Dr Boogey "less gain" mod help, and bias question

[jwyse]

I just completed a Dr Boogey, and man, the thing's a beast. 

I wanted to go ahead and put a "less gain" mod in right from the start.  I couldn't find a good picture or great description, so I hacked it together from the brief descriptions I found.  It seemed simple enough, but I can't tell any difference between the switch positions, so I think I made a mistake in planning it.

From my gain pot (1M) I have a switch (on/on with one side unused) that should be switching a 1M resistor in parallel across the outer lugs of the pot.  It was late and/or I was overconfident when I soldered this piece together, so I didn't test the resistance with a DMM before I attached the rest of the circuit.  When I try to test the resistance now, the reading jumps around.  I'm guessing that's the capacitors+transistors from the rest of the circuit coming into play?

Hopefully my picture is clear enough.  From the pot, the yellow and purple wires go over to the switch, with a resistor in line.  The Y/G/B wires coming down from the pot go over to the board.

Should I have put the switched resistor across different pot lugs in this case?


Also: I biased the JFETs by putting them in (sockets), then measuring DC voltage on the drain and adjusting the corresponding trimpots to ~5.5V for each one.  Is this the proper way to bias the JFETS?  I've seen a lot of comments about "half the input voltage" (4.5V), and various preferences on the bias voltage, but can someone give me a quick explanation of what a lower/higher voltage will translate to?  Less/more gain?  Less/more noise?  What should I be listening for if I want to tweak it?

Thanks!

[jwyse]

Here's the picture I was referring to.

[Govmnt_Lacky]

I know the "resistor across lugs 1 & 3" trick will work with LINEAR (B taper) pots as it is recommended with the MIDS pot on the Boogie build however, I do not know how it will react with a LOG/AUDIO (A taper) pot.

I have read several places that in order to get the 2.5K MID pot all you do is take a 5KB pot and solder a 5K resistor across lugs 1 & 3. But, for the Drive it is an Audio taper so I am not sure if/how it will work.

Can anyone else chime in on this? 

[jwyse]

Thanks for the reply.  Regarding the 2.5k mid pot, you can barely see it at the top of my pic, with a resistor across the outer lugs.  Again, I didn't measure it before I connected the rest of the circuit, but it seems to work fine.

I've been trying to read up on this. I'm fairly new to this (especially the theory stuff) so this is probably stating the obvious to most of you.  From what I can tell, the gain pot acts as a voltage divider (as opposed to a rheostat) because of the way it is connected (ground/"input" on outer lugs; "output" on the wiper).  Does the resistor-across-outer-lugs trick work the same way for pots in this scenario?  Should I put the switched resistor across the "input" and "output" lugs instead of "input" and "ground"?   

[John Lyons]

The 1M in parallel will with a 1M audio ptater pot will make the pot a
500K with a slight reverse taper but mostly linear.
So anything fater about half way up will not make much difference
in gain increase... I'd try some lower, say 470K. It'll still have the same
taper issue but the gain will be lower.

[Govmnt_Lacky]

Is there a HUGE difference between the 1M and 500K Gain pots? Is the gain too overpowering with the 1M? Is this a personal preference?

I just dont see why you cant use the 1M and adjust the pot down half way. Am I missing something in the design that would make the Boogie sound better with the 500K?

[Quackzed]

no, same.
if you want lower gain, then cut with a voltage divider/cap combo. less signal=less signal... here there wherever.

[John Lyons]

The Gain pot is a voltage divider. (I know you know this.)  
You could always put a 500K resistor in series with the gain pots hot lug (3).
This would be with a 500K pot. Same as a 1M turned halfway down.
So you would have from half the range of the 1M, from half to zero.