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coupling factor??.....

Discussion in 'Electronic Design' started by [email protected], Apr 27, 2005.

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  1. Guest


    I would like to know why coupling factor for IF transformer affects the
    centre frequency. I hv a circuit in PSpice with two IF transformers...
    and when I reduce the coupling factor I get the required centre
    frequency which is 455KHz...

    for both IF transformers

    The inductor values are

    L1 = 680uH
    L2 = 97uH

    and original coupling factor is 0.9

    but after simulations the centre frequency obtained is 451KHz

    when the coupling factor is reduced to 0.75 I get the required centre

    anyone with an answer...

  2. The k for an IF-type of coupled transformer is normally
    taken as << 1. With a coupling factor of 0.9 the two
    tuned circuits are probably going to merge into looking
    like one circuit, which could have a resonant frequency
    that is slightly offset by the leakage inductance... but
    exactly how far offset depends on the definition of the
    centre frequency. Possibilities could be, the frequency
    at which the circuit looks resistive to the generator, or
    where the maximum voltage occurs on one of the windings,
    and so on.

    So more information would be handy....

    What values of capacitor across the pri and sec?
    Did you assign an estimated Q for each resonant circuit?
    How is the circuit stimulated?
    What did you measure in the circuit to decide on the
    centre frequency?
  3. John - kd5yi

    John - kd5yi Guest

    The mutual inductance (M) between the primary and secondary is M = K *
    sqrt(L1 * L2). As you can see, the mutual inductance changes with the
    coefficient of coupling (K).

  4. Jim Backus

    Jim Backus Guest

    Have you checked the frequency response over a range of (say) +/- 25
    kHz? You may find that the simulation is showing a double peaked
    response. Coupled tuned circuits used to be described as undercoupled,
    critically coupled and over coupled. When circuits are undercoupled,
    the centre frequency is correct but the output amplitude is reduced.
    With critical coupling the maximum output is obtained with a single
    resonant frequency. If overcoupled, the circuit has two peak
    responses, one below the nominal resonant frequency and the other
    above. The output at the nominal resonant frequency is reduced
    relative to the two peaks. Overcoupling can be useful in increasing
    the bandwidth of a coupled circuit.

    Refer to one of the old textbooks for more explanation - I've just
    checked it in a book called "Radio Engineering Handbook" editied by K
    Henney and published in 1950. Terman would cover it but I don't have a
  5. There's a brief description in an old Samuel Seely
    textbook. However, at moderate Q's (20-ish) the k
    would be around 0.05 for optimum coupling, going
    even lower for higher Q's. This is at least a decade
    below the values of k that OP is talking about.

    With a k near unity the circuit might be better imagined
    as a tuned power transformer, with the leakage inductance
    and secondary capacitor transformed over to the primary.
  6. I read in that Tony Williams
    <>) about 'coupling factor??.....',
    I think caution is necessary. Some people use 'k' to mean what other
    people call 'Qk' (Q x k). Critical coupling is achieved at Qk = 1.If the OP's values were not 'Qk' values, he would have seen response
    peaks very far away from his expected centre frequency, not just a small
  7. Sorry for the delay in replying to this post.

    Ok, thank you. Presumably the Q to use is the single
    value calc'd at Fres...... which reminded me of a remark
    made by DNA the other day, about apparently rigorous final
    sums (for transformers) being based early assumptions
    that may not precisely carry through.

    In this case the textbooks develop a sum for the final
    output voltage (against coupling), based on the assumption
    that Xc=Xl at Fres. But what happens to other things, like
    the primary current, at just off Fres (where Xc is not
    equal to Xl)?

    So I LTspice'd it, and the .asc file is at the end of this
    post. The numbers were jigged slightly to ensure that Fres
    and Q were the same on both sides. Fres is 455249Hz, Q is
    92.48, giving a critical K1 of 0.01812.

    Sweep it, look at I[R2], and there is the single peak of
    critical coupling in the secondary current. Fiddle K1 up
    and down, and you can see the effect of over/under coupling.

    But now look at I[R1] when K1 is 0.01812. At critical
    coupling the primary current is already double-peaking, and
    it is not until K1 is significantly reduced that it reduces
    to a single peak, at Fres.

    Perhaps this is what the OP was seeing?

