| skaara |
How can I make a volume control with one pot (or attenuator?) for both channels of bosoz running balanced (now im using two 5k stereo pots of not so good quality)?
I searched forum but I didnt found answer to this question or It was there and I didnt interpreted it right so im sorry if that has been talked before. |
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| stefanobilliani |
| You must find a quadruple pot :bigeyes: |
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| Peter Daniel |
| Instead of placing the pot between output and ground (4 needed) place the pot between positve and negative output of ea balanced channel (2 needed), so one stereo pot would work. This way, depending on impedance of the pot, positive and negative output will cancel each other without conecting to ground. You can find this setup in Pass D1 DAC. |
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| jwb |
You can use Peter's idea but be aware that it can change the character of the circuit. Look at the original BoSoZ article. The distortion is related to gain, and using a pot between the two sides of the circuit changes the gain.
On my BoSoZ I used a four-deck stepped attenuator from GoldPoint. |
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| Gabrielshorn |
Peter,
Is that option for a stereo pot in the BOSOZ schematic as well? I seem to remember that being built in to the design.
Any idea on what resistance value to put there? Any pro/cons to doing it this way?
I am thinking now of getting an Electroswitch 4-pole 23 position switch and making a 5-10 kOhm attenuator and putting it at the output, seeing as all the IC-based pots have built-in opamps, or require an opamp. |
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| Peter Daniel |
| quote: | Originally posted by jwb
You can use Peter's idea but be aware that it can change the character of the circuit. Look at the original BoSoZ article. The distortion is related to gain, and using a pot between the two sides of the circuit changes the gain.
On my BoSoZ I used a four-deck stepped attenuator from GoldPoint. |
The gain pot is the separate pot. The pots I'm talking about are at the output. |
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| skaara |
| Ill check if I can get quad-pot here.. |
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| Peter Daniel |
| I don't think it was discussed in the original article. What I'm suggesting here is a quite clever way of using single stereo pot to control balanced stereo output. I can only see pros (since pots don't even have to be matched perfectly) and none of the cons. Yet, this method is not very popular with DIYers, anybody knows why?:scratch2: |
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| Peter Daniel |
| quote: | Originally posted by Gabrielshorn
I am thinking now of getting an Electroswitch 4-pole 23 position switch and making a 5-10 kOhm attenuator and putting it at the output, |
How much is it? |
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| Gabrielshorn |
| 4 deck 23 position electroswitch $59.95 |
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| Peter Daniel |
| That's not too bad. For everybody considering that approach I suggest using fixed series resistor and change only the resistors to the ground with a switch, like Harry did in his passive pramp.;) |
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| Gabrielshorn |
| Is that what you call a shunt attenuator? Changing the resistance to ground, I mean. |
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| Nelson Pass |
We used a pot across the balanced output in the
DAC, as described by Peter. It works fine except that
it won't give you absolute zero volume. There is always
some residual output at minimum.
By the way, we bought a pile of those Electroswitch 23
position switches and regretted it. I have boxes of them
if anybody wants to make me an offer.
:bawling: |
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| jwb |
| If you use that electroswitch non-shorting model, wont the volume go to maximum between positions? And wouldn't that be Bad(TM)? |
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| Peter Daniel |
| quote: | Originally posted by Nelson Pass
By the way, we bought a pile of those Electroswitch 23
position switches and regretted it. I have boxes of them
if anybody wants to make me an offer.
:bawling: |
There must be something about those switches, because Sonic Frontiers were getting rid of them too. I have only 3 left.;)
jwb, in that location it has to be a shorting one. |
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| jag |
| quote: |
Peter:
Instead of placing the pot between output and ground (4 needed) place the pot between positve and negative output of ea balanced channel (2 needed), so one stereo pot would work. This way, depending on impedance of the pot, positive and negative output will cancel each other without conecting to ground. You can find this setup in Pass D1 DAC. |
Is this what you mean? |
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| Peter Daniel |
| Taken from Denon high end ad. You can place pot either at the input or the output. |
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| skaara |
| Ill try peters suggestion tomorrow.. |
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| Peter Daniel |
| Would be glad to hear your observations, I plan to use it in my upcoming preamp as well. |
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| jag |
| So it should be like this? |
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| Peter Daniel |
Yes.
