you from a single measurement of the
total reactance X at frequency f can solve both the inductance L and
capacitance C.

If you can not do it, then the single measurement of the total
reactance X is not sufficient for solving both L and C :-).

You could also start with the simpler series resonance (X = Xl + Xc)
case in which the measured reactance X is the end to end reactance.

And please, do not try to invent your own formulas such as
L = X / ( 2 pi f) when the components of X are unknown. That formula
is only valid in a special case when X is known to be a pure inductive
reactance Xl.

Paul OH3LWR

This argument could have been obviated, if the parties concerned had agreed
on the definition of inductance.
Is it an inherent property of an arrangement of current-carrying conuctors,
or the result of measuring the current and voltage applied to that
arrangement?

--
igor
_____________________________________________
Acorn RISC OS4
_____________________________________________

This argument could have been obviated, if the parties concerned had agreed
on the definition of inductance.
Is it an inherent property of an arrangement of current-carrying conuctors,
or the result of measuring the current and voltage applied to that
arrangement?

--
igor
_____________________________________________
Acorn RISC OS4
_____________________________________________

The dimensions of both inductance and capacitance are Length.

Length is idependent of frequency.