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Grid-Leak Bias Phase Inverter?

Discussion in 'Amp Tech Center' started by K Teacher, Aug 21, 2020.

  1. Aug 21, 2020 #1
    K Teacher

    K Teacher TDPRI Member

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    Greetings,

    I read several articles on cathodyne phase inverters and how to improve their response to overdrive.


    And then, to my surprise, I bumped on this:



    upload_2020-8-21_6-16-51.jpg


    So, for curiosity, I did plot the cathodyne load-line.


    From the values on the schematic:

    Ebb = 260V

    Eb_k = 188V - 75V = 113V

    R_load_dc = 68K + 68K = 136K

    I_max = Ebb / R_load_dc = 260V / 136K = 1.91 mA


    upload_2020-8-21_6-16-51.jpg



    From the plot, to get 113 V drop across the 12AX7, it is necessary a plate current of 1.08 mA, commanded by a grid voltage bias of - 0.7V… the difference from this voltage to the power supply voltage is divided between the two 68K resistors, that is, (260V - 113V)/ 2 = 74V each (close enough).


    Looking on the net, I found a few more vintage guitar amps (dubbed “bargain bin models” that use similar cathodyne grid-leak bias PI design) …


    I thought that grid-leak bias was OK for input stages, were the signal level is low (approx. 100 mV or so), but I never saw that bias method used on the stage before the output tubes…


    Therefore, I would like to benefit from the experience of you guys that worked on these models, and ask a couple of questions to get a better understanding on this PI design.


    Question 1: In yours experience, is it possible to get 0.7V grid-leak voltage bias on a 12AX7 out of 1M grid resistor? I thought that this would require a grid-leak resistor from 6M8 to 10M… (?)
    As reference, I plotted (in GREEN) the values for a hypothetical 1K5 cathode bias resistor…


    Question 2: I also did think that these hot biased cathodyne PI (Ec1 near zero) are susceptible to draw grid current and produced unpleasant distortion, as pointed out on several cathodyne PI posts…
    However, I heard people playing this amp (and similar) on YouTube and they sound quite nice…
    How is this possible? Do they have enough gain to drive the output tubes into overdrive?


    Any clarification would be much appreciated… Thanks.
     
  2. Aug 21, 2020 #2
    tubeswell

    tubeswell Friend of Leo's

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    You don't actually need much bias voltage on a cathodyne. The load resistors set the plate and cathode idle points and determine the tube current. The grid can be sitting at the cathode voltage and the cathode voltage swing will still follow the grid voltage swing, As long as the load resistors are balanced, the plate will have the same (but opposite) swing. And provided you have a big enough (~1M) grid leak, you will maintain enough input impedance.

    Its a different situation to grid leak bias on an input stage, where the cathode voltage is fixed at ground potential, and you need about 1V bias on the grid to stop the input signal from clipping (too soon).
     
    tubegeek and D'tar like this.
  3. Aug 21, 2020 #3
    K Teacher

    K Teacher TDPRI Member

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    Hello tubeswell,

    Thanks for your response. I always thought that the bias point needed to be centered on the load-line... however, looking at the plotted load-line again, it seems that in this case (Silvertone 1482), the output voltage swing will be enough to drive a pair of 6V6's biased at -17.0V to the max.

    Have you had the opportunity to listen or to play one?
    If yes, how does it sound?

    Thanks again...
     
    Last edited: Aug 21, 2020
  4. Aug 21, 2020 #4
    Ten Over

    Ten Over Tele-Holic

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    So the tube is sitting there at idle with 113Vak and -0.7Vgk with 1.08mA flowing through it. Then an AC signal is applied to the grid. When the AC signal swings positive, the cathode signal follows it. In order for the cathode to swing positive, more current must flow through it which means that Vgk must change to a value closer to zero. Vgk doesn't have to go very far before it reaches zero and grid current increases exponentially.

    Yes, I know that I state the obvious. But the radical increase in grid current doesn't lead to the clipping of the positive signal as may be expected. Instead, the grid current charges the 0.01uF input capacitor and it charges it lickety-split because there is nothing between it and the grid. When the AC signal ceases to be positive, the input capacitor discharges. But the discharge path has a large resistance and the input capacitor discharges very slowly. The result is that the bias moves more negative due to the negative charging of the input capacitor and then a larger output voltage swing is possible.
     
  5. Aug 21, 2020 #5
    K Teacher

    K Teacher TDPRI Member

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    Hi Ten Over,

    Thanks for the detailed explanation...
    I understand that the voltage at the cathode node follows the voltage at the 68K-to-ground resistor and thus, the grid-to-cathode voltage does not get very large...

    Now, what is confusing me, is that there are several articles stating that you need a 470K grid-stopper on the cathodyne to prevent from unpleasant distortion (frequency doubling) when the cathodyne has grid current due to overdrive…

    So, my question is -- how all of this fits together and is the cathodyne grid leak-biased PI more susceptible to this issue than regular cathode biased cathodyne PI?

    Thanks again...
     
  6. Aug 21, 2020 #6
    K Teacher

    K Teacher TDPRI Member

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    I want to apologize to all… I guess I asked the question in the most cumbersome way…

    Putting it in other words:

    Tone-wise, how does a grid-leak biased cathodyne PI compares to a regular cathode bias one when overdriven?
     
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