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9th January 2011, 05:09 PM  #1 
diyAudio Member
Join Date: Aug 2005

Pentode curves  how to find operating point withoud going crazy?
Hi out there,
anyone so bored that he wants to do tubeaudionoobcalculations with me? I need to make a pentode CCS with a current of 8,6mA, negative supply needed should be between 100V and 250V, tube could be EF184/6EJ7 or maybe EF94/6AU6. I have printedout anode curve sheets scattered all around me, and don't seem to find an operating point compatible with supply voltage, needed current and acceptable linearity... I need help, or a strong drink, or both. Greetings, Andreas 
9th January 2011, 05:23 PM  #2 
diyAudio Member
Join Date: Jul 2003
Location: Ann Arbor, MI

heyHey!!!,
What is so complex? Consider also type 6AC7. Very high plate resistance, and can flow enough current to meet your spec. I'll take that one as an example since I have some experience using it... Find the set of plate curves at variable g2 voltage( set with g1=0 V ). IMO, taking the highest g2 voltage will be useful as it will deliver a maximum g1 negative voltage at your desired current flow. You've left off how big your negative rail will be, and that number will be your g2 voltage( you'll attach g2 to signal ground ). With this one, g2@150V it looks like you'll have 2.2V on g1 to flow your 8.6 mA. Your cathode R should then be an unbypassed 255 Ohms, give or take jus' a bit. cheers, Douglas
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9th January 2011, 06:01 PM  #3  
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Join Date: Feb 2010

Quote:
I see that the total current with 2,2 V bias is Ia + Ig2 ( 8,6mA + 2,2mA) = 10,8 mA. So the cathode resistor would be 2,2V/10,8mA=204 ohms. 

9th January 2011, 06:27 PM  #4 
diyAudio Member
Join Date: Aug 2005

Hi there,
thanks for the fast response. I found curves for the 6AC7 and could follow your explanations. My problem is that for all suitable smallsignal pentodes the required 8mA are either above specs (EF86) or in the very lower part of the graphs where the curves for different g1 voltages are not very equally distanced. In Morgan Jones' book it is stated that linearity is also important for current sinks. So I find it a little difficult to meet all requirements... Greetings, Andreas 
9th January 2011, 06:41 PM  #5  
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Join Date: Feb 2010

Quote:
I looked the curves for EF184 since you said you have those tubes. When Ug2 = 160 V, and Ia = 8,6 mA, then Ug1 = 1,95 V and Ig2 = 3,6 mA. So total cathode current is 12,2 mA and required cathode resistor (1,95 V/ 12,2 mA)= 160 ohms. 

9th January 2011, 07:13 PM  #6  
diyAudio Member
Join Date: Jul 2003
Location: Ann Arbor, MI

Quote:
And yes, I totally neglected g2 current in the cathode R calculation... cheers, Douglas
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9th January 2011, 07:38 PM  #7  
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