Mixing two pickups

[David Schwab]

Actually, that is exactly what happens when you connect them together passively (assuming identical pickups) also, except that the loading by external circuit elements is different. That is, connecting two identical pickups together and buffering that combination gives the same frequency response as buffering each and then adding the results.

You can show this using Thevenin's theorem, which Joe Gwinn introduced on another thread. I would be happy to show you how this works if you are interested.

So the major difference between two pickups in parallel, buffered and non-buffered, is from the cable capacitance. When paralleled, the series inductance drops by a factor of two compared to a single pickup. So an important difference between the sound of a single pickup and two in parallel is the increase in high frequencies from shifting the resonance up. The addition of the two different signals accounts for the rest.

I realize that this might seem non-intuitive, but it really is how it works, and you do have to understand the electronics and do the analysis to understand it.

Hi Mike,

Yes I am interested in hearing more about this.

I have to say however that it sounds very different when each pickup is buffered first than mixing the two passively and buffering them, though that sounds very different from them not buffered.

So if you buffer the two pickups, when you combine them the inductance isn't going to drop. So two 8K coils are still 8K and not the combined 4K. So how is it the same as the two pickups in parallel before the buffer?

[Mike Sulzer]

Hi Mike,
So if you buffer the two pickups, when you combine them the inductance isn't going to drop. So two 8K coils are still 8K and not the combined 4K. So how is it the same as the two pickups in parallel before the buffer?

So consider a pickup model composed of a voltage source in series with an inductor (which can have a resistor in series if we like) and a capacitor in parallel with the series combination. We put two of these in parallel. For simplicity, detach the capacitors; we will put them back in later after combining the source and inductors.

Consider one pickup at a time. This means that we short out the voltage source of the inactive one. Now we have a voltage divider consisting of two identical inductors (maybe with resistors in series, but still both are the same). This means that the voltage at the output is half that of the voltage source. This holds true at all frequencies; so if the source varies as a function of frequency, this variation is preserved. If we short out both sources and look back into the network, we have two inductors in parallel which is equivalent to a single inductor of half the value. So the Thevenin equivalent is a voltage source of half the original source and an inductor of half the value of the original. Now we can put the capacitors back in. Note that two capacitors in parallel have half the impedance of a single one, and so the frequency response is preserved. That is, the product of L and C stays the same because the L is halved, and the C is doubled.

So this answers the question above ("So how is it the same as the two pickups in parallel before the buffer?"). One pickup connected into the other has its output halved, but the frequency response does not change because:
1. The voltage divider formed with the inductor of the other pickup is frequency independent.
2. The inductance is halved, but the capacitance is doubled also; so the resonance stays the same.

We have not yet considered the rest of the problem: what happens when both pickups are on.

[David Schwab]

So this answers the question above ("So how is it the same as the two pickups in parallel before the buffer?"). One pickup connected into the other has its output halved, but the frequency response does not change because:
1. The voltage divider formed with the inductor of the other pickup is frequency independent.
2. The inductance is halved, but the capacitance is doubled also; so the resonance stays the same.

We have not yet considered the rest of the problem: what happens when both pickups are on.

Mike, the idea of a buffer is to prevent loading of a signal by the rest of the circuit. So here we have a pickup connected to the high input impedance JFET or op amp buffer amp. Each pickup has its own buffer amplifier. The outputs of these two buffer amplifiers are then mixed together. For the sake of this argument, let's say it's into two inputs of an audio mixer. Now we have a fader and separate channel for each pickup.

How does mixing the two signals effect the inductance or resonant peak of the other signal? I don't see how it's possible. The input impedance of the buffer is setting what load the pickup sees. That doesn't change. Adding a second signal to the output signal from this buffer can't change the pickup's signal, except by phase cancelation. And still, the resonant peak, etc., of the pickup hasn't changed.

I do a lot of recording, and I've never heard of a situation where two signals will effect the source, just the composite you end up with.

