I have a 250k pot, and I want it to work as a 100k.
Assuming (R1 x R2)/(R1 + R2) is the new resistance, I need ~190k ohm resistor to get this?
And it would go across lugs 3 and 2, no?
Thanks!
I believe you place it on the outside lugs...
--Brad
To parallel a resistor, it goes across the two outside legs, in other words 1 and 3. If you put it from wiper to an outside leg, you are making it a tapering resistor, changing the taper of the pot.
Al
Quote from: gutsofgold on March 28, 2008, 08:36:36 PM
I have a 250k pot, and I want it to work as a 100k.
Assuming (R1 x R2)/(R1 + R2) is the new resistance, I need ~190k ohm resistor to get this?
And it would go across lugs 3 and 2, no?
Thanks!
You can accomplish what you're after with tapering resistors as well. Check out http://www.diystompboxes.com/analogalchemy/emh/emh.html. There are calculators there for creating custom value linear and log pots with larger-value pots and resistors.
You didn't say if you were trying to create a linear or log taper pot, but if you want to make a 100K linear pot from a 250K linear pot, you'd need to put an 82K resistor between lugs 1+2, as well as lugs 2+3. This will give you a 99K linear pot. (To get a 100K linear pot, you'd need to use 83K resistors, which I could not locate in a quick search.)
For a 100K log (audio) pot with a 90% taper, you'd need a 250K linear pot. Solder a 320K resistor between lugs 1+2 (or 2+3, your pick), then solder a 10K resistor between the other lugs. If you want a different taper percentage, visit the aforementioned page, pick your taper and it'll tell you what you need to use to get the resistance and taper you're looking for.
Good luck!
So many variables :icon_eek:
I need to turn this pot into a 100k, http://www.allparts.com/store/ep4485-000,Product.asp
I'm assuming it is a linear taper pot so I might as well keep it linear. Thanks!
Quote from: gutsofgold on March 28, 2008, 11:49:11 PM
So many variables :icon_eek:
I need to turn this pot into a 100k, http://www.allparts.com/store/ep4485-000,Product.asp
I'm assuming it is a linear taper pot so I might as well keep it linear. Thanks!
I don't know if you need both sections of that pot to be 100k, but Allparts also has a Dano concentric pot that has one 100k section:
http://www.allparts.com/store/electronics-miscellaneous-pots-ep-4868-000,Product.asp (http://www.allparts.com/store/electronics-miscellaneous-pots-ep-4868-000,Product.asp)
Al
Yup, both need to be 100k (one is the tone pot for a Big muff, the other is a blend control). That 1Meg/100k would make things easier, just need a resistor on the 1Meg than. Thanks !
Ok so using this calculator here...
http://www.diystompboxes.com/analogalchemy/emh/emh.html
(http://www.diystompboxes.com/analogalchemy/emh/custompot.gif)
I am told I will need two 83k Ohm resistors. Where do I place them tho?
Quote from: gutsofgold on April 01, 2008, 11:47:21 PM
Ok so using this calculator here...
http://www.diystompboxes.com/analogalchemy/emh/emh.html
(http://www.diystompboxes.com/analogalchemy/emh/custompot.gif)
I am told I will need two 83k Ohm resistors. Where do I place them tho?
Just like the diagram, the arrow is the wiper (middle lug), the dots at each end are the other two lugs, so the resistors each go from the wiper to the outside lug. In other words, one goes from lug 1 to lug 2, the other goes from lug 2 to lug 3.
Al
Let me see if I can get this right.
(All of the diagrams above are shown as voltage dividers as in a volume pot.)
(http://www.mrdwab.com/john/POT-TAPER.jpg)
(Fig A) With a linear pot and both fixed resistors the same value, the pot will be linear correct?
Won't fig D be linear also, starting with a linear pot?
With the same set up as Fig A the fixed resistors can be altered to make linear, log or anti log correct?
The fixed resistors would alter the curve of the taper and also allow you to make an S taper pot.
Once I get all this straight I need to make a diagram of all the mutations...sheesh!
