Pentode curves - how to find operating point withoud going crazy? - diyAudio
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Old 9th January 2011, 05:09 PM   #1
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Unhappy Pentode curves - how to find operating point withoud going crazy?

Hi out there,

anyone so bored that he wants to do tube-audio-noob-calculations 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 printed-out 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
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Old 9th January 2011, 05:23 PM   #2
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hey-Hey!!!,
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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Old 9th January 2011, 06:01 PM   #3
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Quote:
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.
Should the Ig2 - some 2,2 mA - also taken into account ?
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.
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Old 9th January 2011, 06:27 PM   #4
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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 small-signal 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
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Old 9th January 2011, 06:41 PM   #5
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Quote:
My problem is that for all suitable small-signal 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.
You should use two different graphs only; Ia vs. -Ug1 and Ig2 vs. -Ug1. There are curves with different Ug2-voltage and fixed Ua-voltage.

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.
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Old 9th January 2011, 07:13 PM   #6
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Quote:
Originally Posted by Rundmaus View Post

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
Not quite sure I agree with that. High resistance( DC, as in horizontal plate curves ), and low capacitance to maintain that plate resistance at high frequency seem like the important ones to me.

And yes, I totally neglected g2 current in the cathode R calculation...
cheers,
Douglas
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Old 9th January 2011, 07:38 PM   #7
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And yes, I totally neglected g2 current in the cathode R calculation...
Yes, this happens often to me too, since usually done most circuits with triodes.
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