Bias Tee Circuit Design & Simulation How-To

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  • Опубликовано: 25 ноя 2024

Комментарии • 11

  • @donepearce
    @donepearce Год назад +1

    The "cold" (non-RF) end of the bias inductor needs decoupling. Also, at 6GHz you should consider a quarter wave line instead of a discrete inductor. It is likely to be a better performer at such a high frequency.

  • @ROMAARIGATO
    @ROMAARIGATO Год назад

    Thanks for your videos! Can you consider in future videos the construction of Bias-T in PoC (Power over Coax) filters for powering and transmitting video signals to/from video sensors.
    Manufacturers such as Murata or TDK show options for constructing such filters and it would be interesting to see your point of view on solving this problem.

  • @juandavidmedinatobon9421
    @juandavidmedinatobon9421 Год назад

    Great video! One question. When choosing a particular inductor component, how should the parasitic elements be considered? Should you choose a component with a self-resonant frequency close to the operating frequency? Thanks!

    • @Zachariah-Peterson
      @Zachariah-Peterson Год назад +1

      Self-resonant frequencies should always be above the operating frequency except in some specialized application. This is true for inductors, capacitors, etc... For inductors there are many options, even SMD wirewound options, that are marketed specifically for RF systems. These will have a frequency specification listed in the product documentation.

  • @lunarjournal
    @lunarjournal Год назад +1

    When computing the S-parameters @ 16:39 where did the 0.26 and 100.26 come from? Thanks! @Altium Acadamy

    • @mushymellow2
      @mushymellow2 Год назад

      Not 100% sure but I believe he is using S11=(Zin-Zs/Zin+Zs) where Zin is the 50.26Ω and Zs is the source impedance which is 50Ω (50.26-50=0.26Ω). Same logic applies for the 100.26.

    • @Zachariah-Peterson
      @Zachariah-Peterson Год назад

      Matthew is correct, it is from the difference between the two impedances (absolute value) being 0.26. Since we have 50 Ohms output impedance set by the source, that is interfacing with the input impedance looking into the tee which is 50.26, so those are the two impedances we use to determine the return loss.

  • @kecsrobi6854
    @kecsrobi6854 Год назад

    In your calculator is there a way to make one if the values fixed? asking since I saw 2 non standard values and those are expensive . So for example can I mimit the capacitor range to lets say E24 or E48 values and then get wherever on the inductor side?
    Also is there a reason NOT to use a higher n?

    • @Zachariah-Peterson
      @Zachariah-Peterson Год назад

      You could limit one of the values for a given target output impedance and frequency, but you couldn't do it in the calculator. This might limit the impedance ratio you can achieve. I think the important point is to look at the impedance ratio like a tuning parameter for the bandwidth, that's what I wanted to show when taking n from 1 to 10.

  • @BMM66666
    @BMM66666 Год назад

    Hi Zach, I made a simple simulation on LTSpice and the results is the same as yours video.
    But the question is how could I determine my real ‘n’ factor in the blog calculator?
    I started with n=1 to see the simulation results, then modify n=10, 100, 1000,…
    The passed energy to the load and the rejected energy to the DC will change by frequency. It makes sense.
    How could I determine the n factor in real circuit? Is it just depends on the simulation?

    • @Zachariah-Peterson
      @Zachariah-Peterson Год назад

      From a real circuit, you would have to know the impedance of each leg in the bias tee, then that would be used to calculate the impedance ratio. As long as you know the components in the circuit then you could calculate it.