But if you mix two passive pickups passively, the resistance of one will load the other, and you can hear it. If one pickup is wound a lot hotter than the other pickup, they wont mix evenly. If you buffer them and actively mix them, you can balance them anyway you like, and each pickup's signal sounds as it does soloed. This is easy to demonstrate if you record them to two separate channels and listen.

The two pickups are isolated from each other via the buffers.

The first time I heard this effect was on an aluminum neck Kramer bass that had a preamp with separate channels for each pickup. I could immediately hear the difference when I blended one pickup into the other. It was the two signals combined, and not the mishmash you get with passive basses. It was also the same as when I ran my Rick in stereo into two channels of the amp. Totally different with both pickups on then when I ran it in mono.

Now if the buffer is after the blending, then you will hear the two pickups effecting each other, but wont hear the cable, etc.

[Mike Sulzer]

Let's look more carefully at what I am saying:

1. If you combine two identical (note identical) pickups in parallel and buffer the result, you have the same frequency response as if you buffer them separately and then combine. The buffer input impedance is assumed very high. I did not finish the proof of that, but rather showed how the frequency response of one pickup stays the same when it is loaded by another identical one. Note that the voltage divider has the same ratio at all frequencies. Read the Wikipedia definition of Thevenin's theorem, and you will see that it shows the steps that I performed, including how the voltage divider is used, and how you get the equivalent impedance by looking back into the network with the voltage sources shorted.

2. If you do not buffer, but rather go to a cable, you get a different sound because the cable capacitance lowers the resonant frequency. But note that this resonant frequency is higher than when you use one pickup because the inductance is less with two in parallel. (But note that putting two pickups in series doubles the inductance, and so lowers the resonant frequency.)

[David Schwab]

Let's look more carefully at what I am saying:

1. If you combine two identical (note identical) pickups in parallel and buffer the result, you have the same frequency response as if you buffer them separately and then combine. The buffer input impedance is assumed very high. I did not finish the proof of that, but rather showed how the frequency response of one pickup stays the same when it is loaded by another identical one. Note that the voltage divider has the same ratio at all frequencies. Read the Wikipedia definition of Thevenin's theorem, and you will see that it shows the steps that I performed, including how the voltage divider is used, and how you get the equivalent impedance by looking back into the network with the voltage sources shorted.

2. If you do not buffer, but rather go to a cable, you get a different sound because the cable capacitance lowers the resonant frequency. But note that this resonant frequency is higher than when you use one pickup because the inductance is less with two in parallel. (But note that putting two pickups in series doubles the inductance, and so lowers the resonant frequency.)

I see what you are saying now. But I find it still sounds different. One of my basses currently has a single buffer after the two volume controls and master tone control. Obviously it sounds different than when it didn't have a buffer. But I have done it both ways, i.e., having a separate buffer for each pickup, and that sounds different from the current setup. It also sounds a little different if I use series resistors to isolate the two pickups before the buffer. Where you hear it is when you blend the two pickups.

[Rick Turner]

Mike, I think you're ignoring both load impedance at the amp and cable capacitance among other things. In the real world, two pickups connected in parallel give you a sound that has neither the lows of the neck pickup nor the highs of the bridge pickup. I don't care what your theory is; you're not listening to the real world, you're reading books. Just try this...hang a pickup that is not under the strings onto your under strings pickup. Take it away. Now try to tell me the sound didn't change. I do this when lecturing to show exactly this effect of one pickup loading another when two are connected in parallel. You can get pretty close to simulating it exactly by just hanging a resistor with the same resistance as the DCR of the off-board pickup across signal to ground. This shows that most of the negative effect is resistive loading.

[Mike Sulzer]

Mike, I think you're ignoring both load impedance at the amp and cable capacitance among other things. In the real world, two pickups connected in parallel give you a sound that has neither the lows of the neck pickup nor the highs of the bridge pickup. I don't care what your theory is; you're not listening to the real world, you're reading books. Just try this...hang a pickup that is not under the strings onto your under strings pickup. Take it away. Now try to tell me the sound didn't change. I do this when lecturing to show exactly this effect of one pickup loading another when two are connected in parallel. You can get pretty close to simulating it exactly by just hanging a resistor with the same resistance as the DCR of the off-board pickup across signal to ground. This shows that most of the negative effect is resistive loading.

