Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: markmalin on April 18, 2015, 10:24:15 am
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Hi all,
I'm building an AB763 with KT-66's and was just looking up the pinout. I have my chassis wired per typical Fender the way they did the 6L6GC where they wire the output of the PI to pin 1, then attach the 1.5K resistor from pin 1 to pin 5. The assumption is pin 1 is NC on a 6L6GC. I just looked up the KT-66 tube chart and see conflicting pinout diagrams. It looks like some show a "surpressor grid" at pin 1, while others show it as a no connect and on others it looks like it's tied to a shield of some sort (based on the diagram).
I've decided not to do Fender's thing stretching the 1.5K across the socket, but rather use a terminal strip for the 470 Ohm and 1.5K resistors. I'm not sure what to do about pin 1, do I leave it unconnected for a KT66?
Thanks!
Mark.
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I like to leave pin1 free "just in case".
And having the socket mounted components on tagstrip is a neat looking construction
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KT66 is an overseas copy of 6L6.
Original 6L6 pin 1 is the metal shell. (This is true for all metal octals.)
As metal 6L6 are out of fashion, you might assume pin 1 is free.
But for KT66 both M-O and GEC data show "IC", internal connection (not "surpressor grid"). So in theory you could get in trouble tying stuff there.
I would think all new-made "KT66" will be sold for drop-in as 6L6 replacements. If they did something with pin 1 there would be a LOT of unhappy customers (because pin 1 is often used as a tie-point).
Would be wisest to leave it free.
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I'll leave it free, then.
Thanks guys!
Mark.