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Transmitter Output Impedance
On Tue, 10 May 2011 15:34:36 -0700 (PDT), Richard Fry
wrote: On May 10, 4:37*pm, walt wrote: This is for Richard Fry, Richard, because the output source impedance of a p-network following a tube RF power amp is non-dissipative, no reflected wave enters it to Here is what I believe you missed: When the 111.11w arrives at the mismatch, 11.111w is reflected, leaving 100w dissipated in the load. Walt, Let us start at T=0. The transmitter is keyed, and instantaneously delivers 100W to the input of a lossless transmission line. That 100W travels to the end of the transmission line and encounters a load connected there having a 2:1 mismatch to the Zo of the transmission line. That load will dissipate about 88.889W, and reflect about 11.111W back to the source. Likewise, the load will dissipate only 88.889% of the RE-reflected power from the source for the same reason it dissipated only 88.889% of the power in the original incident wave. So even if the RE-reflection from the source is lossless, and phase- coherent with the original incident power at the load, how does this yield the 100W of load dissipation that you posit? RF 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On Tue, 10 May 2011 15:34:36 -0700 (PDT), Richard Fry
wrote: So even if the RE-reflection from the source is lossless, and phase- coherent with the original incident power at the load, how does this yield the 100W of load dissipation that you posit? The PI network between the tube and the output connector transforms the tube Hi-Z to 50 Ohms (or any other suitable Z); however, the same PI network's input Z (from the perspective of the returning, original reflection) is NOT 50 Ohms, it is the complex conjugate (whatever 2:1 brings to the table) of the load answering phase-coherency, hence it reflects the reflection (by the same mechanism of mismatch) to the dissipative load the antenna, and thus into the æther. The transmission line is bounded at either end by identical mismatches in conjugation. There is a "ringing" in the line, which at only one end there is dissipation (neglecting loss and damage to the tube for exceeding maximum tolerances, of course). 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On May 10, 6:48*pm, Richard Clark wrote:
On Tue, 10 May 2011 15:34:36 -0700 (PDT), Richard Fry wrote: So even if the RE-reflection from the source is lossless, and phase- coherent with the original incident power at the load, how does this yield the 100W of load dissipation that you posit? The PI network between the tube and the output connector transforms the tube Hi-Z to 50 Ohms (or any other suitable Z); however, the same PI network's input Z (from the perspective of the returning, original reflection) is NOT 50 Ohms, it is the complex conjugate (whatever 2:1 brings to the table) of the load answering phase-coherency, hence it reflects the reflection (by the same mechanism of mismatch) to the dissipative load the antenna, and thus into the æther. The transmission line is bounded at either end by identical mismatches in conjugation. *There is a "ringing" in the line, which at only one end there is dissipation (neglecting loss and damage to the tube for exceeding maximum tolerances, of course). 73's Richard Clark, KB7QHC For Richard F. Where does the 88.889w of power delivered to the load come from? I calculate it as 100w. 100/111.11 = 0.9000. Since 11.111w of power reflected at a 2:1 mismatch with a 100 source and lossless line is indisputable, then the mismatched load will dissipate 90% of the forward power. 90% of 111.11w is 100w. So please explain the significance of 88.889w. I can't see it as relevant. Walt |
Transmitter Output Impedance
On May 10, 7:22*pm, walt wrote:
On May 10, 6:48*pm, Richard Clark wrote: On Tue, 10 May 2011 15:34:36 -0700 (PDT), Richard Fry wrote: So even if the RE-reflection from the source is lossless, and phase- coherent with the original incident power at the load, how does this yield the 100W of load dissipation that you posit? The PI network between the tube and the output connector transforms the tube Hi-Z to 50 Ohms (or any other suitable Z); however, the same PI network's input Z (from the perspective of the returning, original reflection) is NOT 50 Ohms, it is the complex conjugate (whatever 2:1 brings to the table) of the load answering phase-coherency, hence it reflects the reflection (by the same mechanism of mismatch) to the dissipative load the antenna, and thus into the æther. The transmission line is bounded at either end by identical mismatches in conjugation. *There is a "ringing" in the line, which at only one end there is dissipation (neglecting loss and damage to the tube for exceeding maximum tolerances, of course). 73's Richard Clark, KB7QHC For Richard F. Where does the 88.889w of power delivered to the load come from? I calculate it as 100w. 100/111.11 = 0.9000. Since 11.111w of power reflected at a 2:1 mismatch with a 100 source and lossless line is indisputable, then the mismatched load will dissipate 90% of the forward power. 90% of 111.11w is 100w. So please explain the significance of 88.889w. I can't see it as relevant. Walt Richard, the point I missed until rereading your post, is that there is total re-reflection of the reflected wave at either the output of the correctly-adjusted pi-network, or at the output of the antenna tuner, not the same mismatch as between the line and the load. Walt |
Transmitter Output Impedance
On May 10, 5:48*pm, Richard Clark wrote:
The transmission line is bounded at either end by identical mismatches in conjugation. *There is a "ringing" in the line, which at only one end there is dissipation (neglecting loss and damage to the tube for exceeding maximum tolerances, of course). If, as you have posted, dissipation occurs only at one end of the transmission line -- which presumably for the most useful benefit would be at the end of the line opposite the source, e.g., the load -- and the incident power generated and delivered to that line by the final r-f stage in the transmitter is a constant, then how could "damage to the tube" occur? Just to note that as a field engineer for RCA Broadcast for many years I have had to troubleshoot and repair failures that occurred in transmission lines, transmitters, and output networks that resulted from reflections -- either from within the transmission line alone, or together with the load connected at its far end. RF |
Transmitter Output Impedance
On May 10, 6:22*pm, walt wrote:
Where does the 88.889w of power delivered to the load come from? I calculate it as 100w. 100/111.11 = 0.9000. Yes, Walt, 100/111.11 = 0.9. But a transmitter generating 100W does not deliver 111.11 watts in the original incident wave arriving at a load VSWR of 2:1 w.r.t a lossless transmission line used to convey that power. It delivers only 100W. Of that 100W, 88.889W is absorbed by the load, and 11.111W is reflected back to the source. THAT is where the 88.889W comes from. Also note that 0,9 does not equal 0.8889. RF |
Transmitter Output Impedance
On Tue, 10 May 2011 16:28:23 -0700 (PDT), Richard Fry
wrote: If, as you have posted, dissipation occurs only at one end of the transmission line -- which presumably for the most useful benefit would be at the end of the line opposite the source, e.g., the load -- and the incident power generated and delivered to that line by the final r-f stage in the transmitter is a constant, then how could "damage to the tube" occur? Hi Richard, Note how you were careful to specify "phase coherent." This is dangerous to speak of power in this way, energy would be better, but I will proceed with caution thrown to the winds: When the reflection hits the transmitter port, it sees the conjugate of the load. As the load was 2:1 (and trusting everyone's math with my own rounding) this means that the transmitter port also presents a 2:1 mismatch at this near end of the coax. Some (10%) of the power in the reflection is re-reflected to the load. Some (90%) of the power in the reflection re-enters the transmitter to add to the next cycle of power going out. In this scenario, that amounts to 9W going into and being coherently impressed upon the tube. The transmitter, then, becomes a buck-boosted 109W source conjugately matched into a 2:1 mismatched load. 9 additional watts at the plate isn't much, but I dare say there are many problems with transmitter tolerances even if you could tune into a short, or an open. Yes, somewhere the SWR meter says everything is hunky-dory, while the smoke is poring out. Another way of putting it is saying the ringing line is holding 10W until it is dissipated. So far, only the antenna qualifies as dissipation. Through the magic of Integral Calculus, all the bits and pieces ringing around the system sum up to the whole numbers we started with. No doubt this will provoke examinations, re-examinations, cross-examinations, innuendo, rants, and raves (or possibly disdainful silence) - something for everyone to enjoy. ;-O 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On May 10, 7:28*pm, Richard Fry wrote:
On May 10, 5:48*pm, Richard Clark wrote: The transmission line is bounded at either end by identical mismatches in conjugation. *There is a "ringing" in the line, which at only one end there is dissipation (neglecting loss and damage to the tube for exceeding maximum tolerances, of course). If, as you have posted, dissipation occurs only at one end of the transmission line -- which presumably for the most useful benefit would be at the end of the line opposite the source, e.g., the load -- and the incident power generated and delivered to that line by the final r-f stage in the transmitter is a constant, then how could "damage to the tube" occur? Just to note that as a field engineer for RCA Broadcast for many years I have had to troubleshoot and repair failures that occurred in transmission lines, transmitters, and output networks that resulted from reflections -- either from within the transmission line alone, or together with the load connected at its far end. RF Yes Richard, I'm familiar with your work at RCA Broadcast Div. However, in my posts I'm concerned only with tube rigs used in the Amateur Service. In these rigs reflected power doesn't cause overheating, or other damage to the tubes. But I'll qualify that statement--if the pi-network is originally resonated into, say, a dummy load, and is then switched to a mismatched line without retuning, the reactance appearing at the input of the mismatched line detunes the network and the result is the same is if the network was left off resonance initially. As you well know, a mis-tuned pi-network results in excessive plate current, which, if high enough will damage the tube. Consequently, it's the mis-tuning that causes the damage, not the reflections per se. Incidentally Richard, I was also an electrical engineer with RCA, from 1949 thru 1980, first at the RCA Labs in Princeton and beginning in 1958 with Astro. Walt |
