[Elecraft] my KPA1500 and 160M

David Gilbert ab7echo at gmail.com
Thu Sep 3 00:12:07 EDT 2020 

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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>>
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