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Conjugate matching and my funky VSWR meter
There has been some discussion in the past months about conjugate matching,
VSWR, and power transfer from source to load. I've come across a puzzle while noodling on this. My main issue here is how the heck my VSWR meter is measuring the way it is. My elementary hook-up is an RF power amp feeding directly into a VSWR meter, and from there into a load consisting of a carbon resistor and a variable capacitor rigged in series. The meter connects to the load via about a foot of 50 ohm coax. The frequency is between 1 and 10 MHz. Model the source impedance as Zs = R + jX, and the load impedance as Zl = r + jx (or use phasors if you prefer :). The following two statements are true: 1) The power dissipated in the load (r) is maximised when x = -X (so-called "conjugate matching"), whatever the value of (r). 2) The classical VSWR is minimised (zero "reflected power") when x = +X, whatever the value of (r). However, my VSWR meter, whch is a conventional 2-diode bridge and short transmission line, indicates that minimum indicated VSWR corresponds to max power dissipated in (r).!! (i.e. at conjugate match, and NOT when reflected power is zero). The equation normally used for VSWR is VSWR = ABS( (1 + |p|) / (1 - |p|) ) where p = (Zl - Zs) / (Zl + Zs) and p is a measure of the amount of power reflected back to the source, called the "voltage reflection coefficient" I plotted something I call "conjugate VSWR" or VSWR*. which is the same expression as above, but with p defined as p = (Zl - Zs*) / (Zl + Zs) where Zs* indicates the complex conjugate of Zs. and the behaviour of this VSWR* thingie absolutely matches what I see on my meter. Aye, there's the rub. Some points to note a) Classical VSWR shows NO minimum for all r, when x has the opposite sign to X b) VSWR* always has a minimum at the same r-value which causes maximum power to be dissipated in r, whatever the value of x. Again, I flat don't understand how my VSWR meter can indicate VSWR* when I know it should indicate VSWR. Here are a couple of links to flesh out the theory. 1. Wade through this at your peril - it's you lot fighting abou this issue and is VERY long http://www.ibiblio.org/pub/academic/...S/20030831.ant 2. This is much more succint - cut to the chase on p47 http://my.ece.ucsb.edu/yorklab/Usefu...%20AN64-1B.pdf Best, Andrew |
ok, bonzo, i'll bite on the troll bait. but only because its early in the
morning and the normal endless discussion of this stuff hasn't taken over yet. "Lord Snooty" wrote in message nk.net... There has been some discussion in the past months about conjugate matching, VSWR, and power transfer from source to load. I've come across a puzzle while noodling on this. My main issue here is how the heck my VSWR meter is measuring the way it is. My elementary hook-up is an RF power amp feeding directly into a VSWR meter, and from there into a load consisting of a carbon resistor and a variable capacitor rigged in series. The meter connects to the load via about a foot of 50 ohm coax. The frequency is between 1 and 10 MHz. Model the source impedance as Zs = R + jX, and the load impedance as Zl = r + jx (or use phasors if you prefer :). The following two statements are true: 1) The power dissipated in the load (r) is maximised when x = -X (so-called "conjugate matching"), whatever the value of (r). wrong. you must transform the R+jX along the transmission line to get back to the load seen by the source. you stipulate a low frequency and short line, so you are close anyway. 2) The classical VSWR is minimised (zero "reflected power") when x = +X, whatever the value of (r). doubly wrong. vswr is on a cable and is independent of the source. it knows nothing of R+jX only the characteristic impedance of the cable. all following calculations are wrong for this reason alone. However, my VSWR meter, whch is a conventional 2-diode bridge and short transmission line, indicates that minimum indicated VSWR corresponds to max power dissipated in (r).!! (i.e. at conjugate match, and NOT when reflected power is zero). The equation normally used for VSWR is VSWR = ABS( (1 + |p|) / (1 - |p|) ) where p = (Zl - Zs) / (Zl + Zs) wrong again, the impedance used must be that of the cable not of the source. its not worth commenting further until you understand this. and p is a measure of the amount of power reflected back to the source, called the "voltage reflection coefficient" I plotted something I call "conjugate VSWR" or VSWR*. which is the same expression as above, but with p defined as p = (Zl - Zs*) / (Zl + Zs) where Zs* indicates the complex conjugate of Zs. and the behaviour of this VSWR* thingie absolutely matches what I see on my meter. Aye, there's the rub. Some points to note a) Classical VSWR shows NO minimum for all r, when x has the opposite sign to X b) VSWR* always has a minimum at the same r-value which causes maximum power to be dissipated in r, whatever the value of x. Again, I flat don't understand how my VSWR meter can indicate VSWR* when I know it should indicate VSWR. Here are a couple of links to flesh out the theory. 1. Wade through this at your peril - it's you lot fighting abou this issue and is VERY long http://www.ibiblio.org/pub/academic/...S/20030831.ant 2. This is much more succint - cut to the chase on p47 http://my.ece.ucsb.edu/yorklab/Usefu...%20AN64-1B.pdf Best, Andrew |
Andrew,
Conjugate matching may be a very interesting subject and is indeed a much discussed topic on these walls. However, amongst the amateur fraternity, a "conjugate match" should be classified as being off topic. It does not exist. Not even when the tuner has been adjusted exactly for a VSWR equal to 1 to 1. Why? Because the internal impedance of the transmitter is not 50 ohms. It is not relevant. What is the internal impedance of YOUR transmitter? You will not find it mentioned in the manufacturer's operating or maintenance handbooks. In all likelihood you will not know what it is even if you designed and built the transmitter yourself! --- Reg, G4FGQ |
From the previous posting, I can guess who is going to jump all over this.
