[pw8888]

HI

I have a basic question regarding global input potentiometer “value”

Let’s say I have a 10K log potentiometer at the input . I would like to be able to “fine tune the level “ … one pot would be the “master pot “ and the other the “fine” tune one .
What should  be the value for the second one ? 10K as well ? Or should I use 2 x 5K ?

I uploaded a little drawing of my question.

Thanks a lot 

[ruffrecords]

What range are you wanting the fine tune pot to work over?

Cheers

Ian

[pw8888]

HI Ian 

very hard to say in fact !
probably 50% ?
I would like to be able to kind of "gently" saturate the input when needed...
but that is ...as well for other board(not only yours) that I am using as well .
thanks !!
P
 

[abbey road d enfer]

HI

I have a basic question regarding global input potentiometer “value”

Let’s say I have a 10K log potentiometer at the input . I would like to be able to “fine tune the level “ … one pot would be the “master pot “ and the other the “fine” tune one .
What should  be the value for the second one ? 10K as well ? Or should I use 2 x 5K ?

I uploaded a little drawing of my question.

Thanks a lot 

Cascading two pots like you show does not result in better resolution.
There are several possible solutions. A simple one consists in putiing in series with the coarse pot a fine pot. Using two identical value pots will give about 6dB range for the fine pot. However, teh tapers should probably not be identical. If the coarse pot is a log taper, the fine pot should be linear or reverse-log.
Another possibility is cascading two identical pots, but adding a padding resistor at the bottom. Making this resistor eaqual to the pot value would give you the same 6dB range.

[ruffrecords]

One way to get a 6dB range fine tune is as follows. The main pot comes first and is a 10K LIN type. The second 'fine' pot is 2K5 LIN in series with a 2K4 resistor. Strap this across the first pot from wiper to ground with the top of the second pot to the wiper of the first and the free end of the 2K4 to ground. Output is from the wiper of the fine pot.

The first pot will now give a rough LOG law with the 50% rotation point being 10dB down. The fine pot should give a further 6dB range with it 50% position being another 3dB down. So with both pots centred the output is about 13dB down and the fine pot can adjust this by plus or minus 3dB.

Edit: this is Abbey's second circuit.

Cheers

Ian

[Winston O'Boogie]

Typically called a "Vernier" yes? 
You see it on some equipment (mostly not audio - but it is there on the UA 175 compressor) where the main control is maybe a stepped attenuator with, say, 2dB per, and the vernier control handles the 2dB in-between swing.

Cool 

[mjrippe]

Great example, Winston.  Also some German modules like the Siemens V276, which has a 6dB/step input control with a 10dB range trim!

[pw8888]

Cascading two pots like you show does not result in better resolution.
There are several possible solutions. A simple one consists in putiing in series with the coarse pot a fine pot. Using two identical value pots will give about 6dB range for the fine pot. However, teh tapers should probably not be identical. If the coarse pot is a log taper, the fine pot should be linear or reverse-log.
Another possibility is cascading two identical pots, but adding a padding resistor at the bottom. Making this resistor eaqual to the pot value would give you the same 6dB range.

Thanks everybody for your answers !
but I think i was not clear enough with my questions (sorry for this)
I am NOT looking for a better resolution . but more like this exemple. let me trying to explain it better

Let’s say I have one pot at the input . 10K LOG . Lets call it 0 (full anticlockwise)  to 10 (full clockwise)(scale)
Lets say the sound, style , tone is  good around 5.
How do I do then to add a second pot that I will use to “fine tune” this 0 to 5 range ?

Ands specially what would be the value ?

Do I have to add this 3rdresistor ?

Thanks a lot and I hope I am a little more clear …I hope …
And yes the “vernier” on the Universal Audio 175 was acting like that :-)  I use to own 2- 175 and 1- 176 ...

[abbey road d enfer]

How do I do then to add a second pot that I will use to “fine tune” this 0 to 5 range ?

I think we presented you with two methods that precisely achieve this.

[pw8888]

I think we presented you with two methods that precisely achieve this.

thank you so much for your time !
will do the 2 pots + resistor option
just to be sure ... I can use 2 pots of same value (let's say 2 x 10K LOG pot and 1 resistor of 10K )
does the resistor has to be 1 watt ? or less ?
thanks again guys 

[abbey road d enfer]

thank you so much for your time !
will do the 2 pots + resistor option
just to be sure ... I can use 2 pots of same value (let's say 2 x 10K LOG pot and 1 resistor of 10K )
does the resistor has to be 1 watt ? or less ?
thanks again guys 

The pot value must be determined in view of the pick-up impedance.
J-bass p-u's should be used with no less than 220kohm pots. I would safely opt for 470k. For a good balance between level loss and interaction, I would use no less that half the pot value for the resistors.
There is no significant power in this circuit, you may use almost anything , even 1/16W if that's all you have. 1/4W being the de facto standard, I would use them. Don't be tempted to use those carbon- comp resistor s..t that cost an eye.

[pw8888]

The pot value must be determined in view of the pick-up impedance.
J-bass p-u's should be used with no less than 220kohm pots. I would safely opt for 470k. For a good balance between level loss and interaction, I would use no less that half the pot value for the resistors.
There is no significant power in this circuit, you may use almost anything , even 1/16W if that's all you have. 1/4W being the de facto standard, I would use them. Don't be tempted to use those carbon- comp resistor s..t that cost an eye.

thanks !

[pw8888]

The pot value must be determined in view of the pick-up impedance.
J-bass p-u's should be used with no less than 220kohm pots. I would safely opt for 470k. For a good balance between level loss and interaction, I would use no less that half the pot value for the resistors.
There is no significant power in this circuit, you may use almost anything , even 1/16W if that's all you have. 1/4W being the de facto standard, I would use them. Don't be tempted to use those carbon- comp resistor s..t that cost an eye.

Me again ... I would like to double check something. in the case of this schematics ... what value would you have for P2 and P3 and R 1 ? .Should I have 10lin instead of log for P2 and P3 ? ...and  then with a resistor of 5K ?
Should I keep "log" ?
THANK YOU SO MUCH again for your time 

[abbey road d enfer]

Just make everything 10k. P2 Log, P3 lin.

[pw8888]

Just make everything 10k. P2 Log, P3 lin.

Thanks !

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