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Can you use a normal volume/tone pot as a pickup blender?
- Thread starter Peeved T-40
- Start date
Assuming you have two pickups you want to control, a single-channel volume pot controls a single channel. There's no way for it to make one pickup quieter while the other pickup gets louder.
You've been posting a few electronics questions here, maybe it's worth starting a thread where you lay out the whole project you're trying to do so people can discuss it in terms of how all the parts are going to work with each other.
You've been posting a few electronics questions here, maybe it's worth starting a thread where you lay out the whole project you're trying to do so people can discuss it in terms of how all the parts are going to work with each other.
I was thinking you might be able to run the hot end of each pickup to the outer lugs of the pot then run the center lug to volume. That wouldn't work?
I've been considering this but honestly, it's a pretty big, complicated circuit with a lot of functionality that I'm still trying to sort out. I'm just trying to take it one piece at a time as the ideas come
I know some people doo this on strats so the pickup that's not selected by the switch can be blended in, but really that's more of a glorified volume knob. I want be able to blend between two coils like a standard blend knob
But that's whan an MN blend pot does. For half the rotation, pickup 1 is at full volume and you're adjusting the volume of pickup 2. For the other half of the rotation, pickup 2 is at full volume, and you're adjusting the volume of pickup 1.
The previously posted circuit achieves exactly the same thing - as it says at the top of the diagram. One pickup is fully on, and you adjust the volume of the other. The switch determines which pickup is being adjusted. Flipping the switch is like passing the 50% point on the blend pot (kinda).
A volume pot isn't making part of the signal magically disappear. What it's doing is splitting it in two, the proportions dependent on the sweeper's position. Guitar pots are wired the way they are to split the signal between output and ground. The quieter the output, the more signal going to ground.
So the pot needs to be able to send signal to ground in order to control volume. If lugs 1 and 3 are connected to pickups and 2 (the sweeper) to output, there's no connection to ground. It won't work the way you expect.
potentiometers are reversible right? meaning you could for instance, wire a volume or tone pot backwards and make the knob work in reverse, or am I wrong about that? if that's the case, it means the knob works both directions and if you think about it like a switch, it seems to me, "off" for one side would be "on" for the other and everything in between would be a fade and you just send the other end of the pickups direct to ground.
I'm not trying to be argumentative. If that's not how it works, I'm happy to be told so. I just don't understand why it wouldn't. I presume there would be more cross talk by doing it like this but it seems like it should work on some level unless I'm just completely wrong
We'll use this pic as a key: (it's from Sweetwater, in case you want to read their how-to):
Lug 1 and Lug 3 are at opposite ends of the resistive strip/carbon strip. Usually you'll connect the pickup + to Lug 3 and the output to Lug 2, or vice versa (StewMac explains). Lug 1 gets connected to ground. (Lugs 1 and 3 might be the other way around depending on how you're looking at the pot)
The position of the wiper/sweeper determines how much of the signal comes in through Lug 3 and gets split between Lug 2 and Lug 1. The sum of the output of Lug 2 and Lug 1 have to equal 100% of the signal that came in at Lug 3. When Lug 1 connects to ground, that share of the signal is removed from the output and the volume goes down.
So if you want to control two pickups' volume at the same time, you have to have a way to divert some of each pickup's output to ground. How do you make that happen if Lugs 1 and 3 are connected to pickups and Lug 2 is connected to output?
No worries. You're trying to learn and that's good.
the resistive strip is just that right? it's the resister. If i put a regular resister in place of the potentiometer, I wouldn't need to send that to ground.
If I sent a signal into lug 1 and out of lug 3, would that be equivalent to having a normal resister? the knob wouldn't do anything and it would just have the full resistance of the pot right?
Bill Lawrence had a schematic that used a pot in this way. It was a jazz bass permanently wired in series and the second pot functioned as a blend knob for only one pickup. His setup has the neck as primary and you could blend in bridge to taste. I’d point you to the schematic, but last I checked it was down.
I have used a standard pot as a pickup blender, but it was in a situation where the bass's owner only wanted to go from all neck to a 50/50 blend of neck and bridge. For that situation, it works great, but if you want to be able to favor or solo either pickup, nope.
You can do it by wiring the two hot leads to the outside lugs and pull your output from the wiper lug. Use a high resistance linear taper pot, not an audio taper - that's a volume control. Your blended signal would be all one pickup until you got almost to the full opposite rotation. Here's an example I just pulled off the web. There are tons out there.
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yeah, this is exactly what I was thinking originally but after considering it for a while, wouldn't your volume sag in the middle?
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It will work...perhaps not very well?
I assume the circuit you are imagining looks something like this.
With a 500K pot, at the mid point where blend is equal, each of the pickups will see 250K between them and the output jack. This is the equivalent circuit:
I would expect the 250K to attenuate the signals considerably. On the plus side, the coils are isolated from each other by 500K.
AFAIK, with the MN pot, the resistance between the pickups and the output jack are 0 ohms at the equal blend position. I do believe the pickups are essentially tied together at equal blend, as if they are in parallel. So the pickups will load each other a bit. But I would expect this to be less of a problem than the attenuation from using a regular volume pot..
You may find this thread useful: https://www.talkbass.com/threads/blend-pot-wiring-its-more-than-meets-the-ear.1635584/post-28090429
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You could use a tractor to commute 30 miles to work... but it's not the best option for a lot of reasons. If you want a blend pot, buy a blend pot.
So basically, rolled all the way to one side, the volume of one pickup would be near 0 but the volume of the other pickup would be 50% of the pots' potential, meaning, if you used a 500k pot, it would be like having a 250k pot on each pickup? Could I composite for the volume loss by using a 1M pot?
How would this relate to an actual volume knob where, as @ardgedee pointed out, you actually send the signal to ground?
As far as I understand it, the way you wire an MN blend pot is basically as two mirrored volume knobs. As you turn one up, you turn the other down and they both send their signal to ground. So in a standard configuration with a blend and a volume, you effectively sending a volume pot through a volume pot right?
I don't actually know the common values people use for blend knobs but presumably it should be the same value as your volume pot right?
With 1M, the series resistance between each pickup and the output jack would be 500K at equal blend. So no.