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impedance matching and dynamic impedance

Discussion in 'Electronic Design' started by rob, May 17, 2004.

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  1. rob

    rob Guest

    Hi to all.
    I am trying to match a 50 ohm source to a parallel tuned cct at
    13.56Mhz.The parallel
    cct consists of 140nH // 984pF . The actual cap would be a fixed value
    ie 820p in parallel with a variable cap.
    The problem is that I don't know the resistance of the parallel tank
    at resonance.The theoretical resistance (perfect inductor , cap) is
    infinite. The DC resistance of the coil is less than 0.1 ohms. This
    would give a dynamic impedance of 2.8K ohms if R is 0.05 ohms. ( Rd =
    L/R.C)
    This does not take into effect skin effect of the wire and other
    parasitics.

    I then thought I would try and measure the impedance of the cct at
    resonance.
    Would the following method give acceptable results or am I barking up
    the wrong tree!!

    1. Connect the tuned cct to a signal generator via a 1K resistor.
    2. Set the signal gen to 1V out at 13.56Mhz
    3. Tune the resonant cct to resonance using a scope (watch for peak)
    4. calculate the dynamic impedance using basic voltage divide rule.
    Peak voltage across
    tank being Vx. Vx = (1v * Rx) / ( Rx + 1K)

    Using this method I get a value of 315 ohms.
    Does this sound at all correct? I have not taken into account the
    capacitance added by the scope probe. I assume this should not make to
    much difference to the result , because once removed the cct can be
    re-tuned.

    I would then match 315 ohms to 50 with a l-match.

    As I said before , would this method give acceptable results.

    Cheers
    Rob
     
  2. I was looking up parallel tanks and unloaded/loaded
    Q's the other day (related to another thread). The
    sums below might give you some ideas.

    +---------+--------+
    | | |
    [Rg] [Rl] |
    | | |
    Vin----+ [L] ===C
    | |
    0v----+---------+--------+

    The sketch above is a parallel tuned circuit.
    L+Rl and C being the primary elements, which are
    driven/loaded by an external generator, of source
    resistance Rg.

    You want to know the unloaded Q of just the tuned
    circuit. Call this Q. When driven/loaded by Rg
    then the circuit has a lower Q, call this Qe.

    Q
    Qe = -------------- is the important sum.
    1 + wo*L*Q/Rg

    Measure Qe, (by using the 0.707 bandwidth points when
    Rg is in circuit). You already know wo, L and Rg, so
    the unloaded Q can be estimated.

    If Q >10 then Z(unloaded) = Q * sqrt(L/C) approximately.

    All sums pinched out of "Electron Tuned Circuits", by
    Samuel Seely.
     
  3. Hi,

    I take that the ||-tuned tank is not working into an open circuit
    as there wouldn't be any point in it otherwise. If so, then its load
    resistance will likely elbow out any series 'R' as the main contender
    for circuit losses.


    Cheers - Joe
     
  4. John Fields

    John Fields Guest

    ---
    See below...
    ---
    ---
    To find out, I tried a couple of experiments using this circuit:

    1.25MHz>--[50R]--+------------>Ein
    |
    [Rs] <--100k, 10k, 1k, 100R
    |
    +-------+---->Eout
    | |
    [100pF] [150µH]
    | |
    GND--------------+-------+---->0V

    with the different series resistors shown, just to see what would
    happen, and here's how it went:


    Considering that the reactances of the L and C in the tank (at
    resonance) are about

    Xc = Xl = 2pi f L

    = 6.28*1.25E6Hz*1.5E-4H ~ 1200 ohms,

    One might think that after redrawing the circuit to look like this:

    Ein
    |
    [Rs]
    |
    +---->Eout
    |
    [Rp]
    |
    GND

    Then setting Rs to 100k ohms, Rp to 1200 ohms (the reactances of the
    tank components) and ignoring the 50 ohm generator resistance that
    Eout would be:

    Eout = (Ein Rp)/Rs + Rp

    = (1*1200)/100k + 1200

    = .012V

    However, because of the Q of the circuit at resonance, the actual
    voltage measured at Eout was 0.15V, so since the tank is purely
    resistive at resonance if we rewrite Eout = (Ein Rp)/Rs + Rp to solve
    for Rp, we get:

    Rp = (Eout Rs)/Ein - Eout

    and solving for Rp gets us:

    Rp = (Eout Rs)/Ein - Eout

    = (0.15*100k)/1 - 0.15

    = 17,647 ohms


    Doing the experiment again with Rs = 10k yields a measured Eout of
    0.63VPP, which corresponds to a tank impedance of 17,027 ohms. Pretty
    close, (about a 3.5% error) so it looks like we're on the right track.

