Understanding Parallel to Series conversion

Ahh....1 should have said Series to Parallel Conversion. Mikek

Hi Mike,

Is this a fantasy antenna?

For the 58 Ohm resistive value, it would have to be about 300 feet tall - not the size of operation one usually comes to expect for a Xtal radio aficionado.

If it is that tall, it would exhibit 200 Ohms Inductive reactance (one fifth of what you report, and the opposite sign).

Something doesn't wash here.

58±j17 Ohms is still a complex impedance, and says nothing of match which can only be expressed in terms of the expected load R.

How that is arrived at is something of a mystery. By the numbers, you describe a 26000:1 mismatch.

Well, what you have described is sufficient mismatch to insure that. The English reading of your sentence also is instructive: the tank is isolated from the antenna, i.e. no signal is passed to it. This seems to be counterproductive in regards to detection.

Haven't we been down this road some months ago?

73's Richard Clark, KB7QHC

Perhaps it's a longwire antenna, of crappy wire, parallel to the ground?

It's an example from an article, I don't like the number either, seems like maybe

2 to 12 ohms would be more realistic. I think the capacitance is ok.

Ok, how about just working with the concept.

Sorry, I missed a math step, the R was converted to 1.5M. see formula below.

Here's what is stated in the article: " The concept is that at any given frequency, a parallel RC network has an equivalent series RC network, and vise versa. We can use this property to transform the real component of an impedance to a much higher or lower value. As long as Xc series >> R series. Xc (parallel) = Xc (series) R (parallel) = XC^2 (series) / R (series) The Xc of a 17pf at 1Mhz is 9368 Ohms. To rewrite Rp= 9368^2 / R = 1.513 Mohms

The article then goes on to say, The utility of this equivalence can be seen by choosing a sufficiently small value of C series (a large Xc series) A "small" resistance can then be transformed into a "large" value.

Mikek

=A0 =A0 l

=A0 =A0 l =A0 =A0 =A0 =A0Resonance

na.

anks, Mikek

Seems like it doesn't much matter whether the antenna is real or imagined; the point is rather what's going on when you couple a low impedance source to a parallel-resonant tank through a small capacitance...

Perhaps it will help you to think first about a low impedance source, let's say 50 ohms at 1.591MHz (1e7 radians/sec), that you want to couple to an (imaginary) inductor, 2.5mH with Qu=3D500 at that frequency, and maximize the energy transfer to the inductor. The inductance and Q implies that the effective series resistance of the inductor is 50 ohms. That means all we need to do is cancel out the inductive reactance, and we can do that with a series capacitance-- that happens to be 4pF. That assumes the ESR of the capacitance is zero, or essentially zero.

But what if the inductor has 1/4 as much inductance, 0.625mH, same Qu? Then its effective series resistance is 1/4 as much, or 12.5 ohms, and to get the same energy dissipated in it, you need the current through it to be 2 times as large as before (constant R*I^2). The total capacitance to resonate the tank is 4 times as much: 16pF. If you divide that into two 8pF caps, one directly across the inductor and the other in series with the 50 ohm source, (very close to) half the tank's circulating current flows in each capacitor. For the same energy delivered to the coil, the voltage at the top of the inductor must be half as much as before. Since the coupling capacitor to the source is twice as large as it was before, the load impedance the source sees must be the same as before (50 ohms, as required for max energy transfer).

Carry that another step, to an inductor with 1/16 the original inductance and the same Qu=3D500, and you need 64pF to resonate it and

16pF to couple to it from the low impedance source. Now you divide up the total capacitance with 3/4 of it (48pF) directly across the inductor and 1/4 of it to the source. 3/4 of the tank's circulating current flows in the 48pF capacitor, and 1/4 of it in the 16pF to the source; the source once again sees a 50 ohm load (because the voltage at the top of the inductor is 1/4 of the original, and the capacitor coupling to the source is 4 times as large -- with the current in the source unchanged).

The key to understanding this coupling, to me, is the division of the tank's circulating current between the two capacitances, one directly across the coil and one in series with the source. The smaller the inductance (at constant Q), the more circulating current required for the same energy dissipation, and the smaller _percentage_of_total_resonating_capacitance_ for coupling to the (constant impedance) source.

Cheers, Tom

At resonance, the LC tank disappears and you are left with a 1.5 Meg equivalent of the tank losses.

Adding a small capacitance in series with your antenna gives you, effectively, an antenna that looks like a 58R in series with a 17 pF capacitor.

So now your circuit looks like:

58R---17pF-----| | 1.5M | | ---- GND

I don't see any conversion at all.

John

Ah! After studying this for a while, I think I understand what you are driving at.

