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Components and Circuits For discussions about component types, alternatives and availability, circuit configurations and modifications etc. Discussions here should be of a general nature and not about specific sets. |
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29th Jul 2009, 12:04 pm | #61 | ||
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Re: Impedance matching theory.
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29th Jul 2009, 2:18 pm | #62 | ||
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Re: Impedance matching theory.
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29th Jul 2009, 2:33 pm | #63 | |||
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Re: Impedance matching theory.
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29th Jul 2009, 3:36 pm | #64 | ||
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Re: Impedance matching theory.
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We seem to have come a long way from pentode audio output stage transformers! |
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29th Jul 2009, 3:48 pm | #65 |
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Re: Impedance matching theory.
Transmitter - transmission line - load; one of my pet subjects.
First of all, let's tidy up the use of the word 'impedance'. Although this word, by definition, includes reactance and resistance, in the context of transmission lines it gets mis-used. e.g. "this transmission line has an characteristic impedance of X ohms". Although I would agree that the line will have some reactance, for all intents & purposes, this reactance can be considered as insignificant - if it's a half-decent transmission line, that is! The load, on the other hand can be an impedance: resistive and reactive. What seems to cause confusion to some people is when a transmitter O/P socket is labelled "X ohms". It does not have an O/P impedance of X ohms! The term means that it is designed to 'see' a resistive load of X ohms for perfect matching; the O/P impedance of the transmitter itself ideally will be very low - and almost certainly contain some reactance as well as a resistive component. Look at it this way: if the 'O/P impedance' was X ohms, only half the available power would go to the load of X ohms, where X is resistive only. (Assuming a transmission line with zero losses). When the load contains a reactive element, some forward power is reflected back to the transmitter and since the transmitter does not have an O/P impedance of X ohms, a further reflection takes place and the wave travels back to the load - again - and so on. This last point is frequently not appreciated, in my experience. Al. / Skywave. |
29th Jul 2009, 3:52 pm | #66 |
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Re: Impedance matching theory.
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29th Jul 2009, 4:26 pm | #67 | |
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Re: Impedance matching theory.
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29th Jul 2009, 5:12 pm | #68 | |||
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Re: Impedance matching theory.
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The maximum power transfer theorem does have a relevance to receivers - the match between the antenna and the input (mediated via the transmission line, if present). You do get best power transfer by conjugate matching the impedances, and the power lost in the source resistance is real: it is re-radiated by the antenna (I only realised this a few years ago). So a well-matched receive antenna will radiate half the power it absorbs from the incoming signal, and send the other half to the receiver. |
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29th Jul 2009, 5:22 pm | #69 |
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Re: Impedance matching theory.
If the source impedance is complex, then maximum power transfer does occur if the load impedance is the conjugate of the source impedance. Having a transmission line between them complicate things, because the impedance you see at your source end is not in general the same as at the load end. But, reflections, mismatches, or not, at a given frequency, it can always be represented as a real and imaginary part.
As has been said, decent transmission lines have a resistive characteristic impedance anyway. I can see this is going to be true, but it is something I had not thought of. Thanks very much Dave! What a great thread! |
29th Jul 2009, 6:00 pm | #70 |
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Re: Impedance matching theory.
Quote; Al. / Skywave:
First of all, let's tidy up the use of the word 'impedance'. Although this word, by definition, includes reactance and resistance, in the context of transmission lines it gets mis-used. . . . . Quote; Dave, G8HQP: No, this is not a mis-use of the term. It is a recognition that, in general, a transmission line does not have to have a resistive characteristic impedance. Point taken. I should have written "gets mis-understood". Al. |
29th Jul 2009, 6:26 pm | #71 |
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Re: Impedance matching theory.
Yes, Kalee20, after that post I did the maths and decided that a constant current source with parallel Ri did indeed loook like a constant source voltage with Ri as the series impedance. I guess a triode might have been a better example of a non-linear source.
Of course what all that ignores is the power consumed by the valve in generating the constant current. Being a class A output, the HT power consumed is a constant so arguably the maximum power transfer does indeed occur with an 8 ohm load rather than the 64 ohms. it's been a fun thread but we seem to have got to the point where we all agree so I guess there's no entertainment value left. |
29th Jul 2009, 9:00 pm | #72 | |
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Re: Impedance matching theory.
