Ok, Richard that wasn't clear to me, I think at resonance the tank is all R, but I put mostly R because I figured you would have an objection to all R. So are you saying it is pure R at resonance?
I want to understand the use of an air variable to match an antenna to the tank of a crystal radio, over the AMBCB frequency range. With that, I found I need to understand the series to parallel conversion, which I now understand, just IS, it's not anything you do. A series RC has a parallel RC equivalent. I'm not sure how it can be both at the same time. But as long as that R is transformed up, and minimally loads my tank, that's all good. Then, I understand I still have C left that I can use as part of the C for resonating my LC tank. Mikek
You've got it now, Mike. The series/parallel equivalent circuits are just a mathematical tool for putting things in a form you can handle easier. Adding the capacitor does not change the circuit from series to parallel or the other way around.
Lets say you have a fixed frequency AC source, a resistor, and a capacitor. The R and C is in a box where you can't see how they are wired. You measure the voltage applied to the box and measure the current (with phase) into the box.
You could calculate the value of the R and C, right? Most people would calculate them as an R in series with a C. But, there is a parallel R (of a different value from the series case) and a parallel C (of a different value) which will give the same measurements as the series case. You cannot tell which way they are wired internally and, because of this, you cannot tell the actual values of the components. But, for analysis or synthesis, it won't matter.
You can mathematically change a circuit around from series (impedance) to parallel (admittance). This is the conversion that you've been hung up on.
Ya, I have a better understanding now. I need to run a few cases and see the minimum and maximum capacitor needed for a proposed situation. I wish it would warm up, I'd like to put up an antenna and measure it, to get a real case to work with. Mikek
What I found to be most instructive, in understanding this (that is, series representations vs. parallel representations) was to drop down a level into the underlying mathematics.
Start with the fact that you have two impedances (let's call them Z1 and Z2) which are in parallel.
Toss in the basic formula for the result:
Ztot = (Z1 * Z2) / (Z1 + Z2) Since you have a resistor R and a capacitor C in parallel, Z1 is a real number (it's just R), and Z2 will be an imaginary number (it's
-i/2piFC, or 1/jWC if you prefer engineering notations and squint at the "W" so it looks like an omega).
Plug these values into the equation above, and simplify according to the rules for complex number mathematics. You'll end up with Ztot being a complex number, equal to the sum of a pure resistance (real) and a pure capacitance (imaginary). These are the impedances of the series network equivalent to your original parallel network.
Alternate route to the same solution: take each of the two impedances and invert them, to determine the admittances of the two components. The admittance of R will be purely real, while the admittance of C will be purely imaginary. Add the two together (since they're in parallel) to get the complex admittance of the parallel combination. Now, invert this complex number according to the usual rules, to get the equivalent complex impedance... this will be a complex number, the sum of a pure resistance and a pure capacitance. The numbers you get will be the same as in the previous work-through.
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Look into getting and learning (free) LTSpice for circuit analysis.
Also look into getting and learning EZNEC (free) for antenna analysis.
You can't beat putting up an antenna and building a circuit for it and measuring results. But, during adverse weather, you can experiment with simulation software and learn a lot. Then you can try your simulated experiments when the wx is good.
Let's say there is absolutely no loss in the Tank (superconduction and perfect dissipation values as it were); then we would have to ask ourselves what happens to energy applied to this Tank at resonance? It can never enter it, thus the Tank is, in effect, infinite in resistance. But what about the circulating currents? The Tank is, in effect, infinite in conductance.
Infinite Ohms & Zero Ohms simultaneously.
Is this the Z of the Tank? Is this the R of the Tank to which you are matching? No, not even close and certainly it has nothing to do with resonance - except the condition is a function of it being at resonance. A low Z Tank or a high Z Tank each evidences the same Infinite Ohms & Zero Ohms simultaneity given my initial condition of absolute losslessness.
For the energy being applied to or drawn from the Tank, the Tank is in parallel operation. For energy in the Tank, the Tank is in series operation. Where is the Q in this duality? Q suffers by the nature of what you call R. Q has two different values by this duality. One is called "Loaded Q" and as you might guess, the second is called "Unloaded Q." Consult Terman for the engineering design rationale for optimal Qs as I suggested.
