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what happens to reflected energy ?
Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated. - 50 w are "reflected", what is their fate ? Are they definitely lost for radiation and just heat the line, the final.... JC |
what happens to reflected energy ?
On 6 jun, 16:22, "JC" wrote:
Basic question (at least for me) for a very poor antenna matching : -100 w reach the antenna and 50 w are radiated. - 50 w are "reflected", what is their fate ? Are they definitely lost for radiation and just heat the line, the final.... JC Hello, If the source has a linear 50 Ohms output impedance (assuming 50 Ohm cable, no loss), all power will go back into the source. Partly in the form of heat, partly in the form of power saving. However a PA is not a 50 Ohms source, so what happens may vary. It is very likely that some reverse power reflects back to the antenna (as the transmitter is not a source with 50 Ohms output impedance). So the reading on an instrument that measures true forward power is the sum of the real incident power plus the reflected power from the transmitter towards the antenna. Changing the load may result in a reduction or increase of total forward power. In case of a real 50 Ohms source, the forward power will not change, no matter the load. The bad match seen from the transmitter may result in a reduction of DC input power (input current reduces), but may also result in an increase of DC input power. So when you have reverse power reading of 50W, you cannot just say that the PA has to dissipate an extra 50W, it can be more, less and even negative. The negative case is when the input power reduces significantly and the active element has to dissipate less (with respect to the best match condition). In particular high efficiency amplifiers (where the active devices are used as switches), show strong variation in supply current versus load change. Under certain load conditions such amplifiers may show a strong decrease in efficiency resulting in high relative increase of device dissipation. Best regards, Wim PA3DJS www.tetech.nl without abc, PM will reach me |
what happens to reflected energy ?
On Jun 6, 9:22*am, "JC" wrote:
Basic question (at least for me) for a very poor antenna matching : -100 w reach the antenna and 50 w are radiated. - 50 w are "reflected", what is their fate ? Are they definitely lost for radiation and just heat the line, the final..... What happens to the 50 joules/second of reflected energy depends upon the phasing between the source wave and the reflected wave at the source impedance. What most amateurs don't understand is there are two mechanisms that can redistribute reflected energy back toward the antenna. Those mechanisms are a re-reflection based on the physical reflection coefficient (what RF engineers understand) and wave interaction resulting in constructive/destructive interference (what most RF engineers don't seem to understand) because, unlike optical physicists, have not been forced to follow the energy flow. If the reflected wave arrives 180 degrees out of phase with the source wave, the two waves undergo destructive interference and all of the reflected power is redistributed as constructive interference energy back toward the antenna. This is what happens at the Z0-match established at the input of an antenna tuner. Thus your conditions of 100w forward power and 50w reflected power could be accomplished with a 50w matched source. If the reflected wave arrives 90 degrees out of phase with the source wave, there is zero interference and the reflected power is dissipated in the source resistor (in a source with a source resistor). If the reflected wave arrives in phase with the source wave, all of the reflected power and more than 1/2 of the source power can be dissipated in the source resistor. Such knowledge is old hat for optical physicists who don't have the luxury of measuring voltages in light waves. They rely on a power density (irradiance) equation. Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A) where A is the phase angle between the electric fields of two waves. The last term is called the "interference term" and that value is what most amateurs are missing in their energy analysis. If the sign of the interference term is negative, the interference is destructive. If the sign of the interference term is positive, the interference is constructive. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie
wrote: However a PA is not a 50 Ohms source Hi Wimpie, You say this like others, with the air of "knowing." However, when I ask in response of those who "know" what the PA is NOT, what IS it? Give me the Z value of your transmitter. Specify all initial conditions. We have had lengthy correspondence with Walt Maxwell's very rigorously measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly that as is practicable (say +/- 20%); and yet your voice was missing from this discussion with evidence to the contrary. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On Jun 6, 3:13*pm, Cecil Moore wrote:
On Jun 6, 9:22*am, "JC" wrote: Basic question (at least for me) for a very poor antenna matching : -100 w reach the antenna and 50 w are radiated. - 50 w are "reflected", what is their fate ? Are they definitely lost for radiation and just heat the line, the final.... What happens to the 50 joules/second of reflected energy depends upon the phasing between the source wave and the reflected wave at the source impedance. *What most amateurs don't understand is there are two mechanisms that can redistribute reflected energy back toward the antenna. Those mechanisms are a re-reflection based on the physical reflection coefficient (what RF engineers understand) and wave interaction resulting in constructive/destructive interference (what most RF engineers don't seem to understand) because, unlike optical physicists, have not been forced to follow the energy flow. If the reflected wave arrives 180 degrees out of phase with the source wave, the two waves undergo destructive interference and all of the reflected power is redistributed as constructive interference energy back toward the antenna. This is what happens at the Z0-match established at the input of an antenna tuner. Thus your conditions of 100w forward power and 50w reflected power could be accomplished with a 50w matched source. If the reflected wave arrives 90 degrees out of phase with the source wave, there is zero interference and the reflected power is dissipated in the source resistor (in a source with a source resistor). If the reflected wave arrives in phase with the source wave, all of the reflected power and more than 1/2 of the source power can be dissipated in the source resistor. Such knowledge is old hat for optical physicists who don't have the luxury of measuring voltages in light waves. They rely on a power density (irradiance) equation. Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A) where A is the phase angle between the electric fields of two waves. The last term is called the "interference term" and that value is what most amateurs are missing in their energy analysis. If the sign of the interference term is negative, the interference is destructive. If the sign of the interference term is positive, the interference is constructive. -- 73, Cecil, w5dxp.com as in the other thread, what is the mechanism of that 'interaction' between waves? i contend there can be no 'interaction' between forward and reflected waves if the device is linear. so in an ideal case of a voltage or current source and ideal source resistance there is no interaction, it is reflected by and/or absorbed in the source depending on the impedance of the line and source. if the source is not linear then you would have to calculate the effect of the sum of the voltages or currents at the source to determine the effect. |
what happens to reflected energy ?
On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction' between waves? i contend there can be no 'interaction' between forward and reflected waves if the device is linear. so in an ideal case of a voltage or current source and ideal source resistance there is no interaction, it is reflected by and/or absorbed in the source depending on the impedance of the line and source. There is no mixing as in multiplication of waveforms. Perhaps I can offer a simple analogy. Instead of two AC waveforms (forward and reflected), use a DC equivalent. Start with two DIFFERENT batteries. Connect the two negative ends together and declare that to be ground. Connect a resistor between the positive terminals. The two voltages most certainly "interact" across the resistor, resulting in the current and power being proportional to the difference between the two battery voltages. Nothing in this crude example is non-linear, so there's no need for mixing in order to get interaction. Similarly, the coax cable acts much in the same way. The two batteries are replaced by the incident and reflected signals. At any time, or position on the transmission line, the model can be frozen and the instantaneous voltages and currents be calculated. if the source is not linear then you would have to calculate the effect of the sum of the voltages or currents at the source to determine the effect. If the source (or load) is non-linear, then the waveforms seen on the transmission line will be distorted. This is unlikely because we usually don't install diodes in antennas, or build HF amplifiers with substantial non-linearities (i.e. distortion). -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
what happens to reflected energy ?
