#1395

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  • Опубликовано: 7 ноя 2024

Комментарии • 41

  • @yakovdavidovich7943
    @yakovdavidovich7943 Год назад +1

    That comment at the end about how high power attenuators are directional was interesting. I hadn't seen that presented clearly before, so thanks!

  • @rickwise9910
    @rickwise9910 Год назад +3

    And the reason the big guys looked right without load is that at 30 dB, the resistors to ground are about 53 ohms

  • @aduedc
    @aduedc Год назад +2

    Great Clip.
    I like the way you redraw the T and PI network for better understanding.
    Also, The Pi Network can be used to change impedance( ie. from 75 ohm to 50 ohm etc. ). The advantage Pi network over transformers for impedance transform is that they can have wider band width, and can work all the way down to DC, where transformers can not.
    Thanks for great clip.

    • @bayareapianist
      @bayareapianist Год назад +2

      However, the resistors solution would waste a lot power which the transformer one would not if your range of frequencies used are fixed. The old pocket radios were full of those transformer because the wasted power was important, but the bigger radios and TVs just full of heavy resistors and the wasted power was not an issue for the designers who didn't care how much power converts to heat, because they weren't battery powered.

    • @aduedc
      @aduedc Год назад +1

      @@bayareapianist For higher frequencies, you can use tapered transmission line to do wide band impedance matching.

    • @bayareapianist
      @bayareapianist Год назад

      @@aduedc you can use Balun which is a kind of transformer. Baluns are used both for RF and audio frequencies. Transformers can give you differential outputs which are used in audio equipments to get ride of ground effects (hamming).

    • @aduedc
      @aduedc Год назад

      @@bayareapianist BALUN (BALanced - UNbalanced) are used to convert signal to/from differential to/from common mode.
      A transformer based BALUN is constructed from two winding on a core one of them with a center tap, which gets connected to the other winding and used as ground. Old Push-Pull stages used to use transformer based balun,
      At very high frequencies Rat-Race structure is used to achieve this.
      At low frequencies opamps are used to achieve this.

  • @byronwatkins2565
    @byronwatkins2565 Год назад +2

    The pi network divides the (open circuit) voltage twice; the incoming line also has 50 Ohm output impedance. At 14:40, your resistance values are backwards. As shown, this has input impedance less than 37 Ohm.

    • @IMSAIGuy
      @IMSAIGuy  Год назад +2

      damn it. you are right

    • @byronwatkins2565
      @byronwatkins2565 Год назад +1

      @@IMSAIGuy Don't sweat the small stuff. Peer review is quite valuable.

  • @Randrew
    @Randrew Год назад +2

    I would like to attenuate the calories of pie, while amplifying the flavor!

  • @rfburns5601
    @rfburns5601 Год назад +1

    Wow! You just stretched my head. Thanks.

  • @4dirt2racer0
    @4dirt2racer0 Год назад +2

    its always been crazy as hell to me that were pushin so much rf thru passive components that they need big ass heatsinks thats a good bit of power goin just to heat

    • @robinbrowne5419
      @robinbrowne5419 Год назад

      Travis - It does seem strange to generate a bunch of power just to waste it in resistors. I think that this could be an old way of doing RF. Much as power supplies used to use linear regulators which wasted a lot of heat. Maybe radio transmitters need to waste some heat as part of a feedback loop to produce a stable output. I'm just guessing though. Cheers.

  • @mariomionskowski6223
    @mariomionskowski6223 Год назад +1

    excellent

  • @BryanTorok
    @BryanTorok Год назад

    I would like to point out that at RF frequencies, one can use a 10 dB or higher attenuator as a dummy load, properly rated for the power to be dissipated. It would be best practice to put a 50 ohm load on the output, but in a practical sense it will work without it. At 10 dB, only one-tenth (1/10) of the power reaches the output. With the output open, that power is reflected back. But, by the time it goes back through the attenuator, the reflected power is one-hundredth (1/100). You are not likely to even see that small of reflected power on the typical SWR meter.

