SWR reading from analyser

Discussion in 'Homework Help' started by ninjaman, May 6, 2015.

  1. ninjaman

    Thread Starter Member

    May 18, 2013
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    hello
    i am trying to simulate an SWR analyser circuit.
    it is two peak detectors with amplifiers
    upload_2015-5-6_21-16-33.png

    the bottom meter shows the forward voltage the top right meter shows reverse voltage
    using the formula, forward + reverse / forward - reverse = ?
    i get this, 4.9 + 2.8 / 4.9 - 2.9 = 3.6
    the load is 150 ohms, with a 50 ohm reference this should give a 3:1 ratio. instead i get 3.6:1
    any ideas why?
    i dont know, i think it may be because i am only using a resistance instead of using capacitors/inductors. the load is supposed to be an antenna. i thought that if it had a 150 ohm it wouldnt matter much as it is an impedance.
    any help would be great!
    thanks
    simon
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I believe with this type of circuit one normally creates a scaling factor number array which embodies the relationship between the observed "reflected" signal values and the expected VSWR values (based on the range of load impedance values (short circuit to open circuit)). The problem with this scaling relationship is that it is (potentially) both non-linear and non-symmetrical for load values above and below the system characteristic impedance (typically 50 ohms). Plus one usually has to "normalize" the scaling, by first setting the "full-scale" reading with the load (antenna?) disconnected - i.e. the condition corresponding to 'infinite' VSWR.
     
  3. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    When you substituted a 50 ohm load where you have the 150, did you get a 1:1 ratio?
     
  4. ninjaman

    Thread Starter Member

    May 18, 2013
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    no, when i put in 50 ohm load i got the same 3.6 answer as the 150 ohm. the load is purely resistive. it has no capacitor or inductor.
    i dont know what is wrong here.
    i have to complete my project soon. i have to do a write up and my tutor has pretty much told me im going to get a fail. i dont think that its fair to be honest. i didnt know how hard this would be or what the problem is.
    what would be acceptable for a write up if the project didnt work?

    thanks for your help joe and tnk
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    What level is the input signal?
    Perhaps U1A is saturating ....
    Which may well be more annoying from your perspective.
    It will be easy to check what's happening with the amplifiers by confirming the apparent gain is consistent with the designed gain value.
     
    Last edited: May 7, 2015
  6. ninjaman

    Thread Starter Member

    May 18, 2013
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    i am using 0.5 vp
    i dont understand what im doing wrong the formula is forward + reverse/forward - reverse
     
  7. bertus

    Administrator

    Apr 5, 2008
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  8. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    My response would be to ask - are you actually measuring the system Forward and Reflected wave voltages? The circuit is a simple bridge arrangement which 'balances' when the load impedance is exactly 50 ohms. Vr is a measure of the bridge imbalance - no more than that. In that situation with a 50 ohm load, the observed VSWR should be unity since the bridge is in balance and Vr approaches zero value, meaning your equation is 'correct' in that instance. So the circuit is a useful device for determining whether the user is approaching a good matched condition. For other load impedances the relationship between true VSWR and the ratio (Vf+Vr)/(Vf-Vr) is not a simple linear one using this circuit topology.
    Check this link http://vk5ajl.com/projects/swrbridge.php
    Measuring actual forward and reflected components usually requires the use of a directional coupler.
     
    Last edited: May 7, 2015
  9. ninjaman

    Thread Starter Member

    May 18, 2013
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    hello
    i have to use what i have already. i cant get the correct setup for this. i have tried a few things. i dont know what is wrong. i thought it was the way the diodes were connected. i dont know how to follow the voltage through the circuit.
    upload_2015-5-7_12-8-41.png i have this circuit that seems to work ok. the amp is doing what it should with the correct gain. i get the DC signal i need. i connected the diodes as they should be according to the schematic that i have and the above part is not giving the same voltage as the first setup. using the formula and results im getting some really weird reading. i changed R8 the unknown to 100 ohms and got a result of 19. i dont know what is going on
     
  10. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    What kind of diodes do you use?
    Silicium diodes will have a to large voltage drop.
    For the application pin or germanium diodes can be used.

    Bertus
     
  11. ninjaman

    Thread Starter Member

    May 18, 2013
    306
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    hello
    i measured the voltage across the diodes and one was almost one peak voltage, while the other was negligible. i think that the problem is with the diode that connects to the unknown value. it seems to short circuit somewhere, i think!
     
  12. t_n_k

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    Mar 6, 2009
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    Which diode are you referring to? Was this with load set to 50 ohms?
     
  13. ninjaman

    Thread Starter Member

    May 18, 2013
    306
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    all resistors were 50 ohm, the diode i was talking about is the diode in the leg with the unknown resistor (here still a 50 ohm load)
     
  14. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    With the load at 50 ohm, the bridge is balanced and one would expect either a very small or negligible voltage across the diode in question.
    Some further thoughts on the concept ...
    I would think to get results close to "ideal" over a significant range of loading conditions, one has to ensure the circuitry doing the detection of the relevant signals from the bridge arrangement imposes as small a loading effect as possible.
    Clearly one can achieve this under simulation by measuring the two signals representing Vf & Vr with high impedance AC probes - one single ended (non load bridge arm center to ground) and one differential (between the bridge arms center points). You can easily verify that this works and the observed VSWR is consistent with expectation - including cases in which the load has a reactive component.
    With AC to DC conversion of the two signals, a rectifier circuit which preserves this high impedance loading would be preferred. Practical circuits of this type involving the bridge arrangement with rectifying detectors, attempt to reduce this loading effect. In particular, one would try to reduce the peak impulse currents into the filtering capacitors which might degrade the linearity of signal conversion. With respect to the rectifier itself, it is difficult to approach ideal behavior when the diode forward drop is substantial. I guess with a sufficiently low source frequency one could employ ideal op-amp rectifiers to overcome the latter problem. However, this idea lacks credibility when considering high frequency circuits extending into the tens of MHz range and beyond.
    In principle idealized concept:

    Ideal  VSWR bridge concept..jpg
     
    Last edited: May 10, 2015
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