On Jun 6, 8:55*pm, Jeff Liebermann wrote:
On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote: as in the other thread, what is the mechanism of that 'interaction' between waves? *i contend there can be no 'interaction' between forward and reflected waves if the device is linear. *so in an ideal case of a voltage or current source and ideal source resistance there is no interaction, it is reflected by and/or absorbed in the source depending on the impedance of the line and source. * There is no mixing as in multiplication of waveforms. *Perhaps I can offer a simple analogy. *Instead of two AC waveforms (forward and reflected), use a DC equivalent. *Start with two DIFFERENT batteries. Connect the two negative ends together and declare that to be ground. Connect a resistor between the positive terminals. *The two voltages most certainly "interact" across the resistor, resulting in the current and power being proportional to the difference between the two battery voltages. *Nothing in this crude example is non-linear, so there's no need for mixing in order to get interaction. it is very misleading to try to make a lumped circuit analogy out of a transmission line problem. the first step in any circuits 101 course would be to simplify the two batteries into one then solve for the single simple current across the resistor... no waves, no back and forth, no interaction between batteries, just a single current and voltage. Similarly, the coax cable acts much in the same way. *The two batteries are replaced by the incident and reflected signals. *At any time, or position on the transmission line, the model can be frozen and the instantaneous voltages and currents be calculated. * they can be, but what does it prove? i want to freeze it just as the first wave gets to the load, there is no reflected wave yet, so what good is that? in order to use our common equations there are many unstated but necessary assumptions. the most restrictive of which is that we normally only solve for the sinusoidal steady state waves... this requires that a long time(in terms of the length of the line) has passed since the source was energized, that it is a single frequency pure sine wave, and that nothing is changing in time in the load or line characteristics. given all that there is no need for instantaneous values, sure they can be calculated or measured, but they are of no use in most cases. if the source is not linear then you would have to calculate the effect of the sum of the voltages or currents at the source to determine the effect. If the source (or load) is non-linear, then the waveforms seen on the transmission line will be distorted. *This is unlikely because we usually don't install diodes in antennas, or build HF amplifiers with substantial non-linearities (i.e. distortion). and obviously if either one is non-linear then all the simple equations can be abandoned and a more complete analysis must be done. -- Jeff Liebermann * * 150 Felker St #D * *http://www.LearnByDestroying.com Santa Cruz CA 95060http://802.11junk.com Skype: JeffLiebermann * * AE6KS * *831-336-2558 |
what happens to reflected energy ?
On Jun 6, 10:22*am, "JC" wrote:
Basic question (at least for me) for a very poor antenna matching : -100 w reach the antenna and 50 w are radiated. - 50 w are "reflected", what is their fate ? Are they definitely lost for radiation and just heat the line, the final..... JC I assume that your numbers were obtained using a directional wattmeter, that is, forward power is 100W, 50W are reflected and by subtraction, it is computed that 50W enter the antenna, some of which is dissipated in the antenna losses and some of which is radiated. The 100W forward and 50W reflected have no relation to actual powers but are simply values that are constructed so that when they are subtracted the result is the average power flowing towards the load. From these numbers alone it is impossible to decide whether the system is operating as it should or to compute where the losses might be, or whether there are any losses at all. With an appropriate antenna tuner it is entirely possible that the transmitter is delivering its design output of 50W to the line and all of this energy is reaching the antenna and being radiated. Thus the whole question of "where does the reflected power go?", is rather misleading. ....Keith |
what happens to reflected energy ?
On Jun 6, 4:55*pm, Jeff Liebermann wrote:
On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote: as in the other thread, what is the mechanism of that 'interaction' between waves? *i contend there can be no 'interaction' between forward and reflected waves if the device is linear. *so in an ideal case of a voltage or current source and ideal source resistance there is no interaction, it is reflected by and/or absorbed in the source depending on the impedance of the line and source. * There is no mixing as in multiplication of waveforms. *Perhaps I can offer a simple analogy. *Instead of two AC waveforms (forward and reflected), use a DC equivalent. *Start with two DIFFERENT batteries. Connect the two negative ends together and declare that to be ground. Connect a resistor between the positive terminals. *The two voltages most certainly "interact" across the resistor, resulting in the current and power being proportional to the difference between the two battery voltages. *Nothing in this crude example is non-linear, so there's no need for mixing in order to get interaction. Similarly, the coax cable acts much in the same way. *The two batteries are replaced by the incident and reflected signals. *At any time, or position on the transmission line, the model can be frozen and the instantaneous voltages and currents be calculated. * if the source is not linear then you would have to calculate the effect of the sum of the voltages or currents at the source to determine the effect. If the source (or load) is non-linear, then the waveforms seen on the transmission line will be distorted. *This is unlikely because we usually don't install diodes in antennas, or build HF amplifiers with substantial non-linearities (i.e. distortion). -- Jeff Liebermann * * 150 Felker St #D * *http://www.LearnByDestroying.com Santa Cruz CA 95060http://802.11junk.com Skype: JeffLiebermann * * AE6KS * *831-336-2558 Looks like I should step in here, as the answer to this question is the main theme in the book Reflections--Transmission Lines and Antennas, the first edition published in 1990, the second in 2001, and the third in just this past month of May, released at Dayton. The notion that ANY reflected power enters the source, such as an RF power amp using tubes and a pi-network, is FALSE!!! The output source resistance of these amps is non-dissipative, and totally re-reflects all reflected power from a mismatched antenna. The same is true when using an antenna tuner. When correctly adjusted the antenna tuner totally reflects all reflected power, resulting in a conjugate match at the antenna-coax mismatch, canceling all reactances in the system to zero, thus tuning the non-resonant antenna to resonance. This action if fundamental, and has been a misunderstood myth for centuries. For proof of the above statements I invite you to read Chapter 23 of Reflections, which you can find on my web page at www.w2du.com. Click on 'Read Chapters from Reflections 2' and then click on Chapter 23. In addition, Chapter 19 gives more insight, and the addition to Chapter 19 can be found by clicking on 'Preview Chapters from Reflections 3'. The addition shows measured data proving that the output source impedance of the RF amp is the conjugate of the complex load impedance when the pi-network is adjusted to deliver all the available power at a given level of grid drive. Furthermore, a completely revised edition of Chapter 23 and the total Chapter 19 appear in Reflections 3, which is now available from CQ Magazine. Walt Maxwell, W2DU |
what happens to reflected energy ?
On Jun 6, 5:27*pm, walt wrote:
On Jun 6, 4:55*pm, Jeff Liebermann wrote: On Sun, 6 Jun 2010 11:01:20 -0700 (PDT), K1TTT wrote: as in the other thread, what is the mechanism of that 'interaction' between waves? *i contend there can be no 'interaction' between forward and reflected waves if the device is linear. *so in an ideal case of a voltage or current source and ideal source resistance there is no interaction, it is reflected by and/or absorbed in the source depending on the impedance of the line and source. * There is no mixing as in multiplication of waveforms. *Perhaps I can offer a simple analogy. *Instead of two AC waveforms (forward and reflected), use a DC equivalent. *Start with two DIFFERENT batteries. Connect the two negative ends together and declare that to be ground. Connect a resistor between the positive terminals. *The two voltages most certainly "interact" across the resistor, resulting in the current and power being proportional to the difference between the two battery voltages. *Nothing in this crude example is non-linear, so there's no need for mixing in order to get interaction. Similarly, the coax cable acts much in the same way. *The two batteries are replaced by the incident and reflected signals. *At any time, or position on the transmission line, the model can be frozen and the instantaneous voltages and currents be calculated. * if the source is not linear then you would have to calculate the effect of the sum of the voltages or currents at the source to determine the effect. If the source (or load) is non-linear, then the waveforms seen on the transmission line will be distorted. *This is unlikely because we usually don't install diodes in antennas, or build HF amplifiers with substantial non-linearities (i.e. distortion). -- Jeff Liebermann * * 150 Felker St #D * *http://www.LearnByDestroying.com Santa Cruz CA 95060http://802.11junk.com Skype: JeffLiebermann * * AE6KS * *831-336-2558 Looks like I should step in here, as the answer to this question is the main theme in the book Reflections--Transmission Lines and Antennas, the first edition published in 1990, the second in 2001, and the third in just this past month of May, released at Dayton. The notion that ANY reflected power enters the source, such as an RF power amp using tubes and a pi-network, is FALSE!!! The output source resistance of these amps is non-dissipative, and totally re-reflects all reflected power from a mismatched antenna. The same is true when using an antenna tuner. When correctly adjusted the antenna tuner totally reflects all reflected power, resulting in a conjugate match at the antenna-coax mismatch, canceling all reactances in the system to zero, thus tuning the non-resonant antenna to resonance. This action if fundamental, and has been a misunderstood myth for centuries. For proof of the above statements I invite you to read Chapter 23 of Reflections, which you can find on my web page atwww.w2du.com. Click on 'Read Chapters from Reflections 2' and then click on Chapter 23. In addition, Chapter 19 gives more insight, and the addition to Chapter 19 can be found by clicking on 'Preview Chapters from Reflections 3'. The addition shows measured data proving that the output source impedance of the RF amp is the conjugate of the complex load impedance when the pi-network is adjusted to deliver all the available power at a given level of grid drive. Furthermore, a completely revised edition of Chapter 23 and the total Chapter 19 appear in Reflections 3, which is now available from CQ Magazine. Walt Maxwell, W2DU Forgot to mention that the output of the RF power amp is LINEAR, even though the input is non-linear. The reason is that the the pi-network tank circuit is not only an impedance transformer, it's an energy- storage device that isolates the output from the input. The linearity of the output is indicated by the sinusoidal shape of the output wave, and that the voltage and current are in phase when the load impedance is resistive. Walt, W2DU |
what happens to reflected energy ?