I think it's better to do it after 221 resistors. Also 100k is too much for the output. Both this resistor's value and 5k to ground should be carefully chosen for best operation.
Any advice Mr. Pass? |
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| Gabrielshorn |
| Alps does have a supposedly very high-quality quad potentiometer but costs about $200-250. |
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| jag |
| quote: |
Peter:
I think it's better to do it after 221 resistors. |
That will essentially put the pot in parallel with the load (speaker). Well, I guess you can control volume that way, but wouldn't that imply that the pot will need to dissipate a lot of power :confused: |
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| jag |
| quote: | | That will essentially put the pot in parallel with the load (speaker). |
Stupid, stupid me! We are talking about the pre-amp here. Nevermind...:rolleyes: |
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| Roberto Amato |
I've had same problem, finding a quadruple pot for BOSOZ.
Resolved with a bit of mechanics, as shown. Pots are Alps Blue,
I was a bit worried of channel levels mismatch using two separate pots... but no, nothing at all... very good quality pots.
Roberto Amato |
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| Peter Daniel |
Roberto,
This is very nice. I was thinking about it too, but couldn't find suitable parts. |
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| fcel |
Roberto,
I find your solution very interesting and I think it probably cost less than $10 but it does require some work. Could you provide some specs on the 3 gears? Spacing between the 3 gears, # of tooth per gear and may be a drawing showing the dimensions? If I were to go to a R/C hobby shop, what should I ask for? Thanks. |
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| Peter Daniel |
The more teeth, the better. Big gear should have diameter similar to the pt width, or bigger. Small gear can be actually any size, but 1/4 of a big one is just about right, the only requirement being compatibility with the other ones.;)
I wonder if the size of a center gear has any inluence on the resistance when rotating the pot? |
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| Roberto Amato |
| quote: | Originally posted by fcel
Roberto,
I find your solution very interesting and I think it probably cost less than $10 but it does require some work. Could you provide some specs on the 3 gears? Spacing between the 3 gears, # of tooth per gear and may be a drawing showing the dimensions? If I were to go to a R/C hobby shop, what should I ask for? Thanks. |
The small gear has 18 teeth, outer diameter 10mm and has a central hole of 3mm, sorrounded by a collar (you don't see it in the photo, is in the back) of 6mm that fits nicely (with a help of a cianocrylate drop) in a ball bearing. 6mm hole is a common size for small ball bearings inner holes. You need this one, otherwise the assemble won't turn smoothly.
Peter, The size of this gear in this case has no sensible influence on friction, but of course the smaller it is the more torque it gets and the faster it runs, that's why the ball bearing... it also offers a nice way to hold firmly this gear, difficult to do otherwise.
Moreover, it's size influence the distance between the pots. With the current arrangement, there is just enough space for the Alps Pots and the bearing to come along together nicely. Distance center-to-center of the big gears with the small in between is 39mm. You have to be VERY precise on drilling the holes for the assembly or you'll end up (working best case ) with "unpleasant" feeling turning volume. Best practise is drilling the pots holes just little more (~1mm or less) than is needed, so to adjust distances precisely. Also, the total "resistence" you feel is twice what would be using just one pot, but is not excessive.
The big gear is 60 teeth, outer diameter 31mm, with a central hole of 6mm, just right for the pot axe.
I found them on rs-components.com. The catalogue codes from the RS site (the italian one, but should be the same in other countries) are 232-1185 for the small gear and 745-258 for the big one. Prices are about 2 euros/dollars each.
ciao,
Roberto Amato |
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| raul_77 |
Hi all,
You can find four gang potentiometers in Selectronic: www.selectronic.fr.
Sfernice p11 and Alps Blue Box.
