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Cathode to ground question

  1. Jun 5, 2016 #1
    There is a cathode to ground resistor on a KT66 power tube on a amp kit. The value is 1 ohm. I ASSUME the only reason for it being there would be to be able to measure the current flow. I ASSUME 1 ohm is not going to change the cathode bias. Is my assumption correct?

    The KT66 is a beam power tetrode and this amp kit is based on a Marshal JTM 45 schematic. The original schematic has the cathode going directly to ground.

    Thanks,

    Billy
     
  2. jcsd
  3. Jun 5, 2016 #2

    Svein

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    The cathode to ground resistor introduces a slight negative feedback. At a cathode current of 100mA, it will change the grid bias by 100mV, which is not very much. The famous Williamson amplifier (http://www.preservationsound.com/?p=3823) uses a complex cathode resistor network, but assume an equivalent of 120Ω per tube, giving a cathode bias (at 80mA) of ≈ 10V.
     
  4. Jun 5, 2016 #3

    jim hardy

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    Do i perhaps remember discussing that amp ?
    If i recall it was for measuring cathode current when you adjust bias ?
    While amp is off , make sure it's still one ohm.
     
  5. Jun 5, 2016 #4
    No Jim, not the same amp. This is a new kit I am try fix for a kid to keep him from killing himself. You would not believe how bad he had it messed up. Solder running everywhere. Really dangerous stuff. I had to re-do everything.

    I am now looking it over to see if the instructions are correct. I really dislike these kit projects. No bleeder resistors. No cover on the AC line. On and On...not good.
     
  6. Jun 5, 2016 #5

    jim hardy

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    I agree with your assumption. See curves at http://www.drtube.com/datasheets/kt66-jj2007.pdf
    100 millivolts isn't much bias at all.
     
  7. Jun 6, 2016 #6
    Thanks Jim

    Billy
     
  8. Jul 28, 2016 #7

    Bandit127

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    I am learning this at the moment but can I assume that a 1 Ohm resistor in the cathode circuit allows the plate current to be measured directly as a voltage because a 1 Ohm resistor gives V=A. Once I have my plate voltage I can work out my plate dissipation and then adjust bias accordingly.

    So if my plate voltage is 400 V, the voltage drop on the 1 Ohm resistor is 25 mV (0.025 V = 0.025 A) then my plate dissipation is 10 W.

    By measuring the current in this way I guess I am ignoring the small current added to the plate by the input signal on the grid. Or am I?
     
  9. Jul 28, 2016 #8

    jim hardy

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    grid current is microamps
    and the electrons came from the cathode so plate current will be a teeny bit less than cathode, negligible when using analog meters of tube days.
     
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