On 6 jun, 19:00, Richard Clark wrote:
On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie wrote: However a PA is not a 50 Ohms source Hi Wimpie, You say this like others, with the air of "knowing." *However, when I ask in response of those who "know" what the PA is NOT, what IS it? Give me the Z value of your transmitter. *Specify all initial conditions. We have had lengthy correspondence with Walt Maxwell's very rigorously measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly that as is practicable (say +/- 20%); and yet your voice was missing from this discussion with evidence to the contrary. 73's Richard Clark, KB7QHC Hello Richard, Try the following experiment: Measure the forward power of your PA at a convenient load. Use a directional coupler for that, not a voltage meter calibrated for power. The "SET" or "CAL" position of a VSWR meter can be used as forward power indicator. Now make some mismatch (for example VSWR=2 at different phase) and read the forward power. Did it change? If so, the output impedance is no (longer) 50 Ohms. If possible, disable automatic protection to avoid changing drive level. I don't know the value for my FT7B, but I know forward power changes with load variations (as I use it as "measuring instrument" sometimes). Virtually all power amplifiers I designed do not have a large signal output impedance of 50 ohms under significant load change. For some I measured it because of a discussion on this between colleges. Measuring method used: change in resistive load, from voltage change you can calculate the current change, hence the output impedance. One note, except two, all where solid state. All high efficiency designs (class E, D) that I did have output impedance far from the expected load impedance. With "far" I mean factor 2 or factor 0.5. I did not measure that (as it is not important in virtually all cases), but know it from the overload simulation/measurement and I did the design myself. The reason for not being 50 Ohms (after matching) is that when you change the load, the active device will go into voltage or current saturation. This is not a hard process, so for small load variation (low VSWR values), forward power will not change much. I think this is especially true for vacuum triode PA where you have significant tube- internal feedback. For large variation (for example VSWR = 2.5, reflected power 18%), you will notice change in forward power for most power amplifiers. High efficiency CW amplifiers use saturated switches (for example half, full bridge or push-pull), so behave (seen at the active device) as a voltage source. Depending on the total phase shift of the filter sections this may convert to a current source behavior (seen at the output). I had to spent much time to avoid destruction of some circuits in case of mismatch. If you can drop me a link to the discussion of the TS830s, I would appreciate that. Best regards, Wim PA3DJS www.tetech.nl Remove abc first in case of PM. |
what happens to reflected energy ?
On 6 jun, 19:00, Richard Clark wrote:
On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie wrote: However a PA is not a 50 Ohms source Hi Wimpie, You say this like others, with the air of "knowing." *However, when I ask in response of those who "know" what the PA is NOT, what IS it? Give me the Z value of your transmitter. *Specify all initial conditions. We have had lengthy correspondence with Walt Maxwell's very rigorously measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly that as is practicable (say +/- 20%); and yet your voice was missing from this discussion with evidence to the contrary. 73's Richard Clark, KB7QHC Hello Richard, Walt did respond and did a solid statement regarding the amplifier with matching section to obtain maximum output (so I know the conditions) This is the point where the tube/transistor is at the edge of current/voltage saturation. At that operating point the output impedance is 50 Ohms (for small load variations). When you change the load significantly (or change drive level), current or voltage saturation will dominate, hence the output impedance is no longer 50 Ohms. This is also the reason that PA intermodulation may occur in close spaced transmitters where some power from transmitter A enters the amplifier of transmitter B and vice versa. This also proves that there is no linear 50 Ohms output impedance. Best regards, Wim PA3DJS www.tetech.nl |
what happens to reflected energy ?
On Jun 6, 6:36*pm, Wimpie wrote:
On 6 jun, 19:00, Richard Clark wrote: On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie wrote: However a PA is not a 50 Ohms source Hi Wimpie, You say this like others, with the air of "knowing." *However, when I ask in response of those who "know" what the PA is NOT, what IS it? Give me the Z value of your transmitter. *Specify all initial conditions. We have had lengthy correspondence with Walt Maxwell's very rigorously measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly that as is practicable (say +/- 20%); and yet your voice was missing from this discussion with evidence to the contrary. 73's Richard Clark, KB7QHC Hello Richard, Walt did respond and did a solid statement regarding the amplifier with matching section to obtain maximum output (so I know the conditions) *This is the point where the tube/transistor is at the edge of current/voltage saturation. At that operating point the output impedance is 50 Ohms (for small load variations). When you change the load significantly (or change drive level), current or voltage saturation will dominate, hence the output impedance is no longer 50 Ohms. * This is also the reason that PA intermodulation may occur in close spaced transmitters where some power from transmitter A enters the amplifier of transmitter B and vice versa. This also proves that there is no linear 50 Ohms output impedance. Best regards, Wim PA3DJSwww.tetech.nl Sorry Wim, I can't agree with some of your statements in your last post. Concerning maximum output, I will agree that saturation will occur when the minimum of the peak AC plate voltage equals the peak AC grid voltage. This condition occurs when the tube is delivering its total maximum possible power. However, when the grid drive level is less than that which brings the plate-voltage minimum down to the grid- voltage level, saturation does not occur. In addition, when the pi-network has been adjusted to deliver all the available power at some drive level less than the maximum possible power, the source resistance at the output of the pi-network will be exactly equal to its load resistance, not somewhat higher or lower. This follows from the Maximum Power Transfer Theorem. This is not speculation, but proof determined by data from many, many measurements. Walt, W2DU |
what happens to reflected energy ?
