Using a NanoVNA to measure the efficiency of an End Fed Half Wave Transformer

Поделиться
HTML-код
  • Опубликовано: 27 авг 2024
  • Hello - the is Ariel NY4G this time going into a deeper dive on how a NanoVNA can be used to measure the efficiency of an end fed half wave transformer.
    End Fed Half Wave (EFHW) antennas have certainly become popular. One thing that is not well understood nor documented is whether one of these autotransformers used in these popular antennas is wasting power that we can begin to care about.
    An explanation of how to compute this efficiency is explained based on the measured S parameters by the NanoVNA.
    Two methods are explored - the resistor method and the back to back method. Both these methods ought to give answers in the same ball park.
    I tend to favor the resistor method because it matches my intuition of how an antenna system is configured.
    In the end several comparisons are made depicting both iinefficient transformer windings against efficient ones.

Комментарии • 14

  • @mareksp9tkw908
    @mareksp9tkw908 Год назад +1

    Thank you!!! This is a great video, very informative and has many detailed explanations what and why. 73 Marek SP9TKW

  • @K9BCM
    @K9BCM Год назад

    Thank you for putting this video and information together! It has been very helpful!

  • @yowsa52
    @yowsa52 6 месяцев назад

    Awesome presentation.

  • @kennymanchester
    @kennymanchester 2 месяца назад +1

    Ariel, excellent presentation Sir. Thank you. My hang up in my quest to understand how to measure the performance of my entire EFHW single band vertical antenna is that there is this "assumed range" of impedance at the feedpoint. . So, picking an arbitrary resistor value is making my head hurt a bit. While I believe the resistor method for effectively measuring the efficiency of the transformer is sound, what if the transformer matching ratio is inaccurate? How do I know I'm not off by a thousand ohms or more between the transformer secondary and the radiator -the real actual feedpoint impedance? I am hoping this makes sense. For some reason, I'm not seeing how I can use my tools and instruments to know I've got the most power possible as radiation resistance, and not heat loss in my efficient but incorrect matching transformer. I'm hoping you can help me figure this out. Best 73, Kenny NZ5i

    • @CaptainRadioAdventures
      @CaptainRadioAdventures  2 месяца назад

      The 64:1 impedance ratio is an estimate based on the secondary to primary turns ratio. 50x64 is 3200 ohms. The actual end fed impedance is band dependent and will have a return loss or VSWR that is also band dependent. The real antenna impedance at the feedpoint will vary and in my experience - as one adjusts the wire length for lowest SWR - this is typically very close to 3200 ohms on 20m. On a wire cut for 40m fundamental - the best SWR is typically on the first harmonic and not the fundamental but that is the way I tend to balance the bands. You will get 3200 ohms on the real antenna by adjusting wire length. You will not be off by a thousand ohms unless you choose to do so. More than likely, you will be off by a few ohms. I am an engineer - and as engineers typically think close is good enough

  • @ingeniousys
    @ingeniousys 9 месяцев назад

    Thanks for the video. Very interesting. I like the explanations, but when I use the calculations prescribed with readings obtained from a NanoVNA, I sometimes get results that are physically impossible - with transformer efficiencies in excess of 100%. Something isn't quite right. The traces on the NanoVNA look good, but the efficiency calculations don't hold up. I have checked and re-checked the calculations, including using the proper sign of the variables, but still get physically impossible results.
    In the case of the back-to-back method, are you sure that the transformer_loss = (insertion_loss/2) - mismatch_loss? Or should it be transformer_loss = (insertion_loss - mismatch_loss)/2? My reasoning behind this is that if you look at both transformers back-to-back as a single transformer, then transformer_loss = insertion_loss - mismatch_loss. Splitting the transformer loss between the two transformers means simply dividing the transformer loss by 2. Hence it seems that it should be transformer_loss = (insertion_loss - mismatch_loss)/2,

    • @Mike-nf5pt
      @Mike-nf5pt 8 месяцев назад

      I've had the same problem with calculated efficiency > 100%. It's caused by inaccuracy in the NanoVNA's measurement of S21 when you take into account all the strays associated with the measurement. You can usually get a reasonable result with a "naked" transformer, short leads and a good test jig, but once you build it into a box it becomes very difficult. Incidentally, you need to add the input impedance of the S21 port (nominally 50 ohms) to the resistor value.

  • @hankhamner3671
    @hankhamner3671 9 месяцев назад

    Great video! How do you calculate or find the through power used in your analysis?

  • @geokuf9679
    @geokuf9679 10 месяцев назад

    Hello Ariel, with a great deal of excitement I watch your video as EFHW antennas are my favorite. My location is overwhelmed by RFI and operating portable is my preferred way of playing radio. Seeing S0 between stations on 40m is a real joy. All of my EFHW transformers are made by myself. The latest is 56:1 based on experiments of Colin MM0OPX. It works very well, I made two of them just to test the efficiency of it.
    Your method of testing them on the single unit is fantastic. I attempted to replicate your spreadsheet, but I got stuck. Would you be willing to share your spreadsheet file with me?Thank you for all the good things you do for us on RUclips.
    Jurek SP6CAL
    73, 44

  • @LA9RKA
    @LA9RKA 11 месяцев назад

    Very interesting. Is the spreadsheet complicated?

    • @CaptainRadioAdventures
      @CaptainRadioAdventures  11 месяцев назад

      Not at all

    • @radioman5466
      @radioman5466 8 месяцев назад

      @@CaptainRadioAdventures Would you share a copy of the spreadsheet? I am getting lost at the calculation of the Through Power. I do not see where the 0.0023 value is coming from.

  • @moondog13gmail
    @moondog13gmail Год назад

    I loved this explanation and the equations. Thank you for all the time and effort