ENERGY AMPLIFICATION

[gyulasun]

Yes Gyula  Maybe it is that they have the same resonance, but I remember the equal energy thing.
Like a huge coil and a tiny pico cap may not be as good as a larger cap, And the same vice versa, a huge cap and a tiny inductor.  where is the middle?   If we know the middle, we can scale our parts equally in either direction for what ever purpose..   Like a pendulum and flywheel,  the pen is just a coil with a cap.  The total weight of the offset flywheel is the inductor and the cap is how much offset there is. Think about the many combination's that would give different results?


Today I am messing with a larger wire/few turns coil. Just to see what I can see. 

I was checking out Core's project.  He says it gives him headaches pretty quick, and goes away as quick as they come when he shuts it off.   I would like to stay away from such projects. Hopefully this is not one of them.  =]

Be back in a bit.

Mags

Hi Mags,

So you are thinking of the so called L/C ratio I guess. It means that the AC impedance of an LC parallel circuit at resonance can be much different in case you use a 10uH - 100pF combination (their resonant frequency is 5.03MHz) or you use for instance a 100uH - 10pF combination for the same 5.03MHz resonant frequency.
The impedance for parallel LC tanks comes from the reactance value of either the L or C (they are the same at resonance) multiplied by the coil's Q or quality factor. (Assuming the Q of the capacitor is much higher than that of the coil, normally this is valid in practice.)
So assuming a moderate Q of say 80 for a practical coil at 5MHz, the impedance, Z for such a 10uH coil + 100pF combination comes out as Z=Q*XL, XL=2PI*f*L=6.28*5*10=314 Ohm so Z=Q*314=80*314=25.12 kOhm. Now if you use the 100uH - 10pF combination and suppose you can make a 100uH coil for 5MHz with the same Q=80 value (still practical), then the impedance for this case comes out as XL=6.28*5*100=3140 Ohm, Z=Q*3140=80*3140=251.2 kOhm

So we have a resonant AC impedance of 251kOhm versus the 25.1kOm just by using a 100uH coil instead of the 10uH (assuming a Q of 80 for both, this is important when you compare) and using a 10pF cap instead of the 100pF capacitor value, respectively.

Now the question is how the stored energy varies in the different cases? The stored energy in a capacitor for instance depends on the second power of the voltage across it and linearly depends on its actual capacitance value: E=C*V²/2  This means if you decrease the cap value 10 times, from 100pF to 10pF, the stored energy gets 10 times less, assuming the voltage would remain the same,  however, for the capacitance case of 10pF the 251kOhm resultant AC impedance inherently involves higher resonant voltages too (still assuming the loaded Q is 80 of course) so the stored energy can easily turn to not a decrease but a gain. The resonant impedance increase is ten times (251kOhm/25kOhm) so the AC voltage across it can also be 10 times higher and the 10 times increase in voltage means a 100 times increase in the stored cap energy.

Considering this for the coil, the stored energy E=L*I²/2 and the ten times linear increase of L from 10uH to 100uH gives a 10 times increase in stored energy and the second power of the current may remain at the same value like in case of the 10uH if the loaded Q is kept at the same Q=80 value.

If these 'ramblings' are correct (I believe they are), then there seems to be no middle value for the L-C combinations you look for but an extreme limit of using the highest value coil with the smallest value cap if you are to maximize stored energy...  This may correspond to Tesla's coil where he did not use any parallel capacitance but the unavoidable self-capacitance and used thick wire for his coils to reduce losses to a minimum.

I have not mentioned yet that if you increase the coil's inductance from 10uH to 100uH and suppose you use a thicker wire for the 100uH coil to get the same DC copper resistance like the 10uH coil has, then the quality factor also increases 10 times because Q=X_L/R where R is the losses that include the DC resistance, the main loss factor for an air core coil. The higher coil Q gives higher AC resonant impedance. So your larger wire tinkering is good direction.

If there is a correct equivalence comparison between a mechanical pendulum and an LC tank circuit, then the above reasonings somehow should reflect back on the pendulum, within reasonable mechanical limits of course.

I hope my 'ramblings' may help you in your progress.

