Curious about the effect of switches and pickup resistance

dpaterson

New member
Hello.

Just curious (and not seen this mentioned anywhere) (and I have looked):

Let's assume you have a bridge pickup with resistance of, say, 15kΩ and a neck pickup with a resistance of, say, 9kΩ. You also have a 3 way switch (bridge pickup, both, neck pickup).

Obviously when only the bridge pickup is selected via the switch then the resistance measured is 15kΩ. And the same applies when only the neck pickup is connected obviously (9kΩ).

BUT what happens when the switch is in the middle position i.e. both pickups selected???

I'm ASSUMING that this results in the combined resistance being (15kΩ + 9kΩ) / 2 = 12kΩ (parallel resistance)???
 
Re: Curious about the effect of switches and pickup resistance

You follow the law of parallel resistance. So your figure is correct.

Keep in mind that’s just the DC resistance of the pickups. It doesn’t mean much tone wise.

When both pickups are on the impedance (AC resistance) drops and the lower resistance pickup loads the higher one. So having both on and mixed passively changes the tone if the pickups a bit.

You can hear this effect if you connect a second pickup to the circuit while that pickup is not mounted under the strings.


Sent from my iPhone using Tapatalk
 
Back
Top