    Version 4
    SHEET 1 880 680
    WIRE -208 272 -208 224
    WIRE -208 224 -64 224
    WIRE 0 224 96 224
    WIRE 96 304 96 336
    WIRE 96 416 -208 416
    WIRE -208 416 -208 352
    WIRE -208 448 -208 416
    WIRE 208 224 352 224
    WIRE 352 224 352 288
    WIRE 208 416 352 416
    WIRE 352 416 352 352
    WIRE 208 304 208 336
    WIRE 352 416 352 448
    FLAG -208 448 0
    FLAG 352 448 0
    SYMBOL ind2 80 208 R0
    WINDOW 0 -64 45 Left 0
    WINDOW 3 -66 81 Left 0
    SYMATTR InstName L1
    SYMATTR Value 679µ
    SYMATTR Type ind
    SYMBOL ind2 192 208 R0
    SYMATTR InstName L2
    SYMATTR Value 97µ
    SYMATTR Type ind
    SYMBOL res 80 320 R0
    WINDOW 0 -58 39 Left 0
    WINDOW 3 -59 75 Left 0
    SYMATTR InstName R1
    SYMATTR Value 21
    SYMBOL res 192 320 R0
    SYMATTR InstName R2
    SYMATTR Value 3
    SYMBOL cap -64 240 R270
    WINDOW 0 32 32 VTop 0
    WINDOW 3 0 32 VBottom 0
    SYMATTR InstName C1
    SYMATTR Value 180p
    SYMBOL cap 336 288 R0
    SYMATTR InstName C2
    SYMATTR Value 1260p
    SYMBOL voltage -208 256 R0
    WINDOW 3 37 87 Left 0
    WINDOW 123 37 115 Left 0
    WINDOW 39 0 0 Left 0
    SYMATTR InstName V1
    SYMATTR Value ""
    SYMATTR Value2 AC 1
    TEXT -76 129 Left 0 !.ac lin 2000 450000 460000
    TEXT 56 200 Left 0 !K1 L1 L2 0.010812
  8. I read in that Tony Williams
    <>) about 'coupling factor??.....',
    Maybe. This double-peaking of the primary current is a little-known
    effect, mainly because it hardly ever matters. But if you thieve off a
    bit of energy from the primary side of a critically-coupled IFT for some
    purpose, you may run into the effect.
  9. There are other apparent anomolies. For example, the
    theoretical value for critical coupling (K1 = 0.010812)
    appears too low. At this K1 there is still a single peak,
    shifted down slightly to 454.7KHz. K1 has to be upped by
    about 6% to get even the first hint of double peaking.
    There is another slight departure from the textbook sums
    when over-coupled and calculating f1 and f2.

    These departures don't matter in practical terms, but it
    is an interesting observation on textbook sums, and the
    approximations that were made to reach a final answer.
  10. /|\
  11. I read in that Tony Williams
    <>) about 'coupling factor??.....',
    The textbooks generally assume that the two circuits have approximately
    the same component values. Here the asymmetry is quite pronounced.
  12. Try identical comps each side, it doesn't make any
    difference, nor should it afaics. The textbooks'
    initial assumptions only seem to require precisely
    equal L*C products and (preferably) similar Q's.

    The OP does seem to have noticed an oddity in the
    behaviour of coupled circuits. As long as that
    primary current is double-peaking then the sec-
    -ondary passband is distorted slightly, resulting
    in a peak below Fres. At about 2/3 critical coupling
    (for my chosen Q) the primary double-peaking more or
    less disappears, and the secondary voltage peak is then
    at about Fres.
  13. I read in that Tony Williams
    <>) about 'coupling factor??.....',
    But the textbooks assume *simplified* circuits, just as your model does.
    The inductors have shunt capacitance and shunt resistance as well as
    series resistance, and the capacitors have both series and shunt losses.

    Considering the thousands of precision theoretical analyses and millions
    of actual measurements that have been made on tuned coupled circuits
    over the years, a claim to have found an asymmetry (other than one
    produced by varying either C or L to 'align' the circuits) has to be
    regarded as an extraordinary claim, requiring extraordinary proof.

    If you actually twiddle trimmers or inductor cores to 'tune up', then
    you DO get asymmetries. This is exhaustively dealt with in two books by
    W Th H Hetterscheid, for example.
  14. Fred Abse

    Fred Abse Guest

    That's what I get, too. Double peaking was the first thing that came to
    mind here.

    BTW, it might be a good idea to switch on "Greek mu conversion to
    lowercase u" for posted netlists. Some news readers (Pan for example)
    display mu correctly, but convert to a quoted string on copy-and-paste.
    Took me a while to figure out what the curly braces that appeared on the
    copy were all about.
  15. Sorry about that...... I didn't notice that
    LTspice changed a 'u' typed-in on the keyboard
    into a mu.
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