Dumb question of the week: Mis-matching & matching

Hi All,

I have been trying to figure out this basic concept for quite some time now, but without much success: When purposefully mis-matching an RF stage (such as a PA) in a 50 ohm system, I sometimes see references alluding to the fact that the next stage will *not* be able to see 50 ohms due to this mismatched stage. The part that really confuses me is: Why not? Why can't the PA's output matching network not only purposefully be designed to conjugately mis-match the PA's output (for max Pout) AND also present exactly 50 ohms at the other end of this same matching network? And if, for some reason, this is not possible, then why not just add another pi (or L) network so as to give the next stage exactly 50 ohms? Am I understanding this correctly, or do I have it completely FUBAR'ed up?

Many thanks!

-Bill

Reply to
billcalley
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You can make it look like 50ohms for one particular frequency. Power amplifiers are for the most part hard voltage sources. The matching network or nowadays broadband transformers are there to match that into a load. But the main goal is not to do an impedance match. It is to maximize the power output while not going too low in the impedance the output transistors see, in order not to smoke them out. For example, an amp stage might feed into a 1:4 xfmr and thus see a load of 12.5ohms. You could get a whole lot more power into the antenna if you stepped it up to 1:8. But then the output stage would sweat a 6.25ohm load and pretty soon there will be a bang and molten solder will splatter about.

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

Hi Joerg,

Thanks for the response! So you are saying that even if we purposfully mismatch the output of the PA for the highest P1dB, that we can still make the next stage, such as a filter, "think" that it is looking into a 50 ohm PA?

-Bill

Reply to
billcalley

Not if there is nothing else between the two stages, then mismatch would not make a lot of sense. If you have a filter in between you'd need make sure both sides of it are terminating with the spec'd impedance.

Why does the next stage have to see 50 Ohms? It'll usually be quite happy with less. Unless, of course, it is built "on the edge" in terms of stability but IMHO a designer of an amp that goes unstable upon small source changes should be tarred, feathered and flogged ;-)

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

Hi Joerg,

But any filter at the output of the PA stage would have to 'see' 50 ohms in order to maintain its as-designed response...

Best Regards,

-Bill

Reply to
billcalley

Why not instead design the filter for the impedance it does see? You suggested a pi network for coupling the load to the PA. Be aware that you will have real trouble getting a lossless linear passive reciprocal network to present, say, the desired 4000 ohm load to the plate of a beam power tube and at the same time transform the 10k ohms you see looking back at that plate to 50 ohms. It ain't going to happen. If you add an L, or change the pi values, or whatever, to get the 10k transformed to 50, then the 50 will present a 10k load to the plate, rather far from the optimum. Same with Joerg's example of a low-impedance output, say less than an ohm, and your 50 ohm load transformed to the 12.5 ohm optimal load he suggested. Depending on the network you use, the 1 ohm will transform to something different from 50 ohms. A transformer would transform in constant impedance ratio; a quarter wave transmission line would transform in a reciprocal fashion: 25 ohm line to transform 12.5 to 50 would transform 1 to 625. A pi behaves more like a quarter wave transmission line than like a transformer, but not the same as a line in general.

Three ways to solve your dilemma: (a) design whatever follows, be it a filter or another stage or whatever, to operate properly with the source impedance it WILL see; (b) use feedback in the amplifier to adjust the source impedance to what you want (which gets at least tricky at RF); and (c) add resistive (dissipative) loading. For example, for (c) applied to the hypothetical valve mentioned above with an optimum plate load of 4k ohms and an effective plate resistance of 10k ohms, put 13.333k ohms shunt from plate to ground (with DC blocking). Then design a pi network to match between 50 ohms and 10k||13.333k. That means the 50 ohm load will "see" a 50 ohm source impedance, and will present 10k||13.333k to the plate; but there's also 13.333k plate load, so the net plate load will be 4k, as desired. But that wastes 30% of the available power in the 13.333k resistance. Much better to use (a) or (b) or a combination.

I thought I had posted a reply this afternoon about this, with a reference to a nice article on reciprocal networks, but I see it didn't make it. You might find the section beginning at "Properties of reciprocal and non-reciprocal networks" at

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to be helpful in understanding why with "normal" networks you can't solve your problem as stated in the basenote of this thread. You'll see there that you may be able to solve your problem using a network with nonisotropic material in it...

