Judging complex output coupling of this RF circuit....

Discussion in 'Wireless & RF Design' started by Himanshoo, Jan 16, 2017.

  1. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    Hi guys...

    In the diagram shows a common impedance coupling between the RF amplifier composed of T2 and mixer the there are two resonant circuits L2 //C2 and L3//C3 which are not directly coupled but coupled via an inductor L4 which controls the bandwidth.
    By introducing the controlled soft grounding of the common reference terminal through L4, the coupling coefficient across the circuit involving T2 can be very carefully controlled by component value. That makes a much more predictable VHF passband.The difference current between L2 and L3 flows in L4 and it is this difference current is effectively doing the mutual coupling. The bandwidth of the RF amplifier needs to cover the whole VHF TV band as it is not tuned with channel changes. The coupling coefficient needs to remain reasonably constant over the entire band.

    The coupling coefficient k = Lm + M / √(Lp + Lm)(Ls + Lm) —————————————1

    VHF passband = 90 - 100 MHz
    XC2 = 300 Ω // reactance of C2
    XC3 = 600 Ω // reactance of C3
    Q = 10
    M = 1
    XL4 = 42 Ω // reactance of L4

    Coupling coefficient k should remains constant throughout the whole passband of 20 Hz as described but if I substitute above values in equation 1. it would provide me different values of k for the entire passband. Then how can I seem to validate the text above…??

    Screen Shot 2017-01-17 at 12.03.49 am.png
     
  2. MrAl

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    Jun 17, 2014
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    Hi,

    It is a little hard to tell what you are asking here.

    On the one hand, you are saying that you have a quantity that you are calculating that depends only on three measurable quantities lets call them x1, x2, x3, and these quantities do not change, yet when you calculate a function of those three somehow the result changes:

    For y=f(x1,x2,x3) and with x1,x2,x3 constant we get a first value for y:
    y1=f(x1,x2,x3)

    and with x1,x2,x3 still constant we somehow get:
    y2=f(x1,x2,x3)

    a different value for y. This doesnt make sense because if x1,x2,x3 are really constant then we cant get a different value for y.

    You must be leaving something out or else not doing something right, or some other error. Maybe go over it again and see if you left something out, such as some dependence on another quantity for x1, x2, or x3, or any two or all three.
     
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  3. Himanshoo

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    Apr 3, 2015
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    As we have a 10 Mhz passband in discussion which needs to coupled to the mixer circuit...so the output stage of the RF circuit needs to bandpass all the frequencies which are lying in the VHF passband.....
    In order to pass the passband frequencies a constant coupling coefficient k is required...now the values of reactance of C2,C3 and L4 along with quality factor and mutual coupling coefficient is given. From these given values the values of C2,C3 and L4 can be calculated...
    as the vhf passband contains a set of 10 frequencies and for each single frequency the values of reactances will be different (as reactance is frequency dependent) so if reactance differ then C2,C3 and L4 will also differ....and as they differ they will change the value of the coupling coefficient k according to relation........... k = Lm + M / √(Lp + Lm)(Ls + Lm)............which is not desired ....

    So how can I prove this numerically that the common impedance coupling circuit (consisting of L2,L3,L4 and C2,C3) will provide the coupling coefficient constant for entire value of vhf passband..??
     
  4. MrAl

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    Jun 17, 2014
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    Hello again,

    It sounds like you are still asking the same question, but let me elaborate.

    You are giving the formula:
    k = Lm + M / √(Lp + Lm)(Ls + Lm).

    and that appears to show that 'k' is the coupling coefficient and depends on Lm, M, Lp, and Ls, but then you are stating that something also called the coupling coefficient depends also on some capacitor values.

    So the logical question then is how are you defining Lm, M, Lp, and Ls?

    Or, is the coupling coefficient you are talking about actually two different coupling coefficients?
    If so, then i would assume you would apply circuit analysis techniques to calculate whatever it is you need to show.
     
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  5. BR-549

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    Sep 22, 2013
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    Is this a textbook problem? Are you trying to solve for k?
    If you vary L4...will the bandpass vary? Yes, but L4 is NOT controlling bandpass. Varying any component in a resonant circuit can vary bandpass.
    If this is a real circuit.......there are TWO alignment strategies. Neither one cares about L4.
     
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  6. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    No it is not a text book problem

    yes I think it should vary...reason being is that by varying L4 we are changing the common ground level, an as the circuit is arranged as a common impedance coupling....and the current that flows across L2 and L3 also follows through L4 ..by varying L4 we actually are varying the reactance of L4 which change the potential at junction of L2, L3 and L4 hence effect coupling...

    Its my thought that k should remain constant because if k vary coupling will vary and it may happen not all the frequencies of the vhf passband will be coupled to next stage so k is essential for me....and I am looking for an experimental evidence to this.....for this is the thread all about...

    Which are those..??
    What difference would it make with and without L4 in the circuit...what is its exact role...??
     
  7. BR-549

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    Sep 22, 2013
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    I guess I look at circuits a little differently than most. To me, L4 just isolates ac from dc. L4 is not variable.

    L2 and L3 are just frequency adjustments. You may set both frequencies to the center bandpass frequency.......OR you could set one to resonate a little higher than center.....and the other to resonate a little lower than center.

    If you tune L2 for center.......then tune L3 for coupling.......you will end up either high or low of center for bandpass.
    It's important to split the difference on both coils. To keep the bandpass centered.

    You can see this easily with a sweep generator, a scope and a tuning wand.

    You start off by setting all coils to center, sweep and look at bandpass. Adjust for symmetry from center.
     
