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Post by perfboardpatcher on Feb 13, 2020 14:27:16 GMT -5
Nice one Perf 🙂 Now if you could just help me out and explain what I’m looking that would be great 😆 my brain melts when I look at schematics. I might sound like I know more than I do but im definitely kindergarten. Excited to learn tho. What sounds am I getting ? How do the switches work ? As I look at it , you can sellect either position 1 or 4 on the five way switch , is that right ? You say there is Switches (a five way and a super switch ) Do you mean two “types of switches” or two switches in total ? As I look at this I’m seeing 3 five way switches (one for each single coil in a SSS setup) and a separate super switch off to the side. Good that Retread and Newey did some of the explaining. The first questions I ask myself are always: "Am I capable to build it?" and "Do I want to build it?" I guess you have the same kind of questions on your mind. If the brainstorming results in a circuit with 15 or more caps and resistors would you then still be determined to build the circuit? I wouldn't blame you if you were not. It's hard to come up with a schematic that meets all your demands, ourclarioncall! Apart from volume switching and pup combinations switching you're asking for pickup blending. My schematic offers 5 different volume levels (sw2) and 5 different pup/blend combinations (sw1). About those examples 1&4. The #4 option is an attempt to simulate pickup blending like on a Gibson LP style guitar. The more the signal is attenuated the more linear - in terms of frequency - the blending. The #1 option does blending but the result is somewhat like one of my graphics in your HOOP topic shows, the blending works only works well for the low end. I don't like this circuit much. Not an elegant design. Giving up #4 would make it easier to come up with a schematic that is more fun to build. I'd keep the 5-way volume control and ditch the part with the superswitch. Instead I would use 3 switches for pickup selection and fine-tuning/mixdown. The idea is to use the 5 way switch for the big steps (e.g. 6dB) and use the 3 switches for smaller increments (e.g. 3dB) or for mixing purposes when 2 or 3 pickups are engaged. 3db attenuation : by means of R2+C2 attenuation as blend option: by means of R1 or R1+C1 or R1*L1 or (R1*L1)+C1. I expect versions with inductor L1 included to give the best results.
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Post by ourclarioncall on Feb 26, 2020 19:24:24 GMT -5
Sorry Perf, I fell off the face of the planet for a while there
I will come back to this at a later date as there is a few things I need to do circuit wise and it’s taking a bit of time to get there , but I’m making progress even if it’s snail pace.
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Post by ourclarioncall on Mar 23, 2020 20:50:52 GMT -5
Back over here again im trying to figure out how to change RT’s module below into a 2/3rds , 1/3rd, full volume . I was in the highest level maths class at school .......about 25 years ago 😬 can anyone lay out a step by step procedure, or point me in the right direction. Treat me as Kindergarten level 🙂 If it is the same thing Ash posted earlier in this thread then I will definitely need help !
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Post by reTrEaD on Mar 26, 2020 12:33:32 GMT -5
can anyone lay out a step by step procedure, or point me in the right direction. Treat me as Kindergarten level 🙂 The 150k resistor in the Treble Bleed network makes it very difficult to define a step-by-step process for selecting standard-value resistors. Maybe JohnH is enough of a mathemagician to do something with that. In my estimation, using these resistors (top to bottom) in the divider section should give you results close to what you're describing: 180k 47k 22k This will emulate a 249k pot with (approximately) these division ratios: 0bB -7dB -14dB
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Post by ourclarioncall on Mar 26, 2020 16:31:26 GMT -5
can anyone lay out a step by step procedure, or point me in the right direction. Treat me as Kindergarten level 🙂 The 150k resistor in the Treble Bleed network makes it very difficult to define a step-by-step process for selecting standard-value resistors. Maybe JohnH is enough of a mathemagician to do something with that. In my estimation, using these resistors (top to bottom) in the divider section should give you results close to what you're describing: 180k 47k 22k This will emulate a 249k pot with (approximately) these division ratios: 0bB -7dB -14dB Is the treble bleed a necessary part of this design or can it be left off ? I will use it but just wondered If the current values for the treble bleed are not accurate , would it make much of a difference? It would also be great to have a 500k and a 1meg option too but no rush for any of that . I was thinking that I use the 2 single coils in series settings a lot so the 500k might be a better option. Why arnt 1 meg pots standard ? Wouldn’t it be better to cut treble etc from the amp at the end instead of cutting something from the pickup?
