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IamSmooth

Tue Dec 12 2006, 02:53PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
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Posts: 1567

I am in the process of tuning my primary and secondary. I was reading in an old RF handbook that when the two are matched and I feed the resonant frequency into the primary with a signal generator there is a peak on the secondary just to the left and right of the resonant frequency. I can pick this up with a meter long antenna attached to my oscilloscope probe. Why is there a double hump and not just a peak at the resonant frequency?

Steve Ward

Tue Dec 12 2006, 07:50PM

Registered Member #146 Joined: Sun Feb 12 2006, 04:21AM
Location: Austin Tx
Posts: 1055

What you are experiencing is what some call "Frequency splitting". Why does this happen? The cause is due to having 2 resonant circuits coupled to eachother (and note, that the coupling is relatively loose so the effects are small, making the 2 humps close together). There are a few ways of explaining this, i will go with the way that i sorta understand. Energy is transfered between 2 coupled circuits at a certain frequency (depending on the coupling). This frequency is generally much lower than the resonant frequency for tesla coils. The effect is the multiplication of 2 cosines, one being the "carrier" and the other the "modulation" if you like. From trigonometry, we see that when 2 cosines are multiplied such as cos(m)*cos(c), the result is 1/2cos(c-m) + 1/2cos(c+m) (using c for carrier, and m for modulation). Notice that you now have 2 frequencies, one slightly above the natural resonance, and the other slightly below, each containing half of the average energy content of the system (i have to say average, because the energy is actually shifting between these 2). Its interesting to note that this "modulation" frequency goes UP when the coupling is increased, thus you will find the 2 peaks get pushed further apart.

There is another way of viewing this phenomenon which involves the secondary circuit seeing some of the primary capacitance, and the primary circuit seeing some of the secondary inductance, causing the resonant frequency of of the secondary to drop, and the primary to raise slightly, thus again, the double peak.

The natural Fo is in the middle of these 2.

Hazmatt_(The Underdog)

Tue Dec 12 2006, 08:04PM

Registered Member #135 Joined: Sat Feb 11 2006, 12:06AM
Location: Anywhere is fine
Posts: 1735

Try to get some simulation software from somewhere. PSpice is pretty good. It has answered a lot of my questions through simulation and evaluation.

What you're looking at is a 2 LCR system, a 4th order differential equation. You can tune it for one peak but its pretty hard to get there. Solving it by by hand would be pretty rough business, I know because I've had a couple of 4th order projects and they were a mess.

I think the double peak is a result of the damping of both LCR circuits. If they had only complex components then you would have an oscillator. And if both systems were in the same resonance you would have a very large peak. But since they have real losses damping comes into play. The primary resistance being a few tenths of an ohm, and the secondary being a few tens of ohms, you can see how the one resonance is shifted away from the other slightly.

What I would recommend is checking out your results in a solvable manner, like 2 second order differentials. Then you can put your secondary resistance in and primary resistance in for both second order equns'. and see how the solutions vary.
Then you could take the secondary's solution and use it to solve for primary L, knowing your C and R values.

Sulaiman

Tue Dec 12 2006, 09:10PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3140

You could have a play with this applet

Hazmatt_(The Underdog)

Tue Dec 12 2006, 09:30PM

Registered Member #135 Joined: Sat Feb 11 2006, 12:06AM
Location: Anywhere is fine
Posts: 1735

I played with the applet to find the critical coupling and I think its really interesting.

With a 4th order spring mass system for our Physics projects we were seeing the fast (carrier) and slow (modulation) frequencies regardless of any coupling being introduced. The fast and slow frequencies were purely characteristic to the values of the springs and masses chosen. And since this is the mechanical analog to an LC circuit, I payed little attention to the coupling of the circuit.

It looks like the instructors generalities have come back to haunt me once again.

IamSmooth

Tue Dec 19 2006, 12:34AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

Ok, so there is a peak above and below the resonant frequency. My coil is ready to go. I have measured the resonant frequency of the coil to be 125KHz and I have tuned the primary to this value. When I connect the secondary and drive the primary with a sinusoidal signal at 125Khz the Fl = 116Khz and Fh = 135. This gives me a coupling coeff = 0.15 with a 3.7% difference between the two (calculation are approximations).

My theoretical question is this: If the primary resonates at 125, and my peak voltages occur at 116 and 135, how do I get high voltage output if the "sweet spots" are 10Khz above and below the system's resonanting frequency?

Steve Conner

Tue Dec 19 2006, 09:00AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

It's a long story. I've tried explaining it to dozens of people over the years, and as far as I can tell, not one of them understood it, except a 70-something retired mechanical engineer who knew all about coupled resonant modes and stuff, from doing vibration analysis of machines.

I'm going to try one last time though:

1) When the primary and secondary coils are brought into coupling, they DO NOT RESONATE AT THE SAME FREQUENCIES THEY DID BEFORE when they were apart.

2) It is now POINTLESS to try and think of the two coils as separate resonators. They work together as a fourth-order resonator with two new resonant frequencies. Both coils respond to both frequencies (and a Tesla resonator produces HV output when driven at either frequency, or even both simultaneously, which is effectively what a spark-gap driver does)

3) Altering the circuit constants of either coil will change both of the coupled resonant frequencies.

There is no mental trick to visualise the system made from simpler building blocks: you just have to learn the concept of a fourth-order resonator. (As a bonus, you'll then understand how bass reflex loudspeakers work too.)

IamSmooth

Tue Dec 19 2006, 05:37PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

Thanks Steve. Actually, if I accept the fact that they *resonate* at these two frequencies as a coupled unit I do understand your answer. I was thinking that they resonated at the frequency when they were apart; you've cleared up that is the Flow and/or the Fhigh to which they resonate.

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