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1/4 vs 1/2 wavelength antenna
Hi
I am building an rf transmitter for a short range data link at 433MHZ and am almost done, but I would like to understand better exactly what I am seeing with regard to antenna performance. All technical notes I have read recommend a 1/4 wave whip over ground plane as offering the best performance, statements like: "Best range is achieved with either a straight piece of wire, rod or PCB track @ 1/4 wavelength over a ground plane", I understand many factors effect performance however I have found that a "bent" 1/2 wavelength length of wire offers better performance. If I use a 1/4 wavelength I need (due to case requirements) to have two 90 degree bends in it (feed - up, across, up). If I use a 1/2 wavelength I need to run it once around the (plastic) case (feed - up, around the case, up). I hope this makes some sense, anyway I have found the 1/2 wave is less effected by polarisation and offers generally better performance. However while more ground plane may help a 1/4 wave it seems to hinder the 1/2 wave, I guess because it shields the loop around the case? Regards |
"Nug" wrote in message
m... Hi I am building an rf transmitter for a short range data link at 433MHZ and am almost done, but I would like to understand better exactly what I am seeing with regard to antenna performance. All technical notes I have read recommend a 1/4 wave whip over ground plane as offering the best performance, statements like: "Best range is achieved with either a straight piece of wire, rod or PCB track @ 1/4 wavelength over a ground plane", I understand many factors effect performance however I have found that a "bent" 1/2 wavelength length of wire offers better performance. If I use a 1/4 wavelength I need (due to case requirements) to have two 90 degree bends in it (feed - up, across, up). If I use a 1/2 wavelength I need to run it once around the (plastic) case (feed - up, around the case, up). I hope this makes some sense, anyway I have found the 1/2 wave is less effected by polarisation and offers generally better performance. However while more ground plane may help a 1/4 wave it seems to hinder the 1/2 wave, I guess because it shields the loop around the case? A 1/4 wavelength antenna really needs to be straight and at right angles to the ground plane. That is probably why the 1/2 wavelength antenna works better in your case. Leon -- Leon Heller, G1HSM http://www.geocities.com/leon_heller |
I would make the grandiose statement that since you are bending the wire
it no longer exhibits the performance of a "standard" 1/4 or 1/2 wave antenna. I would suggest that if you indeed made a 1/4 wave GP that protruded from the box surface (with a suitable counterpoise) it would outperform the 1/2 wave bent one.. Assuming you have to put the antenna inside the box or wrapped around it I suggest you look into tuning it with some C and/or L. In that case you would construct the antenna to fit your case parameters and adjust the matching for best radiation. Keep in mind that the C/L tuning components could be lengths of coax and open feeder/wire. (because of the high operating freq) Cheers Bob VK2YQA (Sydney Australia) Nug wrote: Hi I am building an rf transmitter for a short range data link at 433MHZ and am almost done, but I would like to understand better exactly what I am seeing with regard to antenna performance. All technical notes I have read recommend a 1/4 wave whip over ground plane as offering the best performance, statements like: "Best range is achieved with either a straight piece of wire, rod or PCB track @ 1/4 wavelength over a ground plane", I understand many factors effect performance however I have found that a "bent" 1/2 wavelength length of wire offers better performance. If I use a 1/4 wavelength I need (due to case requirements) to have two 90 degree bends in it (feed - up, across, up). If I use a 1/2 wavelength I need to run it once around the (plastic) case (feed - up, around the case, up). I hope this makes some sense, anyway I have found the 1/2 wave is less effected by polarisation and offers generally better performance. However while more ground plane may help a 1/4 wave it seems to hinder the 1/2 wave, I guess because it shields the loop around the case? Regards |
On Sun, 20 Feb 2005 22:09:15 -0800, Nug wrote:
