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Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
Hi, you need to retune the impedance of the plate circuit, what happens is that the tank circuit looks almost like a voltage shunt and when increasing the voltage, it makes the MO cap resonate with the magnetic shunts in the MOT and correct power factor (the magnetron in a MO runs in a similar way). The second increase of current is because of core saturation.
Do you have a RF bypass capacitor parallel to the doubler diode?
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
Try increasing the number of primary turns, but just a little, go 1 turn at time (basically you are detuning the circuit to get the correct loaded impedance). The spark length should first increase and then drop, you leave the tap where you get the longest sparks. PS. A primary tank circuit Q of 38 and even 19 is too much, I usually go for around 10. The higher Q just heats up the coils and doesn't make sparks any longer.
Q=R/X X=primary reactance R=load impedance of the tube
X=2PI F L_pri F=secondary resonance L_pri=primary inductance
John Freau suggests that with AC (or pulsed DC) supplied coils, the equation should look like R=V/4I R=V/4I V=plate supply voltage I=plate supply current
Note: You need to simply assume what current you want the tank to pull from the plate supply, use the tube spec sheet to find the maximum plate current and go from there. This is not an exact science.
I would like to have confirmation about R (load impedance of the tube) calculation
I my Case, My MOT's output voltage is 2480V With voltage doubler, it makes 4960V According to the 572B datasheet max plate current is 275mA R=V/4I = 4960 / 4 x O,275 = 4509
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
1. the doubled peak voltage is 4960 V * sqrt(2) which is ~7 kV, but this is just theoretical, lets assume 6 kV 2. the 572B has a plate dissipation of 160 watts, it is safe to assume plate input of 4 times the plate dissipation which is 640 watts 3. the peak input power is 8/3 times * average input power which is 1700 watts, so with a Q of 10 you need 17 kVAR of peak reactive power 4. at this peak power the rms voltage on the primary circuit is 6 kV / sqrt(2) which is around 4200 volts, you use this to calculate the C and L (from their reactances)
Registered Member #3806
Joined: Sat Apr 02 2011, 09:20PM
Location: France
Posts: 259
Dr. Dark Current wrote ...
Do you have a RF bypass capacitor parallel to the doubler diode?
I added it
Dr. Dark Current wrote ...
A primary tank circuit Q of 38 and even 19 is too much...
To lower my Q, I changed my primary capacitance from 470pF to 235pF by adding a similar cap in serie and increased number of primary turns to keep the same resonant frequency.
According to Steve's formulas, this should divide my Q by 2, right ?
Then with a signal generator and o'scope, I did the following:
I measured my secondary resonnant frequency (with primary in place but not connected to primary cap). To take into account secondary detuning due to spark, I also put a piece of 15cm copper wire comming out of the top load. It gave me a frequency of 2.171 MHZ
Using primary taps I tuned my primary circuit to the same resonnant frequency
Then I fired the coil, it was working fine but sparks were not longer than without voltage doubler
So I tried moving the primary tap, when decreasing primary L (higher primary frequ) I got less output as expected.
Then I tried to increase primary L (lower primary frequ) as suggested:
Dr. Dark Current wrote ...
Try increasing the number of primary turns, but just a little, go 1 turn at time (basically you are detuning the circuit to get the correct loaded impedance). The spark length should first increase and then drop, you leave the tap where you get the longest sparks.
And this is where weird things happened: I get higher and higher output when detuning the primary to a lower frequency, even past way beyond resonnance point. I got the best output at the last tap, which gives a primary resonnant frequency of 1.660 MHz
But doing this, the secondary began to smoke and I had to quickly turn the coil off:
The area where the secondary heated up corresponds to the center of the primary The copper wire is not broken and the secondary is still working, but I guess it got really hot
So basically I have three questions:
1) How comes I kept getting longer sparks when detuning the primary to a higher L way past resonance point ?
2) Why did the secondary heated up so much under this condition ?
Dr. Dark Current wrote ...
1. the doubled peak voltage is 4960 V * sqrt(2) which is ~7 kV, but this is just theoretical, lets assume 6 kV 2. the 572B has a plate dissipation of 160 watts, it is safe to assume plate input of 4 times the plate dissipation which is 640 watts 3. the peak input power is 8/3 times * average input power which is 1700 watts, so with a Q of 10 you need 17 kVAR of peak reactive power 4. at this peak power the rms voltage on the primary circuit is 6 kV / sqrt(2) which is around 4200 volts, you use this to calculate the C and L (from their reactances)
3) How do you calculate C and L from this ?
(sorry, this is not laziness, I tried but I'm not sure I got it right. Could you please explain further ?)