When the discussion of "matching" seeks to employ R (pure resistance), then the next step is toward a conjugate match and elaborations of efficiency and maximum transfer of power. There is also an alternative discussion called the Zo Match. This second match seems to invite the same elaborations (many who post here try to force them both into the same salad bowl and cover the illogic with dressing).
When you offered the comment about "the tank is a high impedance (mostly R)" it was steering the car off the cliff. Is this a Zo match or a Conjugate match you are seeking? (I can already anticipate this has gone over your head, as well as many readers. This and the questions that follow are rhetorical.)
For instance, and returning to antennas (the purpose of this group's discussion focus), you can have very high Z antennas with very low resistance characteristics. Do you want a Zo Match, or a Conjugate Match? Let me flip the antenna: you can have very high Z antennas with very high resistance characteristics. Do you want a Zo Match, or a Conjugate Match? Let's do this sideways: you can have very low Z antennas with very low resistance characteristics. Do you want a Zo Match, or a Conjugate Match? I could box the compass here, but the I think I will let the reader off.
John had some number issues with Tony's explanation, but the gist of Tony's rational treatment should be your lesson as it provides for your requested "why." It also implies (by my comments of the sudden appearance of two new components) that our (Ham) tuners have been designed to introduce the proper amounts of reactances in the proper parallel/series relationships to enable the necessary transform towards optimal Q and loading balance. The most elaborate of tuners can change from Pi to T topologies, or series L parallel C (or series C parallel L), or series LC, or parallel LC, or parallel C series L (or parallel L series C)... and any of the other combinations I have not enumerated (about 9 in all). Each shines for a particular situation - you have named only one.
It is not a trivial discussion by any means even when we are talking about the addition of only two new components. So, your obtaining an understanding is not going to be achieved at one sitting in front of the "definitive" posting to a thread.
One problem of seeking the "definitive" posting is that it cannot be born from a broken premise that article you were trying to figure out is lame in the extreme. Given everything you have revealed about it, it didn't present a solution to its fantasy antenna. That is why this work of fiction is not understandable.
just throw a wire out the window, plug it in, and see if it works! ANY antenna connected with a hunk of hookup wire will work better than NO antenna with a perfectly designed match!
A lot there, but I didn't get anything out of it. At resonance, does the tank look capacitive, inductive, resistive, or all
Because of the wavelength of the BCB antennas are usually short and capacitive, so simple tuning with a single series cap works. The problem begins when you tune to the high end of the band and you have to much capacitance to get to resonance with your tank. Then the inductor can be added parallel to the series antenna tuning cap. Alternately and you could increase the tank inductor size.
I didn't rewrite the whole article here, there was a solution with three equations, that, using the antenna finds the L with the constrants of highest operating frequency and lowest capacitance of your capacitor, then a program is run that finds values for C (antenna) and C (tank) and I'm not ready to go here yet C (load), he also tunes the diode/earphone load for optimum with yet another capacitor. The funs over for now go to get ready for work, Thanks, Mikek PS, he runs the program with another, what you call fantasy antenna, and I agree...
What you want is the lightest final load sufficient to drive a speaker for a detectable sound matched to the Tank such that it does not degrade its Q which in turn is the highest possible value for supporting the largest amount of signal from the antenna at hand.
There is a chance the Q (loaded) will be high enough to limit audio bandwidth. So (I think) we couple more energy into the tank for more signal and this would lower Q for a wider bandwidth.
Here's a question I have brewing.
I have three circuits to put together, a source, a tank and, a load. I have two scenerios. hmm..seems as though I have three! For now assume they are all resistive. These are all set up for maximum power transfer, just in different order.
Scenerio 1. Let's say the tank is 1 megohm. I drive the tank with a 1meg source, so now I have 500Kohm circuit impedance. Then I load this with 500Kohm load. So.. The 1 megohm tank is loaded with 333,333ohms, 1meg//500k The 1 meg antenna is loaded with 333,333ohms, 1meg//500k The 500Kohm load is drive by 500kohms. 1meg//1meg
Scenerio 2.
1 megohm tank. I put a 1 megohm load I can drive the tank with a 500Kohm source, So.. The 1 megohm tank is loaded with 333,333ohms, 500k//1meg The 500 Kohm antenna is loaded with 500Kohms, 1meg//1meg The 1 megohm load is driven by 500kohms. 1meg//500k
Scenerio 3.