Hi I am building an rf transmitter for a short range data link at 433MHZ and am almost done, but I would like to understand better exactly what I am seeing with regard to antenna performance. All technical notes I have read recommend a 1/4 wave whip over ground plane as offering the best performance, statements like: "Best range is achieved with either a straight piece of wire, rod or PCB track @ 1/4 wavelength over a ground plane", I understand many factors effect performance however I have found that a "bent" 1/2 wavelength length of wire offers better performance. If I use a 1/4 wavelength I need (due to case requirements) to have two 90 degree bends in it (feed - up, across, up). If I use a 1/2 wavelength I need to run it once around the (plastic) case (feed - up, around the case, up). I hope this makes some sense, anyway I have found the 1/2 wave is less effected by polarisation and offers generally better performance. However while more ground plane may help a 1/4 wave it seems to hinder the 1/2 wave, I guess because it shields the loop around the case? A 1/4 wave antenna will match to a low impedance, unbalanced. A 1/2 wave dipole will match to a low impedance, balanced. A 1/2 wave piece of wire fed at the end will match to a high impedance. What kind of circuit are you using for your output? Thanks, Rich |
In article ,
Nug wrote: Hi I am building an rf transmitter for a short range data link at 433MHZ and am almost done, but I would like to understand better exactly what I am seeing with regard to antenna performance. [.. 1/4 wave and 1/2 wave ...] An antenna looks like an LC tuned circuit loaded by the radiation resistance. Your output stage has some impedance that correctly matches to it (there are exceptions we will ignore) and it is this impedance you want the antenna system to have. When the correct matching is done, the antenna works as an impedance mathcing network that matches the output stages impedance to the radiation resistance. The normal (90 degrees to) 1/4 wave whip over a ground plane is one half of a dipole that is 1/2 wave length. The ground plane operates like a mirror. The electrostatic lines of force follow the same path with the mirroring as they would if the other 1/2 of the dipole was there. This lets you use a smaller (1/4 wave) antenna to get the same effect as the 1/2 wave. In your case, you are not using a whip antenna. If I've read what you wrote correctly, the antenna spends more of its length parallel to the surface of the PCB than it does running 90 degrees away from it. You have some circuit with a ground plane and a limitted sized box to work with, so the mechanical shape is constained by the box and not the ideal electronics. Since the box is small: If you have the equipment to do so, I suggest you measure (estimate) the impedance of the longest single loop of wire that will fit within the case. ie: connect to both ends. You have to have the electronics PCB in the case when you do this. If you are very lucky, its impedance will not be too hard to match to the output stage. -- -- forging knowledge |
Nug wrote:
"I am building an rf transmitter for a short range data link at 433 MHz and am almost done, but I would like to understand better exactly what I am seeing with regard to antenna performance." Your wavelength is about 0.69 meter or about 2.3 feet. In antennas everything depends on wavelength. If you use a transmitter housing as the ground plane for your antenna, it needs to be a sizeable part of a wavelength or the salient part of your antenna must be longer to compensate for the small ground plane. If you had an infinite ground plane, a 1/2-wave wire perpendicular to it would produce up to 50% more volts per meter field strength than a 1/4-wave wire perpendicular to the ground plane. It`s not something for nothing. Total radiation is the same in both cases. More of the radiation is perpendicular to the wire in the 1/2-wave and less goes off at some other angle to the wire. 50% more field in some particular direction is realy not very significant in most cases, and there are other consequences of using a 1/2-wave wire instead of a 1/4-wave wire. An end-driven 1/2-wave wire presents a very high impedance. It is equivalent to a parallel-resonant circuit. It would match a direct connection to a parallel resonant tank circuit perhaps. An end-driven 1/4-wave wire presents a very low impedance when worked against a ground plane, maybe about 30 ohms. How well you are able to radiate a signal from a wire is likely to depend on how well it is matched to the transmitter and less about the bends in the wire. In any case the complete antenna must be resonant to eliminate reactance which opposes the signal`s entry into the wire. For a small transmitter operating at a very short wavelength, the size of the antenna is not onerous and it would be possible to use a center-fed 1/2-wave antenna. Each half would be just a little over a half foot in length. Drivepoint impedance is in the 70-ohm range. Another possibility is a full-wave loop, about 2.3 feet in perimeter with a drivepoint impedance of about 120 ohms. Performance of all the suggestions is probably about the same. You can find the best by trying them. Best regards, Richard Harrison, KB5WZI |
Thanks for all your responses, good stuff here.