The Sfernice are a excelent potentiometer and very cheap, about 22 Dolars.
Happy days,
Raúl Couto |
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| skaara |
| I did it, it didnt worked good, because I dont have equal o/p voltages -> I use single ended input (that will change soon..), theres some losses on undriven side as Nelson said and the voltage doesnt get cancelled 100%, so I get pretty loud volume when I have pot at 0 resistance.. I didnt applied current sources yet, because I didnt buy tip31c transistors (these were proposed to me) yet, Ill have more questions then and I guess ill have to get quad pot somewhere.. |
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| fcel |
Roberto,
Thanks for the detail information ... which even includes the part number! I want to check with you to see if this is the right part # for the ball bearing that you use. RS-Components has a part # of 285-0841 which has a 6mm ID and 19mm OD. I'm not sure if 19mm OD is a common size for a drill bit to drill the hole to fit the bearing.
I wonder if RS-Component has any distributors in the United States because USA is not listed in the choice of country to pick on their "start page". |
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| Roberto Amato |
| quote: | Originally posted by fcel
Roberto,
Thanks for the detail information ... which even includes the part number! I want to check with you to see if this is the right part # for the ball bearing that you use. RS-Components has a part # of 285-0841 which has a 6mm ID and 19mm OD. I'm not sure if 19mm OD is a common size for a drill bit to drill the hole to fit the bearing.
I wonder if RS-Component has any distributors in the United States because USA is not listed in the choice of country to pick on their "start page". |
Well, my bearing was one of a few sitting in a box, I didn't get it
from RS so I can't help you there. But I checked in the past and they were ok... RS also has some bearings that have a little collar over one of the two faces. That collar helps a lot to keep the bearing "flat" on mounting. If I would do it again I would use it.
19mm is a common size, but NOT for a drill bit, use a... (don't know the word in english) is a small bit (about 6mm) with a crown of teeth around... (maybe someone will tell me). The small bit in the center works as a pilot for the rest. They came from 10mm upwards, and they do the work nicely.
I noticed that RS do not have a US site, weird. In any case you can of course get it by any other RS site... can't resist a smile here, someone from US getting parts from Europe... world is upside down! ;)
ciao,
Roberto Amato |
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| Roberto Amato |
[QUOTE]Originally posted by fcel
[B]Roberto,
RS-Components has a part # of 285-0841 which has a 6mm ID and 19mm OD.
That's fine. |
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| Keld |
| quote: | | 19mm is a common size, but NOT for a drill bit |
If you dont get hold of a 19 mm drill bit( I would say it is a Common size) you could try with a 3/4" thats 19. 05mm. Some light hammer work and maybe a little Loctite and the hole would be just fine!
Keld |
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| Mad_K |
I tried it, and it seems to work really well! (BZLS & SOZ)
Right now I am using a dual 10K log pot (between 221R and 100K to ground on each output and channel).
Does anybody know the most ideal value of this pot, and should it be a lin or log version?:confused: |
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| skaara |
| 5k is the good value, but 10k isnt bad either.. You should use log pot for best effect.. |
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| Nelson Pass |
The only problem with non-linear taper is that they
don't tend to track as well between channels.
:( |
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| mrothacher |
NP: May I ask why you weren't happy with your Electroswitch's. Are you referring to the "flat" vertical pcb mount design? I ask because I was about to go that route myself. I've been looking for a good pcb mounted, many-position-switch, so I could make up a sort of "universal" pcb.
Sourcing good pots is right behind sourcing heatsinks when it comes to the "pain" factor.
Thanks
Mike |
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| UrSv |
I am sure everybody knows Elma (http://www.elma.ch/) but they make decent switches and the type 4 is available in 4-gang/24-position/gold-plated. Price is about USD 120 (CHF 200) at qty 1 here in Switzerland. That is a shorting switch which is used quite often as shunt attenuator in it's 2-gang version.