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Aside: I phrase this as a Q. since it's not something I've actually had practical experience with, only read about. Al. / Skywave. |
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29th Jul 2009, 10:18 pm | #73 | |
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Re: Impedance matching theory.
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29th Jul 2009, 11:12 pm | #74 | |
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Re: Impedance matching theory.
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Assuming a perfectly matched antenna surely there is no reflected power back to the source. My limited understanding of antennas is that it may aborb the power but how well it raditaes that power is entirely dependent upon the antenna. You make the statement that half the power is radiated and half is relected back to source with a matched antenna. There are antenna designs which use terminating resistors etc ( folded dipole with the ends terminated with a resistor as an example) which give a good broadband low SWr across a very broad frequency range however very little power is radiated because they are poor radiators broadband but good absorbers of power. Mike |
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29th Jul 2009, 11:38 pm | #75 | |
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Re: Impedance matching theory.
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For lower power levels, there are alternative configurations that can be used to 'mop up' reflected power that don't incur quite the high loss levels (or the expense) of a circulator - the 'Engelbrecht' configuration of using two parallel amplifiers combined with a 3dB hybrid being a good example. |
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30th Jul 2009, 9:18 am | #76 | |
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Re: Impedance matching theory.
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Anyone who doesn't have a button marked "tune" will know that there is a definite setting of the output matching that gives highest power output. This is characteristic of the maximum power theorem situation. But I have a transmitter (Syncal 30) that, if any is going to have a low impedance output, this will be it. Designed to drive short aerials and with an internal tuner that can only correct for reactance, surely this will be one with a low output impedance? But on high power it has a 50 ohm impedance. I just measured it. (On low power it seems to be a little higher, but that may just be a calibration issue.) So I think it likely that RF systems do indeed match their outputs as suggested by the documentation. If you disagree then find a transmitter that actually does have a low output impedance while maximising it's output into 50 ohm load, i.e. not just a consequence of it's output matching being set that way. |
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30th Jul 2009, 9:40 am | #77 | |
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Re: Impedance matching theory.
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(I'll just give you some typical figures for a 25kW vision TX. RF output power from TX = 25kW, measured. DC power input to final amplifier, 42kW, measured. Overall efficiency thus about 60%. But if the TX had an output impedance of 50 ohms, there'd also be 25kW being dissipated in the TX: total RF power, 50kW, but input DC power still 42kW. Efficiency, 120%. maybe we're onto something in the perpetual motion line, here...) Last edited by Ray Cooper; 30th Jul 2009 at 9:46 am. |
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30th Jul 2009, 9:56 am | #78 |
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Re: Impedance matching theory.
You are assuming that a thing that varies it's output voltage as if it has an internal resistance actually does contain a resistor and hence dissipates the undelivered power.
It doesn't have to be that way and I suspect that most transmitters do manage to dissipate less power than you would expect from the matching. |
30th Jul 2009, 10:28 am | #79 | |||
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Re: Impedance matching theory.
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Of course there's an internal resistance. It may not be a physical component with wire ends and '50 ohms', or whatever, written upon it, but a source resistance is present in equivalent form in some manner, whether copper losses in tuning circuits, AC load lines of active components or whatever. And that equivalent resistance will dissipate power. If you're maintaining that there is no internal resistance, then of course the thing must have a zero output impedance (let's leave reactive components aside for the moment) annd would dissipate no power at all. Which seems a far cry from your claimed 50 ohms output impedance... |
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30th Jul 2009, 12:01 pm | #80 | ||
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Re: Impedance matching theory.
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What I am saying is that I think it is possible to have a system that appears to have a source impedance, in that as you load it up the voltage goes down, but there are (almost) no losses. This situation occurs when there is some energy limited transfer going on, e.g. like in a switched mode power supply (or class C output stage?). But I would add that the impedance will not appear to be a simple constant in this situation, and (rather usefully) it seems to mirror the load, which is perhaps not so surprising. Having said that I calculate that the example of 42KW in to get 25KW into 50 ohms could have a simple output impedance of 34 ohms - so not really that low at all. |
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