Home |
Search |
Today's Posts |
|
#1
![]() |
|||
|
|||
![]()
OK, if we shine a flashlight at a mirror the light bounces back and what
ever is caught by the reflector will be reflected. Take away the reflector and the reflection just keeps going. If someone can tell me what the hot filament does perhaps I can understand what happens in the finals, or whatever. tnx Hank WD5JFR "Cecil Moore" wrote in message ... Henry Kolesnik wrote: I just want to know the reflection physics in the Tx, no antenna tuner, just a mismatched antenna. I recall no pysics book that tell me how the reflection sees the transmitter. The reflected waves obey the laws of physics. The kicker is that we don't know (and apparently cannot directly measure) the source impedance. What the reflections can do is modify the designed-for load line through superposition of the forward and reflected waves. Modification of the designed-for load line is not desirable and, if unprotected, can cause over-voltage, over-current, or phase problems. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#2
![]() |
|||
|
|||
![]()
Henry Kolesnik wrote:
OK, if we shine a flashlight at a mirror the light bounces back and what ever is caught by the reflector will be reflected. Take away the reflector and the reflection just keeps going. If someone can tell me what the hot filament does perhaps I can understand what happens in the finals, or whatever. The hot filament provides a source of electrons. How many electrons are emitted depends on the instantaneous voltages on the other elements, the plate and grids. Reflected waves have an effect on those instantaneous voltages but IMO, there's not much sense in pursuing the "filament" line of reasoning. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#3
![]() |
|||
|
|||
![]()
I thought filaments produced photons/light waves as well. You took me to
light now you want to leave! Come on I just want a good basic understand on what it is the the Tx reflects the power , how it does it and a little simple math. My dad used to say if you can't explain something you think you know to someone else it might because you don't understand it yourself or lack command of the language. In this case for me it's both. 73 Hank WD5JFR "Cecil Moore" wrote in message ... Henry Kolesnik wrote: OK, if we shine a flashlight at a mirror the light bounces back and what ever is caught by the reflector will be reflected. Take away the reflector and the reflection just keeps going. If someone can tell me what the hot filament does perhaps I can understand what happens in the finals, or whatever. The hot filament provides a source of electrons. How many electrons are emitted depends on the instantaneous voltages on the other elements, the plate and grids. Reflected waves have an effect on those instantaneous voltages but IMO, there's not much sense in pursuing the "filament" line of reasoning. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#4
![]() |
|||
|
|||
![]()
Henry Kolesnik wrote:
I thought filaments produced photons/light waves as well. You took me to light now you want to leave! Well, the light emitted by tube filaments is irrelevant to the RF function. I use light examples because light and RF are both EM waves and a lot more is known about light than about RF. Come on I just want a good basic understand on what it is the the Tx reflects the power , how it does it and a little simple math. My dad used to say if you can't explain something you think you know to someone else it might because you don't understand it yourself or lack command of the language. In this case for me it's both. What I am trying to say is that I don't know what happens inside a transmitter. Under Z0-matched conditions, as with a tuner, I believe that what happens inside a transmitter is pretty much irrelevant. The transmitter sees its designed-for impedance and that is essentially all that matters. What I am willing to discuss in detail is what happens at a Z0-match point (x) in an antenna system with reflections - something like the following: XMTR---51.5 ohm line---x---1/4WL 300 ohm line---1749 ohm load 100W -- 200W -- -- 0W -- 100W For these typical conditions, all voltages and currents are either in-phase or 180 degrees out of phase at the match point (x), which makes a power analysis the most simple analysis of all. -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#5
![]() |
|||
|
|||
![]() "Cecil Moore" wrote in message ... What I am willing to discuss in detail is what happens at a Z0-match point (x) in an antenna system with reflections - something like the following: what you are willing to discuss is irrelevent as it has nothing to do with the original topic which was about what happens in the transmitter. For these typical conditions, all voltages and currents are either in-phase or 180 degrees out of phase at the match point (x), which makes a power analysis the most simple analysis of all. that should read "For these specific conditions", those conditions are hardly 'typical', they are a very exactly contrived example which makes it easy to compare powers. thus making the proper analysis of currents or voltages seem unnecessary. taking simple cases like this and improperly generalizing them is what leads to the worst mis-conceptions and circular arguments in this group. |
#6
![]() |
|||
|
|||
![]()
Dave wrote:
"Cecil Moore" wrote: What I am willing to discuss in detail is what happens at a Z0-match point (x) in an antenna system with reflections - something like the following: what you are willing to discuss is irrelevent as it has nothing to do with the original topic which was about what happens in the transmitter. Uhhhh Dave, the original topic is the Subject: line. If anything, what happens inside a transmitter is the irrelevant subject since appreciable reflections hardly ever reach the typical ham transmitter. For these typical conditions, all voltages and currents are either in-phase or 180 degrees out of phase at the match point (x), which makes a power analysis the most simple analysis of all. that should read "For these specific conditions", those conditions are hardly 'typical', they are a very exactly contrived example which makes it easy to compare powers. No, those are typical conditions, where the ham radio antenna system is tuned to a Z0-match by a tuner, either external or internal. It is not a "very exactly contrived example" at all. It is absolutely typical of any ham radio installation where the final amp sees close to a 1:1 SWR and that is the great majority. At the Z0-match point at the input of every properly tuned transmatch, the voltages and currents are either in phase or 180 degrees out of phase. If you don't know that, it is no wonder that you label my power analysis stuff as "contrived". -- 73, Cecil http://www.qsl.net/w5dxp -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#7
