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On Mon, 03 Nov 2008 04:46:57 GMT, Owen Duffy wrote:
Jeff Liebermann wrote in : ... In the real world, the power that a transmitter delivers to a non ideal load is not so simply predicted, and it is entirely possible that it delivers more power to the mismatched load. Owen I beg to differ somewhat. In order for the reflected power to contribute to the incident power, the reflected power would first be attenuated by the coax loss. It would then require a substantial Are you proposing vector addition of power? Nope. No need to complexicate things. I can safely assume a real 75 ohm load. I can cheat a bit and assume some multiple of 1/2 wave coax cables, thus eliminating any imaginary contributions from the coax. Please note that my purpose was to demonstrate that 75 ohm antennas and 75 ohm coax will work adequately in a 50 ohm system. I think I've done most of that. Including complex impedances to the calculations will yield a more accurate result, but the resultant reflected power that will be added to the forward delivered power, will be LESS than the results produced by my calculations using only the real part of the impedances. mismatch at the transmitter, which is unlikely. However, assuming there is a mismatch at the source, some of the reflected power will be sent back to the load (antenna), after getting attenuated by the coax for a 2nd time. There may be some contribution, but it will very very very very small. Let's try some more or less real numbers. I kinda prefer doing everything in dBm but hams have this thing about using watts... Of course it doesn't matter, which unit system you use, but if you start adding 'forward' and 'reflected' power in dBm because it is real convenient, you have peformed a flawed vector addition of power. Is that valid? It's only valid for the level of accuracy with which you are working. For convenience, perhaps we can just assume that the re-reflected contribution to the forward power is in phase, thus yielding the maximum delivered power. Any phase shifts between the two signals will result in LESS delivered power than the in phase simplistic calculation. I'm sure the accuracy might be useful for academic purposes, but my example demonstrated that only 35 mw was added to 33 watts, an error of 0.1%. Of course, that's ridiculous because the initial measurement of the originating 50 watts is probably only accurate to 2 significant figures. Start with a 50 watt xmitter and 20 meters of LMR-240 coax at 0.09dB/meter for an attenuation of 1.8dB. The power delivered to the antenna is: 50 watts / ((1.8/10)^10) = 50 / 1.5 = 33 watts The 1.5:1 VSWR reflects 4% of 33 watts for 1.3 watts reflected. The 1.3 watts is again attenuated by the 1.8dB coax loss resulting in: 1.3 watts / (1.8/10)^10) = 1.3 / 1.5 = 0.87 watts You start with a limited view of the mismatch, VSWR conveys only one dimension of a two dimensional mismatch. Sure. It's good enough for a back of the envelope estimate of how much power the re-reflected signal can possibly add to the forward power. Your treatment of the forward wave and reflected waves as independently attenuated is an approximation that will lead to significant errors in some cases. True. However, as long as I assume a 1.8dB coax cable loss, the reflected and re-reflected powers will be sufficiently low to be considered negligible. Including the necessary vector arithmetic to include the possibility of random coax cable lengths will improve accuracy, but not affect the result very much. For example, what percentage of the power at the source end of the line is lost as heat in 1m of LMR400 at 1MHz with a) a 5+j0 ohm load, and b) a 500+j0 ohm load. The VSWR is approximatly the same in both cases but the answers are very different, one is almost 100 times the other. I'll work out the exact numbers tomorrow, but I see your point. However, please note that I made an effort to use a REALISTIC example of a typical 2m radio, coax, and coaxial antenna arrangement. Of course, you can conjure a set of numbers that will result in a substantially increased calculation error. I can do the same thing if I take my example and simply reduce the coax cable length to the point where coax attenuation is dramatically smaller. A 100:1 load impedance change is not the same as a 1.5:1 impedance change (from 50 to 75 ohms) Doesn't it stand to reason that as the length of the transmission line approaches zero, that the power lost transmission in this type of line in the high voltage low current load scenario is lower than the low voltage high current load scenario. Ummmm... you lost me there. I've got a headache tonite. I'll see it makes more sense tomorrow morning. Another issue is that the V/I characteristics of a transmitter output stage is not necessarily (or usually for most ham transmitters) a straight line, in other words it does not exibit a constant Thevenin equivalent source impedance with varying loads and the application of some linear circuit analysis techniques to the output stage are inappropriate. I really don't know if that's true for a 2m FM transmitter. I'm not sure it even matters. The V/I characteristic (slope) is just the source impedance of the output stage. Whether it's 50 or 75 ohms is close enough for my simplistic calculation to be accurate without throwing in non-linearities. The source impedance may be different for a 50 watt radio, running at 1 watt, but not enough to make a big difference. If the source impedance were magically 10 times as high, the 35 milliwatts of re-reflected RF would become 350 mw and still be a fairly negligible contribution to the delivered 33 watts. -- Jeff Liebermann 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 |
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