# Does this schematic approximate volume control?

Discussion in 'Pickups & Electronics [BG]' started by skycruiser, Oct 9, 2019.

1. ### skycruiser

Jan 15, 2019
Texas
I'm trying to understand how the volume and tone controls work on a PJ bass.
This is a pretty good schematic I found online, showing how the pots wire up with the pickups.

I think what this is doing is effectively a voltage summing circuit with the two pickup outputs feeding into the summing node. Here's a schematic of a basic summing circuit. The arrow marks the point shown above.

I think you can approximate the pickup coil as a perfect voltage generator in parallel with a resistor (roughly equal to the DC resistance of the coil). I think coil resistance is usually in the 5k to 10k Ohm range, but for this it doesn't really matter.

So if this is correct, the full circuit including the pickups would look something like this. pickup1 might be the P pickup wired in series, and pickup2 to the J pickup.

I don't think R_coils and R_Ls matter too much. The parallel resistance of those two will change as you adjust the pot, but you will always see the pickup voltage at the node between R_U and R_L. From the upper summing circuit equation, each Voltage will be summed in but divided by the series resistor, in this case R_U. So as you turn the volume down on pickup1 for example, R_U1 will become very large, so the summing in of V_pickup1 will be very small. As you turn the volume up, R_U2 will become very small, raising the summed in level of V_pickup1.

Does anyone know if this is more or less correct? Do the standard pots always leave some resistance on the top end (maybe a few kohm or few hundred ohms)? I don't think you want to directly short the two pickups into the summing node - the series resistance is required.

Thinking about the summing circuit a little more, I think the summing equation is only truly correct if the resistance in parallel with the voltage source is very large (like an open circuit). So this relationship will be distorted by the parallel resistance between R_coil and R_L. If anyone knows how to adjust the equations for this I would be interested to learn about it. Something else I could look into I suppose.

Thanks for any help on this.

2. ### Killed_by_DeathSnaggletooth

The output of pickups will not be DC but AC Voltage & Current, and only around 1/4 of a Volt.
Most two-pickups basses with two volume pots do sum directly when both volumes are fully clockwise.

However, in the case of a Blend pot that is A/C taper, there will be about 10% Resistance between the pickups when the blend pot is in the center position.

With two-pickup basses (with two volumes) there's a noticeable mid-scoop when both pickups are on full-volume, which many attribute to overlapping sine-waves of the two pickups in different positions.

I understand that an A/C taper blend-pot solves that issue, but no one has ever confirmed if it is indeed that 10% of Resistance between the pickups that prevents the mid-scoop.

BTW, if both pickups are putting out 1/4 Volt, the total Voltage will still only be 1/4 Volt, but the milliAmps of output will be increased. So essentially, slightly more power.
Pickups wired in Series have noticeably more output, since they're additive Voltage.

3. ### b/o 402

Jul 14, 2015
DC & RVA
You have a key error on your schematic. The wiper of the tone pot, where you have drawn a blue arrow, should also be connected to the output. Where you have the output lead connected now, on the other end of the tone pot, will have no connection.

Spacecase and skycruiser like this.

Also, saying you go to tonepot is misleading. You go to the hot peg of the jack.
The tone pot is a controlled bridge between output signal and ground. You can take it straight out of the circuit and it keeps working.

skycruiser likes this.

Jul 14, 2015
DC & RVA
6. ### skycruiser

Jan 15, 2019
Texas
Thanks for this info. I realize it's not a DC signal we're talking about, but I think since the summation is a linear process it just directly sums (AC and/or DC). That's interesting what you said about the voltage levels not doubling, but if this is a true summation I'm not sure how that can be correct (if the AC signals are in phase). It may be related to the fact that it's not an ideal summation due to the parallel resistances of the coil resistance and the lower pot resistance. Need to think about that a little more.

Last edited: Oct 9, 2019
Killed_by_Death likes this.
7. ### skycruiser

Jan 15, 2019
Texas
Thanks guys - I think you are pointing out similar points. I see now that drawing is missing an important connection. Need to be extra careful with things pulled off the web. And on my schema (the goofy looking one), as you said it needs to say "to hot output peg and tone pot (if used)" or something like that.