04-4133 is the part number. This is the one I think I will be using with 0.1 % matched Welwyn resistors in my P1.7 remembering Nelson's comments on CMRR when pot is out of track. However I will cheat a little and fit a suitable normal pot at the output and adjust the volume in steps I find good. I will measure the pot for each of the steps and then fix the values for the stepped attenuator not thinking about proper LOG or audio taper which I am sure is not what would fit perfectly anyways.
Thel (http://www.thel.de/) sells silver/carbon potentiometers that are 6-gang. They also sell gears that will resemble the set-up mentioned previously where 2 pots where controlled with one knob.
/UrSv |
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| Peter Daniel |
| I'm using Elma type 4, 2 gang switch at the output of my DAC. Besides slight noise when switching, I'm quite happy with it. There are two versions of that switch, this is the better one, with improved contacts. There is one series resistor in 50 ohm value and I'm only switching shunt resistors which range from 500 to 0.5 ohm. This assures very low output impedance. If there is an interest I can post the values.;) |
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| UrSv |
Peter,
Please post your values if you have them.
As for your question about the size of the inner gear affecting the resistance to turn the pot (assuming we are talking force needed and not the resistance of the pot itself :drunk: but I take that for granted since you were the one who asked) it certainly does. The smaller the inner gear the smaller the force needed and vice versa. It should be reasonably linear and an inner gear half the diameter of the outer gear would mean that you would only need half the force compared to direct connection (or the same as before if the gears are running two pots/switches). This of course using provided that you use the same knob in all cases...
/UrSv |
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| Peter Daniel |
| These are the values. The series resistor is 50 ohm and these resistors are switched between the output and the ground. This choice works pretty well. |
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| jwb |
Also using Elma type 4 switches in my gear. If you are in the USA, you can order directly from their facility in Fremont, CA. Double bonus if you live in the San Francisco Bay Area.
I wonder where Arn Roatcap (http://www.goldpt.com/) gets Elma type 4 switches with 6 decks? |
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| Roberto Amato |
| quote: | Originally posted by UrSv
As for your question about the size of the inner gear affecting the resistance to turn the pot (assuming we are talking force needed and not the resistance of the pot itself :drunk: but I take that for granted since you were the one who asked) it certainly does. The smaller the inner gear the smaller the force needed and vice versa. It should be reasonably linear and an inner gear half the diameter of the outer gear would mean that you would only need half the force compared to direct connection (or the same as before if the gears are running two pots/switches). This of course using provided that you use the same knob in all cases...
/UrSv |
Sorry, UrSv, but this is not the case. That would be valid if we are building a train of gears, with the aim to exchange torque for speed, or vice-versa. In our case whatever dimension the central gear is, the force needed for turning both pots is always the same. The central gear has only to connect the two pots. The overral (phisical) resistence is just the sum of the two pots, providing that whatever gear is in between has a negligible resistance of it's own, but whatever it is depends only by it's friction. Passing from the gear we are turning (the one connected to the right pot) to the central one we are speeding it up, exchanging torque for speed. In turn, the small one connected to the gear on the left exchange speed for torque. The balance is even. Since the smaller the central one the faster it turns, we have only to worry about friction. :hypno2:
Roberto Amato |
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| UrSv |
Elma makes them any way you wish with even 7 decks or more.
They are however not standard but custom order...
/UrSv |
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| UrSv |
Roberto, Peter,
Sorry for not checking the picture closely enough. I assumed that the control of the set-up would be done thorugh center gear (in which case my statement is true) which I see it is not. Obviously then the size is transparent to the user and the force is sum of all the individual forces for pots/switches. Did not mean to be confusing...:Pinoc:
/UrSv |
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| skaara |
| Does anyone have plans for physical construction of attenuator in which resistors from peters table can be used? how did you do it peter? |
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| Peter Daniel |
| I did it like that. PCB comes with Elma switch. One end of ea. resistor goes to PCB (and is switched), the other ends are connected together with wire and grounded. The pic is not very detailed, but I don't feel like opening my DAC now.;) |
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| Peter Daniel |
| And to avoid confusion, this is how it's connected.;) |
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| blschuler |
Peter,
It looks to me as if the attenuation values you posted are for a 500 ohm series resistor not a 50 ohm. |
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| Peter Daniel |
| What makes you think that? |
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| skaara |
| I see now:) Didnt looked at that pic closely before.. Attenuator on this pic is for 1 balanced ch. only? |
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| UrSv |
Does anybody know which is preferable (if useful at all in bottom version) of the different versions below? I read something about having the switch as close to ground as possible but in this case it seems maybe a lot of stray capacitance would be added?