Keep up the good work. Tam "Dave" wrote in message ... ok, bonzo, i'll bite on the troll bait. but only because its early in the morning and the normal endless discussion of this stuff hasn't taken over yet. "Lord Snooty" wrote in message nk.net... There has been some discussion in the past months about conjugate matching, VSWR, and power transfer from source to load. I've come across a puzzle while noodling on this. My main issue here is how the heck my VSWR meter is measuring the way it is. My elementary hook-up is an RF power amp feeding directly into a VSWR meter, and from there into a load consisting of a carbon resistor and a variable capacitor rigged in series. The meter connects to the load via about a foot of 50 ohm coax. The frequency is between 1 and 10 MHz. Model the source impedance as Zs = R + jX, and the load impedance as Zl = r + jx (or use phasors if you prefer :). The following two statements are true: 1) The power dissipated in the load (r) is maximised when x = -X (so-called "conjugate matching"), whatever the value of (r). wrong. you must transform the R+jX along the transmission line to get back to the load seen by the source. you stipulate a low frequency and short line, so you are close anyway. 2) The classical VSWR is minimised (zero "reflected power") when x = +X, whatever the value of (r). doubly wrong. vswr is on a cable and is independent of the source. it knows nothing of R+jX only the characteristic impedance of the cable. all following calculations are wrong for this reason alone. However, my VSWR meter, whch is a conventional 2-diode bridge and short transmission line, indicates that minimum indicated VSWR corresponds to max power dissipated in (r).!! (i.e. at conjugate match, and NOT when reflected power is zero). The equation normally used for VSWR is VSWR = ABS( (1 + |p|) / (1 - |p|) ) where p = (Zl - Zs) / (Zl + Zs) wrong again, the impedance used must be that of the cable not of the source. its not worth commenting further until you understand this. and p is a measure of the amount of power reflected back to the source, called the "voltage reflection coefficient" I plotted something I call "conjugate VSWR" or VSWR*. which is the same expression as above, but with p defined as p = (Zl - Zs*) / (Zl + Zs) where Zs* indicates the complex conjugate of Zs. and the behaviour of this VSWR* thingie absolutely matches what I see on my meter. Aye, there's the rub. Some points to note a) Classical VSWR shows NO minimum for all r, when x has the opposite sign to X b) VSWR* always has a minimum at the same r-value which causes maximum power to be dissipated in r, whatever the value of x. Again, I flat don't understand how my VSWR meter can indicate VSWR* when I know it should indicate VSWR. Here are a couple of links to flesh out the theory. 1. Wade through this at your peril - it's you lot fighting abou this issue and is VERY long http://www.ibiblio.org/pub/academic/...S/20030831.ant 2. This is much more succint - cut to the chase on p47 http://my.ece.ucsb.edu/yorklab/Usefu...%20AN64-1B.pdf Best, Andrew |
nope, all done. reg is here and cecil can't be far behind. i've had my
fun, time to do other more productive things than watch them re-hash conjugal matches for the next month or two. "Tam/WB2TT" wrote in message ... From the previous posting, I can guess who is going to jump all over this. Keep up the good work. Tam "Dave" wrote in message ... ok, bonzo, i'll bite on the troll bait. but only because its early in the morning and the normal endless discussion of this stuff hasn't taken over yet. "Lord Snooty" wrote in message nk.net... There has been some discussion in the past months about conjugate matching, VSWR, and power transfer from source to load. I've come across a puzzle while noodling on this. My main issue here is how the heck my VSWR meter is measuring the way it is. My elementary hook-up is an RF power amp feeding directly into a VSWR meter, and from there into a load consisting of a carbon resistor and a variable capacitor rigged in series. The meter connects to the load via about a foot of 50 ohm coax. The frequency is between 1 and 10 MHz. Model the source impedance as Zs = R + jX, and the load impedance as Zl = r + jx (or use phasors if you prefer :). The following two statements are true: 1) The power dissipated in the load (r) is maximised when x = -X (so-called "conjugate matching"), whatever the value of (r). wrong. you must transform the R+jX along the transmission line to get back to the load seen by the source. you stipulate a low frequency and short line, so you are close anyway. 2) The classical VSWR is minimised (zero "reflected power") when x = +X, whatever the value of (r). doubly wrong. vswr is on a cable and is independent of the source. it knows nothing of R+jX only the characteristic impedance of the cable. all following calculations are wrong for this reason alone. However, my VSWR meter, whch is a conventional 2-diode bridge and short transmission line, indicates that minimum indicated VSWR corresponds to max power dissipated in (r).!! (i.e. at conjugate match, and NOT when reflected power is zero). The equation normally used for VSWR is VSWR = ABS( (1 + |p|) / (1 - |p|) ) where p = (Zl - Zs) / (Zl + Zs) wrong again, the impedance used must be that of the cable not of the source. its not worth commenting further until you understand this. and p is a measure of the amount of power reflected back to the source, called the "voltage reflection coefficient" I plotted something I call "conjugate VSWR" or VSWR*. which is the same expression as above, but with p defined as p = (Zl - Zs*) / (Zl + Zs) where Zs* indicates the complex conjugate of Zs. and the behaviour of this VSWR* thingie absolutely matches what I see on my meter. Aye, there's the rub. Some points to note a) Classical VSWR shows NO minimum for all r, when x has the opposite sign to X b) VSWR* always has a minimum at the same r-value which causes maximum power to be dissipated in r, whatever the value of x. Again, I flat don't understand how my VSWR meter can indicate VSWR* when I know it should indicate VSWR. Here are a couple of links to flesh out the theory. 