Gyula

[iflewmyown]

@gyulasun
An excellent explanation!!
Garry

[IWD]

to gyula: great nice facts

actually ..can be easily say: the energy in LC tank will be grater when you use bigger capacitance than inductance. in case 100uh 10pf will be energy smaler than in case 10uh 100pf.
But the tank will also need more energy to delivery (smaler resistance)

In reality, on the coil will be almost same voltage in both cases, but the current will be bigger in case of bigger capacitor.(that means also bigger power on coil)

When you use a thicker wire to coil, then the coil wil also have bigger capacitance because bigger surface of the wire, and Tesla uses a ground which add serious capacitance to the coil but do not decrease his Q.


But the ground ...is already charged, so when coil oscilate every time on the grounding point of the coil is positive is smal amout of charges indrawn to the coil, that create a magnetic field which will add to the magnetic field of the coil and when the magnetic field collapse.... charges up the capacitance to bigger value that before ...and bigger potential will indrawn more charges from earth, so in that case on secondary coil can be bigger movements of current (bigger power) as tesla stated in one of his patent. that of course will have bigger influence on the primary.
if you feed your tesla coil directly from semiconductors or some like that. you will see a serious rise in consumption form source in resonance point. Interesting is... that is not happening  when you use disruptive discharge thru spark gap metod because there is only final amount of energy which can be delivered.

[gyulasun]

Garry and IWD: Thanks for the kind words.

To IWD:

You wrote the energy in an LC tank is greater when using the lower value coil i.e. in my example the 10uH - 100pF values at 5.03MHz, compared to the 100uH - 10pF combination. And you added that for the 10uH - 100pF combination the tank needs a higher input energy to keep up oscillations because that combination has less resonant impedance than 100uH -10pF combination. 
Ok, I agree but this is just the opposite conclusion to what I wrote in my previous post...
However, now I think both of us are right LOL and the explanation for this virtual controversy is found here, see the very bottom text in this link:
http://www.play-hookey.com/ac_theory/ac_lc_parallel.html

"...if we use a large value of L and a small value of C, their reactances will be high and the amount of current circulating in the tank will be small. If we reverse that and use a low value of L and a high value of C, their reactances will be low and the amount of current circulating in the tank will be much greater. Many applications of this type of circuit depend on the amount of circulating current as well as the resonant frequency, so you need to be aware of this factor. In fact, in real-world circuits that cannot avoid having some resistance (especially in L), it is possible to have such a high circulating current that the energy lost in R (p = i²R) is sufficient to cause L to burn up!"

Now I wonder if Magluvin has received an answer for his question... 

Gyula

to gyula: great nice facts

actually ..can be easily say: the energy in LC tank will be grater when you use bigger capacitance than inductance. in case 100uh 10pf will be energy smaler than in case 10uh 100pf.
But the tank will also need more energy to delivery (smaler resistance)

In reality, on the coil will be almost same voltage in both cases, but the current will be bigger in case of bigger capacitor.(that means also bigger power on coil)

When you use a thicker wire to coil, then the coil wil also have bigger capacitance because bigger surface of the wire, and Tesla uses a ground which add serious capacitance to the coil but do not decrease his Q.


But the ground ...is already charged, so when coil oscilate every time on the grounding point of the coil is positive is smal amout of charges indrawn to the coil, that create a magnetic field which will add to the magnetic field of the coil and when the magnetic field collapse.... charges up the capacitance to bigger value that before ...and bigger potential will indrawn more charges from earth, so in that case on secondary coil can be bigger movements of current (bigger power) as tesla stated in one of his patent. that of course will have bigger influence on the primary.
if you feed your tesla coil directly from semiconductors or some like that. you will see a serious rise in consumption form source in resonance point. Interesting is... that is not happening  when you use disruptive discharge thru spark gap metod because there is only final amount of energy which can be delivered.

[Magluvin]

Good stuff guys.  Thanks for your work here.  It is all work when you think about it.  =]

Im tryin to bust a problem on my buddys laptop. Has an antivirus virus. lol   You know, the antivirus that yells and screams at you to purchase their product, that probably doesnt exisit. Every 20 seconds it interupts anything you do.
I should be done in a bit.

I cant wait to get back and absorb it all..  =]


Mags