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#11
July 5th 15, 09:45 PM
posted to rec.radio.amateur.antenna
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An antenna question--43 ft vertical
On 7/5/2015 3:58 PM, Roger Hayter wrote:
Jerry Stuckle wrote: On 7/5/2015 9:23 AM, Ian Jackson wrote: In message , writes Wayne wrote: "Jeff Liebermann" wrote in message ... On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle wrote: Think of it this way, without the math. On the transmitter side of the network, the match is 1:1, with nothing reflected back to the transmitter. So you have a signal coming back from the antenna. You have a perfect matching network, which means nothing is lost in the network. The feedline is perfect, so there is no loss in it. The only place for the signal to go is back to the antenna. Wikipedia says that if the source is matched to the line, any reflections that come back are absorbed, not reflected back to the antenna: https://en.wikipedia.org/wiki/Impedance_matching "If the source impedance matches the line, reflections from the load end will be absorbed at the source end. If the transmission line is not matched at both ends reflections from the load will be re-reflected at the source and re-re-reflected at the load end ad infinitum, losing energy on each transit of the transmission line." Well, I looked at that section of the writeup. And, I have no idea what the hell they are talking about. Looks like a good section for a knowledgeable person to edit. If the termination matches the line impedance, there is no reflection. Both the antenna and the source are terminations. This is a bit difficult to visualze with an RF transmitter, but is more easily seen with pulses. Being essentially a simple soul, that's how I sometimes try to work out what happening. You'll be better off if you killfile the troll. You'll get a lot less bad information and your life will become much easier. The wikipedia entry is correct as written. In the real world, the output of an amateur transmitter will seldom be exactly 50 Ohms unless there is an adjustable network of some sort. I've always understood that the resistive part of a TX output impedance was usually less than 50 ohms. If a transmitter output impedance WAS 50 ohms, I would have thought that the efficiency of the output stage could never exceed 50% (and aren't class-C PAs supposed to be around 66.%?). Also, as much power would be dissipated in the PA stage as in the load. Fixed output amateur transmitters are a nominal 50 ohms. It can vary, but that is due to normal variances in components, and the difference can be ignored in real life. But output impedance has little to do with efficiency. A Class C amplifier can run 90%+ efficiency. It's output may be anything, i.e. high with tubes but low with solid state. But the output impedance can be converted to 50 ohms or any other reasonable impedance through a matching network. A perfect matching network will have no loss, so everything the transmitter puts out will go through the matching network. Of course, nothing is perfect, so there will be some loss. But the amount of loss in a 1:1 match will not be significant. Also, the amplifier generates the power; in a perfect world, 100% of that power is transferred to the load. The transmitter doesn't dissipate 1/2 of the power and the load the other 1/2. It's not like having two resistors in a circuit where each will dissipate 1/2 of the power. BTW - the resistive part of the impedance is not the same as resistance. For a simple case - take a series circuit of a capacitor, an inductor and a 50 ohm resistor. At the resonant frequency, the impedance will be 50 +j0 (50 ohms from the resistor, capacitive and inductive reactances cancel). But the DC resistance is infinity. Again, a simple example, but it shows a point. The resistive part of the impedance is exactly the same as a resistance as far as the frequency you are using is concerned. And if the amplifier output impedance *and* the feeder input resistance were *both* matched to 50 ohms resistive then 50% of the power generated (after circuit losses due of inefficiency of generation) would be dissipated in the transmitter. It would, at the working frequency, be *exactly* like having two equal resistors in the circuit each taking half the generated power. So the amplifier has a much lower output impedance than 50ohms and no attempt is made to match it to 50 ohms. OK, so then please explain how I can have a Class C amplifier with 1KW DC input and a 50 ohm output, 50 ohm coax and a matching network at the antenna can show 900 watt (actually about 870 watts due to feedline loss)? According to your statement, that is impossible. I should not be able get more than 450W or so at the antenna. It is true that if the transmitter is thought of as a fixed voltage generator in series with, say, a 5 ohm resistor then the maximum power transfer in theory would occur with a 5 ohm load. But to achieve this the output power would have to be 100 times higher (ten times the voltage) and half of it would be dissipated in the PA. Not the best way to run things, it is better to have a voltage generator chosen to give the right power with the load being much bigger than its generator resistance. So why do all fixed-tuning amateur transmitters have a nominal 50 ohm output instead of 1 or two ohms? And why do commercial radio stations spend tens of thousands of dollars ensuring impedance is matched throughout the system? The maximum power transfer at equal impedance theorem only applies if you started with a *fixed* output voltage generator. We don't; we start with a load impedance (50 ohm resistive), then we decide what power output we want, and we choose the voltage to be generated accordingly. (Thank you for giving me the opportunity to think about this!) Actually, it doesn't matter if it's a fixed or a variable output voltage - maximum power transfer always occurs when there is an impedance match. -- ================== Remove the "x" from my email address Jerry, AI0K ================== |
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