Using MosFETs or JFETs in a "stock" NPN Fuzz Face

[JDoyle]

[I don't really understand but] Q1 input [base/gate] isn't tied to ground, so maybe it floats up when the source bias is raised, making it difficult to bias Jfet there.

Using Ohm's law, the base of a Q1 JFET is tied to a voltage that is equal to I_E(Q2) * R_E(Q2) (or the Gain pot), so the gate will always be above the source because the source is at ground and the gate is at the same voltage as that of Q2's emitter, which will always be positive. (I'm assuming an NPN as Q2 and an NJFET as Q1 for ease of understanding)

Jay Doyle

[johngreene]

IIRC J.G. had a BJT MOSFET circuit on the web years ago.

Still do, it was the Axis fuzz done with a MOSFET though.

http://www.greene-pedals.com/GEAD/pdf/fuzz001.pdf

--john

[mac]

If the fet is a 2sk117 or j201 then the source must be 0.5v or more above the gate. As the gate is at 0.6v (*) the source should be at least 1.1v. A higher Vgsoff fets will set the drain voltage too high. All this assuming there is a bjt at Q2.

(*) 4.5v/8.67k = 0.52ma ; Ve2 = 0.52ma*1k = 0.52v
Add 0.2v - 0.3v across the 100k.

I´m going to breadboard using 2 fets and trimpots for both drains and sources.

mac

[JDoyle]

If the fet is a 2sk117 or j201 then the source must be 0.5v or more above the gate. As the gate is at 0.6v (*) the source should be at least 1.1v. A higher Vgsoff fets will set the drain voltage too high. All this assuming there is a bjt at Q2.

(*) 4.5v/8.67k = 0.52ma ; Ve2 = 0.52ma*1k = 0.52v
Add 0.2v - 0.3v across the 100k.

I´m going to breadboard using 2 fets and trimpots for both drains and sources.

mac

mac, where do you get the additional voltage across the 100k? I'm curious. By my estimation, if Q1 is FET and a FET's gate doesn't draw current (or very little) the voltage at the emitter of Q2 would be the voltage at the gate.

Also, you could put two diodes in series with the source to ground to bring it up to 1.2V without the addition of any resistors.

How do you like the 2SK117? I've been wanting to try those for a while.

Regards,

Jay Doyle

[earthtonesaudio]

Someone already mentioned this one: http://www.geocities.com/tpe123/folkurban/fuzz/fuzzyfaces.html

But this one goes a little more in-depth (and has sound samples): http://runoffgroove.com/multiface.html

I've breadboarded the ones on Tim E's site, and done some of the runoffgroove mods as well.  I can attest that they all work, with minor adjustment to the bias pot (10K in schematic).  I don't know how you could get a JFET to work in the Q1 position without major mods to the circuit (as evidenced by previous comments here) but there's no harm in trying!

I find the FETs to give it more of a buzzy, metal type of tone.  And one thing I noticed is that you can get some weird oscillations with the FET versions when you roll back the guitar volume.  Most likely an impedance issue, but I never investigated it further. 

Have fun!
-Alex

[mac]

mac, where do you get the additional voltage across the 100k? I'm curious. By my estimation, if Q1 is FET and a FET's gate doesn't draw current (or very little) the voltage at the emitter of Q2 would be the voltage at the gate.

Yeap, you're right. I was thinking standard FF.   Anyway it was a rough estimation.

How do you like the 2SK117? I've been wanting to try those for a while.

They have low Vgsoff and very high gain.

mac

[mac]

Today I breadboarded a q1=2SK117, q2=BD139 Fuzz Face. Standard circuit, only a pot and a big bypass cap at the fet source. It works.

Rsource=3.5k
Vd=1.1v
Vs=0.85v
Vg=0.5v

Lots of highs, I have to put a 0.1uf cap across the 470 resistor, the higher input impedance I guess. The final gain and fuzz are not as the standard FF. Maybe playing with the feedback and drain resistor.
Anyway it deserves some experimenting.

mac

[JDoyle]

Today I breadboarded a q1=2SK117, q2=BD139 Fuzz Face. Standard circuit, only a pot and a big bypass cap at the fet source. It works.

Rsource=3.5k
Vd=1.1v
Vs=0.85v
Vg=0.5v

Lots of highs, I have to put a 0.1uf cap across the 470 resistor, the higher input impedance I guess. The final gain and fuzz are not as the standard FF. Maybe playing with the feedback and drain resistor.
Anyway it deserves some experimenting.

mac

mac, with .85V at the source you may want to try a si-ge diode pair off the source. You will get near the same Vs and also have the dynamic gain changes that occur as the effective source resistance changes with signal level (or more accurately, the current level through the FET and diode pair), just like with a 'real' FF. A bypassed source resistor gives maximum gain available but is also stabilized because the 3.5k resistor swamps the dynamic change in rs. In the FF, Q1's re is a big part of the sound, in my opinion, because it is dynamic and directly controls the gain of Q1 when it is connected straight to ground.