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Post by reTrEaD on Mar 27, 2020 9:45:34 GMT -5
Is the treble bleed a necessary part of this design or can it be left off ? It's no more necessary here than it is on a regular volume pot. If you don't mind having a significant loss of treble when the volume selection is reduced from 0dB, you can omit the treble bleed and recalculate the resistor values needed for your desired division ratios. If the current values for the treble bleed are not accurate , would it make much of a difference? That all depends on how 'inaccurate'. 10~20% ? Not a big difference. -50% or +100% ? Big difference. It would also be great to have a 500k and a 1meg option too but no rush for any of that . I was thinking that I use the 2 single coils in series settings a lot so the 500k might be a better option. Personally I won't rush at all because I have no intention to ever invest any time in it, although others might choose to do so. When using individual volumes for each pickup, the loading of the volume control doesn't change when you put two pickups in series. If you have a master volume, it does. But by compensating for the series configuration, you're presenting less loading when only one pickup is selected or when both pickups are selected in parallel. In those situations, with less load, the sound is overly bright. So you still have a darker sound when you have pickups in series, compared to the 'normal' one pickup or both pickups in parallel selections. Yes, you could redesign to get the best tone the series configuration, but doing so basically amounts to this: Getting a proper 'blend' when using individual volume controls on each pickup is a bit tricky when using conventional pots and the pickups are in parallel with each other. Far more difficult when using volume switches. The division necessary to make a slightly noticeable difference when only one pickup is selected will introduce enough series resistance to cause that pickup to have very little contribution to the parallel combination, if the other pickup is at full volume. So in the case of volume switches, separate volume controls tends to be less useful. Why arnt 1 meg pots standard ? Wouldn’t it be better to cut treble etc from the amp at the end instead of cutting something from the pickup? No, that would be much worse for several reasons. To begin with, the increased brightness comes in the form of a rather sharp peak. The treble control on an amplifier can't be 'tuned' to flatten out just the peak. Then we have the issue of four times series resistance for the same division ratio if using a 1Meg pot rather than a 250k pot. This means the cable capacitance is that much more effective as a high-cut filter when the volume is reduced. You can mitigate that with a treble-bleed network. But the series resistance also allows hum and noise to more easily enter the system through the guitar cable when the volume is reduced. Also, in guitars which have a bass-cut control, a 1Meg volume pot means the bass-cut is mostly ineffective. Finally, circling back to the independent volume controls: Personally, I wouldn't choose to have them. But if I did, 1Meg pots would definitely be a no. I'd use 250k pots (or 300k pots, like Gibson uses on their LPs) and they would be linear, rather than audio taper. That allows fading one pickup slightly out of the mix when both pickups are selected. 1Meg pots, particularly audio taper, are far too abrupt.
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Post by sumgai on Mar 27, 2020 10:43:14 GMT -5
occ,
Everything reTrEaD just said above, but I'll add a bit of history.
Essentially, Leo used 1M Ohm pots in the Fender Jazzmaster, beginning in 1958, because in those days, most players used large diameter flatwound strings. More or less, there was no "sparkle" to be found anywhere, which suited jazz players just fine. In order to keep the JM from sounding too dull, the Volume pot value was quadrupled to 1M. (Most of the rest of the Fender line was using 250K Ohms in that position.)
That's true to this day, as my own JM will attest, even though strings have decreased in size and increased in technology (resisting corrosion and such, to keep the sizzle alive for longer, before replacement), and the move almost entirely away from flatwounds.