/UrSv |
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| Peter Daniel |
UrSv
I would go with a top one, somehow I don't like the idea of so many resistors hanging from the output.
Skaara
It's stereo, I'm not using balanced output there. |
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| grataku |
Peter,
doesn't the type of connection vary the output impedance of the circuit? I don't know if I like that.
Also I was wondering what would happen if you put a 5k fixed resistor to ground after the 50 Ohm as a way of getting rid of the switching noise. The log curve would be changed but that should be about it. |
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| Peter Daniel |
It changes, but 50-550 ohm range difference is negligible in my opinion. Volume control is for convenience, and I usually use 5 position in the last 3/4 of a range. So i can say that my output impedance is fixed and around 200 ohms. I'd say it's not too bad.;)
As to 5k to output, maybe it's worth a try. |
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| blschuler |
Peter,
How do you calculate dB from a simple voltage divider??? I must be doing it wrong.:bigeyes: |
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| Peter Daniel |
| I tried to do it yesterday and couldn't figure out either. Last time I did it was 1996. BTW, the chart comes from David Broudhurst and is recommended for his DAC. I thought it was 20logR1/R1+R2 but somehow the numbers don't come up.;) |
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| Steve Eddy |
| quote: | Originally posted by Peter Daniel
I tried to do it yesterday and couldn't figure out either. Last time I did it was 1996. BTW, the chart comes from David Broudhurst and is recommended for his DAC. I thought it was 20logR1/R1+R2 but somehow the numbers don't come up.;) |
That won't work if R1 is the series resistor and R2 is the shunt resistor. If it is, then it's 20 x log R2/(R1 + R2).
se |
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| Peter Daniel |
| That is interesting, because I built Broadhurst attenuator and it works fine. Didn't bother to check the values before I built it though.;) |
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| Steve Eddy |
| quote: | Originally posted by blschuler
That exactly what I have been doing, still can't duplicate the attenuation values in the Broudhurst chart. |
It works for me.
You're aware by the way that the calculator keeps the series resistor value constant, right? In other words, R<sub>x</sub> stays the same and only R<sub>y</sub> changes for each step.
So let's run through a simple one.
Set the series resistance to 1,000 ohms.
Set the number of steps to 1.
Set the first step attenuation to 20dB.
The calculator returns 1,000 ohms for R<sub>x</sub> (natch) and 111 ohms for R<sub>y</sub>
Now let's check it against the calculation 20 x log R<sub>y</sub>/(R<sub>x</sub> + R<sub>y</sub>).
R<sub>x</sub> + R<sub>y</sub> = 1,000 + 111 = 1,111.
R<sub>y</sub> / 1,111 = 111 / 1,111 = 0.0999
Log 0.0999 = -1
20 x -1 = -20.
So, we get -20dB.
It checks.
se |
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| Steve Eddy |
Oops. Just realized that the calculator you gave the URL for wasn't the Broudhurst chart you mentioned.
Where can I find this Broudhurst chart?
se |
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| Peter Daniel |
| quote: | Originally posted by Steve Eddy
Where can I find this Broudhurst chart?
se |
At the beginning of the previous page. The series resistor is 50 ohm. |
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| Steve Eddy |
| quote: | Originally posted by Peter Daniel
At the beginning of the previous page. The series resistor is 50 ohm. |
Thanks.