John
Not to steal the thread but what does the single taper (as a percentage) value mean in: Tapered Potentiometer, http://www.diystompboxes.com/analogalchemy/emh/emh.html
In http://www.geofex.com/article_folders/potsecrets/potscret.htm is stated that the taper op a pot has a "curve" of the "fraction of total rotation" against "resistance division ratio". This I understand. So can the whole curve of the linear pot with resistors be calculated in: Tapered Potentiometer, http://www.diystompboxes.com/analogalchemy/emh/emh.html
The percentage of the taper is the slope of the pot taper/curve I beleive.
I assume 90% is an extreme log taper and 0% is linear...?
I'm not sure how the percentage relates to Log vs reverse log though.
In anyone can confirm or deny my grasp of the taper diagams above I'd appriciate it.
Thanks
John
"please allow me to bump thee"
anyone guess that lyrical quote?
too much information here for somethin that simple.. just put a 190k resistor between outer lugs ..the taper will remain the same...
Quote from: John Lyons on April 03, 2008, 07:39:51 PM
"please allow me to bump thee"
anyone guess that lyrical quote?
That's from 'The Humpty Dance' by Digital Underground. Do I win some of puretube's free chips for answering first? :)
As for the taper percentage question, I think 50% taper is linear, < 50% is reverse-log and > 50% is log. Somebody please correct me if I'm incorrect, but that's how I grokked it.
So is this taper percentage value, the percentage of the total taper width when the pot is set halfway or something?
So for a log pot this is 10% and for a linear pot this is 50% and for a reverse pot 90%?
Beware weary internet traveler.
There are some rather imaginative theories about parallel resistance here.
No, putting equal values resistors across pins 1+2 and 2+3 does not magically create a smaller value pot with a linear curve. Why wouldn't anybody actually try it to find out (or just think about how parralel resistance works when one resistance is constantly changing) before making a broken "calculator" for this?
For example, this magical calculator tells me that a 50K linear pot can be turned into a 25K linear pot by putting 25K resistors across pins 1+2 and 2+3. LOL. No. That creates a pot with 25K resistance at *center* position, and about 12.5K at the outer pins. Perfect!
welcome to the forum
that's correct, there is some sketchy maths going on here.
cheers
Well, at extremes the pot shorts out one resistor. I make that 17k.
What d'yer think of this?
https://www.diystompboxes.com/brisance/old/?R1=25000&R2=25000&VR=50000
OK, so many pot - resistor in parallel topics here.
Aka, changing 10K pot into an Approx 1K pot for instance.
So I connected a 10k pot (1.20R to 9.52K per DMM) and a 978R to my BB.
(978R in // with 9.52K = 887R)
I put the 978R resistor across 1 and 3.
Measuring across 3 and 2, I get:
1.20R going Up constantly to 2.7K.
Then Drops constantly to 890R.
So this isn't going to be a 10K pot turned into an Approx 1K pot using this method.
Lets try two near equal 978R resistors across 1-2 and 2-3.
Measuring across 3 and 2 :
Goes Up near constantly from 1.20R to 890R.
This looks better.
Now I will try two ~1.2K resistors to get closer to 1K......
Moral of the story? BB it and measure.
As can be seen from using the spreadsheet here, the use of any resistors across the terminals of a linear potentiometer will change the taper to a non-linear one.
http://www.fleetingspider.com/files/CustomPots.xlsx
BTW, this is not my spreadsheet and I don't actually know who created it, I just have it here to allow people to access it.
Here is the resultant taper you will get (in red) with two 1.2k resistors and a 10k potentiometer.
(https://i.postimg.cc/4mJ1Lsvc/Modified-10k-pot.png) (https://postimg.cc/4mJ1Lsvc)
I'm of the opinion that ain't nothin' like the real thing - but if you can't wait, or just can't get one, then trying this might work ok, depending on what you're building. Or try one close in value instead.
I would rather use, say, a 50k volume pot in a Fuzzrite than TRYING to make a "33k" that was only available in the 60s (AFAIK). Same goes for using a 500k (instead of 350k) for the dirt pot. Every time I've done the 'trick', the control feels odd - evidenced by the curves above.
Unfortunately my 10K hack to make a 1K odd ball is necessary.
