| Page 1 of 2 | 1 | 2 |
|
|
Understanding Parallel to Series conversion
Hi All,
Please look at this in fixed font. I'm looking for understanding of a series to parallel conversion for antenna matching. I'm pretty sure I'm missing something, so point it out to me. This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. The antenna: R=58 ohms C=1072 ohms at 1Mhz. The tank: L=240uh C=106pf Q = 1000 Tank Z =~1.5 Mohms Here's my understanding of what I think I'm reading. I put a matching capacitor in series with the antenna. Antenna-- -----R-----C--------Match cap-----tank------ground. and this is supposed to transform the circuit to this. ( Maybe better said, equivalent to this) l------------l l l Antenna--- R C LC---Tank l l l------------l ^ Ground I calculate an 18.5pf cap for the match, making the antenna look like 58R and 17pf. So this; Antenna-- -----58R-----270pf--------Match cap18.5pf-----1.5Mohms------ground. This converts to; l-----------------l l l Antenna---58R 17pf 1.5M---LC Tank at l l Resonance l-----------------l ^ Ground And I now have a 1.5 Mohms source feeding a 1.5 Mohm load. The purpose of which is to cause minimal loading of the tank by the antenna. I don't understand how adding a series capacitor makes a parallel conversion. What do I misunderstand or do just need to believe the numbers. Thanks, Mikek |
Understanding Parallel to Series conversion
Ahh....1 should have said Series to Parallel Conversion.
Mikek |
Understanding Parallel to Series conversion
On Thu, 13 Jan 2011 11:05:38 -0600, "amdx" wrote:
This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. The antenna: R=58 ohms C=1072 ohms at 1Mhz. 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. I calculate an 18.5pf cap for the match, making the antenna look like 58R and 17pf. 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. And I now have a 1.5 Mohms source feeding a 1.5 Mohm load. How that is arrived at is something of a mystery. By the numbers, you describe a 26000:1 mismatch. The purpose of which is to cause minimal loading of the tank by the antenna. 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. I don't understand how adding a series capacitor makes a parallel conversion. Haven't we been down this road some months ago? 73's Richard Clark, KB7QHC |
Understanding Parallel to Series conversion
Richard Clark wrote:
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. Perhaps it's a longwire antenna, of crappy wire, parallel to the ground? |
Understanding Parallel to Series conversion
"Richard Clark" wrote in message ... On Thu, 13 Jan 2011 11:05:38 -0600, "amdx" wrote: This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. The antenna: R=58 ohms C=1072 ohms at 1Mhz. Hi Mike, Is this a fantasy antenna? 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. 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. Ok, how about just working with the concept. I calculate an 18.5pf cap for the match, making the antenna look like 58R and 17pf. 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. Sorry, I missed a math step, the R was converted to 1.5M. see formula below. And I now have a 1.5 Mohms source feeding a 1.5 Mohm load. How that is arrived at is something of a mystery. By the numbers, you describe a 26000:1 mismatch. 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. The purpose of which is to cause minimal loading of the tank by the antenna. Mikek |
Understanding Parallel to Series conversion
On Jan 13, 9:05*am, "amdx" wrote:
Hi All, *Please look at this in fixed font. *I'm looking for understanding of a series to parallel conversion for antenna matching. I'm pretty sure I'm missing something, so point it out to me. This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. *The antenna: R=58 ohms C=1072 ohms at 1Mhz. *The tank: L=240uh C=106pf Q = 1000 *Tank Z =~1.5 Mohms Here's my understanding of what I think I'm reading. I put a matching capacitor in series with the antenna. Antenna-- -----R-----C--------Match cap-----tank------ground. and this is supposed to transform the circuit to this. ( Maybe better said, equivalent to this) * * * * * * * * * * * l------------l * * * * * * * * * * * l * * * * * *l Antenna--- *R C * * * *LC---Tank * * * * * * * * * * * l * * * * * *l * * * * * * * * * * * l------------l * * * * * * * * * * * * * *^ * * * * * * * * * * * Ground I calculate an 18.5pf cap for the match, making the antenna look like 58R and 17pf. So this; *Antenna-- -----58R-----270pf--------Match cap18.5pf-----1.5Mohms------ground. This converts to; * * * * * * * * * * * * l-----------------l * * * * * * * * * * * * l * * * * * * * * l Antenna---58R 17pf * * * *1.5M---LC Tank at * * * * * * * * * * * * l * * * * * * * * l * * * *Resonance * * * * * * * * * * * * l-----------------l * * * * * * * * * * * * * * * *^ * * * * * * * * * * * * * Ground And I now have a 1.5 Mohms source feeding a 1.5 Mohm load. The purpose of which is to cause minimal loading of the tank by the antenna. *I don't understand how adding a series capacitor makes a parallel conversion. What do I misunderstand or do just need to believe the numbers. * * * * * * * * * * * * * * * * * *Thanks, 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=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 |