HTH
sumgai
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Post by Deleted on Apr 23, 2020 15:15:46 GMT -5
> DELETED because the FORMULA is WRONG, and I dont want any one working on ERRORS of my doing . < Spotted by sumgai THANKS great thing doing maths in bulk.. you cook up in one area it spreads
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Post by sumgai on Apr 23, 2020 20:45:33 GMT -5
'bunny, Your circuit and table both look good, with one minor exception - the third column is not the result of 10^(dB/20), it's actually showing the reciprocal of the resultant value. Which of course is the number you want as a multiplier for any given value of R2. Nonetheless, the formula for R1 should read as: R1 = R2 x 1/(10^(dB/20)) But since you do have that result in the third column, no one should get erroneous results if they follow your instructions. Good job. sumgai
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Post by Deleted on Apr 24, 2020 3:50:21 GMT -5
Your circuit and table both look good, with one minor exception - the third column is not the result of 10^(dB/20), it's actually showing the reciprocal of the resultant value. Which of course is the number you want as a multiplier for any given value of R2. Nonetheless, the formula for R1 should read as: R1 = R2 x 1/(10^(dB/20)) But since you do have that result in the third column, no one should get erroneous results if they follow your instructions. Good job. sumgai The THRID Column Formula is 10^(db/20) First db, Second db/20, Third 10^(db/20) Formula should be { R2/(10^[db/20]) } - R2 where R2 is at the TOP and R1 is all the Figures below The formula for some reason a fraction in the spread sheet Vo=(Vs x R2) / (R1 + R2) So if we say Vs is 1v makes life easy Vo is 10^(db/20) R2 is FIX value, whatever resisters fits to get the near perfect match so we can get the db correct as Vs is 1v its out of here 1x Whatever is Whatever now leaves Vo=R2/(R1+R2) ... R2/Vo = R1+R2 .. (R2/Vo)-R2 = R1 then just a matter of changing R2 to find the near perfect match to give a nice db drop rate
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Post by sumgai on Apr 24, 2020 13:36:32 GMT -5
Your circuit and table both look good, with one minor exception - the third column is not the result of 10^(dB/20), it's actually showing the reciprocal of the resultant value. Which of course is the number you want as a multiplier for any given value of R2. Nonetheless, the formula for R1 should read as: R1 = R2 x 1/(10^(dB/20)) sumgai The THRID Column Formula is 10^(db/20) I think you want to use a Lifeline right about now. Look at Row 24, the -20dB calculation. Column B has dB/20 = 1, and there's no room for argument there, eh? But in Column C, 10^1 is not 0.100, it's 10. And the reciprocal of 10 is.... 0.1, just as you show in the Column C of that row. (Sorry, but that proof was just begging to be called out.) But hey, don't take the word of your calculator for it, simply use any search engine, and enter "10 to the 0.05th power" (from Row 5, the first decimal drop), and watch what pops up. That's the correct answer for the portion of your formula under discussion. What you show in Column C is the reciprocal of this answer. To prove my assertion, simply go back to your search engine, and type in "reciprocal of 10 to the 0.05th power", and Voila! there's the number in your column. Which is the necessary number in order for the correct formula to work as desired. You need to change the whole formula to read as "R1 = R2/( 1/10^(dB/20))-R2", and change the Header on Column C to read 1/10^(dB/20. Yes, I'm harping here, but the fact is, we can't be seen as a website that dis-respects actual mathematical formulas. If we're going to use them (and we must do so, for things like the topic of this thread), then we need to be spot-on accurate, not just close enough for government work. In reality, your table prints the formulaic values for calculating R1, for exactly one value of R2. (And reiterated in powers of 10.) Viewers are going to want to use other values for whatever reason, and when they go to use the formula that you describe, instead of just looking up the correct value in Column C, then they're going to be in for a surprise, to say the least. I trust I've made my point. HTH sumgai
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Post by Deleted on Apr 24, 2020 14:35:19 GMT -5
I think you want to use a Lifeline right about now. Look at Row 24, the -20dB calculation. Column B has dB/20 = 1, and there's no room for argument there, eh? But in Column C, 10^1 is not 0.100, it's 10. And the reciprocal of 10 is.... 0.1, just as you show in the Column C of that row. (Sorry, but that proof was just begging to be called out.) True 10^1 is 10, but 10^-1 is 0.1!
