In response to offline requests and suggestions, here is the detailed
calculation for the toroidal coil. I would like to proceed by making a very
clear statement of the problem to be analyzed - many of the disputes that
occur here are a simple result of fuzzy or missing assumptions that should
always preceed any analysis of a model.
In a typical tank circuit, PI, PI-L or even parallel-link coupled, there will
be a significant high RF voltage across the coil. It's peak-to-peak value is
typically in the range of VCC. For a 4KVDC amp this results in a peak of 2KV
and and RMS of about 1400V. These values are all somewhat approximate and
depend upon the details of the amp and its operating mode. I do not want this
analysis to be at all dependent on those unknowns, for sake of simplicity and
generality. I am just going to state the problem in terms of RMS voltage
across the coil. For purposes of evaluating the coil and/or the ferrite
material from which it is made, the coil voltage is the only variable
parameter.
The question is simply stated: For a given voltage across the coil, what
power will it dissipate due to RF losses (I am going to neglect the ohmic
wire losses - by comparison, they are negligible). My calculations will show
the exact values, limited only by the accuracy of the mfrs. data My
introductory estimate of the actual RF voltage is only used as a guideline in
knowing where on the data curves to look for answers (more on this shortly.)
The calculation to be performed here is basic EM field theory, which I will
review as succinctly as possible. This calculation must find the peak value
of B (the magnetic flux density) during an RF cycle, given the RMS voltage
across the coil. With B(pk) known, the power loss is simply read off of
graphical data from mfr. This graph specifically gives us the power loss per
unit volume of core material (mw/cm(cube) ) and we multiply this by the
volume of the core, known from its geometry.
The core in question is a MM T400-2 which has the following physical
parameters:
OD=4.0 ID=2.25 Ht=.65 area=3.46cm(sq.) VOL=86.4cm(cube) A-sub-L=18nH.
The manufacturers data for power loss is only listed for freq. up to 2.5MHZ,
so I am going to assume a 2.5MHZ tank circuit. Using a typical PINET
calculation I arrive at
a target L of 11.5uH (assuming 2000 ohm Rplate). To achieve this inductance
we use 25 turns of wire on the T400-2 core:
L=25*25*18=11250nH=11.25uH
The next step is to determine the peak flux, and here it gets as bad as it's
gonna. In a rigorous calculation, which would have to calculate the peak flux
by finding the time-integral of the voltage across the coil (Faraday's
law/Maxwell's second ea.) Rather than marching out the dreaded integral signs
here, I will use the fact that this problem has been solved so many times
that its generic answer is also an EE101 formula. It is known as the
'transformer' equation, the same one that is used to design the primary of a
power transformer (same problem -different frequency). I repeat that equation
here in my favorite units;
B(pk)=V*22.5/(N * f(MHZ) *A(cm-sq) )
V is the RMS voltage across the coil terminals, N is the number of turns, A
is the cross section area of the core. Plugging in the numbers for our case:
B(pk)=V*22.5/(25 * 2.5 *3.56) = V * .104
We are almost there now. Lets pick a voltage, say 500V. This gives us a peak
flux of 500 * .104 = 50Gauss (there is no need for more than 2 SF of
accuracy here)
Looking in the mfrs data book for their type-2 material, we find that at
2.5mhz and a peak flux of 50G. the power loss is 160mw/cm(cube). For our core
this results in:
P(core)=160 * 86.4 = 13,700mw=13.7Watts.
This is the final answer - 13.7 Watts loss at 2.5MHZ with 500VRMS applied. I
repeated this for several power levels relavent to kw amplifiers, with the
following results. I state the results in the format VoltsRF/B(pk)/k/P,
where k is the data read from the mfrs graph (unless you have their data
book, you will have to trust me on this one):
500/50/160/13.7
1000/104/680/58.75
1500/156/1600/138.2
2000/208/3000/259
There has been some confusion about what the core WC6W had for sale actually
is, and I cannot answer that. He stated it as a T400-2A, but there is no such
core listed in the Micrometals catalog. They do list a T-400-2D which is the
double height version of the T400-2, so I repeated all of these calculations
for that core. There was a surprise in the results, which are as follows. The
calculations are similar, but the area doubles, the volume doubles, and
A-sub-L doubles. To accommodate the latter, I decreased the turns count to 18
to maintain a constant inductance value of 11.5 uH. Here are the results:
500/36/80/13
1000/72/360/62
1500/108/850/147
2000/144/1500/259
I would like to re-iterate that these calculations were for f=2.5MHZ, forced
by lack of more data from the mfr. One thing is for certai, however, the
losses increase very rapidly with frequency. At 3.5MHZ, the above loss
figures could possibly double, even after decreasing N to scale up the PINet.
In other words, these results are seriously optimistic for 80M, somewaht
pessimistic for 160.
These calculations show that this core, or its alternate, would produce an
enormous amount of heat/loss at voltages above 500-1000. On that basis alone
I can't see its use in typical amplifiers.
Here is another dirty,little secret. This RF saturation effect is really
another form of distortion, when viewed as a circuit element. I predict,
without yet having made the calculations (which are possible but quite
involved) that this could also be a significant factor in rejecting this
core.
The first person who finds a significant flaw in this analysis, as alleged by
WC6W, will receive from me a reward consisting of a steak dinner at a
restaurant in his area.
73
Eric von Valtier K8LV
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