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wideband RF impedance matching

Discussion in 'Electronic Design' started by steve, Sep 11, 2005.

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  1. steve

    steve Guest

    Hi,

    I'm trying to impedance match a 250-450MHz VCO (designed for a 50 Ohm
    load) to an AD8343 mixer which has a 2.7+6.8j input impedance.

    I'm fairly new to impedance matching in the wild (have only got limited
    pure-resistance based impedance matching from university), so I'm
    getting a bit stuck here.

    I've read a bit about LC matching networks but it seems that they're
    only for narrowband applications, unsuitable for my setup. I've also
    come across using transformers for wideband matching applications, but
    I'm unsure how to apply this. From what I've read, the impedance
    relationship of a transformers primary and secondary coils is:

    Zs/Zl = (Ns/Nl)^2

    So for my application I need a transform with a turns ratio which
    satisfies the following:

    50/(2.7+6.8j) = 2.52-j6.35 = (Ns/Nl)^2

    But this gives me a complex transformer ratio, which doesn't seem
    possible.

    I'm a little stuck at the moment, can anyone explain to me where I
    should go from here?

    Steve
     
  2. I read in sci.electronics.design that steve <>
    Wind some turns at right-angles to the others.(;-)
    Is it really true that the j term in your AD8343 input impedance is
    constant over your bandwidth?

    If, instead, the input looks like R (= 2.7 ohms) and L in series, you
    can connect a CR series combination across the input. If the added
    resistor is 2.7 ohms, and LC = 2.7^2, the input will magically be
    converted to 2.7 + j0 at all frequencies. Now you can match your 50 ohm
    source to that with a transformer or equivalent, with turns ratio
    sqrt(50/2.7) = 4.3.

    The added CR is called a Zobel network. More complicated networks can be
    used to convert more complicated impedances to pussy-cat resistive.
     
  3. Tim Wescott

    Tim Wescott Guest

    First, that input impedance is for the signal, not the LO input. The LO
    input is designed for a fairly low power 50 ohm source.

    If the input impedance is complex then its also frequency dependent.
    The value given sounds right for the high end of the frequency range but
    is quoted at 50MHz -- which contradicts the Smith chart in figure 9 of
    the data sheet.

    I'm not sure what frequency range you want on the input, but you should
    probably just consider a simple balun and live with the inductive reactance.
     
  4. Hi,
    The impedance you quote is for the signal input at 50MHz only
    and not for the local oscillator which is presumably non-critical
    since it is squared-off internally. You should only need to
    transform your VCO's output to a simple differential signal just
    as in the data sheet. If the VCO is choosey impedance-wise, then
    maybe a simple resistive isolating pad between it and the balun
    would help.


    Cheers - Joe
     
  5. steve

    steve Guest

    Tim,

    Yes, the impedance I gave was for the signal input. I'm trying to
    create a 0-500MHz frequency generator and I've got a 250-500MHz VCO, so
    I was going to mix the VCO output with a 250MHz LO signal from a
    MAX2620 fixed frequency oscillator.

    Thanks to everyone for their replys, they've been really helpful!

    Steve
     
  6. Chris Jones

    Chris Jones Guest

    The signal inputs go straight to the emitters of the switching transistors,
    by the look of it. The impedance they give would be for small-signal
    operation e.g. in a receiver. Do you mind if there are harmonics in the
    input current? You could drive it hard, if you want lots of output power.
    The impedance would be different then but it doesn't matter, all that
    matters is the input current will come out the top of the mixing cell
    chopped up by the LO waveform. If you just use a centre-tapped
    mini-circuits transformer or something like that, with a resistor from the
    centre tap to ground to sink 20mA or so out of the centre tap, that ought
    to do. You wouldn't need much swing at the mixer inputs to fully switch
    the current from one side to the other, 100-200mV ought to do it. You will
    however need plenty of current so a transformer which steps down the
    voltage and steps up the current should help. The only problem is that if
    you drive it hard like this, then you will get other mixing products such
    as 3LO +/- 3RF, which would be hard to get rid of. If you don't want these
    then you could put resistors in between the transformer and the input pins
    so that the current is sinusoidal again. The 3LO, 5LO etc. frequencies
    will still be present in the LO however, because the limiting action of the
    LO buffer will create them. For that reason, with your VCO, I would try
    using a 500MHz fixed oscillator, so that 3LO+/- fVCO will be so high that
    you can filter it out effectively.

    You might want to reconsider the oscillator frequencies, I don't know if it
    matters for you if there is some VCO leakage in the output. If you used a
    750-1250MHz VCO and a 750MHz fixed oscillator, then it would be easier to
    get rid of any LO and VCO leakage.

    Chris
     
  7. Joerg

    Joerg Guest

    Hello Steve,
    Just a word of caution: If you do this and your output needs to be, say,
    249.9MHz you will also have 250.1MHz. Unless you employ an I/Q scheme.

    Regards, Joerg
     
  8. Another hint :
    With the IQ, where the LO and the IF are required as 0 and 90 degrees,
    you'll get together with your 249.9 also 250.0 plus 250.1, but
    attenuated, say -30dB. They'll be there !
    While the fixed frequency is doable with some effort as 0 and 90
    degrees, the sweepable is rather hard to get as 0 and 90 degrees
    broadband.

    Rene
     
  9. Joerg

    Joerg Guest

    Hello Rene,
    You can get better than -30dB but yes, they'll be there.
    Unless the VCO itself runs at 4F. Which would make it more expensive.

    I would consider DDS. The AD9858 may be able to serve a range of
    0-400MHz but there should be higher clock speed chips somewhere. Of
    course, it'll cost.

    Regards, Joerg
     
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