8. ### skycruiser

Jan 15, 2019
Texas
corrected diagrams

9. ### Killed_by_DeathSnaggletooth

Start plugging numbers into the formula from your own post above & you'll see it's true:

Let's use 3 Volts, for simplicity:

(3+3+3)/3 = 3 Volts

10. ### b/o 402

Jul 14, 2015
DC & RVA
Still not quite right. If you look at the pictorial I posted, you can see the other lug on the tone pot - the one you have connected to the output - is left disconnected. This is standard. The tone control will still work the way you show it, but it leaves the full resistance of the pot across the output, which loads it down unnecessarily.
Ok?

skycruiser likes this.
11. ### b/o 402

Jul 14, 2015
DC & RVA
As I'm reading your comments, I'm seeing that you're attempting to work out the principles of electronics with insufficient information. I certainly don't want to discourage you from exploring this field, but I will say you will make much faster progress if you get acquainted with the basics of AC. You need to know at least Ohm's law and all its applications, as well as the different functions of resistance, impedance, reluctance, and reactance. Good luck!

12. ### skycruiser

Jan 15, 2019
Texas
Ha of course - it's summing but averaging at the same time. How did I miss that. Thanks for pointing out the oversight.

13. ### Leo Thunder

Sep 27, 2018
There is no summing involved. These voltages are in parallel.

thetragichero likes this.
14. ### skycruiser

Jan 15, 2019
Texas
Well I'm not quite trying to work out basic principles. I have been doing electronics work for quite awhile and have an EE degree, though analog is not my forte (my focus was digital design), and I know pretty much nothing about audio. Probably could use a thorough refresher on AC circuits. As for the schematic, I didn't look carefully at the drawing and as I said earlier, the top schematic wasn't mine. I pulled it off the web and didn't do a pin-to-pin comparison to check that it's right. My mistake. I see the other error -the top end of the tone pot needs to be disconnected.

For completeness here's the corrected drawing.

Last edited: Oct 9, 2019
15. ### skycruiser

Jan 15, 2019
Texas
Voltages can't be directly in parallel - a node can only be at one voltage. If the volume pots are basically behaving like the summing circuit, which schematically they appear to be, then it is averaging the signals from the two pickups. It behaves like a weighted average, with the "weight" being set by the position of the pot - basically by setting the series resistance going into the summing node. It's starting to make sense to me anyway.

16. ### crazyBassClown

Sep 25, 2007
sydney, australia
A couple of quick points,
The last 2 versions of your circuit are effectively identical, you have a zero resistance in parralel withthe upper part of the tone pot, so whether the top is connected or not the values will be the same to the centre wiper.
Secondly the pickups are not just voltage sources but are also a load that the other pickup will drive,
Thirdly the pickups are not perfect voltage sources so as the lower resistance decreases on the volume pots the current draw from the pickups will increase dragging the output voltage down.

And finally, i may not be 100% correct because I just play the thing rather than think about it because that circuit has worked well enough for the last 60+ years.

17. ### keyboardguySupporting Member

May 11, 2005
...Isn't this an easier diagram to visualize?

Ric5 likes this.
18. ### Leo Thunder

Sep 27, 2018
Definitely not.

19. ### skycruiser

Jan 15, 2019
Texas
Good catch on your point #1. Honestly my original post and questions didn't really involve the tone pot so I didn't give it much attention. The reason I updated the drawing was just so it would match the actual wiring. I don't think anyone would hook up that 3rd terminal even though as you point out it has no effect on the circuit. Points 2 and 3 are helpful - thanks for adding them to the mix. It's clearly much more complex that my simplified representation of the circuit.

On the last point - I didn't bring this up to see if I could improve it or anything like that. I'm interested in how the components affect the output just to better understand how/why the output of my bass changes as I fiddle with the knobs. I just started playing bass again after 20+ years without one (played a little back in high school). I'm finding that I'm a bit overwhelmed by all the knobs on my bass and amp. Seems like I endlessly cycle through various settings on knobs trying to find satisfying sounds. Probably need to start taking notes. But anyway, I think if I better understand how they work, the settings will start to make more sense. I think the way the EQ knobs and Gain/Volume are clear enough. Still having a little trouble understanding the notch filter, and then the interaction of the 3 knobs on my bass need some clarification in my mind. This discussion has helped with that.

20. ### crazyBassClown

Sep 25, 2007
sydney, australia
For someone with no background in electronics it probably is.