Please go back and read what I wrote again. I am not ignoring the cable capacitance; I say what the effect is a couple of times, but the detailed analysis is yet to come. Yes, I am temporarily ignoring the effect of the amp load; it is often a megohm and not a big effect.

I am dividing up the problem into steps; the first step is to consider the loading of one pickup by another in the absence of an external load. You can think of the pickup(s) as operating into a high impedance buffer. That is the analysis I have given so far. I have not analyzed what happens when you turn on the second pickup, but of course the sound changes; the voltage source of the other pickup has a different frequency function.

Now, in response to what you have written:

The pickup is resistive only at low frequencies; the inductive reactance dominates over much of the frequency range (although the details can be quite complicated); as the frequency rises the impedance becomes very large at the resonance and then becomes capacitive above that.

When you connect two pickups together, it is not correct to say that it is like loading with a resistor. It cannot be because the pickup doing the loading is not resistive over most of the frequency range. As I explained above, when you load one pickup with another identical to it, you make a frequency independent voltage divider. If you do this when connected to a cable, the sound changes because you have lowered the inductance, and so the resonant frequency rises.

I listen to what I make a lot, of course. What I hear agrees with the theory. I can make mistakes, of course, but I do know this material well. If you think I am wrong, please be more specific. But we do need to agree on the basic facts about pickup impedance before we can make any progress.

[Rick Turner]

Mike, please work in the real world if you want to talk about pickup sound.

Try my experiment. Try it with a Strat pickup just for simplicity. Listen to a single Strat pickup. Now put a second pickup in parallel with it, but with that pickup NOT under the strings. Do so into any real world load you choose from the typical 1 meg on up to 10 megs. Now replace the load pickup with a resistor of reasonably similar value. If you don't hear the difference both ways, I just don't know what to tell you. Hell, I can hear the difference when I simply remove two pots...a volume and tone control pot...from being shunts to ground whether they're 250 K or 500 K. Shunt loading pickups affects the sound, simple as that, no matter what your textbook learning is appearing to tell you.

The sound of two Strat pickups together...positions 2 & 4 on a 5 way...is often incorrectly described by musicians as being "out of phase". If wired correctly, of course, it's not, but there is that pinched quality about it where it's not so hard to understand why musicians would describe it as such. What it is is simply one pickup loading the other. Just do the experiment I suggest, and you'll hear it.

I don't give a rat's a.. about the theory of pickups into an permanently infinite load, besides which, if you put two high impedance pickups passively in parallel you no longer have any such thing. I don't care what the capacitance or inductance of anything else in or not in the circuit is; you have a shunt to ground unless you buffer each pickup independently and go from there. That shunt to ground drastically changes your LCR factor, and you now have neither the tone of one or the other pickup, you have a band pass filtered tone that may or may not be useful. Folks have gotten used to it, and it therefore is now part of the lexicon of electric guitarists' tonal world.

I made my early career by listening first and searching for explanations after. Along the way I've been lucky enough to work with some of the best audio engineers in the biz, and you know, the funny thing is that they all rely first on their ears and then go to the technical explanation. Along the way I picked up enough of the tech side to understand what I hear. Sure, knowing theory can aid in design, but when it gets in the way of hearing, I have to object.

[Mike Sulzer]

Of course pots affect the sound of the pickup, as does putting a "dummy" pickup in parallel. The question is why. When you determined the reason for this you made a mental model of how a pickup works. You assumed that it acts very much like a resistor across much of the frequency range. That is wrong. Until you deal with that, there is no point to further conversation with you.

[RedHouse]

Don't get too fired-up Rick, it's his (Mike) modus operandi, there are a few here on this forum that _if_ you get engaged, will argue till the cows come home.
(no disrespect intended)

IIRC it wasn't too long ago there was this big'ol thread about cable capacitance (actually it was supposed to be about treble bleed circuits) some members feel that the ears can't tell anything unless the head understands it.

