• 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, 250k VS 500k issue

chalie

chalie

Nov 3, 2004
632
7
70
Bangkok, Thailand
#1
Hi Guys,
From time to time we were told that a 500k blen pot is appropriate for blending passive pickups but I wonder why we hardly found a 500k blend pot as OEM in bass guitar, both with and without preamplify.
Can anyone help me with this issue.
Thanks,
Chalie
 
#4
chalie said:
Hi Guys,
From time to time we were told that a 500k blen pot is appropriate for blending passive pickups but I wonder why we hardly found a 500k blend pot as OEM in bass guitar, both with and without preamplify.
Can anyone help me with this issue.
Thanks,
Chalie
Click to expand...

500k is the usual correct value. The thinking is that the usual pickup loading in a passive bass is 250k. Modding to 500k is usually done to make the tone brighter.

But the deal with a blend pot is that it is really TWO pots in parallel. Thus the pickup loading with a 500k blend pot will be the standard 250k. If you think that blend should be 250k because that's the standard value, in truth you'll be loading with 125k which will make the tone warmer.

Which leads us into part two of your question. When you choose a blend pot value for an OEM bass, the choice isn't made by theory as above. The choice is made LISTENING to the tone the bass puts out. Then a decision is made based upon the tone the factory wants the bass to have. The final tone, of course depends on MANY things besides the blend value, including pickups, bridges, etc.. Which is why the final choice is made by listening to the tone, rather than by some "electronic rule".
 
  • Like
Reactions: fruitlabor
#7
bassbenj said:
But the deal with a blend pot is that it is really TWO pots in parallel.
Click to expand...

How so? In a real blend pot, it's two pots which are ganged together physically but not electrically. The resistance strip for one gang of the pot extends from the center detent position to one lug, but there's no resistance from center to the other lug. The other gang is reversed.

3284625829_ba5bc22451.jpg


The pots are NOT in parallel unless you wire them that way, which would defeat the purpose of the blend pot.

However, the issue with a blend pot is that you really have three volume pots in most circuits. In addition to the two pots that make up the blend pot, you also have the master volume pot. In that sense yes it IS two pots in parallel. So, using a 500K/500K blend pot with a 500K master volume would get you to the 250K.

John

John
 
#8
Blend pots ARE in parallel. Look at your own drawing. Each end of each deck of the pot is grounded and the other end is connected to the hot. The wipers with the pickups are inconsequential since they are connected to the pickups. Both pickups see the load of both pots, just as they do on a Jazz Bass.

So you have two 500K resistors wired in parallel between the hot and ground. That parallel resistance is 250K. Add in a volume pot, and you have another resistance in parallel.

You might be thinking that blend pots are no-load pots, but they are not. They are like two volume controls with one wired in reverse. Only the taper goes to full on at the detent. But the hots and grounds are connected together.
 
#9
JTE said:
In a real blend pot, it's two pots which are ganged together physically but not electrically.
Click to expand...

The pots are indeed running parallel to each other. On both gangs, one side is grounded, and the other side goes to the output. This means you have two resistors running parallel to the output.

JTE said:
The pots are NOT in parallel unless you wire them that way, which would defeat the purpose of the blend pot.
Click to expand...

Quite the contrary. That is indeed the purpose of a blend pot. You have two volume pots in inverse parallel.
 
#10
line6man said:
What? The resistance from one side to the wiper is always equal to the total resistance minus the resistance from the other side to the wiper. The only way for a 0 Ohm continuity to occur between the wiper terminal and one lug, when the wiper is not all the way to one extreme of the resistive track is if you have shorted the wiper to one end of the resistive track.
Click to expand...
??

M/N blend pots do indeed have no (or almost no) resistance from center to one side. all the taper happens over one half of each pot, with both middle lugs shorting to opposite sides when it's centered.
 
#11
walterw said:
??

M/N blend pots do indeed have no (or almost no) resistance from center to one side. all the taper happens over one half of each pot, with both middle lugs shorting to opposite sides when it's centered.
Click to expand...

Ah! You're right about the M/N tapers reaching 0 Ohms at halfway on one side. If the pot is not set to the center detent, one of the gangs should function as a standard voltage divider, though.
 
#12
Are you saying that a real blend pot is fundamentally different from the two volume pots in a typical Jazz Bass-style dual volume control? I don't see that at all. If the 250K works as separate pots in what way are they different when mounted on the same shaft, assuming one useba real blend with the resistive part only over half of the pit's travel?

John
 
#14
DavePlaysBass said:
A 250K Blend pot is no different than two 250K volumes in terms of blending assuming you never trie to run both volume / volume at a cut position. So for practical purposes for most people they are the same.
Click to expand...


No its not, because the two 250K volume pots are wired in parallel, so that equals 125K load on the pickups.

A 250K blend pot would also equal 125K, but you also have a volume control, and if that's 250K you end up with an 88.33K load. That's 250/250/250 = 88.333.

With a 500K blend, you have two 500K resistors in parallel for a total of 250K. Then if you add a 250K volume you end up with a 125K load. That's the same as two 250K volumes.

If you use a 500K volume, the pickups will see a 250K load.

The pickups see a load even if they are on ten!
 
#15
SGD Lutherie said:
No its not, because the two 250K volume pots are wired in parallel, so that equals 125K load on the pickups.
Click to expand...
That's assuming you're at the center detent. It's a constant 125K resistance parallel to the output. If you adjust the blend pot, that resistance between each pickup and ground will change. If you turn the blend all the way to one extreme, for example, the parallel resistance against one of the pickups will drop to 0 Ohms.
 
#16
line6man said:
That's assuming you're at the center detent. It's a constant 125K resistance parallel to the output. If you adjust the blend pot, that resistance between each pickup and ground will change. If you turn the blend all the way to one extreme, for example, the parallel resistance against one of the pickups will drop to 0 Ohms.
Click to expand...

I doubt it if LineMan6 knows what he is talking about...
 
#20
line6man said:
That's assuming you're at the center detent. It's a constant 125K resistance parallel to the output. If you adjust the blend pot, that resistance between each pickup and ground will change. If you turn the blend all the way to one extreme, for example, the parallel resistance against one of the pickups will drop to 0 Ohms.
Click to expand...

No, the 125K is always there because it's always between hot and ground. The pickup is not isolated from the output until you turn it down, and then it's load increases because the resistance to ground decreases.

The zero resistance is series resistance between the pickup and the output. Burt you still have a load resistance of 125K to ground.

Draw out the schematic and see. As you turn the pickup down, the series resistance increases, and the resistance to ground decreases.
 

Latest posts

Top Bottom
Back