Understanding Parallel to Series conversion
On 1/13/2011 11:05 AM, amdx wrote:
Hi All, Please look at this in fixed font. I'm looking for understanding of a series to parallel conversion for antenna matching. I'm pretty sure I'm missing something, so point it out to me. This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. The antenna: R=58 ohms C=1072 ohms at 1Mhz. The tank: L=240uh C=106pf Q = 1000 Tank Z =~1.5 Mohms Here's my understanding of what I think I'm reading. I put a matching capacitor in series with the antenna. Antenna-- -----R-----C--------Match cap-----tank------ground. and this is supposed to transform the circuit to this. ( Maybe better said, equivalent to this) l------------l l l Antenna--- R C LC---Tank l l l------------l ^ Ground I calculate an 18.5pf cap for the match, making the antenna look like 58R and 17pf. So this; Antenna-- -----58R-----270pf--------Match cap18.5pf-----1.5Mohms------ground. This converts to; l-----------------l l l Antenna---58R 17pf 1.5M---LC Tank at l l Resonance l-----------------l ^ Ground And I now have a 1.5 Mohms source feeding a 1.5 Mohm load. The purpose of which is to cause minimal loading of the tank by the antenna. I don't understand how adding a series capacitor makes a parallel conversion. What do I misunderstand or do just need to believe the numbers. Thanks, Mikek 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 |
Understanding Parallel to Series conversion
On 1/13/2011 11:05 AM, amdx wrote:
Hi All, Please look at this in fixed font. I'm looking for understanding of a series to parallel conversion for antenna matching. I'm pretty sure I'm missing something, so point it out to me. This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. The antenna: R=58 ohms C=1072 ohms at 1Mhz. The tank: L=240uh C=106pf Q = 1000 Tank Z =~1.5 Mohms Here's my understanding of what I think I'm reading. I put a matching capacitor in series with the antenna. Antenna-- -----R-----C--------Match cap-----tank------ground. and this is supposed to transform the circuit to this. ( Maybe better said, equivalent to this) l------------l l l Antenna--- R C LC---Tank l l l------------l ^ Ground I calculate an 18.5pf cap for the match, making the antenna look like 58R and 17pf. So this; Antenna-- -----58R-----270pf--------Match cap18.5pf-----1.5Mohms------ground. This converts to; l-----------------l l l Antenna---58R 17pf 1.5M---LC Tank at l l Resonance l-----------------l ^ Ground And I now have a 1.5 Mohms source feeding a 1.5 Mohm load. The purpose of which is to cause minimal loading of the tank by the antenna. I don't understand how adding a series capacitor makes a parallel conversion. What do I misunderstand or do just need to believe the numbers. Thanks, Mikek 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 |
Understanding Parallel to Series conversion
On Thu, 13 Jan 2011 13:30:52 -0500, "Greg Neill"
wrote: Richard Clark wrote: 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. Perhaps it's a longwire antenna, of crappy wire, parallel to the ground? 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 |
Understanding Parallel to Series conversion
On Thu, 13 Jan 2011 12:42:01 -0600, "amdx" wrote:
"Richard Clark" wrote in message .. . On Thu, 13 Jan 2011 11:05:38 -0600, "amdx" wrote: This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. The antenna: R=58 ohms C=1072 ohms at 1Mhz. Hi Mike, Is this a fantasy antenna? 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. Hi Mike, Then your intuition is firing on all cylinders, great. 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. This is classically true. The frill of "at any given frequency" can be discarded. We can use this property to transform the real component of an impedance to a much higher or lower value. 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). 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 Well, what looks like hand-waving is probably close to the numbers one could expect. 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. 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? What happens if Xc series R series? What happens if Xc series = R series? 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 |
Understanding Parallel to Series conversion
On Thu, 13 Jan 2011 12:41:53 -0800, Richard Clark
wrote: No doubt there is more to be extracted from this article. 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 |
Understanding Parallel to Series conversion
Richard Clark wrote:
If by being close to ground you mean the 58 Ohms finds itself invested in the dirt, well, yes that might be the case. :-) |
Understanding Parallel to Series conversion
