Does a log pot's resistance drop to zero faster than a linear pot? And if you put a 1M resistor across a 1M linear pot you get a kind-of log 500K pot, but in reverse (ohms drop slower)? I did some quick math on this, just wanted confirmation. I want my tone control to go from treble to bass (hi to lo ohms) quicker than the 1M linear pot I'm using. Thanks! And PS I don't mind if its 500K, just want quicker response in the right direction...
Thanks, what a gold mine of info. Hadn't thought about a fixed resistor on the wiper.....And the graphs tell you what to expect. Taper pot here I come!
Its actually not possible to add resistors to make a linear pot into a log one. But you can do better adding a resistor to a log pot to reduce its taper nearer towards linear. In guitars, this is sometimes done as a treble bleed circuit (including a cap). The resistor (about 150k on a 500k log volume pot) goes from hot outer to centre lug, and will reduce the taper from about 10% to 30% at mid turn.
Same here.
In fact, you can make a linear pot behave exactly like a log one .... on "5" and on "5" only 🙁
As useful as a stopped clock which gives you the exact time twice a day. 😉
Hi,
Some serious nonsense being talked about linear pots and log faking.
ESP - A Better Volume Control
It has the advantage in stereo of far better channel gain matching.
rgds, sreten.
Some serious nonsense being talked about linear pots and log faking.
ESP - A Better Volume Control
It has the advantage in stereo of far better channel gain matching.
rgds, sreten.
Yes agreed my mistake. A linear pot with a resistor to pull it down makes a good log taper, so long as it is in an active circuit that can deal with low impedance from the pot when at max. I was thinking more of passive guitar circuits which suffer from loading.
I was wondering what the application was for this pot. Is it for a tone pot in a guitar or in an amplified circuit?
Either one, guitar or amp. You are correct saying the impedance of the pot is important, and since I need 1 megohm in my amp tone circuit, I am just going to buy a log taper pot....
As you've got to buy a pot, and it's only mono, then it makes sense to buy a log one 😀
If nothing else, it saves the cost of a resistor.
Oh yes it does make a BIG difference.
A 1M Log pot shows:
a) 20dB attenuation when on "5" (while a linear one would show just 6dB attenuation, way less range)
b) at any pot position it shows 1M load to the guitar pickup. (good).
a 1M Lin pot, with a 60k resistor wired from wiper to ground shows:
c) ~20dB attenuation when on "5" , which in principle "looks" good, semi-Techs and Amateurs become happy and stop the analysis there.
Pot on 5 now shows 550k load to pickup, , let's say it's acceptable ... sort of, although in a tone control it's shifting crossover frequency higher by one octave, not a minor matter. 🙄
d) on 10 it loads the pickup, which was expecting 1M, with a mere ~50k , 20X lower than expected.
Pickup output is now down by at least 6 to 10dB and you lost all treble.😱
Not a minor thing either. 🙄
In a nutshell: use a Log pot where it's needed.
do they really use the main gain pot right at the front?
seems a bad idea from the get go. but I reckon most use a active pedal before that so your points about source loading are moot.
loading pickups is another area the OP hasn't really asked about. specific to the app
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Perhaps you should try READING my post 😀
Everything you posted had no bearing at all on what I posted.
Dear Nigel, I not only read your post, I also quoted it. 🙂
it read:
What am I missing? 😕
As of:
Yup, as up front as can be: straight inside the guitar:
An externally hosted image should be here but it was not working when we last tested it.
fully agree, it's 40's , 50's Technology.
Try making Musicians change that 🙄
the guitar volume pot is before any active pedal you can throw in the path, so rather than moot, it's the key of the problem.
did you read the OP post?
he asked about his guitar tone control, which does load the pickup.
So much so that some guitars now use a specially made "no load" 😱 potentiometer, with a split track which opens when on 10, precisely to remove its loading from the pickup.
Results are quite audible, contrary to many weird ideas posted around here.
No placebo effect here 😛
If our friend Nigel allows me, I'll analyze what a so called "Log converted" pot would do in the OP situation:
the classic Guitar tone control works by loading the high impedance, mostly inductive pickup with a large value cap, to short Highs to ground.
Typical value : ".05 mfd" (written in Oldspeak 😀 ) or .047uF/47n today.
Treble cut is brutal, so it's tamed at will by adding a series resistor, in this case pot wiper to ground .
So (sorry for introducing Math 🙁 ) let's see what both pots would do, the real one and the fake.
Real one: 500k Log
Fake: 500k Lin with, say, 50k from wiper to ground (10:1 resistance to approximate a classic Log taper pot)
1) 500k Log:
a) on 0: 0 ohms (duh!), cap shorts highs in full.
b) on 5: 50k ... treble cut is noticeable but neither full not minimum, a reasonable pot response as heard by the Musician.
Not a surprise that 500k Log a popular value, used by millions.
