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## [Amps] Re: [Amps] Re: [Amps] BirdŽ 43 Manual

 To: [Amps] Re: [Amps] Re: [Amps] BirdŽ 43 Manual w8ron@stratos.net (Ron) Fri, 29 Mar 2002 17:18:11 -0500
 ```What a nice discussion ......I feel I have to pipe in what I think I know. During the discussion of transmission lines , the subject is started by assuming an endless line of some characteristic impedance. Because a step or an impulse transforms to the frequency domain as all frequencies of equal amplitude , one can use a step to analyze without concern of frequency and it gives it a less mathematical explanation. When you apply a step voltage , the line draws current at the ratio of the characteristic impedance where the current is charging the line capacitance and is limited by the series inductance. Because it is an endless line , as long as you apply the step voltage , the line draws current charging the line further at the front and the series current must flow from the source. The activity of charging proceeds according to the velocity factor of the line times the speed of light. All sections of line behind the propagating front have the line capacitance charged and the series inductance is carrying a constant current of the step voltage divided by the line impedance. So where is the power ? It is stored in the line as a charge line and a series current. The next step is to cut the line and replace it with a non reactive load of the characteristic impedance that looks like the remainder of the endless line. Now the line carries the current and delivers it to the load and holds some stored energy. So the next step is to remove the load to an open, Now, voltage in the line shunt capacitor does not change instantly and current in the series inductance does not change instantly so what happens at the open load is that the current continues to flow into the end but has nowhere to go so it accumulates in the line capacitance at the line end where current is charge / time and voltage is charge / capacitance. The current is slowed to zero and the voltage rockets up to a value higher than the step voltage applied . Then the capacitance at the end has a very high stored charge as well as a high voltage and the line further away has a lower voltage so an electric field exists in the end and current starts to flow back towards the source. In the line there exists a shunt resistance that is high but is there and so some of the charge is leaked through this shunt resistance and there is loss and some power consumed. The current bounds back and forth in the line loosing power each trip except that when it reflect off the supply end , it sees a low resistance with the step voltage. Instead of the current bunching up as it did on the open , it runs back through the supply with it's resistance consuming some power and out through the shield. When you use the shield as a reference , it looks like the current is flowing on the center wire in the wrong direction. Sort of like being in a car and think you are moving forward but the guy next to you is going backward and your sitting still. Eventually , it damps out and there is just a charged line with no current flowing. So where is the power going? Some lost on each trip , some lost in the source impedance. This seems to be the easiest situation that one could visualize . Adding a moving source signal makes it more complicated but does not change the answer , just the ratio of where the power goes. --- Ron Ian White, G3SEK wrote: > Gary Schafer wrote: > > > > There is no 100 watts > >of power coming out of the radio. The meter in this mode is not > >functioning as a watt meter. There is no current flowing in the > >line as there is no termination. No current no power. If there is no > >power going out there is no power to be reflected back to the > >transmitter. You are only seeing voltage that the watt meter is telling > >you there is power. It is not telling you correctly. > > > >Yes your final may be drawing current from the power supply when you > >turn up the drive. Yes the final is dissipating power. Is it > >dc power or rf power. Probably some of both as you will have some > >circulating currents in the tank circuit due to the mismatch and > >the efficiency will probably be way down. It is all happening internally. > > > I agree with Gary - zero RF power is being generated in that idealized > worst case of exactly zero or infinite impedance. > > It may be more helpful to get away from those zeros and infinities, and > think what happens with a very severe (but not quite infinite) load > mismatch. We may be pouring in lots of DC power, but the RF power being > delivered to the load is almost zero. In other words, the more severe > the load mismatch, the less RF power gets delivered to the load, and the > closer the efficiency gets to zero... which is actually something we > already knew. > > The question about the validity of "forward and reflected power" came up > in the rec.radio.amateur.antenna discussion. As someone pointed out, > power is a scalar quantity so it cannot have any direction. In other > words, any statements about "forward" or "reflected" power can not be > completely correct, and will not stand up to detailed examination. > > -- > 73 from Ian G3SEK Editor, 'The VHF/UHF DX Book' > 'In Practice' columnist for RadCom (RSGB) > New e-mail: g3sek@ifwtech.co.uk > New website: http://www.ifwtech.co.uk/g3sek > _______________________________________________ > Amps mailing list > Amps@contesting.com > http://lists.contesting.com/mailman/listinfo/amps ```
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