1. Wade through this at your peril - it's you lot fighting abou this issue and is VERY long http://www.ibiblio.org/pub/academic/...S/20030831.ant 2. This is much more succint - cut to the chase on p47 http://my.ece.ucsb.edu/yorklab/Usefu...%20AN64-1B.pdf Best, Andrew |
Dave wrote:
nope, all done. reg is here and cecil can't be far behind. i've had my fun, time to do other more productive things than watch them re-hash conjugal matches for the next month or two. I guess I need to say this again. My take on discussions of conjugate matching in ham antenna systems is that it is a waste of time. If reflected energy is not allowed to reach the source, e.g. typical ham Z0-matched systems, the source impedance is irrelevant and doesn't affect anything in the system except for efficiency. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
Lord Snooty wrote:
Indeed so, because the whole idea of characterising my SWR meter is towards the goal of measuring the output impedance of my RF amp! I agree that it isn't 50 ohms - it's R + jX, and I want a bulletproof procedure to find R and X. Got one? R + jX is a linear function. As your amp is probably push-pull AB1 class, you could only ever obtain the averaged impedance of two non-linear devices. Is that what you want? -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Sun, 23 May 2004 14:17:51 +0000 (UTC), "Reg Edwards"
wrote: Because the internal impedance of the transmitter is not 50 ohms. It is not relevant. Old Son, Lord Kelvinator would be saddened by such a dismissal of actually delving into "knowing" by your meagre understanding that fails to offer an actual value that it IS. Such superstition that dominates this paucity would lead us to believe the Z is as slight as an angel wafting past in a dream (would that be in English Units, or metric?). Definitions by negatives is an amusing troll however. I do enjoy participation in kind: Let's see, the Z of a transmitter is NOT a dead parrot, nor five farthings, nor 10 inches, nor a polka dot dress, nor the gross national income of Lithuania, nor the combined weight of all your dead white scientists. I'm sure I missed a few.... ;-) 73's Richard Clark, KB7QHC |
On Sun, 23 May 2004 16:39:19 GMT, "Lord Snooty" wrote:
Indeed so, because the whole idea of characterising my SWR meter is towards the goal of measuring the output impedance of my RF amp! I agree that it isn't 50 ohms - it's R + jX, and I want a bulletproof procedure to find R and X. Got one? Cheers, Andrew G3UHD Hi Andrew, You have asked an inescapable question that will lead to a deluge of scribbling commemorating the best attempts of Houdini. There are several many methods to determine exactly what you want to know. The simplest and certainly the one that contains as much information necessary is called "load pulling." To even mention this time and bench proven method will result in hoots from those who would be the last to offer you a fixed answer; however, we shall proceed. This requires that you have access to known, but non-standard value loads capable of sustaining the power you will perform your measurement at. This is not a trivial requirement. It also requires that you can in some way defeat your ALC which will attempt to offset the pull of the non-standard load. It is simplicity itself that only demands you consider the elements of a Thevenin model and how to determine the model's source Z (or likewise, the Norton model's source Z). You will need a means to measure the voltage across the load, or the current through it. Even here, proportionality is all that is required as long as the Load is characterized and thus the tools can be rather spartan. In the long run, this will mean you have to construct and verify your own non-standard loads. Take care that through your verification you confirm their value across all power applications (resistors are very susceptible to drift with temperature). You should also take care to insure that all paths and leads are as short as possible. Loading directly at the terminals will save grief of complex compensation math (and reduce introducing other errors). However, you can choose to employ remote loads if you take care to characterize the lines through which they are attached (this means you should be adept at the Smith Chart). There is more to be said, but this enough to offer you a significant lead to find that, yes, the source exhibits nearly 50 Ohms (the common Ham transmitter running at rated power will fall between 30 and 70 Ohms) - as specified and designed. 73's Richard Clark, KB7QHC |
Wise words. The true load-pulling technique advocated professionally uses a