Rich, I am using TX modules from Liapac which specify a 50 ohm load. So putting it all together (I think) by bending the antenna (both 1/4 and 1/2) I am messing with its impedance and subsequent matching? So I would be better off with a straight 1/4 over ground plane if I had the choice. From reading I've done I think that a bend from vertical to horizontal (closer to ground) would be decreasing the antenna impedance, does that sound correct? Would I benefit from adjusting (extending) the antenna length slightly (I don't think so because it would no longer be at wavelength, but I am not sure?). Ken said ... "If I've read what you wrote correctly, the antenna spends more of its length parallel to the surface of the PCB than it does running 90 degrees away from it"... Actually it's about half and half (feed up 15mm , loop around the case, up - {clear of case} 130mm). Note the horizontal loop does not cross over itself. Sort of like this (~~ is horizontal loop) side view. | | | ~~~~ | feed I understand that more of the radiation is perpendicular to the wire in the 1/2-wave (than 1/4 wave), broadly speaking how will the bend('s) in the aerial effect radiation pattern (I understand this is tough to answer)? I can't fit a full loop inside the case. I have no rf test equipment so can only use trial and error, thanks again to all who responded. As I said it actually works fine it's just I don't like not understanding the reason its working, and would like to make any small changes that may improve performance. Thanks Again |
Ken Smith wrote:
In article , Nug wrote: Hi I am building an rf transmitter for a short range data link at 433MHZ and am almost done, but I would like to understand better exactly what I am seeing with regard to antenna performance. [.. 1/4 wave and 1/2 wave ...] An antenna looks like an LC tuned circuit loaded by the radiation resistance. Your output stage has some impedance that correctly matches to it (there are exceptions we will ignore) and it is this impedance you want the antenna system to have. When the correct matching is done, the antenna works as an impedance mathcing network that matches the output stages impedance to the radiation resistance. RF transmitters are not impedance matched to antennae in the sense of maximum transfer of power. "Maximum transfer of power" is a small signal (ideal linear parameters) issue, not a large signal issue. That is, the antenna/load are not conjugately matched. What is said, is that a TX'er will deliver some given power into, for example, 50 ohms. This says nothing about the output impedance of the PA. Power amplifiers are concerned with DC input power to RF output power efficiency, thus they are load-line "matched," not impedance matched. The concept of "output impedance" breaks down for large signal devices. For example, what is the output impedance of a class C or D amp taken when the transistor is on or off? I suppose one could consider the time-averaged impedance, but I'm not sure of the utility (to be fair, the time-averaged reactive output component is tuned out as best possible). The vague output impedance is a problem even for large signal class A devices. Again, RF PA's should be load-line matched. Output-Z is irrelevent. |
On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote:
RF transmitters are not .... Sorry OM, This was all nonsense. 73's Richard Clark, KB7QHC |
G. White wrote:
"Output Z is irrelevant." This is an old argument in this newsgroup. I became convinced long ago that there are cases in which impedance is very important. "Transmission Lines, Antennas, and Wave Guides" by King, Mimno, and Wing make a clear and concise case for the princuple of conjugates in impedance matching on page 43: "If a dissipationless network is inserted between a constant-voltage generator of impedance Zg, and a load of impedance ZR such that maximum power is delivered to the load, at every pair of terminals the impedance looking in opposite directions are conjugates of each other. To secure maximum power output from a generator whose emf and internal impedance are constant the load must have an impedance equal to the conjugate of the generator`s internal impedance." Radio transmitters don`t produce significant harmonics. It`s the law. They are linear power sources. We can and do tune them for all the power they will produce under their particular operating conditions of drive and d-c power supply. They operate at more than 50% efficiency which means that they don`t take power 100% of the time, but are switched-off during part of the r-f cycle. Output impedance is thus an average over the entire cycle. It`s OK. We have no harmonics. Gaps are filled by the tank circuit and other filters. The radio is a proper source. The impedance added by off-time is called "dissipationless resistance" because no power is lost in the radio while it is switched-off. Best regards, Richard Harrison, KB5WZI |