![]() |
|||
|
|||
![]() "Cecil Moore" wrote in message ... Dave wrote: "Cecil Moore" wrote: What I am willing to discuss in detail is what happens at a Z0-match point (x) in an antenna system with reflections - something like the following: what you are willing to discuss is irrelevent as it has nothing to do with the original topic which was about what happens in the transmitter. Uhhhh Dave, the original topic is the Subject: line. If anything, what happens inside a transmitter is the irrelevant subject since appreciable reflections hardly ever reach the typical ham transmitter. here is the original: I know that any power not dissipated by an antenna is reflected back to the transmitter. Then the transmitter "reflects" this reflection back to antenna, ad nauseum until its all gone. I also know that a short or an open is required to reflect power and I'm searching for which it is, an open or a short. I'm inclined to think it's a virtual open but I'm at a loss to understand that and I wonder if someone has a good explanation or analogy and some math wouldn't hurt. tnx Hank WD5JFR obviously not about what is going on at some contrived transmission line joint. For these typical conditions, all voltages and currents are either in-phase or 180 degrees out of phase at the match point (x), which makes a power analysis the most simple analysis of all. that should read "For these specific conditions", those conditions are hardly 'typical', they are a very exactly contrived example which makes it easy to compare powers. No, those are typical conditions, where the ham radio antenna system is tuned to a Z0-match by a tuner, either external or internal. It is not a "very exactly contrived example" at all. It is absolutely typical of any ham radio installation where the final amp sees close to a 1:1 SWR and that is the great majority. At the Z0-match point at the input of every properly tuned transmatch, the voltages and currents are either in phase or 180 degrees out of phase. If you don't know that, it is no wonder that you label my power analysis stuff as "contrived". of course it is contrived. no one uses loads of those exact impedances, or lengths of coax like you do. |
#8
![]() |
|||
|
|||
![]()
On Sun, 23 May 2004 17:18:52 GMT, "Henry Kolesnik"
wrote: OK, if we shine a flashlight at a mirror the light bounces back and what ever is caught by the reflector will be reflected. Take away the reflector and the reflection just keeps going. If someone can tell me what the hot filament does perhaps I can understand what happens in the finals, or whatever. tnx Hank WD5JFR Hi Hank, Well I see in correspondence following this as "replies," that they are far from satisfactory. All part of the arcana that precedes the convolutions of math you would have had to endure. The abandonment of this lead is simply a matter of Cecil's lack of experience in the metaphor of light. To answer your question above. Removing the reflector is unnecessary as it is part of the initial condition and has nothing to do with it serving as the correlative to a tuner that you want to remove from the argument (which is a perfectly acceptable imposition of conditions). Your question also goes to the heart of the matter. We begin with a hot filament which emits radiation (and yes, no reflector is required so we will skip that as one of your conditions). The amount of radiation is directly correlated to the amount of heat. This will simplify matters, but in the end it will yield a failure of metaphors (which always occur if you cannot bridge the logic). The radiation strikes a reflection (immaterial whether complete or partial) and that portion which returns, impinges upon the filament, the source. The filament absorbs the power, which in turn raises its temperature (everyday experience proves the heat of such radiation). This will, in turn, cause a higher radiation (given the quid-pro-quo of heat and radiation). In a sense, this means the reflected power is re-radiated. The confirmation of this is that if you achieved full reflection, you then define total insulation of the radiation (no heat escapes) and temperature rises accordingly, and this may lead to catastrophic failure of the filament (a very bright illumination if you could see it, and consequent fusing current - electric kilns use this principle but tolerate the current by under rating the source). You can imagine the correlative to transmitter failure for the same conditions. The failure of the metaphor? RF is not heat (common light is) and the return of power to be rendered into heat does not result in a higher RF output. I will anticipate the sophomore's comments that RF reflections do not become heat, in and of itself: The returned power (either through wave mechanics or lumped circuitry) must result in either a higher potential across the source, or a higher current through it. Elevated potentials yield the everyday experience of an arc (heat). Elevated currents yield the everyday experience of current density through the same element (much like the filament of our metaphor - heat). One failure mode comes with the peak power snap, followed by the muttering of "Oh ****!" Or it comes through the more progressive thermal runaway, followed by the muttering of "what's that funny smell?" 73's Richard Clark, KB7QHC |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
Rho = (Zload-Zo*)/(Zload+Zo), for complex Zo | Antenna | |||
Derivation of the Reflection Coefficient? | Antenna | |||
Length of Coax Affecting Incident Power to Meter? | Antenna |