On Sun, 6 Jun 2010 15:21:59 -0700 (PDT), Wimpie
wrote: Try the following experiment: -sigh- Hi Wimpie, This is not what I asked for. You said what the source was NOT, but you cannot say what the source IS. The source is not a bipedal reptile. The source is not an elongated diphthong. The source is not the resurrection of a deity (some may argue that more than I would care to follow). The source is not.... There are many ways to say what the source is NOT, and that will never inform us about the source. I see many draw deuces to this question and try to convince everyone that the card is a pair of winning aces. Now make some mismatch (for example VSWR=2 at different phase) and read the forward power. Did it change? If so, the output impedance is no (longer) 50 Ohms. What IS it now? If you could measure it once, you should be able to tell us what it is this time too. I did this for years to methods set by the National Bureau of Standards. You have drawn a deuce, not two aces. Measuring method used: change in resistive load, from voltage change you can calculate the current change, hence the output impedance. One note, except two, all where solid state. Yes, that is called a load pull, but you offer no data - another deuce. I have done a lot of load pulls. All high efficiency designs (class E, D) that I did have output impedance far from the expected load impedance. With "far" I mean factor 2 or factor 0.5. I did not measure that (as it is not important in virtually all cases), but know it from the overload simulation/measurement and I did the design myself. You didn't measure it, but you can state the value - interesting state of guessing. So, what did the simulation/non-measurement give you as a value? What IS the value? Another deuce. The reason for not being 50 Ohms Reasons abound. BP is giving us reasons why the Gulf Coast shouldn't worry. Data has proven that reasons don't work and neither does their well. That is a Joker draw. High efficiency CW amplifiers The TS830S is not such an animal. Another Deuce. If you can drop me a link to the discussion of the TS830s, I would appreciate that. Google "Plate Resistance" in this group. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On 7 jun, 01:51, walt wrote:
On Jun 6, 6:36*pm, Wimpie wrote: On 6 jun, 19:00, Richard Clark wrote: On Sun, 6 Jun 2010 07:42:31 -0700 (PDT), Wimpie wrote: However a PA is not a 50 Ohms source Hi Wimpie, You say this like others, with the air of "knowing." *However, when I ask in response of those who "know" what the PA is NOT, what IS it? Give me the Z value of your transmitter. *Specify all initial conditions. We have had lengthy correspondence with Walt Maxwell's very rigorously measured Kenwood TS830s that demonstrates a Z of 50 Ohms, or nearly that as is practicable (say +/- 20%); and yet your voice was missing from this discussion with evidence to the contrary. 73's Richard Clark, KB7QHC Hello Richard, Walt did respond and did a solid statement regarding the amplifier with matching section to obtain maximum output (so I know the conditions) *This is the point where the tube/transistor is at the edge of current/voltage saturation. At that operating point the output impedance is 50 Ohms (for small load variations). When you change the load significantly (or change drive level), current or voltage saturation will dominate, hence the output impedance is no longer 50 Ohms. * This is also the reason that PA intermodulation may occur in close spaced transmitters where some power from transmitter A enters the amplifier of transmitter B and vice versa. This also proves that there is no linear 50 Ohms output impedance. Best regards, Wim PA3DJSwww.tetech.nl Hello Walt, I think the problem is in the initial conditions that we use as starting point. Your starting point is matched to maximum power output in all cases, also under bad antenna VSWR. My starting point was a fixed tuned amplifier, made for 50 Ohms and experiencing large mismatch (50W reflected, 100W forward). Sorry *Wim, I can't agree with some of your statements in your last post. Concerning maximum output, I will agree that saturation will occur when the minimum of the peak AC plate voltage equals the peak AC grid voltage. This condition occurs when the tube is delivering its total maximum possible power. However, when the grid drive level is less than that which brings the plate-voltage minimum down to the grid- voltage level, saturation does not occur. In that case the final tube will more or less behave like a current source (LF appraoch, there will be feedback via capacitance), so the impedance will change from the value for maximum power. There are two saturation conditions: voltage saturation, this lowers the plate impedance, current saturation (below maximum voltage swing), this increases the plate impedance (especially for pentodes). We start with a maximim power matched situation (so ZL = Zout) and don't touch the tuning. When you now change the load (for example VSWR =2), you will probably have a strong voltage saturation condition or strong current saturation condition, so voltage source or current source behavior dictates, but not both. Therefore the output impedance at the plate will rise (current saturation) or decrease (voltage saturation). For triodes the effect is less then for pentodes as the transition for current saturation to voltage saturation is relatively smooth for triodes. In addition, when the pi-network has been adjusted to deliver all the available power at some drive level less than the maximum possible power, the source resistance at the output of the pi-network will be exactly equal to its load resistance, not somewhat higher or lower. This follows from the Maximum Power Transfer Theorem. This is not speculation, but proof determined by data from many, many measurements. Fully agree with this, but is this always the case under practical circumstance? When you change one of the paramers (drive level or VSWR, but not the matching), you will divert from the optimum setting, hence impedance will change. You can try the VSWR dependence by measuring the forward power under varying VSWR. I am sure it will change (keeping drive level and pi filter matching the same). Most solid state amplifiers do not have the possibility of matching when you change VSWR, so with bad VSWR, you will be in the current or voltage saturating range of the active device and your output impedance will change. Virtually all high efficient amplifiers work in voltage saturated mode and are therefore not operated at maximum available power, therefore their output impedance doens't match the expected load. Except for CW, most SSB amplifiers run in the current saturated mode (so below maximum voltage swing), so they do not work at their maximum power point given the instantaneous drive level (or you need to modulate your supply also). In real world the situation is more complex because of phase changes in VSWR and parasitic feedback that may change the gain of the active device (worst case spoken, you may get instability). I discovered (when I was younger) that with a class C fixed tuned amplifier, small changes in VSWR resulted in change of forward power, but significant change in DC supply current (and heatsink temperature). So what happens with the reflected power depends on many factors. Walt, W2DU Wim PA3DJS www.tetech.nl |
what happens to reflected energy ?
On Jun 6, 3:21*pm, Wimpie wrote:
Measuring method used: change in resistive load, from voltage change you can calculate the current change, hence the output impedance. One note, except two, all where solid state. Of course, since the source impedance may be anywhere in the complex plane, you need to change more than just the resistive part of the load, I believe, to get an accurate picture... All high efficiency designs (class E, D) that I did have output impedance far from the expected load impedance. With "far" I mean factor 2 or factor 0.5. I did not measure that (as it is not important in virtually all cases), but know it from the overload simulation/measurement and I did the design myself. What I've seen in similar situations is that the source impedance is likely to be strongly reactive, depending on the filter network you use to get sinusoidal output. In any event, the source impedance is likely to have a reflection coefficient magnitude that is quite close to unity. That is exactly what you should expect: there's nothing to absorb reflections. You could (theoretically at least) use feedback to make the output look like 50 ohms, but just as you say, Wim ... why?? There's really almost never any point in doing so. .... Cheers, Tom |
what happens to reflected energy ?
On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote: The 100W forward and 50W reflected have no relation to actual powers From HP Journal V.7n.2: If an incident wave is applied at the left as shown in the diagram, the wave passes down the main arm. In the region where the lines are cou pled, a wave 20 db below the inci dent wave will be coupled to the "Forward" terminal, while a second wave 20 db below the incident wave will be coupled to the resistive ter mination in the "Reverse" arm. Since the combined power in the two split-off waves amounts to only 2% of the power in the main wave, the main wave is essentially unalt ered and continues to the right-hand terminal. A wave applied at the right end of the coupler is coupled in an analo gous manner. Waves 20 db below the left-traveling wave will be coupled to the "Reverse" terminal and to the resistive termination in the "For ward" arm, while the main part of the wave continues to the left-hand terminal. The couplers thus provide equal fractions of right-traveling and left traveling waves at separate termin als. The ratio of these waves will be equal to p *, the magnitude of the reflection coefficient of any device connected to the output of the cou pler. This ratio can be measured (Fig. 4) by applying the outputs of the "Forward" and "Reverse" ter minals to the -hp- 416A Ratio Meter, using suitable detectors to demodu late the amplitude-modulated power which must be used with the system. In an ideal directional coupler, no power from a forward wave would be received at the reverse terminal and no power from a reverse wave would be received at the forward ter minal. In practice some undesired power is received at these terminals, although it has been possible to de sign the couplers so that this unde sired power is at least 46 db below the parent wave, i.e, at least 26 db below the desired wave at the oppo site terminal. In other words the di rectivity of the couplers is at least 26 db (30 db in the lower frequency couplers) over the complete fre quency range [Fig. 3(a), (b)]. The coupling mechanism itself consists of quarter-wavelength sec tions of the conductors placed suit ably near one another to achieve the desired degree of coupling. The combined effects of electrical and magnetic coupling impart directiv ity to the coupled wave. The unused terminal of each of the auxiliary arms is terminated in a special widerange low-reflection resistor to ab sorb any power coupled to that ter minal. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On Jun 6, 10:06*pm, K7ITM wrote:
On Jun 6, 3:21*pm, Wimpie wrote: Measuring method used: change in resistive load, from voltage change you can calculate the current change, hence the output impedance. One note, except two, all where solid state. Of course, since the source impedance may be anywhere in the complex plane, you need to change more than just the resistive part of the load, I believe, to get an accurate picture... All high efficiency designs (class E, D) that I did have output impedance far from the expected load impedance. With "far" I mean factor 2 or factor 0.5. I did not measure that (as it is not important in virtually all cases), but know it from the overload simulation/measurement and I did the design myself. What I've seen in similar situations is that the source impedance is likely to be strongly reactive, depending on the filter network you use to get sinusoidal output. *In any event, the source impedance is likely to have a reflection coefficient magnitude that is quite close to unity. *That is exactly what you should expect: *there's nothing to absorb reflections. *You could (theoretically at least) use feedback to make the output look like 50 ohms, but just as you say, Wim ... why?? *There's really almost never any point in doing so. ... Cheers, Tom Again Wim, we're not on the same page, so let me quote from your eariler post: "Virtually all high efficient amplifiers work in voltage saturated mode and are therefore not operated at maximum available power, therefore their output impedance doens't match the expected load." I have not been talking about MAXIMUM available power, only the power available with some reasonable value of grid drive. In the real world of amateur radio operations, of which I'm talking, when we adjust the pi-network for that given drive level, we adjust for delivering all the AVAILABLE power at that drive level. When all the available power is thus delivered, the output source resistance equals the load resistance by definition. We're now not talking about changing the load, phase or SWR--we're talking about the single condition arrived at after the loading adjustments have been made. And Tom, why would the source be reactive when the pi-network is tuned to resonance? And because the source resistance of the (tube) power amp is non-dissipative, its reflection coefficient is 1.0 by definition, and so it cannot absorb any reflected energy, and therefore re-reflects it. Walt, W2DU |
what happens to reflected energy ?
