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#1 |
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diyAudio Member
Join Date: Apr 2005
Location: denmark
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Hi
I have a problem. I have a china amp that i am modding. It is a williamson copy, but with Kt88 pp as triode and fixed bias. The second grid have a 2k7 2w to the anode. Why that big value. Is there a way to calculate that resistor. B+ is 430 v. Thanks for reading. Soren |
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#2 |
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diyAudio Member
Join Date: Jul 2005
Location: Leverkusen
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Hi Soren,
when converting a tetrode, pentode or beam power valve as in the case of KT88, to triode, the screen grid stopper resistor should be chosen in values as low as possible, while maintaining that no parasitic oscillation takes place. In your case, a resistor in the range of 100 to 270 ohms should do nicely (but make sure that there are no parasitics anyway). If you want to learn more on how to convert tetrodes and pentodes back to triode operation correctly, you might have a look at this script. Regards, Tom
__________________
If in doubt, just measure. |
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#3 |
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diyAudio Member
Join Date: Apr 2005
Location: denmark
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Thank you Tom.
/Soren |
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#4 |
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diyAudio Member
Join Date: Apr 2005
Location: denmark
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Can you please tell me, why the g2 resistor have to be 2w type.
Thanks Soren |
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