But hey, don't take the word of your calculator for it, simply use any search engine, and enter "10 to the 0.05th power" (from Row 5, the first decimal drop), and watch what pops up. That's the correct answer for the portion of your formula under discussion. What you show in Column C is the reciprocal of this answer. To prove my assertion, simply go back to your search engine, and type in "reciprocal of 10 to the 0.05th power", and Voila! there's the number in your column. Which is the necessary number in order for the correct formula to work as desired. You need to change the whole formula to read as "R1 = R2/( 1/10^(dB/20))-R2", and change the Header on Column C to read 1/10^(dB/20. Im not sure where the 1/10^(db/20) is coming from dB = 20 × log (Vout / Vin) .. dB/20 = Log (Vout/Vin) .. 10^(dB/20)= Vout/Vin F**K Vin=1 to make it easy on life, So .. 10^(dB/20)/1 = Vout = (Vin x R2)/(R1+R2) Vout = (1xR2)/(R1+R2) = R2/(R1+R2) R2/Vout = R1+R2 .. (R2/Vout)-R2=R1
R1 = (R2/(10^(dB/20)/1))-R2
Yes, I'm harping here, but the fact is, we can't be seen as a website that dis-respects actual mathematical formulas. If we're going to use them (and we must do so, for things like the topic of this thread), then we need to be spot-on accurate, not just close enough for government work. In reality, your table prints the formulaic values for calculating R1, for exactly one value of R2. (And reiterated in powers of 10.) Viewers are going to want to use other values for whatever reason, and when they go to use the formula that you describe, instead of just looking up the correct value in Column C, then they're going to be in for a surprise, to say the least. I trust I've made my point. HTH sumgai So use that as the GOD FORMULA R1 = (R2/(10^(dB/20)/1))-R2
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Post by sumgai on Apr 24, 2020 16:22:42 GMT -5
'bunny, I see your point - yes indeed, 10^-1 does indeed equal 0.1. However, in my defense, dB (deciBels) represent a ratio, and a ratio is an absolute number. That's why I don't "drop down a negative 3dB", I "drop down 3dB", or I "reduce it by 3dB", or I "subtract 3dB" when I reduce the level by half. Nor does anyone else I've ever spoken with - we all use the absolute form of the term, and prefix it with "reduce, drop down, subtract", or some other synonym. I suppose I could say "add -3dB to the level" (where the minus sign is spoken as 'negative'), and that would be grammatically correct, but a misleading choice of words, don't you think? And I can't say "subtract -3dB" for the same reasons. I dunno, maybe they speak differently over there.... Do you normally say "reduce it by negative 3dB"? Nonetheless, your table stands as is, no correction needed. But only because you included a negative sign in Column B. sumgai
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Post by Deleted on Apr 25, 2020 3:19:29 GMT -5
Gotta remember when in passive can only take away. These is no gain no boost
Kirchoff's law what goes in comes out If you gain more than is going it is a act of God
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Post by sumgai on Apr 25, 2020 11:53:07 GMT -5
Gotta remember when in passive can only take away. These is no gain no boost Kirchoff's law what goes in comes out If you gain more than is going it is a act of God Nope, don't gotta remember that at all. A ratio has no recognition of such things as gains or losses, it's only a measurement of a difference in level. By using a sign in front of that ratio, the reader is clued in as to the direction of that ratio, up or down, in comparison to the initial level. And herein, I speak of the Electrical/Electronic Engineering's discipline overall, not just one tiny subset. Were I to open it up to include the Physics discipline, we'd really be in trouble, trying to limit what a dB is, and what it can represent. (Physics would be the more appropriate ground for discussing acoustic phenomena, such as how loud a sound is perceived, and then changed in loudness, up or down.) And for the record, Kirchhoff's Laws concern themselves with the distribution of current or voltage through multiple pathways within a circuit, not the fact that current out equals current in. For that matter, Kirchhoff explicitly states that: "The sum of all currents leaving a node in an electrical network always equals zero." This particular statement can be found on numerous