passive volume control without losing dBs?

right. the thing is, I have wired the L side to a pot, without any resistor, and the right terminal of the pot to ground. it works but the response curve is very weird. i tried different pots. any log pot will go from 0 to 100 almost instantly, and i tried the linear pots that I have (10k and 50k) and they kinda work but the curve starts turning vol up at around 60% of the pot travel.

I just wanted to know what kind of pot I should use for a more linear response curve
 
Moderator
Joined 2011
Can you be more specific about your purpose?
You want a volume control that only works on one channel, and the other channel is always at maximum?
Or do you just want a lossless balance control?

With a switched discrete resistor control, you can have literally any taper that you want.
 
  • Like
Reactions: 1 users
Yeah, basically a volume control for one channel, the other always at max.

Only two important things for this project, it's a cheap DYI so nothing off the shelf, I want to use basically a pot and resistors if needed. And also that when the vol is maxed out, it's basically at the same volume as the other channel, at least no noticeable losses.
 
right. the thing is, I have wired the L side to a pot, without any resistor, and the right terminal of the pot to ground. it works but the response curve is very weird. i tried different pots. any log pot will go from 0 to 100 almost instantly, and i tried the linear pots that I have (10k and 50k) and they kinda work but the curve starts turning vol up at around 60% of the pot travel.

I just wanted to know what kind of pot I should use for a more linear response curve

It sounds like you may be wiring the pot backwards. If you're looking at the shaft, ground would be the left pin. The middle pin is the wiper(output) and the rightmost pin would be input.
 
  • Like
Reactions: 2 users
It sounds like you may be wiring the pot backwards. If you're looking at the shaft, ground would be the left pin. The middle pin is the wiper(output) and the rightmost pin would be input.
I was wiring them backwards yeah, but that just changed the pot direction, response curve is still the same. should a linear 250k pot result in a more linear volume increase throughout the shaft travel? I kinda just want to know what pot value should I buy before I go buying one of each and trying them individually.
 
Moderator
Joined 2011
If your sources are all solid state, use a mono 10k log pot.
If some are sources are tube, a 100k or 250k mono log pot would be necessary.

Perhaps a linear pot with a selected resistor from the wiper to ground would work better for you.
Start with a resistor of around 20% of the pot value, and adjust the value up or down as you wish.
 
I was wiring them backwards yeah, but that just changed the pot direction, response curve is still the same.
Please wire 1,2 and 3 on your logarithmic pot accordingly, you will know what thirdicomplex meant.

Screenshot 2023-06-27 at 19-27-20 IMG_2386.jpg (JPEG Image 635 × 683 pixels) — Scaled (70%).png
Screenshot 2023-06-27 at 19-49-41 Scheme-it Free Online Schematic and Diagramming Tool DigiKey.png
 
  • Like
Reactions: 1 user
If your sources are all solid state, use a mono 10k log pot.
If some are sources are tube, a 100k or 250k mono log pot would be necessary.

Perhaps a linear pot with a selected resistor from the wiper to ground would work better for you.
Start with a resistor of around 20% of the pot value, and adjust the value up or down as you wish.
It's all solid state. I wired properly a 10k log pot, but the curve seems to do nothing from 0% up to 80 or 90% and then it starts increasing the volume.

I tried a wide range of resistors wired from wiper to ground, but the only ones below 10k I've got with me right now are 4.7k and 10ohms. 4.7k and anything above 10k did nothing. 10ohms lowered the volume a bit at 100%, but the curve seems the same
 
Moderator
Joined 2011
Using a pot lower than 10k is not necessary. The value of the pot does not determine the behavior.
The electrical taper of the pot, log or linear, and an added wiper resistor, do.

The resistor values you tried are inappropriate. To approximate a log pot with a linear pot,
the added resistor from the wiper to ground should be between 10% and 20% of the pot's ohm value.
For a 10k pot, that is between 1k and 2k. Other values will not be useful.
 
Moderator
Joined 2011
Certainly there is a resistor after the pot in the circuit already, but with a 10k pot,
that resistor is likely to be large enough to not have a substantial effect compared
to the recommended values for the added resistor, and probably is around 10k to 50k.
 
there's basically nothing there, just whatever pot I'm testing, between the (L channel from the) stereo out jack and the headphones jack

R channel is connected directly from stereo out to headphones

Ground too, and also to the pot

The only resistors I wired were temporarily the ones from wiper to ground


If I remove the pot from the equation and wire the L channel directly I get the same output level as the pot maxed out without any resistors

In fact I'm gonna eliminate the R channel entirely for testing since what I want is basically a volume pot for a mono channel.
 

Attachments

  • 20230627_223602.jpg
    20230627_223602.jpg
    223.2 KB · Views: 49