Like I said, you're refusing to understand the physics of dielectric
loss in a ferrite. The heat is caused by dielectric currents induced by
the high RF E-field from the VSWR, not the emf caused by the current in it.
I'm done with trying to educate you.
Dave AB7E
On 9/2/2020 8:12 PM, Adrian wrote:
You neglect to fill in the missing part where voltage is causing the
current which is causing the heat.
The use of a magnetic core can increase the strength ofmagnetic field
<https://en.wikipedia.org/wiki/Magnetic_field>in anelectromagnetic
coil <https://en.wikipedia.org/wiki/Electromagnetic_coil>by a factor
of several hundred times what it would be without the core. However,
magnetic cores have side effects which must be taken into account.
Inalternating current
<https://en.wikipedia.org/wiki/Alternating_current>(AC) devices they
cause energy losses, calledcore losses
<https://en.wikipedia.org/wiki/Core_losses>, due tohysteresis
<https://en.wikipedia.org/wiki/Hysteresis_loss>andeddy *currents*
<https://en.wikipedia.org/wiki/Eddy_current>in applications such as
transformers and inductors. "Soft" magnetic materials with
lowcoercivity <https://en.wikipedia.org/wiki/Magnetic_coercivity>and
hysteresis, such assilicon steel
<https://en.wikipedia.org/wiki/Silicon_steel>, orferrite
<https://en.wikipedia.org/wiki/Ferrite_(magnet)>, are usually used in
cores.
Sudden high swr issues at high power are caused by insulation breakdown.
The heat is directly proportional to the current producing it. You
half the current and therefore halve the heat.
However this does not apply to voltage as current flow cannot be taken
for granted. talking only about voltage, as voltage can in a current
equation.
You can halve the voltage and may have little heat due to the voltage
breakdown no longer in place.
Magnetism is created by current, Magnetism cutting a conductor
produces current which produces heat.
Voltage only defined, never has and never will be responsible for heat
production P =I^2 R .
*Dielectric loss*quantifies adielectric material
<https://en.wikipedia.org/wiki/Dielectric_material>'s inherent
dissipation of electromagnetic energy (e.g. heat).^[1]
<https://en.wikipedia.org/wiki/Dielectric_loss#cite_note-1> It can be
parameterized in terms of either the*loss angle*/δ/or the
corresponding*loss tangent*tan /δ/. Both refer to thephasor
<https://en.wikipedia.org/wiki/Phasor>in thecomplex plane
<https://en.wikipedia.org/wiki/Complex_plane>whose real and imaginary
parts are theresistive
<https://en.wikipedia.org/wiki/Electrical_resistance>(lossy) component
of an electromagnetic field and itsreactive
<https://en.wikipedia.org/wiki/Reactance_(electronics)>(lossless)
counterpart
It is the *current* induced by this electromagnetic field, not
cancelled by back emf, that causes heat.
On 3/9/20 12:25 pm, David Gilbert wrote:
OK ... I'm going to make this simple for you. Picture a material
(like a capacitor) with a lossy dielectric, and then apply a high RF
voltage across it. The dielectric passes a current as the result of
the voltage, and the lossiness of the dielectric generates heat. With
a lot of voltage the heat generated can be considerable.
Now then, whether you want to acknowledge it or not, a ferrite core
IS a lossy dielectric and can get hot when you put a high enough RF
voltage across it INDEPENDENT OF THE CURRENT FLOWING THROUGH THE
WINDING AROUND THE CORE. You can find innumerable references to the
dielectric losses of ferrite materials if you just bother to do some
internet searching. This is NOT an insulation breakdown issue ...
not at all.
All of this can happen as soon as power is applied to the system
containing the ferrite. Certainly the rate of temperature rise will
be dependent upon how much voltage is applied, the frequency of it,
and the dielectric loss characteristics of the particular ferrite,
but VSWR is the voltage we're talking about here and that becomes
relevant immediately upon application of power.
I suspect that you will dig in your heels and continue to dispute
this basic physics, but at least I hope others here will understand
things better than you do.
Dave AB7E
On 9/2/2020 5:52 PM, Adrian wrote:
I say that your response is completely false and you are missing
basic electricity facts. The high voltage becomes an issue when
insulation breaks down, and then *current *starts
to flow through the fault path converting to emf & heat directly and
via induced current resulting ; P = E X I*. *Without the current
the heat does not occur, it is basic physics, and
the heat is directly proportional to the current. Voltage can exist
without current, but current cannot exist without voltage. Heat
produced is directly proportional to the current whether
it be in the intentional circuit path, or fault path caused by high
voltage insulation breakdown..
In addition, your statement that only current in the balun circuit
can produce heat is completely false. High voltage RF can create
major core heating due to dielectric losses in the ferrite core
independent of the magnitude of current flow in the tuning circuit.
Several discussions on the TowerTalk reflector have pointed this
out over the years for baluns and common mode chokes in ham radio
applications.
Dave AB7E
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