Bottlehead Forum
Bottlehead Kits => Crack => Topic started by: karl on February 18, 2011, 11:39:39 AM
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I have found the volume to be too loud in the right channel at the point at which the left channel comes out of the "dead zone." Would padding the volume control fix this problem? If so, how can I do it?
Regards,
Karl
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There are several ways to shed some of the input voltage to bring your listening setting off the bottom.
VoltSecond's site will help some:
http://www.siteswithstyle.com/VoltSecond/12_posistion_shunt/12_Position_Pure_Shunt.html
Figure 1.0 shows how attenuators work. The top of figure 1.0 shows three ways to wire a volume pot. The two shunt modes attenuate the range. This should help you.
Sorry, I don't have the schematic handy to give you values. Start with a resistor the same value as the pot.
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Grainger -
What about simply installing a resistor for each channel after the RCA connector?
Regards,
Karl
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Karl,
Yes, that is what you can do. I would put the resistor in series with the signal wire coming in at the volume pot but it probably makes no difference where it is. Try something the same resistance as the volume pot first.
This wouldn't be exactly as any of the pots are drawn on VoltSecond's site. But a quick jumper added from the top (input side) of the pot to the wiper (output) of the pot will give you the "Shunt mode pot easiest" wiring. The jumper can be temporary to see if you like it that way.
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Grainger -
I tried the resistors on the input and like the result. The channel imbalance is no longer an issue. It is much more comfortable to listen now.
Thanks for the help.
Regards,
Karl
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What value resistor did you end up using?
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I'm glad it worked!
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Forte -
I used 100K Ohm resistors purchased from Radio Shack.
Regards,
Karl
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I'm planning to make a go of padding my input, but I want to make sure I'm not misunderstanding any of the above.
The simplest way of gaining some headroom, as I understand it, is to simply insert two resisters in the left and right channels of the signal path running in from the RCA jacks. So for example, I could disconnect the L and R wires from the input terminals of my pot, solder in one resister for each channel, attach the resisters to the the pot inputs, and call it a day? The other four terminals on the pot remain untouched, correct?
I'm working with a 100k TKD-2511 series pot and have a couple of Shinkoh tantalum 75k resisters on hand - so I think this is probably my best option in terms of giving my pot a little bit of extra headroom? [Presently, unless I digitally reduce the volume going into my dac (which I would like to avoid given that it means I'm no longer getting a bit-perfect data stream), I'm only able to turn my pot from 6:00 (zero volume) to 7:30 or so before my HD800's become uncomfortably loud.]
I'd really love to better understand the various "shunt mode" wiring options, but I'm finding myself fairly confused by the above-link (http://www.siteswithstyle.com/VoltSecond/12_posistion_shunt/12_Position_Pure_Shunt.html ), largely due to the fact that it doesn't clearly show input and output terminals on the pot in the schematics. Here's a crop of the image I'm referring to:
(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fcdn.head-fi.org%2F9%2F9d%2F9d2dc935_Shuntmodes.JPG&hash=584df4027c66d91eb50b136a1193838ae5d58dab)
Is shunt mode "typical" where you essentially rewire the whole pot - as depicted in the picture below and discussed here: http://www.world-designs.co.uk/forum/showthread.php?t=5795 ?
(https://forum.bottlehead.com/proxy.php?request=http%3A%2F%2Fhomepage.ntlworld.com%2Frichard.maile%2FWD%2FShuntMod2.gif&hash=d27a903ddd0253620d58f0bbb4a01b78a6ee1db7)
Alternatively, in shunt mode "easiest" - are you basically inserting resisters in the input path (as described above) and then bridging the input and output terminals of the pot with an additional wire, in each channel? I can't wrap my head around how this works, but it seems to be what the author is saying (as quoted below) in section 1.5 of http://www.siteswithstyle.com/VoltSecond/12_posistion_shunt/12_Position_Pure_Shunt.html
1.5 The Easy Way to make the Stock Foreplay Volume Control into a Shunt Mode Volume Control:
1. Remove the wire connection from the top (right terminal) of the volume pot to the selector switch at the volume pot end.
2. Insert a resistor between the volume pot top (right terminal) and the loose wire going to the selector switch.
This resistor is typically 47.0K (3.3 dB minimum attenuation) to 200K (9.5 dB minimum attenuation) metal film when using a 100K volume pot.
3. Run a short wire from the wiper (middle terminal) to the top (right terminal) of the volume pot.
Am I understanding this correctly? The selector switch in the foreplay is simply the input right? Contrary to the above, Grainger's comment regarding using a resister as a bridge between the input and output terminals on the pot makes a lot more sense to me because this would at least give you parallel resistance. If you are placing the resisters in series with the input of the pot though, why would you then bridge the input and output terminals of the pot?
Thanks for any guidance you can offer me!
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Sorry, but there is a lot of information in your post.
The drawing is not right. If you want "shunt mode typical" then you need terminal 3 to be open, the ground still connected to terminal 1 and both the input resistors and the wire to the grid of the tube on terminal 2 of the pot.
The wiring you show has the wiper of the pot shorted to input of the pot then grounded. The ground of your pot is wired so it will be at maximum volume fully counter clockwise (opposite of volume controls) and it will always be at maximum volume.
So don't wire it that way.
To say this a different way, what is shown on VoltSecond's site the input is at the top. That would be where the wiper is shorted to the lug when the pot is fully clockwise. Usually lug 3 as you drew it. The to the amp lug is most often the center lug, lug 2 as you drew it. And the ground lug is usually lug 1 as you drew it.
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Thanks Grainger! I'll make sure to steer clear of that proposed way of wiring it =)
Is there any reason I shouldn't just adopt the simplest method and drop my 75k ohm resisters in in series with the inputs, in front of the pot?
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Nope, no reason to shy away from a series resistor in each channel. That is the way it is most often done.
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Perhaps no reason to 'shy away', but I believe there may be a reason to goto the shunt mode typical. As I understand it, your signal will be going through the resistor AND the pot. If you go with shunt mode, then the signal will go through the resistor only. I'm not sure how much the resistor will contribute to the overall sound, but since you would only need to buy two, you could roll some boutique varieties through and see.
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Most of your topics are above me too. However, a suggestion. Try your resistors at the input first, to see if it pads enough (jumpers off the end of the Interconnect cable to the RCA jacks internally). I had a similar problem and needed to pad. (not Crack though put this worked well) I used a pot (a cheapo 100k) wired this way to pad down a signal. I tinkered with this until I found the perfect balance I wanted. Then measured the resistance at the cheapo pot and used resistors close to that value. Beats trying to calculate everything out and gives you real, operational padding value that should work (hopefully not high enough to put the signal in the noise floor)
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Perhaps no reason to 'shy away', but I believe there may be a reason to goto the shunt mode typical. As I understand it, your signal will be going through the resistor AND the pot. If you go with shunt mode, then the signal will go through the resistor only.
I've read this many times about Shunt mode now, but I'm not convinced. Just because the pot portion goes to ground doesn't mean it won't affect the signal. The same fraction of the signal goes through the shunt section as it does the amp circuitry. Everything that is connected to the signal line (whether it be in-line or branches off) will affect sound quality. Whether or not it's an audible difference is up to debate.
Additionally, a shunt config also makes the volume control even more sensitive in the first half of the pot. You'll have an overall attenuation, but your low volume control will be worse off.
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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.
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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.
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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.
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williaty ... sounds like you found this but in case you didnt, I post the link again ... http://www.goldpt.com/mods.html
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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!
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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.
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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.
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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.
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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?
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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