On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote: The 100W forward and 50W reflected have no relation to actual powers From HP Journal V4n5-6: The multi-hole coupler design has now been extended to a 3 db coupler, that is, a coupler in which half the power entering the main guide is coupled into the auxiliary guide. .... The -hp- multi-hole couplers consist of two sections of wave guide mutually coupled by two rows of coupling holes (Fig. 2). Power entering the input arm of the coupler flows down the main guide and divides at the coupling mechanism. Part of the power con tinues down the main guide where it will be incident on any device connected at the end of the guide. The other part of the power is coupled into the auxiliary guide. It is a prop erty of directional couplers that the power coupled into the auxiliary guide flows essen tially in only one direction. In the -hp- coup ler, this power flow is in the same direction as the power in the primary guide. .... R E F L E C T O M E T E R S E T - U P .... Two directional couplers are connected back-to-back as shown. ******* I will leave it to the readership to imagine why. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On Jun 6, 9:34*pm, Richard Clark wrote:
On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart wrote: The 100W forward and 50W reflected have no relation to actual powers From HP Journal V4n5-6: The multi-hole coupler design has now been extended to a 3 db coupler, that is, a coupler in which half the power entering the main guide is coupled into the auxiliary guide. ... The -hp- multi-hole couplers consist of two sections of wave guide mutually coupled by two rows of coupling holes (Fig. 2). Power entering the input arm of the coupler flows down the main guide and divides at the coupling mechanism. Part of the power con tinues down the main guide where it will be incident on any device connected at the end of the guide. The other part of the power is coupled into the auxiliary guide. It is a prop erty of directional couplers that the power coupled into the auxiliary guide flows essen tially in only one direction. In the -hp- coup ler, this power flow is in the same direction as the power in the primary guide. ... R E F L E C T O M E T E R * S E T - U P ... Two directional couplers are connected back-to-back as shown. ******* I will leave it to the readership to imagine why. 73's Richard Clark, KB7QHC Walter was the first to place things in a book and a book is the first point of reference in this newsgroup Nothing personal Richard, but it is obvious that engineers are totally split as to what theorem is correct and the one to be used. There is no way you will get agreement if they all insist their position is correct and thus other positions are not worth consideration.You may be correct in your position but the idea of change is not in your favor. |
what happens to reflected energy ?
On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote: The 100W forward and 50W reflected have no relation to actual powers From HP Journal V.3n7-8 DIRECTIONAL couplers have been used widely in wave guide applications for such purposes as monitoring power, meas uring reflections, mixing, and for isolation of signal sources. All of these applications make use of the property that power flowing in one direction in the main branch of the coupler induces a power flow in only one direction in the auxiliary circuit. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote: The 100W forward and 50W reflected have no relation to actual powers From HP Journal V.6n.1-2: Power from the source flows down the main arms of the two couplers (Fig. 2) and impinges on the load. The power split off by the 20 db forward coupler is all passed to the forward detector, since the directiv ity characteristic of the multi-hole directional couplers prevents any but a negligible amount of the splitoff power from turning back and being absorbed in the coupler's in ternal termination. The power split off the incident wave by the 10 db reverse coupler, however, is essen tially all absorbed in that coupler's internal termination because of the reversed direction of connection of that coupler (Fig. 2). If the magnitude of the reflection coefficient ~.T. of the load is, say, 0.1, Çf of the incident voltage will be reflected back toward the source. As this reflection passes back through the main arm of the reverse coupler, a 10 db split occurs and is applied to the reverse detector. The remain der of the reflection will proceed back toward the generator where it will be absorbed in the generator impedance and in the termination in the forward coupler. ******** Of course, in regard to this last sentence, HP engineers didn't know jack-**** about power reflections, especially what could be absorbed by the source. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On 7 jun, 04:29, walt wrote:
On Jun 6, 10:06*pm, K7ITM wrote: On Jun 6, 3:21*pm, Wimpie wrote: Measuring method used: change in resistive load, from voltage change you can calculate the current change, hence the output impedance. One note, except two, all where solid state. Of course, since the source impedance may be anywhere in the complex plane, you need to change more than just the resistive part of the load, I believe, to get an accurate picture... All high efficiency designs (class E, D) that I did have output impedance far from the expected load impedance. With "far" I mean factor 2 or factor 0.5. I did not measure that (as it is not important in virtually all cases), but know it from the overload simulation/measurement and I did the design myself. What I've seen in similar situations is that the source impedance is likely to be strongly reactive, depending on the filter network you use to get sinusoidal output. *In any event, the source impedance is likely to have a reflection coefficient magnitude that is quite close to unity. *That is exactly what you should expect: *there's nothing to absorb reflections. *You could (theoretically at least) use feedback to make the output look like 50 ohms, but just as you say, Wim ... why?? *There's really almost never any point in doing so. ... Cheers, Tom Again Wim, we're not on the same page, so let me quote from your eariler post: "Virtually all high efficient amplifiers work in voltage saturated mode and are therefore not operated at maximum available power, therefore their output impedance doens't match the expected load." I have not been talking about MAXIMUM available power, only the power available with some reasonable value of grid drive. In the real world of amateur radio operations, of which I'm talking, when we adjust the pi-network for that given drive level, we adjust for delivering all the AVAILABLE power at that drive level. When all the available power is thus delivered, the output source resistance equals the load resistance by definition. We're now not talking about changing the load, phase or SWR--we're talking about the single condition arrived at after the loading adjustments have been made. Walt, this restriction was not given in the original posting, it is added by you and you exclude many other practical amplifier solutions (like push pull 3...30 MHz no-tune amplifiers, fixed tuned single band amplifiers, emerging high efficiency designs for constant envelope modulation schemes and/or AM with supply voltage modulation). I fully agree with your statement on optimum matching given certain drive and output impedance in case of (pi-filter) matching, no doubt about this. However there are more flavors as I tried to explain. Even during SSB modulation with an optimally tuned amplifier without power supply modulation, the amplifier is most of the time operated into current saturation mode (instead of optimally tuned output, given a certain drive). For tetrode/pentode, assuming no voltage saturation, the RF plate impedance is seldom equal to the conjugated load impedance (load impedance will be lower). I think for a triode in common grid operation, this will apply also because of the cathode is not grounded for AC, hence plate impedance increases. I have to be careful now to avoid that I have to do a lot of work to fulfill Richard's requests. I dismantled my PL519 common grid amplifier (22 years ago), so a cannot measure it anymore. All amateurs that do not have a tuner inside their PA, have to live with non-optimum VSWR (hence amplifier not operating at optimal tuning, even under CW) or have to insert a tuner. For the latter case, the problem now reduces to a cable and tuner loss problem, as after successful tuning, there will be now power reflected to the PA, hence the PA's output impedance doesn't matter. Maybe the JC could provide some background behind his question. And Tom, why would the source be reactive when the pi-network is tuned to resonance? And because the source resistance of the (tube) power amp is non-dissipative, its reflection coefficient is 1.0 by definition, and so it cannot absorb any reflected energy, and therefore re-reflects it. Walt, W2DU Best regards, Wim PA3DJS www.tetech.nl without abc, PM will reach me |
what happens to reflected energy ?
