replacing log pot with stepped switch

[briomusic]

I can't help feeling this has been answered before, but I am not getting anywhere with searching the forum or indeed the internet.
I am trying to calculate the steps needed to replace a 100k log pot with a stepped switch (24 steps). Every spreadsheet or java calculator that I find, asks me how much attenuation in dB I would like to achieve. Like this one:
http://www.quadesl.com/attenuator.html
But I don't know! Maybe my log pot isn't even used as an attenuator! I just want roughly the same distribution of resistance values that a pot provides from fully CCW to fully CW spread out over the 24 steps of my switch. Can anyone point me to an online calculator or spreadsheet?
thanks!

 

[pucho812]

check out goldpoint switches.


http://www.goldpt.com/diy.html

 

[mitsos]

This may not be 100% accurate, but it looks like it might be close enough.

http://gaussmarkov.net/wordpress/parts/resistors/resistors-5-potentiometers/#moreinfo

down the page he gives this formula:
VOUT/VIN = xlog 10/log 2 ≈ x^3.3

X is the % of rotation, so if you use a 24 position switch, divide 100% by 24 (unless you mean 23 steps): 24 positions = 4.17% per click (hint, use 0.0417), but you should do 23 steps if you want a zero in the first click, no? 

Anyway once you make a list of the percent rotation per click, you take each to the power of 3.3 and that will be the percent of resistance per position, so just multiply each by the total, in your case 100K and you'll have the restitance at each position.

run it in Excel and see what you get. This is what I got for 24 positions, 100K. Obviously these aren't

1 3
2 28
3 105
4 271
5 566
6 1034
7 1719
8 2671
9 3940
10 5578
11 7639
12 10180
13 13258
14 16931
15 21259
16 26305
17 32131
18 38801
19 46381
20 54935
21 64532
22 75239
23 87127
24 100264

 

[Kingston]

Your original thread still applies http://www.groupdiy.com/index.php?topic=44553

a "log pot" is not something set in stone. Most log pots don't match at all and in fact aren't even anywhere near logarithmic. They are simply close enough to get a way with calling it log or "audio". Manufacturers all seem to have their own idea about this and most cheap ones simply can't be trusted.

You can use this same dB step calculator to replace just about any log pot for audio use: http://homepages.tcp.co.uk/~nroberts/atten.html

For optimal results you should first use a standard carbon pot to find what resistance area is the most interesting range, and where you want most resolution (1dB vs. 3dB steps for example). Another helpful rule is that there is probably not a single audio application that would benefit from more than about 60dB of attenuation. Unless you absolutely require "off". That should significantly narrow down your calculator settings and to get you the most optimal rotary stepped switches.

 

[moamps]

Hi,
Some 4 years ago I wrote an article on stepped attenuators (excell file included)
which you may find helpful: http://www.moxtone.com/O_stupnjevitom_atenuatoru_2.htm

Regards,
Milan

 

[briomusic]

Hi all, and thanks for your help.
I don't know what the dB of attenuation is in my circuit. (It's the Peak Reduction of an LA2A)
But it shouldn't really matter, should it?
Like, if I go to farnell and ask for a 100k log pot, they don't ask me "how much dB attenuation", it's just a 'resistance' divider with resistance x to one side of the wiper and 100-x to the other, right?
I really don't understand how those online calculators come up with different rotary switches to replace a run-of-the-mill 100k log pot depending on the desired attenuation as it wouldn't be a consideration if a pot was used.
It think mitsos example is closest to what I am looking for, but I am worried as his values are very different to all the other online calculators.
As I am completely useless at maths, can someone just give me a formula to calculate 0-100% of 100k resistance on a logarithmic scale VERSUS 0-100% of travel from fully CCW to fully CW? (Vin, Vout or dB, shouldn't play a part in this)
Thanks!

 

[Kingston]

But it shouldn't really matter, should it?
Like, if I go to farnell and ask for a 100k log pot, they don't ask me "how much dB attenuation", it's just a 'resistance' divider with resistance x to one side of the wiper and 100-x to the other, right?

The manufacturers are only able to give vague approximations of their run of the mill log pots. Making carbon log pots is a highly inaccurate process, and even today after half a century of carbon pots almost no manufacturers care to do it better. There is no market for it. The accuracy they provide is mostly "enough".

the only difference between an ideal 10-log pot and something with dB scale is the shape of the logarithmic curve. And dB always wins for audio use. It corresponds best to our ears. This also applies to LA2A peak reduction.