Since the 1K pot is not available in the form I need for my next build.
An Axis Face with Attack and Smooth control.
(Built and tested but not boxed)
Fortunately I built the same effect a few years ago with a 1K.
(It used a "Standard" 16mm pot)
The hack works as well.
Thanks for the comments and graph.
The graph shows what to listen for in its rotation.
Posted comments on this Attack control mention that it really "kicks" in at 1 to 3.
That is what I hear in both cases.
Quote from: FSFX on May 18, 2023, 12:32:02 PM
Here is the resultant taper you will get (in red) with two 1.2k resistors and a 10k potentiometer.
(https://i.postimg.cc/4mJ1Lsvc/Modified-10k-pot.png) (https://postimg.cc/4mJ1Lsvc)
Now that would be quite handy to avoid needing the stupid pot taper used in the TubeScreamer tone control. Can't think of of much else to do with it, but if it fixes the poor control response of gyrator tone controls, that's actually a pretty big plus. Compare with this:
(https://www.electrosmash.com/images/tech/tube-screamer/g-potentiometer.png)
(From https://www.electrosmash.com/tube-screamer-analysis (https://www.electrosmash.com/tube-screamer-analysis))
Please note I'm not claiming this as an original thought! I'm quite sure it's been done a load of times already. Only that I haven't see it or thought about it till now, so I'm playing my usual game of catch-up ;)
QuoteNow that would be quite handy to avoid needing the stupid pot taper used in the TubeScreamer tone control. Can't think of of much else to do with it, but if it fixes the poor control response of gyrator tone controls, that's actually a pretty big plus
Putting two parallel resistors across a 10k pot creates the wrong taper for a tube screamer, it has an ever larger deadband in the center. Near the ends the resistance change speeds up.
The G taper used on the Tube screamer has the inverse tape. It creates an S shaped which slows down the resistance change towards the ends. There's simple no way to get that with external resistors.
The only way to improve the adjustment of the tube screamer near the ends of adjustment with a linear pot is to use a lower pot value, but that changes the response slightly. A 5k pot is nice to use, a 10k is on the edge of improvement but affects the response less.
The main idea behind adding parallel resistors is to shift the voltage divider ratio of the pot in the center. However the taper itself (ie adjustment per degree of rotation) isn't a good match with a real pot with the correct taper.
Difficult to know the difference in tone sound / control of said tone using G20K , B25K , A20K or modified // resistor pot .
Hard knowing not witnessing Tube Screamer with these different pot setups.
What is the biggest difference?
QuoteDifficult to know the difference in tone sound / control of said tone using G20K , B25K , A20K or modified // resistor pot .
Hard knowing not witnessing Tube Screamer with these different pot setups.
What is the biggest difference
As far as tone goes:
- 25K vs 20k is a small change.
- taper changes don't affect tone.
Tapers affects usability of the control.
(For TS9 and Gyrator EQ's) With a linear pot nothing happens in the center then everything is cramped up in the last 10% of rotation. If you set everything to 12 O'Clock you won't notice anything. If you like to fine tune around 2 to 3 O'Clock on G pot the linear pot will be a pain because it will require very fine adjustment around 4 to 5 O'clock.
Quote from: Rob Strand on May 18, 2023, 05:40:49 PM
QuoteNow that would be quite handy to avoid needing the stupid pot taper used in the TubeScreamer tone control. Can't think of of much else to do with it, but if it fixes the poor control response of gyrator tone controls, that's actually a pretty big plus
Putting two parallel resistors across a 10k pot creates the wrong taper for a tube screamer, it has an ever larger deadband in the center. Near the ends the resistance change speeds up.
The G taper used on the Tube screamer has the inverse tape. It creates an S shaped which slows down the resistance change towards the ends. There's simple no way to get that with external resistors.
Ok, so that diagram on Electrosmash is misleading? It should be the other way round? More S, not Z?
Quote from: ElectricDruid on May 18, 2023, 07:50:27 PM
Ok, so that diagram on Electrosmash is misleading? It should be the other way round? More S, not Z?
I see the bigger problem now. I actually thought the plot was for a funky taper pot but marked G. It just looks wrong.
