better clean up for fuzz

[Unlikekurt]

I've got a simple fuzz on a breadboard.  It's NPN common emitter with  a feedback resistor biasing the base.
The clean up from the guitar volume pot isn't bad, but it's fairly limited in the sweep.  For kicks, I whipped up an op amp buffer put it between the guitar and the fuzz input.  As one would have inspected, the nice bloomy tone vanished. However, the volume pot cleanup seemed to operate over more of the range. 
I searched the forum a little bit as well as the net, it's clear that the low output impedance of the buffer isn't what the input of the fuzz wants to see.
That being said, if one were to calculate the input impedance of the 1st stage in the fuzz, could one set up an op amp buffer to drive the fuzz and maintain the tone; obviously not a 1/10 of the input but perhaps matching?

Thanks

[Rob Strand]

In some cases you can tweak the tone by just adding a series resistor between the buffer and the fuzz circuit.  Going a little further would be to tweak the input cap at the input of the fuzz.  (Beyond that is to add some sort of pickup impedance emulation but matching a pickup might not be best you might need to match a pickup with the impedance scaled down.  This type of stuff is pretty open ended.)

[mozz]

Couple questions,
what kind of pickups are you using
what kind of transistors are you using
what are all your voltages

[j_flanders]

The clean up from the guitar volume pot isn't bad, but it's fairly limited in the sweep. 
op amp buffer put it between the guitar and the fuzz input, the volume pot cleanup seemed to operate over more of the range. 

Without a buffer, the guitar volume control reduces pickup output voltage AND increases input resistance and thereby reduces the gain of the fuzz face circuit.
With a buffer, the guitar volume control it only reduces the pickup output voltage.

Maybe overly simplified, but if you model the Fuzz Face circuit as an inverting opamp with a similarly low input impedance I find it easier to understand the difference:

Without a Buffer, volume at 10 or 8:


With a buffer, volume at 10 or 8:


The above does not take into account the resistance of the pickup itself which already reduces the FF gain compared to a buffer in between:


All of the above should make clear why adding a pot at the input of the FF circuit to control the input resistance, and thus gain, helps if you want to use a buffer between the guitar and the FF.
The FF no longer loads down the pickups, and you cannot control its gain with the volume knob, but at least you can set its 'normal, standard gain'

[Unlikekurt]

Wow, thank you so much for such a great explanation as well as illustrations!
It makes a lot of sense now for me.  But also, as one might expect, brings forth a follow up inquiry.
If the output of the opamp buffer had a coupling cap and then a shunt resistor to bleed off the coupling cap, how would that R come into play?

[antonis]

If the output of the opamp buffer had a coupling cap and then a shunt resistor to bleed off the coupling cap, how would that R come into play?

Just like R = 0.159/(C*f), where f = HPF cut-off frequency..

[Steben]

Might be strange, but always noticed best clean up usually comes from silicon fuzz(face)s.
High current gain usually has slightly higher input impedance, which means brighter tone.

[Steben]

Without a Buffer, volume at 10 or 8:

Pickups indeed usually have impedance on their own of 5k to 10k pure resistance for single coils, yet with an induction as well which means treble cut with low input impedance. The addition of pure resistance when rolling off volume completely changes the treble cut (as in eliminates). This changes the calculations a bit of the volume roll off. The results given are low frequencies and it shows how much the gain of bass goes down, giving a "brighter" clean up.
The first situation gives output "bass" level actually a bit lower than 50% (impedance of guitar itself is usually a tad higher than input impedance of FF) = 5k / (5k+4,9k) and gain of 100k / (5k + 5k) = 10.
The second situation gives output "bass" level actually a bit lower than 8.3% (impedance of guitar itself is usually a tad higher than input impedance of FF) = 5k / (5k +54.9k)

Strat single coil usually has about 2H inductance and this gives simplified a eq R of 2*pi*2*f, or for 1khz about 12,6k ohms.
This means first situation 1khz level of 5k / (5k+4,9k + 12.6k) = 22 % ! at full volume.
This means first situation 1khz level of 5k / (5k+54,9k + 12.6k) = 6.4 % ! at volume at 8. which is not much less than 8.3% for bass frequencies.
Hence the "bright" clean up