Convert series 58R and 17 pF to a parallel equivalent. That is, take the reciprocal of Z = 58 - 9368j to get Y = 661e-9 + 106.8e-6.

Now, what is the reciprocal of the real part of Y? That's 1/661e-9 or about 1.5 Meg. So, the real part of the parallel equivalent now looks like the resistance you are interested in.

Does this help?

Cheers, John

Hi Greg,

No, that would more likely result in a characteristic Z of 600 Ohms. If by being close to ground you mean the 58 Ohms finds itself invested in the dirt, well, yes that might be the case.

The long and short of it means that to investigate the problems of an antenna "loading" a tank coil means that you really must understand the antenna - not a simple thing as this thread will undoubtedly reveal.

73's Richard Clark, KB7QHC

Hi Mike,

Then your intuition is firing on all cylinders, great.

This is classically true. The frill of "at any given frequency" can be discarded.

This concept is not a "property," however; more a transformation (as should be apparent in the original statement). In other words concept equivalent property transform use four words to describe one thing - transform (plain and simple).

Well, what looks like hand-waving is probably close to the numbers one could expect.

The reason why I say hand-waving (and this is probably your gut reaction as to "why") is that the five lines of operations you quote starts with a presumed requirement and then proves it has been met.

What happens if Xc series R series?

This somewhat clouds the mystery for you of understanding Parallel to Series conversion. I wonder too. Are we dropping in a new component and stepping back with a wave and Voila! to find the Parallel circuit has suddenly been transformed? Yeah, that WOULD be a mystery.

And what is this Xc(parallel) and Xc(series) stuff?

Xc is stricty a function of pi, capacitance, and frequency.

And what is this R(parallel) and R(series) stuff?

R is a function of its, well, resistance. No variables to be found.

No doubt there is more to be extracted from this article.

73's Richard Clark, KB7QHC

Upon review of other correspondents to your plight, ditch the article and pick up an EE sophomore book on circuit analysis. The coverage should be encompassed in the section (from my own copy) called "Transform Driving Point Impedance and Admittance (Immittance)."

73's Richard Clark, KB7QHC

:-)

The author said it was an antenna in his attic. That's all I know. Mikek

The article's focus is on matching a crystal radio tank to the antenna, So as a general statement, the tank is a high impedance (mostly R) and the antenna has a low R and a C in the 100's of ohms. So I don't know, is that (>>) ? The question is retorical in nature.

I took a little liberty there, The article had a schematic diagram showing a series RC, then the radio showing the Parallel equivalent antenna connected. The the actual labels were Xs, Xp, Rs, and Rp. Sorry if I muddied it, the attempt was to make it clearer.

Seems like it doesn't much matter whether the antenna is real or imagined; the point is rather what's going on when you couple a low impedance source to a parallel-resonant tank through a small capacitance...

Perhaps it will help you to think first about a low impedance source, let's say 50 ohms at 1.591MHz (1e7 radians/sec), that you want to couple to an (imaginary) inductor, 2.5mH with Qu=500 at that frequency, and maximize the energy transfer to the inductor. The inductance and Q implies that the effective series resistance of the inductor is 50 ohms. That means all we need to do is cancel out the inductive reactance, and we can do that with a series capacitance-- that happens to be 4pF. That assumes the ESR of the capacitance is zero, or essentially zero.

But what if the inductor has 1/4 as much inductance, 0.625mH, same Qu? Then its effective series resistance is 1/4 as much, or 12.5 ohms, and to get the same energy dissipated in it, you need the current through it to be 2 times as large as before (constant R*I^2). The total capacitance to resonate the tank is 4 times as much: 16pF. If you divide that into two 8pF caps, one directly across the inductor and the other in series with the 50 ohm source, (very close to) half the tank's circulating current flows in each capacitor. For the same energy delivered to the coil, the voltage at the top of the inductor must be half as much as before. Since the coupling capacitor to the source is twice as large as it was before, the load impedance the source sees must be the same as before (50 ohms, as required for max energy transfer).

Carry that another step, to an inductor with 1/16 the original inductance and the same Qu=500, and you need 64pF to resonate it and

16pF to couple to it from the low impedance source. Now you divide up the total capacitance with 3/4 of it (48pF) directly across the inductor and 1/4 of it to the source. 3/4 of the tank's circulating current flows in the 48pF capacitor, and 1/4 of it in the 16pF to the source; the source once again sees a 50 ohm load (because the voltage at the top of the inductor is 1/4 of the original, and the capacitor coupling to the source is 4 times as large -- with the current in the source unchanged).

The key to understanding this coupling, to me, is the division of the tank's circulating current between the two capacitances, one directly across the coil and one in series with the source. The smaller the inductance (at constant Q), the more circulating current required for the same energy dissipation, and the smaller _percentage_of_total_resonating_capacitance_ for coupling to the (constant impedance) source.

