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Posted by Maxamillion (A) on July 01, 2009 at 09:00:23

This may be a dumb question, but it's confusing me.

When reading average plate characteristic charts, which have plate volts on the x-axis, plate current on the y-axis and a series of curves labeled "Grid Volts", are the grid volts shown on the curves referenced to ground, or to the cathode? For example, does the line corresponding to grid volts = 0 mean that the voltage is actually 0 relative to ground, or that the voltage is equal to the cathode voltage? I need this info for a cathode-biased preamp, and obviously it makes a big difference in the selection of operating points. I have attached the tube data sheet I'm looking at, specifically chart #92C-6436 in the attachment.

Any help is greatly appreciated!!

• 2C22 Tube Data   (Open in New Window)

• RE: Reading Plate Characteristic Charts - Question - Tre' 10:24:52 07/01/09 (3)
  In Reply to: RE: Reading Plate Characteristic Charts - Question posted by Maxamillion on July 01, 2009 at 09:00:23

  
  
  

  
  
  The grid voltage is grid to cathode. When the signal is present on the control grid the grid voltage changes. The operation of the tube follows the black load line (22K load in this case).
  
  This 6sn7 example has an idle current of 11.3ma and a idle plate voltage of 188vdc. The grid is 4 volts negative with respect to the cathode.
  
  If you place a 4 volt peak to peak signal on the grid, the plate voltage will fall to 155vdc on the positive half (as the grid voltage is forced by the input signal from -4 to -2vdc) of the input wave form and rise to 220vdc on the negative half (as the grid voltage is forced by the input signal from -4 to -6vdc) of the input wave form.
  
  The tube and operating point is not totally linear but not far from it. The plate voltage increase is 32vdc (220-188)while the plate voltage decrease is 33vdc (188-155).
  
  This difference in the plate voltage change from the positive half of the input wave form to the negative half of the wave form causes harmonic distortion.
  
  
  The trick is to find a tube and a operating point where the positive plate voltage change and the negative plate voltage change are equal.
  
  Tre'
  Have Fun and Enjoy the Music
  "Still Working the Problem"
  
  

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  • "The trick is"... - slownlo 15:00:10 07/02/09 (0)
    In Reply to: RE: Reading Plate Characteristic Charts - Question posted by Tre' on July 01, 2009 at 10:24:52

    ..."to find a tube and a operating point where the positive plate voltage change and the negative plate voltage change are equal".
    
    Tre',
    That statement just blew my mind...when I read that, it "unlocked" a flood of realizations! Its amazing how a simple statement like that helped me to "create" the missing peices of a mental puzzle of discovery...
    
    Thanks for that.
    
    slownlo
    

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  • RE: Reading Plate Characteristic Charts - Question - Maxamillion 10:35:52 07/01/09 (1)
    In Reply to: RE: Reading Plate Characteristic Charts - Question posted by Tre' on July 01, 2009 at 10:24:52

    Wow - Great info - Thanks!!
    

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    • RE: Reading Plate Characteristic Charts - Question - Michael Samra 22:41:19 07/01/09 (0)
      In Reply to: RE: Reading Plate Characteristic Charts - Question posted by Maxamillion on July 01, 2009 at 10:35:52

      Max
      Always remember that every voltage to any tube is referenced to the cathode except filaments..The reason that people somtimes think voltages are referenced to ground like in fixed bias output tubes is because they usually use a 10 ohm resistor cathode to ground which is so close to chasis potential that people usually measure from ground.
      
      If the power supply waveform isn't pretty,neither is the sound in most cases.
      
      

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• Relative to the cathode - Caucasian Blackplate 09:28:50 07/01/09 (4)
  In Reply to: RE: Reading Plate Characteristic Charts - Question posted by Maxamillion on July 01, 2009 at 09:00:23

  Avoid that 0 line on these little triodes...
  

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  • RE: Relative to the cathode - Maxamillion 10:05:50 07/01/09 (3)
    In Reply to: RE: Relative to the cathode posted by Caucasian Blackplate on July 01, 2009 at 09:28:50

    Thanks for the reply! Is that also true of Plate Volts as well?
    

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    • RE: Relative to the cathode - Caucasian Blackplate 10:15:10 07/01/09 (2)
      In Reply to: RE: Relative to the cathode posted by Maxamillion on July 01, 2009 at 10:05:50

      Yes indeed!
      
      In regards to dialing in plate voltage, maybe I can offer an easier solution. Put two HLMP-6000 diodes under your cathode. That fixes your cathode voltage right around 3v, which is a decent place for a preamp.
      
      Now put a 100k pot in as your plate load, dial in plate voltage, and replace with a resistor.
      
      The diodes offer low impedance to ground, and they don't require bypass caps. Also, you only end up having to juggle one part.
      

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      • RE: Relative to the cathode - Maxamillion 10:35:18 07/01/09 (1)
        In Reply to: RE: Relative to the cathode posted by Caucasian Blackplate on July 01, 2009 at 10:15:10

        Hmm, yes, I've been reading about diode biasing (so much to learn on this subject!). The cathode resistor on my preamp is bypassed with a relatively large Oscon cap (220uF, I think I recall), which can't be too good for the sonics, even if the Oscon is considered a pretty good electrolytic.
        
        If I did this, would I lose any of the advantages of having a bypassed cathode resistor, such as increased gain and no negative feedback?
        

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        • RE: Relative to the cathode - Caucasian Blackplate 14:07:26 07/01/09 (0)
          In Reply to: RE: Relative to the cathode posted by Maxamillion on July 01, 2009 at 10:35:18

          You may lose a little gain, but I don't think you will pick up and local feedback from this arrangement.
          

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