Re: [TCML] tank circuit power

[jhowson4@xxxxxxxxxxx]

Thank you for the explanation Jim, 
I never though to just approximate it as multiple sine half pulses at differing amplitude, that simplifies the integral considerably. 
I shall have to apply that approximation to my rms current integral tomorrow and see what happens. 

So you have addressed the power within the tank circuit , and what is directly happening to it. 
I still don't see where the sqrt(Q) * Irms nst = Irms tank = 200ma comes from though. 

I modeled the waveform of the current through the primary coil, as a cos(a t)*sin(b t)*exp(c t) type function, where a,b,c are parameters relating to the total energy transfer time, what notch am I expecting it to quench at, etc. then numerically calculate the rms integral with Irms = bps * sqrt(1/T integral(function^2,t,0,T)) where T is on the order of 0.5 microseconds. 
Before I made some changes to the program (which I am still trying to work the bugs out of) I was getting to be expected rms currents in the range of 20-150 amps from small and large system imputs respectively . 




Thanks, 
John "Jay" Howson IV 


"Why thank you, I will be happy to take those electrons off your hands." 

----- Original Message -----
From: "Jim Lux" <jimlux@xxxxxxxxxxxxx> 
To: "Tesla Coil Mailing List" <tesla@xxxxxxxxxx> 
Sent: Tuesday, May 15, 2012 11:46:01 PM 
Subject: [TCML] tank circuit power 

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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