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Transmitter Output Impedance
This group has presented members with valuable lessons in antennas and
transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. My curiosity was piqued today as I took some baby steps into EZNEC. A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. Doesn't seem like it. My point: Using 75-ohm cable to improve the match at the antenna won't help me ... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. (I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. Thanks. "Sal" (KD6VKW) |
Transmitter Output Impedance
"Sal M. Onella" wrote in message ... This group has presented members with valuable lessons in antennas and transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. My curiosity was piqued today as I took some baby steps into EZNEC. A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. Doesn't seem like it. My point: Using 75-ohm cable to improve the match at the antenna won't help me ... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. (I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. Thanks. "Sal" (KD6VKW) A transmitter output impedance is designed for maximum power transfer at a specific impedance. Most of the the older tube transmitters impedance was tunable within a range. In simple terms the impedance of the transmitter tube is the plate voltage devided by the current. This impedance is then transformed to the nominal 50 ohms of the antenna system. If the transmitter has the usual tune and load controls, the exect impedance will not mater as you adjust for maximum transmitter output. Most of the transistor transmitters are not adjustable so the output impedance is usually fixed at 50 ohms for maximum power transfer. If the impedance of the antenna system is not 50 ohms, then the output power will be less than the designed output. You can use the antenna tuner to adjust for a match. |
Transmitter Output Impedance
Sal M. Onella wrote:
This group has presented members with valuable lessons in antennas and transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. My curiosity was piqued today as I took some baby steps into EZNEC. A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. Doesn't seem like it. My point: Using 75-ohm cable to improve the match at the antenna won't help me ... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. (I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. Thanks. "Sal" (KD6VKW) I suspect that most ham transmitters do NOT have a 50 ohm output impedance. What they do have is a specification that they will adequately drive a 50 ohm load (and some sort of internal circuitry that detects an "unacceptable" output condition and turns down the drive). After all, your transmitter could have an output impedance of zero ohms (a "stiff" voltage source), and adequately drive your transmission line and antenna at 50 ohms (yes, this is not the optimum power transfer, but nobody ever said that ham transmitters are designed for optimum power transfer... maybe they're perfectly happy with less transfer, but still operating within their safe area) ON9CVD made some simple measurements using a couple of resistors and foudn that a TS440 has a Zout somewhere around 15-40 ohms (depending on frequency and output power). http://sharon.esrac.ele.tue.nl/~on9c...impedantie.htm Grant Bingeman also has words on this: http://www.km5kg.com/loads.htm |
Transmitter Output Impedance
On Mon, 25 Apr 2011 17:35:29 -0700 (PDT), "Sal M. Onella"
wrote: Something I haven't seen is a discussion of the source impedance of the transmitter. sigh.... My point: Using 75-ohm cable to improve the match at the antenna won't help me ... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. Hi OM, Look at the prospective SWR and how much is lost/reflected/absorbed/what-have-you? More heat comes from a less than optimal system efficiency than what your computation will reveal. So much that it will swamp it. But the trick here is that the reflected "power" (arguments turn on this word) doesn't always mean heat and it could actually cool - however hot or cold it may alter the situation, that same "power" never got out into the air. Now, as for source impedance, that is a subject fraught with denial in the face of the obvious: Those fins in the back of your rig are to help bookend your QSL cards into groups (the heat bears no relation to efficiency nor match loss). A standard definition (courtesy of Wikipedia) for Return Loss is: where Zs is the impedance toward the source and Zl is the impedance toward the load. and we find from the values you supply that it is 0.20 Of course, such a definition is utterly useless when the concept of Zs is replaced with (in most cases) "it ain't 50 Ohms, thet's fur shure"). If, perchance, some brave soul steps into the breach of NOT 50 Ohms to suggest what Zs is, then we can give it the acid test of engineering (an act that I am usually reminded is beyond the understanding of readers and the province of discussion here). Let's be gentle and go only by an order of two (which is reasonably available and can be coaxed out of my TS-440). Return loss for a rig exhibiting an Zs of 25 Ohms into the 75 Ohm line (presuming it is infinite in length) would give us: 0.50 that doesn't look good, so let's try Zs of 100 Ohms: 0.14 that looks better all 'round. Even intuition agrees. Let's press intuition to the proximal limit and say that Zs is 74 Ohms (yes, my thumb is on the scale): 0.01 What does intuition affirm? What is preferable? 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On Apr 25, 7:35*pm, "Sal M. Onella" wrote:
Wisdom in any form would be appreciated. *Thanks. Have you seen these? http://www.w2du.com/QEXMayJun01.pdf http://www.w2du.com/Appendix12.pdf http://www.w2du.com/r3ch19a.pdf -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On 26 abr, 02:35, "Sal M. Onella" wrote:
This group has presented members with valuable lessons in antennas and transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. *My curiosity was piqued today as I took some baby steps into EZNEC. *A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. *Doesn't seem like it. My point: *Using 75-ohm cable to improve the match at the antenna won't help me *... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. *My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. *(I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. *Thanks. "Sal" (KD6VKW) Hello Sal, Make yourself up for a long discussion (as we had such a thread earlier). Most amplifiers are designed just to provide the desired amount of power into a certain load. The actual output impedance of the amplifier is not important in many cases. Changing the load impedance (for example in case of a solid state amplifier) may result in less or more heat generated in the active device(s). You can see this by changing the load (for example with an external matcher between power meter and PA) and watching the output power and current consumption. If you have an amplifier with tunable output section (vacuum tube PA), you are within the range of the tuner, and you tune it for maximum output power (given certain drive), the output impedance equals the load impedance (or conjugated value in case of non-ohmic load). If you change the drive (so adjust the output power), the output impedance may change (due to saturation issues). Same is valid for the load, if you change the load, voltage across and current through the active device may saturate, hence changing the output impedance. Non-tunable amplifiers (for example a 3…30 MHz balanced amplifier) will mostly not present 50 Ohms to your load (unless specially designed for that using feedback). Virtually all high-efficient switching amplifiers do not show 50 Ohms to the load. If you add an external tuner and match to maximum output power, you will very likely destroy the amplifier in case of no supervisory circuits present. You can do some experiments with your own amplifiers. Just change the load impedance and see what the forward power indicator on your reflectometer/VSWR meter does. If it doesn't change, your output impedance is very close to 50 Ohms. Measuring the output impedance (for relative small change in load) is possible, but is not a simple task. Very likely other people will comment on this. With kind regards, Wim PA3DJS www.tetech.nl without abc, PM will reach me very likely. |
Transmitter Output Impedance
Ralph Mowery wrote:
"Sal M. Onella" wrote in message ... This group has presented members with valuable lessons in antennas and transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. My curiosity was piqued today as I took some baby steps into EZNEC. A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. Doesn't seem like it. My point: Using 75-ohm cable to improve the match at the antenna won't help me ... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. (I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. Thanks. "Sal" (KD6VKW) A transmitter output impedance is designed for maximum power transfer at a specific impedance. Most of the the older tube transmitters impedance was tunable within a range. In simple terms the impedance of the transmitter tube is the plate voltage devided by the current. This impedance is then transformed to the nominal 50 ohms of the antenna system. If the transmitter has the usual tune and load controls, the exect impedance will not mater as you adjust for maximum transmitter output. Most of the transistor transmitters are not adjustable so the output impedance is usually fixed at 50 ohms for maximum power transfer. If the impedance of the antenna system is not 50 ohms, then the output power will be less than the designed output. You can use the antenna tuner to adjust for a match. Not exactly.. A "match" provides the optimum power transfer from generator to load, but that is NOT the maximum load power, nor is it either the maximum or minimum power dissipated in the source. Say I have a zero output impedance on my source and I'm putting out 7 Volts RMS into a 50 ohm load. That's about 1 watt into the load. Now.. if I reduce the load impedance to 25 ohms, and since I've got zero output impedance, I'm now putting out 2 Watts. The source impedance is zero, so I'm not dissipating any extra power in the source, either. It is true that a "matched load" to my zero ohm source would, in fact, be zero ohms, and would have infinite power. Any other load impedance would have less power into the load, so the Thevenin theorem is satisfied. Now.. if my generator had a fixed output impedance, it's true that the load impedance that will get the most power out is the conjugate of the output Z. For resistive sources/loads, here's an example.. You also have to be careful about looking at Thevenin equivalent sources (e.g. a ideal voltage in series with a Z, or a ideal current in parallel with a Z), because just because *the model* has a resistor in series does NOT mean that you're actually dissipating any power in the source. (If I had a very efficient op amp, I could simulate any arbitrary output impedance, without dissipating any power in the source) Say my generator is 40 ohms, and I'm putting out 7 Volts into a 40 ohm load. OK, that means that the imaginary voltage source is putting out 14 V. I'm getting about 1.23 Watts into my load. Now, if I decrease my load Z to 20 ohms, what do I get? Now, I have 4.67 (=14/3) Volts instead of 7, and I get 1.1 Watts. Yep, less.. Thevenin works. Let's try increasing the Z to 60 ohms.. Now the voltage on my load is 8.4 V, and I'm dissipating 1.18W, again, less than my 1.23. But here's some weird stuff.. let's look at how much power is dissipated in that imaginary resistor (i.e. our source *really is* a ideal voltage source in series with a resistor) At 40 ohm load, we've got 7 volts on the load and 7 volts across the resistor, so they both dissipate the same 1.225 Watts. Pload/Pgen = 1 In the 20 ohm load case, we've got 1.1 dissipated in the load and 2.2 dissipated in the generator. Pload/Pgen = 0.5 In the 60 ohm load case, we've got 1.18 dissipated in the load, and 0.78 dissipated in the generator. Pload/Pgen = 1.5 (i.e. we dissipate more in the load than in the generator... how about that!) And let's look at "efficiency" of the system (assuming that the total power in is the sum of what's dissipated in the generator and the load) 20 ohm load, 33% 40 ohm load, 50% (what you'd expect) 60 ohm load, 60% (hey.. it's more efficient, too) - take home message... a "good match" is sort of an artificial thing from a power transfer standpoint.. it depends on what you're trying to optimize for. - Where you get bitten is when "match" varies with frequency... now, all of a sudden, you have a system that has a response that varies with frequency, which is generally undesirable. When you get up into the microwave region, where a transmission line is often many wavelengths long, that mismatch can result in remarkably wild fluctuations in gain with respect to frequency. |
Transmitter Output Impedance
Measuring the output impedance (for relative small change in load) is possible, but is not a simple task. Very likely other people will comment on this. ON9CVD's website I linked to has a very simple technique.. 50 ohm dummy load and a 220 ohm resistor you can switch in. At 100W (into 50 ohms), the 220 ohms would only dissipate 22W. You could get some of those non-inductive resistors from Caddock and series them up to do something like this. BTW, this is a simplified version of what's called a "load pull" test... which makes me wonder if one could cobble up a quick test set that could be controlled by a computer to do automated output Z measurements of an HF transceiver over a reasonably wide range... One approach would be to use a RS-232 controlled antenna tuner and, maybe, a antenna relay box with several different load resistances). The challenge (having actually looked at doing this with a LDG AT200PC) is that the Z of the tuner isn't very well defined. It's a pretty big calibration project in itself. Maybe, though, one could build a few test dummy loads.. say a 25 ohm and a 75 or 100 ohm, along with your vanilla 50 ohm, and the antenna switch (like an RCS-8V). Basically, you're building a "high power resistor substitution box" You'd want some sort of nice inline watt meter (like an LP100) to make the measurements. |
Transmitter Output Impedance
Jim Lux wrote:
Ralph Mowery wrote: "Sal M. Onella" wrote in message ... This group has presented members with valuable lessons in antennas and transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. My curiosity was piqued today as I took some baby steps into EZNEC. A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. Doesn't seem like it. My point: Using 75-ohm cable to improve the match at the antenna won't help me ... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. (I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. Thanks. "Sal" (KD6VKW) A transmitter output impedance is designed for maximum power transfer at a specific impedance. Most of the the older tube transmitters impedance was tunable within a range. In simple terms the impedance of the transmitter tube is the plate voltage devided by the current. This impedance is then transformed to the nominal 50 ohms of the antenna system. If the transmitter has the usual tune and load controls, the exect impedance will not mater as you adjust for maximum transmitter output. Most of the transistor transmitters are not adjustable so the output impedance is usually fixed at 50 ohms for maximum power transfer. If the impedance of the antenna system is not 50 ohms, then the output power will be less than the designed output. You can use the antenna tuner to adjust for a match. Not exactly.. A "match" provides the optimum power transfer from generator to load, but that is NOT the maximum load power, nor is it either the maximum or minimum power dissipated in the source. Say I have a zero output impedance on my source and I'm putting out 7 Volts RMS into a 50 ohm load. That's about 1 watt into the load. Now.. if I reduce the load impedance to 25 ohms, and since I've got zero output impedance, I'm now putting out 2 Watts. The source impedance is zero, so I'm not dissipating any extra power in the source, either. It is true that a "matched load" to my zero ohm source would, in fact, be zero ohms, and would have infinite power. Any other load impedance would have less power into the load, so the Thevenin theorem is satisfied. Now.. if my generator had a fixed output impedance, it's true that the load impedance that will get the most power out is the conjugate of the output Z. For resistive sources/loads, here's an example.. You also have to be careful about looking at Thevenin equivalent sources (e.g. a ideal voltage in series with a Z, or a ideal current in parallel with a Z), because just because *the model* has a resistor in series does NOT mean that you're actually dissipating any power in the source. (If I had a very efficient op amp, I could simulate any arbitrary output impedance, without dissipating any power in the source) Say my generator is 40 ohms, and I'm putting out 7 Volts into a 40 ohm load. OK, that means that the imaginary voltage source is putting out 14 V. I'm getting about 1.23 Watts into my load. Now, if I decrease my load Z to 20 ohms, what do I get? Now, I have 4.67 (=14/3) Volts instead of 7, and I get 1.1 Watts. Yep, less.. Thevenin works. Let's try increasing the Z to 60 ohms.. Now the voltage on my load is 8.4 V, and I'm dissipating 1.18W, again, less than my 1.23. But here's some weird stuff.. let's look at how much power is dissipated in that imaginary resistor (i.e. our source *really is* a ideal voltage source in series with a resistor) At 40 ohm load, we've got 7 volts on the load and 7 volts across the resistor, so they both dissipate the same 1.225 Watts. Pload/Pgen = 1 In the 20 ohm load case, we've got 1.1 dissipated in the load and 2.2 dissipated in the generator. Pload/Pgen = 0.5 In the 60 ohm load case, we've got 1.18 dissipated in the load, and 0.78 dissipated in the generator. Pload/Pgen = 1.5 (i.e. we dissipate more in the load than in the generator... how about that!) And let's look at "efficiency" of the system (assuming that the total power in is the sum of what's dissipated in the generator and the load) 20 ohm load, 33% 40 ohm load, 50% (what you'd expect) 60 ohm load, 60% (hey.. it's more efficient, too) - take home message... a "good match" is sort of an artificial thing from a power transfer standpoint.. it depends on what you're trying to optimize for. - Where you get bitten is when "match" varies with frequency... now, all of a sudden, you have a system that has a response that varies with frequency, which is generally undesirable. When you get up into the microwave region, where a transmission line is often many wavelengths long, that mismatch can result in remarkably wild fluctuations in gain with respect to frequency. Vgen Rgen Rload V load I load P load Pgen Pload/Pgen Ptotal Pload/Ptotal 10 40 20 3.33 0.17 0.56 1.11 0.50 1.67 0.33 10 40 40 5.00 0.13 0.63 0.63 1.00 1.25 0.50 10 40 60 6.00 0.10 0.60 0.40 1.50 1.00 0.60 |