  8. MrAl

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    You need to read ALL replies if you want help. You did not answer these important questions:

    So the logical question then is how are you defining Lm, M, Lp, and Ls?
    Or, is the coupling coefficient you are talking about actually two different coupling coefficients?
     
  9. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    Lp is actually L2 , Ls is L3 and Lm is L4 and M is the mutual coefficient between Lp and Ls...you can compare this from this figure

    Screen Shot 2017-01-18 at 8.56.22 am.png
     
  10. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    dc from where? L4 is an inductor, it passes dc if you are talking about dc from 15 V supply...

    what I know is that if we couple two resonant circuit and individually set both the resonant circuit at different resonant frequencies will give prominent output at both the frequencies dropping other non dominant frequencies...

    Don't you consider it as a "bottom inductive mutual coupling"..?? where a common impedance is provided by L4 where the voltage drop across it is reflected to next stage...
     
  11. BR-549

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    Sep 22, 2013
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    When you sweep a circuit, the scope will display the bandpass. You can see the adjustment change in real time. You can watch and shape the peaks. You can see the coupling.

    The sweep part starts at 18 min.

     
  12. MrAl

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    Hi again,

    I had a feeling that's what you are talking about but wanted to be sure.

    If you look at the attachment, you can see that the factor k is for the circuit shown in figure A and is not valid for circuit B or C.
    It's only valid as a 'coupling' factor (if you want to call it that) for the circuit in figure A because when we have other components those have to be considered in the analysis of the transfer function from input to output as well as the normal k coupling factor that is associated with the three inductors alone.

    The circuit shown in B is your circuit but you probably have to compute that using a current source as input.

    The circuit in C is the circuit with the usual input and output impedances shown as resistors, which you may be able to use. You have to find a way to estimate the input and output impedances in order to use that circuit, or you might get away with using some arbitrary set of impedances like 50 Ohms just to get an idea how this circuit works.

    So to put it another way, the coupling factor associated with the three inductors alone shown in A does not change with frequency, but the transfer function amplitude will probably change with frequency for the other two circuits because we have capacitors included in those circuits. But if you only need to consider the coupling coefficient for the three inductors, then the formula shown in A works ok.
    The formula does not work, in general, for the circuits shown in B and C, that's why i put an unknown function under them for now.

    You could choose which circuit you want to use and then evaluate it, or we could do that next if you like.
     
  13. BR-549

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    Sep 22, 2013
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    Can we see some photos of the circuit?
     
  14. MrAl

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    Jun 17, 2014
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    Here's an example using the circuit shown in figure C.

    With R2=R1=1, C2=C1=1, L2=L1=1, L3=1,

    we see a bandpass like filter with w from 0 to 4, with peak amplitude at w=sqrt(sqrt(37)+1)/(3*sqrt(2))
    which is about w=0.627, but on either side of that 'w' we get a reduced amplitude which indicates the 'coupling' had changed.

    Also important is we have to know how we are going to define what is the same amplitude and what is different.
    For example, if we have an amplitude at w=1 of 0.9990 and an amplitude at w=2 of 9991 will we still consider that the same amplitude? In most cases we would consider the bandwidth which means that we dont consider that we have seen an amplitude change until we see a decrease of about 30 percent (down 3db) in the amplitude. So it partly depends on what the objective is.
     
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  15. BR-549

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    Sep 22, 2013
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    I'm old school. To me "mutual coupling" means the field of a coil interacting with another coil.

    Unless a photo shows me different, the only mutual coupling I see is at the ant. transformer and the output transformer.

    I would think L2 and L3 are shielded coils.

    Course we don't know if he's trying to mod an existing circuit or build from scratch.
     
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  16. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    I have found a nice article discussing the type of coupling in the original circuit...but none of the description seems valid with my understanding..
    It would be better if we understand this in terms of simple voltages and current across the component....I have rearranged the circuit..

    equ_circuit.png

    Now suppose the two resonant circuits are at resonance as a result they will show high impedance at resonant frequency...both the circuits will have the same impedance due to resonance...so point X and Y will at same potential hence no current will flow across the path and the inductor L4 and the signal voltage is coupled directly due to mutual induction to the next resonant circuit as well....but the text claim opposite of it...as it says..
    " ..signal voltage across L4 (Lm) is coupled from one tuned circuit to the other."
    Since if there would be some signal voltage developed across L4 there would be some signal current flowing through it, but as no current flows through L4 then how signal voltage coupled through L4...

    impedance coupling.PNG
     
    Last edited: Jan 22, 2017
  17. BR-549

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    Sep 22, 2013
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    Do you have a photo of the circuit in post #1?

    I do not believe the circuit in post #1.......is the same as all the other posted examples in this thread.
    I believe that you are using one circuit to explain another. BUT......they are DIFFERENT circuits.

    Mutual inductance shares magnetic flux, not current. L2/C2 are series tuned. L3/C3 are series tuned. That means peak current at resonance.
    But all of the examples you have used to understand this are parallel tuned.

    We can only answer on and with the information given. I am not positive on my comments because I can't see the circuit.
    I look at circuits in a mechanical sense. If you need a mathematical explanation.......MrAl will be glad to help you.....if he has the needed information.

    I would highly recommend a sweep generator. Just a 5 MHz scope with detector can be used to sweep up to UHF and GHz circuits.

    These rf sweep generators can cost. Dick Cappels has provided a solution.

    http://www.cappels.org/dproj/functsweep/functionswp.html

    And here is something more elementary.

    rf-sweeper.jpg

    http://www.vk2zay.net/article/93

    Good luck.
     
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