All I could do is:
I = 17000 kvar / 4200V = 4A Needed primary Z = 4200/4 = 1050 ohms
With my actual L (16µH) and C (235pF) values: Reactance of my primary L at 2.171MHz is 218 ohms Reactance of my primary C at 2.171MHz is 311 ohms This gives a primary reactance of 1/(1/218-1/311) = 729 ohms
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
Hi 1. Detuning alters the primary loaded impedance (and Q-factor) to better match the tube 2. Eddy current losses combined with IIR losses 3. I = V/X (X.. Xl, Xc) = Q/V (Q as in reactive power) You can not calculate the actual primary loaded impedance and Q. The reactances of L and C have nothing to do with primary loaded impedance.
Registered Member #3806
Joined: Sat Apr 02 2011, 09:20PM
Location: France
Posts: 259
Hi Dr DC and thanks for your time.
Dr. Dark Current wrote ...
Hi 1. Detuning alters the primary loaded impedance (and Q-factor) to better match the tube
I think I Got this. This somehow means that impedance matching, and correct Q-Factor are more important than proper primary tuning to resonnant frequency, right ?
Dr. Dark Current wrote ... 2. Eddy current losses combined with IIR losses
If I make a higher primary with interturn spacing, would this reduce this effect by spreading the magnetic field over a longer portion of the secondary ?
Dr. Dark Current wrote ... 3. I = V/X (X.. Xl, Xc) = Q/V (Q as in reactive power)
Aren't X, Xl and Xc reactances of tank circuit, tank L and tank C in your formula ?
Dr. Dark Current wrote ... You can not calculate the actual primary loaded impedance and Q. The reactances of L and C have nothing to do with primary loaded impedance.
Then I'm completely lost ... What do you call "Primary loaded imedance" ? And how do you properly calculate the Q-Factor of a VTTC ?
As it seems increasing primary L gives better results, I guess I have to increase it even more and decrease my C value accordingly to keep the same resonnant frequency. Do you agree with this ?
Registered Member #152
Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384
Hi, you say "correct Q-Factor are more important than proper primary tuning to resonnant frequency, right ?" Yes
"If I make a higher primary with interturn spacing, would this reduce this effect by spreading the magnetic field over a longer portion of the secondary ?" Yes, exactly.
"Aren't X, Xl and Xc reactances of tank circuit, tank L and tank C in your formula" Xl .. inductive reactance = 2*pi*f*L Xc .. capacitive reactance = 1/(2*pi*f*C)
"What do you call "Primary loaded imedance" ?" It is the impedance seen externally by the tube, or you might even say "damping resistance". The capacitive and inductive reactances cancel out and are not seen in any way by the tube. The loaded impedance depends on the spark loading (which you can not calculate) and other power losses.
"And how do you properly calculate the Q-Factor of a VTTC ?" I think there is no formula for this as the optimal primary tank Q-factor depends mainly on spark loading (which you can not calculate) and coupling to the secondary. The primary tank Q of 10 usually works well.
"As it seems increasing primary L gives better results, I guess I have to increase it even more and decrease my C value accordingly to keep the same resonnant frequency. Do you agree with this ?" No, you calculate the C for proper Q-factor and then just tune L for the biggest sparks.
Registered Member #3806
Joined: Sat Apr 02 2011, 09:20PM
Location: France
Posts: 259
Hi Jan, thanks for all these explanations, much appreciated
Dr. Dark Current wrote ...
"If I make a higher primary with interturn spacing, would this reduce this effect by spreading the magnetic field over a longer portion of the secondary ?" Yes, exactly.
That's a good point !
I now understand what you call Primary Loaded Impedance and why we cannot calculate it, thanks for the detailed explanation.
Dr. Dark Current wrote ...
I think there is no formula for this as the optimal primary tank Q-factor depends mainly on spark loading (which you can not calculate) and coupling to the secondary. The primary tank Q of 10 usually works well.
Ok, I think I made a confusion between Primary Q and Q-Factor (I thought they were the same thing)
If I get it right, Primary Q is what we also call selectivity or bandwith of an RLC circuit where R is here tube load impedance in serie with the LC circuit. Is that right ?
Q = R/(2*pi*f*L) = R*2*pi*f*C with R = tube plate voltage / 4 * tube plate current (for AC or pulsed DC) or R = tube plate voltage / 2 * tube plate current (for smoothed DC)
Do you agree with Steve's formulas ?
And then, my next question is, what do you call "Q-Factor". All researches I made about it lead me to LC circuit selectivity or Q, but according to what you said it is a different thing that applies to the whole coil and not only to the primary.
Dr. Dark Current wrote ...
No, you calculate the C for proper Q-factor and then just tune L for the biggest sparks.
How can I do this if Q-Factor cannot be calculated ? Maybe you meant "you calculate the C for proper Q and then just tune L for the biggest sparks" ?
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My first coil : 572B VTTC
