What size & type resistor to reduce series output?

[wiscoaster]

The bass I built has a humbucking pickup that I've wired to switch between parallel and series output. My problem is that the series output overdrives the amplifier and I'd like to reduce the output in that mode. In series mode, the pickup resistance is about 8.5 ohms, so if I add a resistor in parallel to that, a resistor of about 28 ohms will reduce the series output net resistance to about six ohms, which I'm figuring should be about right to reduce the output signal to the amp to where I want it to be.

My question is what type and watt rating resistors are appropriate for bass guitar electronics? 1/4 watt carbon film be OK for this use?

Thanks!!

 

[wiscoaster]

Oh, I suppose somebody's going to want a pic of the bass:

Please be gentle. It's my first. I know there are members here who are really, really good at this. But not bad for 1x12 maple board lumber from Home Depot, and no special skill, if I say so myself.

 

[dwizum]

1/4w carbon film will be more than adequate. The smallest, weakest, least stable resistor you could get your hands on will be more than adequate. Don't lose sleep over it - go ahead and throw it in.

I'm assuming you left the "k" out of your numbers, i.e. you're planning on a 28k ohm resistor not a 28 ohm resistor? If I were in your shoes I'd pick a handful of resistors and try them, the "by the math" answer may end up sounding different than the math tells you, considering the EQ changes that come along with series wiring. For instance, I might compare your 28k ohm resistor to two others - one in the 5-10k range, and one in the 50k range.

 

[wiscoaster]

...I'm assuming you left the "k" out of your numbers, i.e. you're planning on a 28k ohm resistor not a 28 ohm resistor? ....

Yes, you're right, I had the multimeter on the "20K" scale so my reading of 8.5 was not 8.5 ohms, but 8.5K ohms. Thanks for reminding me of that. My bad.

 

[E2942]

That's not how the DC resistance of the pickups works. It's not a measure of output, it' a very rough indication of how many turns of wire are on the pickup coil.
If you want lower output, do the obvious thing & roll back the volume knob a little, or turn down the gain on the amp.

 

[wiscoaster]

...
If you want lower output, do the obvious thing & roll back the volume knob a little, or turn down the gain on the amp.

I know; it's rather an irritation to have to do that. Volume controls have a tendency to "walk". Then you have to tweak them every time you play. I'd rather just get the desired output the same all the time with everything turned all the way up.

 

[wiscoaster]

That's not how the DC resistance of the pickups works. It's not a measure of output, it' a very rough indication of how many turns of wire are on the pickup coil.

Yes, but adding resistance in parallel to reduce the output resistance does reduce the voltage presented to the output since the current produced by the turns of wire on the pickup coil is the same for a given string vibration.

V = I * R.

That's essentially what you're doing with a volume control.

Strictly technical speaking I know we're dealing with AC and impedance but I have no way of measuring that. So I'm just SWAGing it. Hopefully gets me in the ballpark.

 

[E2942]

Use a trimpot, adjustable with a screwdiver. That way you won't need to keep taking in & out different values.

 

[rumblesaust]

I will 2nd a trim pot
Buy a 50k or 100k one and fine tune.
They are tiny, cheap, infinitely adjustable and will fit easily in the control cavity

 

[wiscoaster]

Use a trimpot, adjustable with a screwdiver. That way you won't need to keep taking in & out different values.

Great idea!! Thanks for that.

 

[iiipopes]

Yes, all the above math is correct. Now, real world: the smallest resistor wattage rating will suffice, since the output of a passive pickup is measured in millivolts. I have actually done this to balance the output of a neck humbucker with a bridge humbucker in a guitar, although it was years ago. I believe I used a 500 ohm 1/8 watt resistor in series from the hot lead to selector switch. I do like the idea of the trim pot, but I would go 500 ohms linear, to make sure there is enough room to trim to match the outputs. That should be physically small enough to work with the switch used to for the series/parallel wiring. The problem is finding a place to wire it since most conventional push/pull series/parallel wiring diagrams show the hot lead from the pickup to one lug on the switch and then from that lug to the jack, leaving the rest of the DPDT switch to accomplish the series/parallel.

 

[mike buzzetti]

Get a 10 turn trim pot they are VERY stable and cheap !!

 

[JeezyMcNuggles]

The bass I built has a humbucking pickup that I've wired to switch between parallel and series output. My problem is that the series output overdrives the amplifier and I'd like to reduce the output in that mode. In series mode, the pickup resistance is about 8.5 ohms, so if I add a resistor in parallel to that, a resistor of about 28 ohms will reduce the series output net resistance to about six ohms, which I'm figuring should be about right to reduce the output signal to the amp to where I want it to be.

