• TalkBass has been independent since 1998. Add your voice.
    Create a free account to reply to discussions, view embedded media, and browse with fewer display ads.
    Join free Log in
    Want zero display ads or expanded classifieds tools? Compare plans.

Blend Pot Problems

MustangWally

MustangWally

Inactive
Feb 5, 2019
921
1,551
San Marcos TX
#1
I recently put a blend pot in my Mustang PJ in place of my pickup selector switch, and while it generally works as planned there is a problem: all of the blend action takes place in a tenth each side of the midpoint, the rest of the rotational travel does virtually nothing. Suggestions would be very welcome.

Blend pot is an Alpha 250K linear taper, wiring is the usual crossover arrangement: [Invalid or Expired Link Removed]

Should I be using an audio taper pot or somesuch?
 
#3
When I started reading your post, my first thought was that you must be using log taper pot, but then I've read that it's linear, which I find unexpected. Other than checking your pot's taper (maybe they sent you a log pot by mistake), the only other thing i can think of is that all the guys who spoke of ungrounded balance controls were using, IIRC, 100k controls. Perhaps 250k is too much.
 
#8
Blend pots on passive basses are always a compromise. You're trying to mix two sources that are highly reactive (their impedances are wildly different depending on what frequency you're talking about) with resistive components (who's impedance is flat over frequency. A VVT circuit has the same issue, by the way.
 
  • Like
Reactions: Jazz Ad
#9
I'm just trying to understand if I can 'widen the sweet spot' on my blend pot to the point where I can use most of the rotational travel instead of just the middle 3/10. I lack the electrical background to work this out but I am hoping that TBers with greater knowledge than I can help me with this one. Thanks in advance.
 
Last edited:
#10
Here are a few choices of MN taper pots. If you look around you should be able to find some log/antilog taper stacked pots. Either of those should get you closer to what you want. The StewMac description seems to indicate theirs is linear taper which could cause what you’re describing.

Invalid Link Removed
 
#11
They are MEC. They came with the basses. Buying direct from Germany is relatively expensive, especially with shipping. I am happy with them. They work better than my two volume situation on a jb4.

Really I think you are running into a situation where not every log taper is the same.
 
#17
Thanks for that link Matt. I have read through it all and it seems that if I swap to a 100K linear MN pot that is about as good as it's gonna get. Walter W is really helpful!

Also found the difference between MN and AC pots - MN gives you the 100% neck and 100% bridge at middle, AC is 50/50.
 
#18
MustangWally said:
Thanks for that link Matt. I have read through it all and it seems that if I swap to a 100K linear MN pot that is about as good as it's gonna get. Walter W is really helpful!

Also found the difference between MN and AC pots - MN gives you the 100% neck and 100% bridge at middle, AC is 50/50.
Click to expand...

That’s what I was trying to say about AC vs MN, I just couldn’t remember the right numbers. :rollno:

Let us know how it works out.
 
#19
MustangWally said:
Thanks for that link Matt. I have read through it all and it seems that if I swap to a 100K linear MN pot that is about as good as it's gonna get. Walter W is really helpful!

Also found the difference between MN and AC pots - MN gives you the 100% neck and 100% bridge at middle, AC is 50/50.
Click to expand...
This is incorrect. A blend pot should give 100% in the middle position for both signals for both AC and MN pots. If it doesn't, I would argue it isn't really a blend pot.

What makes a blend pot different than a standard stacked pot is that in the middle position you are at the the maximum amount resistance between the common terminal and the outer terminal on one side. As you rotate the potentiometer, resistance decreases to ground for one pole and stays the same for the other. Something like this where the green marking is the amount of resistance between terminals in the middle position:

image001.png


If these are perfectly linear pots, resistance to ground is 50% in the middle position in the standard pot and 100% in the blend pot. The shape of the curve doesn't have to be linear though. It could be any curve.

The A, B, C, M, and N refer to the shape of the taper. A is a log taper. B is a linear taper. C is a reverse log taper. M and N are a type of sigmoid taper that are mirror images of one another.

Every potentiometer company has slight differences in how even the same taper is applied. Those difference can impact how well the pot works greatly.
 
Last edited:
#20
This site gives a good brief overview of why this is and should make your problem make sense.

https://e2e.ti.com/blogs_/archives/b/thesignal/archive/2012/10/22/logarithmic-potentiometers

Logarithmic tapers are often made by putting together two or more linear tapers on one pot. A manufacturer can determine what the two linear halves of the curve are that approximate a logarithmic curve. Every company does this a little differently. If they made it so the first part is too shallow a slope, you would have nothing happen then hit the steep slope and get a lot of change.
 

Latest posts

Top Bottom
Back