Transmitter Output Impedance
On May 10, 7:00*pm, Richard Clark wrote:
Another way of putting it is saying the ringing line is holding 10W until it is dissipated. *So far, only the antenna qualifies as dissipation. Apparently you have never had to repair the components of a transmitter PA, an output network, or a transmission line that arced over and/or melted down due to load reflections (coherent, or not). RF |
Transmitter Output Impedance
On Tue, 10 May 2011 17:14:56 -0700 (PDT), Richard Fry
wrote: On May 10, 7:00*pm, Richard Clark wrote: Another way of putting it is saying the ringing line is holding 10W until it is dissipated. *So far, only the antenna qualifies as dissipation. Apparently you have never had to repair the components of a transmitter PA, an output network, or a transmission line that arced over and/or melted down due to load reflections (coherent, or not). Not more than a quarter million Watts, fur shure. Does it show? Seriously, if that was your only rant, based on that snippet, like Wim, you have just indicted your own witness (experience?). RF 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On Tue, 10 May 2011 16:27:00 -0700 (PDT), walt wrote:
Richard, the point I missed until rereading your post, is that there is total re-reflection of the reflected wave at either the output of the correctly-adjusted pi-network, or at the output of the antenna tuner, not the same mismatch as between the line and the load. Walt Hi Walt, The mismatch, of the reflection wave approaching the transmitter, is seen at the antenna/tuner connection. Hence we have a reflection there back to the load. In regard to solving the transmission line problem where the mismatched load acts as the source of energy (found in the reflection); then that energy must travel to meet a matching load where the load (our transmitter) complete takes it in, or a mismatch where some energy passes into the load (our transmitter) and some is reflected according to the conventional mechanics of mis-matching. I take this step given the evidence of physics and reflection at even the most specular of surfaces. There is some portion of the energy that exists within a quarter wave behind/beneath the surface of a reflector (I am speaking of light, but lower frequencies doesn't alter the situation). That is to say, there is no infinitesimally thin boundary through which energy does not pass - even for a perfect reflector. In antennas, we call this zone the reactive field. Hence, the reflection seeing the mismatch of the tuner/tuned output, finds its way beyond the connector, back into the internals, passing through transform circuitry. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On May 10, 8:40*pm, Richard Clark wrote:
On Tue, 10 May 2011 17:14:56 -0700 (PDT), Richard Fry wrote: On May 10, 7:00*pm, Richard Clark wrote: Another way of putting it is saying the ringing line is holding 10W until it is dissipated. *So far, only the antenna qualifies as dissipation. Apparently you have never had to repair the components of a transmitter PA, an output network, or a transmission line that arced over and/or melted down due to load reflections (coherent, or not). Not more than a quarter million Watts, fur shure. *Does it show? Seriously, if that was your only rant, based on that snippet, like Wim, you have just indicted your own witness (experience?). RF 73's Richard Clark, KB7QHC To Richard C: Richard, I must totally disagree that the same degree of mismatch occurs at the tx output as at the dissipative load, the antenna. As I have said many times, with the non-dissipative output source impedance of the RF amp (tubes) the reflection at the junction of the line input and the pi-network output is total. In other words, all of the reflected power incident on the output of the pi-network is TOTALLY re- reflected. Can't be anything less! To Richard F. And I disagree that the load mismatched at 2:1 absorbs 88.8889% of the forward power. It is 90% !!! We are discussing steady- state values, in which arrival of the first forward wave is irrelevant. If you are so confident that the number is 88.8889%, please derive the conditions that yield that number. I never thought I'd be having this discussion. Walt |
Transmitter Output Impedance
On Tue, 10 May 2011 19:35:55 -0700 (PDT), walt wrote:
To Richard C: Richard, I must totally disagree that the same degree of mismatch occurs at the tx output as at the dissipative load, the antenna. As I have said many times, with the non-dissipative output source impedance of the RF amp (tubes) the reflection at the junction of the line input and the pi-network output is total. In other words, all of the reflected power incident on the output of the pi-network is TOTALLY re- reflected. Can't be anything less! Hi Walt, There's an active load behind that 2:1 mismatch seen by the reflected power returning from the load. That is my part of the discussion. However, if you are going to neglect that interpretation, then there must be some explanation for TOTAL reflection. Fortunately there is, it is called an open, or a short. Now, we come to a philosophical quandry of the "Conjugate Match." If we tune to a load, and we guarantee that the tune is the conjugate of the load, then at any point along the transmission line in between there exists looking at either end a conjugate of the opposite end. In other words, putting a network analyzer into the line looking at the load it sees R +jX, and looking back at the source it sees R -jX. This is conjugation. Moving the analyzer along the length of the line will transform R and X, but they will always track to a solution of the conjugate. You've been arguing this for years. And, as the load is neither a short, nor an open (which is a special solution of conjugation); then it stands to reason it is something other which necessarily demands a solution in the form of: R ±jX from the source, where R has some non-zero, and non-infinite value. For trivial antennas, that value of R falls somewhere between the 10s of Ohms and the 100s of Ohms. For trivial, retail transmitters available to Hams, similar values of R abound but most often fall closest to 50 Ohms, or they would have a hard time getting optimal power into the 50 Ohm pipe. Vendors are not going to shoot themselves in the foot before that race. A TOTAL reflection is available courtesy of the "active" part of this special load. The "active" load (the transmitter sinking the reflected power from the dissipative load) can offer one of two mechanisms to provide the necessary short or open. When the source wave coming out meets the returning wave coming in, and they are phase coherent at 0 degree difference, then there is no potential difference through the connector - no current flow. No current flow = open circuit. Open circuit satisfies TOTAL reflection mechanics (but violates conjugation). If they are phase coherent, but 180 degrees different, then there is maximum current = short circuit. Short circuit satisfies TOTAL reflection mechanics (but violates conjugation). However, in either of these scenarios, both powers (actually energies) must be equal. Obviously that isn't the case with a nominal 100W source looking at a 10W reflection. Hence the analysis draws out slightly more along the lines I've already posted and won't repeat here. However, for the sake of argument, the condition of tune (conjugation) and the state of coherency (a slippery slope, but here we are), combine through a ringing in the line to create a virtual TOTAL reflection at the connector of the Tuner/tuned amplifier. So, Walt, I am not denying your case, but in the earlier posting I do preserve conjugation in the explanation. No transmitter, even with the best of tuning attempting to match will forgive a short or open as load however. And, yes Richard, I have experienced such conditions with 10s of W to 100s of kW and not all of them accidental. Another issue of moving sensors along the transmission line between high mismatches is that there will be considerable error in the determination of power - I presume this is where Jim's bookkeeping explanation will lead, but that is still hanging on the line (no pun there). I have experience the difficulty of those conditions too when making precision power measurements. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On May 10, 9:35 pm, walt wrote:
If you are so confident that the number is 88.8889%, please derive the conditions that yield that number. Ã = R(load) - R(source) / R(load) + R(source) SWR = (1 + |Ã|) / (1 - |Ã|) Power Accepted by the Load = Incident Power * (1 - Ã^2) For a 50 ohm source connected to a 100 ohm load: Ã = (100 - 50) / (100 + 50) = 50 / 150 = 0.333333... SWR = (1 + 0.333333) / (1 - 0.333333) = 1.333333 / 0.666666 = 2:1 Load Power = Incident Power * (1 - 0.333333^2) = Incident Power * 0.888889 (or 88.8889%) The answers are the same for a 50 ohm source with a 25 ohm load. RF |
Transmitter Output Impedance
On 11/05/2011 12:21, Richard Fry wrote:
On May 10, 9:35 pm, wrote:9.542 If you are so confident that the number is 88.8889%, please derive the conditions that yield that number. Ã = R(load) - R(source) / R(load) + R(source) SWR = (1 + |Ã|) / (1 - |Ã|) Power Accepted by the Load = Incident Power * (1 - Ã^2) For a 50 ohm source connected to a 100 ohm load: Ã = (100 - 50) / (100 + 50) = 50 / 150 = 0.333333... SWR = (1 + 0.333333) / (1 - 0.333333) = 1.333333 / 0.666666 = 2:1 Load Power = Incident Power * (1 - 0.333333^2) = Incident Power * 0.888889 (or 88.8889%) The answers are the same for a 50 ohm source with a 25 ohm load. RF Or expressed another way a 2:1 vswr equates to a return loss of 9.542dB and so a mismatch loss of -0.512dB which equates to 0.888 or 88.8% . Jeff |
Transmitter Output Impedance
On May 10, 5:34*pm, Richard Fry wrote:
Let us start at T=0. *The transmitter is keyed, and instantaneously delivers 100W to the input of a lossless transmission line. *That 100W travels to the end of the transmission line and encounters a load connected there having a 2:1 mismatch to the Zo of the transmission line. That load will dissipate about 88.889W, and reflect about 11.111W back to the source. Likewise, the load will dissipate only 88.889% of the RE-reflected power from the source for the same reason it dissipated only 88.889% of the power in the original incident wave. So even if the RE-reflection from the source is lossless, and phase- coherent with the original incident power at the load, how does this yield the 100W of load dissipation that you posit? It's what happens while going from T0 to steady-state. Pload = 88.889w -- 98.8w -- ... -- 100w Pfor = 100w -- 111.11w -- ... -- 112.5w Pref = 11.11w -- 12.3w -- ... -- 12.5w -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On May 10, 9:35*pm, walt wrote:
Richard, I must totally disagree that the same degree of mismatch occurs at the tx output as at the dissipative load, the antenna. May I suggest that we simplify the example by removing the tuner from the discussion? 100w--+----1/4WL 100 ohm feedline----200 ohm load Same conditions as before. The voltage reflection coefficient at the load is 0.3333. The power reflection coefficient at the load is 0.1111. A 50 ohm Z0-match exists at point '+' without a tuner. No reflected energy reaches the source. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On 10 mayo, 23:37, walt wrote:
[deleted] Now Wim, your math is very challenging: You state 100w delivered by the source, but at one point you also state 111w is delivered to the 100-load--at another point you state that 100w is delivered to the 100- ohm load. Which is it? You also state that voltage out of the source is 212.1v--sum ting wong here. 212.1v across 50 ohms yields 899.73w, and 212.1v across 100 ohm yields 449.86w. These power values are nowhere near the values appearing in you statements. In my calculations, with 100w the voltage across a 50-ohm line is 70.71v, and across a 100-ohm line the voltage is 100v. With a 50-ohm line terminated in 100 ohms, the vrc is 0.3333 as you stated, thus the power-reflection coefficient is 0.1111. This means, as I've stated continually, that with 100w delivered by the source, the power reflected at the 2:1 mismatch is 11.111w, which when added to the 100w supplied by the source, makes the forward power 111.11w. Now with 11.111w of power reflected at the mismatch, this leaves 100w delivered to the 100-ohm load, not 111w. Please tell me where the 111w came from. You also haven't told us why you calculate 130w forward and 15w reflected. Someone else may have made these calculations, but it was you I asked for an explanation, because you repeated those calculations. When using the correct physics and math in this example, how can the results *be so different? To close, let me present the procedure I use to calculate the total forward power--I guess it is related to Ohm's Law: With the power-reflection coefficient as prc, the forward power PF = 1/ (1 - prc). So Wim, can you clarify the confusion you appear to have made? Walt Hello Walt, The source (that is the PA in this case) produces an EMF (as mentioned before) of 212.1Vrms, Source impedance is 100 Ohms (so forget the "conjugated match" thing for this case). When terminated according to the numbers above the socket ("100W into 50 Ohms"), this results in 100W ( Vout = 212.1*50/(50+100) = 70.7V, just a voltage divider consisting of 100 Ohms and 50 Ohms ). 70.7V into 50 Ohms makes 100W. Now we remove the 50 Ohms load and create a mismatch (referenced to 50 Ohms), by connecting a 100 Ohms load (VSWR = 100/50 = 2) Now the output voltage will be: Vout = 212.1*100/(100 +100) = 106V 106V into 100 Ohms makes 112W (forgive me the one Watt difference). This all without any transmission line theory. Now the forward and reflected power balance (this is exactly the same as in my previous posting, except for using 112W instead of 111W): 100 Ohms equals |rc| = 0.3333 (50 Ohms reference). Hence "reflected power" = 0.333^2*"incident power" = 0.111*"incident power". "net delivered power" = "incident power" – "reflected power" = 112W. "net delivered power" = (1-0.11111)*"incident power" = 112W. "incident power"= 126W, "reflected power" = 14 W As far as I can see, there is nothing new with respect to the previous calculation, except for the truncation/rounding. Similar reasoning can be applied to my class-E PA. It is designed for 500W into 4.5 Ohm, but the output impedance is below 1 Ohms. If you try to achieve conjugate match, you fry the active devices. Same is valid for most audio amplifiers, they may mention: "80W into 8 Ohm", but that does not mean that Zout = 8 Ohms. Awaiting your comment, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On 11 mayo, 02:05, walt wrote:
On May 10, 7:28*pm, Richard Fry wrote: On May 10, 5:48*pm, Richard Clark wrote: The transmission line is bounded at either end by identical mismatches in conjugation. *There is a "ringing" in the line, which at only one end there is dissipation (neglecting loss and damage to the tube for exceeding maximum tolerances, of course). If, as you have posted, dissipation occurs only at one end of the transmission line -- which presumably for the most useful benefit would be at the end of the line opposite the source, e.g., the load -- and the incident power generated and delivered to that line by the final r-f stage in the transmitter is a constant, then how could "damage to the tube" occur? Just to note that as a field engineer for RCA Broadcast for many years I have had to troubleshoot and repair failures that occurred in transmission lines, transmitters, and output networks that resulted from reflections -- either from within the transmission line alone, or together with the load connected at its far end. RF Yes Richard, I'm familiar with your work at RCA Broadcast Div. However, in my posts I'm concerned only with tube rigs used in the Amateur Service. In these rigs reflected power doesn't cause overheating, or other damage to the tubes. To Walt: (Steady state) Load conditions can be specified in many forms (complex impedance, complex reflection coefficient, VSWR with phase, etc), but depending on the actual integral of the Vp*Ip product, the plate's dissipation may decrease or increase (due to a load change). So changing the load (without retuning as you mentioned below) may put your tubes at risk. Your qualification below seems to contradict your statement above. But I'll qualify that statement--if the pi-network is originally resonated into, say, a dummy load, and is then switched to a mismatched line without retuning, the reactance appearing at the input of the mismatched line detunes the network and the result is the same is if the network was left off resonance initially. As you well know, a mis-tuned pi-network results in excessive plate current, which, if high enough will damage the tube. Consequently, it's the mis-tuning that causes the damage, not the reflections per se. Incidentally Richard, I was also an electrical engineer with RCA, from 1949 thru 1980, first at the RCA Labs in Princeton and beginning in 1958 with Astro. Walt Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On May 11, 10:57*am, Wimpie wrote:
On 11 mayo, 02:05, walt wrote: On May 10, 7:28*pm, Richard Fry wrote: On May 10, 5:48*pm, Richard Clark wrote: The transmission line is bounded at either end by identical mismatches in conjugation. *There is a "ringing" in the line, which at only one end there is dissipation (neglecting loss and damage to the tube for exceeding maximum tolerances, of course). If, as you have posted, dissipation occurs only at one end of the transmission line -- which presumably for the most useful benefit would be at the end of the line opposite the source, e.g., the load -- and the incident power generated and delivered to that line by the final r-f stage in the transmitter is a constant, then how could "damage to the tube" occur? Just to note that as a field engineer for RCA Broadcast for many years I have had to troubleshoot and repair failures that occurred in transmission lines, transmitters, and output networks that resulted from reflections -- either from within the transmission line alone, or together with the load connected at its far end. RF Yes Richard, I'm familiar with your work at RCA Broadcast Div. However, in my posts I'm concerned only with tube rigs used in the Amateur Service. In these rigs reflected power doesn't cause overheating, or other damage to the tubes. To Walt: (Steady state) Load conditions can be specified in many forms (complex impedance, complex reflection coefficient, VSWR with phase, etc), but depending on the actual integral of the Vp*Ip product, the plate's dissipation may decrease or increase (due to a load change). So changing the load (without retuning as you mentioned below) may put your tubes at risk. *Your qualification below seems to contradict your statement above. But I'll qualify that statement--if the pi-network is originally resonated into, say, a dummy load, and is then switched to a mismatched line without retuning, the reactance appearing at the input of the mismatched line detunes the network and the result is the same is if the network was left off resonance initially. As you well know, a mis-tuned pi-network results in excessive plate current, which, if high enough will damage the tube. Consequently, it's the mis-tuning that causes the damage, not the reflections per se. Incidentally Richard, I was also an electrical engineer with RCA, from 1949 thru 1980, first at the RCA Labs in Princeton and beginning in 1958 with Astro. Walt Wim PA3DJSwww.tetech.nl To Richard C.: It seems to me that you're overlooking the fact that the source resistance appearing at the output terminals of the pi-network is non-dissipative, and thus cannot absorb the energy in the reflected wave incident on it. The resistance is E/I and therefore non-dissipative, has a real value and is thus a part of the conjugation. The result is total re-reflection, no transfer of energy into the amp. I thought I'd made this very clear over the years. To Richard F. My humble apology for my moronic error. My bad!!! I can't believe I was so stupid as to not know where the 0.888889 came from. Thanks for spelling it out for me. I got totally sidetracked with the 100/111,11 = 0.90, and didn't see any farther. Wim: I'm working on your response--it's finally beginning to make sense--gimme a few more minutes. Walt |
Transmitter Output Impedance
Not replying to anyone in particular - Because I can post graphics