1 megohm tankThe I drive the tank with 2 megohm source and load it with a 2 megohm load. So.. The 1 megohm tank is loaded with 1 Mohm, 2Mohm//2Mohm The 2 meg antenna is loaded with 666,666 ohms, 1meg//2meg
2 megohm load is drive by 666,666 ohms. 1meg//2meg
I have no clue where maximum power is delivered from the antenna to the load.
I do not understand where your -2947j figure comes from. I see it appearing nowhere in my calculations.
The antenna mentioned by Mikek has an impedance of 58R-1,072j which, according to my spreadsheet, corresponds (at 1 MHz) to the parallel of 19862R and -1075j (that is a 148,1 pF capacitor).
In any case, parallel series trasformations never result in a change of the reactance sign; therefore it is not possible that a -2957j (negative) reactance is transformed into a +2948j (positive) reactance.
Well, it's not mine and it appears in you post as 54 pF. Isn't 58R in series with 54 pF equal to 58-2947j? And isn't the parallel equivalent of that equal to 149k ohms of resistance in parallel with -2948 ohms of reactance (~54 pF)?
I'm pointing out that you slipped a decimal point or you would have seen that 54 pF is too much it results in the parallel equivalent resistance of 149k rather than 1.49M. Mike's figure of about 18 pF (17 pF series combination) will do the job.
You are correct. I allowed the sign of the suseptance to creep through.
Yes, at one o' clock in the morning, I slipped the decimal point. So, the total antenna series capacitance should have been about 17 pF, not 54 pF. This requires putting a 19-pF capacitance in series with the antenna, not 85 pF.
And the inductance resonating the residual parallel capacitance becomes 1,490 uH instead of 470 uH.
Hasn't anyone pointed out that this a problem made for using a Smith Chart?
(Since no one really seems capable of doing the math :-) ...Jim Thompson
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I love to cook with wine. Sometimes I even put it in the food.
How? (It would be a sad day for us all if more input to a Tank lowered its Q were true.)
Actually, that remains to be seen. We must first establish that we have the same available power in all three which I will give the arbitrary value of 1 to simplify the math. Also, you mix your source loading in these models, so in the sense of Norton/Thevenin sources I describe the power being supplied by a parallel current source to keep the units uniform.
This presumes a parallel current source by your description of the input and tank appearing as a 500K circuit. This is why set the initial condition of there being a current source for all scenarios.
The parallel current source then sees 250K Ohm for the same power available to all scenarios. Pavailable = 1 = i²·250K i = sqrt(1/250K) As the current does not divide evenly, then we will work to find the power to the load through voltage sharing. Obviously, there is the same voltage across the three components, hence: e = i·250K = 500 Pload = e²/500K = 0.50
This does not qualify either a parallel current nor series voltage source, but as both are fungible to design with the same value resistance, then I will proceed as before with all three resistors in parallel to a parallel current source: Pavailable = 1 = i²·250K i = sqrt(1/250K) Obviously, there is the same voltage across the three components, hence: e = i·250K = 500 Pload = e²/1000K = 0.25
This does not qualify either a parallel current nor series voltage source, but as both are fungible to design with the same value resistance, then I will proceed as before with all three resistors in parallel to a parallel current source: Pavailable = 1 = i²·500K i = sqrt(1/500K) Obviously, there is the same voltage across the three components, hence: e = i·500K = 707 Pload = e²/2000K = 0.25
Any clues now? Barring any math or conceptual error on my part, then by one account more power (that is one measure of success) is delivered to the load when its resistance is lowest.
My thought was, if we couple more energy from the antenna, it loads the tank more lowering Q. The thought might need more work....
Scenerio 1 is how I have always thought about the system. Nice to know where max power transfer is.
I'm still at design highest Q tank circuit then transform antenna to match Z of tank. That's about as I want to go for now. Not ready to get into that diode thing again. Unless you've been studying :-) running for cover..... Thanks, Mikek
Well, that is nominally considered as "matching" in that the coupling from the antenna is minimal so as to not load the parallel resonant circuit; a tradeoff actually between energy transfer and loading. A better way might be to have a low Z tap on the inductor; a method used in radios since the 60's i think.
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