On Jun 6, 7:29*pm, walt wrote:
.... And Tom, why would the source be reactive when the pi-network is tuned to resonance? And because the source resistance of the (tube) power amp is non-dissipative, its reflection coefficient is 1.0 by definition, and so it cannot absorb any reflected energy, and therefore re-reflects it. Walt, W2DU If you allow that the thing driving the pi network has an effective source impedance different from the load that the pi network presents to it, then clearly the output impedance seen at the other end (the "50 ohm" end) of the pi network won't be 50 ohms. Try it with some numbers; for example, assume a pi network that transforms your 50 ohm load to a 4k ohm load to the amplifier output, and assume an amplifier output stage that looks like a 20k ohm source. Design the pi network for a loaded Q of 10. I believe you'll find that the source impedance seen by the 50 ohm load is about 11+j18 ohms. As Wim has pointed out, requiring an amplifier to be loaded and driven in a very particular way unnecessarily dismisses some very important classes of amplifier. What do you do, for example, with a linear amplifier? What do you do with an amplifier that drives a voltage very hard (and for which a simple pi network is inappropriate for matching to a load)? Perhaps an even more basic question is: why exactly do we tune a pi network to present a particular load to an RF amplifier stage? Why should we operate a 6146 with, say, a 3000 ohm plate load? Why not 1000, or 6000? And what if I set up a tube and pi network for operation such that the apparent output source impedance is 50 ohms (while driving a 50 ohm load), and then I add feedback to the amplifier in such a way that the operating conditions are not changed, but the impedance looking back into the plate is changed? How did we get to the source resistance of "the (tube) power amp" being non-dissipative? I know there are some of us who don't buy into that... Cheers, Tom |
what happens to reflected energy ?
Hello Walt,
I did some simulations with EL34. Conditions: Common Grid circuit, Valve connected as triode, 3.6 MHz. Vsupply, 600V, Vgrid -58V, bias current 40mA (yes I know that it will run very hot). Maximum available input power: 9.2W out of 600 Ohms, sinusoidal. Anode loaded with 20uH, 96pF, 3800 Ohms (parallel circuit), QL= 8.4, so the anode voltage has sinusoidal shape. Without changing the drive (9.2W), 3800 Ohms with 96pF gives highest output power of 64W (Vanode = 698Vp). So power gain is 7.0. Output impedance: Change to 4100 Ohms load gives Zout = 1826 (voltage saturation), Change to 3500 Ohms load gives Zout = 4350 (current saturation), So for small change in Zload, Zout will be close the load resistor that gives maximum power. This supports that under well-matched conditions, given a certain drive level, Zout = ZLoad. Input power reduced with 3 dB, output becomes 32.6W (Vanode = 497.5Vp, gain = 7.08). There is a slight increase in gain showing that at 64W the amplifier is not into deep voltage saturation. Output impedance at 32.6W output: Change from 3800 to 3500 Ohms gives Zout = 8100 Ohms. This shows that reducing the drive level (without changing the matching) results in Zout not equal to Zload for this triode common grid circuit. Also shown that small changes in load (reduction or increase seen at the anode) results in changes in output impedance. That is why I used three resistance values for determining Zout at well-matched output. I used low frequency to avoid endless matching because of parasitic capacitances and I also skipped the output matching (I have to work sometimes, so time is limited). It is not strange to see an RF swing above supply voltage as this is a common grid circuit with cathode grounded for DC. If you like, I could do same exercise for modern amplifier topologies, but be prepared to see unexpected results (As Tom, K7ITM already noted). Best regards, Wim PA3DJS www.tetech.nl remove abc first in case of PM. |
what happens to reflected energy ?
On Jun 6, 1:01*pm, K1TTT wrote:
as in the other thread, what is the mechanism of that 'interaction' between waves? *i contend there can be no 'interaction' between forward and reflected waves if the device is linear. David, you are preaching to the choir. I explained before that there is NO interaction between forward and reflected waves because they are traveling in different directions. The only time that coherent, collimated waves can interact is when they are traveling in the same direction in a transmission line. Keith's argument requires that forward waves and reflected waves interact. I say they cannot interact in a constant Z0 environment. You say they cannot interact. We are on the same side of this argument. The mechanism of the interaction of two coherent, collimated waves traveling in the same direction is that the superposition process is irreversible. The source photons and the reflected photons are indistinguishable. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On Mon, 7 Jun 2010 09:19:59 -0700 (PDT), K7ITM wrote:
As Wim has pointed out, requiring an amplifier to be loaded and driven in a very particular way unnecessarily dismisses some very important classes of amplifier. This is the standard rebuff with "the rest of the world works differently" distractions to Walt having stated a premise, described initial conditions, taken measurements, and having shown the data supports his hypothesis. The hypothesis is dismissed through shifting initial conditions to suit an avoidance of committing an honest answer to Walt's specific and very explicit observation. Walt, if this community were pressed for an "up or down" vote: "Does Walt's data support the evidence of a Conjugate Match?" then you would be out in the weeds with your only supporting vote from me (and maybe from others who would do this by email). I doubt this will set off a stampede to the ballot box, but what few votes are stuffed in, I bet they will have the "up or down" stapled to a dissertation of "however...." To this last, if I sinned in that regard, I used only Walt's data, his equipment references, and his citation sources. As no one else seems to tread that narrow path, much less commit beyond grandiose statements, I don't feet too bad. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On Jun 7, 1:37*am, Richard Clark wrote:
All of these applications make use of the property that power flowing in one direction in the main branch of the coupler induces a power flow in only one direction in the auxiliary circuit. Power flow??? Heaven forbid. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
On 7 jun, 19:46, Richard Clark wrote:
On Mon, 7 Jun 2010 09:19:59 -0700 (PDT), K7ITM wrote: As Wim has pointed out, requiring an amplifier to be loaded and driven in a very particular way unnecessarily dismisses some very important classes of amplifier. * This is the standard rebuff with "the rest of the world works differently" distractions to Walt having stated a premise, described initial conditions, taken measurements, and having shown the data supports his hypothesis. *The hypothesis is dismissed through shifting initial conditions to suit an avoidance of committing an honest answer to Walt's specific and very explicit observation. Walt, if this community were pressed for an "up or down" vote: * "Does Walt's data support the evidence of a Conjugate Match?" then you would be out in the weeds with your only supporting vote from me (and maybe from others who would do this by email). * I doubt this will set off a stampede to the ballot box, but what few votes are stuffed in, I bet they will have the "up or down" stapled to a dissertation of "however...." To this last, if I sinned in that regard, I used only Walt's data, his equipment references, and his citation sources. *As no one else seems to tread that narrow path, much less commit beyond grandiose statements, I don't feet too bad. 73's Richard Clark, KB7QHC Hello Richard, I think, most people that are willing to read will support Walt's statements on output impedance of an amplifier under matched output power (given a certain drive condition). However there are many practical circumstances where Walt's conditions are not met. As soon as you change the drive (for example during an AM or SSB transmission), your matching is no longer guaranteed, especially when using circuits with current behavior (tetrode/pentode, common grid, etc). See my EL34 posting. Note that under practical circumstances, this is no problem. As soon as you change the load (without changing the matching), you may run into current or voltage saturation, of course depending on phase of VSWR. As you know many people use non-tune solid-state amplifiers, so they don't have the possibility to tune/match for maximum output given certain drive. That means, live with it, or use an external tuner where you don't tune for maximum power, but for minimum VSWR presented to the PA. Amplifiers for constant envelope modulation use saturation to increase efficiency (and accept the loss in gain). These are deliberately used under mismatch, therefore the gain is less with respect to a non- saturating approach. You can also see this from the Pout versus Pin curve for FM transistors. I know that severak CB owners retuned (or even modified) the output stage to get 1 dB more power, but they did forget that the final transistor had dissipate 2dB more. Result: some japanese transistors became very popular (2SC1306, 1307, 1969, etc). High efficiency circuits are the extreme case and are entering the amateur world. Active devices behave like switches, output impedance can have every value as long as it is close to the edge of the passive Smith Chart. Amplitude modulation with these topologies can only be done via supply voltage modulation. Tuning for maximum power with an external tuner will surely destroy the amplifier if no protection is present. For me it was a surprise how this thread developed. Best regards, Wim PA3DJS www.tetech.nl |
what happens to reflected energy ?