Should be like this, (just change word slider to % pot rotation)
(https://i.stack.imgur.com/4DmFg.png)
The wiper to end resistance is a low fraction of the total but changes slowly.
Quote from: gutsofgold on April 01, 2008, 11:47:21 PM
Ok so using this calculator here...
http://www.diystompboxes.com/analogalchemy/emh/emh.html
(http://www.diystompboxes.com/analogalchemy/emh/custompot.gif)
I am told I will need two 83k Ohm resistors. Where do I place them tho?
It will be good for the task at hand.
^ According to my BB test, to make 10k into 1K you need two 1.2K .
Lugs 1 to 2 and 2 to 3.
With those Pot measures 1.2R to 1039.
That calculator tells me to use two 555R ?
I trust my DMM.
Quote from: Phend on May 19, 2023, 07:41:53 AM
^ According to my BB test, to make 10k into 1K you need two 1.2K .
Lugs 1 to 2 and 2 to 3.
With those Pot measures 1.2R to 1039.
That calculator tells me to use two 555R ?
If you were to have used the spreadsheet that I posted a link to then you would have got more accurate results.
Using a 10k pot and values of -0.92 for the A/R factor and B/R factor produces resistor values of 1.202k for A and B and a resultant resistance across the pot of 1.073k.
(https://i.postimg.cc/PLM84XGF/pot-view.png) (https://postimg.cc/PLM84XGF)
Unfortunately in my current electronics bag there is no accuracy.
I usually if not always measure the component and try to find the one closest to what is specified.
Mostly never find an "accurate" one.
Hence these guitar effect circuits are like "rubber".
To a degree, values can deviate quit a bit from the specification.
Which can often make an even better circuit than the original.
Because everything is out of spec and your lucky.
When I first started making simple effects, I mentioned that I "Blue Printed" the circuit I was building.
At that time I had a stash of stuff so as I could measure and find close or "~exact" values.
I heard the word "Blue Printed" when I was a teen and some guy had a muscle car that was BPed.
Meaning that the parts in the engine were all selected to be dead nuts tolerance.
Of course there are times when transistors need to be matched...etc..
So in this effect business 1K can be 950 or 1090, and still work, just don't turn the knob as far.!!
I have never seen any tolerance call out on an effect circuit except the Phase 45.
But nothing wrong with all that, effects sound good built with what they are built with. !
Quote from: Phend on May 18, 2023, 03:38:38 PMAn Axis Face with Attack and Smooth control.
Posted comments on this Attack control mention that it really "kicks" in at 1 to 3.
Need a pot lesson (again)
So the B1K Attack Control "kicks" in at 1 to 3 o'clock.
Would changing this to a C1K pot solve this and make it more "useful" (or work better thru its range) ?
(https://i.postimg.cc/7bD7p2Rp/axisfacesi3schematic.png) (https://postimg.cc/7bD7p2Rp)
I don't know my ABCs.
The emitter is like 30 Ohms. So 100 ohms in the pot is a significant drop from max. Taking habitual CW=more, you want a Reverse Audio, or even more extreme.
Or wire an Audio backward. You will learn. People who "pick up" pedals won't like it and may leave it behind some day.
Is it possible to make a C taper pot from a B taper? :icon_rolleyes:
I found an answer...
The following is from General Guitar Gadgets in their Fuzz Face instructions:
"Our kits include a 1k C (reverse log taper) for the Fuzz control instead of the
1k B that the originals have. It is a known fact that the reverse log taper gives
a much better rotation for the fuzz control. We wouldn't have made this
upgrade if we weren't absolutely sure this is a worthwhile improvement. You
can use a 1k B and it will work fine, but we like the 1k C better. We have left
the schematics and diagrams to show a 1k B since that is the original value."
