radiance

Making a log pot with a rotary
« on: July 14, 2006, 07:16:22 AM »
For doing this I use this website.
Still I find it very hard to get a smooth response with  a 12 step rotary switch. For example what is a sensible value to use as "Total attenuation" in this calculator? I use something like 40dB but I was wondering, does a stepped attenuator differ in any way from a stepped log pot? That is: do they use different slopes?

Another question: with a stepped log pot, the direction from step 1 to 2 should be clockwise right? Where as with a stepped attenuator the direction is reversed,  that is: when going ccw you lower the volume right?
"Knowing that you are dreaming, however, does not automatically guarantee full rationality.
Then again, being awake doesn't ensure good thinking, either." -  Lynne Levitan


radiance

Making a log pot with a rotary
« Reply #1 on: July 15, 2006, 11:32:00 AM »
Any comments?



 :cry:
"Knowing that you are dreaming, however, does not automatically guarantee full rationality.
Then again, being awake doesn't ensure good thinking, either." -  Lynne Levitan

Kit

Making a log pot with a rotary
« Reply #2 on: July 15, 2006, 12:56:12 PM »
Im not sure i understand what you mean by a stepped log pot.....
Do you mean like a "stud" type where you cant feel the clicks?
"Relaxing on the axis of the wheel of life."

synthi

Making a log pot with a rotary
« Reply #3 on: July 15, 2006, 09:13:52 PM »
Sometimes there is a kind og telepathy between diy members  :shock:

I was about to ask exactly the same question!

I haver a bunch of 12 pos. rotary switches and I want to replace some potentiometer with those. The linear ones are easy just dividing the total resistance for the original pot/ steps in the rotary switch, but is there a easy calculator for the log curve?

Synthi

spacewig

Making a log pot with a rotary
« Reply #4 on: July 15, 2006, 10:02:29 PM »
Just finished making a balanced stepped attenuator for my active monitors. If you can 'lock' your switch positions the following values will work for 9 steps:

5.1K Ohms
2.4
1.2
.510
.240
.120
.060
.030
0

radiance

Making a log pot with a rotary
« Reply #5 on: July 16, 2006, 07:18:48 AM »
Kit, what I ment was a 12 position rotary switch...




Quote from: "synthi"
...., but is there a easy calculator for the log curve?

Synthi


See top post.....It has worked quite well for me altough it took some fiddling around with the total attenuation setting and I'm still not sure which value works best.
"Knowing that you are dreaming, however, does not automatically guarantee full rationality.
Then again, being awake doesn't ensure good thinking, either." -  Lynne Levitan

synthi

Making a log pot with a rotary
« Reply #6 on: July 16, 2006, 07:23:58 AM »
hmmmm... I guess 100dB attenuation?

About the wiring, I think you could wire it as  the calculator say and then just reverse the conctions to the pcb where the pot goes so it will work in reverse... right?

Synthi

synthi

Making a log pot with a rotary
« Reply #7 on: July 16, 2006, 07:25:04 AM »
the lower value need to be 0 resistance and the max value the total resistance you want for the pot, so 100dB seems to work...

Synthi

radiance

Making a log pot with a rotary
« Reply #8 on: July 16, 2006, 11:18:25 AM »
Quote from: "synthi"
the lower value need to be 0 resistance and the max value the total resistance you want for the pot, so 100dB seems to work...

Synthi


What do you mean by this?
"Knowing that you are dreaming, however, does not automatically guarantee full rationality.
Then again, being awake doesn't ensure good thinking, either." -  Lynne Levitan

synthi

Making a log pot with a rotary
« Reply #9 on: July 16, 2006, 11:26:36 AM »
for example a 10k pot goes from 0 to 10k resistance, ok?
Now go to that page with the calculator and fill it for 10000 ohms total resistance, 12 steps and 40dB attenuation: the first steep will be 53.78 ohms, not "0"...
now try the same values but using 100dB attenuation, the first step is 0.19ohms, enough for leaving this step with a wire for 0 resistance.

well thats that my brain tell me...

Synthi


radiance

Making a log pot with a rotary
« Reply #10 on: July 16, 2006, 11:47:30 AM »
Well, I asked this before and  PRR sugested the 40dB value. I tried bigger values like 60dB but that gave me a rotary switch with very big steps. For example when using 100dB value each step is about 9 db. That's way to big for me...
"Knowing that you are dreaming, however, does not automatically guarantee full rationality.
Then again, being awake doesn't ensure good thinking, either." -  Lynne Levitan

WJS

Making a log pot with a rotary
« Reply #11 on: July 16, 2006, 03:24:48 PM »
If you have 12 steps to go down 60 db (60/12) thats about 5db per step. if you want to go all the way down to 100... (100/12) thats 8 1/3 per step.
Do you really need 100db of attenuation? unless you have VERY strong signals coming in, 40db should be quite fine. whats your application for this?
If you need more presicion and a higher db value, you might have to get a switch with more positions than 12.
"CLUELESSNESS.
There Are No Stupid Questions, But There Are A Lot Of Inquisitive Idiots."

syn

Making a log pot with a rotary
« Reply #12 on: July 16, 2006, 04:31:28 PM »
Quote from: "synthi"

About the wiring, I think you could wire it as  the calculator say and then just reverse the conctions to the pcb where the pot goes so it will work in reverse... right?

Synthi


Sorry to hijack.

Does the same logic applies for getting the a rev-log pot out of a log pot?

Thanks :thumb:

synthi

Making a log pot with a rotary
« Reply #13 on: July 16, 2006, 06:26:15 PM »
Quote
Do you really need 100db of attenuation? unless you have VERY strong signals coming in, 40db should be quite fine. whats your application for this?


Thats the question, I don`t need it for attenuating signals, but as replacement for a normal log potentiomter in a circuit! So I`m sure I need 0 resistance at one end and max resistance at the other...

About the reverse log pot, No I think you`ll need a different aproach since the slope is reversed (ie. smaller increments at the final stages).

Synthi

Freq Band

Making a log pot with a rotary
« Reply #14 on: July 16, 2006, 10:08:43 PM »
Is it possible to:

-take out the orig pot (since you know it works),
-temporarily connect it with alligator clips,
-create some way of marking the pot's rotation ,
-make 12 marks at key positions you find useful,
-measure those marks with a DVM to find resistor values for your rotary switch?

(of course they will not be dB specific.)  
((can I find more ways to avoid math ?? :cool: ))

=FB=
Facebook is an unfortunate way to receive news, and a good place to receive rumors.

radiance

Making a log pot with a rotary
« Reply #15 on: July 17, 2006, 06:56:12 AM »
Quote from: "Freq Band"
Is it possible to:

-take out the orig pot (since you know it works),
-temporarily connect it with alligator clips,
-create some way of marking the pot's rotation ,
-make 12 marks at key positions you find useful,
-measure those marks with a DVM to find resistor values for your rotary switch?

(of course they will not be dB specific.)  
((can I find more ways to avoid math ?? :cool: ))

=FB=


YES  :thumb:  I think I'll do just THAT, just to see how the outcome copares to the one from the calculator.....
"Knowing that you are dreaming, however, does not automatically guarantee full rationality.
Then again, being awake doesn't ensure good thinking, either." -  Lynne Levitan


 

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