On Jun 7, 2:09*pm, Wimpie wrote:
As soon as you change the drive (for example during an AM or SSB transmission), your matching is no longer guaranteed, ... AM should be the easiest mode to analyze since it requires linear finals. -- 73, Cecil, w5dxp.com |
what happens to reflected energy ?
I haven't followed this thread since it's been beaten to death so many
times here before. But there's an interesting fact that might have been overlooked, and might (or might not) be relevant: If you put a directional coupler such as a Bruene circuit at the output of a transmitter, and use its "forward" output to control the transmitter power to keep the "forward" directional coupler output constant, you'll find that the power output vs load resistance characteristic is exactly the same as if the transmitter had 50 ohm output impedance. This is assuming that the directional coupler is designed for a 50 ohm system, and that the load is purely resistive. It also assumes that any load is left in place long enough for the feedback circuit to stabilize. The effective source impedance to a very rapidly varying load (that is, one changing so fast that the ALC feedback system doesn't have time to respond) would be the open-loop output impedance which could be quite different. I haven't taken the time to analyze how it behaves with a complex load. I stumbled across this some time ago when designing a rig using this ALC method and found it interesting. I believe many if not most modern solid-state transmitters use this ALC method. Roy Lewallen, W7EL |
what happens to reflected energy ?
On Jun 7, 5:00*pm, Cecil Moore wrote:
On Jun 6, 1:01*pm, K1TTT wrote: as in the other thread, what is the mechanism of that 'interaction' between waves? *i contend there can be no 'interaction' between forward and reflected waves if the device is linear. David, you are preaching to the choir. I explained before that there is NO interaction between forward and reflected waves because they are traveling in different directions. The only time that coherent, collimated waves can interact is when they are traveling in the same direction in a transmission line. Keith's argument requires that forward waves and reflected waves interact. I say they cannot interact in a constant Z0 environment. You say they cannot interact. We are on the same side of this argument. The mechanism of the interaction of two coherent, collimated waves traveling in the same direction is that the superposition process is irreversible. The source photons and the reflected photons are indistinguishable. -- 73, Cecil, w5dxp.com This is what YOU said: What happens to the 50 joules/second of reflected energy depends upon the phasing between the source wave and the reflected wave at the source impedance. What most amateurs don't understand is there are two mechanisms that can redistribute reflected energy back toward the antenna. Those mechanisms are a re-reflection based on the physical reflection coefficient (what RF engineers understand) and wave interaction resulting in constructive/destructive interference (what most RF engineers don't seem to understand) because, unlike optical physicists, have not been forced to follow the energy flow. There is no interaction between the waves, they may be superimposed, and maybe their photons are indistinguishable, but there is no 'interaction'. interaction implies that one wave affects the other, energy is transferred or fields are affected, such things occur in non- linear media, but not in linear ones. |
what happens to reflected energy ?
On Jun 7, 12:19*pm, K7ITM wrote:
On Jun 6, 7:29*pm, walt wrote: ... And Tom, why would the source be reactive when the pi-network is tuned to resonance? And because the source resistance of the (tube) power amp is non-dissipative, its reflection coefficient is 1.0 by definition, and so it cannot absorb any reflected energy, and therefore re-reflects it. Walt, W2DU If you allow that the thing driving the pi network has an effective source impedance different from the load that the pi network presents to it, then clearly the output impedance seen at the other end (the "50 ohm" end) of the pi network won't be 50 ohms. *Try it with some numbers; for example, assume a pi network that transforms your 50 ohm load to a 4k ohm load to the amplifier output, and assume an amplifier output stage that looks like a 20k ohm source. *Design the pi network for a loaded Q of 10. *I believe you'll find that the source impedance seen by the 50 ohm load is about 11+j18 ohms. As Wim has pointed out, requiring an amplifier to be loaded and driven in a very particular way unnecessarily dismisses some very important classes of amplifier. *What do you do, for example, with a linear amplifier? *What do you do with an amplifier that drives a voltage very hard (and for which a simple pi network is inappropriate for matching to a load)? *Perhaps an even more basic question is: *why exactly do we tune a pi network to present a particular load to an RF amplifier stage? *Why should we operate a 6146 with, say, a 3000 ohm plate load? *Why not 1000, or 6000? And what if I set up a tube and pi network for operation such that the apparent output source impedance is 50 ohms (while driving a 50 ohm load), and then I add feedback to the amplifier in such a way that the operating conditions are not changed, but the impedance looking back into the plate is changed? How did we get to the source resistance of "the (tube) power amp" being non-dissipative? *I know there are some of us who don't buy into that... Cheers, Tom Hello Tom, Well, the reason I chose to discuss only the tube amp with a pi- network filter and impedance transformer is that that arrangement is the only one I've measured, and that I'm not sufficiently acquainted with other arrangements to discuss them. As to your question, why not use plate loads of 3000, 6000 or 1000? Because the transceivers I measured using two 6146s in parallel are Kenwood TS-830S and Heathkit HW-100. I don't remember the exact plate load with the HW-100, close to 1400 ohms, while the TS-830S was 1400 ohms. I had no control over those plate loads because the plate and grid voltages were preset, and 1400 ohms is the RL I measured at the input of the pi-network when the grid drive and the network were adjusted to deliver precisely 100w to the load 50-ohm load. Now concerning the non-dissipative source resistance of the tube-type power amp. There are two separate resistances in the amp, the cathode- to-plate resistance, Rpd, that accounts for all the dissipation in the tube; and the output-source resistance that is non-dissipative. It is a common myth that an RF power amp cannot have an efficiency greater than 50% when conjugate matched to the load, because half the RF power is dissipated in the source resistance. This is not true, because when the amp is operating properly, resistance Rpd is less than the output source resistance, thus allowing more power delivered to the load than that dissipated in the plate-to-cathode resistance. The output source resistance is derived from the voltage-current ratio E/I that appears at the output terminals of the pi-network. A ratio, as such, cannot dissipate energy, but the load it feeds does. Using an example from Terman's Radio Engineer's Handbook I explain this phenomenon in great detail in Chapter 19 in Reflections, and further in Chapter 19A, an addition to Chapter 19. Both are available on my web page at www.w2du.com. Chapter 19 appears in 'Read Chapters from Reflections 2', and 19A appears in 'Preview Chapter from Reflections 3'. The entire Chapter 19 appears in Reflections 3, which is now available from CQ. I invite you to review these Chapters. Walt, W2DU |
what happens to reflected energy ?