    Dropping Rs down to 1000 ohms and then down to 100 ohms yields just
    about 1V out for 1V in which is consistent with the tank being a much
    higher impedance than Rs, so I'd have to say that if that's the way
    you're doing your measurements, you're doing them right.
     
  5. Tom Bruhns

    Tom Bruhns Guest

    To Tony's good advice, I'd add:

    Can you make an estimate of the coil's unloaded Q from its
    construction? If it's an air-core single-layer solenoid coil, you
    should be able to get quite close. Same if it's wound on a powdered
    iron toroid core of known characteristics. If nothing else, you can
    put an upper bound on it by considering the RF resistance of the wire
    its wound with, assuming you can determine that. This would be a
    double-check on the measurement you make. You can make it per Tony's
    diagram, but you can also do it by coupling lightly to the tank
    circuit (with a tiny capacitor, say 1pF, from a signal generator) and
    measuring the 3dB bandwidth.

    Another way to do it is just make an educated guess what L-network
    values you'll need and tune them until you get a match to 50 ohms,
    assuming you have a way to measure that match. Then you have the
    values you need for the L-network, and if you can measure them, you
    can calculate the load you've matched to 50 ohms. (But read the down
    to the last suggestion in this message for a much easier way!)

    By the way, unless your coil is quite small, I'd expect an unloaded Q
    of 100 or above at 13MHz. It should only take a coil less than 1cm
    (less than 3/8 of an inch) diameter to get Q above 100, unless you
    have lossy material around its field. So I'd expect a tank resistance
    at resonance of more like 100*12 = 1200 ohms or above. Your 315 ohms
    seems very low. Is it a very small inductor?

    An L network will have its own losses, too, of course. You may find
    that you can match to the tank more easily some different way. One
    way that used to be quite common in radio transmitters, and is still a
    fine way to couple, is a link of a couple turns near the "cold" end of
    the tank coil. If you can make it adjustable coupling (by varying the
    distance and/or angle) you should be able to adjust for a match. You
    can also do it by putting a variable capacitor in series with the
    link.

    Yet another way is to realize that your L matching network will have a
    reactance shunt across your tank. It can be either an inductance or a
    capacitance. Then to complete the match you put a series capacitance
    or inductance back to the source. But if you say the L network will
    have an inductance shunt across your tank, that's JUST the same as
    decreasing the capacitance that's already in the tank. So you save a
    coil, and just use a capacitor to the top of the tank. Assuming that
    your tank is 1000 ohms at resonance, it takes a 54pF series cap back
    to the source, and reducing your tank capacitance by about the same
    amount, to about 933pF. Beware: tuning will be quite sharp! If
    you're right about the 315 ohms at resonance, then it would be more
    like 102pF series, and reduce the tank capacitance to 898pF. If you
    have a high-Q coil and R=3000 ohms, the series cap is 30.5pF and the
    tank shunt is reduced to 954pF. So you can see, with variables for
    both the tank and the series cap, you can adjust over a rather large
    range of coil Qs. Just be sure the frame of the series variable is
    connected to the low-impedance point...the generator...and isolate it
    from ground as best you reasonably can. (It will add a small amount
    of capacitance to ground at that generator output node.)

    Hope that helps...

    Cheers,
    Tom
     
  6. Tam/WB2TT

    Tam/WB2TT Guest

    Why don't you measure the real part when the reactance has been tuned out,
    and the loop is *coupled to the load*. BTW, I don't think you have a 50 Ohm
    source; you have a source that is meant to drive a 50 Ohm load.

    Tam
     
  7. Tom Bruhns

    Tom Bruhns Guest

    As one who is surrounded by network analyzers and signal generators
    that really do have 50 ohm sources, I'd never assume that someone
    doesn't actually have a 50 ohm source at hand, but in any event, does
    it matter in this case? (I didn't get any sense from the OP whether
    the excitation was going to be a few milliwatts or a few kilowatts.)

    Cheers,
    Tom
     
  8. Tam/WB2TT

    Tam/WB2TT Guest

    Tom,

    I think 13.56 is the industrial heating frequency (among ather things). So,
    there is probably a class C or class D amp putting out kilowatts. He doesn't
    say what his loop is coupled to. Is he melting aluminum, treating patients,
    or what...

    Tam
     
  9. rob

    rob Guest



    Thanks for all the tips guys. The application is for a wireless tag
    reader system. Only low power.Few watts at most.
    Have got a working system with a wound coil. Put in a 47 ohm resistor
    to bring down the Q and make matching easier. Works fine , but I want
    to experiment a bit.
    The "Load" is the inductivly coupled Tag.

    chers
    Rob
     
  10. Just a tip: read the application notes of Texas Instruments (Tag-it).
    It has some (practical) info on antenna matching.

    Reinier
     
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