I really don't understand how those online calculators come up with different rotary switches to replace a run-of-the-mill 100k log pot depending on the desired attenuation as it wouldn't be a consideration if a pot was used.

It's not a consideration with pots, no. But that's only because most of the time we simply ignore the fact the pots aren't logarithmic at all, and match all over the place. Equipment is usually designed with this in mind.

Do some googling on the topic and you'll quickly find out most carbon pots actually stick two linear carbon tracks together, perhaps three. And then call it "logarithmic" or "audio", sometimes it's not even specified. To recap, when you buy a "log pot", it's not in any way guaranteed it will even be 10-log. Any calculator churning out resistor values will be much more accurate and predictable in comparison, but a dB scale will probably handle closest to what you actually hear.

The rest of your worries sound to me more like superstition, especially since LA2A schematic/design is extremely non-critical. It was designed for something like 10-20% parts variance.

 

[mitsos]

It think mitsos example is closest to what I am looking for, but I am worried as his values are very different to all the other online calculators.

I see I left off mid-sentence  ??? in the last post, sorry, must have been drinking.    I guess what I was going to say is that obviously that isn't the resistor at each position but the total resistance you want AT that position. To get the actual values, I'm really bad at explaining things, but if R(x) is the resistance I gave you, and r(x) is the actual resistor you need and x is the switch position it's in (like X=22 would be the second to last, X=23 would be the last one, since they are soldered in between positions on the switch), r(x)=100K-R(x)-[100K-R(x+1)].  I have no idea if I wrote that properly, its been ages since any math class... so here is the list I get doing this step:

1 24.7
2 77.4
3 166.2
4 295.1
5 467.3
6 685.4
7 951.8
8 1268.7
9 1638.0
10 2061.5
11 2540.8
12 3077.5
13 3673.1
14 4328.8
15 5046.0
16 5826.1
17 6670.0
18 7579.1
19 8554.3
20 9596.8
21 10707.5
22 11887.5
23 12873.4

If you add resistors one and two and look at the first list, you'll see its roughly equal to the third resistance. Add one more and you get the fourth reistance. If you add them all up, you get 100K. And if you plot these you get a lot pot curve. Now the important thing as Kingston says is to figure out what range you really want, and divide that range up by the number of positions you have available, to maximize resolution where you need it and forget parts you don't really use.

hope that's a bit clearer.

 

[abbey road d enfer]

But it shouldn't really matter, should it?
Like, if I go to farnell and ask for a 100k log pot, they don't ask me "how much dB attenuation", it's just a 'resistance' divider with resistance x to one side of the wiper and 100-x to the other, right?

That would define a 40dB attenuator, but you wou wouldn't know what happens in between. It may have a linear taper. Resistance at mid-position is what defines the taper.
Many pot manufacturers offer different variation of log. Alpha offers as standard 6 variations of Log
http://www.alphapotentiometers.net/html/taper_curves.html
Log tapers are defined by three points: CCW = 0, CW = nominal value, and center position, as a percentage of nominal.

 

[moamps]

one example:

 

[Kingston]

Great example, already compensated for standard resistors. I've used a similar scale for preamps.

Since briomusic has a specific case of LA2A he will want more resolution in the attenuated end, for both gain and peak reduction. Basically 4dB vs. 2dB steps reversed compared to the above. The reason is that LA2A has a stupid amount of gain for a compressor; it was designed for a different era. Most of the time you are compensating in around the -50 to -30dB area for modern material. Very rarely, if at all it will sit at full blast. Never needed the 0dB setting personally, even as a preamp.

Similarly, peak reduction just about looses its effectiveness at very high settings. It'll just turn to distorted mush and full blast, not the good kind of mush. Again, something like -50 to -30dB area needs most resolution.

As an empirical example, if you try playing around using standard log pots, you will quickly find out they rarely -if ever - go past halfway in LA2A.

 

[radiance]

Somehow, all info given here is for attenuators. But what if I want a stepped gain pot in (for example) a pultec EQ?
DO the same rules apply?
All calculators I've found so far are for attenuators.

I have to say, I have made many stepped gain pots in the past with varying results. Somehow, it always came down to trial and error.
For example, measuring different LOG pots mounted on a piece of carton with a 24 position scale on it.
Then draw out a curve etc. etc. 
Maybe I should buy a high quality log pot for reference?

I want to get it right this time...cause soldering 24 steps rotary switches is not my idea of fun :-(