The other popular value, 250k Log, works in a similar way, is preferred for single coil pickups which have roughly half the impedance of Humbuckers.
c) on 10: 500k in series with the cap ... it still loads the pickup somewhat but is accepted by most players.
For those who don't, the no load pot was invented (it just opens circuit on 10).
2) Fake: 500k Lin + 50k resistor across wiper to ground
d) on 0: 0 ohms (duh!), cap shorts highs in full. As before.
e) on 5: 50k ... treble cut is noticeable but neither full not minimum, a reasonable pot response as heard by the Musician.
Hey!!!! looks good !!!! (says the guy who didn't want to wait to get a true Log pot)
f) on 10: STILL 50k in series with the cap, as strong a treble cut as if it were on 5, in fact no practical change from 5 to 10 ... pissed off Musician now thinks:
* WTF? .... is my tone pot broken? 😕
* WTF? ... why is this guitar so dull?
In a nutshell: get the proper 500k tone pot, even if it's special order (many shops don't carry Log pots 😡 )
As a side note: I suggest rereading Rod Elliotts's page again, he suggests tweaking pots no doubt, but he clearly thinks it a second rate kludge, that's why he subjects it to important "if" clauses:
You can work around it by double buffering each pot section, (before and after) , in any case more than "just adding a resistor" .
As a side note, not that it gives me any more credentials about the subject but simply a relevant anecdote: I have actually made pots, with my own hands (to be more precise, running the machines) at a pot Factory (Seiwa in Buenos Aires) .
I was friends with the owner and he let me make my own weird tracks.
Reverse Log? 10% 30% Log? S taper? (great for EQs) Different value tracks? ... no problem 😉
Only slight inconvenient was that the minimum run on any value was 500 tracks each, and in some values I didnt use that many in, say, a year, but I made the tracks anyway and then assembled the completed pots when needed.
he never said guitar in the OP? infact in post 12 when answered to a clarification he indicated more as a "general rule". so pardon me
Good post JMF
In the guitar tone circuit, where we just use the pot with two of its lugs, as a variable resistor, we just cant make a lin pot any way nearer to log.by adding a resistor. As you demonstrated, the net resistance at half turn becomes much more than half of the net resistance at full turn. This is the opposite of what is wanted in a guitar tone control where we want a less than half value at mid settings.
For a volume pot, I think the idea of a linear pot with pull-down resistor per sreten's link, is good to know about particularly for use in active circuits. Actually, I think I have that in my mixer.
I was thinking about this concept for a guitar volume control. To keep a clear sound without excessively loading the pickups (assuming passive), the lowest overall volume pot resistance that we usually want is 250k. This could be approximated with a 1M linear pot with 330k resistor from centre lug to ground. This will give close to 250k at max. At half way, we have an overall resistance of 700k comprising 500k (half the pot) and about 200k (500k and 330k in parallel) that will provide a voltage divider of 28% which might be a nice overall taper. I think that may be worth considering further, possibly with a very tiny treble bleed cap.
I had some success 'stretching' pots by strategically scraping along the edges of the track. It can be very precise because yoy can clip a meter to the track and monitor resistance as you work. It may be possible to take a linear pot and work on it to add 100k or two to the upper part of its track resistance to make a slightly steeper initial taper than linear.
In the guitar tone circuit, where we just use the pot with two of its lugs, as a variable resistor, we just cant make a lin pot any way nearer to log.by adding a resistor. As you demonstrated, the net resistance at half turn becomes much more than half of the net resistance at full turn. This is the opposite of what is wanted in a guitar tone control where we want a less than half value at mid settings.
For a volume pot, I think the idea of a linear pot with pull-down resistor per sreten's link, is good to know about particularly for use in active circuits. Actually, I think I have that in my mixer.
I was thinking about this concept for a guitar volume control. To keep a clear sound without excessively loading the pickups (assuming passive), the lowest overall volume pot resistance that we usually want is 250k. This could be approximated with a 1M linear pot with 330k resistor from centre lug to ground. This will give close to 250k at max. At half way, we have an overall resistance of 700k comprising 500k (half the pot) and about 200k (500k and 330k in parallel) that will provide a voltage divider of 28% which might be a nice overall taper. I think that may be worth considering further, possibly with a very tiny treble bleed cap.
I had some success 'stretching' pots by strategically scraping along the edges of the track. It can be very precise because yoy can clip a meter to the track and monitor resistance as you work. It may be possible to take a linear pot and work on it to add 100k or two to the upper part of its track resistance to make a slightly steeper initial taper than linear.
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I am sorry 🙁 , I'm into Music Electronics and visit lots of such Forums, I tend to forget that this is DIYAudio and although this is the MI section, the regular Forum member usually sees things from another perspective. 🙂
They say that he who has a hammer,tends to see problems as nails.
Ogh well .
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