Richard Clark wrote:
On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote: RF transmitters are not .... Sorry OM, This was all nonsense. Nice articulation. I don't know who OM is, but RF transmitter power amps are not "impedance matched." Neither are audio power amps for that matter. |
In article , gwhite wrote:
Richard Clark wrote: On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote: RF transmitters are not .... Sorry OM, This was all nonsense. Nice articulation. I don't know who OM is, but RF transmitter power amps are not "impedance matched." Neither are audio power amps for that matter. "OM" is an amateur radio term. It is short for "Old Man". It is a respectful term for all other males that is quick to transmit via Morse code. Richard Clark appears to be an amateur radio operator or the like. RF transmitter power amps are certainly "impedance matched" to the intended load. Take a look in the ARRL "The radio amateur's handbook". If you have the 1944 addition, you will need to start reading at page 96 in the lower right column. If you don't have that, try Motorola's AN-721. As for audio amp, you are 1 for 3 my friend. -- -- forging knowledge |
On Fri, 25 Feb 2005 03:17:12 GMT, gwhite wrote:
RF transmitter power amps are not "impedance matched." Neither are audio power amps for that matter. Hi OM, You seem to be shy of facts and long on claims. Got any experience at the bench, or is this all arm-chair philosophy? 73's Richard Clark, KB7QHC |
Dear gwhite [no call, no location]:
Notwithstanding the clear limitations on making conclusions about what happens inside of a circuit that has been modeled using Thevenin's theorem, it is part of Religion that the least important theorem in circuit theory is applicable. Debates about Faith are a waste of energy. Avoid the tar-baby. 73 Mac N8TT -- J. Mc Laughlin; Michigan U.S.A. Home: "gwhite" |
Ken Smith wrote:
RF transmitter power amps are certainly "impedance matched" to the intended load. Take a look in the ARRL "The radio amateur's handbook". If you have the 1944 addition, you will need to start reading at page 96 in the lower right column. If you don't have that, try Motorola's AN-721. A CMOS Class-E amp is in full saturation (0.5v at 2a) for 10% of a cycle and off (12v at 0a) for the other 90% of a cycle. The tank circuit changes the digital energy to analog energy by filtering out everything except the fundamental frequency component. How in the world does one determine the steady-state impedance of the CMOS source? Isn't the best one can do with a digital switch is to keep it within specified parameters? The CMOS device dissipates 2 watts for 10% of the time - therefore 0.2 watts steady-state. -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Cecil Moore wrote:
The CMOS device dissipates 2 watts for 10% of the time - therefore 0.2 watts steady-state. Sorry, should have been: The CMOS device dissipates one watt for 10% of the time - therefore 0.1 watts. ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
Ken Smith wrote:
RF transmitter power amps are certainly "impedance matched" to the intended load. Take a look in the ARRL "The radio amateur's handbook". If you have the 1944 addition, you will need to start reading at page 96 in the lower right column. If you don't have that, try Motorola's AN-721. It appears that I may have canceled an earlier posting by accident so will repeat it. A certain Class-E CMOS amp is in full saturation for 10% of a cycle, 0.5v at 2a. For the rest of the time it is off. The supply voltage is 12v. What is the steady-state impedance of the source at the fundamental frequency? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
On Fri, 25 Feb 2005 09:47:16 -0600, Cecil Moore
wrote: Ken Smith wrote: RF transmitter power amps are certainly "impedance matched" to the intended load. Take a look in the ARRL "The radio amateur's handbook". If you have the 1944 addition, you will need to start reading at page 96 in the lower right column. If you don't have that, try Motorola's AN-721. It appears that I may have canceled an earlier posting by accident so will repeat it. A certain Class-E CMOS amp is in full saturation for 10% of a cycle, 0.5v at 2a. For the rest of the time it is off. The supply voltage is 12v. What is the steady-state impedance of the source at the fundamental frequency? Now, now, Cecil! Don't sully the thread with facts !-) ...Jim Thompson -- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | | http://www.analog-innovations.com | 1962 | I love to cook with wine. Sometimes I even put it in the food. |
