Pots and HF rolloff - explain?

[Groover]

Can someone explain the physics behind HF rolloff when the vol pot in a guitar is turned down.

(I know this is the AMP forum, but it seems to me that as soon as a guitar is plugged in the guitar electronics become part of the amp, buffer amps notwithstanding.)

I have a pretty good sense of how HF rolloff works in a RC network, but I don't see a C in the guitar [Lets stipulate that it has no tone control connected.].

Is the frequency-dependent response due to the inductance of the pickup somehow?

I have seen/heard that this effect is more pronounced with higher value potentiometers. Is this the case and if so why?

Or to put it another way: Why is the voltage divider have more pronounced effect n HF the greater the total R is?

[Robert M. Martinelli]

I have a pretty good sense of how HF rolloff works in a RC network, but I don't see a C in the guitar

Are you sure there's no C(apacitance) to be taken into account?

While it's true pickups' dominant factor is inductance, you should remember a strat pickup has a stray ( or parasistic ) capacitance of some 100 pf....( not to mention its some 6K equivalent resistance, but this is not the point you raised, so I leave that sleeping dog lie )

( Good quality ) Guitar cords' stray capacitance is slightly over 100 pf/mt ( or some 35 pf/ft if you like it more ), so a 6 mt good quality cord behaves like a 650 pf capacitor shunting some HFs to ground....let alone a "bad quality" cord...

Hope this helps

Best regards

Bob

[Steve Conner]

The C is the capacitance of the guitar cord, and the R is the Thevenin equivalent resistance (google it) of the volume control. You get the worst rolloff when the volume control is at 50%. Note due to the log taper, this is not "5" on the volume dial, more like 8 or 9.

With the volume up full, the C of the cord resonates with the L of the pickup, to form a 2nd-order lowpass filter with a resonant peak. The peak is damped by the volume control resistance and amp's input resistance in shunt, and the pickup's AC resistance in series. (AC resistance is DCR plus dielectric losses, and eddy current losses in covers and other metalwork.)

This is why different pickups sound different, single coils brighter than humbuckers, and so on. Pickups have other characteristics than inductance and AC resistance, but I believe they can all be ignored to a first order. You can consider the self-capacitance if you like, but it's probably swamped by the cord capacitance.

With the tone knob at zero, the pickup inductance is now resonated by the tone capacitor, in parallel with the cord capacitance. The resonant peak moves down in frequency and gets bigger: a hot pickup sounds like a parked wah under these conditions.

At all other tone and volume settings, the resonant peak is damped more or less, and shifted around in frequency. Someone once published a family of graphs of it here.

And so on. What I take from this is that guitar cord capacitance is a vital part of tone, and you should find a cord you like and stick to it

[Robert M. Martinelli]

Hi Steve,
( BTW - how are you doin' ? )

That was exactly the point I wanted to raise....cords are often neglected as they weren't part of the equation, while they're vital in preserving ( hmm...actually trying to preserve, as they're not perfect ) the signal in its journey to the amp.

As said, a Strat pickup has a stray capacitance around 100 pf, a "hot" humbucker reaches 250 pf, but a 6 mt. cord easily exceeds 650 pf ( which have to be summed to the pickup's stray capacitance, being the equivalent total stray capacitance a parallel of the two ).

Groover,
The resonant peak of the equivalent circuit is located at 1/2pi(sqrroot(L/C)), where L is the pickup's inductance and C is the overall stray capacitance resulting from pickup windings, guitar cord and other parts' ( jack plugs and sockets, pickup switches etc. ) negligible stray capacitance.

The Q factor can be found with this equation: Q= (1/R)*(sqrroot(L/C)) and the bandwidth is Bw=Resonant peak frequency/Q.

Things change when the guitar is connected to different loads, but I think this is enough to give you an idea...

I think you'll look at your guitar's cord with new respect from now on...

Hope this helps

Best regards

Bob

[tubeswell]

FWIW the Anderton mod (that was in an '80s Guitar Player Mag), has the guitar's vol control bypassed by a 'treble bleed cap' (1nF-5nF) across the wiper to the input of the guitar volume pot. This counteracts the effects of the RC network formed by the pot and cable (hey! - maybe that should be called a 'PC' network ;-).

[black_labb]

speaking of instrument leads and capacitance, notice those instrument leads that are marketed towards bass guitar. those are the bad batches of cable with too much capacitance, so they say it "increases the bass". just a rolloff of the highs. i have a tone pot for that thank you very much.

[Groover]

hey!!!!

I am completely cognizant of the fact that the cable has capacitance - what I want to know is the way this effect changes with a change in potentiometer value.

[6267]

The pot has internal capacitance leading to a rolloff (in many cases) as the pot is turned down. Especially true inside the amp as well. Try putting a cathode follower in front of your volume control and see

[Groover]

Yes but is it more pronounced proportionally to the pot value?

Say a 250k pot set at 80% vs a 1meg pot set at 80%

[ETR]

The biggest high end suck in a guitar is the tone cap. Even tho there's 250-500K in series with it to ground with the tone control on 10 it's still quite effective with the 100-400K in line resistance youre adding in series when you roll your volume pot back. It's an RC network, adding R in series to the equation is the same as adding C in parallel. Of course theres the C from the cord as previously mentioned but even the small capacitance of a triode has the same effect in an amp when you get the volume control rolled back. R is a multiplier here in the RC network. That's why a lot of amps use bright caps on the volume controls. Just the few pf of capacitance of the tubes plates is enough when you have hundreds of K series resistance before it.

[Groover]

PLEASE

I love all this information but could someone please answer my real question which is

IS IT WORSE WITH HIGHER VALUE POTS?

[black_labb]

yes it will be.

[ETR]

I love all this information but could someone please answer my real question which is

IS IT WORSE WITH HIGHER VALUE POTS?

YES

More series resistance = bigger time constant = more HF rolloff

[Groover]

Thank you!

[Robert M. Martinelli]

I am completely cognizant of the fact that the cable has capacitance - what I want to know is the way this effect changes with a change in potentiometer value.

Groover, that was exactly what we were trying to guide you to.

Now that you have the picture, to understand the way this effect changes with the pot value, you have to draw some equivalent circuits.

Draw the pickup as an inductor with a series resistance and a parallel capacitor, then draw a potentiometer as a fixed resistor ( 500 K, then 1M so you can answer your question yourself ) between the signal output and GND. Now draw a capacitor between the output and GND. You have just drawn the situation with the pot on "10". Use real world values and do some calculation with the equations we were talking 'bout.
( cables have negligible resistance and inductance, their capacitance being their "dominant" factor ).

The volume pot acts like a variable voltage divider, so repeat the drawing e.g. with the pot at 90%, this means drawing two resistors, the smaller one between the pickup and the output ( pot's wiper ), the other between
the output ( pot's wiper ) and GND. Draw the cable's capacitance between the wiper and GND. This RC network's time constant changes with the pot's position.

Go on with other values to get an idea....or....you could ( of course ) use a simulation software to analyze all the cases...

BTW the trick tubeswell highlighted ( bypass capacitor across the volume pot ) is also used in some guitars : the Ibanez JS series has a switchable one.

and BTW, The answer to your question is YES, as you will be able to verify using the above methodologies.

Hope this helps

Best regards

Bob