directly to QRZ, I have started a thread on the antenna forum at: http://forums.qrz.com/showthread.php...er-Brainteaser It is my contention that reflections are eliminated by a Z0-match which, in a tuner system, occurs at the *input* to the tuner. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On May 11, 2:40*pm, Cecil Moore wrote:
Not replying to anyone in particular - Because I can post graphics directly to QRZ, I have started a thread on the antenna forum at: http://forums.qrz.com/showthread.php...er-Brainteaser It is my contention that reflections are eliminated by a Z0-match which, in a tuner system, occurs at the *input* to the tuner. -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Cecil, you are correct!!! Walt |
Transmitter Output Impedance
On May 11, 3:09*pm, walt wrote:
Cecil, you are correct!!! Thanks Walt, it would be interesting to see what else about which I am correct. W7EL has a food-for-thought publication with one section about reflections from the source. His source has a 50 ohm source impedance, by design, so any dynamic load pulling experiment should result in zero reflections from the source. Question is, does it? http://eznec.com/misc/Food_for_thought.pdf The "forward and reverse power section" is at the bottom of the page. Using the dynamic load pulling method, do the results always indicate 50 ohms for the source impedance no matter what is the load and/or the transmission line length? If not, the method yields invalid results. IMO, it is a gross error to presume that all redistribution of energy is a result of reflections. In W7EL's example, there are no reflections from the source yet it is obvious that energy is being redistributed from the source back toward the load in some cases but not in other cases. IMO, destructive/constructive interference must be taken into account in order to explain the results. Yet, no one except yours truly has even mentioned interference effects as a method of redistributing energy. Anyone interested in understanding the role of interference at impedance discontinuities in transmission lines is welcome to read my article at: http://www.w5dxp.com/energy.htm -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK |
Transmitter Output Impedance
On May 13, 4:09*pm, Cecil Moore wrote:
On May 11, 3:09*pm, walt wrote: Cecil, you are correct!!! Thanks Walt, it would be interesting to see what else about which I am correct. W7EL has a food-for-thought publication with one section about reflections from the source. His source has a 50 ohm source impedance, by design, so any dynamic load pulling experiment should result in zero reflections from the source. Question is, does it? http://eznec.com/misc/Food_for_thought.pdf The "forward and reverse power section" is at the bottom of the page. Using the dynamic load pulling method, do the results always indicate 50 ohms for the source impedance no matter what is the load and/or the transmission line length? If not, the method yields invalid results. IMO, it is a gross error to presume that all redistribution of energy is a result of reflections. In W7EL's example, there are no reflections from the source yet it is obvious that energy is being redistributed from the source back toward the load in some cases but not in other cases. IMO, destructive/constructive interference must be taken into account in order to explain the results. Yet, no one except yours truly has even mentioned interference effects as a method of redistributing energy. Anyone interested in understanding the role of interference at impedance discontinuities in transmission lines is welcome to read my article at: http://www.w5dxp.com/energy.htm -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Right on, Cecil. The article you URL'ed is the most clear and accurate paraphrasing of my presentation of the subject in 'Reflections'. Nowhere else is the subject clearly defined concerning destructive and constructive interference in relation to impedance matching and 100% re-reflection. I hope RF engineers everywhere can realize that optics and RF are indeed using the same principles of electromagnetics, and therefore follow exactly the same rules. However, I would like for you to add one more reference to your article, an article I published in QEX that proves Steve Best's QEX three articles wrong. It appeared in the Jul/Aug 2004 issue, entitled, "A tutorial Dispelling Certain Misconceptions Concerning Wave Interference In Impedance Matching." Let me know here whether you have a copy of my QEX article. If not, I'll email you a copy. Walt, W2DU If anyone else on this thread would like a copy of my QEX article please let me know by email. My email address is shown above. |
Transmitter Output Impedance
On May 13, 5:39*pm, walt wrote:
On May 13, 4:09*pm, Cecil Moore wrote: On May 11, 3:09*pm, walt wrote: Cecil, you are correct!!! Thanks Walt, it would be interesting to see what else about which I am correct. W7EL has a food-for-thought publication with one section about reflections from the source. His source has a 50 ohm source impedance, by design, so any dynamic load pulling experiment should result in zero reflections from the source. Question is, does it? http://eznec.com/misc/Food_for_thought.pdf The "forward and reverse power section" is at the bottom of the page. Using the dynamic load pulling method, do the results always indicate 50 ohms for the source impedance no matter what is the load and/or the transmission line length? If not, the method yields invalid results. IMO, it is a gross error to presume that all redistribution of energy is a result of reflections. In W7EL's example, there are no reflections from the source yet it is obvious that energy is being redistributed from the source back toward the load in some cases but not in other cases. IMO, destructive/constructive interference must be taken into account in order to explain the results. Yet, no one except yours truly has even mentioned interference effects as a method of redistributing energy. Anyone interested in understanding the role of interference at impedance discontinuities in transmission lines is welcome to read my article at: http://www.w5dxp.com/energy.htm -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Right on, Cecil. The article you URL'ed is the most clear and accurate paraphrasing of my presentation of the subject in 'Reflections'. Nowhere else is the subject clearly defined concerning destructive and constructive interference in relation to impedance matching and 100% re-reflection. I hope RF engineers everywhere can realize that optics and RF are indeed using the same principles of electromagnetics, and therefore follow exactly the same rules. However, I would like for you to add one more reference to your article, an article I published in QEX that proves Steve Best's QEX three articles wrong. It appeared in the Jul/Aug 2004 issue, entitled, "A tutorial Dispelling Certain Misconceptions Concerning Wave Interference In Impedance Matching." Let me know here whether you have a copy of my QEX article. If not, I'll email you a copy. Walt, W2DU If anyone else on this thread would like a copy of my QEX article please let me know by email. My email address is shown above. Sorry guys, my error, my email address is . Walt |
Transmitter Output Impedance
On 13 mayo, 22:09, Cecil Moore wrote:
On May 11, 3:09*pm, walt wrote: Cecil, you are correct!!! Thanks Walt, it would be interesting to see what else about which I am correct. W7EL has a food-for-thought publication with one section about reflections from the source. His source has a 50 ohm source impedance, by design, so any dynamic load pulling experiment should result in zero reflections from the source. Question is, does it? http://eznec.com/misc/Food_for_thought.pdf The "forward and reverse power section" is at the bottom of the page. Using the dynamic load pulling method, do the results always indicate 50 ohms for the source impedance no matter what is the load and/or the transmission line length? If not, the method yields invalid results. IMO, it is a gross error to presume that all redistribution of energy is a result of reflections. In W7EL's example, there are no reflections from the source yet it is obvious that energy is being redistributed from the source back toward the load in some cases but not in other cases. IMO, destructive/constructive interference must be taken into account in order to explain the results. Yet, no one except yours truly has even mentioned interference effects as a method of redistributing energy. Anyone interested in understanding the role of interference at impedance discontinuities in transmission lines is welcome to read my article at: http://www.w5dxp.com/energy.htm -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Hello Cecil, When a source behaves like a 50 Ohms source, it will not rereflect the reflected power back to the load. In other words the forward power generated by the source will not change when changing the load. The reflected power that is absorbed by the source may result in increase or reduction of power dissipation of the active device. It all depends on the change of the integral of V*I (for the active device). With regards to load pulling, output impedance of an amplifier may change depending on the rate of change of the load. As I mentioned earlier, manual load pulling may result in change of supply and bias voltage and current. Average Voltage and current have time to settle to the new load condition. In case of load pulling where the load changes relatively fast, but well within the matching section's bandwidth (as is the case for the off-carrier signal injection method) supply and bias voltage change may be less due to decoupling capacitance. In simulation this can easily be seen by applying a load step change and observe the envelope and phase response versus time. Of course this requires you to model your power supply and bias circuit correctly. You may see a slow step response that cannot be explained by the bandwidth of the matching network. To walt: I didn’t see your announced comment to my last post (regarding the forward/reverse power mathematics), did I missed it? With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On May 13, 6:51*pm, Wimpie wrote:
On 13 mayo, 22:09, Cecil Moore wrote: On May 11, 3:09*pm, walt wrote: Cecil, you are correct!!! Thanks Walt, it would be interesting to see what else about which I am correct. W7EL has a food-for-thought publication with one section about reflections from the source. His source has a 50 ohm source impedance, by design, so any dynamic load pulling experiment should result in zero reflections from the source. Question is, does it? http://eznec.com/misc/Food_for_thought.pdf The "forward and reverse power section" is at the bottom of the page. Using the dynamic load pulling method, do the results always indicate 50 ohms for the source impedance no matter what is the load and/or the transmission line length? If not, the method yields invalid results. IMO, it is a gross error to presume that all redistribution of energy is a result of reflections. In W7EL's example, there are no reflections from the source yet it is obvious that energy is being redistributed from the source back toward the load in some cases but not in other cases. IMO, destructive/constructive interference must be taken into account in order to explain the results. Yet, no one except yours truly has even mentioned interference effects as a method of redistributing energy. Anyone interested in understanding the role of interference at impedance discontinuities in transmission lines is welcome to read my article at: http://www.w5dxp.com/energy.htm -- 73, Cecil, w5dxp.com "Halitosis is better than no breath at all.", Don, KE6AJH/SK Hello Cecil, When a source behaves like a 50 Ohms source, it will not rereflect the reflected power back to the load. In other words the forward power generated by the source will not change when changing the load. The reflected power that is absorbed by the source may result in increase or reduction of power dissipation of the active device. It all depends on the change of the integral of V*I (for the active device). With regards to load pulling, output impedance of an amplifier may change depending on the rate of change of the load. As I mentioned earlier, manual load pulling may result in change of supply and bias voltage and current. *Average Voltage and current have time to settle to the new load condition. In case of load pulling where the load changes relatively fast, but well within the matching section's bandwidth (as is the case for the off-carrier signal injection method) supply and bias voltage change may be less due to decoupling capacitance. In simulation this can easily be seen by applying a load step change and observe the envelope and phase response versus time. Of course this requires you to model your power supply and bias circuit correctly. You may see a slow step response that cannot be explained by the bandwidth of the matching network. To walt: I didn’t see your announced comment to my last post (regarding the forward/reverse power mathematics), did I missed it? With kind regards, Wim PA3DJSwww.tetech.nl Hello Wim, I hadn't yet given you my final understanding of it. I understand the voltage divider action, but I don't yet understand the following quote from you: "When terminated according to the numbers above the socket ("100W into 50 Ohms"), this results in 100W ( Vout = 212.1*50/(50+100) = 70.7V, just a voltage divider consisting of 100 Ohms and 50 Ohms )." The portion I don't understand is the mathematical basis for: "( Vout = 212.1*50/(50+100) = 70.7V," And Wim, on your last post above I agree that if the source is 50 ohms with a real resistor, as in the classical generator used in text books. However, are you not aware that the output resistance of the RF power amp is R = E/R that appears at the output. Since this R is non- dissipative it re-reflects all the reflected power. I have proved this to be true in the experiment I presented with the Kenwood TS-830S presentation that I'm sure you have a copy of. Consequently, the RF power amp is not going to absorb the reflected power. At this point my email server is down, but I'll forward a copy of my QEX article ASAP. Walt |
Transmitter Output Impedance
On May 11, 4:21*am, Richard Fry wrote:
On May 10, 9:35 pm, walt wrote: If you are so confident that the number is 88.8889%, please derive the conditions that yield that number. Ã = R(load) - R(source) / R(load) + R(source) SWR = (1 + |Ã|) / (1 - |Ã|) Power Accepted by the Load = Incident Power * (1 - Ã^2) For a 50 ohm source connected to a 100 ohm load: Ã = (100 - 50) / (100 + 50) = 50 / 150 = 0.333333... SWR = (1 + 0.333333) / (1 - 0.333333) = 1.333333 / 0.666666 = 2:1 Load Power = Incident Power * (1 - 0.333333^2) = Incident Power * 0.888889 (or 88.8889%) The answers are the same for a 50 ohm source with a 25 ohm load. RF Richard F is, of course, correct. You can just as well do it for DC: 2V O.C., 1 ohm source will deliver 1A, 1V, 1 watt to a 1 ohm load. For a 2 ohm load, the current is 2/3 amp; the power is 8/9 watt. For an 0.5 ohm load, the current is 4/3 amp; the power is 8/9 watt. It's not terribly difficult to show, for the AC case, that the same is true for any 2:1 load, regardless of phase angle--trivial, in fact, if you accept that the SWR along a lossless TEM line is constant. Cheers, Tom |
Transmitter Output Impedance
On May 13, 7:07*pm, K7ITM wrote:
On May 11, 4:21*am, Richard Fry wrote: On May 10, 9:35 pm, walt wrote: If you are so confident that the number is 88.8889%, please derive the conditions that yield that number. Ã = R(load) - R(source) / R(load) + R(source) SWR = (1 + |Ã|) / (1 - |Ã|) Power Accepted by the Load = Incident Power * (1 - Ã^2) For a 50 ohm source connected to a 100 ohm load: Ã = (100 - 50) / (100 + 50) = 50 / 150 = 0.333333... SWR = (1 + 0.333333) / (1 - 0.333333) = 1.333333 / 0.666666 = 2:1 Load Power = Incident Power * (1 - 0.333333^2) = Incident Power * 0.888889 (or 88.8889%) The answers are the same for a 50 ohm source with a 25 ohm load. RF Richard F is, of course, correct. *You can just as well do it for DC: 2V O.C., 1 ohm source will deliver 1A, 1V, 1 watt to a 1 ohm load. For a 2 ohm load, the current is 2/3 amp; the power is 8/9 watt. *For an 0.5 ohm load, the current is 4/3 amp; the power is 8/9 watt. *It's not terribly difficult to show, for the AC case, that the same is true for any 2:1 load, regardless of phase angle--trivial, in fact, if you accept that the SWR along a lossless TEM line is constant. Cheers, Tom My apology, Wim, the equation I quoted in my last post is incorrect--I mean't to say R = E/I, not E/R. Sorry. Walt |
Transmitter Output Impedance
On May 11, 7:15*am, Wimpie wrote:
On 10 mayo, 23:37, walt wrote: [deleted] Now Wim, your math is very challenging: You state 100w delivered by the source, but at one point you also state 111w is delivered to the 100-load--at another point you state that 100w is delivered to the 100- ohm load. Which is it? You also state that voltage out of the source is 212.1v--sum ting wong here. 212.1v across 50 ohms yields 899.73w, and 212.1v across 100 ohm yields 449.86w. These power values are nowhere near the values appearing in you statements. In my calculations, with 100w the voltage across a 50-ohm line is 70.71v, and across a 100-ohm line the voltage is 100v. With a 50-ohm line terminated in 100 ohms, the vrc is 0.3333 as you stated, thus the power-reflection coefficient is 0.1111. This means, as I've stated continually, that with 100w delivered by the source, the power reflected at the 2:1 mismatch is 11.111w, which when added to the 100w supplied by the source, makes the forward power 111.11w. Now with 11.111w of power reflected at the mismatch, this leaves 100w delivered to the 100-ohm load, not 111w. Please tell me where the 111w came from. You also haven't told us why you calculate 130w forward and 15w reflected. Someone else may have made these calculations, but it was you I asked for an explanation, because you repeated those calculations. When using the correct physics and math in this example, how can the results *be so different? To close, let me present the procedure I use to calculate the total forward power--I guess it is related to Ohm's Law: With the power-reflection coefficient as prc, the forward power PF = 1/ (1 - prc). So Wim, can you clarify the confusion you appear to have made? Walt Hello Walt, The source (that is the PA in this case) produces an EMF (as mentioned before) of 212.1Vrms, Source impedance is 100 Ohms (so forget the "conjugated match" thing for this case). When terminated according to the numbers above the socket ("100W into 50 Ohms"), this results in 100W *( Vout = 212.1*50/(50+100) = 70.7V, just a voltage divider consisting of 100 Ohms and 50 Ohms ). 70.7V into 50 Ohms makes 100W. Now we remove the 50 Ohms load and create a mismatch (referenced to 50 Ohms), by connecting a 100 Ohms load (VSWR = 100/50 = 2) Now the output voltage will be: Vout = 212.1*100/(100 +100) = 106V 106V into 100 Ohms makes 112W (forgive me the one Watt difference). This all without any transmission line theory. Now the forward and reflected power balance (this is exactly the same as in my previous posting, except for using 112W instead of 111W): 100 Ohms equals |rc| = 0.3333 (50 Ohms reference). Hence "reflected power" = 0.333^2*"incident power" * = 0.111*"incident power". "net delivered power" = "incident power" – "reflected power" = 112W.. "net delivered power" = (1-0.11111)*"incident power" = 112W. "incident power"= 126W, "reflected power" = 14 W As far as I can see, there is nothing new with respect to the previous calculation, except for the truncation/rounding. Similar reasoning can be applied to my class-E PA. It is designed for 500W into 4.5 Ohm, but the output impedance is below 1 Ohms. If you try to achieve conjugate match, you fry the active devices. *Same is valid for most audio amplifiers, they may mention: *"80W into 8 Ohm", but that does not mean that Zout = 8 Ohms. Awaiting your comment, Wim PA3DJSwww.tetech.nl Of course, with pretty much any decently designed power amplifier, no matter the frequency, included protection circuits will limit the dissipation and voltage in the critical areas (e.g., the output active devices). Loads become disconnected or shorted sometimes; winds blow stray wires or tree limbs across antennas. Through a piece of transmission line, such a change can reflect back to become any possible phase angle, and any of a very wide range of magnitudes, at the amplifier's output. Though it's a red herring typical of audio-speak, most modern high fidelity audio amplifiers have a very low output impedance, a small fraction of an ohm, so they can claim a high damping factor. Off topic: the reason it's a red herring is that the impedance of the speaker connected to the amplifier must be included to figure the damping, and that impedance (even just the DC resistance of voice coils) changes by considerably more than the amplifier's output impedance (resistance) just because of heating on audio peaks. Especially when reactive components (inductors and capacitors) couple power between a source and a load, you can get stresses--voltages and/ or currents--well beyond what's safe when you try to operate the source into a load it's not intended to handle. That's true even when the net power delivered to the load is considerably LESS than the rated output power of the source. Wim's example of the class-E amplifier is true enough, but it's not necessary to ask the source to deliver more net load power than it's rated to deliver, to establish conditions that cause trouble. Thus, even sources that have an output impedance at or very close to the rated load impedance will have circuits to protect against loads that could destroy things inside the amplifier. Cheers, Tom |