Cheers, Tom

Reply to
Tom Bruhns

Bill, I'm not sure I understand your question but if I do , then the answer is NO you cannont do that using LOSSLESS matching becasue lossless matching is reciprocal. i.e. if you desgin a 2:1 matching circuit it will look like 1:2 the other way. Now that is true for lossless networks like LC matching networks and transformers. If you can use a PAD i.e. lossy resistors, then you can create any combination of in/ out match you want, but you must burn some of the power in the resistors.

Is that your question?

Mark

Reply to
Mark

No, the other way around. You have to design the filter so the PA feels alright. In ham radio it was the usual game. Bigger tube would be nice but the budget ain't there. So, let's tweak the input impedance of the filter down a bit. Plates glow dark red. Ah, maybe they can do a little more. An orange glow begins. POOF.

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

Several kohm to 50ohm is tough to achieve with a Pi-filter. Unless you can splurge and put in a fancy vacuum variable cap, maybe. But even then I wouldn't do it. Another architecture that works better with such extreme ratios is the tapped parallel resonant circuit. Just don't touch anything with the fingers when it's running ;-)

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

Hi Joerg, Tom, and Mark,

Thanks so much for the informative responses! It is becoming a bit more clear to me now, but the practical talk of high output impedance vacuum tubes is getting me confused again, since I am only interested in the theory of exactly how matching networks can work when a mis-match MUST be seen by a certain stage; but when 50 ohms

*must* be presented to a prior stage. For instance, let's say we have a receiver chain that looks like this: ANTENNA-BPF-LNA-MIXER-IF-DETECTOR. Now, the discrete low noise transistor for the LNA stage, in this case, must see 10-j10 for a perfect noise match, so we then design a PI or 'T' or 'L' input matching network that permits the LNA's transistor to see this exact value; yet the (ceramic) BPF filter that will be attaching to this completed LNA circuit's input *must* see exactly 50 ohms. What I'm asking is: Is this possible? Can we do this conjugate mis-match between the filter and LNA stages to satisfy the low noise transistor, while still presenting *exactly* 50 ohms to the BP filter?

Thanks again guys!

-Bill

Reply to
billcalley

As Tom hinted, a network consisting of only "lossless" L and C is reciprocal and can't really translate between a fully resistive port and a complex port. Time for making compromises, I guess. I'd start at the filter, see if it can live with slightly non-resistive port behavior. If it really falls off the rocker and you need all that noise figure of the LNA plus steep filtering you may have to go back to ye olde LC filtering between antenna and LNA.

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

Bill et al,

Ther are several reasons why a non 50 ohms termination or drivin point impdance is used. The best case is in low noise applications. The lowest noise in a devcie usually is not at 50 ohms, therfore to optimize noise noise a purposeful non 50 ohm is created at the drving point.

Also, the actvie devcies are NO unilateral. What this mean, there is a small interelectrode capcitance from the out put to the input, te "miller capacitance". This parastic element will chnage the effective inout (or output) match depending on the loading, bias and drvie levels of any given device in the chain.

The effects of interstage mis-match ad changes with operating point, can be ameliorated by:

feedback interstage "swaping" resisotrs and the configuration of the amplifier. Some amplifiers chains aremore susuceptable to miller capctiance than others. For exmaple, the cascode (two transistors, first common emmiter second common base) recues the effect of the miller capacitance, by lowering the voltage opn the first stages collector, therby redcuing the effective "size" of the miller capacitnace"

swapming resitors 10-27 ohms, in series with the amplifier chain, provides matching stablity a the expense of gain and effeciency as the swamping resitor are pure loss, and substract proportionate with swamping resisotr size.

Finally pure 50 ohms is needed where electrically long(> 1/32 lambda) signals are used. so from a connector and cable to generator the 50 ohms is required. and from the output to a connector, where a cable is used requires 50 ohms too. However, interstage and with electrically small interconnections(as is possible on a single substrate) does not at all require 50 ohms interfaces. IN fact, with interstage and filtering the impedance will not be 50 ohms, but is selected to obtain the "q" selectivity of the interstage filter(s).

Best Regards,

Marc

Reply to
LVMarc

Although a network made only of lossless L and C components is indeed reciprocal, it's not a limitation that it be lossless.

