on the basic functioning of the Fuzz Face.

[Steben]

Reading RG's article about the FF (technology of...), I always believed the negative feedback biasing resulted in a musical flavoured response. It is, however, negative feedback. I looked closely and I really think the feedback makes the response more linear than otherwise. The circuit is used massively in phono and tape recorder preamps by the way. Since the logarithmic response of Q1 is countered by a opposite changing biasing, the assymetry is lessened. The remaining "touch responsive" aspect is AFAIK the follow-up of assymetric clipping by symmetric.

I suspect the change in tone between Germanium and Silicon is more determined by Q2 than 1. A hybrid FF would be better by using a Germ for the second tranisistor.

[R.G.]

Good thought workout. It's always possible to think of how circuits work in many ways as an aid to understanding them.

Here are my thoughts on that topic. Negative feedback does indeed linearize things, so there is always some degree of that going on. How much it linearizes depends on lots of things - forward gain and the impedances involved among others.

The voltage feedback stage is indeed used in many older hifi and phono stages. It was one way they got consistent response without having opamps to rely on.

The classical way you would use a FF style voltage feedback stage is with an input resistor. The equivalent input circuit is then the input resistor in series with the input voltage, running into the very low input impedance of Q1 base. The feedback is taken from effectively Q1 collector (buffered by Q2 emitter), so you really have a current summing junction at Q1 base. Under those circumstances, the circuit begins to resemble an opamp - a lot - if there is enough forward gain to linearize the input, or alternatively, enough fed-back output current to make the forward gain much lower than the open loop gain. It all depend on how much input impedance, how much feedback impedance and how much forward gain.

In the case of the FF, the input "resistor" is the impedance of the incoming signal. For a guitar, that's the pickup voltage through a resistance of maybe 6K to 20K, depending on the pickup, and a rising inductive impedance from the pickup inductance. Then there's that emitter resistor and capacitor on Q2 emitter. As that is changed, you get from substantially all of the collector signal fed back through the 100K to substantially none of it, depending on the pot setting. The less AC feedback you get, the less the feedback through the 100K resistor can linearize the response of Q1. That's the fundamental method the FF has of changing how much distortion you get - it changes the forward gain of the whole mess by varying how much AC feedback gets back to the input.

What makes for clipping being asymmetrical or not is how and when the signal bangs into cutoff or saturation on the transistors. In this case, we have two very different clipping methods. Q1 as a voltage feedback stage offers the advantage of having the highest possible signal swing out of the given power supply, plus the fact that the voltage feedback stage may cut off sharply, but has a mushy, soft saturation. Positive going signals run Q1 into saturation and this soft, mushy clipping. This is reflected in Q2 as soft, mushy clipped tops on signals (tops in the sense of "near the power supply"; it's really inverted from this since there is a negative power supply). However, when Q1 swings toward cutoff, it pushes Q2 into saturation. Q2 will saturate without the advantages of voltage feedback to soften it, so it's saturation clipping is harder than Q1's. Whether Q2 ever hits saturation or cutoff depends on the bias on Q2 and the values of the Q1 and Q2 collector resistors. This is one reason that changing those resistors and the collector bias point on Q2 has such an effect on tone - it determines whether you get to see Q1's cutoff or Q2's cutoff and at what signal level.

Which one controls the change in tone?? I don't think one can say that apriori. At least I can't. It depends on how you set up the biasing.

On the log response of Q1 - the collector current of a transistor is an exponential function of the base voltage.

[Paul Perry (Frostwave)]

I wonder if it is possible to make the overload a bit more 'squashy' by putting non-linear elements (eg diodes) in series or parallel with the collector resistors?

[Steben]

Good thought workout. It's always possible to think of how circuits work in many ways as an aid to understanding them.

Here are my thoughts on that topic. Negative feedback does indeed linearize things, so there is always some degree of that going on. How much it linearizes depends on lots of things - forward gain and the impedances involved among others.

The voltage feedback stage is indeed used in many older hifi and phono stages. It was one way they got consistent response without having opamps to rely on.

The classical way you would use a FF style voltage feedback stage is with an input resistor. The equivalent input circuit is then the input resistor in series with the input voltage, running into the very low input impedance of Q1 base. The feedback is taken from effectively Q1 collector (buffered by Q2 emitter), so you really have a current summing junction at Q1 base. Under those circumstances, the circuit begins to resemble an opamp - a lot - if there is enough forward gain to linearize the input, or alternatively, enough fed-back output current to make the forward gain much lower than the open loop gain. It all depend on how much input impedance, how much feedback impedance and how much forward gain.