On 7 jun, 22:36, Roy Lewallen wrote:
I haven't followed this thread since it's been beaten to death so many times here before. But there's an interesting fact that might have been overlooked, and might (or might not) be relevant: If you put a directional coupler such as a Bruene circuit at the output of a transmitter, and use its "forward" output to control the transmitter power to keep the "forward" directional coupler output constant, you'll find that the power output vs load resistance characteristic is exactly the same as if the transmitter had 50 ohm output impedance. This is assuming that the directional coupler is designed for a 50 ohm system, and that the load is purely resistive. It also assumes that any load is left in place long enough for the feedback circuit to stabilize. The effective source impedance to a very rapidly varying load (that is, one changing so fast that the ALC feedback system doesn't have time to respond) would be the open-loop output impedance which could be quite different. I haven't taken the time to analyze how it behaves with a complex load. I stumbled across this some time ago when designing a rig using this ALC method and found it interesting. I believe many if not most modern solid-state transmitters use this ALC method. Roy Lewallen, W7EL Hello Roy, If you have sufficient headroom and under the conditions you mentioned, you mimic a 50 Ohms source. I think it also works for any (complex) loads (I couldn’t find why not). The difference between a real 50 Ohms circuit may be that the phase of the belonging EMF may change, in many amplifiers phase shift is somewhat excitation dependent, but who bothers? I expect such scheme in combination with VSWR measurement also, as several PAs have a reverse power indicator present (the other half of a the Bruene circuit). You need the reverse power indication to avoid destroying active devices and/or intermodulation distortion due to voltage saturation. Imagine that you have full reflection |RC| = 1 and it appears at your active device as RC=-1. You want to maintain the original forward power. Your active device has to deliver in that case double the current at zero collector/plate voltage to maintain same forward power as under matched condition. The actual power delivered to the load is zero (as the active device supplies current, but no voltage, RC=-1 means a short circuit). This will result in massive dissipation in the active device. In case of RC=+1, it has to provide double the voltage with no current. In other circumstances you will have a significant phase shift between current and voltage resulting also in increased device dissipation and inconvenient combinations of instantaneous voltage and current. So above some value for VSWR, you may have to reduce the forward power I had a discussion recently about the power control scheme for TETRA terminals, but we couldn't find the answer to what is happening under high VSWR (so we have measure it). It only states VSWR2. Best regards, Wim PA3DJS www.tetech.nl without abc, PM will reach me. |
what happens to reflected energy ?
Wimpie wrote:
. . . I expect such scheme in combination with VSWR measurement also, as several PAs have a reverse power indicator present (the other half of a the Bruene circuit). You need the reverse power indication to avoid destroying active devices and/or intermodulation distortion due to voltage saturation. . . . Yes, I was referring to the operating region at which normal power is maintained. As you pointed out, rigs employing this method will only maintain the same "forward" power over some range of SWR beyond which they'll begin reducing it, otherwise they'd self-destruct in the attempt. In the reduced power region, the apparent 50 ohm output impedance of course no longer holds. Roy Lewallen, W7EL |
what happens to reflected energy ?
On Mon, 07 Jun 2010 13:36:41 -0700, Roy Lewallen
wrote: I haven't followed this thread since it's been beaten to death so many times here before. But there's an interesting fact that might have been overlooked, and might (or might not) be relevant: If you put a directional coupler such as a Bruene circuit at the output of a transmitter, and use its "forward" output to control the transmitter power to keep the "forward" directional coupler output constant, you'll find that the power output vs load resistance characteristic is exactly the same as if the transmitter had 50 ohm output impedance. This is assuming that the directional coupler is designed for a 50 ohm system, and that the load is purely resistive. It also assumes that any load is left in place long enough for the feedback circuit to stabilize. The effective source impedance to a very rapidly varying load (that is, one changing so fast that the ALC feedback system doesn't have time to respond) would be the open-loop output impedance which could be quite different. I haven't taken the time to analyze how it behaves with a complex load. I stumbled across this some time ago when designing a rig using this ALC method and found it interesting. I believe many if not most modern solid-state transmitters use this ALC method. Roy Lewallen, W7EL Hi Roy, This application that you describe was written up in exactly the same terms within the recent HP Journals I have posted extracts here. HP used Directional Couplers (the Bruene circuit, also called a bridge, qualifies too but uses a non-wave design) to separate out the forward from the reverse power reflected from the mismatch to create a reference power. Later, HP and others strapped the signals back into the source in much the manner you describe. The rudimentary version can be found in HP Journal v.6 n.1-2. HP Journal v.12 n.4 strengthens the concept with hard copy sweeps of the reflection coefficient of a load. By HP Journal v.16 n.6, we have the description of automatic level control. For the 45 years beyond that last article, more refinements. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On Mon, 7 Jun 2010 12:09:23 -0700 (PDT), Wimpie
wrote: I think, most people that are willing to read will support Walt's statements on output impedance of an amplifier under matched output power (given a certain drive condition). However there are many practical circumstances where Walt's conditions are not met. So, that is one vote for (with bulky provisos), and one vote against (with bulky provisos) going into the ballot box. Walt, if this community were pressed for an "up or down" vote: "Does Walt's data support the evidence of a Conjugate Match?" It seems writing the phrase "Conjugate Match is shown by Walt's data" is a fearsome (loathsome) step to take, or can be taken with qualifiers and other soporifics to deaden the pain. Walt's data shows evidence of a Conjugate Match. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote: The 100W forward and 50W reflected have no relation to actual powers From a current NARDA specification: GENERAL COUPLER OPERATION A coaxial directional coupler has the general appearance of a section of coaxial line, with the addition of a second parallel section of line and with one end terminated (see Figure 9). These two sections are known as the main and auxiliary lines. The two lines are internally separated from each other; the amount of spacing between lines determines the amount of RF energy that may be transferred from the main line to the auxiliary line. In operation, assume that energy is fed into port A of the main line. Most of this energy will appear at output port B of the main line. However, a fraction of this energy (determined by coupling value) will also appear at the coupled port C, of the auxiliary line. A dual-directional coaxial coupler, such as the reflectometer coupler, consists essentially of two single- ended couplers connected back-to-back. Perhaps the most important characteristic of the directional coupler (and the one from which its name originates) is its directivity. .... For reflectometry applications, the dual directional coupler, incorporating two auxiliary outputs, permits the simultaneous sampling of incident and reflected power. .... RF power applied to the load is reflected to some degree depending on load characteristics, thereby resulting in a voltage standing wave ratio (VSWR) which is reflected back to the main line output port. this reflected power is coupled out of the reflected output port at a level 10 dB down from the reflected power level at the load. 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On Mon, 07 Jun 2010 15:49:04 -0700, Richard Clark
wrote: The rudimentary version can be found in HP Journal v.6 n.1-2. HP Journal v.12 n.4 strengthens the concept with hard copy sweeps of the reflection coefficient of a load. By HP Journal v.16 n.6, we have the description of automatic level control. For the 45 years beyond that last article, more refinements. HP Journal Nov. 1970 is dedicated to all system elements going into one box, 8620A, 8632A (with options for an external Directional Coupler and Power Meters). 73's Richard Clark, KB7QHC |
what happens to reflected energy ?
On Jun 7, 4:21*pm, K1TTT wrote:
There is no interaction between the waves, they may be superimposed, and maybe their photons are indistinguishable, but there is no 'interaction'. *interaction implies that one wave affects the other, energy is transferred or fields are affected, such things occur in non- linear media, but not in linear ones. Note that I was not talking about forward and reflected waves in a constant Z0 transmission line. I am talking about the four wavefront components that are generated at an impedance discontinuity. It has been proven in experimentally that those waves indeed interact. In fact, the transfer of destructive interference energy from wave cancellation to the areas that permit constructive interference is obviously interaction since the canceled waves disappear in their original direction of travel. But, as the FSU web page says, they are not annihilated - their energy components simply change direction. How can you possibly argue that wave cancellation doesn't require wave interaction? Those two waves completely disappear in the direction of destructive interference. Dr. Best argued that those two waves don't interact and continue propagating (completely devoid of energy) forever in the direction of original travel. I asked him to prove that his phantom waves exist but he could not. Is there such a proof available? At a 1/4WL thin-film coating on non-reflective glass, when the internal reflected wave arrives with equal magnitude and 180 degrees out of phase with the external reflection, wave cancellation occurs. That is an *obvious* effect that one wave has on the other. Wave cancellation is an obvious interaction. In the s-parameter equation: b1 = s11*a1 + s12*a2 = 0 the s11*a1 wavefront has obviously interacted with the s12*a2 wavefront to accomplish wave cancellation. -- 73, Cecil, w5dxp.com |
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