Start here...
Treble bleed alternatives
Treble bleed alternatives
Treble bleed alternatives
Treble bleed alternatives

I really had to laugh when after all the "debate" came this post "you have to be practical at some level"...

Treble bleed alternatives

No offense intended to Mike, but sometimes I've had to wonder if (Mike) even has a guitar, or winds pickups, or wires instruments or amps. It's never the practical stuff everybody with ears should know (by listening of course) it's always the so-n-so theorum or who'sits equation.

You really wanna read some "discussion" here Rick, do a search on Eddy Currents.

[David Schwab]

Until you deal with that, there is no point to further conversation with you.

It doesn't matter. Not in the context that by buffering the pickups, that doesn't happen. Period, the end. After you buffer each pickup independently, they no longer have any loading effects on one another, dummy coil or not, and are now two signals being mixed together that share a common source. Because of which there will be some comb filtering, just as two microphones picking up the same speaker in an amp will. But not the loading, which as Rick described, is a pinched tone. You never get the full tone of both pickups. You get much closer to that with buffered pickups.

I mean look at how the resistance of the coils change in series and parallel. When they are buffered, that doesn't happen. The 4K coil is still 4K, not 2K in parallel.

[Mike Sulzer]

It doesn't matter. Not in the context that by buffering the pickups, that doesn't happen.

What does not matter?

By the way, do you agree with Rick that putting a pickup away from the strings in parallel with another is like putting a resistor in parallel with the pickup?

What about you, Brad? Do you agree with that?

[Rick Turner]

Mike, the question is not "why". The question is whether you have the frigging ears to hear it. From your totally "up in the head; totally wrong text book" read on what is being discussed, I'd say that you're not qualified to ask "why" because you've totally ignored the truth of hearing which is ALL that pickup discussions are about. You're the classic "listen with your eyes" at a situation that is utterly beyond your perception as a listener. You're trying to make hearing line up with your misinterpretation of book learning. Good freakin' luck...

Thank God I usually make guitar pickups for musicians...

[Rick Turner]

OK, I've been winding pickups...and looking at them as scientifically as I could...since 1969. The one absolute truth I've learned is that pickup design is at least 50% art...and the remainder...usually a minor part...is science. And if you can't hear what you're doing and take that over whatever the Gaussmeter or LCR meter or VOM tells you, you're screwed.

And this is very little different from microphone design, by the way, and I'm lucky to have one of the planet's great mic designers about one block from my own shop...

If there were a scientifically perfect pickup or microphone, one or the other would be the only one acceptable; does not science and perfection reign supreme over mere taste? We know that's not true, so we have to listen learnedly, carefully...we have to apply our hopefully trained aesthetic sense of hearing and active listening to approving of one design over another...in a particular setting, whether that be a stage, a venue, or a particular phrase of music.

And if you can't hear the effect of an 8 K Hz shunt applied across a Strat pickup, you should probably retire from commenting on this forum...

[David Schwab]

What does not matter?

By the way, do you agree with Rick that putting a pickup away from the strings in parallel with another is like putting a resistor in parallel with the pickup?

I haven't tried putting a resistor in parallel, but having another coil in parallel changes the tone.... unless it's buffered, which was the original point I made. Clearly sticking a dummy coil into the system alters the tone. Doesn't matter if it's in series or parallel. Buffer it and you don't hear the effect anymore.

You said it's the same, buffered or not, and I think that's not what one hears when mixing pickups. In the real world we are never dealing with two identical pickups in parallel, and even in the case of a humbucker, adding the two coils together in parallel sounds different from a single coil or in series, right? If you buffer both coils and sum them, they don't sound like the two coils in parallel passively.

That was the original concern raised by Marko. He didn't like the results. A lot of people don't like the results, including me. Active blending fixes that issue by eliminating the pickups loading each other.