Transmitter Output Impedance
On Apr 26, 10:44*am, Wimpie wrote:
The actual output impedance of the amplifier is not important in many cases. Yes, consider a source with 70.7 volts at its output terminals connected to a 50 ohm load through a 50 ohm feedline. Except for source efficiency, the source impedance simply doesn't matter. Any 70.7 volt source will deliver 100 watts to the feedline no matter what the source impedance. There's a simple method called load-pulling. Keeping everything the same at the source, if the power to the load increases when the load is changed away from 50 ohms, the source doesn't have a 50 ohm internal impedance. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On Apr 26, 7:20*am, Cecil Moore wrote:
On Apr 25, 7:35*pm, "Sal M. Onella" wrote: Wisdom in any form would be appreciated. *Thanks. Have you seen these? http://www.w2du.com/QEXMayJun01.pdf http://www.w2du.com/Appendix12.pdf http://www.w2du.com/r3ch19a.pdf -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Well, I have _now_! 'Twill take just a bit of time to digest them. Thanks. "Sal" |
Transmitter Output Impedance
On 26 abr, 02:59, "Ralph Mowery" wrote:
"Sal M. Onella" wrote in ... This group has presented members with valuable lessons in antennas and transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. *My curiosity was piqued today as I took some baby steps into EZNEC. *A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. *Doesn't seem like it. My point: *Using 75-ohm cable to improve the match at the antenna won't help me *... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. *My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. *(I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. *Thanks. "Sal" (KD6VKW) A transmitter output impedance is designed for maximum power transfer at a specific impedance. Most of the *the older tube transmitters impedance was tunable within a *range. In simple terms the impedance of the transmitter tube is the plate voltage devided by the current. *This impedance is then transformed to the nominal 50 ohms of the antenna system. *If the transmitter has the usual tune and load controls, the exect impedance will not mater as you adjust for maximum transmitter output. Most of the transistor transmitters are not adjustable so the output impedance is usually fixed at 50 ohms for maximum power transfer. *If the impedance of the antenna system is not 50 ohms, then the output power will be less than the designed output. *You can use the antenna tuner to adjust for a match. Hello Ralph, The actual output impedance can be anything, but is mostly not 50 Ohms. If you want it close to 50 Ohms, you have to spend money in components and design time. As 50 Ohms isn't mostly required, one will not design for that. Just as an example, take a hard-driven totem pole or push pull stage with only a series tuned circuit to suppress harmonics (so the LC circuit shows zero ohms at the carrier frequency). As the active devices are used a switches, the output impedance of this arrangement is almost zero (at least far below 50 Ohms). When you connect a 50 Ohms quarter-wave cable between the output and the 50 Ohms load, the amplifier-cable combination has very high output impedance (quarter wave transformer formula). For power amplifiers, there is no relation between actual output impedance and efficiency. When an amplifier is designed for 50 Ohms, it only means that the amplifier will work correctly when terminated with 50 Ohms. When you deviate from that, output power may decrease or increase. This may result in more or less stress on the amplifier's components. With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On Apr 26, 1:59*am, Richard Clark wrote:
On Mon, 25 Apr 2011 17:35:29 -0700 (PDT), "Sal M. Onella" wrote: Something I haven't seen is a discussion of the source impedance of the transmitter. sigh.... My point: *Using 75-ohm cable to improve the match at the antenna won't help me *... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. Hi OM, Look at the prospective SWR and how much is lost/reflected/absorbed/what-have-you? *More heat comes from a less than optimal system efficiency than what your computation will reveal. So much that it will swamp it. But the trick here is that the reflected "power" (arguments turn on this word) doesn't always mean heat and it could actually cool - however hot or cold it may alter the situation, that same "power" never got out into the air. * Now, as for source impedance, that is a subject fraught with denial in the face of the obvious: *Those fins in the back of your rig are to help bookend your QSL cards into groups (the heat bears no relation to efficiency nor match loss). A standard definition (courtesy of Wikipedia) for Return Loss is: * * * * where Zs is the impedance toward the source and * * * * Zl is the impedance toward the load. and we find from the values you supply that it is * * * * 0.20 Of course, such a definition is utterly useless when the concept of Zs is replaced with (in most cases) "it ain't 50 Ohms, thet's fur shure"). If, perchance, some brave soul steps into the breach of NOT 50 Ohms to suggest what Zs is, then we can give it the acid test of engineering (an act that I am usually reminded is beyond the understanding of readers and the province of discussion here). *Let's be gentle and go only by an order of two (which is reasonably available and can be coaxed out of my TS-440). *Return loss for a rig exhibiting an Zs of 25 Ohms into the 75 Ohm line (presuming it is infinite in length) would give us: * * * * 0.50 that doesn't look good, so let's try Zs of 100 Ohms: * * * * 0.14 that looks better all 'round. *Even intuition agrees. Let's press intuition to the proximal limit and say that Zs is 74 Ohms (yes, my thumb is on the scale): * * * * 0.01 What does intuition affirm? *What is preferable? * * * * 73's Richard Clark, KB7QHC Thanks Richard. Intuition is that the Zs is near 50 ohms for as many frequencies as the designer can manage. I am on record (including in this group) of letting intuition lead me down the path to ruin. I get from you that there's a presupposition that I know the source impedance or can easily establish it. Hm-m-m ... not so. One big problem I see is the need to try to measure power delivered in a non-50-ohm system with my existing instruments that depend on a 50-ohm system. I don't have a nominal 75-ohm power meter. Won't putting a 50-ohm meter into a 75-ohm circuit not only read wrong but introduce reflection losses, too? I think I'd need a collection of non-inductive load resistors and an accurate rf ammeter. I'd need to connect them and calculate power at a few points in every band. Maybe the papers that Cecil cited for me will fill in the gaps or suggest other approaches. "Sal" |
Transmitter Output Impedance
On Tue, 26 Apr 2011 13:12:23 -0700, Jim Lux
wrote: (If I had a very efficient op amp, I could simulate any arbitrary output impedance, without dissipating any power in the source) I can see why this is parenthetical, because it covers a lot of sins of omission. First we bang up against the wall of Gain Bandwidth Product. If you are talking about resistive loads at low power DC, then your statement is trivially valid. Second, the ability to "simulate" any arbitrary output (or input for that matter) impedance for an OP AMP is well defined in the closed loop gain (which robs from the open loop gain for frequency by proportion to GBP). Taking the conventional RF Power Deck of any consumer (Ham) product, the similarity to an OP AMP is wholly foreign, and for good and commercial reason. If one were motivated to engineer in the necessary noise amplifier (a term coined by H.W. Bode who defined this topic of source Z and applied it to the negative gain or feedback path); then we would find that the exact same loss is exhibited in the exact same component(s). However, by virtue of OP AMP characteristics we would benefit to vastly better distortion figures, far less spurious content, and virtually no need for either the conventional impedance transformer, nor the bandwidth filter that follows the same power deck (provided, of course, that the drive input is sinusoidal - which it never is, unfortunately, for this scenario). This novel OP-AMP/Power-Deck redesign would also confer considerable power supply rejection (that voltage could sag or rise without appreciable effect) and noise rejection (the internal noise from other circuitry would not migrate into the signal output). ALL such benefits are strictly derived from the amount of negative feedback (not to be confused, as are many readers to this topic, with the rather ordinary compensation cap in the last stage). Why isn't this done as a service to the customer? Cost. Again, OP AMP design merits are paid for in lost gain and bandwidth. The price is found in the amount of negative feedback that goes to lower the overall amplification. Would you pay for this improved cool performance to run 10W in the 80M band from a formerly crackly and hot 100W 10-80M band capable source? 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On 26 abr, 22:21, Jim Lux wrote:
Measuring the output impedance (for relative small change in load) is possible, but is not a simple task. Very likely other people will comment on this. ON9CVD's website I linked to has a very simple technique.. 50 ohm dummy load and a 220 ohm resistor you can switch in. At 100W (into 50 ohms), the 220 ohms would only dissipate 22W. *You could get some of those non-inductive resistors from Caddock and series them up to do something like this. BTW, this is a simplified version of what's called a "load pull" test... which makes me wonder if one could cobble up a quick test set that could be controlled by a computer to do automated output Z measurements of an HF transceiver over a reasonably wide range... One approach would be to use a RS-232 controlled antenna tuner and, maybe, a antenna relay box with several different load resistances). The challenge (having actually looked at doing this with a LDG AT200PC) is that the Z of the tuner isn't very well defined. *It's a pretty big calibration project in itself. Maybe, though, one could build a few test dummy loads.. say a 25 ohm and a 75 or 100 ohm, along with your vanilla 50 ohm, and the antenna switch (like an RCS-8V). *Basically, you're building a "high power resistor substitution box" You'd want some sort of nice inline watt meter (like an LP100) to make the measurements. Hello Jim, Other method is injecting a slightly off-carrier frequency signal into the amplifier (this emulates a constant small VSWR shown to the PA (wtih 50 Ohms load), but with continuous varying phase). Because of the difference in frequency, one can measure the forward (towards the PA) and reverse (reflected by PA) signal with a two channel VSA. This will give you the PA's complex output impedance. Tom (K7ITM from my head) did this with a HP89410 with couplers. With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On Tue, 26 Apr 2011 14:37:58 -0700 (PDT), "Sal M. Onella"
wrote: Thanks Richard. Intuition is that the Zs is near 50 ohms for as many frequencies as the designer can manage. I am on record (including in this group) of letting intuition lead me down the path to ruin. Well, if you miss the path, you are certain to be reminded where it is. I get from you that there's a presupposition that I know the source impedance or can easily establish it. Hm-m-m ... not so. It is printed in the specifications. There are other ways to derive it, of course, and they would merely confirm that number. One big problem I see is the need to try to measure power delivered in a non-50-ohm system with my existing instruments that depend on a 50-ohm system. I don't have a nominal 75-ohm power meter. Won't putting a 50-ohm meter into a 75-ohm circuit not only read wrong but introduce reflection losses, too? It is easier to measure voltage and current and use your calculator. Now having said that, measuring voltage and current is damned hard at HF. It is achievable with care, but now we are back into your same question with many hands pointing at that path to ruin. As both are difficult (power or voltage times current), you could trust authority (which confirms what is intuitive), or you could listen to argument (which at the distance of time and recall becomes murky and opaque). I think I'd need a collection of non-inductive load resistors and an accurate rf ammeter. I'd need to connect them and calculate power at a few points in every band. Bravo! This reduces complexity because RF Power resistors of high accuracy and bandwidth are commercially available. You will need to practice your skill at mounting to a heat sink, however (another non-trivial achievement). Maybe the papers that Cecil cited for me will fill in the gaps or suggest other approaches. Walt and I have been corresponding over these matters just these past two weeks. Skimming that content will once again confirm intuition. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
Richard Clark wrote:
On Tue, 26 Apr 2011 13:12:23 -0700, Jim Lux wrote: (If I had a very efficient op amp, I could simulate any arbitrary output impedance, without dissipating any power in the source) I can see why this is parenthetical, because it covers a lot of sins of omission. yep.. not possible to build such a thing, anymore than one can build a zero ohm output impedance RF source with any signficant power. Suggested more as an example that the power dissipation in the source doesn't necessarily correlate with match, load Z, or anything else in general. (You can get pretty darn close at powers less than a watt and HF, though..) However, by virtue of OP AMP characteristics we would benefit to vastly better distortion figures, far less spurious content, and virtually no need for either the conventional impedance transformer, nor the bandwidth filter that follows the same power deck (provided, of course, that the drive input is sinusoidal - which it never is, unfortunately, for this scenario). This novel OP-AMP/Power-Deck redesign would also confer considerable power supply rejection (that voltage could sag or rise without appreciable effect) and noise rejection (the internal noise from other circuitry would not migrate into the signal output). ALL such benefits are strictly derived from the amount of negative feedback (not to be confused, as are many readers to this topic, with the rather ordinary compensation cap in the last stage). One can also do a lot of this with various clever schemes if the input to your PA is coming out of some signal processing. Generically, predistortion, but it can be so much more. People have literally spent their lives working out ever more sophisticated approaches Why isn't this done as a service to the customer? Cost. Like race cars... how fast do you want to go..just bring money |
Transmitter Output Impedance
Wimpie wrote:
On 26 abr, 22:21, Jim Lux wrote: Measuring the output impedance (for relative small change in load) is possible, but is not a simple task. Very likely other people will comment on this. ON9CVD's website I linked to has a very simple technique.. 50 ohm dummy load and a 220 ohm resistor you can switch in. At 100W (into 50 ohms), the 220 ohms would only dissipate 22W. You could get some of those non-inductive resistors from Caddock and series them up to do something like this. BTW, this is a simplified version of what's called a "load pull" test... which makes me wonder if one could cobble up a quick test set that could be controlled by a computer to do automated output Z measurements of an HF transceiver over a reasonably wide range... One approach would be to use a RS-232 controlled antenna tuner and, maybe, a antenna relay box with several different load resistances). The challenge (having actually looked at doing this with a LDG AT200PC) is that the Z of the tuner isn't very well defined. It's a pretty big calibration project in itself. Maybe, though, one could build a few test dummy loads.. say a 25 ohm and a 75 or 100 ohm, along with your vanilla 50 ohm, and the antenna switch (like an RCS-8V). Basically, you're building a "high power resistor substitution box" You'd want some sort of nice inline watt meter (like an LP100) to make the measurements. Hello Jim, Other method is injecting a slightly off-carrier frequency signal into the amplifier (this emulates a constant small VSWR shown to the PA (wtih 50 Ohms load), but with continuous varying phase). Because of the difference in frequency, one can measure the forward (towards the PA) and reverse (reflected by PA) signal with a two channel VSA. This will give you the PA's complex output impedance. Yes.. that's another way to do it. I think one would want variable load impedances for the testing in any case, because I'll bet that most ham rigs have a load dependent Z. I've occasionally kicked around the idea of what would it take to do it simply, especially with nifty devices becoming available to help with the measurements.. $500 VNAs, $400 power meters that directly read current and voltage, etc. What I'm not sure about is whether it is "useful" to know. Consider a ham with a manual or auto tuner. They'll adjust for minimum reflected power, which is probably as good as anything else. Hams, as a class, don't much care about "DC power to RF radiated" efficiency (because the regulatory requirements are imposed at the "RF output" measurement plane.. Power dissipation in the PA is only a second order concern.. can I plug it into a standard outlet? Will it get too hot in my car? backpack QRPers are concerned about DC power consumption, but I'm not sure they're worried about whether their PA is 30% or 35% efficient. The people designing battery powered tracking transmitters ARE concerned about this, as are high power broadcasters (since they're both directly paying for the supply power and have a "radiated RF power" requirement to meet their need). |
Transmitter Output Impedance
On Tue, 26 Apr 2011 15:27:28 -0700, Jim Lux
