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grid stoppers, open circuits, voltage divider, ground reference

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  • grid stoppers, open circuits, voltage divider, ground reference

    Considering the grid circuit of a typical capacitor-coupled common-cathode triode stage, the AC signal runs through the grid load resistor to ground, but I am trying to envision what effect the grid stopper has on the AC signal amplitude at the grid.

    Since the grid is not internally connected to anything there would be no current flow through the grid stopper except when the grid voltage approaches the cathode voltage and grid current flows, and so according to Ohms law there would be no voltage drop and the AC amplitude would be the same on both ends of the grid stopper.

    But my intuition tells me you can't have a resistor in the path without giving something up. So therefore it seems like the grid stopper and grid load would interact in some way to act as a voltage divider, and the AC voltage at the grid would be lower than the AC voltage at the grid stopper and grid load junction.

    Who is right, me or Senor Ohm?

    The same question might apply to screen grid stoppers. I think it is also why I am a little puzzled about how to inject bias tremolo into the output stage of a cathode-biased push-pull amp.


    This is a specific subset of my general lack of understanding about ground reference circuits, which also seems to come into play in figuring out

  • #2
    Check it:
    http://music-electronics-forum.com/t16346/
    -Mike

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    • #3
      Originally posted by Groover View Post
      Since the grid is not internally connected to anything there would be no current flow through the grid stopper except when the grid voltage approaches the cathode voltage and grid current flows, and so according to Ohms law there would be no voltage drop and the AC amplitude would be the same on both ends of the grid stopper.

      But my intuition tells me you can't have a resistor in the path without giving something up. So therefore it seems like the grid stopper and grid load would interact in some way to act as a voltage divider, and the AC voltage at the grid would be lower than the AC voltage at the grid stopper and grid load junction.

      Who is right, me or Senor Ohm?
      Your intuition about losing something is correct, but it is also correct that there is little or no loss in the grid stopper - at low frequencies.

      What you're missing is that every conductor has some small but finite capacitance to every other conductor in the universe. The grid has a capacitance to the cathode, and to the plate. The plate capacitance is effectively multiplied by the voltage gain of the circuit, so there is an effective capacitance to ground at the grid of Cg-k + G*Cg-p where these are the grid-cathode capacitance, voltage gain, and grid-plate capacitances.

      At DC, no current flows until the grid goes positive. You have that part right. However, the series resistor causes a lowpass filter with the effective capacitance that rolls off high frequency gain. The grid stopper resistor is called that because it often stops high frequency oscillations by rolling off high frequency gain.

      The same question might apply to screen grid stoppers. I think it is also why I am a little puzzled about how to inject bias tremolo into the output stage of a cathode-biased push-pull amp.
      Screen resistors are not doing the same thing, exactly. Screen voltage effectively changes the gain of the tube. Screens pull a lot more current than grids because they're biased positive with respect to the cathode instead of negative like the control grid, so any electron that hits, sticks. Screens can also melt down if they pull too much current. Series resistances lower the screen voltage automagically if the screen current gets high-ish, and can stop runaway or melting down. Sometimes. Sometimes the screen current blows the screen resistor too.

      You can make a tremolo by modulating the screen voltage. At least one guy (Eric Barbour, as I remember; it's been a while) made a voltage controlled amplifier out of a signal pentode this way. Takes some power to do it. You can also modulate the grid voltage of a cathode biased stage. Just cause it's cathode biased doesn't mean you can't wiggle the grid, too.
      Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

      Oh, wait! That sounds familiar, somehow.

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      • #4
        Miller effect. That's the whole point (well, one important point) of a grid stopper, it forms a low-pass filter with the Miller capacitance of the grid, keeping RF out of the stage.
        "Enzo, I see that you replied parasitic oscillations. Is that a hypothesis? Or is that your amazing metal band I should check out?"

        Comment


        • #5
          Originally posted by Steve Conner View Post
          Miller effect. That's the whole point (well, one important point) of a grid stopper, it forms a low-pass filter with the Miller capacitance of the grid, keeping RF out of the stage.
          Actually, it's of some value even in followers. Followers can get into oscillation easily enough, although it's more of a problem with high-transconductance devices like FETs and bipolars. The series resistance both forms a lowpass filter on the front and damps any resonances there that would help form the voltage gain needed to get over unity.

          But for most cases, yes, the resistor in concert with the multiplication of the grid-place capacitor by Miller effect (G*Cgp) is the biggie in keeping RF out.
          Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

          Oh, wait! That sounds familiar, somehow.

          Comment

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