Posted by
Adrian-3 on
Sep 03, 2020; 3:12am
URL: http://elecraft.85.s1.nabble.com/OT-WWV-Time-Transmissions-tp7664561p7664640.html
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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