Nope, they don't add up. Of course the basic calculation assumes a zero impedance source and an infinite impedance load. These numbers may have been calculated for specific source/load impedances.
se |
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| bigswede |
Where can one find a suitable 2-point volume pot as mentioned by Daniel at the beginning of this thread, to be located between positive and negative of the two channels of BoSoZ?
...at reasonable price. No freak gear needed in my first DIY project... |
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| mefinnis |
Resurrecting an old thread, which was brought to my attention by a friend.
Clearly there are a number of options as to exactly where a single pot/switch may be placed.
Which of the following (A/B/C) do we think would be the best option?
Thanks,
mark |
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| Brian Donaldson |
| I would say A. Putting the attenuator as late as possible lets it also attenuate as much noise as possible. |
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| promitheus |
| I built a test board with 6 relays using the ADC0804 like in the pass preamp. Should I use a linear or log pot? I tried a linear pot 100K because there is a 10K resistor across middle pin (moving pin) and earth. This makes it act logarythmic. Also the relays switch in this fashion for linear movement. But I think in the begining the volume doesn´t go up so fast as at the end. I have to say I used it at the output of the BLS with full volume. So it could be that the volume pot interacts with the relay volume control. Just wanted to know if you guys used linear or log pot. |
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| till |
I use a ADC0804 volume controll with my BOZ. For me the range is not enough, and this volume control makes a lot of noise, clicks and pops when volume is tuned. Also its not absoltely stable, sometimes relays change without me moving the pot knob. I use a 100k lin pot.
The better way with more range and less noise, absolutely stable, is this very simple microcontroller relays volume controll i made for my BZLS :
http://home.arcor.de/dddddd-/relais/relais.html
Its easy to use for SE only, mount only half the number of relays. It needs less parts than the ADC0804 version. Its very cheap. |
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| promitheus |
| I did see the same problems with instability. The problem with uP is that you have to programm them. I don´t really know how to do that. When I will need a few more for friends or to repair it that would be a problem. |
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| till |
| quote: | | The problem with uP is that you have to programm them |
no problem, contact me. |
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| promitheus |
what do you use to programm the pics?
do you use the programmers with basic? |
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| Petter |
| quote: | Originally posted by mefinnis
Resurrecting an old thread, which was brought to my attention by a friend.
Clearly there are a number of options as to exactly where a single pot/switch may be placed.
Which of the following (A/B/C) do we think would be the best option?
Thanks,
mark |
Don't forget changing the value of R15, except it works in the opposite direction. Higher resistance is lower gain.
Petter |
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| till |
| quote: | | what do you use to programm the pics? |
assembler, the code lays on the page i linked above. |
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| promitheus |
| what kind of programmer do you have? |
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| till |
| quote: | | what kind of programmer do you have? |
these informations could be found at the page i started to make some documentation for my PGA2310 volume controll
In university there is also a professional programmer avaiable, but i had no problems with the diy one from sprut.de until now.
in case you need only some for private usage i´m able to send them to you ready programmed.
http://home.arcor.de/dddddd-/PGA/index.html |
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| macka |
I used the values as stated in Aleph L attenuator for the ouptut attenuators and the Goldpoint calculated values for the 10K input attenuators.
The attenuation has a very smooth action for the X Bosoz.
The 24 position Electroswitch referred to by Mr Pass works pefectly for both input and output.
macka:) |
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| Brian Guralnick |
My passive, wireless remote control, battery powered preamp is coming! It should be puny in size & price compared to what I’ve seen so far in this topic.
I'm now doing the IR receiver / processor & display PCB, schematic of the attenuator module is attached.
-It runs on 2 lithium coin cells. They should last at least a year.
-Consumption: 100ma for 2 seconds just after a volume change, 30ma for 3 more seconds while the 4-digit display is on, less than 0.003ma (3ua) while nothing is happening. Processor is completely stopped here, ABSOLUTELY NO EM signals being generated at this time. Even the microcontroller oscillator is shut down.
-Supply range, 5v to 16v dc.