known model of the output stage using S parameters. Yuk. Not for this soldier. I'd rather sniff underwear in a geriatric clinic thank you. Whatever one did in this direction, as an amateur, would doubtless be either wrong or so inaccurate as to be not even wrong as Dr. Pauli was wont to say. The amp is single-ended out of an MRF136, so I presume it's Class A. The amp's designation is H-10 (I bought it surplus). It's rated at around 15W, 0.1 - 30 MHz. The circuit diagram shows no hint of current limiting circuitry. If one is serious about proper design of a matching network - a network, I might add, which attaches *directly* (near as dammit) to the Tx output - then one is all at sea without a proper knowledge of source impedance. See my comments in the other thread about this. I tried to use a technique advised by K7ITM to measure source impedance (R + jX), and it produces a negative value of R by calculation. This is the technique, based on a load setting of (r + jx) ohms. And once again, there is no cable, folks. And it wouldn't matter if I did have a few inches of it either. 1. Determine X. a) Set r ~= R based on a best guess (which would be R=50 ohms nominally in many systems). b) Monitor the voltage, current, or power in the load (r). c) Adjust (x) to maximise the monitored value. This setting corresponds to x = -X, and we have determined X. 2. Determine R a) Leave x = -X set as in step 1,so now the circuit is pure resistive. b) Monitor the voltage across the resistor. c) Set r to R(nominal) plus and minus a small percentage, and measure the monitored voltage at both values of r. The voltage across r is given by V/V0 = r / (R + r) Solving the 2 simultaneous equations to eliminate V0 shows that R is determined by the equation R = r1*r2*(V2 - V1) / (V1*r2 - V2*r1) ------------------------------------- My problems ----------- 1. My measurements of V1,V2 lead to the inescapable conclusion that the above model fails, because the calculated value of R comes out negative. Let us assume that we set (r2 r1) and we obtain (V2 V1), which is predicted from the model, and is also the case for my measurements. Under these conditions, a negative value of R can only be obtained from the equation if V2/V1 r2/r1, which in my case is true. I undertook a full (and rather exhaustive and tedious!) calculation of the expression for R when the value of (x) was not set correctly to -X, thinking that perhaps this was the cause of the discrepancy. It turns out in this case that, if the calculated value of R is negative, it has nothing to do with the setting of (x), and depends ONLY on the condition V2/V1 r2/r1. Since we know that the actual value of R cannot be negative, this implies a failure of the model. How then can the model fail? Since we are maintaining frequency constant, any collection of resistances and reactances, however complicated, can be modelled as (R + jX), so it cannot be that. The only assumption left to question is the constancy of V0, and this is what the failure must be. This leaves me with more questions than answers, because the way forward is now completely unclear. 2. I should also mention another, less serious, problem I had, and that is with the determination of X. The value of (x) I determine from measurement would be expected to be constant at a given frequency. It is in my case not so. The derived value of X appears to depend on a) the power level setting of my amplifier b) the value of (r) I set in the circuit when determining X. Quite probably this second problem relates to the first problem's identification of the failure of the model, and can probably be subsumed under that category. Best, Andrew "Richard Clark" wrote in message ... On Sun, 23 May 2004 16:39:19 GMT, "Lord Snooty" wrote: Indeed so, because the whole idea of characterising my SWR meter is towards the goal of measuring the output impedance of my RF amp! I agree that it isn't 50 ohms - it's R + jX, and I want a bulletproof procedure to find R and X. Got one? Cheers, Andrew G3UHD Hi Andrew, You have asked an inescapable question that will lead to a deluge of scribbling commemorating the best attempts of Houdini. There are several many methods to determine exactly what you want to know. The simplest and certainly the one that contains as much information necessary is called "load pulling." To even mention this time and bench proven method will result in hoots from those who would be the last to offer you a fixed answer; however, we shall proceed. This requires that you have access to known, but non-standard value loads capable of sustaining the power you will perform your measurement at. This is not a trivial requirement. It also requires that you can in some way defeat your ALC which will attempt to offset the pull of the non-standard load. It is simplicity itself that only demands you consider the elements of a Thevenin model and how to determine the model's source Z (or likewise, the Norton model's source Z). You will need a means to measure the voltage across the load, or the current through it. Even here, proportionality is all that is required as long as the Load is characterized and thus the tools can be rather spartan. In the long run, this will mean you have to construct and verify your own non-standard loads. Take care that through your verification you confirm their value across all power applications (resistors are very susceptible to drift with temperature). You should also take care to insure that all paths and leads are as short as possible. Loading directly at the terminals will save grief of complex compensation math (and reduce introducing other errors). However, you can choose to employ remote loads if you take care to characterize the lines through which they are attached (this means you should be adept at the Smith Chart). There is more to be said, but this enough to offer you a significant lead to find that, yes, the source exhibits nearly 50 Ohms (the common Ham transmitter running at rated power will fall between 30 and 70 Ohms) - as specified and designed. 73's Richard Clark, KB7QHC |