    • @IMSAIGuy
      @IMSAIGuy  Год назад

      I did a video: ruclips.net/video/WakbPENJHXA/видео.html

    • @BryanTorok
      @BryanTorok Год назад

      @@IMSAIGuy Tnx. I went to watch the video and found that not only had I watched it, I had made a comment on it. I just went and joined your Patreon. I watch enough of your videos that I thought I should contribute.

  • @subramanianr7206
    @subramanianr7206 Год назад +1

    Hi, a nice attempt to explain the PADs. But there are many confusions.
    There are symmetrical and non-symmetrical configurations as there are balanced and unbalanced types.
    They can be pi, L and T shapes.
    Apart from the attenuators there are 50 ohm ( or 75 ohm) terminators that would have only one connector.
    The nearest value commercially available for the -6dB pad is 150 ohm and 39 ohm.
    I used 5w carbon resistors in my 4×SD2933 MOSFET transistor amplifier.
    By simple oversight you drew the value of resistors swapped for the -6dB pad at 14.26, though it was drawn correctly in the start.
    The thin film resistors on the BeO ceramic substrate mounted on the aluminium finned heat sink are real engineering Marvel.
    De VU2RZA

    • @yakovdavidovich7943
      @yakovdavidovich7943 Год назад +1

      Sounds like your plan would make a much longer video. All good info, but content creators have to decide where the boundaries for a single video are, and I think this one is fine. The nit about the values being swapped is fair, but other than that I don't think there are "errors" (or "confusions") so much as it's just not treating all kinds of attenuators in maximum depth.

    • @subramanianr7206
      @subramanianr7206 Год назад +1

      @@yakovdavidovich7943 yes, a longer video is not new to Imsai guy and I would always appreciate and enjoy them.
      As he has shown it cannot be considered as a simple voltage divider but can be explained only by the mathematical formula.
      Only the symmetrical types would pose 50 ohm (or say 75 ohm) at the input and the output.
      Asymmetric devices especially the L type would present different values and by adopting a 50 ohm you can measure the same.
      Moreover it doesn't require a 50 ohm termination at the output; it's designed with the assumption that the circuit would present the characteristic impedance both at the input and the output.
      The brunt of the power is borne by the first shunt component and must be rated to handle it. Other components on the downstream can be derated accordingly. The value of the resistor (usually the shunt value) would be very different the 50 ohm.
      It's like the LC low pass filter design where the passive resistors are replaced by the passive LC components to achieve certain degree of attenuation while maintaining certain characteristics impedance value.
      De VU2RZA

  • @dennisqwertyuiop
    @dennisqwertyuiop Год назад

    great information

  • @johnwest7993
    @johnwest7993 Год назад +1

    I'm looking for a source of SMT resistors that I can use to build attenuators into my RF circuits. I can't be plugging $20 attenuators into everything I build.

  • @JanJeronimus
    @JanJeronimus Год назад +1

    Indeed a nice attempt for explanation. However several things are presented simple. E.g. To the attenuators that measured a 50 ohm resistance no load was added. And impedance is not the same as resistance.

    • @aduedc
      @aduedc Год назад +1

      Indeed, you need to connect it to network analyzer to find out how good match it is.

    • @bayareapianist
      @bayareapianist Год назад +1

      @@aduedc I totally agree. I mentioned to him in one of his posts while ago but I think this guy's doesn't know the difference. He keeps measuring everything with his ohmmeter.
      A VNA would be a better idea.
      If I say in a stupid way, ohmmeter is to measure resistance in DC. VNA measure impedances and phases in ac which are frequency dependent. A good attunator has diagrams which show these values within its frequency range.

    • @IMSAIGuy
      @IMSAIGuy  Год назад +1

      the pads are rated to DC but you are right about S11 here is video: ruclips.net/video/WakbPENJHXA/видео.html

    • @bayareapianist
      @bayareapianist Год назад +1

      @@IMSAIGuyYes. A good way to see if they work down to DC is to measure them with ohmmeter ;)

  • @pravardhanus
    @pravardhanus Год назад +2

    I am waiting for the teardown of one of the high power attenuator.

