Hi Jim..
Hey, I'm just gonna put something out here.
The main question I gathered from the person inquiring was do you need
large copper on the primary? For SG, usually not, but you may need
adequate spacing between turns simply because the turn to turn voltage
is higher for an SG coil. For a solid state coil (SSTC, DRSSTC, etc),
your better to go big on the tubing, but narrow spacing is usually
okay (low turn to turn voltage). Heating depends on I^2R losses over
time and the heat that will occur on the primary at that inner turn or
two if very different for SG compared to SS. We can attempt rms
numbers in the primary, but if your gonna do that, don't leave out
proximity effects that occur from circulation currents from turn to
turn and whatever else. There is much that goes into the heating of a
primary coil (time of running is a big factor) and it is not a simple
calculation. For most SG coilers, power is low enough to not require a
large primary tubing (1/4" to 3/8" suits most everyone). I would bet
that most coilers would say "i never realized any heat in my primary".
But there are other coilers that have run higher power on SG's and
also many SS coilers that would say "damn, primary got real hot". Of
course, one of the side affects to solid state coils is that if the
primary support doesn't melt, we tend to not pay attention to
temperature. Then when it melts, we freak out. This is much less of a
concern to SG coilers, but it can and does happen.
Take care,
Bart
5/15/2012 8:46 PM, Jim Lux wrote:
The question comes up of what is the RMS current in the primary tank.
This comes up when looking at losses, and when looking at RF safety
calculations for the magnetic field from the primary.
A few basic assumptions:
The loaded Q of a tesla coil is approximately 10.
The equation for Q = 2*pi * Energy stored/(energy lost per cycle)
or, Energy lost = Energy stored * 2 *pi/Q
For Q = 10, that's 6.28/10 or .628...
That is, the tank loses about 62.8% of its stored energy in every
cycle. If you look at the voltage, it goes as the square root (so
each cycle is about 60% the amplitude of the previous)
Previous analyses have shown that one of the larger losses is in the
spark gap (which has a "cathode drop" of around 100V per gap. (why a
rotary or blast gap is more efficient than a static multi gap)
From that, we can calculate what the stored energy in the primary
capacitor is (1/2 C *V^2) and from the stored energy in the cap, we
can calculate the peak current in the primary (= energy in cap, to a
first order)
the next question is "how much energy is dissipated in the resistance
for a half sine with peak value X". For this, I assume that there's
no loss during the actual half cycle, so the current is perfectly
symmetrical around the peak:
So what we want is
integral[sin(omega t)^2] for t=[0, pi/omega]
which is
(omega t)/2 - sin(2omega t)/4 +C
This is zero for t=0
(omega pi/omega)/2 - sin(2* omega pi/omega)/4
or
pi/2 - sin (2pi)/4 = pi/2
So the energy dissipated in the primary resistance, R, if the peak
current is I, is
R * pi/2 * I^2 * thalfcycle
Then you build up a little spreadsheet and sum things up..
( stored is how much energy is in the cap at the peak, the loss is
the joules lost to the resistance, to sec is the energy transferred
to the secondary)
Here's an example:
L 3.70E-05 Q 10
C 6.80E-08 r 0.1
f 1.00E+05
V I
t stored loss to sec
0 21000 900 14.99 0.63 7.30
0.5 14406 618 7.06 0.30 3.44
1 9883 424 3.32 0.14 1.62
1.5 6779 291 1.56 0.07 0.76
2 4651 199 0.74 0.03 0.36
2.5 3190 137 0.35 0.01 0.17
3 2189 94 0.16 0.01 0.08
3.5 1501 64 0.08 0.00 0.04
4 1030 44 0.04 0.00 0.02
4.5 707 30 0.02 0.00
Total joules 1.20 13.78
With twice the Q
L 3.70E-05 Q 20
C 6.80E-08 r 0.1
f 1.00E+05
V I
Cycle stored loss to sec
0 21000 900 14.99 0.63 3.70
0.5 17703 759 10.66 0.45 2.63
1 14924 640 7.57 0.32 1.87
1.5 12581 539 5.38 0.23 1.33
2 10605 455 3.82 0.16 0.94
2.5 8940 383 2.72 0.11 0.67
3 7537 323 1.93 0.08 0.48
3.5 6353 272 1.37 0.06 0.34
4 5356 230 0.98 0.04 0.24
4.5 4515 194 0.69 0.03
Total joules 2.12 12.21
with 0.4 ohms and Q=10
L 3.70E-05 Q 10
C 6.80E-08 r 0.4
f 1.00E+05
V I
Cycle stored loss to sec
0 21000 900 14.99 2.54 5.40
0.5 14406 618 7.06 1.19 2.54
1 9883 424 3.32 0.56 1.20
1.5 6779 291 1.56 0.26 0.56
2 4651 199 0.74 0.12 0.26
2.5 3190 137 0.35 0.06 0.12
3 2189 94 0.16 0.03 0.06
3.5 1501 64 0.08 0.01 0.03
4 1030 44 0.04 0.01 0.01
4.5 707 30 0.02 0.00
Total joules 4.79 10.19
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