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Re: Ideal Magnifier Model - PSPICE (Antonio?)
Original poster: Rob Maas <robm-at-nikhef.nl>
> > Why is the "self capacitance of L3" included in C2:
>
>We usually model a coil that has distributed capacitance and one side
>grounded as a grounded coil in parallel with a "self-capacitance",
>grounded too.
>In a magnifier, the third coil has both ends floating. If we assume that
>the capacitances from both ends of the coil to ground are identical,
>the simplest model for a floating coil has then a capacitor to ground at
>each end:
>
>+--L3--+
>| |
>C3a C3b
>| |
>+------+-o ground
>
>If one end of the coil is grounded, if we assume that the
>capacitances don't change due to this, the two capacitors must be
>each identical to the "self-capacitance" of the coil.
>And really, a measurement (or calculation) of the free-space
>capacitance of a cylinder results in a capacitance that is quite
>close to twice the Medhurst capacitance for a coil with the
>shape of the cylinder.
>Example: A coil with 32 cm of length and 4.4 cm of radius has a
>Medhurst capacitance of 5.55 pF.
>The free-space capacitance of an open cylinder with these dimensions
>results as 10.15 pF.
>
> > Isn't C3 = 'self capacitance of L3' + 'capacitance of topload' on top of
>L3 ?
>
>Yes, it is, using the "normal" copy of the self-capacitance at the
>output side of L3.
>
>Antonio Carlos M. de Queiroz
Antonio,
Thanks for the above explanation.
Recently there was some discussion in this group, related to the announcement
of the 'Green Monster', about coupling values (k_12) between L1 and L2.
A high coupling speeds up energy transfer between C1 and C3, but can only
be realized by close proximity of L1 and L2. This causes serious isolation
problems. So the practical value of k_12 has to be a compromise; most
practical
values seem to be in the region 0.35 < k_12 < 0.50 .
In order to get a feeling what that means, I wrote a small program that
basically just calculates eq's (5), (4) and (8) of your paper "Designing a
Tesla Magnifier"; the list is ordered in ascending order of k12, and discards
C3/C2 values smaller than a chosen input value:
--------------------------------------------------------------------------
table for k12-values between 0.350 and 0.500
and C3/C2-values > 0.500
k12( 6: 7:16) = 0.357 L3/L2 = 6.849 C3/C2 = 0.560
k12( 6: 7:18) = 0.365 L3/L2 = 6.525 C3/C2 = 0.744
k12( 5: 6:13) = 0.384 L3/L2 = 5.776 C3/C2 = 0.564
k12( 5: 6:15) = 0.395 L3/L2 = 5.411 C3/C2 = 0.802
k12( 5: 6:17) = 0.402 L3/L2 = 5.192 C3/C2 = 1.074
k12( 4: 5:10) = 0.417 L3/L2 = 4.741 C3/C2 = 0.540
k12( 4: 5:12) = 0.434 L3/L2 = 4.303 C3/C2 = 0.857
k12( 4: 5:14) = 0.444 L3/L2 = 4.075 C3/C2 = 1.231
k12( 4: 5:16) = 0.450 L3/L2 = 3.940 C3/C2 = 1.663
k12( 9:12:23) = 0.470 L3/L2 = 3.533 C3/C2 = 0.585
k12( 8:11:20) = 0.487 L3/L2 = 3.220 C3/C2 = 0.543
k12( 3: 4: 9) = 0.488 L3/L2 = 3.204 C3/C2 = 0.889
------------------------------------------------------------------
The three items in the list, k12, L3/L2 and C3/C2 depend only on
the the three mode numbers k:l:m, as given as 'argument' of k12.
The relation between L2 and L3 does not seem to be problematic:
L3 is 3-7 times larger than L2; since L scales with turns squared,
this is easily accomplished. But for most of the given k12-values,
C3 is significantly smaller than C2. Suppose we write
(1) C3 = R*C2
splitting C3 in a 'Medhurst' part C3(M), and a topload part C3(T),
and C2 in its 'own' Medhurst part and the part of C3 (as you explained
above), and neglecting the transmission line contribution, we get
C3(M) + C3(T) = R*(C2(M) + C3(M))
for a small 30cm x 10 cm topload, C3(T) = 14 pF, and for a
small 10 cm x 40 coil C3(M) = 6 pF, so rewriting (1):
(2) 20 = R*(C2(M) + 6)
Even for an R-value of 1 (a full table reveals that most R-values
are smaller than 1), one still needs C2(M) = 14 pF
(corresponding to an L2 of 30cm diameter, height 45 cm).
In case R = 0.5, C2(M) = 34 pF, which makes a secundary of 70 cm
diameter and more than a metre high. Moreover, these larger R-values
only occur for high mode numbers.
>From the above list, only mode 3:4:9 seems attractive;
modes (5:6:15,17) and (4:5:12,14,16) also produce reasonable
C3/C2 ratios and realistic values for k12.
Aloowing k12 to be larger, gives more possibilities:
-------------------------------------------------------------------
table for k12-values between 0.500 and 0.600
and C3/C2-values > 0.500
k12( 8:11:22) = 0.500 L3/L2 = 2.994 C3/C2 = 0.707
k12( 3: 4:11) = 0.502 L3/L2 = 2.963 C3/C2 = 1.436
k12( 3: 4:13) = 0.510 L3/L2 = 2.840 C3/C2 = 2.092
k12( 3: 4:15) = 0.515 L3/L2 = 2.768 C3/C2 = 2.857
k12( 7:10:19) = 0.523 L3/L2 = 2.658 C3/C2 = 0.666
k12( 7:10:21) = 0.536 L3/L2 = 2.485 C3/C2 = 0.870
k12(10:15:26) = 0.542 L3/L2 = 2.398 C3/C2 = 0.557
k12( 6: 9:16) = 0.547 L3/L2 = 2.341 C3/C2 = 0.600
k12( 2: 3: 6) = 0.565 L3/L2 = 2.133 C3/C2 = 0.833
k12( 6: 9:18) = 0.565 L3/L2 = 2.133 C3/C2 = 0.833
k12( 9:14:25) = 0.572 L3/L2 = 2.052 C3/C2 = 0.642
k12( 6: 9:20) = 0.577 L3/L2 = 2.006 C3/C2 = 1.094
k12( 8:13:22) = 0.590 L3/L2 = 1.873 C3/C2 = 0.579
k12( 2: 3: 8) = 0.591 L3/L2 = 1.862 C3/C2 = 1.698
k12( 5: 8:15) = 0.599 L3/L2 = 1.792 C3/C2 = 0.766
--------------------------------------------------------------------
but the oscillator part is probably very difficult to build.
The conclusion seems to be that C3 should not become too large in
order to avoid a huge L2-C2 system - unless a 'lumped' capacitor
is added to the L2-C2 system, but I don't know whether this is really
practical due to the high voltages - or am I missing something here?
I have also a question about k12: is k12 independant from the
presence of L3-C3. What I mean is this: suppose one has a magnifier
design, with a particular k12 = k12(k:l:m) value. Is it sufficient
to model (e.g. using your Inca program) the L1-L2 system for that
particular value of coupling?
Any comment about this is appreciated.
best regards, Rob Maas