(https://i.postimg.cc/cKXMxQyh/ggg-ff5-sc-pnp-231010-071952.jpg) (https://postimg.cc/cKXMxQyh)
Quote from: Baran Ismen on October 10, 2023, 01:20:31 AMIs it possible to make a C taper pot from a B taper? :icon_rolleyes:
As possible as is to make a A taper from a B taper.. :icon_wink:
The Secret Life of Pots (http://geofex.com/Article_Folders/potsecrets/potscret.htm)
Am I too stupid to comprehend this matter or my pots are just weird?
https://www.elby-designs.com/webtek/documents/tailoringpotentionometers.pdf
According to this PDF file's 5th Figure, I need to solder a 100k resistor between 2 active lugs of my 1m linear pot and it becomes logarithmic, right?
Mine becomes however something completely different, it reads around 80k or something. For Rev.Log, what shall I do? Add a resistor of 10m?
Thing is, the circuit I deal with right now is Small Stone and there are only 2 lugs connected. 3rd one is empty, or shorted to 2nd.
Quote from: Baran Ismen on October 10, 2023, 01:12:50 PMAm I too stupid to comprehend this matter or my pots are just weird?
https://www.elby-designs.com/webtek/documents/tailoringpotentionometers.pdf
According to this PDF file's 5th Figure, I need to solder a 100k resistor between 2 active lugs of my 1m linear pot and it becomes logarithmic, right?
Mine becomes however something completely different, it reads around 80k or something. For Rev.Log, what shall I do? Add a resistor of 10m?
Thing is, the circuit I deal with right now is Small Stone and there are only 2 lugs connected. 3rd one is empty, or shorted to 2nd.
May be (you're dumby,a little). ;D
It is not clear what value of the pot you want to get.
PDF link is not work.
Post the circuit with the pot marked, if it's not too much trouble.
Quote from: Dormammu on October 10, 2023, 01:47:39 PMQuote from: Baran Ismen on October 10, 2023, 01:12:50 PMAm I too stupid to comprehend this matter or my pots are just weird?
https://www.elby-designs.com/webtek/documents/tailoringpotentionometers.pdf
According to this PDF file's 5th Figure, I need to solder a 100k resistor between 2 active lugs of my 1m linear pot and it becomes logarithmic, right?
Mine becomes however something completely different, it reads around 80k or something. For Rev.Log, what shall I do? Add a resistor of 10m?
Thing is, the circuit I deal with right now is Small Stone and there are only 2 lugs connected. 3rd one is empty, or shorted to 2nd.
May be (you're dumby,a little). ;D
It is not clear what value of the pot you want to get.
PDF link is not work.
Post the circuit with the pot marked, if it's not too much trouble.
Thanks for the compliment, mate :)
http://web.archive.org/web/20210203023928/https://www.elby-designs.com/webtek/documents/tailoringpotentionometers.pdf
I need a C1M pot and got B1M pot at hand that needs to be converted into C.
It's irrelevant as it won't work in the rheostat (variable resistor, 2 lugs connected) application.
Quote from: FiveseveN on October 10, 2023, 02:47:58 PMIt's irrelevant as it won't work in the rheostat (variable resistor, 2 lugs connected) application.
I've just noticed that after reading the secret life of pots couple of times.
Time to order it from Tayda, then.
Quote from: Baran Ismen on October 10, 2023, 01:20:31 AMIs it possible to make a C taper pot from a B taper? :icon_rolleyes:
Not a real potentiometer (3 terminals), but you can make a C taper variable resistor (2 terminals) from a linear pot or variable resistor. See Antonis' answer above.
Quote from: GGBB on October 10, 2023, 04:24:09 PMQuote from: Baran Ismen on October 10, 2023, 01:20:31 AMIs it possible to make a C taper pot from a B taper? :icon_rolleyes:
Not a real potentiometer (3 terminals), but you can make a C taper variable resistor (2 terminals) from a linear pot or variable resistor. See Antonis' answer above.
Im truly lost here.
In theory, i need to connect a b2m pots 1st and 2nd lugs with a 400k resistor and use these lugs for board connection, right? 3rd one is not connected anywhere. So that ill get a c1m pot curve?
Do a breadboard. Wire up your pot and connect to breadboard. Select resistors and insert.
Use your meter on resistance and watch how the pot functions as you turn it.
Functions, meaning how, and when you see how ohms change.
I can't help thinking that it's a lot easier to buy the correct value pot with the correct taper in the first place - runs for cover!!
