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Dumb question of the week: Mis-matching & matching

Discussion in 'Electronic Basics' started by billcalley, Apr 10, 2007.

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  1. billcalley

    billcalley Guest

    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!

  2. Joerg

    Joerg Guest

    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.
  3. billcalley

    billcalley Guest

    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?

  4. Joerg

    Joerg Guest

    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 ;-)
  5. billcalley

    billcalley Guest

    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,

  6. Tom Bruhns

    Tom Bruhns Guest

    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 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...

  7. Mark

    Mark Guest

    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

    Is that your question?

  8. Joerg

    Joerg Guest

    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.
  9. Joerg

    Joerg Guest

    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 ;-)
  10. billcalley

    billcalley Guest

    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!

  11. Joerg

    Joerg Guest

    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.
  12. LVMarc

    LVMarc Guest

    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:

    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,

  13. Tom Bruhns

    Tom Bruhns Guest

    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."

  14. Joerg

    Joerg Guest

    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.
  15. Tom Bruhns

    Tom Bruhns Guest

    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.

  16. Joerg

    Joerg Guest

    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.
  17. Tom Bruhns

    Tom Bruhns Guest

    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.

  18. Tom Bruhns

    Tom Bruhns Guest

    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.

  19. Mark

    Mark Guest

    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..

  20. billcalley

    billcalley Guest

    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,


    *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.
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