"gwhite" wrote in message ... Richard Clark wrote: On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote: RF transmitters are not .... Sorry OM, This was all nonsense. Nice articulation. I don't know who OM is, but RF transmitter power amps are not "impedance matched." Neither are audio power amps for that matter. My stereo amp has a spec on output impedance. As I recall, it was around 0.16 Ohms. Intended load is 4 - 16 Ohms. Tam |
Ken Smith wrote:
RF transmitter power amps are certainly "impedance matched" to the intended load. I'm sorry, but they are not. Nor are any power amps that I know of. Efficiency (and thus necessarily output swing) is what matters for power amps. To maximize swing requires load line matching, not impedance matching. If you want to study RF PA's I suppose Cripps is one of the best I know of for a modern text: http://www.amazon.com/exec/obidos/tg.../-/0890069891/ |
Cecil Moore wrote:
What is the steady-state impedance of the source at the fundamental frequency? Exactly Cecil. That's exactly why I said the concept of impedance is dubious for large signal devices and thus provides a second argument, albeit weaker, against the "impedance matching" idea for power amps. "The concept of 'output impedance' breaks down for large signal devices. For example, what is the output impedance of a class C or D amp taken when the transistor is on or off?" The strongest argument for dropping the impedance matching concept is PA efficiency, and therefore maximum signal swing. Obtaining maximum swing is a load line issue. |
In article , Cecil Moore wrote:
Ken Smith wrote: RF transmitter power amps are certainly "impedance matched" to the intended load. Take a look in the ARRL "The radio amateur's handbook". If you have the 1944 addition, you will need to start reading at page 96 in the lower right column. If you don't have that, try Motorola's AN-721. A CMOS Class-E amp is in full saturation (0.5v at 2a) for 10% of a cycle and off (12v at 0a) for the other 90% of a cycle. The tank circuit changes the digital energy to analog energy by filtering out everything except the fundamental frequency component. How in the world does one determine the steady-state impedance of the CMOS source? Isn't the best one can do with a digital switch is to keep it within specified parameters? The CMOS device dissipates 2 watts for 10% of the time - therefore 0.2 watts steady-state. For what you say here really to be true the transistors must switch very fast. About 25pS switching speed is needed at about 400KHz. If we take that to be the case however, I think you will see why matching still applies. Lets take the reactive component first. If there is a reactive component to the loading, the current in the switch will have a higher RMS value without that increase in RMS increasing the radiated power of the system. So the reactive component of the matching is fairly obvious. Imagine that you have a well designed Class-E circuit loaded with the load the designer optimized it for. Now imagine that you slightly increase the resistance slightly. When you do so, the current into the load will decrease but the voltage will not increase enough to compensate for this. Now lets assume that you slightly decrease the resistance. Since we are assuming that this is a well designed case, we can assume that the designer took steps to ensure that the output devices would be protected from excess currents. This could be done by reducing the operating voltage of the output section, for example. In any case, the voltage on the load will decrease by a larger factor than the current will increase. So it is obvious that the reactive part is matched and the resistive part is matched just as it would be in a non-class-E output section. -- -- forging knowledge |
In article , Cecil Moore wrote:
Ken Smith wrote: RF transmitter power amps are certainly "impedance matched" to the intended load. Take a look in the ARRL "The radio amateur's handbook". If you have the 1944 addition, you will need to start reading at page 96 in the lower right column. If you don't have that, try Motorola's AN-721. It appears that I may have canceled an earlier posting by accident so will repeat it. I answered your other copy of the posting. -- -- forging knowledge |
In article ,
