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On 7/8/2015 10:09 AM, John S wrote:
On 7/2/2015 1:38 PM, rickman wrote: On 7/2/2015 1:56 PM, Ralph Mowery wrote: "Jerry Stuckle" wrote in message ... Try this - connect the output of an HF transmitter to an SWR bridge. Now connect a piece of 75 ohm coax such as RG-59 to the output of the SWR meter, and connect that to a 75 ohm resistive load. Do you think the SWR bridge will show a 1:1 SWR? Not a chance. It will be 1.5:1. What you have described is a case of using the wrong swr bridge. You are trying to use a 50 ohm bridge on a 75 ohm system. If a 75 ohm bridge is used it will show a 1:1 SWR. The real SWR is 1:1. With a 75 ohm line and 75 ohm load there is no reflected power. My knowledge of antenna systems is limited, but I do know that this is correct, there will be no reflection from the antenna. If there is no reflections from the antenna, how can there be a loss in the source end? There is NO power returned according to your own statement. I don't see any contradiction. The power comes from the source through the source impedance. The source impedance will create a loss, no? If the transmitter output is 50 ohms there will be a loss in this matching that will result in less power being delivered to the feed line, but that will not result in reflections in the feed line. Why? What causes the loss? The transmitter output resistance? So that would mean that one can never achieve more that 50% efficiency at the transmitter's OUTPUT! And that would mean that a 1000W transmitter is dissipating 500 watts under the BEST circumstances. Good luck on getting that to work to your satisfaction. Maybe "loss" isn't the right term then. The output of a 50 ohm source driving a 75 ohm load will deliver 4% less power into the load than when driving a 50 ohm load. That comes to -0.177 dB. Is there any part of that you disagree with? Nothing in this analysis addresses the theoretical maximum possible efficiency of an arbitrary transmitter and an arbitrary load. In particular I posted the results of a simulation that showed very clearly that the loss in the transmitter output impedance can be well below 50% of the total power drawn from the PSU. Just set the load impedance and make your output impedance as low as you would like. It is when you set the output impedance of the transmitter to a fixed value that a matched load impedance will draw the maximum power from the transmitter while the loss in the transmitter output will be 50%. -- Rick |
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