Sorry everyone, but I just retested with no cable and the results I obtain are
precisely the same. The coax cable was only 26" long anyway. So forget about that transmission line stuff. It's irrelevant here. What I want to know is why the VSWR indications are the way they are. If anyone's interested, I can email a small spreadsheet that deals with this simple circuit (V0-R-jX-r-jx) and allows you to set a) R,X and r, and vary x b) R,X and x, and vary r. I plot side by side on the two corresponding graphs - modulus of total load voltage - modulus of load resistor voltage - modulus of load reactance voltage - power dissipated in load resistor - VSWR between source and load - "conjugate VSWR" between source and load. One more time with feeling - What I want to know is why the VSWR indications are the way they are. Best, Andrew "Cecil Moore" wrote in message ... Dave wrote: nope, all done. reg is here and cecil can't be far behind. i've had my fun, time to do other more productive things than watch them re-hash conjugal matches for the next month or two. I guess I need to say this again. My take on discussions of conjugate matching in ham antenna systems is that it is a waste of time. If reflected energy is not allowed to reach the source, e.g. typical ham Z0-matched systems, the source impedance is irrelevant and doesn't affect anything in the system except for efficiency. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
probably because the meter is not measuring what you think it is. remember,
vswr meters are meant to be showing what the vswr is in a transmission line of a given characteristic impedance. they are not impedance meters, nor are they proper power meters even though they are often calibrated in watts... they are only accurate in the specific characteristic impedance system they were 'calibrated' for... and then only roughly in most cases. if you want to make proper measurements give up on the vswr meter and measure the voltage or current with an oscilloscope or properly calibrated rf voltmeter. "Lord Snooty" wrote in message nk.net... Sorry everyone, but I just retested with no cable and the results I obtain are precisely the same. The coax cable was only 26" long anyway. So forget about that transmission line stuff. It's irrelevant here. What I want to know is why the VSWR indications are the way they are. If anyone's interested, I can email a small spreadsheet that deals with this simple circuit (V0-R-jX-r-jx) and allows you to set a) R,X and r, and vary x b) R,X and x, and vary r. I plot side by side on the two corresponding graphs - modulus of total load voltage - modulus of load resistor voltage - modulus of load reactance voltage - power dissipated in load resistor - VSWR between source and load - "conjugate VSWR" between source and load. One more time with feeling - What I want to know is why the VSWR indications are the way they are. Best, Andrew "Cecil Moore" wrote in message ... Dave wrote: nope, all done. reg is here and cecil can't be far behind. i've had my fun, time to do other more productive things than watch them re-hash conjugal matches for the next month or two. I guess I need to say this again. My take on discussions of conjugate matching in ham antenna systems is that it is a waste of time. If reflected energy is not allowed to reach the source, e.g. typical ham Z0-matched systems, the source impedance is irrelevant and doesn't affect anything in the system except for efficiency. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Sun, 23 May 2004 20:42:59 GMT, "Lord Snooty" wrote:
The amp is single-ended out of an MRF136, so I presume it's Class A. The amp's designation is H-10 (I bought it surplus). It's rated at around 15W, 0.1 - 30 MHz. The circuit diagram shows no hint of current limiting circuitry. If one is serious about proper design of a matching network - a network, I might add, which attaches *directly* (near as dammit) to the Tx output - then one is all at sea without a proper knowledge of source impedance. See my comments in the other thread about this. Hi Andrew. Specification sheets respond to these issues quite well. The MRF136 is a 400MHz device, normally offering 15W max with about 16dB gain at 28Vdc (although rated higher) in a class A configuration (showing about 60% efficiency). As would be expected, it covers a lot of turf. In the HF, the output Z runs easily near 50 Ohms in push-pull circuit configurations; otherwise it is simpler to describe it in the teens to tens of Ohms across any number of variables you do not disclose (like frequency, the actual configuration, additional interface components). 1 - 10 MHz does not bode well towards the best implementation of an UHF device. Load pulling is vastly simpler than S-parameters (also specified in the data sheets for this device, down to 1MHz), why you want to marry the two is a mystery. 73's Richard Clark, KB7QHC |