"A reciprocal network is one in which the power losses are the same between any two ports regardless of direction of propagation (scattering parameter S21=S12, S13=S31, etc.) A network is known to be reciprocal if it is passive and contains only isotropic materials. Examples of reciprocal networks include cables, attenuators, and all passive power splitters and couplers."

Cheers, Tom

Reply to
Tom Bruhns

True. I was just assuming that Bill didn't want any resistive pads in there because that would ruin the noise figure of the whole setup.

--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

Huh? It's done all the time. e.g., my 13.33k||10k example, 5714 ohms. Let's say 5MHz. 70pF at the plates, 15.09uH, 400pF at the 50 ohm output. Ql about 14. Qu of the coil pretty easy to make 30 or more times Ql. 4000 ohms lets you go to lower Ql if you wish, but 14 is, by most folk, considered quite reasonable. For lower Ql, you can add poles. A second inductor at the output of the example above lets you very considerably lower the Ql.

If you insist on a parallel resonant circuit, use link coupling...coupled resonators.

Cheers, Tom

Reply to
Tom Bruhns

Sure, I've done it as well. The last one was a gorilla amp. Two QB5/1750, 5kV on the plates. Ok up to 20MHz or so but at 30MHz I was unable to get low enough in primary capacitance. I had to shell out big bucks for a vacuum variable capacitor to make that work.

That's the really classic approach. I remember when that was "the" method used by all hams around me. Ok, now I gave away the fact that I am over the hill...

BTW, to keep the Q up there we'd regularly polish the coil (made from

3/8" copper pipe) with Wenol metal polishing paste. Boy was I glad when I found that stuff again after moving to the US. In a kitchen store, bought all they had.
--
Regards, Joerg

http://www.analogconsultants.com
Reply to
Joerg

Why *must* the BPF be loaded with exactly 50 ohms? What happens if it isn't? I post these as rhetorical questions, suggesting you go answer them yourself. Next: what is that filter's purpose? Next: Can you use a different filter there, and/or add a filter after the LNA, so that you achieve the desired overall response? What IS the impedance the filter sees, looking toward the LNA's input, including the matching network, which will almost certainly NOT look like 10-j10 transformed by the network.

There are reasons to put a filter there, but you need to understand exactly what those reasons are, and to what degree you need the filter to perform like it does in an "exact" 50 ohm environment. You may be surprised to find that the filter does a pretty respectable job working into a different load impedance. You may be able to design a filter (L-C, microstrip, ... or even a different ceramic one) that does the job you need in the environment you have. You may find that you can use a different LNA transistor that works better. You may find a compromise between the LNA noise, its gain, and the filter response, that works for you.

Cheers, Tom

Reply to
Tom Bruhns

...

Shoot, in what I just posted I forgot to suggest: add feedback to the LNA so that the impedance you see looking back into it is close (closer) to the impedance the LNA device wants to see for optimal noise performance. This may be difficult at your operating frequency, but it's something we do at moderate RF frequencies. Note that you can add reactances as simple passives. Just avoid resistors that dissipate precious RF: those will degrade the noise figure, almost certainly worse than operating the LNA device a ways off its optimal source impedance.

Cheers, Tom

Reply to
Tom Bruhns

yes but you may need a dissipative element that will brun some power... you probably can't do it with just L, C and transofermer..

you may need resistors...

So you may want to re-consdier if you really want to do it this way..

Mark

Reply to
Mark

Hi Guys,

From what I have been able to absorb now from all of your really terrific responses is that it is totally and completely impossible to LC match, even theoretically, a purposefully mis-matched active device*, and then look back into that mis-matched network and see a perfect 50 ohms. However, the purposefully mis-matched active device WILL get to see the exact impedance it wants to see. And all this is due to the reciprocal nature of LC matching networks. Thus, it would be wise, as you have all mentioned, to design any LNA or PA to be as close to 50 ohms as possible, or to design the next connecting stage so that it properly works with something other than 50 ohms -- 'cause there is no way to "fix" this mis-match issue with any LC matching networks. Is that correct, or have I misunderstood something? (I can't believe I didn't know -- or didn't understand -- this stuff from the get-go! A major glitch in my knowledge-base, that's for sure.)

Many Thanks,

-Bill

*The mis-match created so as to optimize an LNA transistor's input for NF, or a PA transistor's output for P1dB, to name two common reasons.
Reply to
billcalley

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