Anyway, just a thought.

Jay Doyle

Anyway, just an idea.

[mac]

Jay,
Today I'm going to try the diodes.
I have a 3 fet circuit with diodes at the source bypassed by caps. IIRC when I posted the circuit someone said that when the diode is on by a static current, it is like a small resistor of about 100ohm or so. I guess this is true when the dc current is enough to turn the diode on in the linear region, but with a smaller dc current one can put the diode in the non-linear zone and let the ac current do the transition.

Anyway, with either diodes or resistor at the source, the high input impedance of the fet makes a difference, it "sounds like" the boss' Rocket at low-to-med gain settings. It is like the fet is both the buffer and the driving transistor.

I forgot to say that removing the cap from q1 source resistor cleans the distortion better than q2 drive pot.

mac

[JDoyle]

Jay,
Today I'm going to try the diodes.
I have a 3 fet circuit with diodes at the source bypassed by caps. IIRC when I posted the circuit someone said that when the diode is on by a static current, it is like a small resistor of about 100ohm or so. I guess this is true when the dc current is enough to turn the diode on in the linear region, but with a smaller dc current one can put the diode in the non-linear zone and let the ac current do the transition.

I don't really think that was accurate advice. The resistance of a diode varies according to whatever current is passing through it at a particular time and the resistance can be determined by:

r_d = V_T / I_d

where:

- r_d is the effective resistance of the diode
- V_T is the thermal volatage (26mV at room temp)
- I_d is the current through the diode

So at 1mA the effective resistance the diode is 26 ohms, at 0.1mA: 260 ohms, at 0.01 mA: 2600 ohms, etc. (these numbers are 'calculated' and will always be different when actually measured because the value is so intimately linked to temp., but the idea will be the same).

So if your JFET has a I_dss of 2mA and you bias the stage to a quiescient current of 1mA, the resistance of the diode will vary from 26 ohms at rest, to 13 ohms at the negative clipping point (when V_G = V_S and the JFET is passing I_dss), to well over 100 MEG at the positive clipping point when the JFET's channel is only passing leakage current (it isn't infinity due to the diode's own leakage current, 1 x 10^-12 or 1 pA, is a number I see often in texts, but whatever the absolute value, it is really high and might as well be infinity).

BUT can you see how it is possible to make sure that the positive clipping point doesn't really 'clip'? As the signal swings up at the drain, the current through the JFET decreases, at some point the current through it decreases to a point where the effective resistance of the diode is MORE then that of the load resistor on the drain, reducing the gain to less than one. So it is possible to bias it so as the output swings positive instead of it 'clipping' on the positive swing, the gain incrementally reduces, 'rounding off' the signal instead of 'clipping' it.

You can also see that the negative swing is 'relatively' linear as the effective resistance of the diode only changes by 13 ohms over the entire negative swing.

Hope this makes sense,

Jay

PS - For germanium diodes (and Si diodes running at substantially larger currents) the formula is almost the same: r_d = 2V_T / I_d

[mac]

BUT can you see how it is possible to make sure that the positive clipping point doesn't really 'clip'? As the signal swings up at the drain, the current through the JFET decreases, at some point the current through it decreases to a point where the effective resistance of the diode is MORE then that of the load resistor on the drain, reducing the gain to less than one. So it is possible to bias it so as the output swings positive instead of it 'clipping' on the positive swing, the gain incrementally reduces, 'rounding off' the signal instead of 'clipping' it.
You can also see that the negative swing is 'relatively' linear as the effective resistance of the diode only changes by 13 ohms over the entire negative swing.

I tried the diodes, two series 1n4148 bypassed with a 47uf electro + 6.8uf poly. I put a pot at q2 collector to set 4.5v.
Rc=5.43k
Vd=1.37v
Vs=1.09v
Vg=0.76v

Then I replaced the diodes with a pot at the source.
Rc=5.43k
Rs=4.83k
Vd=1.36v
Vs=1.09v
Vg=0.75v

Both sound very close. I tested before reading your post, I did not tried the diodes without the bypass cap.

mac

[petemoore]

Rc=5.43k
Vd=1.37v
Vs=1.09v
Vg=0.76v

Then I replaced the diodes with a pot at the source.
Rc=5.43k
Rs=4.83k
Vd=1.36v
Vs=1.09v
Vg=0.75v

   Both sound very close.
  So..how is the FF responding / sounding, does guitar volume rolloff gain, and does it sound like it's biased [looks like Jfet's biased ok] and not gating, nice fuzz?

[mac]

pete,
Bias are ok, it is not gating. Maybe tweaking the 33k and the 100k may improve the sound. This is just a proof that it works.
Guitar volume rolls off gain very well, it is brighter than a std ff but not so gainy. It sounds thinner than a ff and not so fuzzy. I should record a sample.

mac