See the September 1988 copy of Home & Studio Recording which has the circuit of a Dual Fader/Gate/Panner, and look at the current mirror section so described in the article!!
Quote from: StephenGiles on October 10, 2023, 05:30:51 PM...easier to buy the correct value pot with the correct taper in the first place
I believe Baran Ismen is in Turkey/Türkiye. Now, there are "no" electronics dealers in my part of the USA. But even mail-order may not be robust in much of the world, including I suppose Turkey. I see that iPhones get hit with heavy taxes. The Ministry of Trade says e-trade is brisk, and I believe it may be, relative to the past. Stuff I read online suggests that our Paypal/VISA/AmEx card payments don't work in Turkey; you call the seller, haggle the price, and wire the money to his account.
Quote from: PRR on October 10, 2023, 10:39:11 PMQuote from: StephenGiles on October 10, 2023, 05:30:51 PM...easier to buy the correct value pot with the correct taper in the first place
I believe Baran Ismen is in Turkey/Türkiye. Now, there are "no" electronics dealers in my part of the USA. But even mail-order may not be robust in much of the world, including I suppose Turkey. I see that iPhones get hit with heavy taxes. The Ministry of Trade says e-trade is brisk, and I believe it may be, relative to the past. Stuff I read online suggests that our Paypal/VISA/AmEx cgard payments don't work in Turkey; you call the seller, haggle the price, and wire the money to his account.
It's Turkey, not North Korea. :icon_lol: Indeed we have some problems but not that bad, for now at least... PayPal has been blocked for a long time, but for the rest, there's no problem, only thing is international trade for such small items is costly..
We have plenty of suppliers all around the country yet C taper is simply not available -let alone the A tapers-. The last chance is Tayda. I'll pay %20 import tax over the total price inc. the shipping, so 0.50$ worth of C1M pot will cost me about 10 bucks :)
But seems like I have no other choice, right?
Quote from: Baran Ismen on October 10, 2023, 02:02:43 PMThanks for the compliment, mate :)
http://web.archive.org/web/20210203023928/https:/ (http://web.archive.org/web/20210203023928/https:/)/www.elby-designs.com/webtek/documents/tailoringpotentionometers.pdf
I need a C1M pot and got B1M pot at hand that needs to be converted into C.
The link still doesn't work.
But based on the data presented — having a 1M pot cannot be fully converted into a different curvature in the described way.
You need to decide on an acceptable resistance value or the functionality of the regulator — and act on this basis.
Quote from: Dormammu on October 11, 2023, 02:31:47 AMQuote from: Baran Ismen on October 10, 2023, 02:02:43 PMThanks for the compliment, mate :)
http://web.archive.org/web/20210203023928/https:/ (http://web.archive.org/web/20210203023928/https:/)/www.elby-designs.com/webtek/documents/tailoringpotentionometers.pdf
I need a C1M pot and got B1M pot at hand that needs to be converted into C.
The link still doesn't work.
But based on the data presented — having a 1M pot cannot be fully converted into a different curvature in the described way.
You need to decide on an acceptable resistance value or the functionality of the regulator — and act on this basis.
Try this? (http://web.archive.org/web/20210203023928/https://www.elby-designs.com/webtek/documents/tailoringpotentionometers.pdf) It's an old & dead link, can only be reachable via Wayback Machine.
I also have B2M in stock, I can use it if the total resistance will be decreased.
Quote from: Baran Ismen on October 11, 2023, 03:35:58 AMI also have B2M in stock, I can use it if the total resistance will be decreased.
Oh damn, what a misunderstanding.
Having googled the Small Stone diagram, I found 1 pot (1М) connected according to the rheostat circuit — there is no reason to use a pot with a certain curvature there. Perhaps you should just reduce its value using any available method, including simply shunting it.
Quote from: Dormammu on October 11, 2023, 05:45:31 AMQuote from: Baran Ismen on October 11, 2023, 03:35:58 AMI also have B2M in stock, I can use it if the total resistance will be decreased.
Oh damn, what a misunderstanding.