Transmitter Output Impedance
Hello Walt,
We have just a voltage source of 212Vrms with 100 Ohms in series. This is all in a black box and called my PA. That black box has a UHF socket and that mentions "100 W into 50 Ohms". When you connect 50 Ohms to the SO 239 socket, you get a voltage divider. 2/3 of the EMF is across the internal 100 Ohms and 1/3 is across the 50 Ohms load. This results in 70.7V across the load (so 100W). Does this make sense? Wim PA3DJS |
Transmitter Output Impedance
On 14 mayo, 01:28, K7ITM wrote:
On May 11, 7:15*am, Wimpie wrote: On 10 mayo, 23:37, walt wrote: [deleted] Now Wim, your math is very challenging: You state 100w delivered by the source, but at one point you also state 111w is delivered to the 100-load--at another point you state that 100w is delivered to the 100- ohm load. Which is it? You also state that voltage out of the source is 212.1v--sum ting wong here. 212.1v across 50 ohms yields 899.73w, and 212.1v across 100 ohm yields 449.86w. These power values are nowhere near the values appearing in you statements. In my calculations, with 100w the voltage across a 50-ohm line is 70.71v, and across a 100-ohm line the voltage is 100v. With a 50-ohm line terminated in 100 ohms, the vrc is 0.3333 as you stated, thus the power-reflection coefficient is 0.1111. This means, as I've stated continually, that with 100w delivered by the source, the power reflected at the 2:1 mismatch is 11.111w, which when added to the 100w supplied by the source, makes the forward power 111.11w. Now with 11.111w of power reflected at the mismatch, this leaves 100w delivered to the 100-ohm load, not 111w. Please tell me where the 111w came from. You also haven't told us why you calculate 130w forward and 15w reflected. Someone else may have made these calculations, but it was you I asked for an explanation, because you repeated those calculations. When using the correct physics and math in this example, how can the results *be so different? To close, let me present the procedure I use to calculate the total forward power--I guess it is related to Ohm's Law: With the power-reflection coefficient as prc, the forward power PF = 1/ (1 - prc). So Wim, can you clarify the confusion you appear to have made? Walt Hello Walt, The source (that is the PA in this case) produces an EMF (as mentioned before) of 212.1Vrms, Source impedance is 100 Ohms (so forget the "conjugated match" thing for this case). When terminated according to the numbers above the socket ("100W into 50 Ohms"), this results in 100W *( Vout = 212.1*50/(50+100) = 70.7V, just a voltage divider consisting of 100 Ohms and 50 Ohms ). 70.7V into 50 Ohms makes 100W. Now we remove the 50 Ohms load and create a mismatch (referenced to 50 Ohms), by connecting a 100 Ohms load (VSWR = 100/50 = 2) Now the output voltage will be: Vout = 212.1*100/(100 +100) = 106V 106V into 100 Ohms makes 112W (forgive me the one Watt difference). This all without any transmission line theory. Now the forward and reflected power balance (this is exactly the same as in my previous posting, except for using 112W instead of 111W): 100 Ohms equals |rc| = 0.3333 (50 Ohms reference). Hence "reflected power" = 0.333^2*"incident power" * = 0.111*"incident power". "net delivered power" = "incident power" – "reflected power" = 112W. "net delivered power" = (1-0.11111)*"incident power" = 112W. "incident power"= 126W, "reflected power" = 14 W As far as I can see, there is nothing new with respect to the previous calculation, except for the truncation/rounding. Similar reasoning can be applied to my class-E PA. It is designed for 500W into 4.5 Ohm, but the output impedance is below 1 Ohms. If you try to achieve conjugate match, you fry the active devices. *Same is valid for most audio amplifiers, they may mention: *"80W into 8 Ohm", but that does not mean that Zout = 8 Ohms. Awaiting your comment, Wim PA3DJSwww.tetech.nl Of course, with pretty much any decently designed power amplifier, no matter the frequency, included protection circuits will limit the dissipation and voltage in the critical areas (e.g., the output active devices). *Loads become disconnected or shorted sometimes; winds blow stray wires or tree limbs across antennas. *Through a piece of transmission line, such a change can reflect back to become any possible phase angle, and any of a very wide range of magnitudes, at the amplifier's output. Though it's a red herring typical of audio-speak, most modern high fidelity audio amplifiers have a very low output impedance, a small fraction of an ohm, so they can claim a high damping factor. *Off topic: *the reason it's a red herring is that the impedance of the speaker connected to the amplifier must be included to figure the damping, and that impedance (even just the DC resistance of voice coils) changes by considerably more than the amplifier's output impedance (resistance) just because of heating on audio peaks. Especially when reactive components (inductors and capacitors) couple power between a source and a load, you can get stresses--voltages and/ or currents--well beyond what's safe when you try to operate the source into a load it's not intended to handle. *That's true even when the net power delivered to the load is considerably LESS than the rated output power of the source. * Hello Tom, I agree with your statement. If you have a CLC LPF section that introduces about 90 degrees phase delay (see it as a quarter wave section) and it looks into an open load (so zero power is delivered to this open load), this will result in a low impedance at the input of the CLC network. If the plate is connected at this CLC network, the plate's dissipation will greatly increase. Wim's example of the class-E amplifier is true enough, but it's not necessary to ask the source to deliver more net load power than it's rated to deliver, to establish conditions that cause trouble. *Thus, even sources that have an output impedance at or very close to the rated load impedance will have circuits to protect against loads that could destroy things inside the amplifier. I also fully agree on this one. Cheers, Tom With kind regards, Wim PA3DJS www.tetech.nl |
Transmitter Output Impedance
On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote:
Though it's a red herring typical of audio-speak, most modern high fidelity audio amplifiers have a very low output impedance, a small fraction of an ohm, so they can claim a high damping factor. Off topic: the reason it's a red herring is that the impedance of the speaker connected to the amplifier must be included to figure the damping, and that impedance (even just the DC resistance of voice coils) changes by considerably more than the amplifier's output impedance (resistance) just because of heating on audio peaks. Especially when reactive components (inductors and capacitors) couple power between a source and a load, you can get stresses--voltages and/ or currents--well beyond what's safe when you try to operate the source into a load it's not intended to handle. That's true even when the net power delivered to the load is considerably LESS than the rated output power of the source. Wim's example of the class-E amplifier is true enough, but it's not necessary to ask the source to deliver more net load power than it's rated to deliver, to establish conditions that cause trouble. Thus, even sources that have an output impedance at or very close to the rated load impedance will have circuits to protect against loads that could destroy things inside the amplifier. I'm going to slip a mickey into this by a careful, editorial change of focus BACK to the subject line. Though it's a red herring typical of Ham-speak, most modern retail 100W RF transmitters for amateur service have a very low output impedance, a fraction of an ohm [ editorial: until, of course, it goes to the Z transformer that precedes the bandpass filter]. It must be a rare condition for which there is a transition frequency below which audio amps source rated at sub Ohm Z feeding loads up to 10 times the sub-Ohm design work - and above which retail Amateur transmitters at loads up to 10 times the sub-Ohm design do not work [editorial: without that Z transformer]. Curious logic. I wonder if that works (sic) backwards? Would the RF deck's finals feed a speaker with audio with equal power performance of the Audio amp? Both are sub-Ohm sources, and the RF deck certainly has the bandwidth. I await the specification of this transition frequency and the analysis of how the 100W RF final deck would fail to supply 100W audio (however crummy it may sound). Any problems that might arise for the retail Ham transmitter are already handled by protection circuitry so that's a wash. 73's Richard Clark, KB7QHC |
Transmitter Output Impedance
On May 13, 8:57*pm, Richard Clark wrote:
On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote: Though it's a red herring typical of audio-speak, most modern high fidelity audio amplifiers have a very low output impedance, a small fraction of an ohm, so they can claim a high damping factor. *Off topic: *the reason it's a red herring is that the impedance of the speaker connected to the amplifier must be included to figure the damping, and that impedance (even just the DC resistance of voice coils) changes by considerably more than the amplifier's output impedance (resistance) just because of heating on audio peaks. Especially when reactive components (inductors and capacitors) couple power between a source and a load, you can get stresses--voltages and/ or currents--well beyond what's safe when you try to operate the source into a load it's not intended to handle. *That's true even when the net power delivered to the load is considerably LESS than the rated output power of the source. *Wim's example of the class-E amplifier is true enough, but it's not necessary to ask the source to deliver more net load power than it's rated to deliver, to establish conditions that cause trouble. *Thus, even sources that have an output impedance at or very close to the rated load impedance will have circuits to protect against loads that could destroy things inside the amplifier. I'm going to slip a mickey into this by a careful, editorial change of focus BACK to the subject line. Though it's a red herring typical of Ham-speak, most modern retail 100W RF transmitters for amateur service *have a very low output impedance, a fraction of an ohm [ editorial: until, of course, it goes to the Z transformer that precedes the bandpass filter]. It must be a rare condition for which there is a transition frequency below which audio amps source rated at sub Ohm Z feeding loads up to 10 times the sub-Ohm design work - and above which retail Amateur transmitters at loads up to 10 times the sub-Ohm design do not work [editorial: without that Z transformer]. *Curious logic. I wonder if that works (sic) backwards? *Would the RF deck's finals feed a speaker with audio with equal power performance of the Audio amp? *Both are sub-Ohm sources, and the RF deck certainly has the bandwidth. * I await the specification of this transition frequency and the analysis of how the 100W RF final deck would fail to supply 100W audio (however crummy it may sound). Any problems that might arise for the retail Ham transmitter are already handled by protection circuitry so that's a wash. 73's Richard Clark, KB7QHC Yes Wim, I understand your black box, and the voltage-divider action that occurs when terminated into 50 ohms. But you didn't answer my question, as you normally don't do. So I'll ask it again, as I did in my previous post: "( Vout = 212.1*50/(50+100) = 70.7V," I asked for the mathematical basis for this equation. I agree that it's mathematically correct, but 212.1 is voltage, or E, but where does 212.1*50 come from? The equation appears to be of the form E/R = E, which is absurd. In what equational form is this equation? I'm trying to learn here. Walt " |
Transmitter Output Impedance
On 5/13/2011 8:08 PM, walt wrote:
On May 13, 8:57 pm, Richard wrote: On Fri, 13 May 2011 16:28:52 -0700 (PDT), wrote: Though it's a red herring typical of audio-speak, most modern high fidelity audio amplifiers have a very low output impedance, a small fraction of an ohm, so they can claim a high damping factor. Off topic: the reason it's a red herring is that the impedance of the speaker connected to the amplifier must be included to figure the damping, and that impedance (even just the DC resistance of voice coils) changes by considerably more than the amplifier's output impedance (resistance) just because of heating on audio peaks. Especially when reactive components (inductors and capacitors) couple power between a source and a load, you can get stresses--voltages and/ or currents--well beyond what's safe when you try to operate the source into a load it's not intended to handle. That's true even when the net power delivered to the load is considerably LESS than the rated output power of the source. Wim's example of the class-E amplifier is true enough, but it's not necessary to ask the source to deliver more net load power than it's rated to deliver, to establish conditions that cause trouble. Thus, even sources that have an output impedance at or very close to the rated load impedance will have circuits to protect against loads that could destroy things inside the amplifier. I'm going to slip a mickey into this by a careful, editorial change of focus BACK to the subject line. Though it's a red herring typical of Ham-speak, most modern retail 100W RF transmitters for amateur service have a very low output impedance, a fraction of an ohm [ editorial: until, of course, it goes to the Z transformer that precedes the bandpass filter]. It must be a rare condition for which there is a transition frequency below which audio amps source rated at sub Ohm Z feeding loads up to 10 times the sub-Ohm design work - and above which retail Amateur transmitters at loads up to 10 times the sub-Ohm design do not work [editorial: without that Z transformer]. Curious logic. I wonder if that works (sic) backwards? Would the RF deck's finals feed a speaker with audio with equal power performance of the Audio amp? Both are sub-Ohm sources, and the RF deck certainly has the bandwidth. I await the specification of this transition frequency and the analysis of how the 100W RF final deck would fail to supply 100W audio (however crummy it may sound). Any problems that might arise for the retail Ham transmitter are already handled by protection circuitry so that's a wash. 73's Richard Clark, KB7QHC Yes Wim, I understand your black box, and the voltage-divider action that occurs when terminated into 50 ohms. But you didn't answer my question, as you normally don't do. So I'll ask it again, as I did in my previous post: "( Vout = 212.1*50/(50+100) = 70.7V," I asked for the mathematical basis for this equation. I agree that it's mathematically correct, but 212.1 is voltage, or E, but where does 212.1*50 come from? The equation appears to be of the form E/R = E, which is absurd. In what equational form is this equation? I'm trying to learn here. Walt " I apologize for stepping in, Walt, but, maybe this will help: Vout = 212.1 * (50/(50+100)) In other words, Vout is equal to Vsource times the divider ratio. I know you know this. I think you just misread it. Cheers, John |
Transmitter Output Impedance
On 14 mayo, 03:08, walt wrote:
On May 13, 8:57*pm, Richard Clark wrote: On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote: Though it's a red herring typical of audio-speak, most modern high fidelity audio amplifiers have a very low output impedance, a small fraction of an ohm, so they can claim a high damping factor. *Off topic: *the reason it's a red herring is that the impedance of the speaker connected to the amplifier must be included to figure the damping, and that impedance (even just the DC resistance of voice coils) changes by considerably more than the amplifier's output impedance (resistance) just because of heating on audio peaks. Especially when reactive components (inductors and capacitors) couple power between a source and a load, you can get stresses--voltages and/ or currents--well beyond what's safe when you try to operate the source into a load it's not intended to handle. *That's true even when the net power delivered to the load is considerably LESS than the rated output power of the source. *Wim's example of the class-E amplifier is true enough, but it's not necessary to ask the source to deliver more net load power than it's rated to deliver, to establish conditions that cause trouble. *Thus, even sources that have an output impedance at or very close to the rated load impedance will have circuits to protect against loads that could destroy things inside the amplifier. I'm going to slip a mickey into this by a careful, editorial change of focus BACK to the subject line. Though it's a red herring typical of Ham-speak, most modern retail 100W RF transmitters for amateur service *have a very low output impedance, a fraction of an ohm [ editorial: until, of course, it goes to the Z transformer that precedes the bandpass filter]. It must be a rare condition for which there is a transition frequency below which audio amps source rated at sub Ohm Z feeding loads up to 10 times the sub-Ohm design work - and above which retail Amateur transmitters at loads up to 10 times the sub-Ohm design do not work [editorial: without that Z transformer]. *Curious logic. I wonder if that works (sic) backwards? *Would the RF deck's finals feed a speaker with audio with equal power performance of the Audio amp? *Both are sub-Ohm sources, and the RF deck certainly has the bandwidth. * I await the specification of this transition frequency and the analysis of how the 100W RF final deck would fail to supply 100W audio (however crummy it may sound). Any problems that might arise for the retail Ham transmitter are already handled by protection circuitry so that's a wash. 73's Richard Clark, KB7QHC Yes Wim, I understand your black box, and the voltage-divider action that occurs when terminated into 50 ohms. But you didn't answer my question, as you normally don't do. So I'll ask it again, as I did in my previous post: "( Vout = 212.1*50/(50+100) = 70.7V," *I asked for the mathematical basis for this equation. I agree that it's mathematically correct, but 212.1 is voltage, or E, but where does 212.1*50 come from? The equation appears to be of the form E/R = E, which is absurd. In what equational form is this equation? I'm trying to learn here. Walt " Walt, First you should change language, because if you don't understand something, it is not always the other party who is wrong. For a voltage divider: Vout = Vin*R1/(R1+R2). R1 is the lower resistor , here 50 Ohms, R2 is the upper resistor, here 100 Ohms (inside the black box). I can't make it easier. Vin = 212.1 Vrms. Wim |
Transmitter Output Impedance
On May 13, 9:17*pm, Wimpie wrote:
On 14 mayo, 03:08, walt wrote: On May 13, 8:57*pm, Richard Clark wrote: On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote: Though it's a red herring typical of audio-speak, most modern high fidelity audio amplifiers have a very low output impedance, a small fraction of an ohm, so they can claim a high damping factor. *Off topic: *the reason it's a red herring is that the impedance of the speaker connected to the amplifier must be included to figure the damping, and that impedance (even just the DC resistance of voice coils) changes by considerably more than the amplifier's output impedance (resistance) just because of heating on audio peaks. Especially when reactive components (inductors and capacitors) couple power between a source and a load, you can get stresses--voltages and/ or currents--well beyond what's safe when you try to operate the source into a load it's not intended to handle. *That's true even when the net power delivered to the load is considerably LESS than the rated output power of the source. *Wim's example of the class-E amplifier is true enough, but it's not necessary to ask the source to deliver more net load power than it's rated to deliver, to establish conditions that cause trouble. *Thus, even sources that have an output impedance at or very close to the rated load impedance will have circuits to protect against loads that could destroy things inside the amplifier. I'm going to slip a mickey into this by a careful, editorial change of focus BACK to the subject line. Though it's a red herring typical of Ham-speak, most modern retail 100W RF transmitters for amateur service *have a very low output impedance, a fraction of an ohm [ editorial: until, of course, it goes to the Z transformer that precedes the bandpass filter]. It must be a rare condition for which there is a transition frequency below which audio amps source rated at sub Ohm Z feeding loads up to 10 times the sub-Ohm design work - and above which retail Amateur transmitters at loads up to 10 times the sub-Ohm design do not work [editorial: without that Z transformer]. *Curious logic. I wonder if that works (sic) backwards? *Would the RF deck's finals feed a speaker with audio with equal power performance of the Audio amp? *Both are sub-Ohm sources, and the RF deck certainly has the bandwidth. * I await the specification of this transition frequency and the analysis of how the 100W RF final deck would fail to supply 100W audio (however crummy it may sound). Any problems that might arise for the retail Ham transmitter are already handled by protection circuitry so that's a wash. 73's Richard Clark, KB7QHC Yes Wim, I understand your black box, and the voltage-divider action that occurs when terminated into 50 ohms. But you didn't answer my question, as you normally don't do. So I'll ask it again, as I did in my previous post: "( Vout = 212.1*50/(50+100) = 70.7V," *I asked for the mathematical basis for this equation. I agree that it's mathematically correct, but 212.1 is voltage, or E, but where does 212.1*50 come from? The equation appears to be of the form E/R = E, which is absurd. In what equational form is this equation? I'm trying to learn here. Walt " Walt, First you should change language, because if you don't understand something, it is not always the other party who is wrong. For a voltage divider: Vout = Vin*R1/(R1+R2). R1 is the lower resistor , here 50 Ohms, R2 is the upper resistor, here 100 Ohms (inside the black box). *I can't make it easier. Vin = 212.1 Vrms. Wim Thank you for the clarification, Wim. I was unaware of that equation relative to voltage dividers. I'm going to have to do some thinking on that equation before I'll understand the relevance of the numerator, or how it relates to the calculation. Walt |
Transmitter Output Impedance
On May 13, 9:36*pm, walt wrote:
On May 13, 9:17*pm, Wimpie wrote: On 14 mayo, 03:08, walt wrote: On May 13, 8:57*pm, Richard Clark wrote: On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote: Though it's a red herring typical of audio-speak, most modern high fidelity audio amplifiers have a very low output impedance, a small fraction of an ohm, so they can claim a high damping factor. *Off topic: *the reason it's a red herring is that the impedance of the speaker connected to the amplifier must be included to figure the damping, and that impedance (even just the DC resistance of voice coils) changes by considerably more than the amplifier's output impedance (resistance) just because of heating on audio peaks. Especially when reactive components (inductors and capacitors) couple power between a source and a load, you can get stresses--voltages and/ or currents--well beyond what's safe when you try to operate the source into a load it's not intended to handle. *That's true even when the net power delivered to the load is considerably LESS than the rated output power of the source. *Wim's example of the class-E amplifier is true enough, but it's not necessary to ask the source to deliver more net load power than it's rated to deliver, to establish conditions that cause trouble. *Thus, even sources that have an output impedance at or very close to the rated load impedance will have circuits to protect against loads that could destroy things inside the amplifier. I'm going to slip a mickey into this by a careful, editorial change of focus BACK to the subject line. Though it's a red herring typical of Ham-speak, most modern retail 100W RF transmitters for amateur service *have a very low output impedance, a fraction of an ohm [ editorial: until, of course, it goes to the Z transformer that precedes the bandpass filter]. It must be a rare condition for which there is a transition frequency below which audio amps source rated at sub Ohm Z feeding loads up to 10 times the sub-Ohm design work - and above which retail Amateur transmitters at loads up to 10 times the sub-Ohm design do not work [editorial: without that Z transformer]. *Curious logic. I wonder if that works (sic) backwards? *Would the RF deck's finals feed a speaker with audio with equal power performance of the Audio amp? *Both are sub-Ohm sources, and the RF deck certainly has the bandwidth. * I await the specification of this transition frequency and the analysis of how the 100W RF final deck would fail to supply 100W audio (however crummy it may sound). Any problems that might arise for the retail Ham transmitter are already handled by protection circuitry so that's a wash. 73's Richard Clark, KB7QHC Yes Wim, I understand your black box, and the voltage-divider action that occurs when terminated into 50 ohms. But you didn't answer my question, as you normally don't do. So I'll ask it again, as I did in my previous post: "( Vout = 212.1*50/(50+100) = 70.7V," *I asked for the mathematical basis for this equation. I agree that it's mathematically correct, but 212.1 is voltage, or E, but where does 212.1*50 come from? The equation appears to be of the form E/R = E, which is absurd. In what equational form is this equation? I'm trying to learn here. Walt " Walt, First you should change language, because if you don't understand something, it is not always the other party who is wrong. For a voltage divider: Vout = Vin*R1/(R1+R2). R1 is the lower resistor , here 50 Ohms, R2 is the upper resistor, here 100 Ohms (inside the black box). *I can't make it easier. Vin = 212.1 Vrms. Wim Thank you for the clarification, Wim. I was unaware of that equation relative to voltage dividers. I'm going to have to do some thinking on that equation before I'll understand the relevance of the numerator, or how it relates to the calculation. Walt Thank you for stepping in, John, as from you and Wim I understand that that equation represents the voltage-divider circuit. However, I'm still agonizing over the numerator in the ratio. I just can't fit it into the voltage-divider concept. Gimme a little time and it will probably sink in. Walt |
Transmitter Output Impedance
On Fri, 13 May 2011 17:57:27 -0700, Richard Clark
wrote: I await the specification of this transition frequency and the analysis of how the 100W RF final deck would fail to supply 100W audio (however crummy it may sound). :-) That was way too hard. |
Transmitter Output Impedance
On May 13, 5:57*pm, Richard Clark wrote:
On Fri, 13 May 2011 16:28:52 -0700 (PDT), K7ITM wrote: Though it's a red herring typical of audio-speak, most modern high fidelity audio amplifiers have a very low output impedance, a small fraction of an ohm, so they can claim a high damping factor. *Off topic: *the reason it's a red herring is that the impedance of the speaker connected to the amplifier must be included to figure the damping, and that impedance (even just the DC resistance of voice coils) changes by considerably more than the amplifier's output impedance (resistance) just because of heating on audio peaks. Especially when reactive components (inductors and capacitors) couple power between a source and a load, you can get stresses--voltages and/ or currents--well beyond what's safe when you try to operate the source into a load it's not intended to handle. *That's true even when the net power delivered to the load is considerably LESS than the rated output power of the source. *Wim's example of the class-E amplifier is true enough, but it's not necessary to ask the source to deliver more net load power than it's rated to deliver, to establish conditions that cause trouble. *Thus, even sources that have an output impedance at or very close to the rated load impedance will have circuits to protect against loads that could destroy things inside the amplifier. I'm going to slip a mickey into this by a careful, editorial change of focus BACK to the subject line. Though it's a red herring typical of Ham-speak, most modern retail 100W RF transmitters for amateur service *have a very low output impedance, a fraction of an ohm [ editorial: until, of course, it goes to the Z transformer that precedes the bandpass filter]. It must be a rare condition for which there is a transition frequency below which audio amps source rated at sub Ohm Z feeding loads up to 10 times the sub-Ohm design work - and above which retail Amateur transmitters at loads up to 10 times the sub-Ohm design do not work [editorial: without that Z transformer]. *Curious logic. I wonder if that works (sic) backwards? *Would the RF deck's finals feed a speaker with audio with equal power performance of the Audio amp? *Both are sub-Ohm sources, and the RF deck certainly has the bandwidth. * I await the specification of this transition frequency and the analysis of how the 100W RF final deck would fail to supply 100W audio (however crummy it may sound). Any problems that might arise for the retail Ham transmitter are already handled by protection circuitry so that's a wash. 73's Richard Clark, KB7QHC FWIW, I've built audio amplifiers with output transistors with f-sub-t around 50MHz. Those transistors certainly would provide decent power gain at the lower frequency HF ham bands. They were used in the audio amp to achieve very low distortion across the audio spectrum (a very few PPM at full rated power output). One big difference between (typical) audio amplifiers and (typical) RF power amplifiers is in the use of negative feedback. In an audio amplifier, voltage-derived negative feedback yields a very low amplifier output impedance. In almost all RF power amplifiers, little or no negative feedback is used, so the output impedance is generally much higher than with an audio amplifier running similar power and supply voltage to the output devices. But again, the source impedance of a transmitter is seldom important in the application of the amplifier. What's important are things like the optimal load impedance and the rated power output. I'd put things like distortion specs for a linear amp far above source impedance in importance. If you (the lurking reader) think source impedance is important, please explain in detail _why_. Cheers, Tom |
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