Jim Thompson wrote: On Fri, 25 Feb 2005 09:47:16 -0600, Cecil Moore wrote: [...] A certain Class-E CMOS amp is in full saturation for 10% of a cycle, 0.5v at 2a. For the rest of the time it is off. The supply voltage is 12v. What is the steady-state impedance of the source at the fundamental frequency? Now, now, Cecil! Don't sully the thread with facts !-) This needs a 12V transistor that switches in about 25pS. Would that be likely on a current chip? -- -- forging knowledge |
In article , gwhite wrote:
Cecil Moore wrote: [...] "The concept of 'output impedance' breaks down for large signal devices. Motorola seems to disagree with you. Perhaps more importantly, I disagree with you. The large signal impedance of a transistor working into a tuned load still applies quite well. For example, what is the output impedance of a class C or D amp taken when the transistor is on or off?" When dealing with an RF output section you don't deal with just one part of the cycle. You can tell if the real part of the impedance is matched if: (1) If you increase the resistance of the load does the power decrease? AND (2) If you decrease the resistance of the load does the power decrease? The strongest argument for dropping the impedance matching concept is PA efficiency, and therefore maximum signal swing. Obtaining maximum swing is a load line issue. What do you mean by "maximum signal swing" in this context. I can get a bigger swing by leaving the output completely unloaded and hence causing the actual efficiency to be zero. The reactive component issue is still there too. Reactive loads cause increased currents in the output stage without delivering any power to the load so they still need to be reduced as much as practical. -- -- forging knowledge |
In article , gwhite wrote:
Ken Smith wrote: RF transmitter power amps are certainly "impedance matched" to the intended load. I'm sorry, but they are not. Nor are any power amps that I know of. Efficiency (and thus necessarily output swing) is what matters for power amps. To maximize swing requires load line matching, not impedance matching. I still say they are. Motorola AN-721 takes on the theory. AN-758 does a practical example matching 12.5 Ohms into 50 Ohms -- -- forging knowledge |
"Tam/WB2TT" wrote in message ... "gwhite" wrote in message ... Richard Clark wrote: On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote: RF transmitters are not .... Sorry OM, This was all nonsense. Nice articulation. I don't know who OM is, but RF transmitter power amps are not "impedance matched." Neither are audio power amps for that matter. My stereo amp has a spec on output impedance. As I recall, it was around 0.16 Ohms. Intended load is 4 - 16 Ohms. Tam Tam An audio amplifier can have an improved capability of minimizing a speaker's deviation from its desired displacement if it has a low output impedance. That is, the speaker's cone tends to swing past its desired displacement when its terminals are loaded to a high impedance. Jerry |
Ken Smith wrote:
In article , gwhite wrote: Ken Smith wrote: RF transmitter power amps are certainly "impedance matched" to the intended load. I'm sorry, but they are not. Nor are any power amps that I know of. Efficiency (and thus necessarily output swing) is what matters for power amps. To maximize swing requires load line matching, not impedance matching. I still say they are. Motorola AN-721 takes on the theory. AN-758 does a practical example matching 12.5 Ohms into 50 Ohms Sure are. A 30 year ago 500 level course I took called Non-Linear Transistor Design said so. The way you handle class C etc. is by handling each harmonic separately in your analysis of the transistor plus tank circuit. You match to the harmonic you want. It may be the fundamental, or the third for a tripler, etc. tom K0TAR |
Jerry Martes wrote:
"Tam/WB2TT" wrote in message ... "gwhite" wrote in message ... Richard Clark wrote: On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote: RF transmitters are not .... Sorry OM, This was all nonsense. Nice articulation. I don't know who OM is, but RF transmitter power amps are not "impedance matched." Neither are audio power amps for that matter. My stereo amp has a spec on output impedance. As I recall, it was around 0.16 Ohms. Intended load is 4 - 16 Ohms. Tam Tam An audio amplifier can have an improved capability of minimizing a speaker's deviation from its desired displacement if it has a low output impedance. That is, the speaker's cone tends to swing past its desired displacement when its terminals are loaded to a high impedance. Jerry Correct - in the case of an audio amplifier damping factor is critical to accurate reproduction of transients. In an RF amp, it is normally not required or desired. tom K0TAR |
On Sat, 26 Feb 2005 00:40:09 +0000, Ken Smith wrote:
In article , gwhite wrote: Ken Smith wrote: RF transmitter power amps are certainly "impedance matched" to the intended load. I'm sorry, but they are not. Nor are any power amps that I know of. Efficiency (and thus necessarily output swing) is what matters for power amps. To maximize swing requires load line matching, not impedance matching. I still say they are. Motorola AN-721 takes on the theory. AN-758 does a practical example matching 12.5 Ohms into 50 Ohms Evidently, the guy's never tuned up a 40 meter pi-net output transmitter. ;-) If that's not impedance matching, I don't know what it is! (Oh, "Load line" matching? What are the two parameters of the load line? Voltage and Current, right? What's the slope of the load line? Impedance!) Cheers! Rich |
On Fri, 25 Feb 2005 17:59:56 -0500, Tam/WB2TT wrote:
"gwhite" wrote in message ... Richard Clark wrote: On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote: RF transmitters are not .... Sorry OM, This was all nonsense. Nice articulation. I don't know who OM is, but RF transmitter power amps are not "impedance matched." Neither are audio power amps for that matter. My stereo amp has a spec on output impedance. As I recall, it was around 0.16 Ohms. Intended load is 4 - 16 Ohms. That works because the transmission line is less than 0.01 wavelength. So impedance matching becomes moot. If the speaker line were 1/4 wavelength long, there would be almost no signal transferred at all. Cheers! Rich |
Arguments about whether the power ammplifier is matched or not matched to 50
ohms arise due to misunderstandings about the meaning of "matched". Meaning 1. ------------- The PA has been designed for maximum, linear, undistorted power output when loaded with Ro ohms and the load reistance has actually been adjusted to equal Ro. Ro is usually 50 ohms. (There may be additional criteria to define what constitutes an optimum match.) Meaning 2. ------------- The load impedance Z = R+jX has been adjusted to equal the conjugate of the internal impedance resistance of the PA. (The internal impedance of the PA is usually unknown but the circuit is assumed to behave as if a conjugate match exists.) The two meanings are entirely different from each other. If there is danger of confusion then the meaning should be stated. Some people already use the descriptions "Zo match" and "Conjugate match". ---- Reg, G4FGQ |
In article ,
Rich Grise wrote: [... audio matching ..] That works because the transmission line is less than 0.01 wavelength. So impedance matching becomes moot. If the speaker line were 1/4 wavelength long, there would be almost no signal transferred at all. ( Unless you use the specially tapered euphonic cable! ) Actually, if you used a thick enough wire, you could go 1/4 wavelength at audio frequencies. If you connect a speaker straight onto the mains, chances are you are going further than that from the generator. -- -- forging knowledge |
In article ,
Reg Edwards g4fgq,regp@ZZZbtinternet,com wrote: Arguments about whether the power ammplifier is matched or not matched to 50 ohms arise due to misunderstandings about the meaning of "matched". Meaning 1. ------------- The PA has been designed for maximum, linear, undistorted power output when loaded with Ro ohms and the load reistance has actually been adjusted to equal Ro. Ro is usually 50 ohms. (There may be additional criteria to define what constitutes an optimum match.) Meaning 2. ------------- The load impedance Z = R+jX has been adjusted to equal the conjugate of the internal impedance resistance of the PA. (The internal impedance of the PA is usually unknown but the circuit is assumed to behave as if a conjugate match exists.) Meaning 3: The PA has been designed to deliver the maximum power at that load impedance and the distortion is not an issue. The two meanings are entirely different from each other. If there is danger of confusion then the meaning should be stated. Some people already use the descriptions "Zo match" and "Conjugate match". Actually meanings 2 and 3 are effectively equal in the case of the tuned system, if you define the Zo based on the change in output power vs connected impedance for small changes. Since a lot of such systems aren't linear, this is the way you end up having to define the impedance. You can't use open circuit voltage and short circuit current. Remember that this all started with the OP having a "transmitter". This would include any needed filtering. He was just connecting a 1/4 and 1/2 wave lengths of bent up wire. His output filter, I assume is just a bunch of LC sections. -- -- forging knowledge |
Ken Smith wrote:
"Motorola AN-721 takes on the theory. AN-758 gives a practical example matching 12.5 Ohms into 50 Ohms." As King, Mimno, and Wing said, a conjugate match is required to get maximum power out of the radio. Terman says the same on page 76 of his 1955 edition of "Electronic and Radio Engineering." There is also an Eimac application note on matching their tubes to a load on the final amplifier. If Larry, Moe, and Curly published an application note, we`d have a crowd of skeptics here. See Terman`s Fig. 3-21 on page 77 of his 1955 edition. Best regards, Richard Harrison, KB5WZI |
"Rich Grise" wrote in message ... On Fri, 25 Feb 2005 17:59:56 -0500, Tam/WB2TT wrote: "gwhite" wrote in message ... Richard Clark wrote: On Wed, 23 Feb 2005 19:08:20 GMT, gwhite wrote: RF transmitters are not .... Sorry OM, This was all nonsense. Nice articulation. I don't know who OM is, but RF transmitter power amps are not "impedance matched." Neither are audio power amps for that matter. My stereo amp has a spec on output impedance. As I recall, it was around 0.16 Ohms. Intended load is 4 - 16 Ohms. That works because the transmission line is less than 0.01 wavelength. So impedance matching becomes moot. If the speaker line were 1/4 wavelength long, there would be almost no signal transferred at all. Cheers! Rich There is nothing wrong with driving a transmission line/antenna from a zero impedance source. It does NOT change the SWR. The point is that an audio amplifier with a damping factor of 50 is NOT conjugate matched. Somebody mentioned Motorola Application note 721. This is what it says: ************************************************** ************************************** " ..the load, in first approximation, is not related to the device, except for VCE(sat). The load value is primarily dictated by the required output power and the peak voltage; it is not matched to the output impedance of the device. " ************************************************** ***************************************** When device people talk about "matching", they mean matching the load to what the transistor wants to see, which is not the conjugate of the output impedance. The way this is done is to build an amplifier, and vary the load until maximum output power is reached. The transistor is then removed, and the impedance looking into the coupling network is measured. The conjugate of this is sometimes listed as "output impedance" on data sheets. Newer data sheets will have an asterisk * next to that, and a note explaining what it means. If you look at Philips literature, you will see exactly the same explanation. This whole thing has been hashed out here about 5 times during the past year. Tam |
"Ken Smith" wrote in message ... In article , gwhite wrote: Ken Smith wrote: RF transmitter power amps are certainly "impedance matched" to the intended load. I'm sorry, but they are not. Nor are any power amps that I know of. Efficiency (and thus necessarily output swing) is what matters for power amps. To maximize swing requires load line matching, not impedance matching. I still say they are. Motorola AN-721 takes on the theory. AN-758 does a practical example matching 12.5 Ohms into 50 Ohms Ken, Motorola 721 specifically says the device IS NOT MATCHED. Tam -- -- forging knowledge |
If a conjugate match matters two hoots, or is involved in the slightest way
with the design of power amplifiers or any other sort of amplifier, why don't tube manufacturers state the internal resistance or impedance in tube date sheets? |
In article ,
Tam/WB2TT wrote: [...] Ken, Motorola 721 specifically says the device IS NOT MATCHED. It say " Strictly speaking..." is that the section you are refering to? Yes they do say that but then if you follow the design through, I think you will find that they try to cancel the reactive component. If you then put in the output device protection they didn't include, you end up with the matching as I explained elsewhere. As the impedance moves away from what the designer intended, the radiated power decreases. The point is that the OP was taking about hooking a "transmitter" to a length of wire. I was talking about the matching from a complete transmitter which I assume contains such protection to this length of wire. -- -- forging knowledge |
gwhite wrote:
The strongest argument for dropping the impedance matching concept is PA efficiency, and therefore maximum signal swing. Obtaining maximum swing is a load line issue. So what impedance does the reflected wave encounter? -- 73, Cecil http://www.qsl.net/w5dxp ----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups ---= East/West-Coast Server Farms - Total Privacy via Encryption =--- |
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