Padding the Pot

karl · 14832

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Offline Grainger49

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Reply #15 on: June 23, 2011, 11:15:55 AM
I'm with David.  The signal comes through the shunt resistor and then into the grid of the tube.  The pot to ground should be linear with temperature changes but is not "classically" in the signal path.  But on the opposite side of the coin a cathode resistor and bypass capacitor are not in the signal path.  However, upgrading them makes a difference.

Obviously not conclusive since I have argued both sides.  But information to throw out there anyway.



Offline williaty

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Reply #16 on: June 23, 2011, 06:25:08 PM
So I'm off on my own pot-padding adventure tonight. It's a little difficult since i don't know how to calculate this stuff, so I'm trying to follow someone else's chart and then make that work with the resistors I actually have.

1) A 10dB L-pad consisting of a 68k and a 47k wasn't nearly enough. Still stuck down on the first 45* of the pot.

2) A 30-ish (again, don't have the math to calculate this) dB l-pad consisting of a 100k and a 3.3k seems just slightly too much. "Wide open" on the pot is now basically exactly the loudest I'd ever be willing to listen to the quietest music I have. However, it'd be nice to have a little bit of headroom. However, highly compressed pop music (Lilian by Depeche Mode) causes me to turn the pot down almost 180*.


So that leads me to a question: How much is the pot attenuating? Is there a way to measure? My long term goal is to make a stepped attenuator, so knowing how many dB I personally want from top to bottom would be good. Obviously, I just proved that the "wide open" setting needs to be about -25dB but I have no idea how much I need to make it able to "turn down" without knowing how many dB the pot's dropping at intermediate volume settings.



Offline Grainger49

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Reply #17 on: June 23, 2011, 11:57:26 PM
Sounds to me like #2 is what you want.  There will always be a low level CD or LP that you will need a little more turn on the volume for.  I had a CD that I found I could play straight from my CDP into a Krell KST250S with no attenuation, no preamp at all.  It wasn't that loud because it was recorded/mastered at a low level.

The pot attenuates from zero dB to infinite dB.  All the way up the pot attaches the input directly to the wiper.  There is no attenuation.  All the way down the wiper is shorted to ground, so there is infinite attenuation.  Even stepped attenuators have the same two end points, zero and infinite attenuation.



Offline Laudanum

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Reply #18 on: June 24, 2011, 02:50:03 AM
williaty ... sounds like you found this but in case you didnt, I post the link again ...  http://www.goldpt.com/mods.html

Desmond G.


Offline dubiousmike

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Reply #19 on: June 28, 2011, 09:36:52 PM
williaty ... sounds like you found this but in case you didnt, I post the link again ...  http://www.goldpt.com/mods.html

Thanks for the link Laudanum! 

Mike M.


Offline dubiousmike

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Reply #20 on: July 06, 2011, 09:36:18 AM
So the goldpoint pre-attenuation circuit definitely looks like the way to go if you want to pad your input signal by more than a few decibels (to say nothing of preserving input impedance).  It has been about a decade since I took any class relating to E&M, so I enlisted my physicist father to help me with some napkin math while we put back a couple of Newcastles on the 4th. 

To recap from page 1, I'm using a 100k TKD pot in my crack, and was planning on simply adding a pair of 75k ohm resisters in series with the inputs to pad the volume.  However, after calculating it out, it turns out that 75k ohm resisters in series with a 100k pot will actually only reduce the input signal by about 5 db.  By contrast, adding in the 75k ohm resisters and also a second set of resisters (R2 as shown in  http://www.goldpt.com/mods.html), of approximately ~33k ohms, bridging the input and ground terminals of the pot, will reduce the input signal by nearly 12 db, while also preserving the impedance. 

This should be about perfect in my setup since I am often lowering the volume in Foobar by ~10 db if I have my pot in the vicinity of 9:00.


Mike M.


Offline Doc B.

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Reply #21 on: July 06, 2011, 10:22:11 AM
Quote
so I enlisted my physicist father to help me with some napkin math while we put back a couple of Newcastles on the 4th. 

Thus supporting my theory that the cocktail napkin was the most significant invention of the 20th century. Moon landings and cloud computing probably wouldn't exist without it.

Dan "Doc B." Schmalle
President For Life
Bottlehead Corp.


Offline Armaegis

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Reply #22 on: July 15, 2011, 10:18:49 AM
Has anyone here tried running a resistor in parallel with a pot output to ground? I've seen it done before and it pulls the attenuation curve into a nicer shape (compared to a linear pot that is), though it does mess with your input impedance.



Offline Grainger49

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Reply #23 on: July 15, 2011, 10:50:03 AM
Check the link in the second post, reply #1.  One of the drawings shows just that.  I haven't tried it but maybe VoltSecond might chime in?



Offline Armaegis

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Reply #24 on: July 22, 2011, 10:42:10 AM
Check the link in the second post, reply #1.  One of the drawings shows just that.  I haven't tried it but maybe VoltSecond might chime in?

Hmm nope, didn't see a picture that matched what I was talking about...

Actually I found a link: http://sound.westhost.com/project01.htm