"Greg Neill" wrote in message m... Richard Clark wrote: 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. Perhaps it's a longwire antenna, of crappy wire, parallel to the ground? The author said it was an antenna in his attic. That's all I know. Mikek |
Understanding Parallel to Series conversion
"Richard Clark" wrote in message ... On Thu, 13 Jan 2011 12:42:01 -0600, "amdx" wrote: "Richard Clark" wrote in message . .. On Thu, 13 Jan 2011 11:05:38 -0600, "amdx" wrote: This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. The antenna: R=58 ohms C=1072 ohms at 1Mhz. Hi Mike, Is this a fantasy antenna? 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. Hi Mike, Then your intuition is firing on all cylinders, great. 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. This is classically true. The frill of "at any given frequency" can be discarded. We can use this property to transform the real component of an impedance to a much higher or lower value. 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). 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 Well, what looks like hand-waving is probably close to the numbers one could expect. 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. 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? What happens if Xc series R series? What happens if Xc series = R series? 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. 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. And what is this Xc(parallel) and Xc(series) stuff? 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. 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 |
Understanding Parallel to Series conversion
"K7ITM" wrote in message ... On Jan 13, 9:05 am, "amdx" wrote: Hi All, Please look at this in fixed font. I'm looking for understanding of a series to parallel conversion for antenna matching. I'm pretty sure I'm missing something, so point it out to me. This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. The antenna: R=58 ohms C=1072 ohms at 1Mhz. The tank: L=240uh C=106pf Q = 1000 Tank Z =~1.5 Mohms Here's my understanding of what I think I'm reading. I put a matching capacitor in series with the antenna. Antenna-- -----R-----C--------Match cap-----tank------ground. and this is supposed to transform the circuit to this. ( Maybe better said, equivalent to this) l------------l l l Antenna--- R C LC---Tank l l l------------l ^ Ground I calculate an 18.5pf cap for the match, making the antenna look like 58R and 17pf. So this; Antenna-- -----58R-----270pf--------Match cap18.5pf-----1.5Mohms------ground. This converts to; l-----------------l l l Antenna---58R 17pf 1.5M---LC Tank at l l Resonance l-----------------l ^ Ground And I now have a 1.5 Mohms source feeding a 1.5 Mohm load. The purpose of which is to cause minimal loading of the tank by the antenna. I don't understand how adding a series capacitor makes a parallel conversion. What do I misunderstand or do just need to believe the numbers. Thanks, 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=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 |
Understanding Parallel to Series conversion
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 |
Understanding Parallel to Series conversion
On Thu, 13 Jan 2011 16:18:35 -0600, "amdx" wrote:
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) 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. and the antenna has a low R and a C in the 100's of ohms. This is the sad (and useful) fate of short antennas, yes. So I don't know, is that () ? The question is retorical in nature. 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. And what is this Xc(parallel) and Xc(series) stuff? 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. 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 |
Understanding Parallel to Series conversion
On 1/13/2011 6:05 PM, Antonio Vernucci wrote:
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. Antonio - I think you slipped a decimal point. The parallel equivalent of the series combo 58R-2947j is actually 149k+2948j. 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 |
Understanding Parallel to Series conversion
On Fri, 14 Jan 2011 01:05:17 +0100, "Antonio Vernucci"
wrote: 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). 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 |
Understanding Parallel to Series conversion
On Jan 13, 2:27*pm, "amdx" wrote:
"K7ITM" wrote in message ... On Jan 13, 9:05 am, "amdx" wrote: Hi All, Please look at this in fixed font. I'm looking for understanding of a series to parallel conversion for antenna matching. I'm pretty sure I'm missing something, so point it out to me. This is in regard to a crystal radio, so the match is for a low impedance antenna to a high impedance tank circuit. The antenna: R=58 ohms C=1072 ohms at 1Mhz. The tank: L=240uh C=106pf Q = 1000 Tank Z =~1.5 Mohms Here's my understanding of what I think I'm reading. I put a matching capacitor in series with the antenna. Antenna-- -----R-----C--------Match cap-----tank------ground. and this is supposed to transform the circuit to this. ( Maybe better said, equivalent to this) l------------l l l Antenna--- R C LC---Tank l l l------------l ^ Ground I calculate an 18.5pf cap for the match, making the antenna look like 58R and 17pf. So this; Antenna-- -----58R-----270pf--------Match cap18.5pf-----1.5Mohms------ground. This converts to; l-----------------l l l Antenna---58R 17pf 1.5M---LC Tank at l l Resonance l-----------------l ^ Ground And I now have a 1.5 Mohms source feeding a 1.5 Mohm load. The purpose of which is to cause minimal loading of the tank by the antenna. I don't understand how adding a series capacitor makes a parallel conversion. What do I misunderstand or do just need to believe the numbers. Thanks, 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=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 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 |
Understanding Parallel to Series conversion
"Richard Clark" wrote in message ... On Thu, 13 Jan 2011 16:18:35 -0600, "amdx" wrote: 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) 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). 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? ......which, then leads me to ask "What do you really want?" 73's Richard Clark, KB7QHC 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 |
Understanding Parallel to Series conversion
On 1/13/2011 7:51 PM, amdx wrote:
"Richard wrote in message ... On Thu, 13 Jan 2011 16:18:35 -0600, wrote: 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) 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). 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? ......which, then leads me to ask "What do you really want?" 73's Richard Clark, KB7QHC 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. Does this make any sense? Cheers, John |
Understanding Parallel to Series conversion
"John - KD5YI" wrote in message ... On 1/13/2011 7:51 PM, amdx wrote: "Richard wrote in message ... On Thu, 13 Jan 2011 16:18:35 -0600, wrote: 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) 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). 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? ......which, then leads me to ask "What do you really want?" 73's Richard Clark, KB7QHC 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. Does this make any sense? Cheers, John 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 |
Understanding Parallel to Series conversion
In article ,
amdx wrote: A series RC has a parallel RC equivalent. I'm not sure how it can be both at the same time. 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. -- Dave Platt AE6EO Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior I do _not_ wish to receive unsolicited commercial email, and I will boycott any company which has the gall to send me such ads! |
Understanding Parallel to Series conversion
On 1/13/2011 8:33 PM, amdx wrote:
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 Mike - 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. Cheers, John |
Understanding Parallel to Series conversion
On Thu, 13 Jan 2011 20:10:46 -0600, John - KD5YI
wrote: Does this make any sense? Up, up, and away, in my beautiful, my beautiful balun! |
Understanding Parallel to Series conversion
On Thu, 13 Jan 2011 19:51:21 -0600, "amdx" wrote:
"Richard Clark" wrote in message .. . On Thu, 13 Jan 2011 16:18:35 -0600, "amdx" wrote: 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) 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). 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? Hi Mike, 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. ......which, then leads me to ask "What do you really want?" 73's Richard Clark, KB7QHC 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 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. 73's Richard Clark, KB7QHC |
Understanding Parallel to Series conversion
On Thu, 13 Jan 2011 21:43:55 -0800, GoldIntermetallicEmbrittlement
g wrote: On Thu, 13 Jan 2011 20:10:46 -0600, John - KD5YI wrote: Does this make any sense? Up, up, and away, in my beautiful, my beautiful balun! That's even worse than your favourite. You know, the one that goes "Your mother should be arrested for ......" **** off, you pathetic dullard. |
Understanding Parallel to Series conversion
On Jan 14, 2:33*am, "amdx" wrote:
"John - KD5YI" wrote in ... On 1/13/2011 7:51 PM, amdx wrote: "Richard *wrote in message . .. On Thu, 13 Jan 2011 16:18:35 -0600, *wrote: 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) 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). * 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? ......which, then leads me to ask "What do you really want?" 73's Richard Clark, KB7QHC * 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. Does this make any sense? Cheers, John *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 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! |
Ahh....1 should have said Series to Parallel Conversion.
