Yeah, this is where support will get a little lean.
I understand. So the current fact is I now have 3 LM431’s in possible play.
LM431 -(1), is on the 9 pin socket, going from the center post, (ground) to B3 cathode.
LM431 -(2), was already on the driver board “A†side and is, before any changes, going to B8 cathode.
LM431 -(3), I added to the driver board “B†side along with the 2.49K @ R3 and 3.5K @ the outboard end of R4 & the inboard larger + pad. The new lead goes from the smaller, outboard + pad.
The power supply regulator has the 431 with fixed resistors. The "K" pad from this portion of the PCB, and the "O" pad that feeds the "+" linked to that 431 (should be "OA") need to hit the same half of the 6CG7 (1,2,3 or 6,7,8).
If I understand correctly, this is LM431 (2). Then this circuit remains unchanged and the wire from KA remains on the cathode B8.
The driver stage will take the lower current "O" pad (OB most likely) and the K from the adjustable half of the board.
So here is where we need to check my understanding. The added LM431 (3), 2.49K & 3.5K resistors are to change the driver cathode bias from 2.5v to 6.0v. If that is correct, then the new lead from "B" side K via "+" would go to B3.
Do I then need to remove the currently installed LM431 (1), (from the center post to B3)? Then wire the new lead from the new circuit with LM431 (3) to B3 to provide the 6.0v bias?
Or? Leave LM431 (1) in place, as is and add the new circuit lead to B3 and the combined old and new LM431 (1+3) will give me the desired 6.0v cathode bias?
OR? Just grab the closest bottle of Scotch and enjoy the Paramour 45's and try thinking again tomorrow!?
Thanks for hanging in with me on this!
Cheers,
Geary
« Last Edit: March 21, 2014, 03:32:10 PM by galyons »
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