My question is what type and watt rating resistors are appropriate for bass guitar electronics? 1/4 watt carbon film be OK for this use?

Thanks!!

You're hooked to a variable resistor. Lower the volume when in series mode.

 

[micguy]

A couple of things:

1) The DC resistance of your pickups is in KOhms (thousands of ohms). If you load your pickup with 28 ohms, you will have very little output.

2) Loading a pickup by shunting it is a poor way to reduce volume - the pickup is not resistive, its impedance varies with frequency. What that means is, if you load it with a resistor, the tone changes - a lot. A P bass does not shunt its pickups with its volume controls. A Jazz does, though it's a bit more complicate than that. Why does almost every Jazz Bass player play with one (or both) pickups full up? Because, if you put both volume controls halfway up, the thing simply doesn't sound very good.

If you want to reduce the output of your bass, I recommend using your volume control. If it's a P bass, that control is relatively decent in how it works. I do know how to make it work a bit better, but that's a subject for another time and place.

 

[mike buzzetti]

add a smaller resistor to the bottom of the POT this will lower overall output because it splits the signal , IF you have separate pots for each pickup ???

 

[Plain Old Barry]

When installing your choice of resistor or trim pot, I suggest running jumper wires out from under the pickguard. You can swap parts or adjust trim to your hearts content without having to disassemble anything. When you get where you want, remove the wires and install the part inside.

 

[micguy]

OK, a suggestion. You want to have different sounds in your bass (why you have series and parallel connections), but you don't like the volume jump, and you're trying to haul down the output in that mode (which will change the sound quite a bit, and not for the better).

The reason the series mode of your bass sounds different is that the high frequency resonance of the pickup's inductance with the cable capacitance goes down in frequency the pickup (since it has a higher impedance in series mode) is loaded more by the cable, so it sounds different. There is another way to achieve the "series" sound, and completely avoid the volume jump: load the pickup capacitively.

I do thing kind of thing in my basses all the time, and it works very well - multiple sounds, all with the same level. If you like the sound of your bass in both modes, all you need to do is, instead of switching the pickup from parallel to series, use a switch to add a capacitor to the circuit, but leave the bass wired parallel. Since the impedance of your pickup goes up by a factor of 4 when you put it in series, what you need to do is increase the capacitive load by a factor of 4.

Most cables (though there is a lot of variation) have somewhere in the neighborhood of 600 picofarads. You want 4 times that much, you need 2400 picofarads total. 2400 minus 600 (what's already there) means an 1800 picofarad capacitor will do the trick. You connect that capacitor from the top of the volume control (the lug where the pickup hooks up) to ground (the back of the volume pot), but with a switch in the top leg - between the pickup and the capacitor, so you can put it in and out of the circuit.

The only other thing you need to know is that switching a capacitor in and out of the circuit, you can get a little pop when switching - a high value resistor across the capacitor will keep the capacitor from staring charge, so you get silent switching.

 

[E2942]

Don't the pickups coils in series also lower the resonant frequency?
It seems like just rolling back the tone control a smidge would accomplish something similar to inserting a 1.8 nF capacitor.

 

[wiscoaster]

You're hooked to a variable resistor. Lower the volume when in series mode.

Please read the whole thread. I've already responded to this opinion.

 

[micguy]

Don't the pickups coils in series also lower the resonant frequency?
It seems like just rolling back the tone control a smidge would accomplish something similar to inserting a 1.8 nF capacitor.

The pickup coils, when put in series, add their inductance. In parallel, their inductance is halved by the combination. A pickup by itself does have a resonant frequency (because the coils do have some interwinding capacitance), but that's very much a "tree falling the forest, nobody there to hear it" thing - when loaded by a cable, the resonant frequency is lowered much more than in the theoretical case where....you can't hear it, because you haven't hooked it up to anything.

Rolling back a tone control a "smidge" damps the resonant peak - it lowers its amplitude a bit - you lose some upper mids - say at 2 kHz or so. Above and below that, it does essentially nothing to the response - if the pickup has too much 10kHz, and you roll off the tone a bit, it still does the SAME thing at 10 kHz, even with the tone rolled back a bit. Loading the pickup capacitively MOVES the resonant peak (downwards) in frequency, not in amplitude. In doing that, you increase the lower mids below resonance some, and you lower the high frequencies above the peak - it can be a considerable amount, depending on the values of the pickup and load capacitor. Quadrupling the load capacitor will lower the treble stuff above resonance by 6dB, and move the peak down about an octave (the aforementioned interwinding capacitance makes the math not exact).

A lot of people don't really understand what a tone control on a passive bass does. It's a very common misconception that "it just turns down the treble". That's kinda true, but there is a lot of nuance that that statement misses.