Mikek __________________ |
Understanding Parallel to Series conversion
"Richard Clark" wrote in message ... On Thu, 13 Jan 2011 19:51:21 -0600, "amdx" wrote: "Richard Clark" wrote in message . .. On Thu, 13 Jan 2011 16:18:35 -0600, "amdx" wrote: 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) 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). 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? Hi Mike, 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. A lot there, but I didn't get anything out of it. At resonance, does the tank look capacitive, inductive, resistive, or all to the antenna you connect to it. (this makes the assumption attaching the antenna didn't change anything, it did, let's say I adjusted back to resonance) And then my original question, Are you saying it is pure R at resonance? A yes or no will be fine, then I can try to reprocess the above that didn't get anything out of. 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). I'll bite, I want maximum transfer of power, I'm still working with the tank as a large pure R so I want the antenna to look like the same large R. I realize there is still capacitance from the antenna to deal with. Now are getting to conjugate :-) 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. ......which, then leads me to ask "What do you really want?" 73's Richard Clark, KB7QHC 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 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. 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. 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. 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... 73's Richard Clark, KB7QHC |
Understanding Parallel to Series conversion
On Fri, 14 Jan 2011 07:20:20 -0600, "amdx" wrote:
I want maximum transfer of power, I'm still working with the tank as a large pure R so I want the antenna to look like the same large R. No you don't. 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. Am I wrong? 73's Richard Clark, KB7QHC |
Understanding Parallel to Series conversion
"Richard Clark" wrote in message ... On Fri, 14 Jan 2011 07:20:20 -0600, "amdx" wrote: I want maximum transfer of power, I'm still working with the tank as a large pure R so I want the antenna to look like the same large R. No you don't. What you want is the lightest final load sufficient to drive a speaker We haven't got to the load yet! it's coming :-) 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. Am I wrong? 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. This aught to be fun :-) Mikek |
Understanding Parallel to Series conversion
Antonio - I think you slipped a decimal point. The parallel equivalent of the
series combo 58R-2947j is actually 149k+2948j. Hi John, 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. 73 Tony I0JX |
Understanding Parallel to Series conversion
On 1/14/2011 10:50 AM, Antonio Vernucci wrote:
Antonio - I think you slipped a decimal point. The parallel equivalent of the series combo 58R-2947j is actually 149k+2948j. Hi John, I do not understand where your -2947j figure comes from. I see it appearing nowhere in my calculations. 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. 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. You are correct. I allowed the sign of the suseptance to creep through. 73 Tony I0JX |
Understanding Parallel to Series conversion
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. 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. Sorry for mistake! 73 Tony I0JX |
Understanding Parallel to Series conversion
On Fri, 14 Jan 2011 19:35:43 +0100, "Antonio Vernucci"
wrote: 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. 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. Sorry for mistake! 73 Tony I0JX 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 -- | James E.Thompson, CTO | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona 85048 Skype: Contacts Only | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food. |
Understanding Parallel to Series conversion
On Jan 14, 6:41*pm, Jim Thompson To-Email-Use-The-Envelope-I...@On-My-
Web-Site.com wrote: On Fri, 14 Jan 2011 19:35:43 +0100, "Antonio Vernucci" wrote: 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. 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. Sorry for mistake! 73 Tony I0JX 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 -- | James E.Thompson, CTO * * * * * * * * * * * * * *| * *mens * * | | Analog Innovations, Inc. * * * * * * * * * * * * | * * et * * *| | Analog/Mixed-Signal ASIC's and Discrete Systems *| * *manus * *| | Phoenix, Arizona *85048 * *Skype: Contacts Only *| * * * * * * | | Voice:(480)460-2350 *Fax: Available upon request | *Brass Rat *| | E-mail Icon athttp://www.analog-innovations.com| * *1962 * * | I love to cook with wine. * * Sometimes I even put it in the food. just plug it in and try it... if the volume isn't high enough get a real radio! |
Understanding Parallel to Series conversion
On Fri, 14 Jan 2011 10:47:50 -0600, "amdx" wrote:
So (I think) we couple more energy into the tank for more signal and this would lower Q for a wider bandwidth. Hi Mike, How? (It would be a sad day for us all if more input to a Tank lowered its Q were true.) 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 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. 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. 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. Then I load this with 500Kohm load. 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 Scenerio 2. 1 megohm tank. I put a 1 megohm load I can drive the tank with a 500Kohm source, 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 Scenerio 3. 1 megohm tankThe I drive the tank with 2 megohm source and load it with a 2 megohm load. So.. 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 I have no clue where maximum power is delivered from the antenna to the load. 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. How does this impact design priorities? 73's Richard Clark, KB7QHC |
Understanding Parallel to Series conversion
"Richard Clark" wrote in message ... On Fri, 14 Jan 2011 10:47:50 -0600, "amdx" wrote: So (I think) we couple more energy into the tank for more signal and this would lower Q for a wider bandwidth. Hi Mike, How? (It would be a sad day for us all if more input to a Tank lowered its Q were true.) My thought was, if we couple more energy from the antenna, it loads the tank more lowering Q. The thought might need more work.... 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 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. 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. 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. Then I load this with 500Kohm load. 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 Scenerio 2. 1 megohm tank. I put a 1 megohm load I can drive the tank with a 500Kohm source, 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 Scenerio 3. 1 megohm tankThe I drive the tank with 2 megohm source and load it with a 2 megohm load. So.. 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 I have no clue where maximum power is delivered from the antenna to the load. 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. Scenerio 1 is how I have always thought about the system. Nice to know where max power transfer is. How does this impact design priorities? 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 73's Richard Clark, KB7QHC |
| All times are GMT +1. The time now is 08:31 AM. |
|
Powered by vBulletin® Copyright ©2000 - 2025, Jelsoft Enterprises Ltd.
RadioBanter.com