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13th April 2008, 06:38 PM  #11 
diyAudio Member
Join Date: Jul 2005
Location: Leverkusen

Hi Darius,
please notice that the curves for positive Eg values are just in parallel to those of Eg = 0, 5 and 10 Volt respectively, so at least for the straight parts, the three determining coefficients mu, gm and ra coefficients have the same values as with normal (negative) g1 operation, for all practical purposes. IOW, what you see is, that 6SN7 simply keeps its linear mu _and_ linear gm behaviour also when g1 is held positive. Hence no plate resistance anomalies, either. So, yes, I still think your argumentation chain is nonsense, because the premise you made it is based on a wrong assumption. Regards, Tom
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13th April 2008, 08:13 PM  #12 
diyAudio Moderator

very nice disscusion here....
keep it going guys, and SY, can we make this thread a sticky to address all G2 drive dicussion issues please? thanks you.....
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13th April 2008, 09:32 PM  #13 
diyAudio Moderator

If the thread continues to evolve, then sure, we'll put some glue on it.
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14th April 2008, 12:53 AM  #14 
diyAudio Member
Join Date: Nov 2005
Location: SoCal

Sorry I couldn't take the time to read the details. But it seems like you are describing space charge operation (G2 driven, G1 biased positive relative to cathode). If so, this is how they got tubes to run off 12V for automotive back in the day.

14th April 2008, 06:40 AM  #15 
diyAudio Member
Join Date: Mar 2008

#5 #10 #11
Hi Tom,
hope this graph helps understanding. Output curve 1 shows a linear resistor. Curve 2 shows a diode or a triode at negative grid voltages. Curve 3 shows a pentode, tetrode etc., transistor FET BJT or a triode at positive grid voltages. You can adjust the output curve of a triode from diode to transistor by adjusting the grid voltage. Ideally curve 3 is horizontally and starts at 0V. Kind regards, Darius 
14th April 2008, 06:48 AM  #16  
diyAudio Member
Join Date: Mar 2008

#14
Quote:
D. 

14th April 2008, 09:38 AM  #17 
diyAudio Member
Join Date: Mar 2008

#15
http://frank.pocnet.net/index.html
Here are the output curves of the AL1 pentode and the 811A Triode positive grid. Kind regards, Darius 
24th June 2008, 01:36 PM  #18  
diyAudio Member
Join Date: Jul 2005
Location: Leverkusen

Re: #5 #10 #11
Hi Darius,
Quote:
Actually, you get: µ = 15 gm = 3,75mA/V rp = 4 kOhm over a wide range, which is quite close (factor less than 2) to 6SN7 coefficients at standard (negative g1) operation. Now compare this to any real "pentode like" coefficients. Again, my points is: The curves may look pentodelike, but don´t get fooled by scaling. For any _practical_ purpose you still get triodelike _coefficients_ when driving g1 of a triode positive, and _that_ is what counts, but not misleading optics by changed scales. Now please fix your signature in this regard, which is needlessly provoking. Regards, Tom
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If in doubt, just measure. 

24th June 2008, 02:44 PM  #19  
diyAudio Member
Join Date: Mar 2008

funny...
Hi Tom,
Quote:
"Resistor like" means differential and static resistance do not change with the voltage, see Curve 1. "Transistor like" means differential resistance increases at increasing DC voltages, see curve 3. "Triode like" means differential resistance decreases at increasing DC voltages, see curve 2. Quote:
You are telling in other words what I wrote in my signature, hi hi. Technical forum rules: µ = hfe_ hie = 1/hoe_ triodes don't change their characteristic at positiv grid voltages _ anode current is independent from g2 current in a pentode_ calculating the gain of a triode cathode follower you have to ignore the influence of µ_ ignore Loftin White_ ignore the fourth circuit topology_ tubes should be replaced by mosfets_ neutralisation is forbidden_ continued ... I don't like these rules, I have to accept them here in the forum. Kind regards, Darius 

24th June 2008, 05:55 PM  #20  
diyAudio Member
Join Date: May 2006
Location: Lancashire

Re: Re: #5 #10 #11
Quote:
Anyway, the way I see it is, oldeurope is saying that at some G1 voltage which is more positive than 0V, you will find a point where the grid curve is very neary linear (resistive). This is of limited use in a real triode, but a pentode/tetrode has the screen grid to play with. Therefore you could fix the G1 voltage at the 'ideal' positive value, and input the audio signal to G2 instead (which would be referenced to ground). The valve can therefore operate over a nearly linear transfer characteristic. Mu at the screen grid is very low of course, so what you gain in linearity you lose in input sensitivity nothing for nothing. Or have I got this all wrong? 

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