wrote: (If I had a very efficient op amp, I could simulate any arbitrary output impedance, without dissipating any power in the source) I can see why this is parenthetical, because it covers a lot of sins of omission. yep.. not possible to build such a thing, anymore than one can build a zero ohm output impedance RF source with any signficant power. Suggested more as an example that the power dissipation in the source doesn't necessarily correlate with match, load Z, or anything else in general. (You can get pretty darn close at powers less than a watt and HF, though..) OP AMPs are a constant of my admiration in the possibilities offered. That and the signal processing you suggest (plus digital oscillators) "could" change the playing field - if conventional design weren't so universally fallen back upon. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On Apr 26, 4:49*pm, Wimpie wrote:
Other method is injecting a slightly off-carrier frequency signal into the amplifier (this emulates a constant small VSWR shown to the PA (wtih 50 Ohms load), but with continuous varying phase). Because of the difference in frequency, one can measure the forward (towards the PA) and reverse (reflected by PA) signal with a two channel VSA. This will give you the PA's complex output impedance. Unfortunately, the impedance encountered by the off-carrier frequency signal is probably not the same as the impedance encountered by the carrier frequency so the results don't correlate and are not very useful. The carrier frequency has interference components that the off- carrier signal doesn't encounter. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On Apr 25, 6:07*pm, Jim Lux wrote:
Sal M. Onella wrote: This group has presented members with valuable lessons in antennas and transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. *My curiosity was piqued today as I took some baby steps into EZNEC. *A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. *Doesn't seem like it. My point: *Using 75-ohm cable to improve the match at the antenna won't help me *... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. *My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. *(I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. *Thanks. "Sal" (KD6VKW) I suspect that most ham transmitters do NOT have a 50 ohm output impedance. *What they do have is a specification that they will adequately drive a 50 ohm load (and some sort of internal circuitry that detects an "unacceptable" output condition and turns down the drive). After all, your transmitter could have an output impedance of zero ohms (a "stiff" voltage source), and adequately drive your transmission line and antenna at 50 ohms (yes, this is not the optimum power transfer, but nobody ever said that ham transmitters are designed for optimum power transfer... maybe they're perfectly happy with less transfer, but still operating within their safe area) ON9CVD made some simple measurements using a couple of resistors and foudn that a TS440 has a Zout somewhere around 15-40 ohms (depending on frequency and output power).http://sharon.esrac.ele.tue.nl/~on9c...impedantie.htm Grant Bingeman also has words on this:http://www.km5kg.com/loads.htm I agree with Jim. While it's true that if a source (transmitter) is tuned for maximum output, the output impedance must necessarily be the conjugate of the load impedance, it is NOT generally the case that the transmitter is tuned for maximum output. Rather, the transmitter is tuned for an output that won't destroy the output devices and will result in acceptable distortion (in the case of a linear amplifier). There are plenty of cases of sources designed to be loaded with an impedance far different from their output (source) impedance: the AC power line, audio amplifiers, ... . A while back, I set up a couple precision high power directional couplers so I could measure the output impedance of a couple different ham rigs. In the case of the rig with vacuum tube output stage, if I operated the output stage with limited grid drive and tuned the plate tank for maximum output power, indeed the output impedance was 50 ohms, within the tolerance of my ability to adjust the output for maximum. But if I increased the grid drive for solid class-C operation and tuned for the rated output power (which is no longer the maximum possible power), the impedance seen at the output dropped. If you work through the pi-network transformation back to the vacuum tube plates, it's apparent that the plates under those operating conditions represent a considerably higher source impedance than when things are tuned for maximum available power (as first described). But coming back to "Sal's" original question, it's always made sense to me given the availability of inexpensive 75 ohm line with low loss to go ahead and use it to feed antennas that have a feedpoint impedance closer to 75 ohms than to 50 ohms. If you need to provide a bit of matching at the transmitter end so that the transmitter is operating correctly, it should be straightforward to do that. But whether the actual source impedance of the transmitter is one value or another is really of very little importance. The only time I can think that it would matter is if you're trying to transmit a very broadband signal and you don't want power that's reflected at the transmission-line:antenna interface to re-reflect from the transmitter:transmission-line interface and go back to the antenna, delayed by enough to cause a "ghost" (in a television picture), for example. In such a case, you'll be well served by insuring that the antenna is well matched to the transmission line so there is an insignificant reflection there anyway. Cheers, Tom |
Transmitter Output Impedance
On Apr 26, 2:49*pm, Wimpie wrote:
On 26 abr, 22:21, Jim Lux wrote: Measuring the output impedance (for relative small change in load) is possible, but is not a simple task. Very likely other people will comment on this. ON9CVD's website I linked to has a very simple technique.. 50 ohm dummy load and a 220 ohm resistor you can switch in. At 100W (into 50 ohms), the 220 ohms would only dissipate 22W. *You could get some of those non-inductive resistors from Caddock and series them up to do something like this. BTW, this is a simplified version of what's called a "load pull" test.... which makes me wonder if one could cobble up a quick test set that could be controlled by a computer to do automated output Z measurements of an HF transceiver over a reasonably wide range... One approach would be to use a RS-232 controlled antenna tuner and, maybe, a antenna relay box with several different load resistances). The challenge (having actually looked at doing this with a LDG AT200PC) is that the Z of the tuner isn't very well defined. *It's a pretty big calibration project in itself. Maybe, though, one could build a few test dummy loads.. say a 25 ohm and a 75 or 100 ohm, along with your vanilla 50 ohm, and the antenna switch (like an RCS-8V). *Basically, you're building a "high power resistor substitution box" You'd want some sort of nice inline watt meter (like an LP100) to make the measurements. Hello Jim, Other method is injecting a slightly off-carrier frequency signal into the amplifier (this emulates a constant small VSWR shown to the PA (wtih 50 Ohms load), but with continuous varying phase). Because of the difference in frequency, one can measure the forward (towards the PA) and reverse (reflected by PA) signal with a two channel VSA. This will give you the PA's complex output impedance. Tom (K7ITM from my head) did this with a HP89410 with couplers. With kind regards, Wim PA3DJSwww.tetech.nl Yep, 'twas me. Using the 89410, I can resolve a signal removed from the transmitter's output by a very small frequency offset, so it's well within the bandwidth of the transmitter. With synchronous averaging, the injected signal could be one or two Hz away from the transmitted carrier, though it helps a lot to have a transmitter running from a really "clean" (low phase noise) source if you're going to do that. The injected signal can be many tens of dB lower in amplitude than the transmitted signal. As I recall, I was using a signal a kHz or so away from the carrier, still well within the bandwidth of the RF power amplifier. One thing you have to be careful about is either disabling ALC, or operating well outside the ALC loop bandwidth; you don't want the ALC screwing up your results. As Wim points out, this setup presents a load to the RF power amplifier that's indistinguishable from an R+jX load that's continuously varying, tracing a path around a little circle on a Smith chart (a very tiny circle, when using signals that are very small compared with the transmitted power). The rate the circles are traced out is just the frequency offset between the transmitter and the test signal. Cheers, Tom K7ITM |
Transmitter Output Impedance
On 27 abr, 03:44, Cecil Moore wrote:
On Apr 26, 4:49*pm, Wimpie wrote: Other method is injecting a slightly off-carrier frequency signal into the amplifier (this emulates a constant small VSWR shown to the PA (wtih 50 Ohms load), but with continuous varying phase). Because of the difference in frequency, one can measure the forward (towards the PA) and reverse (reflected by PA) signal with a two channel VSA. This will give you the PA's complex output impedance. Unfortunately, the impedance encountered by the off-carrier frequency signal is probably not the same as the impedance encountered by the carrier frequency so the results don't correlate and are not very useful. The carrier frequency has interference components that the off- carrier signal doesn't encounter. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Helo Cecil, Depending on the frequency resolution of your VSA, the frequency of the injected signal can be well within 1 kHz of the carrier, so LC filters in the PA will not distort the measurement. In case of a 100W PA and injection of about 100 mW, the difference in wanted signal and signal to be rejected is 30 dB (not that large). If you don't have a VSA, you can still do it (with some more calculations) by using a diode detector, this will give you the difference frequencies directly and you can observe the phase differences on an oscilloscope. There can be a difference between a very slow load variation (for example manually changing loads and noticing current and voltage [including phase]) and the VSA method. This is because of decoupling capacitors in the power supply or bias circuits (for example RC combination in the grid circuitry to limit the grid current). If you vary the load about 10..100 times/s, bias and supply voltages don't have the time to settle to their steady state. If the VSA method is basically wrong, I would love to hear why. With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On Apr 25, 9:07*pm, Jim Lux wrote:
Sal M. Onella wrote: This group has presented members with valuable lessons in antennas and transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. *My curiosity was piqued today as I took some baby steps into EZNEC. *A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. *Doesn't seem like it. My point: *Using 75-ohm cable to improve the match at the antenna won't help me *... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. *My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. *(I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. *Thanks. "Sal" (KD6VKW) I suspect that most ham transmitters do NOT have a 50 ohm output impedance. *What they do have is a specification that they will adequately drive a 50 ohm load (and some sort of internal circuitry that detects an "unacceptable" output condition and turns down the drive). After all, your transmitter could have an output impedance of zero ohms (a "stiff" voltage source), and adequately drive your transmission line and antenna at 50 ohms (yes, this is not the optimum power transfer, but nobody ever said that ham transmitters are designed for optimum power transfer... maybe they're perfectly happy with less transfer, but still operating within their safe area) ON9CVD made some simple measurements using a couple of resistors and foudn that a TS440 has a Zout somewhere around 15-40 ohms (depending on frequency and output power).http://sharon.esrac.ele.tue.nl/~on9c...impedantie.htm Grant Bingeman also has words on this:http://www.km5kg.com/loads.htm Being aware that this was the nature of my solidstate transceiver I attempted to use a tuner with one to improve my match to my antenna system. While I didnt damage my transmitter I did notice that the best settings of the tuner for TX and RX did not coincide. I was wondering if anyone else has observed this . Jimmie |
Transmitter Output Impedance
On Apr 27, 10:30*am, Wimpie wrote:
Depending *on the frequency resolution of your VSA, the frequency of the injected signal can be well within 1 kHz of the carrier, so LC filters in the PA will not distort the measurement. *In case of a 100W PA and injection of about 100 mW, the difference in wanted signal and signal to be rejected is 30 dB (not that large). Would any competent optical physicist suggest that it is valid to study the conditions associated with interfering coherent light waves inside an interferometer by introducing an incoherent light source into the system? Why would any competent RF engineer suggest that the system source conditions associated with interfering coherent RF waves can be studied by introducing an incoherent test signal? -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On Wed, 27 Apr 2011 09:40:14 -0700 (PDT), JIMMIE
wrote: Being aware that this was the nature of my solidstate transceiver I attempted to use a tuner with one to improve my match to my antenna system. Hi Jimmie, This is somewhat cryptic, you would be expected to need a tuner. While I didnt damage my transmitter This transcends cryptic. It's like saying that as you came to a stop sign, you used your brakes and you didn't damage your car. I did notice that the best settings of the tuner for TX and RX did not coincide. I was wondering if anyone else has observed this . Two different circuits inside, hence two different loads. Two different loads, hence two different matches. Some receive paths might be 50 Ohms, but that value is not as essential as for transmission. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
Hello Cecil,
On 27 abr, 20:13, Cecil Moore wrote: On Apr 27, 10:30*am, Wimpie wrote: Depending *on the frequency resolution of your VSA, the frequency of the injected signal can be well within 1 kHz of the carrier, so LC filters in the PA will not distort the measurement. *In case of a 100W PA and injection of about 100 mW, the difference in wanted signal and signal to be rejected is 30 dB (not that large). Would any competent optical physicist suggest that it is valid to study the conditions associated with interfering coherent light waves inside an interferometer by introducing an incoherent light source into the system? Why would any competent RF engineer suggest that the system source conditions associated with interfering coherent RF waves can be studied by introducing an incoherent test signal? As this slightly off-carrier frequency signal behaves like a load with very low VSWR with a cable in between that extends with constant speed. In other words, the amplifier sees a constant VSWR, but with changing phase. Small frequency difference results in slow phase change of VSWR. Maybe you should read the postings from Tom also (K7ITM) Wim PA3DJS www.tetech.nl 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On 27 abr, 18:40, JIMMIE wrote:
On Apr 25, 9:07*pm, Jim Lux wrote: Sal M. Onella wrote: This group has presented members with valuable lessons in antennas and transmission lines, like how to measure, how to match, etc. Something I haven't seen is a discussion of the source impedance of the transmitter. *My curiosity was piqued today as I took some baby steps into EZNEC. *A particular antenna had such-and-such VSWR if fed with a 50-ohm cable and a different value if fed with a 75-ohm cable. While this is hardly news, it got me wondering whether a 75-ohm cable will load the transmitter the same. *Doesn't seem like it. My point: *Using 75-ohm cable to improve the match at the antenna won't help me *... IF ... I suffer a corresponding loss due to mismatch at the back of the radio. *My HF radios, all solid state, specify a 50 ohm load. As necessary, I routinely use an internal autotuner and either of two external manual tuners. *(I'm aware of the published 1/12 wavelength matching method.) Wisdom in any form would be appreciated. *Thanks. "Sal" (KD6VKW) I suspect that most ham transmitters do NOT have a 50 ohm output impedance. *What they do have is a specification that they will adequately drive a 50 ohm load (and some sort of internal circuitry that detects an "unacceptable" output condition and turns down the drive). After all, your transmitter could have an output impedance of zero ohms (a "stiff" voltage source), and adequately drive your transmission line and antenna at 50 ohms (yes, this is not the optimum power transfer, but nobody ever said that ham transmitters are designed for optimum power transfer... maybe they're perfectly happy with less transfer, but still operating within their safe area) ON9CVD made some simple measurements using a couple of resistors and foudn that a TS440 has a Zout somewhere around 15-40 ohms (depending on frequency and output power).http://sharon.esrac.ele.tue.nl/~on9c...impedantie.htm Grant Bingeman also has words on this:http://www.km5kg.com/loads.htm Being aware that this was the nature of my solidstate *transceiver I attempted to use a tuner with one to improve my match to my antenna system. While I didnt damage my transmitter I did notice that the best settings of the tuner for TX and RX did not coincide. I was wondering if anyone else has observed this . Jimmie Hello Jimmie, I noticed this also when I was experimenting with CB equipment and simple antenna experiments. For several CB transceivers I could get more output by slightly mismatching the load as seen by the PA (but in many cases with too much increase in current consumption). I tried to use a matcher/tuner (and later a high Q resonator) to reject other stations close by and then I figured out that when applying mismatch to the receiver, the S-meter moved further. The above isn't strange. As PA's are mostly not designed to show 50 Ohms, many receivers are also not designed to show 50 Ohms. I am not discussing wide band receivers (for example digital or analog video). Most active devices have lowest noise figure when driven from a source impedance that is far from the input impedance of the active device. If you want them to be equal, you need to use feedback and that complicates the design. Also filters with significant pass band ripple show, even when designed for 50 Ohms, significant input reflection when referenced to 50 Ohms With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On Apr 27, 1:43*pm, Wimpie wrote:
In other words, the amplifier sees a constant VSWR, but with changing phase. Small frequency difference results in slow phase change of VSWR. From the IEEE Dictionary: "impedance - (1)(A) The corresponding impedance function with p replaced by jw in which w is real. Note: Definitions (A) and (B) are equivalent. (1)(B) The ratio of the phasor equivalent of a steady- state sine wave voltage ... to the phasor equivalent of a steady-state sine wave current ... (1)(C) A physical device or combination of devices whose impedance as defined in definition (A) or (B) can be determined. Note: This sentence illustrates the double use of the word impedance ... Definition (C) is a second use of 'impedance' and is independent of definitions (A) and (B)." The pinging experiment seems to be measuring a physical impedance (1) (C) the nature of which is unclear. When the amplifier is outputting power, it seems that the source impedance would be a V/I ratio (1)(B) which doesn't respond to incoherent signals. Seems to me, you guys are pinging something other than the source impedance. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
JIMMIE wrote:
Being aware that this was the nature of my solidstate transceiver I attempted to use a tuner with one to improve my match to my antenna system. While I didnt damage my transmitter I did notice that the best settings of the tuner for TX and RX did not coincide. I was wondering if anyone else has observed this . Not surprising at all. Consider the whole black art of adjusting the input match for lowest noise figure, which may or may not coincide with the largest output signal for a given input. All those theorems about matching always have an asterisk about the assumption that they're reciprocal linear devices with constant impedances, etc. Start putting nonideal active devices in the mix, and life gets interesting. |
Transmitter Output Impedance
Cecil Moore wrote:
On Apr 27, 10:30 am, Wimpie wrote: Depending on the frequency resolution of your VSA, the frequency of the injected signal can be well within 1 kHz of the carrier, so LC filters in the PA will not distort the measurement. In case of a 100W PA and injection of about 100 mW, the difference in wanted signal and signal to be rejected is 30 dB (not that large). Would any competent optical physicist suggest that it is valid to study the conditions associated with interfering coherent light waves inside an interferometer by introducing an incoherent light source into the system? Why would any competent RF engineer suggest that the system source conditions associated with interfering coherent RF waves can be studied by introducing an incoherent test signal? Interestingly, it's NOT an incoherent test signal. It's a carefully chosen coherent test signal with a frequency difference. It's the same idea as having an interferometer with dithering piezo transducer on a mirror or a modulator in one of the paths. |
Transmitter Output Impedance
Wimpie wrote:
Hello Cecil, On 27 abr, 20:13, Cecil Moore wrote: On Apr 27, 10:30 am, Wimpie wrote: Depending on the frequency resolution of your VSA, the frequency of the injected signal can be well within 1 kHz of the carrier, so LC filters in the PA will not distort the measurement. In case of a 100W PA and injection of about 100 mW, the difference in wanted signal and signal to be rejected is 30 dB (not that large). Would any competent optical physicist suggest that it is valid to study the conditions associated with interfering coherent light waves inside an interferometer by introducing an incoherent light source into the system? Why would any competent RF engineer suggest that the system source conditions associated with interfering coherent RF waves can be studied by introducing an incoherent test signal? As this slightly off-carrier frequency signal behaves like a load with very low VSWR with a cable in between that extends with constant speed. In other words, the amplifier sees a constant VSWR, but with changing phase. Small frequency difference results in slow phase change of VSWR. to the device under test, this isn't much different than a electrically controlled line stretcher (a classic automated load pull setup... see the stuff from Maury Microwave, for instance) It's a very clever technique. A variant is used in antenna ranges where you have a probe with a mismatch in it. |
Transmitter Output Impedance
On 27 abr, 23:18, Cecil Moore wrote:
On Apr 27, 1:43*pm, Wimpie wrote: In other words, the amplifier sees a constant VSWR, but with changing phase. Small frequency difference results in slow phase change of VSWR. From the IEEE Dictionary: "impedance - (1)(A) The corresponding impedance function with p replaced by jw in which w is real. Note: Definitions (A) and (B) are equivalent. (1)(B) The ratio of the phasor equivalent of a steady- state sine wave voltage ... to the phasor equivalent of a steady-state sine wave current ... (1)(C) A physical device or combination of devices whose impedance as defined in definition (A) or (B) can be determined. Note: This sentence illustrates the double use of the word impedance ... Definition (C) is a second use of 'impedance' and is independent of definitions (A) and (B)." The pinging experiment seems to be measuring a physical impedance (1) (C) the nature of which is unclear. When the amplifier is outputting power, it seems that the source impedance would be a V/I ratio (1)(B) which doesn't respond to incoherent signals. Seems to me, you guys are pinging something other than the source impedance. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Hello Cecil, You may try to figure out how the signal injection method functions (it is a form of active load pulling). Can you agree with: it doesn't matter whether: -power reflect towards the amplifier is caused by load mismatch, or -power is sent towards the amplifier by means of a phase synchronized source. This source is phase synchronized with the PA's exciter, so we have a steady state system. We assume small load mismatch (or low injected power towards the PA) so that the operating point of the PA just changes slightly (to allow linear approximation). Now we insert a coupler between the amplifier and the load. This coupler will measure the forward voltage generated by the PA, plus the reflected part of the voltage that originates from the phase synchronised source. Depending on the phase relationship, it can be more or less then the forward voltage of the PA. If the PA shows 50 Ohms, the coupler's output would not change due to the signal injection (as no signal is reflected by the PA). We note the forward coupler's output voltage (both phase and amplitude). Now we change the phase relationship between the exciter and the source that transmits some power toward the PA. Lets change 180 degrees and keep the amplitude the same. We again note the coupler's output voltage (both phase and amplitude). The voltage that is reflected by the PA equals half the complex voltage change because of the phase change. Off course you have to correct the readings because of the coupler loss. If you know the signal that is send toward the PA, you can now calculate the complex output impedance of the PA for small load change around 50 Ohms. Instead of changing the phase of the source manually, you can do that continuously and note the couplers output continuously. If you change the phase of a signal continuously (with certain constant rad/s), the result is a decrease or increase of frequency. Assuming some reflection by the PA, the complex output from coupler rotates around a certain point. That certain point is the result of the PA's output power and the rotating vector is the result from the injected signal that is reflected by the PA (back towards the load). With a VSA you can discriminate between the voltage component from the PA itself and the reflected component (with slightly different frequency). With a normal spectrum analyser, you can only determine the magnitude of the PA's reflection coefficient (or VSWR as you like). Given the dynamic range of today's equipment, you can inject a very low level signal that may mimic load mismatch well below VSWR = 1.1. With respect to the impedance concept, we as amateurs do not use steady state signals, as they contain no information. We modulate them and are still using the impedance concept, despite the definitiones you showed. As long as the signal that is injected is well within the pass band of the PA and it sufficiently low to allow linear approximation, the concept of superposition and concept of impedance still holds. But if you feel more confident with the manual phase change, or using two known loads with known slight mismatch, I have nothing against it. But if you have a VSA, some couplers and signal source at hand, it may save lots of time. With kind regard, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
Wimpie wrote:
With a VSA you can discriminate between the voltage component from the PA itself and the reflected component (with slightly different frequency). With a normal spectrum analyser, you can only determine the magnitude of the PA's reflection coefficient (or VSWR as you like). Given the dynamic range of today's equipment, you can inject a very low level signal that may mimic load mismatch well below VSWR = 1.1. One needs to have an analyzer with a narrow band detector for this, though. Inexpensive analyzers like the TAPR VNA have untuned detectors, so the PA's main signal will screw things up. The N2PK uses a form of direct conversion detector, so your test signal would have to be far enough away from the main signal so that the LPF in the detector filters it out. The original N2PK design uses, I think, a 100 Hz filter, and the adc samples at 15.36 kHz with a digital filter. The overall BW is something like 5 Hz, so putting your test signal 1kHz away would probably work. |
Transmitter Output Impedance
On 4/26/2011 7:52 PM, Richard Clark wrote:
On Tue, 26 Apr 2011 15:27:28 -0700, Jim wrote: (If I had a very efficient op amp, I could simulate any arbitrary output impedance, without dissipating any power in the source) I can see why this is parenthetical, because it covers a lot of sins of omission. yep.. not possible to build such a thing, anymore than one can build a zero ohm output impedance RF source with any signficant power. Suggested more as an example that the power dissipation in the source doesn't necessarily correlate with match, load Z, or anything else in general. (You can get pretty darn close at powers less than a watt and HF, though..) OP AMPs are a constant of my admiration in the possibilities offered. That and the signal processing you suggest (plus digital oscillators) "could" change the playing field - if conventional design weren't so universally fallen back upon. 73's Richard Clark, KB7QHC I wonder if this should be applied to these sorts of projects - https://www.kb9yig.com/ I have a 1 watt kit that I haven't put together yet. Several others in our club have and they love them. Especially nice for CW. tom K0TAR |
Transmitter Output Impedance
On 4/27/2011 6:33 PM, Wimpie wrote:
it doesn't matter whether: -power reflect towards the amplifier is caused by load mismatch, or -power is sent towards the amplifier by means of a phase synchronized source. This source is phase synchronized with the PA's exciter, so we have a steady state system. We assume small load mismatch (or low injected power towards the PA) so that the operating point of the PA just changes slightly (to allow linear approximation). Thanks Wimpie. It sometimes takes a bit to get me to realize this type of thing isn't hard, it's actually simple. My old brain tries to obfuscate things from itself sometimes. It Calc 100. Make the difference small. tom K0TAR |
Transmitter Output Impedance
On Apr 27, 6:33*pm, Wimpie wrote:
With respect to the impedance concept, we as amateurs do not use steady state signals, as they contain no information. We modulate them and are still using the impedance concept, despite the definitiones you showed. Trouble is, the impedance in IEEE definition (1)(B) doesn't *cause* reflections. If the actual source impedance matches IEEE definition (1) (B), the presumption that source impedance will *cause* a reflection is invalid. Walter Maxwell argues that the actual source impedance of an RF amplifier is in reality a V/I ratio, i.e. it agrees with IEEE definition (1)(B). -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On Apr 27, 2:13*pm, Cecil Moore wrote:
On Apr 27, 10:30*am, Wimpie wrote: Depending *on the frequency resolution of your VSA, the frequency of the injected signal can be well within 1 kHz of the carrier, so LC filters in the PA will not distort the measurement. *In case of a 100W PA and injection of about 100 mW, the difference in wanted signal and signal to be rejected is 30 dB (not that large). Would any competent optical physicist suggest that it is valid to study the conditions associated with interfering coherent light waves inside an interferometer by introducing an incoherent light source into the system? Why would any competent RF engineer suggest that the system source conditions associated with interfering coherent RF waves can be studied by introducing an incoherent test signal? -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Cecil suggested reading Chapter 19A in Reflections to view the results of my extensive measurements of the output resistance (impedance) of RF power amps, but except for Jim and Richard, it appears that the others have not. Actually, Chapter19A is an addition to Chapter 19, which when taken completely will provide some information that will hopefully change some minds concerning the maximum power delivered. It should be understood that 'maximum' power delivered is that power delivered with a specified level of drive. For example, if the drive level is set to deliver a maximum of 100w, and the pi-network is adjusted to deliver that maximum power into its load, the source resistance (impedance) will be the (complex) conjugate of the load impedance. We're not talking here about the very maximum power that the amp can deliver, with max drive, max plate current, etc. If you review the 19A portion of you will see beyond a doubt that the conjugate match exists between the output of the pi-network and its complex load impedance, and that the maximum power delivered at the drive level that allows only 100w to be delivered as the maximum. Further review of all the data presented there will also show that the output resistance of the amp is non-dissipative, while the dissipative resistance is that between the cathode and plate. The reason the efficiency of the amps can exceed 50 percent is because the cathode to- plate resistance is less than the non-dissipative output resistance, where that R = E/I appearing at the output of the pi-network. Walt, W2DU I hope the review of my measured data will clear up some of the confusion concerning the output resistance (impedance) of the RF power amp. Walt, W2DU |
Transmitter Output Impedance
On Apr 27, 2:13*pm, Cecil Moore wrote:
On Apr 27, 10:30*am, Wimpie wrote: Depending *on the frequency resolution of your VSA, the frequency of the injected signal can be well within 1 kHz of the carrier, so LC filters in the PA will not distort the measurement. *In case of a 100W PA and injection of about 100 mW, the difference in wanted signal and signal to be rejected is 30 dB (not that large). Would any competent optical physicist suggest that it is valid to study the conditions associated with interfering coherent light waves inside an interferometer by introducing an incoherent light source into the system? Why would any competent RF engineer suggest that the system source conditions associated with interfering coherent RF waves can be studied by introducing an incoherent test signal? -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Cecil suggested reading Chapter 19A in Reflections to view the results of my extensive measurements of the output resistance (impedance) of RF power amps, but except for Jim and Richard, it appears that the others have not. Actually, Chapter19A is an addition to Chapter 19, which when taken completely will provide some information that will hopefully change some minds concerning the maximum power delivered. It should be understood that 'maximum' power delivered is that power delivered with a specified level of drive. For example, if the drive level is set to deliver a maximum of 100w, and the pi-network is adjusted to deliver that maximum power into its load, the source resistance (impedance) will be the (complex) conjugate of the load impedance. We're not talking here about the very maximum power that the amp can deliver, with max drive, max plate current, etc. If you review the 19A portion of you will see beyond a doubt that the conjugate match exists between the output of the pi-network and its complex load impedance, and that the maximum power delivered at the drive level that allows only 100w to be delivered as the maximum. Further review of all the data presented there will also show that the output resistance of the amp is non-dissipative, while the dissipative resistance is that between the cathode and plate. The reason the efficiency of the amps can exceed 50 percent is because the cathode to- plate resistance is less than the non-dissipative output resistance, where that R = E/I appearing at the output of the pi-network. The earlier portion of Chapter 19, that appears in Reflections 2, can be downloaded from my web page at www.w2du.com, click on 'Read Chapters from Reflections 2', and select Chapter 19. I hope the review of my measured data will clear up some of the confusion concerning the output resistance (impedance) of the RF power amp. Walt, W2DU |
Transmitter Output Impedance
On 28 abr, 14:39, Cecil Moore wrote:
On Apr 27, 6:33*pm, Wimpie wrote: With respect to the impedance concept, we as amateurs do not use steady state signals, as they contain no information. We modulate them and are still using the impedance concept, despite the definitiones you showed. Trouble is, the impedance in IEEE definition (1)(B) doesn't *cause* reflections. If the actual source impedance matches IEEE definition (1) (B), the presumption that source impedance will *cause* a reflection is invalid. Are you familiar with the concept of S-parameters where you determine impedance by measuring of reflection coefficient? Walter Maxwell argues that the actual source impedance of an RF amplifier is in reality a V/I ratio, i.e. it agrees with IEEE definition (1)(B). -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Wim |
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