    • @KG4JYS
      @KG4JYS Год назад

      I'd also like to see a teardown.

    • @mikesradiorepair
      @mikesradiorepair Год назад

      That would be a bad idea since they contain hazardous materials like BeO (beryllium oxide) which is carcinogenic.

    • @subramanianr7206
      @subramanianr7206 Год назад +2

      @@mikesradiorepair there's nothing to worry in opening it up. All RF power transistors have the BeO ceramic caps on them. You can touch or even try to taste it. You won't die. If you powder it and inhale you have trouble. The big attenuator has Hex screws and can be removed. What imsai guy showed in the picture you would see inside; yes, I have opened it and seen the inside. It was a 100W (continuous) 30dB attenuator. By mistake I pumped 300W for a second or so; it went qrt. On opening I could see one of the resistors was open.
      If you inhale the dust (BeO) on several occasions for a considerable time you will end up with a condition called Berylliosis; it would present like other occupational diseases like asbestosis, silicosis etc. Lung cancer, squamous cell carcinoma (or other kinds) will develop after say 20 years or more of being exposed to the dust.
      De VU2RZA

  • @BryanTorok
    @BryanTorok Год назад

    Why do the large attenuators show 50 ohms both ways WITHOUT a 50 ohm load on the output?

    • @IMSAIGuy
      @IMSAIGuy  Год назад

      run the calculator for 30dB and you will see

  • @robinbrowne5419
    @robinbrowne5419 Год назад

    This is interesting. Thank you. I can see how the Pi attenuator can be used to match the impedance while reducing the power. But it still borders on black magic for me. Especially when Dbs are involved. Why not just say 0.5 instead of -3db ??? Weird. But there must be some reason from the days of the ancient wizards at Marconi or somewhere :-)

    • @IMSAIGuy
      @IMSAIGuy  Год назад +1

      using decimals of gain does not work well do to the large dynamic range of radios. +50dBm transmit and -127dBm receive. you would end up with lots of zeros

    • @robinbrowne5419
      @robinbrowne5419 Год назад

      @@IMSAIGuy True there would be lots of 0s. Also I read somewhere that people used dB because it is the Log and they could just add instead of multiply which is easier. And they had books of log tables.

  • @georgesampson4714
    @georgesampson4714 Год назад

    OK, but ...... If I go to all the trouble and expense to generate 30 watts of RF power why would I want to burn 90% in an attenuator? Why not just generate 30 mw to begin with, save lots of money and call it a day.

    • @IMSAIGuy
      @IMSAIGuy  Год назад

      you need to test transmitters at full power to look for harmonics and distortion

  • @bayareapianist
    @bayareapianist Год назад +2

    I think you should stop using your expensive ohmeter to measure high frequency stuff and stick to you cheap VNA to measure them.

  • @markusm.lambers8893
    @markusm.lambers8893 Год назад

    ... forget about 'swaping' the resistor-values at first, in the 'PI-equitation', ... !
    Is there not a relation between the 'T' and 'Pi', to the 'star' and 'triangle', in the 'normal electricity' circuits? Are the mathematical rules are not the same, or can be substituted?
    Do you remember the damned "elecrical-cube" made of resistor? Each single resistor of that cube, was 10KOhms. There were 12 resistors in the cube. It drove me crazy, when I saw the picture of the cube, and didn't know, to 'solve' the problem, ...! Not correct for parallel or serial, ... even a 'drawing'-solution was not possible. - At first! - Then the teacher showed us something to do, ... !
    I Think there was a kind of 'formula' or equitation for the 'star/trangle'-problem, as I remember, ... !
    (I am a graduated and certified electician, since more then 4 decades. - A very long time ago, ... !)
    I could be wrong, nearly half of a century ago I learned something like this!
    Sorry for my bad English, but it's not my native tounge.
    73 de Markus - db9pz (JN39fq - 5km, 3miles east of LX )