websites dealing with engineering disciplines, let alone countless textbooks across the nearly 18 decades since he formulated the rule. (Like, in 1845!) On the face of it, your statement that "current out = current in" would seem to fly against the grain here, wouldn't it? In fact, no, it doesn't. But making a correct statement, in proper context, requires giving a full analysis and explanation to the reader, not just a throwaway bon mot. Now you know (or should know) why I'm such a stickler for these kinds of things. In my personal life, I let people get away with lots of things, because in the big scheme of things, it won't matter. But here in The NutzHouse, it does make a difference. Readers come here from all walks of life, with all manner of experiences. We get them from actual full-blown Engineers all the way down to "how does electricity work?", and everything in between. For those tending towards the latter category, I can't abide telling them half-truths, or without explaining shortcuts that can hurt them in some way if they don't understand the underlying concepts. Call it 'too many years of wearing a college professor's hat' if you wish, but so far in my sojourn here, nobody's complained that I'm too full of hogwash. Pedantic, yes, but not arrogantly stubborn, even when I'm wrong. (You will recall that you have twice successfully extracted apologies from me, yes?) Now, let's put this to bed, shall we? I believe that your original intentions with the 6-position switch have been well serviced here in this thread, and we shouldn't distract too greatly from a future reader who'd like to build your module into their scheme. HTH sumgai
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Post by Deleted on Apr 25, 2020 13:20:58 GMT -5
Gotta remember when in passive can only take away. These is no gain no boost Kirchoff's law what goes in comes out If you gain more than is going it is a act of God And for the record, Kirchhoff's Laws concern themselves with the distribution of current or voltage through multiple pathways within a circuit, not the fact that current out equals current in. For that matter, Kirchhoff explicitly states that: "The sum of all currents leaving a node in an electrical network always equals zero." sumgai YOU ARE NOT GOD AND YOU CAN NOT MAKE POWER SO I CAN NEVER BE IN A +dB THAT was the point .. PASSIVE NO ADDED POWER you where having a HARD TIME knowing that the - MEANS - and im puzzled why you had that hard time understanding that. i am ELECTRONICS and COMPUTERS college and uni, i didnt want to go to HALHAM so i ended up at Nottingham Trent Uni.. ALSO i did Electrics, Fully qualified electrician. also was a Design Tech for a few years on Current based Amps (hearing aids). i know i have faults with dyslexia but you have a problem if miss reading things too, but you dont seem to know you have this problem. That is some thing i cant get my head around .
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Post by sumgai on Apr 25, 2020 15:39:27 GMT -5
ASB,
I'm going to leave your diatribe up, but I'm not going to dignify it with a response.
Besides which, I asked you politely to stop with this bickering, and let other people read this thread in peace. This is the last time I'll ask you to do so.
sumgai (Chief Cook & Bottle Washer)
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Post by perfboardpatcher on Apr 26, 2020 8:49:39 GMT -5
So use that as the GOD FORMULA R1 = (R2/(10^(dB/20)/1))-R2 Good job, ASB! And when the formula gets too complicated you can always make a table and use vlookup(). So in your case you would get columns for the resistances (160, 1.6k etc) with incremental steps. It's a bit hard to explain, hopefully one of my spreadsheets (attenuation with a 5-way switch) will clarify. Just look at the yellow cells. In my spreadsheet it was too complicated to calculate resistance R_dbov (D8:D10) for Vuit (=-6, -12, -18dB! -> A8:A10) Therefore I've created a Table -> A14:E64 Rotat% in steps of 2%, R_dond goes in steps of 2% from 0 to 220000 (R_deler ->C3) 5th column, R_dbov, is what I'm after. D10 contains the formula: =VLOOKUP(ROUND((A10+0.005),3),$D$14:$E$64,2) (0.005 is to make the rounding off work correctly) 0.125 (A10) is searched for in first column of D14:E64 and the value from second column (,2 from formula) is printed. cheers, PP
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