Cheers, Tom

Tom I need to print this out and work with it. Did you change the inductor on purpose or just miss a factor of ten from the numbers I posted? Thanks, Mikek

Mikek,

on of the main problem of this newsgroup is that people write answers without carefully reading the question.

Everyone tends to "convert" the question toward issues he is able to write something about, Unfortunately those issues often have little to do with the original question.

I am offering you my answer, which may be clear, so and.so, or hardly understandable ... I do not know. But be sure that at least I paid maximum efforts to appreciate your question (though my numbers do not always match yours). I'll try to explain the issue in the easiest way I can, assumimng that all components behave in an ideal manner.

Let us first resume your hypotheses:

- the antenna has an impedance equal to the SERIES of a 58-ohm resistance and a capacitive reactance of -1,072 ohms which, at 1 MHz, corresponds to a 148.5-pF capacitance

- your aim is to transform that complex impedance into a 1.5-Mohm purely-resistive impedance, that would match that of your tank circuit.

That said, you must first appreciate that your antenna can be visualized in two ways:

- as the SERIES of R=58ohm, C=148.5pF (as said above)

- or, by applying the series-to-parallel transformation formula, as the PARALLEL of R=19,862ohm, C=148.1pF

These are just two fully equivalent ways of describing the same physical antenna. You can freely use the one which suits you best. For our purposes let us here visualize your antenna as the parallel of R=19,862ohm, C=148.1pF.

That said, assume for a moment that you are able to eliminate in some way (i'll tell you after how) the 148.1-pF parallel capacitance. What would then remain is a 19,862-ohm resistance, a value which unfortunately does not match the 1.5-Mohm figure you wish to get.

So, how to get just 1.5Mohm instead?

Playing with the transformation formulas you would realize that, if the SERIES representation of your antenna would hypothetically be R=58 ohm, C=54 pF (instead of R=58 ohm, C=148.5 pF as it is in the reality), the corresponding PARALLEL representation would then become R=1.5Mohm, C=53.9 pF. Just the resistance value you wish to get!

But modifying the SERIES representation of your antenna according to your needs is very easy: if you put an 85-pF capacitance in series with the antenna, its total capacitance would change from C=148.1 pF to C=54pf. And the antenna SERIES representation would then become R=58 ohm, C=54 pF, as you were aiming at.

Once you have put such 85-pF capacitance in series with your antenna, its PARALLEL representation becomes R=1.5 MHohm, C=53.9 pF, as said earlier.

For removing the 53.9-pF residual parallel capacitance, just resonate it with a

470-uH parallel inductance. The trick is then done: what remains is just the 1.5-Mohm resistance you wanted to get!

In summary:

- put a 85-pF in series (i.e. in between your antenna and the tank circuit)

- put a 470uH inductance in parallel to the tank (in practice this just means to increase the tank inductance by 470uH with respect to its nominal value).

73

Tony I0IX Rome, Italy

Hi Mike,

There's your first mistake. Tank Z is never, ever "mostly R," or you wouldn't be able to make the Q claim of 1000 (or even 10).

Please note the distinction between Z (which includes R) and R in isolation. Further, read Terman's material on Tank Circuits in his classic "Electronic and Radio Engineering" to clear up the clouds that obscure the view of their design rationale.

This is the sad (and useful) fate of short antennas, yes.

It is usually rhetorical, yes, insofar as not being enumerated. Often in technology it is a shorthand for a 10:1 ratio. Perhaps 100:1.

I could follow Xc(parallel/series) easily enough, but what you neglected to mention was there are two schematics embodied in the single one you brought to the discussion.

Making questions simpler often leads to Byzantine answers.

Another problem in this simplification is that there is more than one agenda being served, and they sometimes conflict with any single solution. As often happens, when I ask "What do you really want?" I get a response, I offer a solution, and I am immediately rebuffed that it does not serve the other agenda - which, then leads me to ask "What do you really want?"

73's Richard Clark, KB7QHC

Antonio - I think you slipped a decimal point. The parallel equivalent of the series combo 58R-2947j is actually 149k+2948j.

to

Hi Tony,

Great walk-through, excellent solution. And the appearance of Two additional, unstated components. Are they found in the original text of the article? Possibly not and thus the source of mystery (and a cautionary tale about what might be found as knowledge on the Internet).

73's Richard Clark, KB7QHC

ce

8R

Mike, I don't care what your values are. I'm interested in showing you in the generalized _concept_: how it works. Once you understand that, you can work with whatever component values you want. If you want to actually build something that works, you better pick values you can realize.

Cheers, Tom