5F1 Circuit Question - V1 Cathode Resistors

Discussion in 'Shock Brother's DIY Amps' started by brackp, Feb 22, 2019.

  1. brackp

    brackp TDPRI Member

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    Having managed to build the 5F1, with a lot of help from people here, I feel like I've got some kind of understanding of what's happening in most of the circuit.

    One thing that I've been wondering about is what's going on with the cathode resistors on V1. Can someone explain to me where the 1.5DC is coming from on Pins 3 and 8 of the 12AX7?

    My understanding is that the filaments heat up the cathodes which gives them a negative charge? Would I be right in saying that the cathode resistor acts like a valve to dump some of that charge to ground in order to get the right level of negative charge on the cathodes? Or am I way off?

    Thanks again!
     
  2. D'tar

    D'tar Friend of Leo's

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    First off.... Im not a theory expert and hope to not add confusion...

    The resistor holds the cathode above ground potential and sets the bias for the triode. If it were not there the cathode would be at zero. The grid is at zero therefore negative in respect to the cathode. This alows us to control the flow of electrons. As you apply signal to the grid positive flows more electrons and negative suppresses electron flow. The more positive the grid to cathode runs the tube into saturation. The more negative the grid to cathode runs the tube into cutoff.

    These old military videos are some of my favorites. You can imagine the alternator being guitar signal on the control grid.

     
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  3. DADGAD

    DADGAD Friend of Leo's

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    With one milliamp of current through the tube, there is a voltage drop across the cathode resistor. E=I*R.

    So 1 ma. through 1.5K ohms is 1.5 volts.
     
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  4. Luthier Vandros

    Luthier Vandros Tele-Holic

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    GREAT videos and I enjoy those, as well. Simple enough for my lizard brain.
     
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  5. D'tar

    D'tar Friend of Leo's

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    Oh sure, thats the easy answer:)
     
  6. D'tar

    D'tar Friend of Leo's

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    Me too.

    They had to teach young impressionable teenagers to be radio warriors!
     
  7. DADGAD

    DADGAD Friend of Leo's

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    No. It's the law. :lol::lol::lol:
     
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  8. Whatizitman

    Whatizitman Friend of Leo's

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    The resistor brings the cathode voltage (zero potential at ground) to 1.5DC at the cathode pins.

    Valve Wizard to the rescue. Chaper 1.9 on biasing.

    http://www.valvewizard.co.uk/Common_Gain_Stage.pdf

    Cathode is ground potential, so zero. It can be negative with respect to the grid, but it is zero. A cathode resistor (for cathode biasing) makes the cathode voltage stay within a range positive with respect to grid, as opposed to 'fixing' the grid negative with respect to cathode (fixed bias). Putting resistance in series with the cathode allows for the tube to self-bias with anode current changes.

    At least this is how I understand it. Someone far more knowledgeable please correct me if I ain't got it right.
     
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  9. robrob

    robrob Poster Extraordinaire Ad Free Member

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  10. elpico

    elpico Tele-Holic

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    pretty much right

    And then not so right :D

    The filament heats up the cathode material which causes it to "boil" electrons off into the space around it. They form a cloud around it called the space charge. Electrons have a a negative charge. Since they have a negative charge they only ever want to flow towards a more positively charged point. They came from the ground, attracted by the positive voltage on the plate. They can't flow back the way they came because ground isn't more positive than the cathode, so no, the cathode resistor isn't "acting like a valve to dump charge to ground". It's a one way street in the opposite direction. The negatively charged electrons come from ground, travel up through the cathode resistor to the cathode, and then want to cross the vacuum to the plate because it's strongly positive. The grid is put in between them to throttle that flow.

    The purpose of the cathode resistor is to raise the cathode voltage above the grid voltage so the grid will be an effective throttle. The grid is at ground potential, so the bigger you make the cathode resistor the higher the cathode voltage climbs and the more negative the grid looks to the cathode. Which makes the grid a more effective throttle. At some point you will make the throttle effective enough to cut off current flow completely.
     
    Last edited: Feb 25, 2019
  11. jsnwhite619

    jsnwhite619 Friend of Leo's

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    I like your way of explaining it. Adds a nice mental picture to the process.
     
  12. brackp

    brackp TDPRI Member

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    Thanks a million guys! That's all making a lot more sense to me now. I actually stripped down my 5F1 build last night and rebuilt it with some better components (switchcraft jacks, belton tube sockets and a Fender light bulb socket). It was a much nicer experience having some idea of what each connection was doing.
     
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