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Where does the power go
On Wednesday, November 2, 1994 at 6:52:22 PM UTC-5, Guy Fletcher wrote:
I have followed the thread on reflected power in a transmission line with interest and curiosity, hoping to learn something. I have also thought deeply about the question on buses and in the shower! For any new readers who didn't follow the earlier discussions, the question is about reflected power from a mismatched load (antenna) driven through a transmission line from a high power tube amplifier, though it applies to any power level and any amplifier type. What happens to the reflected power? Is it dissipated in the amplifier? Where? Or is it reflected again at the possible mismatch looking back into the amplifier from the line? The power will all go to the load if the source and the transmission line had no resistive components I believe the problem lies in the question being wrongly posed, though it seems a perfectly reasonable question to ask. A transmission line is a wondrous invention which, if infinitely long to avoid reflections, has the property that the input voltage and forward current are in phase and have a definite ratio - the characteristic impedance, usually 50 ohms for coax. The infinite line can then be terminated at any point by 50 ohms without the input seeing any difference - the line is matched. Please forgive the long and well-known introduction; I am getting to the point! If the load is not 50 ohms (and real), a forward pulse is partly reflected and returns to the input, where it may be reflected again depending on the load it sees looking back into the source. The speed on the line is about 1ns/foot in air, and about 2/3 of that in solid line, so for any reasonable line the round trip time is sub-microsecond. Unless our transmitter is modulated at a frequency above 1MHz, so that the amplitude can vary significantly in less than a microsecond, the returning wave will see the original amplitude still coming out, but several cycles later depending on the line length. This applies to keying too unless you want to splatter over the whole band (or HF spectrum). So we are realistically talking about a steady key-down state of the amplifier, except perhaps for UHF TV signals. In a mismatched line (or a matched one) you can open the line at any point and measure the signal. The ONLY things you can measure are the voltage, the current, and their phase relationship. At the load end, this defines the complex load impedance, and if this were the power frequency of 50 or 60 Hz, we would mutter about power factor to calculate the absorbed power, but we would NOT talk about reflected power back down the wires. A mismatched transmission line has the property that the apparent load impedance varies along the line, and we have a standing wave. The apparent impedance is complex except at the nodes and antinodes where it is real but smaller (50 / SWR) at a node, or larger (50 x SWR) at an antinode. We find it MATHEMATICALLY convenient to analyze this standing wave in terms of a forward wave and a refected wave, though the definition of SWR refers only to the voltage ratio at antinode and node, and doesn't need the formal idea of 2 opposing waves. We have even designed an instrument to measure SWR, which we tend to think of as measuring forward and reflected power because we have put a scale on it marked in this way! What it actually measures is the in-phase (forward) component of current, and the out-of-phase (reverse) component, into a presumed 50-ohm load. There is only one power flow in the line, and that is forward (hopefully). The voltage, current, and phase difference at any point in the line define that power flow (in either direction depending on the phase). Amazingly the magnitude and direction of the power flow are the same everywhere on the line apart from losses. So at the line input a fast pulse always sees 50 ohms, whatever the distant load, but a steady sine wave sees a complex load calculated in the usual way. The tube amplifier is usually designed to deliver its rated power into 50 ohms real. It will probably deliver more into a lower load (which may damage it), and it will certainly deliver less into a higher real load. It will also deliver less into a complex load unless you tune out the imaginary part. The efficiency of the amplifier depends on the load. If into a certain (complex) mismatched load its efficiency is 70%, then 30% of the dc input power is lost as heat, mostly in the plate of a tube amplifier. That 30% was never delivered as power down the line in the first place, so the question of what happens when it comes back down the line never arises. Nor does the output impedance of the amplifier matter - it may be 50 ohms, or it may be much less as with most solid state amplifiers. The output impedance is merely part of the amplifier design. What matters is the rated load resistance, the power delivered into that rated load, and the way the efficiency varies with other loads. This is a rather different way of looking at the question, hence the new thread. Considered comments welcome. Guy Fletcher VK2BBF The power will all get dissipated in the load if there is no resistive component in the source or transmission line. This is true regardless of frequency. In real systems the source and T line both have a real R component which will dissipate the reflected energy as heat, high SWR means less power to the load. In systems where the T line has very little restive component then huge power can be transmitted. For example: a small optical fibers are used to carry 100KW of power without issue. |
#2
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Where does the power go
On Thu, 31 Dec 2015 05:33:17 -0800, woody369 wrote:
On Wednesday, November 2, 1994 at 6:52:22 PM UTC-5, Guy Fletcher wrote: I have followed the thread on reflected power in a transmission line with interest and curiosity, hoping to learn something. big snip Guy Fletcher VK2BBF The power will all get dissipated in the load if there is no resistive component in the source or transmission line. This is true regardless of frequency. In real systems the source and T line both have a real R component which will dissipate the reflected energy as heat, high SWR means less power to the load. In systems where the T line has very little restive component then huge power can be transmitted. For example: a small optical fibers are used to carry 100KW of power without issue. You're responding to a 21 year old post. The OP is long gone. -- Jim Mueller To get my real email address, replace wrongname with dadoheadman. Then replace nospam with fastmail. Lastly, replace com with us. |
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