-Supply is completely isolated from the analog audio path.
-Software selectable linear pot emulation, or, volumetric pot emulation.
-Software controls learns remote signals from 1 of your existing remotes.
-(modules only) Exclusive special price for DIYaudio forum members.
Everything should be ready in around 2-3 months:
ATTENUATOR MODULE:
-Fully passive. Break points at the ins & outs to allow for custom drive sections.
-Emulates 2 individual mono POTs allowing balance control.
-Selectable 100k, 50k, 25k, or 12k models.
-Super matching of channels, even when using all 16 channels.
-Encased in a copper shield.
-Fits vertically along the rear panel of a 1U rack-mount case leaving the rest of the case empty.
-Attached is a 128 step volume module with 6 line ins. It's finished.
-A 32 step, 3 line in econo module will also be available.
-Optional 0.1% resistors, or 0.5% resistors versions.
-1 module is stereo unbalanced, or mono balanced.
PROCESSOR MODULE:
-It will run up to 8 modules simultaneously. That's 16 unbalanced channels, or 8 balanced channels.
-Fits vertically along the front panel of a 1U rack-mount case leaving the rest of the case empty.
-Built in power supply regulation circuitry.
-Connects to attenuator modules via 1mm 6 pin flat flex cable, no other wires anywhere. |
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| jewilson |
Brian,
Why don't you post the schematic, my eye are not good enough to see what your doing. |
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| Brian Guralnick |
Sorry, waiting for the first PCB before I potentially embarrass myself.
Also, I guess I might as well wait for the processor schematic to be finished & post everything in it's own thread.
I gues you could call this a preview... |
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| jam |
Brian,
Looks good..............
Will the gain stage be on a seperate module?
How many inputs will you have?
Regards,
Jam |
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| Brian Guralnick |
I'm not supplying the gain stage, however, I left optional jumpers so you may add your own either right before the relay based pot, or right after depending on your preference.
For a balanced system, I would recomend the bosoz design, for unbalanced & extra simple, I would use a simple JFET buffer. However, these do require a power supply. Since it's so small, I may just squeese it in without placing any of the parts.
Each attenuator module will have 4 line ins, it was originally 6. If you don't want my processor board, these modules have an I2C interface. |
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| jam |
Brian,
With four lines in, does that mean two balanced inputs?
Regards,
Jam |
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| nowater |
Peter Daniel said in post #3 and #22 about the single-pot balanced volume control--| quote: | | Instead of placing the pot between output and ground (4 needed) place the pot between positve and negative output of ea balanced channel (2 needed), so one stereo pot would work. This way, depending on impedance of the pot, positive and negative output will cancel each other without conecting to ground. You can find this setup in Pass D1 DAC. |
| quote: | | Yes.I think it's better to do it after 221 resistors. Also 100k is too much for the output. Both this resistor's value and 5k to ground should be carefully chosen for best operation. |
Please help my ignorance on the question of volume control impedance values.
1. How do you calculate the optimal value for a (normal) volume control at the input to your preamp? I have heard 25-50K mentioned for the BZLS input. How/why?
2. How do you calculate the optimal value for a (normal) volume control at the output to your preamp? I have heard 5-10K mentioned for the BZLS input. How/why?
3. With the single-pot method of volume control of a balanced amp, no part of the pot is grounded, or placed across the amp's input or output. How do you work out the right value for this pot?
Thanks,
Grant |
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| Fox |
Hi Grant,
The options to adjust the volume of the BZLS really freak you out, don't they? Apparently, the simplicity of the line-stage makes it more difficult to decide upon the 'right' level-control ;)
Anyway here are my observations:
| quote: | | 1. How do you calculate the optimal value for a (normal) volume control at the input to your preamp? I have heard 25-50K mentioned for the BZLS input. How/why? |
The maximum value of a pot at the input is restricted by the input capacitance of the MOSFETs. A high value pot together with this reasonable capacitance (in nano-farads) forms an RC-filter that causes a high frequency roll-off. For this reason, Nelson advices not to exceed 10k. In the Aleph P he just omitted the input pot to increase the bandwith of the line-stage.
