Ok, looking at a Helical Resonator, we have a 1/4 WTL driven by a voltage source at one end and open at the 1/4 wave length end. As we know the signal propagates to the open end then reflects back in phase and adds to the transmitted wave to increase the amplitude times two (perfect reflection). Now the reflected wave propagates back to the transmitting end which is a short ( low Z) and reflects back out of phase with no addition amplitude to the standing waves. That causes a VSWR of two! Is it possible to have a VSWR much greater than 2? Given a low Z, (compaired to the reactance of the line) we can only have one in phase reflection with a voltage gain of 2. Is this true or am I missing something? Thanks, Harry
Yes, you're missing lots. There's lots more to a helical resonator than a 1/4 wave transmission line.
If the system, any system, is lossless... If you TRY to put power in and nothing comes out the other end... Then the standing wave ratio can be infinite.
You can have every VSWR in an actual system. In your system (assuming losless quarter wave TL), an open termination (voltage reflection coeffcient of +1, current reflection coefficient = -1) results in a voltage reflection coefficient of -1 at the quarter wave distance from the open termination (that is fully left on the edge of the Smith Chart).
A voltage reflection coefficient of -1 equals zero Ohms and VSWR = infinite. Are you familiar with the Smith Chart?
If the VSWR is >2 implies that the waveform is reflecting in phase (additive) off of both ends of the TL but one end (driven) is a short, which will cause phase reversil and non-addivite reflection. ?? Thanks Harry
Brent, Thanks for the neat tutorials. This TL theroy is different from what I propose. This setup is all back terminated with the line charateristic Z (50R). I my case I have a very low Z voltage sourse, driving 1KW, into the coil. As such I believe that my VSWR must be <
You're mixing apples and oranges. Take a black box. Put 1KW into it. It MUST come out somewhere...electromagnetic radiation, heat, light, something must come out...eventually...or it'll explode...which is just another form of something coming out. You just can't put 1KW into a region of space without it coming out somewhere in some form...eventually. Don't tell me about black holes...you ain't got one.
If nothing is coming out, you can't be putting anything in, so your vswr is infinite.
A VSWR of < 2 implies that the load impedance is inside the 2.0 VSWR circle on the smith chart. VSWR is just a scalar measure of "quality" normalized to transmission line characteristic impedance. It says nothing about whether the source can actually deliver power into the load impedance. The number is useful in many applications if you don't try to extrapolate it into something it isn't. In many applications, the objective is to make VSWR == 0. When you get there, you then know quite a lot about whether the source can deliver power into the matched load.
Maybe you can post a circuit diagram of the setup, together with some properties of the helical resonator and the source.
In addition, when your signal isn't narrow band (so steady state condition isn't reached), you may specify this also. If so, the calculus becomes significantly more elaborate.
Mike, you make a great point and clearly stated. Ok when I incorrectly stated 1KW, I was trying to emphasize the low Z of the transmitting end. Basically I have what looks like a Tesla transformer with the primary loosely coupled to this secondary. The primary is driving lots of power and it all goes somewhere. To get max voltage output, we will resonate the primary and secondary and also try to get a 1/4 wavelength standing wave for added voltage gain. My belief is that the standing wave can add a factor of 2X voltage but no more, due to the low Z driving end. Regards, Harry
OK, no problem. Instead of the Smith Chart, I will use a graph that shows the relation between complex impedance and complex reflection coefficient (or phase and modulus)....
Still too few details to be much help. There's a lot of ART to making a tesla coil. And I know little about it. Once you get past the obvious issues of controlling voltage breakdown, I think it's all about turns ratio and coupling coefficient.
Because of the voltages involved, I don't think there's much you can do about resonating the secondary. Because of the L and distributed C, there will be a natural resonance. You just have to take what you get as the system frequency. You can resonate the primary to match.
In theory, you can tap the primary and drive it anywhere you want. As a practical matter, there will be an optimum tap point that balances losses in the primary against losses in the drive circuitry. So, there will be some impedance at the tap point. For a tesla coil, the stray circuit elements and losses that don't show up on a simple schematic have more effect on performance than those that do. An accurate circuit model can get complex.
You can't drive a resonant circuit with a voltage source. Well, you can, but it isn't helpful. The zero impedance hanging off the tap damps the primary resonance to nothing. If you want high coupling coefficient, the secondary resonance gets damped too.
A chunk of transmission line provides impedance transformation. On the smith chart, you plot the load impedance of your system. Draw a circle around the center. That's the SWR circle. Depending on the length of the cable, its input impedance will be somewhere on that circle. You pick a point on the circle that maximizes the ability of your source to deliver power. That determines the optimum cable length. The transmission line impedance can be anything you want and give you a wide range of theoretical impedance transformations. As a practical matter, with easily achievable impedances, the range of transformation is VERY much more limited.
At the frequency of interest, the cable can be LOOOOOONG! Losses in the cable may overshadow other considerations.
The VSWR concept is leading you astray. Dump the concept.
You have some source. Its ability to deliver power into various loads can be determined. Take the impedance at the primary tap and build a matching circuit that gets the maximum power transferred from the source to the secondary output of your coil. If a transmission line is part...or all...of that circuit, so be it. The only reason you'd care about VSWR is that the line might arc at some node along its length.
Remember, for high voltage circuits, the "load" can be a nonlinear function of drive; e.g. corona discharge. Just one more of those pesky elements that doesn't show up on a simple model. So, maximum power may not coincide exactly with maximum voltage out, but it'll be a good start.
On 7/10/2011 11:12 AM, Harry D wrote: > Ok, looking at a Helical Resonator, we have a 1/4 WTL driven by a > voltage source at one end and open at the 1/4 wave length end. As we > know the signal propagates to the open end then reflects back
Okay so far.
By the time it reaches the transmitting end again, the voltage is 180 degrees out of phase with the input signal.
Yes.
No. Adding 1 at zero degrees plus 1 at 180 degrees gives zero.
SWR = Vmax/Vmin. Vmin is zero due to the out-of-phase summation of the incoming and reflected waves. Therefore, SWR = infinite.
You are missing a great deal. You have already been given excellent advice and chose to ignore it. You will have a hard time getting help here if you leave the impression that "it isn't what I wanted to hear, so I'll just ignore it."
Since the SWR is mostly irrelevant in this situation, having a faulty SWR analysis should not affect your results.
As long as you accurately model your system and optimize appropriately, the transmission line effects will get tuned out.
There are lots of successful products that have unnecessary stuff in 'em. Gives incentives for competitors to enter the market with lower cost products. Capitalism at it's finest. Everybody wins.
I once patched up a troubled instrumentation design using a technique I'd used in another product. Worked great. Took about two-dozen components.
The product shipped and made some money.
I later learned that a competitor had taken a fresh look and solved the same problem with a single resistor. Boy, was my face red...which was surprising. Shoulda been brown based on the place my head was when I implemented the original design.
Your explanation has Vin=3D0 with an output on Vmax so you get infinite gain (SWR). My Vin =3D Vmin=3D voltage source and it has a driven voltage on it. The Zc =3D 40K so the Vmin node is easy to drive. Does this make any sense? Harry
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