Having googled the Small Stone diagram, I found 1 pot (1М) connected according to the rheostat circuit — there is no reason to use a pot with a certain curvature there. Perhaps you should just reduce its value using any available method, including simply shunting it.
There's an imbalance between the rate pot value and LFO cap.
If I decrease the pot value, I need to increase the LFO cap value (which is 22uf at the moment) and vice versa.
After trying a couple of options, I've seen that 1m / 22uf sounds good on both the slowest and fastest settings. The problem is the range and accumulation in between. At the moment, the first quarter (maybe more) of the B1M pot doesn't feel to react precisely and increase the rate proportionally.
Quote from: Baran Ismen on October 10, 2023, 02:02:43 PMThanks for the compliment, mate :)
I need a C1M pot and got B1M pot at hand that needs to be converted into C.
Baran - here is a question for you, based on the information you have provided. if you have a 1M resistor and you want to put a parallel resistor across it, what is the value of the added resistor such that the end result resistance is 1M?
or - 1M // Rx = 1M , Rx = ??
Quote from: duck_arse on October 11, 2023, 09:48:40 AMQuote from: Baran Ismen on October 10, 2023, 02:02:43 PMThanks for the compliment, mate :)
I need a C1M pot and got B1M pot at hand that needs to be converted into C.
Baran - here is a question for you, based on the information you have provided. if you have a 1M resistor and you want to put a parallel resistor across it, what is the value of the added resistor such that the end result resistance is 1M?
or - 1M // Rx = 1M , Rx = ??
It's 1 ohm ? ::)
Okay, I got B2M in stock as well :D
:icon_biggrin: :icon_biggrin: :icon_biggrin: :icon_biggrin:
(sorry but couldn't restrain myself) :icon_redface:
Beni yanlış anlama dostum..
@Baran: ANY resistance set in parallel with another one dominates the value of the later.. :icon_wink:
Quote from: antonis on October 11, 2023, 01:55:36 PMANY resistance set in parallel with another one dominates the value of the later.. :icon_wink:
Not ANY resistance will dominate, but the total resistance will be less than the least resistance.
Quote from: antonis on October 11, 2023, 01:55:36 PM:icon_biggrin: :icon_biggrin: :icon_biggrin: :icon_biggrin:
(sorry but couldn't restrain myself) :icon_redface:
Beni yanlış anlama dostum..
@Baran: ANY resistance set in parallel with another one dominates the value of the later.. :icon_wink:
Thats Ok Antonis, i dont.
I am not as pro as you all people when it comes to theorical information, sorry.
I really just need to know if i can make a rev log conversion with the linear pots i have and be able to use as rate pot or not, otherwise ill put an order to tayda :-[
Quote from: Dormammu on October 11, 2023, 02:31:47 AMThe link still doesn't work.
The Archive[.]org link, "with two URLs", confuses this forum's auto-parser.
Manually linked:
http://web.archive.org/web/20210203023928/https://www.elby-designs.com/webtek/documents/tailoringpotentionometers.pdf (http://web.archive.org/web/20210203023928/https://www.elby-designs.com/webtek/documents/tailoringpotentionometers.pdf)
In some browsers, you can highlight the
whole link, right-click, and "Open in new tab".
Quote from: PRR on October 11, 2023, 09:52:33 PMIn some browsers, you can highlight the whole link, right-click, and "Open in new tab".
Nevermind.
I was inattentive and thought that this was a link to the Small Stone diagram. ;D
Take your B2M pot and put a 500K resistor across 2 and 1 or 3. Now connect your dmm to 2 and 1, turn pot and see what you get.
Quote from: Phend on October 12, 2023, 07:57:01 AMTake your B2M pot and put a 500K resistor across 2 and 1 or 3. Now connect your dmm to 2 and 1, turn pot and see what you get.
It's the same response as in B1M, no rev.log response whatsoever. Rate response is like a deep logarithmic one, rate starts to decrease way too slowly, after midway its getting higher a bit and in the near end it goes up rapidly. Not a balanced distribution I can say..
Well, it's all a mechanical thing, B1M = C1M in resistance. Either will work.
A larger diameter knob will help reduce the sense of logs. Lol