In the case of the FF, the input "resistor" is the impedance of the incoming signal. For a guitar, that's the pickup voltage through a resistance of maybe 6K to 20K, depending on the pickup, and a rising inductive impedance from the pickup inductance. Then there's that emitter resistor and capacitor on Q2 emitter. As that is changed, you get from substantially all of the collector signal fed back through the 100K to substantially none of it, depending on the pot setting. The less AC feedback you get, the less the feedback through the 100K resistor can linearize the response of Q1. That's the fundamental method the FF has of changing how much distortion you get - it changes the forward gain of the whole mess by varying how much AC feedback gets back to the input.

What makes for clipping being asymmetrical or not is how and when the signal bangs into cutoff or saturation on the transistors. In this case, we have two very different clipping methods. Q1 as a voltage feedback stage offers the advantage of having the highest possible signal swing out of the given power supply, plus the fact that the voltage feedback stage may cut off sharply, but has a mushy, soft saturation. Positive going signals run Q1 into saturation and this soft, mushy clipping. This is reflected in Q2 as soft, mushy clipped tops on signals (tops in the sense of "near the power supply"; it's really inverted from this since there is a negative power supply). However, when Q1 swings toward cutoff, it pushes Q2 into saturation. Q2 will saturate without the advantages of voltage feedback to soften it, so it's saturation clipping is harder than Q1's. Whether Q2 ever hits saturation or cutoff depends on the bias on Q2 and the values of the Q1 and Q2 collector resistors. This is one reason that changing those resistors and the collector bias point on Q2 has such an effect on tone - it determines whether you get to see Q1's cutoff or Q2's cutoff and at what signal level.

Which one controls the change in tone?? I don't think one can say that apriori. At least I can't. It depends on how you set up the biasing.

On the log response of Q1 - the collector current of a transistor is an exponential function of the base voltage.

thanks RG fot the reply.
I guess I still don't understand why voltage feedback introduces softer clipping. I only see voltages, maybe the current-aspects of BJT's are still vague to me?

Q1/Q2 clipping: I guess the introduction of larger pots (2k - 2k5) on Q2's emitter is ment to reduce Q2's clipping?

[Steben]

A last annoying question in trying to fine tune: is putting a trimmer on Q1's collector as well as on Q2's collector the best way to fine tune an get some ideal biasing voltages, especially when swpping transistors? Iow: is only one trimmer ok to make it work, but two superb to make it sound ideal too?

[MartyMart]

A last annoying question in trying to fine tune: is putting a trimmer on Q1's collector as well as on Q2's collector the best way to fine tune an get some ideal biasing voltages, especially when swpping transistors? Iow: is only one trimmer ok to make it work, but two superb to make it sound ideal too?

I found that a Q1 trimmer was useful for Ge FF's fine tune but with Si's more pedictable response
just a Q2 trimmer was enough.
I suppose that a "known good" and stable Ge in Q1 would be fine once the `'fixed" resistor value is
calculated .....
I do quite like having Q2 as trim/external pot for "mis-bias" fun !

[Steben]

A last annoying question in trying to fine tune: is putting a trimmer on Q1's collector as well as on Q2's collector the best way to fine tune an get some ideal biasing voltages, especially when swpping transistors? Iow: is only one trimmer ok to make it work, but two superb to make it sound ideal too?

I found that a Q1 trimmer was useful for Ge FF's fine tune but with Si's more pedictable response
just a Q2 trimmer was enough.
I suppose that a "known good" and stable Ge in Q1 would be fine once the `'fixed" resistor value is
calculated .....
I do quite like having Q2 as trim/external pot for "mis-bias" fun !

The resistor to ground from emitter of Q2 is also quite fun as external, it is a fine tune "gate" control, really!
Instead of the fuzz control, unless you make a parallel pot in series with the bypass cap.

[mac]

A last annoying question in trying to fine tune: is putting a trimmer on Q1's collector as well as on Q2's collector the best way to fine tune an get some ideal biasing voltages, especially when swpping transistors? Iow: is only one trimmer ok to make it work, but two superb to make it sound ideal too?

I like to go this way, two trimmers. Also with some transistors one can make both base currents equal. In this case ic2 = current through fuzz pot, and current through Rc1 = ie1, assuming low leakage. I can't find my notes but there is a relationship between resistor, vbe1 and hfe1 to satisfy this curious result.

I do the following: I first adjust Rc1 starting from a low value, say 10k, and Rc2 near 8.2k. It sounds unbiased or no sound at all. Then I increase Rc1 until I find the most focused sound, then tweak Rc2, measure Vc2 and adjust rc1 again a little, and so on. My results are that Rc1 can go as low as 20k in some cases.

IMHO, the 33k is a value which is set to fit many transistors at the factory, ie hfe and leakage, but it maybe tweaked to get the most of a particular pair of Ge. BUt as RG said above, the circuit is very susceptible to resistors changes, so what you get at the end of this fine tune process may be a little away from the original tone.

mac