Richard Clark wrote:
Load pulling is vastly simpler than S-parameters (also specified in the data sheets for this device, down to 1MHz), why you want to marry the two is a mystery. A laboratory bench setup is simpler than a calculator and a couple of sheets of paper? Egads Richard, s-parameter analysis was invented in order to make things "vastly simpler" - and it does! -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
On Sun, 23 May 2004 21:19:23 GMT, Richard Clark
wrote: [snip |in a class A configuration (showing |about 60% efficiency). Huh? |
Cecil wrote, Richard Clark wrote: Load pulling is vastly simpler than S-parameters (also specified in the data sheets for this device, down to 1MHz), why you want to marry the two is a mystery. A laboratory bench setup is simpler than a calculator and a couple of sheets of paper? Egads Richard, s-parameter analysis was invented in order to make things "vastly simpler" - and it does! -- 73, Cecil http://www.qsl.net/w5dxp S-parameters are supposed to work best on small-signal analysis, where everything is fairly linear. In this case, the MRF 136 is a power fet and Motorola says you can use s-parameters as a first approximation. In the data sheet, they provide a list of common source scattering parameters from 2.0 Mhz up to 800 Mhz. I suppose that's useful to someone familiar with s-parameter techniques. 73, KA6RUH |
Wes wrote,
On Sun, 23 May 2004 21:19:23 GMT, Richard Clark wrote: [snip |in a class A configuration (showing |about 60% efficiency). Huh? The data sheet doesn't say it gets 60% efficiency in a class A configuration. It just says, under one bullet, that "Efficiency = 60%." Further down the column it says, "Ideally Suited For Class A Operation." 73, Tom Donaly, KA6RUH |
Tdonaly wrote:
S-parameters are supposed to work best on small-signal analysis, where everything is fairly linear. Therefore, should work well on transmission-line analysis, where everything is linear. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
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Richard Clark wrote:
S-params may be useful for comparing devices, but are inadequate for power application design - unless, of course, you can tolerate 400% error and inverted reactances. The actual subject of the s-parameter analysis was a linear Z0-match network point, not an amplifier, so your statements are irrelevant. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
"Richard Clark" wrote in message
... Hi Guys, I trust both of you appreciate the difference between a vaguely described system and trying to fit a spec sheet to it. Far easier to proceed direct from one or the other, but making a fit between the two is like stepping from one boat to another with a high chance of standing on neither. 73's Richard Clark, KB7QHC There are uncharacterised components surrounding my MRF136, and I'd rather determine their lumped behaviour with the transistor than unsolder them one by one and try and predict the answer. But that's because I think it's possible to determine what's going on. One reason I *don't* think this is the following little bench test this afternoon - with the load jammed directly on the amp output; no cable, no VSWR meter. At 8.000 Mhz and a load consisting of 50 ohms carbon 10W 10% in series with a capacitance trimmer bank, at an output power level of about 2W, a load capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a minimum in the total voltage across the load. Also, as capacitance was increased over the range 100-700pF, the voltage across the load resistor increased monotonically. The latter is easy to explain (it means the source reactance is positive, and smaller than +j28.4 ohms), but the former is beyond my ken. Best, Andrew |
On Mon, 24 May 2004 05:46:01 GMT, "Lord Snooty" wrote:
At 8.000 Mhz and a load consisting of 50 ohms carbon 10W 10% in series with a capacitance trimmer bank, at an output power level of about 2W, a load capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a minimum in the total voltage across the load. Also, as capacitance was increased over the range 100-700pF, the voltage across the load resistor increased monotonically. The latter is easy to explain (it means the source reactance is positive, and smaller than +j28.4 ohms), but the former is beyond my ken. Best, Andrew Hi Andrew, You got me confused too. Is the "load" the resistor, or the resistor-cap combination when you measure these voltages? You describe a voltage minimum across the load for a cap setting of 250pF; but you also maintain that the voltage across the load increases for the variation in capacitance from 100 to 700pF which contradicts the first measurement. Further, the construction of a high power semiconductor does not lend itself to supporting inductive reactances (the junctions are quite manifestly capacitive in structure). By specification the device is characterized as exhibiting 27pF @ 1 MHz (for 28Vdc although there is not much variation until below 10Vdc). There are a world of other variables to consider, but none portray inductance within the device. This alone should provoke you to re-examine your premise. 73's Richard Clark, KB7QHC |
Comment imbedded:
"Dave" wrote in message ... ok, bonzo, i'll bite on the troll bait. ... "Lord Snooty" wrote in message nk.net... ...RF power amp feeding directly into a VSWR meter, ...into a load consisting of a carbon resistor and a variable capacitor rigged in series. The meter connects to the load via about a foot of 50 ohm coax. The frequency is between 1 and 10 MHz. Model the source impedance as Zs = R + jX, and the load impedance as Zl = r + jx (or use phasors if you prefer :). The following two statements are true: 1) The power dissipated in the load (r) is maximised when x = -X (so-called "conjugate matching"), whatever the value of (r). wrong. you must transform the R+jX along the transmission line to get back to the load seen by the source. you stipulate a low frequency and short line, so you are close anyway. The "transform along the coax" part is correct, but the "power is maximised" part can be VERY misleading folks. P.S. Get this MPT theorm blockage out of your minds... It is a synthetic restriction. The "maximum power therom" (ZL=Zs) ONLY applies to ONE special case, NOT all cases. That case is where the source's output power (or if you like current) capability is limited ONLY by the two resistances. That is, the case is when the source can put out all the power needed by these resistors and no other internal limit dominates. A common circuit can be shown to give maximum power at other than Zs=ZL (aparently violating the above referred-to therom). There are things other than these resistances that limit the output power of a practical source.... (see how long this thread goes..... 2) The classical VSWR is minimised (zero "reflected power") when x = +X, whatever the value of (r). doubly wrong. vswr is on a cable and is independent of the source. Without plodding through the rest, it appears Dave has a handle on the error. -- Steve N, K,9;d, c. i My email has no u's. it knows nothing of R+jX only the characteristic impedance of the cable. all following calculations are wrong for this reason alone. However, my VSWR meter, whch is a conventional 2-diode bridge and short transmission line, indicates that minimum indicated VSWR corresponds to max power dissipated in (r).!! (i.e. at conjugate match, and NOT when reflected power is zero). The equation normally used for VSWR is VSWR = ABS( (1 + |p|) / (1 - |p|) ) where p = (Zl - Zs) / (Zl + Zs) wrong again, the impedance used must be that of the cable not of the source. its not worth commenting further until you understand this. and p is a measure of the amount of power reflected back to the source, called the "voltage reflection coefficient" I plotted something I call "conjugate VSWR" or VSWR*. which is the same expression as above, but with p defined as p = (Zl - Zs*) / (Zl + Zs) where Zs* indicates the complex conjugate of Zs. and the behaviour of this VSWR* thingie absolutely matches what I see on my meter. Aye, there's the rub. Some points to note a) Classical VSWR shows NO minimum for all r, when x has the opposite sign to X b) VSWR* always has a minimum at the same r-value which causes maximum power to be dissipated in r, whatever the value of x. Again, I flat don't understand how my VSWR meter can indicate VSWR* when I know it should indicate VSWR. Here are a couple of links to flesh out the theory. 1. Wade through this at your peril - it's you lot fighting abou this issue and is VERY long http://www.ibiblio.org/pub/academic/...S/20030831.ant 2. This is much more succint - cut to the chase on p47 http://my.ece.ucsb.edu/yorklab/Usefu...%20AN64-1B.pdf Best, Andrew |
"Richard Clark" wrote in message
... On Mon, 24 May 2004 05:46:01 GMT, "Lord Snooty" wrote: At 8.000 Mhz and a load consisting of 50 ohms carbon 10W 10% in series with a capacitance trimmer bank, at an output power level of about 2W, a load capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a minimum in the total voltage across the load. Also, as capacitance was increased over the range 100-700pF, the voltage across the load resistor increased monotonically. The latter is easy to explain (it means the source reactance is positive, and smaller than +j28.4 ohms), but the former is beyond my ken. Best, Andrew Hi Andrew, You got me confused too. Is the "load" the resistor, or the resistor-cap combination when you measure these voltages? You describe a voltage minimum across the load for a cap setting of 250pF; but you also maintain that the voltage across the load increases for the variation in capacitance from 100 to 700pF which contradicts the first measurement. Further, the construction of a high power semiconductor does not lend itself to supporting inductive reactances (the junctions are quite manifestly capacitive in structure). By specification the device is characterized as exhibiting 27pF @ 1 MHz (for 28Vdc although there is not much variation until below 10Vdc). There are a world of other variables to consider, but none portray inductance within the device. This alone should provoke you to re-examine your premise. 73's Richard Clark, KB7QHC To clarify a) "Load" in my context means "load resistor (r) and load capacitor (reactance jx) in series" b) The output transistor feeds to the output port through an inductor. One would therefore expect X, the source reactance, to be positive. Andrew |
On Wed, 26 May 2004 02:07:50 GMT, "Lord Snooty" wrote:
To clarify a) "Load" in my context means "load resistor (r) and load capacitor (reactance jx) in series" b) The output transistor feeds to the output port through an inductor. One would therefore expect X, the source reactance, to be positive. Hi Andrew, a load capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a minimum in the total voltage across the load. What was the voltage? Also, as capacitance was increased over the range 100-700pF, the voltage across the load resistor increased monotonically. What were the voltages? The latter is easy to explain (it means the source reactance is positive, and smaller than +j28.4 ohms), but the former is beyond my ken. as the capacitive reactance falls, you note the voltage climbs, this hardly requires an inductance to explain this. Simple divider action serves quite well. You have since revealed the inductor buffered output, but the data is still pretty skimpy to bless it as the major contributor to source Z. What are you using to measure this voltage? 73's Richard Clark, KB7QHC |
"Richard Clark" wrote in message
... On Wed, 26 May 2004 02:07:50 GMT, "Lord Snooty" wrote: Hi Andrew, a load capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a minimum in the total voltage across the load. What was the voltage? Load voltage lies in the 5 to 20 volt peak range, depending on the power setting. This minimum represented about a 15% dip. Also, as capacitance was increased over the range 100-700pF, the voltage across the load resistor increased monotonically. What were the voltages? Again, a nominal value between 5 and 20 V pk. The latter is easy to explain (it means the source reactance is positive, and smaller than +j28.4 ohms), but the former is beyond my ken. as the capacitive reactance falls, you note the voltage climbs, this hardly requires an inductance to explain this. Simple divider action serves quite well. You have since revealed the inductor buffered output, but the data is still pretty skimpy to bless it as the major contributor to source Z. Agreed, but if I keep increasing the load C (decreasing the capacitative reactance ), I will see a peak in the voltage across the load resistor, which will only happen if a conjugate match is occurring. What are you using to measure this voltage? A scope probe set to 10x, which has an unmeasurably high DC resistance and a capacitance of 22 pF (measured). It's a good idea to use the 10x, not 1x, setting, since the latter contributes about 80 pF, and will change the AC response of the circuit. Best, Andrew |
On Thu, 27 May 2004 16:49:26 GMT, "Lord Snooty" wrote:
"Richard Clark" wrote in message .. . On Wed, 26 May 2004 02:07:50 GMT, "Lord Snooty" wrote: Hi Andrew, a load capacitor value of 250 +/- 10pF (-j80 ohms @ 8 MHz) was found to produce a minimum in the total voltage across the load. What was the voltage? Load voltage lies in the 5 to 20 volt peak range, depending on the power setting. This minimum represented about a 15% dip. Pick ONE power setting. What was the voltage across the load. Also, as capacitance was increased over the range 100-700pF, the voltage across the load resistor increased monotonically. What were the voltages? Again, a nominal value between 5 and 20 V pk. Pick the ONE and SAME power setting. What was the voltage across the load resistor for: 100pF 200pF 300pF .... 700pF The latter is easy to explain (it means the source reactance is positive, and smaller than +j28.4 ohms), but the former is beyond my ken. as the capacitive reactance falls, you note the voltage climbs, this hardly requires an inductance to explain this. Simple divider action serves quite well. You have since revealed the inductor buffered output, but the data is still pretty skimpy to bless it as the major contributor to source Z. Agreed, but if I keep increasing the load C (decreasing the capacitative reactance ), I will see a peak in the voltage across the load resistor, which will only happen if a conjugate match is occurring. This is a violation of terms. Just what constitutes the generator? At one time you say the combination of the cap-resistor is the load, hence the source is described ACROSS this series. THEN you isolate the resistor which pushes the cap back into the source. You originally asked for the complex impedance of the source, but if the source contains a variable cap, this makes determination rather a moving target. In the world of metrology (folks who measure this stuff for a living), you have an immutable boundary called the plane of the source. On one side is everything that can be attributed to the source and everything on the other side can be attributed to the load. If there is a transmission line between, then you have two planes, the plane of the source, and the plane of the load. Everything to the right of the plane of the load (thinking in a left-right progression) can be attributed to the load. Between the two planes is a transform. The plane of the source is commonly the output connector, the plane of the load is commonly the input connector. Things in between like tuners, SWR meters, Lines, dividers, splitters, duplexers... are transforms. It is perfectly justifiable to make a component like a tuner resident within (and behind) either plane, but once you do that, it is considered bad form to go tweaking it and maintain nothing has happened to the source/load. What are you using to measure this voltage? A scope probe set to 10x, which has an unmeasurably high DC resistance and a capacitance of 22 pF (measured). It's a good idea to use the 10x, not 1x, setting, since the latter contributes about 80 pF, and will change the AC response of the circuit. Perfectly adequate. 73's Richard Clark, KB7QHC |
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