| quote: | | 2. How do you calculate the optimal value for a (normal) volume control at the output to your preamp? I have heard 5-10K mentioned for the BZLS input. How/why? |
The output impedance of the BZLS is directly related to the value of the pot. The lower the better and as Nelson explained 5k seems reasonable
| quote: | | 3. With the single-pot method of volume control of a balanced amp, no part of the pot is grounded, or placed across the amp's input or output. How do you work out the right value for this pot? |
I have simulated this alternative with Simetrix and found out that the value of the pot needs to be 1k to be effective. Be aware however that with unbalanced sources the output level of the two polarities of the signal is not the same, which makes the attenuation of the signal with a balanced attenuator a problem at low volume settings.
Hope this helps,
Fox |
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| nowater |
| quote: | | quote: | | 3. With the single-pot method of volume control of a balanced amp, no part of the pot is grounded, or placed across the amp's input or output. How do you work out the right value for this pot? |
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| quote: | | I have simulated this alternative with Simetrix and found out that the value of the pot needs to be 1k to be effective. |
1k!! That is extremely low. Why? What goes wrong with a higher value? The multi-channel volume control I am thinking of using only comes in 10k and larger sizes.
Grant |
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| Fox |
| quote: | | 1k!! That is extremely low. Why? What goes wrong with a higher value? The multi-channel volume control I am thinking of using only comes in 10k and larger sizes. |
Above 1k the signal won't be attenuated, which means that with a 10k-pot you can only use the first quarter of the pot-range. This results in a very small range to attenuate between very loud and very low sound pressure levels. Sorry but that's life for ya :D
Fox |
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| ljozsef |
Fox!
It was simulated - where? A, B or C (post 72)?
I think that A, but nearby the input (C) would be maybe fine with a higher value, too -?
Could you possibly simulate it?
Laci |
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| Fox |
Hi Laci,
I simulated the circuit at A (post 72) but contrary to the picture, after the 100K resistors. Attenuation before these resistors does not work.
According to my simulations, attenuation at the input (C) should work well with a 10k-pot. A 20k-pot should also work but does result in an increase of the signal in comparison to a 10k-pot. This implies that with a 20k-pot the range between 10k and 20k does not alter the signal level. So the conclusion seems justified that with a 10k-pot the most accurate level-attenuation is possible because the whole range of the pot can be used. :)
Hope this helps,
Fox |
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| ljozsef |
Hi, Fox,
Thanks for the explanations!
10 k much more fine for me. But, if you could, just one premium simulation: I suppose, the rule were to use log pots. Is this the only truth, or linear ones could be applicable, too? - eventually applying supplimentary items?
I think this alternative solution could be interesting for many diyers having lin pots in their drawers.
Regards,
Laci |
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| Fox |
Hi Laci,
Linear-pots can also be used. However, the attenuation will be heard in a counter-intuitve manner because the way we hear the difference between loud and less loud is on a logaritmic scale.
Somewehere I read that putting a resistor parallel to a linear pot results in a good aproximation of a logaritmic one. For instance, when you put a 20K resistor parallel to a 20k-pot, you would have a 10k-pot that behaves sort of logaritmic.
Good luck
Fox |
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| Prune |
Sorry to resurrect an old thread, but...
I'm having a big problem with putting the pot between the two lines of a balanced channel, as the simulation gives me an attenuation curve that is opposite from what audio taper is supposed to be. Most of the attenuation happens for pot settings that are lower than the series resistors it follows.
The attached image shows my situation; R4 is the pot, with the DAC output represented by the sources on the left, and the buffer on the right. Using a pot of just a few K gives a more normal curve, but the maximum volume is a third of the original, and with no voltage gain stages after this it's no good. I know there was a way to use a shunt resistor to fake a log law, but that does not work in this configuration. I can't figure out how to do this, so I hope someone can help. |
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